Vibrations of Rotating Machinery: Volume 2. Advanced Rotordynamics: Applications of Analysis, Troubleshooting and Diagnosis 9784431554523, 9784431554530, 4431554521

Table of contents :
Preface
Contents
1 An Overview of Vibration Problems in Rotating Machinery
1.1 Structure and Feature of Rotating Machinery
1.2 Vibration Problems in Rotating Machinery and Countermeasures
1.3 Guidelines for Health Evaluation
1.3.1 International Standards on Vibrations of Rotating Machinery
1.3.2 Allowable Vibration Criteria
1.3.3 Condition Monitoring of Machines and Certification System of Vibration Diagnostics Engineers
1.4 Vibrations in Rotating Machinery and Diagnosis
1.4.1 Causal Relationship Matrix
1.4.2 Discrimination Between Forced Vibrations and Self-excited Vibrations
2 Basics of Plain Bearings
2.1 Operating Principles of Plain Bearings
2.1.1 Features of Plain Bearings
2.1.2 Oil Film Formation and Generation of Oil Film Pressure
2.2 Hydrodynamic Lubrication Theory for an Oil Film Bearing
2.2.1 Coordinates and Differential Equation of Oil Film Pressure
2.2.2 Derivation of Reynolds Equation
2.3 Steady-State Characteristics of a Plain Bearing Oil Film
2.3.1 Infinitely Short Bearing Approximation Solution of Reynolds Equation (Cylindrical Bearing)
2.3.2 Circumferential Boundary Condition of Pressure
2.3.3 Equilibrium of Journal Center and Journal Center Locus
2.3.4 Steady-State Oil Film Reaction Force and Sommerfeld Number
2.4 Dynamic Characteristics of a Plain Bearing Oil Film
2.4.1 Oil Film Force of a Cylindrical Bearing
2.4.2 Linear Stiffness Coefficients and Linear Viscous Damping Coefficients of a Cylindrical Bearing
2.5 Turbulent Oil Film
2.6 Actual Plain Bearings
2.6.1 Various Plain Bearings
2.6.2 Details of Plain Bearing Specifications
3 Unbalance Vibration of a Rotor in Plain Bearings
3.1 Feature of Unbalance Vibration
3.2 Mathematical Expression for Unbalance Vibration
3.2.1 Expression with X–Y Coordinates
3.2.2 Complex Displacement Expression z = x + jy
3.3 Simplified Expressions for the Dynamic Properties of an Oil Film with Unbalance Vibration
3.4 Effects of Bearing Type and Design Variables
3.5 Effects of Bearing Pedestal Stiffness
4 Stability of a Rotor in Plain Bearings
4.1 Cause and Phenomena of Oil Whip
4.1.1 Cause of Oil Whip
4.1.2 Oil Whip Phenomena
4.2 Stability Chart of Circular Bearing Based on Linear Vibration Analysis
4.2.1 Coordinates and Equation of Motion
4.2.2 Characteristic Equation and Stability Criterion
4.2.3 Stability Chart
4.3 Countermeasures Against Oil Whip in a Cylindrical Bearing
4.4 Stability Charts of Non-circular Bearings and Oil Whip Countermeasures
4.5 Explanation of Mechanism of Oil Whip
4.6 Countermeasures for Actual Journal Bearings and Points of Attention
4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration
5 Vibration of Rolling Element Bearings
5.1 Stiffness of Rolling Element Bearings [58, 59, 60]
5.1.1 Stiffness in the Radial Direction: kr
5.1.2 Thrust Direction (Axial) Stiffness: kz
5.2 Excitation Frequency of a Rolling Element Bearing [60, B46]
5.2.1 Whirling and Rotational Frequencies of Rolling Elements
5.2.2 Exciting Frequency Caused by a Traveling Ball
5.2.3 Race Vibration (Ringing) of Rolling Element Bearings
5.3 Diagnosis of Rolling Element Bearing Vibrations and Signal Processing
5.3.1 Acceleration Vibration Waveform of Bearing Box
5.3.2 Amplitude Modulation
5.3.3 Frequency Band
5.3.4 Envelope Processing
5.3.5 Concepts of Bearing Vibration Diagnosis
5.4 Case Study of Rolling Element-Bearing Vibration Diagnosis [62]
5.4.1 Degree of Degradation of a Rolling Element Bearing
5.4.2 Vibration Waveform and FFT Analysis of a Damaged Rolling Element Bearing
6 Vibration in Magnetic Bearing Rotor Systems
6.1 Functions and Characteristics of Magnetic Bearings
6.1.1 Magnetic Circuit [91]
6.1.2 Size of an Active Magnetic Bearing (AMB) [B48, 92]
6.1.3 Attractive Force of a Magnetic Bearing
6.1.4 Linearization of a Magnetic Force in Magnetic Bearing
6.2 ISO Standards Related to Magnetic Bearings [93]
6.2.1 Terms Related to Active Magnetic Bearings (ISO 14839-1)
6.2.2 Vibration Evaluation Criteria (ISO 14839-2)
6.2.3 Evaluation Criteria of Stability Margin (ISO 14839-3)
6.2.4 Case Study
6.3 Straight Control of an Active Magnetic Bearing Rotor
6.3.1 Controller Transfer Function
6.3.2 Dynamic Characteristics of AMB
6.3.3 Modeling and Control of an Active Magnetic Bearing Rotor (One AMB)
6.3.4 Modeling and Control of an AMB Rotor (Two AMBs + Symmetrical Rotor) [97–99, VB550]
6.4 Cross-Control of an Active Magnetic Bearing
6.4.1 Cross-Control [100]
6.4.2 Whirl Vibration and Stability of Cross-Control
6.4.3 Cross-Control of Unbalance Vibration
7 Case Studies of Forced Vibration Problems of a Rotor
7.1 Approaches to Resonance Problems in Rotating Machinery
7.1.1 Natural Frequencies and Damping Ratios Varying with Rotational Speed or Load
7.1.2 Constant Speed Machine and Variable Speed Machine
7.1.3 Vibration Amplitude at Critical Speed During Acceleration/Deceleration
7.1.4 Bending Vibrations and Torsional Vibrations
7.2 Criteria for Acceptable Vibration Levels of Rotating Machinery
7.2.1 Transition of the Vibration Standards
7.2.2 ISO 10816: Mechanical Vibration—Evaluation of Machine Vibration by Measurements on Non-rotating Parts [142]
7.2.3 ISO 7919: Mechanical Vibration of Non-reciprocating Machines—Measurements on Rotating Shafts and Evaluation Criteria [143]
7.3 Case Studies of Unbalance Vibration Problems
7.3.1 Balancing of a Gas Turbine Rotor [VB017]
7.3.2 Structural Resonance Problems of a Vertical Pump
7.3.3 Vibration Caused by Low Stiffness of a Fan Framework [VB058]
7.3.4 Thermal Bending Vibration
7.4 Case Studies of Forced Vibration in an Asymmetric Rotor
7.4.1 Secondary Critical Speed of an Asymmetric Rotor
7.4.2 Elliptical Deformation of Circular Nyquist Plots Due to Rotor Asymmetry
7.5 Vibrations Induced by Gears
7.6 Vibration Generated by a Cross Joint
7.7 Case Studies of Other Forced Vibration
7.7.1 Electromagnetic Vibration
7.7.2 Rotating Stall
7.7.3 Rotor Blade
7.7.4 Interaction Between the Rotor Blade and the Stator Vane
7.7.5 Four-Cycle Engine
7.7.6 Belt Drive Machines
7.7.7 Reducing Torsional Vibration
8 Case Studies of Self-excited Vibration of Rotor Stability Problems
8.1 Approaches to Self-excited Vibration Problems in Rotating Machinery
8.1.1 How to Identify Self-excited Vibrations and Apply Countermeasures
8.1.2 Examples of Destabilizing Force
8.2 Self-excited Vibration Problems Caused by Oil Film Bearings or Seals
8.2.1 Self-excited Vibration Caused by Oil Film Bearing (Oil Whip)
8.2.2 Self-excited Vibration Caused by Seal (Clearance Flow)
8.3 Self-excited Vibration Due to Fluid Force of an Impeller
8.3.1 Self-excited Vibration Caused by Turbine Blades (Axial Flow)
8.3.2 Self-excited Vibration Caused by Centrifugal Impellers
8.4 Self-excited Vibration Due to Internal Damping (Hysteresis Whip)
8.5 Fluid-Containing Rotor
8.6 Friction Whip
8.7 Bently–Muszynska Model for Clearance Flow
8.8 Stabilization with Squeeze Film Damper
8.8.1 Differences with and Without a Centering Spring
8.8.2 Feeding/Draining of Oil and End Seals
8.8.3 Dynamic Characteristics of a Simple Model of a  Squeeze Film Damper
8.8.4 General Expression for Dynamic Characteristics of Squeeze Film Dampers
9 Torsional Vibration and Related Coupled Vibration
9.1 Analysis and Measurement of Torsional Vibration
9.1.1 Single-Shaft System
9.1.2 Measurement of Torsional Vibration in an Engine Shaft System
9.1.3 Geared Shaft System
9.2 Torsional Vibration of a Turbine Generator Set
9.2.1 Torsional Vibration of a Turbine Generator Shaft System
9.2.2 Measurement of Torsional Vibration [246]
9.2.3 Calculation of Torsional Natural Frequencies
9.3 Blade–Shaft-Coupled Torsional Vibration [246, 247]
9.3.1 Outset of Blade–Shaft-Coupled Torsional Vibration Problems
9.3.2 Equivalent Mass Model of Blade Vibration (Nodal Diameter Number κ = 0) [251, 252]
9.3.3 Example of Blade–Shaft-Coupled Model
9.3.4 Example of Calculation of Blade–Shaft-Coupled Natural Frequencies
9.3.5 Effect of Blade–Shaft Coupling
9.3.6 Verification of Accuracy of Measurement of Blade–Shaft-Coupled Vibration [253, 254]
9.4 Bending–Torsional Coupled Vibration
9.4.1 Equation of Motion for Bending–Torsional Coupled Vibration [262, 263]
9.4.2 Stability of Bending–Torsional Coupled Vibration
10 Signal Processing for Rotor Vibration Diagnosis
10.1 Vector Monitor (Balance Analyzer)
10.1.1 What is a Vector Monitor?
10.1.2 PLL (Phase-Locked Loop)
10.1.3 Bode Diagram and Nyquist Diagram
10.2 Signal Processing for Unbalance Vibrations
10.2.1 Waveform and Orbit of Shaft Center of Rotation for Unbalance Vibration
10.2.2 Vibration Measurement
10.2.3 Block Diagram of Unbalance Vibration
10.2.4 Extraction of Forward Unbalance Vibration Component
10.2.5 Balancing to Decrease the Forward Unbalance Vibration Amplitude
10.3 Fourier Series Expansion [274, 275]
10.3.1 Example: Estimation of an Inherent Cosine/Sine Wave’s Amplitude
10.3.2 Principles of Fourier Series Expansion
10.4 Discrete Fourier Transformation [276]
10.4.1 Principles of the Discrete Fourier Transformation
10.4.2 Mirror (Aliasing) Phenomenon of Complex Amplitudes
10.4.3 Sampling Values
10.4.4 Aliasing Error
10.4.5 Suppression of Noise
10.5 Practicalities of FFT Analysis [276, 60]
10.5.1 Basic Specifications
10.5.2 Signal Processing Procedure
10.5.3 DFT of Synchronous and Asynchronous Waveforms
10.5.4 Fourier Transforms
10.5.5 Resolution
10.5.6 Overall (OA)
10.5.7 Application of FFT Analyzers [277, 278]
10.6 Zooming of a FFT Analyzer
10.6.1 FFT of the Original Signal
10.6.2 Frequency Shifting and LPF
10.6.3 DFT and Half-Spectrum Displays
10.6.4 Spectrum Zoom Displays
10.7 Full Spectrum
10.7.1 Concept
10.7.2 One-Channel FFT
10.7.3 Two FFTs for Full Spectrum
10.7.4 Example of Vibration Diagnosis Using a Full Spectrum
11 Our Latest Topics Relating to Simplified Modeling of Rotating Systems
11.1 Practical Techniques Recommended for Model Order Reduction (MOR)
11.1.1 Procedures for Guyan Reduction
11.1.2 Mode Synthesis Technique
11.1.3 Accuracy Comparison Between Guyan and Mode Synthesis Models
11.1.4 Discrete Modeling for Continuous Medium
11.1.5 Mode Separation
11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported Rotor
11.2.1 Single-Degree-of-Freedom System and Simplified Stability Criterion [B37, 293]
11.2.2 Two-DOF System and Undamped Natural Frequency [294]
11.2.3 Simplified Stability Criterion for Two-Degree-of-Freedom System
11.2.4 Calculation Example for a Rotor System with Three Disks [295]
11.2.5 Example of Numerical Calculation for Muszynska’s Rotor System [293]
11.2.6 Example of Numerical Calculation for Casing Whirl
11.2.7 Anisotropic Properties of Oil Film Bearing Dynamic Force
11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems [296]
11.3.1 Coupling Mass of Mode Synthesis Model
11.3.2 Modeling of Mode Synthesis Method with 3D Finite Element Method (FEM)
11.3.3 Mode Synthesis Model of Blade
12 Exercises of ISO Certification Examination for Vibration Experts
12.1 The First 30 Questions and Multiple Choice Answers
12.2 Additional 70 Questions
12.3 Answers and Hints for the Additional Questions 31–100
Appendix A: Spring and Damping Coefficients of a Cylindrical Bearing Assuming Infinitely Short Width
Appendix B: Elliptical Coordinates
Appendix C: Fourier Transformation [291]
C.1 Definition of Fourier Transformation
C.2 Examples of Fourier Transformations
C.3 Comparison of Amplitude Spectrum of Window Function
Appendix D: PLL Circuit and Synchronized Sinusoidal Wave Generation Circuit
Appendix E: Campbell Diagram
Appendix F: Details for Obtaining Eqs. (11.46) and (11.49)
Appendix G: Details for Obtaining Eq. (11.51)
Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies [299]
H.1 Variables and Definition
H.2 Equation of Motion for an Entire Blading System
H.3 Mode Synthesis Model for Blading
References
R2 References (JSME v_BASE Data Book)
R2 References (Papers)
R2 References (Books)
Index

Citation preview

Mathematics for Industry 17

Osami Matsushita Masato Tanaka Masao Kobayashi Patrick Keogh Hiroshi Kanki

Vibrations of Rotating Machinery Volume 2. Advanced Rotordynamics: Applications of Analysis, Troubleshooting and Diagnosis

Mathematics for Industry Volume 17

Aims & Scope The meaning of “Mathematics for Industry” (sometimes abbreviated as MI or MfI) is different from that of “Mathematics in Industry” (or of “Industrial Mathematics”). The latter is restrictive: it tends to be identified with the actual mathematics that specifically arises in the daily management and operation of manufacturing. The former, however, denotes a new research field in mathematics that may serve as a foundation for creating future technologies. This concept was born from the integration and reorganization of pure and applied mathematics in the present day into a fluid and versatile form capable of stimulating awareness of the importance of mathematics in industry, as well as responding to the needs of industrial technologies. The history of this integration and reorganization indicates that this basic idea will someday find increasing utility. Mathematics can be a key technology in modern society. The series aims to promote this trend by 1) providing comprehensive content on applications of mathematics, especially to industry technologies via various types of scientific research, 2) introducing basic, useful, necessary and crucial knowledge for several applications through concrete subjects, and 3) introducing new research results and developments for applications of mathematics in the real world. These points may provide the basis for opening a new mathematics-oriented technological world and even new research fields of mathematics. To submit a proposal or request further information, please use the PDF Proposal Form or contact directly: Swati Meherishi, Executive Editor. Editor-in-Chief Masato Wakayama (Kyushu University, Fukuoka, Japan) Scientific Board Members Robert S. Anderssen (Commonwealth Scientific and Industrial Research Organisation, Canberra, ACT, Australia) Yuliy Baryshnikov (Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL, USA) Heinz H. Bauschke (University of British Columbia, Vancouver, BC, Canada) Philip Broadbridge (School of Engineering and Mathematical Sciences, La Trobe University, Melbourne, VIC, Australia) Jin Cheng (Department of Mathematics, Fudan University, Shanghai, China) Monique Chyba (Department of Mathematics, University of Hawaii at Mānoa, Honolulu, HI, USA) Georges-Henri Cottet (Joseph Fourier University, Grenoble, Isère, France) José Alberto Cuminato (University of São Paulo, São Paulo, Brazil) Shin-ichiro Ei (Department of Mathematics, Hokkaido University, Sapporo, Japan) Yasuhide Fukumoto (Kyushu University, Nishi-ku, Fukuoka, Japan) Jonathan R. M. Hosking (IBM T.J. Watson Research Center, Scarsdale, NY, USA) Alejandro Jofré (University of Chile, Santiago, Chile) Masato Kimura (Faculty of Mathematics & Physics, Kanazawa University, Kanazawa, Japan) Kerry Landman (The University of Melbourne, Victoria, Australia) Robert McKibbin (Institute of Natural and Mathematical Sciences, Massey University, Palmerston North, Auckland, New Zealand) Andrea Parmeggiani (Dir Partenariat IRIS, University of Montpellier 2, Montpellier, Hérault, France) Jill Pipher (Department of Mathematics, Brown University, Providence, RI, USA) Konrad Polthier (Free University of Berlin, Berlin, Germany) Osamu Saeki (Institute of Mathematics for Industry, Kyushu University, Fukuoka, Japan) Wil Schilders (Department of Mathematics and Computer Science, Eindhoven University of Technology, Eindhoven, The Netherlands) Zuowei Shen (Department of Mathematics, National University of Singapore, Singapore, Singapore) Kim Chuan Toh (Department of Analytics and Operations, National University of Singapore, Singapore, Singapur, Singapore) Evgeny Verbitskiy (Mathematical Institute, Leiden University, Leiden, The Netherlands) Nakahiro Yoshida (The University of Tokyo, Meguro-ku, Tokyo, Japan)

More information about this series at http://www.springer.com/series/13254

Osami Matsushita Masato Tanaka Masao Kobayashi Patrick Keogh Hiroshi Kanki •

•

•

•

Vibrations of Rotating Machinery Volume 2. Advanced Rotordynamics: Applications of Analysis, Troubleshooting and Diagnosis

123

Osami Matsushita The National Defense Academy Yokosuka, Japan

Masato Tanaka The University of Tokyo Tokyo, Japan

Masao Kobayashi IHI Corporation Yokohama, Japan

Patrick Keogh University of Bath Bath, UK

Hiroshi Kanki Kobe University Kobe, Japan

ISSN 2198-350X ISSN 2198-3518 (electronic) Mathematics for Industry ISBN 978-4-431-55452-3 ISBN 978-4-431-55453-0 (eBook) https://doi.org/10.1007/978-4-431-55453-0 Library of Congress Control Number: 2018965430 © Springer Japan KK, part of Springer Nature 2019 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Japan KK, part of Springer Nature. The registered company address is: Shiroyama Trust Tower, 4-3-1 Toranomon, Minato-ku, Tokyo 105-6005, Japan

Preface

This book is a sequel to the previous book “Vibrations of Rotating Machinery Volume 1. Basic Rotordynamics: Introduction to Practical Vibration Analysis.” Volume 1 used simplified models and mathematical expressions to give succinct explanations of various vibration phenomena occurring in rotating machinery. “Volume 2. Advanced Rotordynamics: Applications of Analysis, Troubleshooting and Diagnosis” provides actual vibration problems and fundamental key techniques for troubleshooting the problems by measurement, analysis, and diagnosis of the vibrations. The v_BASE databook provides a database of vibration problems of actual machinery experienced by industry. The latest edition, published in 2011 March, includes 790 cases in total, and more than a half of them are involved with rotating machinery. A wide variety of vibration problems experienced are described, together with elaborate troubleshooting of the problems. For example, when a vibration problem is found in commissioning a rotating machine, full-power operation and/orfull-speed operation cannot be completed until the problem is solved. The same or similar vibration problems occur repeatedly in the field. This book aims to assist engineers in troubleshooting the problems effectively. Rotating shafts are supported by various types of bearings. Chapters 2–4 describe the dynamical properties of oil film bearings, unbalance vibration of rotors supported by the bearings, and self-excited vibrations of rotors caused by the destabilizing action of bearing oil films or seal films, respectively. Chapter 5 describes the vibration diagnosis methods for rotors supported by rolling element bearings. Chapter 6 describes the mechanical and dynamical properties of active magnetic bearings (AMBs) and the rotor vibration control action available from AMBs. The related ISO information and case studies are introduced to accelerate the familiarization of AMB technologies. Chapters 7 and 8 give actual examples of forced vibration of rotors and self-excited vibration of rotors selected from the v_BASE databook, respectively. The chapters mentioned above deal with the rotor lateral vibrations of whirling modes in a shaft axial rotational plane. In addition, Chapter 9 treats torsional vibration analysis of a turbo-rotor train system, together with shaft–blade coupled v

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Preface

torsional resonances in turbine generator sets. Chapter 10 explains the rotor vibration signal processing based on the Fast Fourier transformation (FFT). The same signal processing is so specialized for the combination with rotor whirling motion that it could be convenient for other vibration diagnoses. Chapter 11 deals with our latest topics on modeling techniques and is recommended mainly for system reduction. It also enables prediction of the onset speed of oil film-induced instability without the need for eigenvalue evaluation and explains how to model the complete form of coupled vibration analysis between shafting and blading. Chapter 12, the final chapter, offers 100 selected questions and answers, referring to the ISO curriculum. We hope that they will be effective to re-affirm the knowledge stated in our books for vibration diagnosis. This book is intended to help design engineers of rotating machinery to prevent the recurrence of the problems reported and also to help candidates pass the ISO machine condition monitoring and diagnosis examination. The target readers of this book also include graduate students, practicing engineers, and research scientists who can utilize methods of analysis, the mathematical approach, and the references cited. While the authors have endeavored to eliminate errors and avoid bias, critical comments from readers are welcome. The authors are grateful to the authors of the literature cited in this book, especially Dr Ronald L. Eshleman (VI, USA). Special thanks are also due to Prof Hiroyuki Fujiwara (National Defense Academy of Japan), Mr Hisayuki Motoi (Deputy General Manager, IHI, Japan), and Dr Hidetomo Komura (3DIM, Japan) for providing valuable comments for the 100 Q&A list of Chapter 12. The authors deeply thank Mr Fumito Shinkawa, President of Shinkawa Electric Co., Ltd., for his support for our authoring. The authors wish to thank Springer for its support in publishing this work. Yokosuka, Japan August 2018

Osami Matsushita

Contents

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An Overview of Vibration Problems in Rotating Machinery . . . 1.1 Structure and Feature of Rotating Machinery . . . . . . . . . . . . 1.2 Vibration Problems in Rotating Machinery and Countermeasures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Guidelines for Health Evaluation . . . . . . . . . . . . . . . . . . . . . 1.3.1 International Standards on Vibrations of Rotating Machinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Allowable Vibration Criteria . . . . . . . . . . . . . . . . . . 1.3.3 Condition Monitoring of Machines and Certification System of Vibration Diagnostics Engineers . . . . . . . 1.4 Vibrations in Rotating Machinery and Diagnosis . . . . . . . . . 1.4.1 Causal Relationship Matrix . . . . . . . . . . . . . . . . . . . 1.4.2 Discrimination Between Forced Vibrations and Self-excited Vibrations . . . . . . . . . . . . . . . . . . . Basics of Plain Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Operating Principles of Plain Bearings . . . . . . . . . . . . . . . . 2.1.1 Features of Plain Bearings . . . . . . . . . . . . . . . . . . 2.1.2 Oil Film Formation and Generation of Oil Film Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Hydrodynamic Lubrication Theory for an Oil Film Bearing 2.2.1 Coordinates and Differential Equation of Oil Film Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Derivation of Reynolds Equation . . . . . . . . . . . . . 2.3 Steady-State Characteristics of a Plain Bearing Oil Film . . . 2.3.1 Infinitely Short Bearing Approximation Solution of Reynolds Equation (Cylindrical Bearing) . . . . . . 2.3.2 Circumferential Boundary Condition of Pressure . . 2.3.3 Equilibrium of Journal Center and Journal Center Locus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Unbalance Vibration of a Rotor in Plain Bearings . . . . . . . . . . . 3.1 Feature of Unbalance Vibration . . . . . . . . . . . . . . . . . . . . . . 3.2 Mathematical Expression for Unbalance Vibration . . . . . . . . 3.2.1 Expression with X–Y Coordinates . . . . . . . . . . . . . . 3.2.2 Complex Displacement Expression z = x + jy . . . . . 3.3 Simplified Expressions for the Dynamic Properties of an Oil Film with Unbalance Vibration . . . . . . . . . . . . . . . . . . . . . . 3.4 Effects of Bearing Type and Design Variables . . . . . . . . . . . 3.5 Effects of Bearing Pedestal Stiffness . . . . . . . . . . . . . . . . . .

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Stability of a Rotor in Plain Bearings . . . . . . . . . . . . . . . . . . . . . 4.1 Cause and Phenomena of Oil Whip . . . . . . . . . . . . . . . . . . . 4.1.1 Cause of Oil Whip . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Oil Whip Phenomena . . . . . . . . . . . . . . . . . . . . . . . 4.2 Stability Chart of Circular Bearing Based on Linear Vibration Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Coordinates and Equation of Motion . . . . . . . . . . . . 4.2.2 Characteristic Equation and Stability Criterion . . . . . 4.2.3 Stability Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Countermeasures Against Oil Whip in a Cylindrical Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Stability Charts of Non-circular Bearings and Oil Whip Countermeasures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Explanation of Mechanism of Oil Whip . . . . . . . . . . . . . . . . 4.6 Countermeasures for Actual Journal Bearings and Points of Attention . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Steady-State Oil Film Reaction Force and Sommerfeld Number . . . . . . . . . . . . . . . . . Dynamic Characteristics of a Plain Bearing Oil Film . . . 2.4.1 Oil Film Force of a Cylindrical Bearing . . . . . . 2.4.2 Linear Stiffness Coefficients and Linear Viscous Damping Coefficients of a Cylindrical Bearing . Turbulent Oil Film . . . . . . . . . . . . . . . . . . . . . . . . . . . . Actual Plain Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Various Plain Bearings . . . . . . . . . . . . . . . . . . . 2.6.2 Details of Plain Bearing Specifications . . . . . . .

Vibration of Rolling Element Bearings . . . . . . . . . 5.1 Stiffness of Rolling Element Bearings [58, 59, 5.1.1 Stiffness in the Radial Direction: kr . . 5.1.2 Thrust Direction (Axial) Stiffness: kz .

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Excitation Frequency of a Rolling Element Bearing [60, B46] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Whirling and Rotational Frequencies of Rolling Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Exciting Frequency Caused by a Traveling Ball . . . . 5.2.3 Race Vibration (Ringing) of Rolling Element Bearings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Diagnosis of Rolling Element Bearing Vibrations and Signal Processing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Acceleration Vibration Waveform of Bearing Box . . 5.3.2 Amplitude Modulation . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Frequency Band . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.4 Envelope Processing . . . . . . . . . . . . . . . . . . . . . . . . 5.3.5 Concepts of Bearing Vibration Diagnosis . . . . . . . . Case Study of Rolling Element-Bearing Vibration Diagnosis [62] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Degree of Degradation of a Rolling Element Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.2 Vibration Waveform and FFT Analysis of a Damaged Rolling Element Bearing . . . . . . . . .

Vibration in Magnetic Bearing Rotor Systems . . . . . . . . . . 6.1 Functions and Characteristics of Magnetic Bearings . . . 6.1.1 Magnetic Circuit [91] . . . . . . . . . . . . . . . . . . . 6.1.2 Size of an Active Magnetic Bearing (AMB) [B48, 92] . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Attractive Force of a Magnetic Bearing . . . . . . 6.1.4 Linearization of a Magnetic Force in Magnetic Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 ISO Standards Related to Magnetic Bearings [93] . . . . 6.2.1 Terms Related to Active Magnetic Bearings (ISO 14839-1) . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Vibration Evaluation Criteria (ISO 14839-2) . . 6.2.3 Evaluation Criteria of Stability Margin (ISO 14839-3) . . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 Case Study . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Straight Control of an Active Magnetic Bearing Rotor . 6.3.1 Controller Transfer Function . . . . . . . . . . . . . . 6.3.2 Dynamic Characteristics of AMB . . . . . . . . . . 6.3.3 Modeling and Control of an Active Magnetic Bearing Rotor (One AMB) . . . . . . . . . . . . . . . 6.3.4 Modeling and Control of an AMB Rotor (Two AMBs + Symmetrical Rotor) [97–99, VB550] . . . . . . . . . . . . . . . . . . . . . . .

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Cross-Control of an Active Magnetic Bearing . . . . . . . 6.4.1 Cross-Control [100] . . . . . . . . . . . . . . . . . . . . 6.4.2 Whirl Vibration and Stability of Cross-Control 6.4.3 Cross-Control of Unbalance Vibration . . . . . . .

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Case Studies of Forced Vibration Problems of a Rotor . . . . . . . 7.1 Approaches to Resonance Problems in Rotating Machinery . 7.1.1 Natural Frequencies and Damping Ratios Varying with Rotational Speed or Load . . . . . . . . . . . . . . . . 7.1.2 Constant Speed Machine and Variable Speed Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.3 Vibration Amplitude at Critical Speed During Acceleration/Deceleration . . . . . . . . . . . . . . . . . . . . 7.1.4 Bending Vibrations and Torsional Vibrations . . . . . . 7.2 Criteria for Acceptable Vibration Levels of Rotating Machinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Transition of the Vibration Standards . . . . . . . . . . . 7.2.2 ISO 10816: Mechanical Vibration—Evaluation of Machine Vibration by Measurements on Non-rotating Parts [142] . . . . . . . . . . . . . . . . . . . . . 7.2.3 ISO 7919: Mechanical Vibration of Non-reciprocating Machines—Measurements on Rotating Shafts and Evaluation Criteria [143] . . . 7.3 Case Studies of Unbalance Vibration Problems . . . . . . . . . . 7.3.1 Balancing of a Gas Turbine Rotor [VB017] . . . . . . . 7.3.2 Structural Resonance Problems of a Vertical Pump . 7.3.3 Vibration Caused by Low Stiffness of a Fan Framework [VB058] . . . . . . . . . . . . . . . . . . . . . . . 7.3.4 Thermal Bending Vibration . . . . . . . . . . . . . . . . . . 7.4 Case Studies of Forced Vibration in an Asymmetric Rotor . . 7.4.1 Secondary Critical Speed of an Asymmetric Rotor . . 7.4.2 Elliptical Deformation of Circular Nyquist Plots Due to Rotor Asymmetry . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Vibrations Induced by Gears . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Vibration Generated by a Cross Joint . . . . . . . . . . . . . . . . . . 7.7 Case Studies of Other Forced Vibration . . . . . . . . . . . . . . . . 7.7.1 Electromagnetic Vibration . . . . . . . . . . . . . . . . . . . . 7.7.2 Rotating Stall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.3 Rotor Blade . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.4 Interaction Between the Rotor Blade and the Stator Vane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.5 Four-Cycle Engine . . . . . . . . . . . . . . . . . . . . . . . . .

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Belt Drive Machines . . . . . . . . . . . . . . . . . . . . . . . . . 211 Reducing Torsional Vibration . . . . . . . . . . . . . . . . . . . 213

Case Studies of Self-excited Vibration of Rotor Stability Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Approaches to Self-excited Vibration Problems in Rotating Machinery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 How to Identify Self-excited Vibrations and Apply Countermeasures . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Examples of Destabilizing Force . . . . . . . . . . . . . . . 8.2 Self-excited Vibration Problems Caused by Oil Film Bearings or Seals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Self-excited Vibration Caused by Oil Film Bearing (Oil Whip) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Self-excited Vibration Caused by Seal (Clearance Flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Self-excited Vibration Due to Fluid Force of an Impeller . . . 8.3.1 Self-excited Vibration Caused by Turbine Blades (Axial Flow) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Self-excited Vibration Caused by Centrifugal Impellers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Self-excited Vibration Due to Internal Damping (Hysteresis Whip) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Fluid-Containing Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Friction Whip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Bently–Muszynska Model for Clearance Flow . . . . . . . . . . . 8.8 Stabilization with Squeeze Film Damper . . . . . . . . . . . . . . . 8.8.1 Differences with and Without a Centering Spring . . . 8.8.2 Feeding/Draining of Oil and End Seals . . . . . . . . . . 8.8.3 Dynamic Characteristics of a Simple Model of a Squeeze Film Damper . . . . . . . . . . . . . . . . . . 8.8.4 General Expression for Dynamic Characteristics of Squeeze Film Dampers . . . . . . . . . . . . . . . . . . . . Torsional Vibration and Related Coupled Vibration . . . . . . . 9.1 Analysis and Measurement of Torsional Vibration . . . . . . 9.1.1 Single-Shaft System . . . . . . . . . . . . . . . . . . . . . . 9.1.2 Measurement of Torsional Vibration in an Engine Shaft System . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.3 Geared Shaft System . . . . . . . . . . . . . . . . . . . . . 9.2 Torsional Vibration of a Turbine Generator Set . . . . . . . . 9.2.1 Torsional Vibration of a Turbine Generator Shaft System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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9.3

9.4

9.2.2 Measurement of Torsional Vibration [246] . . . . . . . 9.2.3 Calculation of Torsional Natural Frequencies . . . . . . Blade–Shaft-Coupled Torsional Vibration [246, 247] . . . . . . 9.3.1 Outset of Blade–Shaft-Coupled Torsional Vibration Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Equivalent Mass Model of Blade Vibration (Nodal Diameter Number j = 0) [251, 252] . . . . . . 9.3.3 Example of Blade–Shaft-Coupled Model . . . . . . . . . 9.3.4 Example of Calculation of Blade–Shaft-Coupled Natural Frequencies . . . . . . . . . . . . . . . . . . . . . . . . 9.3.5 Effect of Blade–Shaft Coupling . . . . . . . . . . . . . . . . 9.3.6 Verification of Accuracy of Measurement of Blade–Shaft-Coupled Vibration [253, 254] . . . . . . . Bending–Torsional Coupled Vibration . . . . . . . . . . . . . . . . . 9.4.1 Equation of Motion for Bending–Torsional Coupled Vibration [262, 263] . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Stability of Bending–Torsional Coupled Vibration . .

10 Signal Processing for Rotor Vibration Diagnosis . . . . . . . . . . 10.1 Vector Monitor (Balance Analyzer) . . . . . . . . . . . . . . . . . 10.1.1 What is a Vector Monitor? . . . . . . . . . . . . . . . . . 10.1.2 PLL (Phase-Locked Loop) . . . . . . . . . . . . . . . . . 10.1.3 Bode Diagram and Nyquist Diagram . . . . . . . . . . 10.2 Signal Processing for Unbalance Vibrations . . . . . . . . . . . 10.2.1 Waveform and Orbit of Shaft Center of Rotation for Unbalance Vibration . . . . . . . . . . . . . . . . . . . 10.2.2 Vibration Measurement . . . . . . . . . . . . . . . . . . . 10.2.3 Block Diagram of Unbalance Vibration . . . . . . . . 10.2.4 Extraction of Forward Unbalance Vibration Component . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.5 Balancing to Decrease the Forward Unbalance Vibration Amplitude . . . . . . . . . . . . . . . . . . . . . . 10.3 Fourier Series Expansion [274, 275] . . . . . . . . . . . . . . . . 10.3.1 Example: Estimation of an Inherent Cosine/Sine Wave’s Amplitude . . . . . . . . . . . . . . . . . . . . . . . 10.3.2 Principles of Fourier Series Expansion . . . . . . . . 10.4 Discrete Fourier Transformation [276] . . . . . . . . . . . . . . . 10.4.1 Principles of the Discrete Fourier Transformation . 10.4.2 Mirror (Aliasing) Phenomenon of Complex Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.3 Sampling Values . . . . . . . . . . . . . . . . . . . . . . . . 10.4.4 Aliasing Error . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.5 Suppression of Noise . . . . . . . . . . . . . . . . . . . . .

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10.5 Practicalities of FFT Analysis [276, 60] . . . . . . . . . . . 10.5.1 Basic Specifications . . . . . . . . . . . . . . . . . . . 10.5.2 Signal Processing Procedure . . . . . . . . . . . . . 10.5.3 DFT of Synchronous and Asynchronous Waveforms . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.4 Fourier Transforms . . . . . . . . . . . . . . . . . . . . 10.5.5 Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.6 Overall (OA) . . . . . . . . . . . . . . . . . . . . . . . . 10.5.7 Application of FFT Analyzers [277, 278] . . . 10.6 Zooming of a FFT Analyzer . . . . . . . . . . . . . . . . . . . 10.6.1 FFT of the Original Signal . . . . . . . . . . . . . . 10.6.2 Frequency Shifting and LPF . . . . . . . . . . . . . 10.6.3 DFT and Half-Spectrum Displays . . . . . . . . . 10.6.4 Spectrum Zoom Displays . . . . . . . . . . . . . . . 10.7 Full Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7.1 Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7.2 One-Channel FFT . . . . . . . . . . . . . . . . . . . . 10.7.3 Two FFTs for Full Spectrum . . . . . . . . . . . . 10.7.4 Example of Vibration Diagnosis Using a Full Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 Our Latest Topics Relating to Simplified Modeling of Rotating Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Practical Techniques Recommended for Model Order Reduction (MOR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 Procedures for Guyan Reduction . . . . . . . . . . . . . . . 11.1.2 Mode Synthesis Technique . . . . . . . . . . . . . . . . . . . 11.1.3 Accuracy Comparison Between Guyan and Mode Synthesis Models . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.4 Discrete Modeling for Continuous Medium . . . . . . . 11.1.5 Mode Separation . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported Rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1 Single-Degree-of-Freedom System and Simplified Stability Criterion [B37, 293] . . . . . . . . . . . . . . . . . 11.2.2 Two-DOF System and Undamped Natural Frequency [294] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.3 Simplified Stability Criterion for Two-Degree-ofFreedom System . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.4 Calculation Example for a Rotor System with Three Disks [295] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.5 Example of Numerical Calculation for Muszynska’s Rotor System [293] . . . . . . . . . . . . . . . . . . . . . . . .

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11.2.6 Example of Numerical Calculation for Casing Whirl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.7 Anisotropic Properties of Oil Film Bearing Dynamic Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems [296] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 Coupling Mass of Mode Synthesis Model . . . . . . . . 11.3.2 Modeling of Mode Synthesis Method with 3D Finite Element Method (FEM) . . . . . . . . . . . . . . . . . . . . . 11.3.3 Mode Synthesis Model of Blade . . . . . . . . . . . . . . . 12 Exercises of ISO Certification Examination for Vibration Experts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 The First 30 Questions and Multiple Choice Answers . . . 12.2 Additional 70 Questions . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Answers and Hints for the Additional Questions 31–100

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Chapter 1

An Overview of Vibration Problems in Rotating Machinery

Abstract This book aims to explain various phenomena and mechanisms of vibrations in rotating machinery based on theory and field experiences. It also aims to help engineers in carrying out diagnosis and in implementing countermeasures for various vibration problems in the field. It is normally easy to know the condition of machinery by means of measurement of vibrations and/or noise. However, it is rather difficult to interpret the observed phenomena correctly, derive real causes of the problems, and ascertain effective countermeasures, because a sufficient knowledge of case studies is needed. This chapter provides an overview of vibration problems in rotating machinery and general approaches for the countermeasures, and is intended to be helpful in solving vibration problems at the front line. Condition monitoring of a machine and vibration diagnostics including relevant ISO standards are illustrated. Case studies of vibration problems and the countermeasures in excess of 1000 cases, which have been collected until now by the Vibration Database (v_BASE) Committee under the Japan Society of Mechanical Engineers (JSME), are also referred to in this chapter. Keywords Rotor · Bearing · Stator · Vibration model · Vibration problems · Case study · Forced vibration · Self-excited vibration · ISO Standard · Condition monitoring · Vibration diagnostics · Causal relationship matrix · v_BASE database

1.1 Structure and Feature of Rotating Machinery Figure 1.1 shows a schematic of a turbocharger for application in automotive engines. In this machine, exhaust gas from an internal-combustion engine drives the turbine wheel that in turn drives the compressor wheel via a shaft fixed coaxially. Thus, much more air can be supplied to the combustion chambers and power output can be increased compared with a naturally aspirated engine. This machine consists of three parts, that are, (1) a rotating shaft with two wheels (rotors), (2) a stationary casing or housing (stator), and (3) bearings and seals (inbetween connection parts).

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_1

1

2 Fig. 1.1 Configuration of vehicular turbocharger

1 An Overview of Vibration Problems in Rotating Machinery Bearing

Casing (Stator)

Seal Ring seal

Turbine shaft

Compressor impeller

Rotating shaft (Rotor) Low pressure rotor (LP rotor) (Inside)

(Outside) High pressure rotor (HP rotor)

Fig. 1.2 Rotor configuration of aircraft engine (provided by Japanese Aero Engines Corporation)

Figure 1.2 shows a cutaway of aircraft gas turbine engine with a double spool arrangement, that is, the long LP rotor is placed inside the hollow HP rotor. This structure is complicated, but is the same as a turbocharger from a basic viewpoint. Theoretical modeling of actual rotating machinery is not always easy. In the case of torsional vibration of a rotor in its rotational degree of freedom, the modeling of the rotor only is sufficient because bearing forces and seal forces do not adversely affect the vibration. On the other hand, in the case of rotor bending vibration, the dynamic characteristics of bearings and seals must be incorporated in the model because they exert significant influence on the analytical results. Furthermore, modeling with the dynamic characteristics of a soft casing needs to be incorporated when the critical speeds of the system are influenced by them and come near to or within the operating speed range. From the perspective of a vibrating system, a rotating machine is special and differs widely from a general structure. First, the largest differences in a rotating machine

1.1 Structure and Feature of Rotating Machinery

3

are associated with the dynamic characteristics of the bearings and seals, which are specific to rotating machines. These dynamic characteristics may have sufficiently large influences to characterize the whole system. Consequently, this book intends to explain the features of bearings in detail, focusing on oil film bearings, rolling element bearings, and active magnetic bearings. The dynamic characteristics of bearings and seals change significantly with operating conditions such as rotating speeds, loads, or pressures, and are represented in two-dimensional whirling degrees of freedom (x, y). Stiffness and damping coefficients are expressed with associated (2 × 2) matrices. Sometimes, the magnitudes of cross-coupled stiffnesses (xy and yx components) become much larger than those of diagonal stiffnesses (xx and yy components). In such cases, the rotor-bearing system can become unstable, resulting in self-excited vibration known as oil whip, for example. Other significant differences in a rotating machine are gyroscopic moments acting on a rotor (e.g., giving rise to the standing up effect of a spinning top) that change in magnitude in proportion to rotating speed. As a result, vibrations in the x and y directions become coupled. The gyroscopic moments can be expressed using a skew-symmetric matrix. From these differences, standard general-purpose finite element method (FEM) software, which is based on symmetric matrix calculation, is not applicable for a rotating machine. This is a reason why vibration analysis software specialized for rotordynamics, such as MyROT as described in R1_Chap. 12, plays an important role in this field. Bearing characteristics sometimes cause self-excited vibration, but the damping coefficient of an oil film bearing or a squeeze film damper is much larger relatively than material or structural damping considered in a general structure. Therefore, it can be said for rotating machines that there are many stabilizing elements, yet also many destabilizing elements. It is important for rotating machinery to attain rotational stability throughout the operational range. For this reason, a complex eigenvalue analysis, which calculates not only natural frequencies but also damping ratios, is performed commonly for rotordynamic assessment.

1.2 Vibration Problems in Rotating Machinery and Countermeasures Unexpected vibration problems often become evident at a stage of developing machinery for higher speed, higher load, and higher efficiency. If the causes of the phenomena are known, countermeasures are straightforward. In cases of phenomena not previously encountered, there may be great difficulties to elucidate the causes. However, the efforts to solve these problems will certainly contribute to the progress of machine technology.

4

1 An Overview of Vibration Problems in Rotating Machinery

Historically, in the field of structural vibration engineering, bridge design for wind-resistant stability progressed greatly after the famous problem of the Tacoma Narrows Bridge. Also, airfoil flutter technologies began with the development of high-speed aircraft. In rotordynamics, a typical example is oil whip, a self-excited vibration first reported in 1925, which was encountered in the development of a highspeed rotor supported by oil film bearings. The technology advanced by elucidating the instability mechanism. In this manner, vibration engineering has been making progress in tandem with machine performance. For these reasons, in order to design high-performance machines without repetition of previously experienced problems, it is essential to know what types of vibration phenomena have occurred and also to know what countermeasures have been taken. Figure 1.3 shows a survey of results relating to problematic modes in some rotating machines. In this case “abnormal vibration” 29%, “abnormal sound” 11%, and “fatigue” 8%, these vibration-related cases account for 48% in total. Also, “wear” or “crack” issues often arise due to vibration; therefore, more than half of problems can be attributed to vibrations in rotating machinery. As this survey shows, a rotating machine is commonly associated with vibration phenomena. There are some causes of resulting vibration. A prompt investigation of these causes and remedies would be required for the benefit of the organization. Usually, troubleshooting is carried out with the following process steps shown below. (1) Understanding of machine structure and mechanisms, and collection of information At first, it is essential to know the structure and understand how the machine functions. Viewing drawings and reports from engineers commonly form this process step. Design difficulties concerning vibration technologies should be assessed using proper engineering judgment. It is also important to derive firsthand information from operators and maintenance engineers on-site. (2) Vibration measurement and investigation

Fig. 1.3 Problematic case classifications in rotating machinery

Insulation deterioration Material deterioration Abnormal temperature Rest Oil deterioration 8% Leakage Erosion Abnormal vibration 3% Looseness 3% 29% Fatigue 8% Crack 9%

Abnormal sound 11%

Wear 20%

1.2 Vibration Problems in Rotating Machinery and Countermeasures

5

Next, measurements are usually taken in order to comprehend the phenomena. In the case of a machine, measurements are necessary to evaluate the existing problematic conditions. For some reason, if a vibration level is high as a result, measurements should not necessarily be restricted to the vibration alone. Other possible causes, such as operational condition (pressure or temperature and so on) should be measured. It is also important to monitor the changes with variation of rotational speed. It may seem better to gather more data, but final decisive data usually form only a relatively small subset from which to comprehend the phenomena and to identify the causes. Gathering data is not a final target, but it is important to identify the decisive data as early as possible. (3) Estimation of the causes When an approximate understanding of the phenomena becomes apparent, the next step is to estimate the cause. Study of relevant precedents and the simulation of simple models will take place in order to clarify the mechanism. It may also include large-scale analysis. Table 1.1 shows the approximately classified vibration phenomena and the causes in rotating machinery. We can see that vibration phenomena are separated into forced vibration and self-excited vibration. (4) Implementing countermeasures The next stage involves the implementation of countermeasures. At this stage, the real cause of the vibration may not be ascertained, but the solution process will be urgent. After building up a hypothesis about the mechanism, countermeasures are implemented. If the results coincide with the expectations, the hypothesis may be considered to be proven. At the end of this stage, it is important to file the result of the troubleshooting as a causal relationship for future use.

Table 1.1 Vibration phenomena and cause Forced vibration

Unbalance vibration

Excessive unbalance Thermal unbalance Rotor deformation Misalignment Cracked rotor Bowed rotor Foundation defect

Resonance

Unbalance resonance Harmonic, Sub-harmonic resonance Torsional resonance Selective resonance

Fluid force

Rotating stall Cavitation Organ pipe vibration Combustion vibration

Force by mechanism

Electromagnetic force Ball bearing force Gear force Cross (Hooke's) joint Mechanism (cam /link)

Self-excited vibration Surge Friction whip Internal damping / hysteresis whip Oil whirl /whip Steam whirl Labyrinth whirl Impeller shroud force

6 Fig. 1.4 Ratio of the categorized cause of vibration

1 An Overview of Vibration Problems in Rotating Machinery

Bearing failure

Torsion 3% Electromagnetic 3% Flow-induced

3%

Nonlinear

5%

5%

Rest 12%

Self-excited vibration 10% Rotor resonance 9%

Unbalance 29%

Resonance of casing 21%

Figure 1.4 shows the causes of vibration problems for a specific machine. Another chart would be presented for a different machine. In this case, the effects concerning forced vibration such as unbalanced resonance are experienced in a significant number of cases, and the effects concerning self-excited vibration are unexpectedly a few (10%). Note: 1 Preventive maintenance of rotating machinery Machinery maintenance can be divided broadly into two categories. One is breakdown maintenance (BM), which implements maintenance work after a breakdown event. Another is preventive maintenance (PM), which implements maintenance work before breakdowns occur. One of the PM categories is time-based maintenance (TBM) that repeats inspections, repairs, and component replacement at regular time intervals. With TBM applied, machinery needs to be shut down after operating for a preset time duration, even if it is running smoothly, and some machine components are replaced even when not necessary. Consequently, TBM tends to result in costly maintenance. Furthermore, it may lead to a higher risk of initial failure at the restart of operation. The other PM category is condition-based maintenance (CBM) or predictive maintenance, which improves the possible shortcomings of TBM. Figure 1.5 illustrates the differences between TBM and CBM. With CBM applied, maintenance work is implemented before any predicted breakdown time, based on condition monitoring data obtained and analyzed. Machines are shut down when some representative measurements exceed preset limit values. The vibration amplitudes of bearing housings or rotating shafts are typically used in the monitoring. When vibration data alone are not sufficient to make a decision, data from wear particle sensors, acoustic emission (AE) sensor signals, and infrared thermography may be combined. Note: 2 Online and offline condition monitoring Figure 1.6 shows a typical online condition monitoring system by means of permanently installed and continuously working sensing equipment. It is costly, but can monitor the system condition continuously. Consequently, this system is applied to machinery that is categorized as of high value and is risk critical. It is implemented in risk-based maintenance (RBM),

1.2 Vibration Problems in Rotating Machinery and Countermeasures

7

Stop for periodic inspection (SPI)

Failure rate

SPI

SPI

SPI

......

SPI

SPI

SPI

......

......

Life cycle ( a ) Time-based maintenance (TBM) Breakdown

Breakdown

Failure rate

Early failure Random failure

Wear failure Repair

Acceptable level

..... Repair Life cycle

( b ) Condition-based maintenance (CBM)

Fig. 1.5 Comparison of maintenance cycle between TBM and CBM Fig. 1.6 Online monitoring (ISO 13373-1)

Turbo-machine Bearing vibration Rotation pulse Shaft vibration Driving machine

Monitoring by permanently installed device Preprocessor · Very important facility · Easy to deteriorate facility

Radial direction Axial direction

Analyzing computer

which uses the product of probability of failure and resulting damage as an evaluation index. On the other hand, Fig. 1.7 shows an offline monitoring system to collect data by means of handheld vibration measurement device at specified time intervals, for example, once per day. It is applied to lower risk machinery. Terminal boxes are used for a machine with difficult human access.

8

1 An Overview of Vibration Problems in Rotating Machinery ( a ) Manual method · Approachable facility · Rare to deteriorate facility

Shaft vibration Turbo-machine Radial direction

Rotation pulse Bearing vibration Driving machine

Axial direction

Handheld Link to computer vibration meter

( b ) Terminal box method · Non-approachable facility · Rare to deteriorate facility

Terminal box

Analyzing computer

Fig. 1.7 Offline monitoring (ISO 13373-1)

1.3 Guidelines for Health Evaluation The health condition of rotating machinery is often evaluated based on bearing or shaft vibration signals. Bearing vibration is measured as an absolute vibration (RMS value of vibration velocity) by means of electrodynamic vibration velocimeters or piezoelectric vibration accelerometers, both fixed on stationary bearing pedestals. Shaft vibration is measured as a relative vibration (vibration amplitude of shaft displacement) from stationary casings using displacement transducers. Machinery manufacturers and users often have different criteria in categorizing measured vibration as normal or abnormal. Therefore, mutual agreements are needed in advance to evaluate health condition, inclusive of allowable vibration criteria. For this agreement, the ISO (International Organization for Standardization [1]) establishes and disseminates internationally unified standards for such agreements. Major international standards concerning vibrations of rotating machinery are presented in the following sections.

1.3.1 International Standards on Vibrations of Rotating Machinery ISO consists of more than 200 Technical Committees (TCs), each TC consists of Sub-Committees (SCs), and each SC consists of Working Groups (WGs). Mechanical vibrations are treated in TC108 (mechanical vibrations, shock and condition monitoring). The following are the main SCs and WGs operated by TC108 concerning rotating machinery.

1.3 Guidelines for Health Evaluation

9

SC1 Balancing (currently WG31 under SC2) SC2 Measurement and evaluation of mechanical vibration and shock as applied to machines, vehicles, and structures /WG1 Rotordynamics and vibration of machines /WG7 Vibration of machines with active magnetic bearings /WG9 Vibration of pumps /WG10 Basic techniques for vibration diagnostics SC5 Condition monitoring and diagnostics of machine systems /WG7 Training and accreditation in the field of condition monitoring and diagnostics Many standards related to rotating machinery vibrations have been authorized and released by ISO, based on the discussions between international experts. Furthermore, the IEC [2] (International Electrotechnical Commission, ISO Cooperation Organization) and the API [3] (American Petroleum Institute) publish similar standards. The following standards are related to the content of this book: • Basics ISO 2041 Mechanical vibration, shock and condition monitoring—Vocabulary • Unbalance and balancing ISO 21940-11 (originated from 1940-1) Mechanical vibration – Rotor balancing – Part 11: Procedures and tolerances for rotors with rigid behaviour ISO 21940-12 (originated from 11342) Mechanical vibration – Rotor balancing – Part 12: Procedures and tolerances for rotors with flexible behaviour ISO 21940-31 (originated from 10814) Mechanical vibration – Rotor balancing – Part 31: Susceptibility and sensitivity of machines to unbalance • Measurement and evaluation of mechanical vibrations and shock ISO 10816 Mechanical vibration—Evaluation of machine vibration by measurements on non-rotating parts, Parts 1–7 ISO 7919 Mechanical vibration of non-reciprocating machines—Measurements on rotating shafts and evaluation criteria, Parts 1–5 Note: Recently, some of ISO 10816 and ISO 7919 are now being unified to a new ISO 20816 series. In this book, we mention them by their original series numbers. ISO 8528 Reciprocating internal combustion engine driven alternating current generating sets, Part 9: Measurement and evaluation of mechanical vibrations. ISO 14839 Mechanical vibration—Vibration of rotating machinery equipped with active magnetic bearings, Part 1: Vocabulary, Part 2: Evaluation of vibration, Part 3: Evaluation of stability margin, Part 4: Technical guidelines.

10

1 An Overview of Vibration Problems in Rotating Machinery

ISO 22266 Mechanical vibration—Torsional vibration of rotating machinery • Condition monitoring and diagnostics ISO 13372 Condition monitoring and diagnostics of machines—Vocabulary ISO 13373 Condition monitoring and diagnostics of machines—Vibration condition monitoring, Part 1: General procedures, Part 2: Processing, analysis and presentation of vibration data. ISO 13374 Condition monitoring and diagnostics of machines—Data processing, communication and presentation, Part 1: General guidelines, Part 2: Data processing, Part 3: Communication. ISO 13379 Condition monitoring and diagnostics of machines—General guidelines on data interpretation and diagnostics techniques ISO 17359 Condition monitoring and diagnostics of machines—General guidelines ISO 18436 Condition monitoring and diagnostics of machines—Requirements for qualification and assessment of personnel, Part 1: Requirements for certifying bodies and the certification process, Part 2: Vibration condition monitoring and diagnostics, Part 3: Requirements for training bodies and the training process, Part 4: Field lubricant analysis. • IEC Standards [2] The following standards, defined according to ISO standards, are often used. IEC 60034-14 Rotating electrical machines, Part 14: Mechanical vibration of certain machines with shaft heights 56 mm and higher—Measurement, evaluation and limits of vibration severity. IEC 60994 Guide for field measurement of vibrations and pulsations in hydraulic machines (turbines, storage pumps and pump-turbines). • API Standards [3] The machine-dependent standards establish incoming inspection criteria from the viewpoint of users of rotating machinery. API 610 Centrifugal Pumps for Petroleum, Petrochemical and Natural Gas Industries, API 616 Gas Turbines for Petroleum, Chemical, and Gas Industry Services, API 617 Axial and Centrifugal Compressors and Expander-compressors, API 677 General Purpose Gear Units for Petroleum, Chemical and Gas Industry Services.

1.3 Guidelines for Health Evaluation

11

1.3.2 Allowable Vibration Criteria Among the ISO Standards described above, the following are the guidelines to determine allowable vibration criteria. (1) Bearing vibrations ISO 10816 (Mechanical vibration—Evaluation of machine vibration by measurements on non-rotating parts) determines vibration evaluation criteria in terms of vibration severity, based on RMS value of broadband vibration velocity (normally 10 Hz–1 kHz) measured at non-rotating parts near bearing casings. Note: As mentioned above, ISO 10816 and ISO 7919 series are now being integrated to new ISO 20816 series. Part 1 General guidelines, Part 2 Land-based steam turbines and generators in excess of 50 MW with normal operating speeds of 1500, 1800, 3000, and 3600 r/min, Part 3 Industrial machines with nominal power above 15 kW and nominal speeds between 120 and 15 000 r/min when measured in situ, Part 4 Gas turbine sets with fluid-film bearings, Part 5 Machine sets in hydraulic power generating and pumping plants, Part 6 Reciprocating machines with power ratings above 100 kW, Part 7 Rotodynamic pumps for industrial application, including measurements on rotating shafts. Part 7, published in 2009, includes the evaluation criteria of shaft vibration, corresponding to ISO 7919. (2) Shaft vibration ISO 7919 (Mechanical vibration of non-reciprocating machines—Measurements on rotating shafts and evaluation criteria) determines evaluation criteria of shaft vibration displacement measured by means of non-contacting sensors (contact-type sensors may also be applicable), Part 1 General guidelines, Part 2 Land-based steam turbines and generators in excess of 50 MW with normal operating speeds of 1500, 1800, 3000 and 3600 r/min, Part 3 Coupled industrial machines, Part 4 Gas turbine sets with fluid-film bearings, Part 5 Machines set in hydraulic power generating and pumping plants. These internationally agreed standards present allowable vibration values determined from past field experiences. Design engineers and operation managers need to know the content of the standards fully since they are important to judge machine vibrations as normal or abnormal.

12

1 An Overview of Vibration Problems in Rotating Machinery

1.3.3 Condition Monitoring of Machines and Certification System of Vibration Diagnostics Engineers TC108/SC2/WG10 (Prof. Aly El-Shafei as convenor, Professor at Cairo University, Egypt) focuses its activities mainly on ISO 13373 that prescribes condition monitoring and diagnostic techniques of mechanical vibrations. TC108/SC5/WG7 (Dr. Eshlemann as convenor, President of Vibration Institute, USA) established ISO 18436 in 2003, defining an international training system for vibration diagnostics engineers, and also a certification system. In accordance with the ISO requirements above, the Japan Society of Mechanical Engineers (JSME) started the certification business for professional engineers qualified for condition monitoring and diagnostics of mechanical vibrations in 2004. The qualification examination is staged twice per year, with more than 100 applicants at each time. The examination is explained in detail on the pages of Qualification and Certification in the JSME website [4]. Figure 1.8 shows a wide variety of the industrial fields (manufacturers and users) of the applicants up to the year of 2008. The qualifications are needed to be able to carry out acceptance testing and vibration troubleshooting. The certified qualifications are classified into four categories, I to IV (highest). According to ISO 18436-2, each category is structured broadly as outlined by the following: Category I Being qualified to perform a range of a simple single-channel machinery vibration condition monitoring and diagnostics of machinery activities, in accordance with ISO 17359 and ISO 13373-1. However, it does not qualify the person to be responsible, for example, for the choice of sensor/transducer or for any analysis to be conducted, nor for the assessment of test results, and other judgment.

Fig. 1.8 Industrial fields of applicants for ISO vibration analyst for condition monitoring and diagnostics of machines

Automobile and tire 0.5% Steel and metal 2.7% Rest 7.1% Petroleum and plant 5.2% Measurement 7.4% Electrical and Electronics

Engineering and maintenance 30.1%

7.5% Heavy industry and Machinery 19.4%

Electric power and Gas 20.1%

Japan Society of Mechanical Engineers' data (statistics from 2008)

1.3 Guidelines for Health Evaluation

13

Category II Being qualified to perform industrial machinery vibration measurements and basic vibration analysis using single-channel measurements according to established and recognized procedures, and also being qualified to recommend minor corrective actions, and others. Category III Being qualified to perform and/or direct and/or establish programmes for vibration condition monitoring and diagnostics of machines in accordance with ISO 17359 and ISO 13373-1. Also qualified to measure and perform diagnosis of single-channel frequency spectra, as well as waveforms and orbits, to establish vibration monitoring programmes, to recommend field corrective actions, and to report to management personnel regarding programme objectives, budgets, cost justification and personnel development, and other associated procedures. Category IV Being qualified to perform and/or direct vibration condition monitoring and diagnostics of machines in accordance with ISO 17359 and ISO 13373-1, and all types of machinery vibration measurements and analyses. Also being qualified to apply vibration theory and techniques, to recommend corrective actions and/or design modifications, to provide technical guidance to vibration trainees and to interpret and evaluate published ISO codes of practice, International Standards and specifications. Individuals trying to get certification of Category III or IV are required to possess the certificates of the lower categories.

1.4 Vibrations in Rotating Machinery and Diagnosis Figure 1.9 shows the classification of mechanical vibrations, based on ISO/TC108. The vibrations are classified primarily as deterministic or random. The former is classified further as periodic or aperiodic. For example, vibrations synchronous with shaft rotating frequency are classified as periodic, and cavitation vibration, as random. This classification is based on recording and analyzing time-series waveform of vibration. From the practical viewpoint of rotating machinery, typical vibrations of rotors are classified as follows: (1) Shaft bending vibration (lateral vibration) This vibration corresponds to shaft center whirling in the shaft rotational plane, showing the mode of shaft deflection as it varies in the axial direction. Rigid mode whirling is inclusive. This bending vibration is the most typical in rotating machinery.

14

1 An Overview of Vibration Problems in Rotating Machinery Vibration

Deterministic

Periodic

Sinusoidal (simple harmonic)

Random (stochastic)

Aperiodic

Multiple sinusoidal

Transient

Stationary ergodic

Strongly stationary

Non-stationary

Weakly stationary

Fig. 1.9 Classification of vibrations (ISO 2041:1990 vocabulary)

(2) Shaft torsional vibration This vibration corresponds to torsional relative deflection oscillating with time between each neighboring shaft element in the axial direction. Although rotating machinery has few inherent factors causing torsional shaft vibration compared to reciprocating machinery, oscillation of the electromagnetic torques of electric motors may lead to torsional vibration. (3) Shaft longitudinal (axial) vibration This vibration usually does not have importance because longitudinal stiffness of shaft is relatively high. However, rigid mode vibration or local resonance may cause contacts with stationary parts or excessive dynamic loading on thrust bearings. (4) Rotating structure vibration Turbine blades or impellers of pumps or compressors may cause vibration problems. Eigen frequencies and resonance amplitudes are calculated precisely by means of 3D FEM applied to real configurations of the structures, and in consideration of the stiffness increase due to the centrifugal field effect. The vibration types mentioned above are classified in terms of vibration direction or of vibrating parts. Each specific vibration can take place alone, but coupled vibrations may also occur. Vibration arises in response to causes. To reduce vibration, it is important to find out the cause. To investigate vibration problems and to probe the causes is referred to as vibration diagnosis.

1.4.1 Causal Relationship Matrix ISO 13372 classifies equipment diagnosis procedure into two missions, that is, condition monitoring and diagnostics. Condition monitoring is defined as judging whether

1.4 Vibrations in Rotating Machinery and Diagnosis

15

the vibration is normal or abnormal, based on the trend against allowable vibration levels. Diagnostics involve determining or estimating the probable causes from various vibration data of the equipment in abnormal situations. These diagnostics are the essentially same as undertaken by medical doctors to identify diseases of patients when examined from various bodily conditions obtained. Many causal relationships between vibration phenomena and causes have been made clear up to now, and one can estimate causes from the vibration signatures on the basis of a causal relationship matrix. Table 1.2 shows an example of such a matrix based on related research achievements and field experiences. Double circles represent strong relationships between vibration phenomena and causes. Double circles and single circles are found in both cells of stabilization and destabilization depending on some phenomena. For instance, in the case of hysteresis whip, internal

Table 1.2 An example of cause-and-effect matrix

( : strong relation, Phenomena / Mechanism

: medium relation,

Special term

Self-excited vibration (Natural frequency) Stabilize

Unbalance

Nonlinear bearing External damping Internal damping Fluid film bearing Unsymmetrical shaft

Hysteresis whip Oil whip Oil whirl Asymmetric rotor

Crack

Cracked rotor

Gravity

Secondary critical speed

Blade passing

Blade passing frequency (BPF)

Electromagnetic force Gear

Mesh frequency

Shaft bending

Bowed rotor

Rotating stall Rubbing

Thermal bending Friction whip

Ball bearing Misalignment Cross joint, Joint Non-contact seal

Labyrinth whirl

Impeller shroud Fluid containing

Fluid trapped rotor

Turbine impeller

Thomas force

: low relation)

Destabilize

Forced vibration (Excitation frequency) Proportional to rotation Rotation Ω νΩ component component

16

1 An Overview of Vibration Problems in Rotating Machinery

damping contributes to stabilization at shaft speeds lower than critical speeds, but to destabilization at shaft speeds higher than critical speeds.

1.4.2 Discrimination Between Forced Vibrations and Self-excited Vibrations Table 1.2 shows that mechanical vibrations are broadly divided into two categories. One is forced vibration with a dominant frequency corresponding to that of the exciting force, such as unbalance vibration with shaft rotating frequency. The other is self-excited vibration, corresponding with an eigen frequency of the system. Consequently, the first step of vibration diagnosis of rotating machinery is a decision whether the phenomenon is due to forced vibration or self-excited vibration. With the use of a FFT analyzer, frequency analysis measurements are possible. They enable a judgment to be made on the vibration frequencies, which are proportional to the rotational frequency or the eigen frequencies of the system. A Campbell diagram is similarly useful, with shaft rotational speed as the horizontal coordinate and vibration frequency as the vertical coordinate. Figure 1.10 shows this diagram for the rotor vibration of a turbocharger supported by rolling element bearings with excessive clearance. It shows the unbalance vibration component with the shaft rotating frequency  and subharmonic or superharmonic vibration components with the frequencies in proportion to  (strictly equal to ν , ν  n/3, n/2, n, n  1, 2, 3, 4, …). These are then assumed as forced vibrations caused by the nonlinear mechanical characteristics of the rolling element bearings. Most types of forced vibration change their frequencies in proportion to shaft rotating speeds. Furthermore, the cause of the forced vibration can be estimated from the vibration frequency of the response because it is equal to the frequency

Fig. 1.10 Example of sub/super-harmonic vibration (Campbell diagram)

4

500

Frequency [Hz]

400

36 mP

3

5 2

300

200

5 3 4 3



3 2

2 3 1 2

100

0

2

5 10 Rotational speed  × 10 3 [rpm]

1 3 15

1.4 Vibrations in Rotating Machinery and Diagnosis

17

of excitation. For a countermeasure against forced vibrations, the principle is either to reduce the magnitude of the exciting force or to make the vibration system more insensitive to excitation by modifying the system parameters, or both in combination. In contrast, self-excited vibrations are characterized with eigen frequencies of the system, irrespective of the causes or mechanisms. Consequently, further tests are needed to identify the causes with various parameters being changed, sometimes taking significant time to decide on effective countermeasures. However, self-excited vibrations are well-known to take place because the system damping ratio of a corresponding eigen mode becomes negative. Consequently, even if the vibration mechanisms are not identified, increasing the initial damping of the eigen mode (for example, additional dampers installed) can stabilize the system in operation: (Increased initial damping ratio) − (decrease in damping ratio by destabilizing forces)  (damping ratio in operation) > 0 This is a supportive measure, but it often suffices to stop vibration problems. The true causes or mechanisms should be examined at a convenient later time. Note: 3 Vibration database (v_BASE) Committee [5] This book provides many actual case studies of vibration troubleshooting, mainly quoting the documents of the Vibration Database (v_BASE) Committee under the Dynamics, Measurement and Control Division of the Japan Society of Mechanical Engineers (JSME). Each of the quoted case studies from v_BASE is given by its number with VB___. The Committee was established in 1991, having continued to collect case studies of various troubleshooting for vibration and noise during 27 years and gathered over 1000 data. The Committee distributed a book of 296 data cases in 1994 as a first edition, and in 2002, updated to a database of all 518 data cases collected in a CDROM, with a full-text search tool included. Most recently, the Committee released the new database of all 790 data cases collected untill 2010 in a CD-ROM as a third edition. Figure 1.11 shows a breakdown of the 790 case studies. The data of

Fig. 1.11 Machine categories of v_BASE data (3rd edition, 790 data)

Construction machinery 1% Reciprocating machinery 4% Information machinery

General machinery/ structure,etc. 13%

3% Rotating machinery 49%

Traffic machinery 10% Plant 19%

18

1 An Overview of Vibration Problems in Rotating Machinery

rotating machinery occupy about 49%, and those of plant equipment, mainly flowinduced vibration, about 19%. Vibration problems in these two fields are found to be statistically large in number.

Chapter 2

Basics of Plain Bearings

Abstract Oil-lubricated plain bearings are used widely to support the main rotors of rotating machinery, because the hydrodynamic oil films formed in the bearing clearances can guarantee smooth rotation of shafts and effectively prevent or suppress rotor vibrations. This chapter explains the principles of oil film formation in typical plain bearings, based on conventional hydrodynamic lubrication theory. The circumferential oil film shape at the steady-state equilibrium is determined, so that the hydrodynamic pressure generated in the oil film can balance the oil film reaction force vectorially with the applied bearing load. In other words, the oil film separates the journal from the bearing surface and the journal can float and rotate on the oil film. The oil film shape determines the journal center position represented by the journal eccentricity ratio and the attitude angle in polar coordinates. The steady-state journal center position and in turn the corresponding oil film shape vary with the operating condition of the journal bearing, that is, Sommerfeld number. When the journal center is slightly perturbed at the equilibrium, the oil film is slightly deformed, resulting in the slightly modified pressure distribution. Then, the corresponding oil film force is found to consist of the static reaction force at the equilibrium and also the two mutually perpendicular components of the small dynamic reaction force. Each of the two components is expressed by the linear sum of the two stiffness components and also the two damping components. In total, four stiffness components and four damping coefficients are determined at the equilibrium. These coefficients are the key factors that dominate the vibrational behavior of the rotor supported in the plain bearings. The coefficients obtained for various types of journal bearing are found to vary with the Sommerfeld number. Consequently, journal bearings need to be selected and designed so that the dynamic coefficients of the hydrodynamic oil films can enable satisfactory control of rotor vibration. Keywords Hydrodynamic lubrication · Journal bearing · Bearing clearance · Oil film formation · Sommerfeld number · Oil film reaction force · Journal eccentricity ratio · Attitude angle · Stiffness coefficient of oil film · Damping coefficient of oil film

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_2

19

20

2 Basics of Plain Bearings

2.1 Operating Principles of Plain Bearings 2.1.1 Features of Plain Bearings The basic functions required of bearings are to support the static load of rotor weight firmly and transmit it to stationary pedestals, to minimize friction against shaft rotation, and to prevent or suppress rotor whirling. Rotating machinery uses plain bearings, rolling element bearings, and active magnetic bearings to rotate and function properly. These bearings work based on different principles to carry out the primary functions mentioned above. Oil-lubricated plain bearings can accomplish these required functions more easily than rolling element bearings, by means of the excellent mechanical properties of an oil film formed in the annular clearance between a journal and a bearing. Furthermore, plain bearings are inexpensive compared with active magnetic bearings. Consequently, plain bearings (oil film bearings) are used widely for rotating machinery.

2.1.2 Oil Film Formation and Generation of Oil Film Pressure Figure 2.1 shows a schematic of a horizontal rotor supported in oil film bearings at its both ends, while Fig. 2.2 shows a schematic of the cross-section of a cylindrical bearing with the journal spinning at an angular rate . Although the diametric clearance of a bearing is normally small, typically between 1.5 and 2.5 thousandth of the bearing diameter, it is depicted oversized to facilitate understanding. The orthogonal coordinates X (in the direction of static load W ) and Y (horizontal) are defined with their origin placed on the bearing center. When the journal does not rotate, it rests on the bearing bottom surface in the bearing clearance. Then, the journal center, Os , is located directly under the bearing center at a distance equal to the mean radial bearing clearance C (half of the difference of the bearing diameter and the journal diameter). The local clearance between the journal and bearing decreases monotonically in the circumferential direction, from the maximum clearance at the top of the bearing to the zero clearance at the bottom. When lubricating oil is fed into the clearance and the journal starts to spin, the viscous shearing action of the journal drags oil into the clearance space, forming a continuous oil film. As lubricant flows in the clearance space, which converges in the direction of journal rotation, mass conservation is maintained naturally. Consequently, the oil film pressure increases gradually toward the minimum film thickness position, resulting in an increase in the reverse flow component near the stationary bearing surface. Furthermore, some of the lubricant mass flows out of the clearance space in the Z direction perpendicular to the X–Y plane from the both ends of the bearing. This lubricant flow determines a hydrodynamic pressure distribution in the oil film as shown in Fig. 2.2.

2.1 Operating Principles of Plain Bearings

21

Fig. 2.1 Rotor-bearing system

Y

Ω

Θ

O X oil film bearing

oil film bearing

Fig. 2.2 Oil film pressure distribution P=0 cavi tat e d

e = εC

Y

Os W

ε : eccentricity ratio e : eccentricity C : mean radial clearance

oil film pressure p

X

φ

This pressure distribution can be integrated over the bearing surface, resulting in an oil film reaction force F. The steady-state vector balance of W and F can be obtained when the journal center, Os , moves from the original position just under the bearing center to the equilibrium position shown in Fig. 2.2. This specific journal position determines a corresponding oil film shape, attaining the force equilibrium. In the case of a cylindrical bearing, the center of the spinning journal at equilibrium is found at the distance of e from the bearing center and in the radial direction inclined by the angle φ from the X axis direction. The angle φ is called the attitude angle. The maximum and minimum oil film thickness positions are located relative to each other on opposite ends of the centerline connecting bearing center and Os . Thus the spinning journal rotates, floating on the oil film without direct contact with the bearing surface, resulting in low viscous friction compared to dry contact friction. When the journal center deviates from the equilibrium, the oil film resists the movement and pushes the journal back to the equilibrium position. This action suppresses and prevents rotor whirling.

22

2 Basics of Plain Bearings

2.2 Hydrodynamic Lubrication Theory for an Oil Film Bearing 2.2.1 Coordinates and Differential Equation of Oil Film Pressure This section explains how hydrodynamic pressure can be developed in the oil film, based on hydrodynamic lubrication theory. Figure 2.3 shows a schematic of a bearing surface (lower) and a journal surface (upper) separated by an oil film having uniform viscosity, μ. Both surfaces can be represented approximately by flat surfaces when a very small part of the lubricating domain is analyzed. Orthogonal coordinates x, y, and z are defined on the lower surface, where x is in the journal rotational direction, y is in the cross-film direction, and z is in the bearing axial direction. The oil film thickness, h, remains unchanged with z, but decreases gradually with x, forming a converging oil film. The bearing surface (y = 0) moves at a speed U 1 in the x direction, while the journal surface (y = h), has a speed U 2 in the x direction and a speed V in the y direction. As the oil film is assumed to be a continuum body of fluid, the three equations of motion of the oil film in the x, y, and z directions are derived from the NavierStokes equations. The equations may be simplified with some assumptions normally applied to the hydrodynamic lubrication theory of plain bearings, such as rigid wall boundaries, incompressible Newtonian fluids, isothermal states, isoviscous fluids, isopycnic fluids, laminar flows, negligibly small inertia forces, and predominant cross-film derivatives compared with the other derivatives in the x and z directions. Then, the three equations of motion are expressed as μ

∂p ∂ 2u = 2 ∂y ∂x

(2.1)

∂ 2v =0 ∂ y2

(2.2)

μ

Fig. 2.3 Model of hydrodynamically lubricated surfaces

U2

V

y h x z U1

2.2 Hydrodynamic Lubrication Theory for an Oil Film Bearing

μ

23

∂ 2w ∂p = ∂ y2 ∂z

(2.3)

where μ is the uniform viscosity coefficient of oil film, u, v, and w are the flow velocities in the x, y, and z directions, respectively, and the pressure p generated in the oil film is assumed to be uniform in the y direction, but varies in the x and z directions.

2.2.2 Derivation of Reynolds Equation No-slip conditions on the wall surfaces are assumed here. Then, the wall velocities shown in Fig. 2.3 are given as the boundary conditions of the oil film velocities at the wall surfaces. When Eqs. (2.1–2.3) are integrated twice with respect to y with the boundary conditions applied, three oil film velocities u, v, and w are obtained as     y(h − y) ∂ p h−y + − (2.4) u = U2 + (U1 − U2 ) h 2μ ∂ x y v=V (2.5) h y(h − y) ∂ p w=− (2.6) 2μ ∂z These velocities are substituted into the equation obtained by integrating the equation of continuity (2.7) with respect to y and further, by applying Eq. (2.8) that changes the order of integration and differentiation, ∂v ∂w ∂u + + =0 ∂x ∂y ∂z h(x) 0

∂ ∂ f (y, x)dy = ∂x ∂x

h(x) f (y, x)dy − f (h(x), x) 0

(2.7) ∂h(x) ∂x

(2.8)

As a result, the two-dimensional dynamic Reynolds Equation is derived as follows:     ∂ ∂ ∂ ∂p ∂p ∂h h3 + h3 = 6μ(U1 − U2 ) + 6μh (U1 + U2 ) + 12μV ∂x ∂x ∂z ∂z ∂x ∂x (2.9)

24

2 Basics of Plain Bearings

2.3 Steady-State Characteristics of a Plain Bearing Oil Film 2.3.1 Infinitely Short Bearing Approximation Solution of Reynolds Equation (Cylindrical Bearing) Equation (2.9) is applied to the oil film of a journal bearing of diameter D (= 2R), width L, and mean radial bearing clearance C (half of the difference of inner bearing diameter and journal diameter). In Eq. (2.9), U1 = 0, U2 ≈ U, V = U (∂h/∂ x) + ∂h/∂t for a constant journal surface velocity U (= R, counterclockwise rotation). Furthermore, the second term on the right-hand side can be assumed as negligibly small. Then, for a uniform viscosity coefficient, μ, and a constant bearing load, W, in the vertical downward direction, Eq. (2.9) can be simplified to the following:     ∂ ∂h ∂h ∂p ∂p ∂ h3 + h3 = 6μU + 12μ (2.10) ∂x ∂x ∂z ∂z ∂x ∂t where the second term on the right-hand side, 12μ∂h/∂t, is omitted when static oil film conditions apply. The oil film thickness, h, is expressed approximately in the following Eq. (2.11), because the magnitude error of h/C is of the order of C/R, which is normally in the range 1.5 × 10−3 to 2.5 × 10−3 for actual journal bearings: h = C(1 + ε cos θ )

(2.11)

where θ is the angle measured in the journal rotation direction from the maximum clearance position (therefore x = Rθ ), and ε is the journal eccentricity ratio defined as journal center eccentricity e from the bearing center, divided by the mean bearing radial clearance, C, resulting in 0 ≤ ε ≤ 1. Consequently, Eq. (2.10) can be solved for oil film pressure p(θ, z) when the values of μ, U, C, and ε at equilibrium are given. It is normally difficult to derive the analytical solution of p because Eq. (2.10) is a second-order partial differential equation. However, Eq. (2.10) may be transformed into an ordinary differential equation with the infinitely short bearing approximation applied. In that case, an analytical solution of p can be derived easily, although numerical solutions can be readily obtained nowadays by discretization of Eq. (2.10) by means of, for example, the finite difference method. The short bearing approximation assumes that the bearing width is infinitely short; then, the derivative term with respect to x on the left-hand side of Eq. (2.10) can be deleted because it is regarded as negligibly small compared to the derivative term with respect to z. When Eq. (2.11) is substituted into Eq. (2.10) without the first term on the left-hand side, the following ordinary differential equation with respect to a single variable z applies:

2.3 Steady-State Characteristics of a Plain Bearing Oil Film

ε sin θ d2 p 6μU =− 2 dz 2 C R (1 + ε cos θ )3

25

(2.12)

Equation (2.12) can be integrated with respect to z, with the boundary conditions p = 0 at z = 0, L. The pressure distribution p(θ, z) is obtained as follows: p=

3μU ε sin θ z(L − z) C 2 R (1 + ε cos θ )3

(2.13)

This is the infinitely short bearing approximation solution that agrees well with measurements in the case of small ε and small L/D.

2.3.2 Circumferential Boundary Condition of Pressure Equation (2.13) gives a parabolically symmetric pressure distribution in the direction of bearing width with the maximum value at the midwidth of bearing z = L/2. On the other hand, Fig. 2.4a shows the circumferential pressure distribution with zero pressure at θ = 0, π, 2π due to the term ε sin θ . Furthermore, because ε is always positive, the oil film pressure is always negative (below atmospheric pressure) in the divergent oil film (π < θ < 2π ). This pressure profile corresponds to the Sommerfeld condition that allows negative pressure to exist in the divergent oil film. However, the oil film flow that decreases to a minimum volume at θ = π cannot fill the divergent film region from θ = π and beyond fully, hence oil film rupture must occur. Furthermore, atmospheric air would flow into the divergent oil film with negative pressure from the bearing sides. Consequently, the pressure in the divergent oil film must remain at atmospheric pressure, which has been confirmed by measurements. Therefore, the oil film pressure in the divergent oil film is reasonably assumed to be atmospheric. Thus, Fig. 2.4b shows Gümbel condition that prescribes zero-constant gauge pressure in the divergent oil film. This boundary condition is simple and reasonable to apply to bearings operating in an atmospheric environment.

Fig. 2.4 Boundary conditions of oil film pressure

26

2 Basics of Plain Bearings

Figure 2.4c shows the Reynolds condition (or Swift-Stieber condition) that prescribes both pressure and pressure derivative to be zero at an angle θ1 . It is the most rigorous mathematically since it includes continuity of mass flow at θ1 . The angle θ1 is not always π , but is determined during the process of calculating the pressure distribution numerically.

2.3.3 Equilibrium of Journal Center and Journal Center Locus Figure 2.5a defines a X–Y orthogonal coordinate system and also a ε0 − θ0 polar coordinate system. In the figure, a journal center position Os at equilibrium in the bearing oil film is given by (ε0 , θ0 ) in the polar coordinate system. Here, ε0 is the journal eccentricity ratio. The journal is assumed to spin counterclockwise. When the journal does not spin ( = 0), the journal center, Os , stays on the X axis with ε0 = 1.0. When the journal spins, the journal center moves to its equilibrium position (ε0 , θ0 ) determined by bearing dimensions and operating conditions such as oil film temperature, spin speed, and load. Here, θ0 is the attitude angle, also denoted by φ. The static bearing load and the steady-state oil film reaction force balance vectorially when the journal center is located at its equilibrium position. In other words, the journal center moves its equilibrium position to generate oil film reaction force that balances with the applied bearing load. The equilibrium position moves depending on the operating condition. In the case of cylindrical bearing, the journal center depicts a near semi-circular arc locus, approaching the bearing center with increasing . bearing center eccentricity ratio ε 0 = ∞ 0.2 0.4 0.6 0.8 0

Y bearing center 0

1.0

F0

F θ0

Fθ 0

θ

F

Fε 0

θ

Ft

y 0.6 0.8

θ0

−fx

W

X ( a ) steady-state condition

( ε ,θ ) ( x , y )

ε0

ε = ε 0 + Δε θ = θ 0 + Δθ

y

ε

fr

X

x

ft

+ Fε 0 =−

1.0

=0

Oj

+ 0 = Fθ

− fy

Fr

1.0

( ε 0 ,θ 0 ) O s x

Os equilibrium position (ε 0 ,θ 0 )

Y

θ

θ0

Fε

0.2 attitude = φ angle θ 0 0.4

eccentricity ratio ε 1.0

( b ) dynamic condition

Fig. 2.5 Steady-state equilibrium and dynamic behavior of journal center in bearing clearance

2.3 Steady-State Characteristics of a Plain Bearing Oil Film

27

2.3.4 Steady-State Oil Film Reaction Force and Sommerfeld Number The steady-state oil film reaction force, F 0 , consists of two orthogonal components, namely, the radial component Fε0 and the tangential component Fθ0 . These two components are obtained by integrating Eq. (2.13) with the Gümbel condition in the oil film area θ = 0 − π, z = 0 − L: ˜  Fε0 = ˜(− p cos θ )R dθ dz Fθ0 = ( p sin θ )R dθ dz

(2.14)

The two components Fε0 and Fθ0 and the resultant F 0 given in the form of squareroot of sum of squares of Fε0 and Fθ0 are expressed as follows:  2  3 L ε02 R C D (1 − ε02 )2  2  3 L π ε0 R Fθ0 = W sin φ = 2μU R C D (1 − ε02 )3/2   2  3 ε π 2 + (16 − π 2 )ε2  0 0 L R 2 2 + Fθ0 = 2μU R F0 = Fε0 2 2 C D (1 − ε0 ) Fε0 = W cos φ = 8μU R

(2.15) (2.16)

(2.17)

where μ = oil film viscosity coefficient (Pa s), R = bearing radius (m), D = bearing diameter 2R (m), L = bearing width (m), C = mean bearing radial clearance (m) and U = surface velocity R (m/s). The attitude angle, φ, is defined as follows: tan φ =

Fθ0 = Fε0

 π 1 − ε02 4ε0

(2.18)

F 0 in Eq. (2.17) can be replaced by W, as F 0 is equal to W at equilibrium. Then,  S

L D

2 =



(1 − ε02 )2

(2.19)

π ε0 π 2 + (16 − π 2 )ε02

where the Sommerfeld number, S, is defined as   μN DL R 2 S= W C

(2.20)

where N = journal spin speed (rps, 1/s). The Sommerfeld number is dimensionless because both L/D and the right-hand side term are dimensionless in Eq. (2.19).

28

2 Basics of Plain Bearings 0

100 80

0.2 φ

0.4

40

0.6

20

0.8 1.0 0.001

60

ε0 0.01

0.1

1

attitude angle φ

eccentricity ratio εε0

Fig. 2.6 Variation of eccentricity ratio ε0 and attitude angle φ at equilibrium (short bearing approx.)

0 10

S (L / D) 2

Once S(L/D)2 is given, the journal center eccentricity ratio, ε0 , and attitude angle, φ, at equilibrium are uniquely determined from Eq. (2.19) and then Eq. (2.18), as shown in Fig. 2.6. Consequently, the dimensionless Sommerfeld number represents the operating state of the journal bearing. The Sommerfeld number can be regarded as representing dimensionless journal spin speed in most rotating machinery applications, because the other parameters in Eq. (2.20) can be assumed constant.

2.4 Dynamic Characteristics of a Plain Bearing Oil Film 2.4.1 Oil Film Force of a Cylindrical Bearing When a journal center is perturbed slightly around its equilibrium position, it is assumed to move from the equilibrium Os to Oj . As the oil film shape changes due to the perturbation, additional oil film pressure is generated, changing the oil film reaction force from the static force F 0 , balancing the static bearing load W, to the dynamic force F that works to push the journal center back to the original equilibrium Os . Under the dynamic condition, the moving journal center Oj is assumed to be located at the position (ε, θ ) at an arbitrary point in time. The time derivative of the oil film is calculated from Eq. (2.11) as ∂h/∂t = C(˙ε cos θ − εθ˙ sin θ ). Then, the corresponding Reynolds equation for a cylindrical bearing with the infinitely short bearing approximation can be derived from Eq. (2.10) as follows: ε sin θ ε˙ cos θ 6μ 12μ d2 p =− 2 ( − 2θ˙ ) + 2 2 3 dz C (1 + ε cos θ ) C (1 + ε cos θ )3

(2.21)

This ordinary differential equation can be integrated to obtain the pressure p that is in turn integrated with the Gümbel condition applied, as shown in Eq. (2.14). Then, the radial component F ε and the tangential component F θ of dynamic oil film reaction force can be derived as follows:

2.4 Dynamic Characteristics of a Plain Bearing Oil Film

29

 ⎫ ˙ μR L 3 ε2 ( − 2θ) π ε˙ (1 + 2ε2 ) ⎪ ⎪ Fε = + ⎪ ⎬ C2 (1 − ε2 )2 2(1 − ε2 )5/2 ⎪  μR L 3 π ε( − 2θ˙ ) 2εε˙ Fθ = + C2 4(1 − ε2 )3/2 (1 − ε2 )2

(2.22)

⎪ ⎪ ⎪ ⎪ ⎭

This equation with ε˙ = θ˙ = 0 gives the static oil film reaction force components expressed in Eqs. (2.15) and (2.16).

2.4.2 Linear Stiffness Coefficients and Linear Viscous Damping Coefficients of a Cylindrical Bearing The dynamic properties of an oil film are outlined here. With regard to detailed formula manipulation, Appendix A should be referred to. As shown in Fig. 2.5b, the journal center is perturbed slightly around equilibrium Os (ε0 , θ 0 ), and is assumed to move to Oj (ε, θ ). The symbol  is defined to denote a small perturbation in the following expressions: ε = ε0 + ε, θ = θ0 + θ, ε˙ = 0 + ˙ε , θ˙ = 0 + θ˙

(2.23)

When these expressions are substituted into Eq. (2.22), the following expressions are obtained: Fε = Fε0 + k11 ε + c11 ˙ε + c12 ε0 θ˙ + O(2 ) · · · Fθ = Fθ0 + k21 ε + c21 ˙ε + c22 ε0 θ˙ + O(2 ) · · ·

 (2.24)

These equations are expressed in the polar coordinate system at Oj (ε, θ ) and are transformed into the following equations expressed in the polar coordinate system at Os (ε0 , θ 0 ):  −Fr = Fε0 + k11 ε + k12 ε0 θ + c11 ˙ε + c12 ε0 θ˙ + O(2 ) · · · ≡ Fε0 − fr Ft = Fθ0 + k21 ε + k22 ε0 θ + c21 ˙ε + c22 ε0 θ˙ + O(2 ) · · · ≡ Fθ0 + f t (2.25) The first term of each expression above corresponds to the static oil film reaction force component, while the other terms correspond to the dynamic oil film reaction force components, f r and f t , with regard to small vibrations around equilibrium point Os . Equation (2.25) expressed in the polar coordinate system is transformed into the expression in the x–y orthogonal coordinate system around the equilibrium point Os . Furthermore, if terms of the second and higher order are deleted, the linear (first-order) oil film reaction force components are expressed as

30

2 Basics of Plain Bearings



fx fy





k k = xx xy k yx k yy

   x x˙ cx x cx y + c yx c yy y y˙

(2.26)

where x and y are small displacements from equilibrium, and x˙ and y˙ are small velocities at equilibrium. Thus, the dynamic oil film force components are expressed in terms of eight linear coefficients, that is, four stiffness coefficients and four viscous damping coefficients. Each coefficient can be made dimensionless as follows: Ki j =

C C ki j , C i j = ci j (i, j = x, y) W W

(2.27)

where W is static bearing load equal to half the weight of a symmetrical rotor placed horizontally. The eight dimensionless coefficients for a cylindrical bearing are obtained as follows:   4 π 2 + (32 + π 2 )ε02 + 2(16 − π 2 )ε04 Kxx = (1 − ε02 )K a   2 π π + (32 + π 2 )ε02 + 2(16 − π 2 )ε04 Kxy = ε0 (1 − ε02 )0.5 K a   2 4 2π + (16 − π 2 )ε02 K yy = Ka   2 π π − 2π 2 ε02 − (16 − π 2 )ε04 K yx = − (2.28) ε0 (1 − ε02 )0.5 K a   2π π 2 + 2(24 − π 2 )ε02 + π 2 ε04 Cx x = ε0 (1 − ε02 )0.5 K a   2 8 π − 2(8 − π 2 )ε02 Cx y = Ka   2π(1 − ε02 )0.5 π 2 − 2(8 − π 2 )ε02 C yy = ε0 K a C yx = C x y 1.5  where K a = π 2 + (16 − π 2 )ε02 . Each dimensionless oil film coefficient is expressed as a function of the journal eccentricity ratio ε0 at equilibrium Os . Consequently, the value of each coefficient varies with ε0 that varies with S(L/D)2 as shown in Fig. 2.6. Hence, each dimensionless coefficient is uniquely determined, corresponding to S(L/D)2 in Fig. 2.7, as is the analogous with the equilibrium coordinates shown in Fig. 2.6. The databook [B13] explains and illustrates both static and dynamic properties of oil film bearings in a general sense.

2.4 Dynamic Characteristics of a Plain Bearing Oil Film

31

Dimensionless stiffness coeff. Kij

100 50 Kxx 10 5

Kxy

Kyy 1 −Kyx

0.5 Kyx

Dimensionless damping coeff. Cij

100 50 Cxx 10 5 Cxy=Cyx 1 0.5

0.001

Cyy

0.01

0.1 S (L / D) 2

1

10

Fig. 2.7 Oil film coefficients of a cylindrical bearing (short bearing approx.)

Currently, numerical solutions of the oil film properties may be obtained by discretization of the Reynolds equation (Eq. 2.10) by means of, for example, the finite difference method, instead of using the infinitely short bearing approximation method alone. The circumferential angle corresponding to the Reynolds condition p = dp/dθ = 0 can be determined easily and automatically in executing numerical calculations repeatedly. Consequently, the Reynolds condition is applied widely in numerical convergence calculations for hydrodynamic oil film pressure instead of the Gümbel condition. Various experimental methods to determine oil film coefficients have been established, as shown in [B13, Sect. 4.2].

2.5 Turbulent Oil Film The steady-state and dynamic characteristics of oil film reaction force components mentioned previously were derived on the assumption that oil film flow is laminar. When the Reynolds number, Re (reference velocity U × reference length C/reference

32

2 Basics of Plain Bearings

kinematic viscosity ν), exceeds a critical value, oil film flow undergoes a transition from a laminar state to a superlaminar/turbulent state. In the case of journal bearings, it is usual to define the Reynolds number (Re = UC/ν) in terms of journal velocity, U, mean bearing radial clearance, C, and effective kinematic viscosity of the oil film, ν. When Re exceeds 1500–3000, the oil film flow is presumed to become turbulent. Therefore, turbulent oil film flows are often found to take place in large journal bearings for utility steam turbines and journal bearings lubricated with low viscosity fluids such as water. Then, the mechanical properties of such bearing oil films may be calculated based on turbulent lubrication theories. Instead, turbulent bearing properties can be obtained easily by compensating the laminar properties results with the Reynolds number applied. For example, the laminar property results with the ordinary Sommerfeld number may be rearranged with the modified Sommerfeld number, Se, defined by Se = S(1 + 0.0007768Re0.9 )

(2.29)

Figure 2.8a [B13, Sect. 4.6] shows theoretical results for the variation of equilibrium eccentricity ratio with Se for a tilting pad journal bearing (4-pad, L/D = 1.0, preload factor 0, pad angle 80°, center pivot, LBP) corresponding to three values of Reynolds number, that is, 0 (laminar), 3000 (turbulent) and 8000 (turbulent). The predictions for the three different Reynolds numbers are represented reasonably by one curve by means of the modified Sommerfeld number, Se. Figure 2.8b [B13, Sect. 4.6] shows similar results with regard to the stiffness coefficient K xx and the viscous damping coefficient C xx . Figure 2.9 [6] shows a case of measured variation of journal eccentricity ratio, ε, for Reynolds numbers of 4500 and 11,000, which can be represented approximately  by one curve with the modified Sommerfeld number S defined by Eq. (2.30) in the case of a water-lubricated cylindrical bearing (diameter 200 mm, L/D = 1.0, C/R = 0.0025): 1.0

Re = 0 Re = 3 000 Re = 8 000

Re = 0 Re = 3 000 Re = 8 000

10

Cxx

0.6

Kxx , Cxx

eccentricity ratio ε

0.8

0.4

1

0.2 0

0.1

Se

1

(a) eccentricity ratio

Kxx

0.1

Se

1

(b) stiffness and damping coeff.

Fig. 2.8 Eccentricity ratio and oil film coefficient variation with modified Sommerfeld number Se

2.5 Turbulent Oil Film

33

Fig. 2.9 Variation of eccentricity ratio with modified Sommerfeld number S 

1.0 Re measured 11 000 4 500

eccentricity ratio ε

0.8 0.6 0.4 0.2 0 0.01

0.1

S'

1.0

10

Cd / 2 b d/4 d

hmin

0.8b

W ( a ) configuration

( b ) oil groove dimensions

Fig. 2.10 Configuration of cylindrical bearing

S  = S(1 + 0.00133Re0.9 )

(2.30)

2.6 Actual Plain Bearings 2.6.1 Various Plain Bearings The cylindrical bearings mentioned so far are assumed to have had a fully cylindrical inner surface for simplified theoretical calculation purposes. As shown in Fig. 2.10, actual plain bearings have inner surfaces with oil holes and/or oil grooves manufactured, without fully cylindrical inner surfaces. Actual cylindrical bearings used in rotating machinery are normally installed by assembling two horizontally split pads (upper half pad and lower half pad). An axial oil groove is manufactured on each of the two mating faces of the two pads. Such cylindrical bearings with two axial oil grooves are standard for rotating machinery. In addition, Fig. 2.11a–d shows various multi-lobe bearings consisting of some fixed pads. In contrast with cylindrical bearings, these bearings do not form circum-

34

2 Basics of Plain Bearings

Ω

Ω

Ω Oj

Oj

Oj

W

W

W

( a ) two-lobe

( c ) four-lobe

( b ) three-lobe pivot pad

Ω Oj W ( d ) offset Oj journal center bearing center

Ω

y

y Oj

Ω

W

Oj W

x ( e ) 4-pad, LBP

x ( f ) 5-pad, LOP

( e, f ) tilting pad bearings

Fig. 2.11 Configurations of various journal bearings

ferentially uniform clearances when a journal center is placed at the bearing center, because each pad arc center of the multi-pad bearings is normally placed at an offset position so that each pad is placed somewhat closer to the bearing center. This type of pad placement is called geometrical preload, or simply preload (Eq. 2.31). Thus, each individual pad of a multi-pad bearing can be operated at an eccentricity ratio larger than that of cylindrical bearings under the same operating condition, resulting in static and dynamic oil film properties different from those of cylindrical bearings. In addition, Fig. 2.11e and f shows tilting pad bearings. Each pad is not fixed, but supported on the back surface by a pivot that enables each pad to tilt. Each pad is normally preloaded similarly to multi-pad bearings. As described in Chap. 4, cylindrical bearings have the lowest stability limit for oil whip, that is, a self-excited vibration of the rotor supported in oil film bearings, compared to other types of plain bearings. As a countermeasure, cylindrical bearings are often replaced by multi-pad bearings or tilting pad bearings of the type shown in Fig. 2.11. These bearings, with clearances fundamentally different from that of a cylindrical bearing, have different mechanical properties of the oil film, resulting in higher stability limits for oil whip than that for a cylindrical bearing. Consequently, these bearings are used widely in high-speed rotating machinery in spite of their higher manufacturing cost. In particular, tilting pad bearings have no cross-coupled oil film stiffness coefficients, as shown in Fig. 2.12 [B13, pp 83–178], resulting in the highest stability margin against oil whip.

2.6 Actual Plain Bearings

35

dimensionless stiffness coeff. K ij

100 50

10 5

Kxx = Kyy (1) (2)

1 K yy

0.5

(1) 4-pad, LBP, L/D = 0.5, m p = 0.5, center pivot (2) 5-pad, LOP, L/D = 0.5, m p = 0.5, center pivot 0.001 100

dimensionless damping coeff. C ij

K xy = K yx = 0

K xx

0.01

0.1

1 C xy = C yx = 0

50

10 5

C xx C xx = C yy

(1) (2)

1 0.5

0.001

10

C yy

0.01

(1) 4-pad, LBP, L/D = 0.5, m p = 0.5, center pivot (2) 5-pad, LOP, L/D = 0.5, m p = 0.5, center pivot 0.1 Sommerfeld number S

1

10

Fig. 2.12 Oil film coefficients of tilting pad bearings

2.6.2 Details of Plain Bearing Specifications Basic design variables of plain bearings are bearing diameter, bearing width and mean bearing clearance. Pad preload factor and number of pad are included for multi-pad or tilting pad bearing designs. Furthermore, load direction, pivot configuration, and pivot position are also included in the case of tilting pad bearing designs. Here, Eq. (2.31) defines the pad preload factor, mp , with C p and C b shown in Fig. 2.13: mp = 1 −

Cb Cp

(2.31)

Figure 2.13a and b illustrates how to manufacture and assemble a two-lobe bearing with a preload factor mp . The upper pad and the lower pad are fixed in one piece with a thin plate of uniform thickness sandwiched on each mating face. Then, the cylindrical bore of a specified machined radial clearance, C p , is machined. When the

36

2 Basics of Plain Bearings Cb

Cp

D

shim thickness

circular C b arc r

r

R

bearing

R1

R2

r

R =2

pivot (line contact)

pad

journal carrier ring (a) manufactured

(b) assembled

journal

(c) tilting pad bearing (rolling pivot)

Fig. 2.13 Primary dimensions of journal bearings

two shims are removed, a two-lobe bearing with an assembled radial clearance, C b , is obtained (C b < C p ). Figure 2.13c shows a five-pad tilting pad journal bearing. A ring of outer radius R2 is split into a specified number of pad elements. Those elements, placed and fixed on the inner surface of a cylindrical jig, are machined into pads with inner surface of radius R1 . The inner radius of the carrier ring of the assembled tilting pad bearing is smaller than that of the jig. Consequently, the machined pads placed in the carrier ring are in the preloaded condition. Normally, the number of pads is 3 for a small bearing, 4–6 for a medium-to-large bearing, and 12 or more for a guide bearing used in very large vertical rotors, as in hydro power turbine generator sets. There are two types of pad arrangement (load direction) for bearings applied to horizontal rotors. Figure 2.11e shows Load Between Pads (LBP), which means that the static bearing load vector is directed from the bearing center to the direction of the mid angle between two adjacent pivots, while Fig. 2.11f shows Load On Pad (LOP), which means that the static bearing load vector is directed from the bearing center to the direction of a pivot position. In the case of a four-pad tilting pad bearing with a LBP arrangement shown in Fig. 2.12, the two direct stiffness coefficients, K xx and K yy , are always equal each other. The two viscous damping coefficients, C xx and C yy , are also equal. The isotropic oil film coefficients are characteristics of a bearing with a LBP arrangement. On the other hand, anisotropic oil film coefficients are characteristics of a five-pad tilting pad journal bearing (LOP) as shown in the same figure. There are several types of pivot configuration. A rolling pivot enables a pad to roll slightly on the inner surface of a cylindrical carrier ring, as in a rocking chair, resulting in circumferential tilt motion of the pad when the radius of the pad back surface is smaller than that of the inner radius of the carrier ring. A key pivot with a rectangular section is manufactured in the axial direction on the pad back surface. Pads can tilt only circumferentially (pitching motion against main shear flow of the oil film). These pivots are suitable for rotating machinery of small-to-medium size.

2.6 Actual Plain Bearings

37

A ball and flat pivot consists of a spherical pivot against a flat plate, while a ball and socket pivot consists of a spherical pivot in a spherical concave. These two pivots enable pads to tilt in the both circumferential and axial directions (pitching and rolling motion). These two types of pivot are suitable for large rotating machinery that need secure bearing alignment in the axial direction of the shaft. A center pivot is placed at the circumferential mid-span of a pad (50%), while an offset pivot is that placed somewhat toward the trailing edge from the circumferential mid-span position (about 60%). Tilting pad bearings with offset pivots can achieve larger load-carrying capacity and lower temperature of the pad surface at the cost of higher unbalance response of a rotor, compared with the same bearing with center pivots. In the case of reverse rotation of shaft, center pivot bearings can maintain the same performance, while offset pivot bearings show higher pad surface temperature. Steel backings are normally lined with soft tin-based alloy or aluminum alloy, to form the pad surface. Most recently, sophisticated plastic surface pads are in practical use because of its higher strength, higher thermal resistance and higher creep resistance. For actual oil film bearings, the oil feed method, oil-supply temperature, and supply oil flow rate are selected so that pad surface temperature does not exceed a prescribed upper limit, for example, 110 °C for tin-based alloy. Bearings are often operated, so that oil temperature measured at discharge ports does not exceed 75 or 80 °C. The minimum oil film thickness calculated from the journal eccentricity ratio in operation should not fall below a prescribed lower limit that is often set to be more than three times the surface roughness of the journal and bearing pad surfaces. The minimum oil film thickness in operation should be checked carefully in the low-speed and high-load operation conditions. Oil film bearings need be designed so as to prevent or control various rotor vibrations satisfactorily, with tribological integrity being secured. Bearing tribology is explained in detail in various books [B28, B29, B44, B60]. References [7–22] present recent results of the static and dynamic properties of various journal bearings. Example 2.1 Calculate oil film coefficients for the following conditions from Fig. 2.7: Rotor mass m = 51.8 kg, Journal diameter D = 40 mm, L/D = 1.0, C/R = 0.003, Bearing load W = half of rotor weight, Oil viscosity μ = 28 cP, Shaft rotational speed N = 50, 90 rps. Answer In the case of N = 50 rps, Sommerfeld number S=

28 × 10−3 (Pa s) 50 (s−1 ) 0.04 (m) 0.6 × 0.04 (m) = 0.588 51.8/2 × 9.8 (N) × 0.0032

⇒ S(L/D)2 = 0.21 ⇒ ε0 = 0.35 (see Fig. 2.6)

38

2 Basics of Plain Bearings

⇒ K xx = 2.013, C xx = 7.43 (see Eq. (2.28) or Fig. 2.7) 25.9 × 9.8 = 851 × 104 (N/m) 0.003 × 0.02 25.9 × 9.8 = 10 × 104 (N s/m) = 7.43 0.003 × 0.02 × 2π 50

⇒ k x x = 2.013 ⇒ cx x

The following results are obtained when similar calculations are undertaken:

rps 50 90

kx x 851 669

kx y 1772 2188

k yy 1000 1042

k yx −848 −1572

cx x 10.0 7.2

cx y 3.2 1.8

c yy 6.7 6.0

c yx 3.2 1.8

C/R = 0.003 ×104 ×104

Example 2.2 Calculate oil film coefficients for the following conditions from Fig. 2.7: C/R = 0.001, Oil viscosity μ = 27 cP The other variables are assumed to be same as shown in Example 2.1. Answer

rps 50 90

kx x 1637 1622

kx y 23, 449 41, 680

k yy 3225 3230

k yx −23, 015 −41, 438

cx x 149 147

cx y 10 6

c yy 147 147

c yx 10 6

C/R = 0.001 ×104 ×104

Example 2.3 Calculate oil film coefficients for the following conditions from Fig. 2.7: C/R = 0.01, Oil viscosity μ = 45 cP. The other variables are assumed to be same as shown in Example 2.1. Answer

rps 50 90

kx x 748 539

kx y 583 533

k yy 248 262

k yx 28 −25

cx x 2.9 1.5

cx y 0.8 0.5

c yy 0.6 0.5

c yx 0.8 0.5

C/R = 0.010 ×104 ×104

Example 2.4 Estimate the journal eccentricity ratio at equilibrium for a case of turbulent lubrication from Fig. 2.8a and Eq. (2.29) with the following operating conditions:

2.6 Actual Plain Bearings

39

Journal diameter D = 600 mm, L/D = 1.0, 2C/D = 0.0025, Specific bearing load pm = 2.0 MPa, Oil viscosity μ = 20 mPa s (Kinematic viscosity ν = 24×10−6 m2 /s), Shaft rotational speed N = 60 rps. Answer The Sommerfeld number, S, is calculated from Eq. (2.20) as follows: S=

    μN DL R 2 μN R 2 20 × 10−3 × 60 = = = 0.096 W C pm C 2.0 × 106 × 2.52 × 10−6

Figure 2.8a gives the laminar eccentricity ratio over 0.9 corresponding to Se(S) = 0.096 when the Sommerfeld number is not modified with the Reynolds number, Re. The Reynolds number, Re, is calculated as follows: Re =

0.6 × π × 60 × 0.3 × 2.5 × 10−3 UC = = 3533 ν 24 × 10−6

Then, the bearing is found to operate in the turbulent lubrication region, with Re greater than 3000, and the modified Sommerfeld number, Se, is obtained from Eq. (2.29) as follows:   Se = S 1 + 0.0007768Re0.9 = 0.096 × (1 + 1.21) = 0.212 Then, Fig. 2.8a gives the turbulent eccentricity ratio of 0.75, which is much lower than the laminar eccentricity ratio because of the higher load-carrying capacity resulting from the high effective viscosity of the turbulent oil film. Example 2.5 Estimate the journal eccentricity ratios at equilibrium for the case of turbulent lubrication with the same operating conditions as Example 2.4, except with the lower oil viscosity μ = 10 mPa s (Kinematic viscosity ν = 12 × 10−6 m2 /s). Answer

S=

    μN DL R 2 μN R 2 10 × 10−3 × 60 = = = 0.048 W C pm C 2.0 × 106 × 2.52 × 10−6 0.6 × π × 60 × 0.3 × 2.5 × 10−3 UC = = 7066 ν 12 × 10−6   Se = S 1 + 0.0007768Re0.9 = 0.048 × (1 + 2.26) = 0.157 Re =

Then, Fig. 2.8a gives the laminar eccentricity ratio to be greater than 1.0 and the turbulent eccentricity ratio to be 0.8. Both eccentricity ratios are higher than those corresponding with Example 2.4 because of the lower oil viscosity of 10 mPa s. On the other hand, the difference of the laminar and turbulent eccentricity ratios at 0.2

40

2 Basics of Plain Bearings

for this case is found to be larger than that of Example 2.4 because of the higher Reynolds number. Example 2.6 Estimate K xx and K yy at equilibrium from Fig. 2.12 (Case (2), five-pad tilting pad bearing LOP) for the following operating conditions: C/R = 0.0025, Specific bearing load pm = 2.5 MPa, Oil viscosity μ = 15 mPa s, Shaft rotational speed N = 75 rps. Answer The Sommerfeld number, S, is calculated from Eq. (2.20) as follows:     μN DL R 2 μN R 2 15 × 10−3 × 75 S= = = = 0.072 W C pm C 2.5 × 106 × 2.52 × 10−6 K xx and K yy at equilibrium obtained from Fig. 2.12 (Case (2)) are 9 and 1.2, respectively, showing strong anisotropy of the two stiffness coefficients at the lower Sommerfeld number. Example 2.7 Estimate K xx and K yy at equilibrium from Fig. 2.12 (Case (2)) for the following operating conditions: C/R = 0.0020, Specific bearing load pm = 1.0 MPa, Oil viscosity μ = 20 mPa s, Shaft rotational speed N = 200 rps. Answer The Sommerfeld number, S, is calculated from Eq. (2.20) as follows: S=

    μN DL R 2 μN R 2 20 × 10−3 × 200 = = = 1.0 W C pm C 1 × 106 × 22 × 10−6

Both of K xx and K yy at equilibrium obtained from Fig. 2.12 (Case (2)) are 11, showing isotropic stiffness coefficients at the higher Sommerfeld number.

Chapter 3

Unbalance Vibration of a Rotor in Plain Bearings

Abstract Oil-lubricated plain bearings used widely for large-sized rotating machinery can effectively suppress unbalance vibration amplitude of a rotor. This chapter outlines the nature of unbalance vibration amplitude of a rotor supported in plain bearings. The mathematical expression of unbalance vibration of a rotor is given in the rotating plane of the shaft. The two equations of motion given in the x–y-plane are converted into a single equation of motion derived with the complex expression of the displacement of the rotor. Then, the rotor is found to make an elliptical whirl orbit with major and minor axes around its equilibrium. The eight rotordynamic coefficients of a bearing oil film can be converted into the four effective coefficients expressed with the two coordinates of the major and minor axes. The vibration response and the whirl orbit are calculated and shown with varying shaft speed. The effects of bearing type and design variables on the unbalance vibration response are also calculated and compared. Furthermore, the effect of bearing pedestal characteristics on the unbalance vibration response is shown. Finally, it is shown how rotor-bearing systems can be designed to reduce the maximum amplitude of unbalance vibration when passing critical speeds. Keywords Unbalance vibration · Elliptical whirl orbit · Vibration amplitude variation with shaft speed · Effect of bearing type · Effect of bearing pedestal

3.1 Feature of Unbalance Vibration The stiffness and viscous damping of oil films in plain bearings are characterized as follows: (1) Higher damping capability than that of rolling element bearings. (2) Both stiffness coefficients and viscous damping coefficients have cross-coupled terms in most plain bearings. (3) Both stiffness coefficients and viscous damping coefficients are anisotropic with respect to the vertical direction (X) and the horizontal direction (Y) in most plain bearings, depending on the magnitude and direction of the static bearing load. © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_3

41

42

3 Unbalance Vibration of a Rotor in Plain Bearings

Consequently, rotors supported in plain bearings have the following vibration characteristics: (4) When rotors and bearings are properly designed, the magnification factor at critical speeds can be kept small, enabling passage through critical speeds safely and operation near critical speeds or even at critical speeds. (5) Because of the X–Y anisotropy of oil film coefficients, rotor-bearing systems have different critical speeds with respect to the vertical and horizontal directions. (6) Rotor-bearing systems may exhibit decreased damping ratios at high shaft speeds due to the increasing magnitude of the cross-coupled stiffness coefficients, eventually resulting in oil whip (Chap. 4). (7) Peak amplitudes of unbalance vibration of rotor at critical speeds can be suppressed well because of the high damping capability that can be attained if plain bearings are designed to align with the vibration characteristics of the rotor. (8) If bearings are supported by pedestals with low stiffness, oil films in plain bearings may not produce a sufficient oil film damping effect, resulting in excessive vibration amplitude of a rotor.

3.2 Mathematical Expression for Unbalance Vibration Vibration of a rotor in rotating machinery is indicative of whirling of the rotor around its equilibrium position in the rotating plane of the shaft where the horizontal coordinate x and the vertical coordinate y are defined (the y-axis direction is 90° from the x-axis in the direction of shaft rotation). Then, the whirl amplitude consists of two components, x(t) and y(t), where t is time. Alternatively, when the x–y-plane is regarded as an Argand diagram complex plane, the complex whirl amplitude, z(t), can be defined as z(t)  x(t) + j y(t)

(3.1)

where j corresponds to the imaginary unit. This section gives two different expressions of equation of motion and whirling amplitude of a rotor.

3.2.1 Expression with X–Y Coordinates The equation of motion for the generalized rotor system with unbalance mass (Fig. 3.1) can be expressed in the x–y coordinates as follows: M x¨ + G y˙ + K x x x + K x y y + C x x x˙ + C x y y˙  U 2 cos(t + θ ) M y¨ − G x˙ + K yx x + K yy y + C yx x˙ + C yy y˙  U 2 sin(t + θ )

 (3.2)

3.2 Mathematical Expression for Unbalance Vibration

43

Fig. 3.1 Rotor motion with unbalance

y(t ) Unbalance U=m ε

G y

O

θ

ε

S

Ωt

Displacement x(t )

x Pulse

where M is the mass matrix, G is the gyroscopic matrix, K ij is the stiffness matrix, C ij is the damping matrix, (i, j  x, y),  is the rotational speed of shaft, U is the unbalance mass-eccentricity, and θ is the phase angle of the unbalance (angle from rotor pulse position). Equation (3.3) represents the solutions of Eq. (3.2): x  x R cos t − x I sin t y  y R cos t − y I sin t

 (3.3)

where x R corresponds to the amplitude of the cosine component in the x direction, while x I , to that of the sine component also in the x direction. Similarly, y R and y I , are defined in the y direction. Equation (3.3) may be substituted into Eq. (3.2), then the coefficients of each of cos t and sin t on both sides can be equated, respectively. This yields the algebraic matrix equation ⎤⎡ ⎤ ⎡ ⎤ −C x x Kxy −G2 − C x y xR Ux −M2 + K x x ⎥⎢ x I ⎥ ⎢ U y ⎥ 2 ⎢ C x x −M2 + K x x G2 + C x y Kxy ⎥⎢ ⎥  ⎢ ⎥ ⎢ 2 2 ⎦⎣ y R ⎦ ⎣ U y ⎦ ⎣ K yx G − C yx −M + K yy −C yy − G2 + C yx K yx C yy −M2 + K yy yI −Ux (3.4) ⎡

where Ux ≡ U cos θ, U y ≡ U sin θ . Equation (3.4) may be solved for x R , x I , y R and y I . Then, x and y in Eq. (3.3) are expressed as follows: jϕx jt x  x R cos t − x I sin t  ax cos(t + ϕx )  Re [ax e jϕ e jt ] y  y R cos t − y I sin t  a y cos t + ϕ y  Re [a y e y e ]

 (3.5)

44

3 Unbalance Vibration of a Rotor in Plain Bearings

where ax represents the vibration amplitude of the x component, while ϕ x is the phase difference from the pulse signal (Fig. 3.2). Then, the complex vibration amplitude can be expressed as follows: A x  x R + j x I  ax e jϕx A y  y R + j y I  a y e jϕ y

 (3.6)

In this case, the whirl orbit of the unbalance vibration is elliptical (Fig. 3.3). The whirl amplitude can then be expressed with two components, X in the major axis direction and Y in the minor axis direction, as X  a cos(t − α0 + φ0 ) Y  b sin(t − α0 + φ0 )

 (3.7)

where 2a is the major axis length (a is major radius), 2b is the minor axis length (b is minor radius), and α 0 is the inclination angle of the major axis from the x-axis direction. From Appendix B, a, b, a0 , and the phase difference φ0 in the X–Y coordinates may be expressed as ϕx

ax x (t ) ϕy

Ax = ax e jϕx Ay = ay e jϕy

ay y(t )

pulse

Fig. 3.2 Unbalance vibration of rotor Fig. 3.3 Coordinates of elliptical orbit

Y X 2a φ0

y

α0

Ωt

2b

x

xr

3.2 Mathematical Expression for Unbalance Vibration



⎡ a

2 ⎣ xR

+

x I2

+ 2

y R2

+

y I2

+



x R2 + x I2 − y R2 − y I2 4

45

⎤1/2

2

+ (x R y R + x I y I )2 ⎦

b  (x I y R − x R y I )/a

2(x R y R + x I y I ) 1 α0  tan−1 2 2 x R + x I2 − y R2 − y I2

(3.8) (3.9) (3.10)

φ0  tan−1 [(x I + y R )/(x R − y I )]

(3.11)

A positive value of b corresponds to forward whirl, while a negative value corresponds to backward whirl. Equation (3.6) can be solved directly for the complex vibration amplitudes. In this case, the unbalance vibration amplitude solutions for Eq. (3.2) correspond to the real part of the complex solution:     x  Re A x e j t y  Re A y e j t

(3.12)

Similarly, the unbalance forces on the right-hand side of Eq. (3.2) correspond to the real parts of the complex forces:

 

 jθ 

1 U cos( t+θ) ReU e  jθ  Re  U e Im U e jθ U sin( t+θ ) −j

(3.13)

Substituting these two equations into Eq. (3.2), the following equation is obtained:

−M2 + jC x x + K x x j2 G + jC x y + K x y − j2 G + jC yx + K yx −M2 + jC yy + K yy



Ax Ay



1  U 2 e jθ −j (3.14)

Note: 1 Complex eigenvalues expressed in the x and y directions Solutions for free vibration are assumed to be expressible as     x  Re φx est y  Re φ y est

(3.15)

Substituting Eq. (3.15) into Eq. (3.2) yields the matrix-vector relation

Ms 2 + C x x s + K x x Gs + C x y s + K x y −Gs + C yx s + K yx Ms 2 + C yy s + K yy



φx φy

0

(3.16)

When the determinant of the coefficient matrix is set to zero, the characteristic equation is derived and solved for the complex eigenvalues or roots s.

46

3 Unbalance Vibration of a Rotor in Plain Bearings

3.2.2 Complex Displacement Expression z  x + jy1 The equation of motion for the rotor system in Fig. 3.1 can be expressed as M z¨ − jG z˙ + K f z + C f z˙ + K b z¯ + Cb z˙¯  U 2 e j(t+θ)

(3.17)

where

K x x + K yy − Kf  2 K x x − K yy Kb  + 2 C x x + C yy − Cf  2 C x x − C yy + Cb  2

K x y − K yx ⎫ ⎪ j ⎬ 2 K x y + K yx ⎪ ⎭ j 2 C x y − C yx ⎫ ⎪ j ⎬ 2 C x y + C yx ⎪ ⎭ j 2

(3.18a)

(3.18b)

The solutions of the unbalance vibration amplitudes can be expressed with the forward and backward complex whirl amplitudes, Af and Ab , respectively, as z  A f e jt + Ab e− jt

(3.19)

Substituting this equation into Eq. (3.17), the coefficients of e j t on both sides must be equal. Similarly, those of e− j t need to be equal. Then, the complex vibration amplitudes can be obtained by solving the matrix-vector equation

K b + jCb −(M − G)2 + K f + jC f K b + jCb −(M + G)2 + K f + jC f



Af Ab



U 2 e jθ 0 (3.20)

Complex vibration amplitudes follow as A f  a f e jϕ f

Ab  ab e jϕb

(3.21)

Then, the elliptical whirl orbit z can be expressed as follows (R1_Fig. 7.10):   z  e jα0 a f e j(t+θ) + ab e− j(t+θ)

(3.22)

ϕ f +ϕb ϕ f − ϕb , θ , a is the major radius, 2b is the minor radius, and where α0  2 2 α 0 is the inclination angle of the major axis from the x axis direction.

1 Details

in R1_Chap. 7.

3.2 Mathematical Expression for Unbalance Vibration

47

The following expressions may thus be obtained (see similar Eqs. (3.8–3.11)): major radius a  a f + ab

(3.23)

  minor radius b  a f − ab 

(3.24)

inclination angle α0 

ϕ f − ϕb 2

(3.25)

phase difference φ0  θ + α0  ϕ f

(3.26)

The case of af > ab corresponds to forward whirl, that of af < ab to backward whirl, and that of af  ab , to straight line translational oscillation. When the dynamic properties of oil film are isotropic, K x x  K yy ≡ K d , K x y  −K yx ≡ K c ⇒ K f  K d − j K c , K b  0 (3.27) C x x  C yy ≡ Cd , C x y  −C yx ≡ Cc ⇒ C f  Cd − jCc , Cb  0 Consequently, the backward vibration amplitude, Ab , in Eq. (3.20) becomes equal to zero and the forward amplitude, Af , may be obtained from the scalar equation   −(M − G)2 + K f + jC f A f  U 2 e jθ ,

Ab  0

(3.28)

Note: 2 Complex eigenvalues in the case of complex displacement expression z = x + jy Solutions for free vibration are assumed to be expressible as z  φ f est + φ¯ b es¯t

(3.29)

for which

Ms 2 − j Gs + K f + C f s K b + Cb s K b + Cb s Ms 2 + jGs + K f + C f s



φf φb

0

(3.30)

When the determinant of the coefficient matrix is set to be zero, the characteristic equation follows. Note: 3 Relationship of two pairs of vibration amplitude expressions The following relationship is derived from R1_Eq. (6.51): A x  A f + Ab A y 

A f − Ab Ax + j A y ⇔ Af  j 2

Ab 

Ax − j A y 2

(3.31)

Note: 4 Relationship of two dynamic stiffness expressions The reciprocating exciting forces in the X and Y directions are expressed as Fx (t)  Re[Fx e jωt ],

Fy (t)  Re[Fy e jωt ]

(3.32)

48

3 Unbalance Vibration of a Rotor in Plain Bearings

The complex exciting forces, F x and F y , can be expressed with the magnitudes and phases, as in Eq. (3.13). The dynamic stiffnesses, S ij (i, j  x, y), may be used to determine the complex response amplitudes, Ax and Ay , caused by F x and F y through

Sx x Sx y S yx S yy



Ax Ay



Fx Fy

(3.33)

This equation is expressed in a generic form, and Eq. (3.14) is a particular example. Alternatively, the exciting forces can be expressed on the complex whirling plane: Fz (t)  F f e jωt + Fb e− jωt

(3.34)

Corresponding complex amplitudes, Af and Ab (Eq. (3.19)), can be derived from

Sf f Sfb Sb f Sbb



Af Ab



Ff Fb

(3.35)

A particular form of this equation is represented by Eq. (3.20). The following equations give the relationship of the dynamic stiffnesses expressed in the x–y coordinates and those expressed in the complex plane:

Sf f Sb f

Sx y − S yx Sx y + S yx Sx x + S yy Sx x − S yy −j , Sfb  +j  2 2 2 2 Sx y + S yx Sx y − S yx Sx x − S yy Sx x + S yy −j , Sbb  +j  2 2 2 2

⎫ ⎪ ⎬ ⎪ ⎭ (3.36)

3.3 Simplified Expressions for the Dynamic Properties of an Oil Film with Unbalance Vibration When a journal moves in the x or y direction in a bearing oil film, corresponding reaction forces f x and f y are generated. When the journal movement is assumed to be a small perturbation around the equilibrium position, f x and f y can be linearized and expressed with four stiffness coefficients, k ij , and four viscous damping coefficients, cij , (i, j  x, y). Furthermore, inertia coefficients may be added, for example, in the case of water-lubricated bearings, to form the dynamic matrix-vector equation

mxx mxy m yx m yy





x¨ x˙ x cx x cx y kx x kx y fx + +  c yx c yy k yx k yy fy y¨ y˙ y

(3.37)

The coefficients in Eq. (3.37) are determined depending on the operating conditions of the bearing. The coefficients vary with Sommerfeld number, as shown in Fig. 2.7. Alternatively, Fig. 3.4 shows the variation with journal eccentricity ratio. However, eight coefficients are too many to understand the mechanical properties

3.3 Simplified Expressions for the Dynamic Properties …

49

100

m

Kxy Cyx = Cxy

S=

Cyy Kyy

1

1 Kyx < 0

ε− S

Kyx > 0 0.1

0.1

Sommerfeld number

8 dimensionless oil film coefficients

10

μ N ( R / C )2 p

Kxx

Cxx 10

Ω y x

0

0.2

0.4 0.6 Eccentricity ratio ε

0.8

0.01 1.0

Fig. 3.4 Oil film coefficients of a 160° partial arc bearing (L/D  0.75)

of an oil film in an intuitive and easy manner. Consequently, simplified expressions involving the coefficients are discussed. The approach is based on the experimental results given by Hagg and Sankey [23] who enabled incorporation of the dynamic properties of an oil film for the first time in the design of rotor-bearing systems. They derived equivalent stiffness coefficients and equivalent viscous damping coefficients from the measured unbalance vibration response of rotor supported in plain bearings and excited with a known unbalance mass. These equivalent oil film coefficients [24] are now explained theoretically. Equation (3.37) can be interpreted to express the journal vibration response caused by given force components, f x and f y , when these are sinusoidally varying inputs of forward rotating unbalance applied to journal. Then, the equation may be solved for the vibration response amplitudes of the journal in the major axis X direction and also in the minor axis Y direction of its elliptical orbit (Fig. 3.3). Then, two equivalent stiffness coefficients and two equivalent viscous damping coefficients may be obtained in the X and Y directions, respectively. These four coefficients can be used to calculate critical speeds and estimate the vibration response amplitudes in an approximate manner. The equivalent coefficients can be obtained by performing a number of steps. As the frequency of the excitation forces is equal to the shaft rotating frequency, , the forces and the responses can be written as

50

3 Unbalance Vibration of a Rotor in Plain Bearings

f x  Fx e jt ,

f y  Fy e jt ⇒ x  A x e jt ,

y  A y e jt

(3.38)

For simplicity, the inertia matrix of the oil film coefficients is assumed to be negligibly small in Eq. (3.33). Then, the matrix-vector equation reduces to 

kx x kx y k yx k yy



 cx x cx y Ax Fx + j  c yx c yy Ay Fy

(3.39)

Then, both sides of the equation are multiplied by C/W (the radial clearance divided by the bearing load) to make the coefficients dimensionless, as shown in Eq. (2.28), resulting in

K x x + jC x x K x y + jC x y K yx + jC yx K yy + jC yy



Ax Ay



C W



Fx Fy

(3.40)

If CF x /W is assumed to be equal to unity, CF y /W must be equal to −j in Eq. (3.40), in other words, unity force excitation is applied to the journal in the x and y directions, respectively. Then, Eq. (3.40) is solved for vibration response amplitudes Ax and Ay . The dimensionless dynamic stiffness coefficients of the oil film, S x and S y , are obtained in the x and y directions, respectively, as Sx 

1 [m] 1 , Sy  [−] A x [m] Ay

(3.41)

However, as shown in Fig. 3.3, the x–y coordinate system does not match up precisely with the X–Y coordinate system defined on the elliptical whirl orbit plane because of the cross-coupled effects of the oil film, and it is not easy to understand the oil film action in a straightforward manner. Consequently, four equivalent dimensionless oil film coefficients K max , K min , C max, and C min are defined in the X–Y coordinate system. K max and C max correspond to the minor Y axis, while K min and C min correspond to the major X axis. When A x  x R + j x I and A y  y R + j y I obtained from Eq. (3.39) are substituted into Eqs. (3.8–3.11), a, b, φ0 , α0 are determined. Then, K max , K min , C max, and C min are obtained as K max  1 [m] × cos φ0 /b [m], Cmax  1 × sin φ0 /b [−] Cmin  1 × sin φ0 /a [−] K min  1 × cos φ0 /a,

 (3.42)

The vibration amplitudes and phase angles calculated with these four equivalent coefficients are identical to those obtained with the original eight coefficients. For example, Fig. 3.5 shows the variation of the four equivalent dimensionless coefficients with journal eccentricity ratio for the 160° partial arc bearing in Fig. 3.4. Figure 3.5 shows clearly that K max increases, but C max decreases with increase in

3.3 Simplified Expressions for the Dynamic Properties … 4 dimensionless effective oil film coefficients for forward circular orbit excitation

Fig. 3.5 Effective oil film coefficients of a 160° partial arc bearing (L/D  0.75)

51

10

C max

K max

1

K min C min

0.1 0

0.2

0.4

0.6

0.8

1.0

Eccentricity ratio ε

journal eccentricity ratio, which would result in larger vibration amplitudes. It is easy to derive design guidelines and to avoid operating the bearing at excessively large journal eccentricity ratios. Example 3.1 Figure 3.4 gives the oil film coefficients at journal eccentricity ratio ε  0.6 as follows: ε 0.6

Kxx 3.39

Kxy 3.6

K yy 1.59

K yx − 0.35

Cx x 7

Cx y 1.74

C yy 1.59

C yx 1.78

Calculate the equivalent K max , K min , C max, and C min . Answer Equation (3.40) is expressed as follows and solved for Ax and Ay :

K x x + jC x x K x y + jC x y K yx + jC yx K yy + jC yy



Ax Ay



1 −j

⇒

Ax Ay



0.478 − j0.048 −0.561 − j0.613 (3.43)

Thus, x R , x I , y R , and y I in Eq. (3.6) are now determined. Substituting these values into Eqs. (3.8–3.11) results in the major radius a  0.89, the minor radius b  0.36, the angle of major axis α0  −67◦ , and the phase delay φ0  −29◦ . The resultant of the spring and damping forces of the oil film balances with the unity excitation forces in the X or Y direction, respectively, as shown in Fig. 3.6. The following equation is derived: 1 [m] × e jφ  (K + jC)min × a [m]  (K + jC)max × b

52

3 Unbalance Vibration of a Rotor in Plain Bearings

Fig. 3.6 Force balance in the major and minor axis directions

1.0 2 Minor axis direction C max K max 1 φ0

×ε

φ0 = −29°

A

B

1

C min

K min 1 Major axis direction y 1.0 × ε

−1.0 α 0 = −67°

0.5 x

Consequently, K max  2.43,

K min  0.98

Cmax  1.36, Cmin  0.54 Another answer When the oil film properties at a journal eccentricity ratio ε  0.6 in Fig. 3.4 are expressed in the forward or backward whirling modes, respectively, Eqs. (3.18a, b) give the following results: ε 0.6

Kf 2.49 − j1.98

Kb 0.9 + j1.63

Cf 4.295 + j0.02

Cb 2.705 + j1.76

When the mass and gyroscopic terms are dropped in Eq. (3.20), the following equation is obtained:

K f + jC f K b + jCb K b + jCb K f + jC f



Af Ab





  1 0.625 − 29◦ Af  (3.44) ⇒ Ab 0.266 105◦ 0

Then, the same values of a, b, φ0 , α0 are obtained, that is, the major radius a  0.89, the minor radius b  0.36, the angle of major axis α0  −67◦ , and the phase delay φ0  −29◦ , giving in the same results. The calculation process thereafter is omitted.

3.4 Effects of Bearing Type and Design Variables This section discusses unbalance vibration responses of a Jeffcott rotor (Table 3.1) supported in various types of plain bearing. The dynamic properties of those bearings are obtained from reference [6].

3.4 Effects of Bearing Type and Design Variables

53

Table 3.1 Basic calculation model Shaft Rotor mass

m [kg]

Shaft bending stiffness

70 k s [N/m]

Shaft elasticity index α = (mg/C)/k s [ - ] Note : L/D = 0.5

m

ks

φ 27 mm L264 mm

1.399 × 107 0.907

Plain bearings

mg/C =1.27 × 107 [N/m]

Dimensionless major vibration amplitude A / ε

100 50

F

F = Forward whirl

F

B = Backward whirl

F 10

F

Ω

5

1 0.5

0.1 0.2

B

0.4

0.6 0.8 Dimensionless shaft speed Ω / ω s

1.0

1.2

Fig. 3.7 Variation of unbalance response and whirl orbit with shaft speed for a two-lobe bearing

Figure 3.7 shows a typical variation of unbalance vibration response and whirl orbit with dimensionless shaft speed. With bearings of anisotropic oil film properties, the first peak response corresponds to horizontal resonance because the horizontal stiffness of the oil film is normally lower than the vertical stiffness. With further increase in shaft speed, the vertical resonance appears. These critical speeds and shapes of whirl orbit change, depending on the types of bearing. In the case of a four-pad tilting pad bearing (LBP) with inherently isotropic oil film properties, only one peak resonance appears. Figure 3.8 shows the effect of shaft stiffness (through the shaft elasticity index, α) on the unbalance response of a rotor supported in cylindrical bearings. The horizontal coordinate in Fig. 3.8a is the dimensionless shaft speed, /ωs (ωs is the pinned rotor natural frequency), while that in Fig. 3.8b is the shaft speed, . With decreasing shaft stiffness (increasing shaft elasticity index, α), the natural frequency decreases, resulting in increase in peak response. In contrast, highly rigid rotors show very low peak responses.

3 Unbalance Vibration of a Rotor in Plain Bearings

(b)

Dimensionless major vibration amplitude A / ε

(a)

Dimensionless major vibration amplitude A / ε

54 100 50

α = ( mg / C ) / ks

α= 4 1

10 5 0.25

1 0.5

0.1 0.2

0.4

0.6 0.8 Dimensionless shaft speed Ω / ω s

1.0

1.2

100

α = ( mg / C ) / ks

50

α=4

1

10 0.25

5 1 0.5

0.1 0

50

100

150

Shaft speed Ω [rps]

Fig. 3.8 Effect of shaft stiffness on unbalance vibration amplitude of rotor in a cylindrical bearing

Figure 3.9 shows the effect of bearing type on unbalance vibration response of rotor of α  0.907 as shown in Table 3.1. Figure 3.9a compares three different responses corresponding to a cylindrical bearing, a two-lobe bearing and a four-pad tilting pad bearing. In the case of the two-lobe bearing with strong anisotropy of oil film properties, two separate peaks are observed clearly. In the case of the four-pad tilting pad bearing (LBP), only one peak is observed. Figure 3.9b compares three different responses corresponding to a cylindrical bearing, and five-pad tilting pad bearings (LBP or LOP). In the case of the five-pad tilting pad bearing, the response changes with load direction. Figure 3.9c compares all five responses in the linear vertical coordinate, showing the differences in detail. In general, oil film stiffness increases with increasing preload factor of a bearing pad, resulting in a slight increase of the natural frequency and the peak amplitude. Thus, critical speeds and peak amplitudes of the same rotor vary significantly, depending on bearing type, load direction, and preload factor. Satisfactory response of a rotor can be attained by means of optimum design of the bearing. As shown in

Dimensionless major vibration amplitude A / ε

Dimensionless major vibration amplitude A / ε

Dimensionless major vibration amplitude A / ε

3.4 Effects of Bearing Type and Design Variables

55

100

α = 0.907

50

(2) Two-lobe (1) Cylindrical

(2) (1)

10 5

(3) 4-pad tilting LBP

1 0.5

0.1 0.2

100

0.4

0.6 0.8 1.0 Dimensionless shaft speed Ω / ω s (a) Comparison of unbalance response for cylindrical bearing, two-lobe bearing and 4-pad tilting pad bearing

1.2

α = 0.907

50

(5) 5-pad tilting LOP (1)

(1) Cylindrical

10 5

(4) 5-pad tilting LBP 1 0.5

0.1 0.2

0.4

0.6 0.8 1.0 Dimensionless shaft speed Ω / ω s (b) Comparison of unbalance response for cylindrical bearing and 5-pad tilting pad bearing (LOP, LBP)

20 (1) Cylindrical (2) Two-lobe (3) 4-pad tilting LBP (4) 5-pad tilting LBP (5) 5-pad tilting LOP

15

10

α = 0.907

(5) (2) (1)

1.2

(1)

(4) (3)

(2) 5 (5) 0 0.6

0.7

0.8 0.9 1.0 Dimensionless shaft speed Ω / ω s

1.1

(c) Comparison of unbalance response for cylindrical, two-lobe, 4-pad and 5-pad

Fig. 3.9 Unbalance vibration response of rotor in various plain bearings

56

3 Unbalance Vibration of a Rotor in Plain Bearings

Fig. 3.8, the most important design variable of rotor is the shaft elasticity index α, which needs be determined properly, based on R1_Fig. 7.15.

3.5 Effects of Bearing Pedestal Stiffness Figure 3.10 shows a rotor supported in plain bearings that in turn are supported by bearing pedestals. In particular, large-sized rotating machinery tends to have larger values of dimensional stiffnesses of shaft and bearing oil films, resulting in relative decrease in bearing pedestal stiffness, k p . Because the rotor, the bearing, and the pedestal are connected in series, any component of relatively low stiffness tends to dominate the total vibration response of the rotor-bearing system. The effect of pedestal stiffness can be summarized as follows: (1) In the case of higher k p , oil film properties are predominant, and the effect of pedestals is negligibly small. (2) In the case of k p with the magnitude comparable to the bearing stiffness, the effect of oil film properties is weakened. (3) In the case of lower k p , the damping effect of bearing oil film is impeded. Instead, the dynamic properties of the bearing pedestal (lower cp compared with bearing oil film) are much more predominant, resulting in excessive vibration response. Figure 3.11 shows the effect of bearing pedestal stiffness on the vibration response of the rotor supported in cylindrical bearings (Table 3.1). The damping force of a bearing pedestal structure made of concrete or steel is normally lower than that of a bearing oil film and is assumed to be 5% of pedestal stiffness force here. The calculation results show that peak response may increase excessively due to lower system damping with decreasing, k p . References [25–28] present recent results of unbalance vibration of rotating machinery.

Rotor ks

m

Plain bearing

kp

cp

Bearing pedestal

Fig. 3.10 Rotor-bearing model with bearing pedestal

Dimensionless major vibration amplitude A / ε

3.5 Effects of Bearing Pedestal Stiffness

57

100

α p = (mg/C) / kp 50 20

αp = 4 2

α =1 cp Ω = 0.05kp

1

0.2 0

10 5

2 0 0.2

0.4

0.6

0.8

1.0

Dimensionless shaft speed Ω / ω s

Fig. 3.11 Effect of bearing pedestal stiffness on unbalance vibration amplitude

1.2

Chapter 4

Stability of a Rotor in Plain Bearings

Abstract This chapter explains the self-excited nature of vibration of a rotor caused by the destabilizing oil film force (oil whip or oil whirl); another self-excited vibration of a rotor caused by the destabilizing fluid force in seals and impellers (flowexcited vibration); and how to prevent the vibrations effectively by selecting appropriate specifications of plain bearings. Both of the unstable vibrations show twodimensional whirl orbits of the rotor around the steady-state equilibrium position in the shaft rotating plane. The dominant component of the shaft whirl orbit is found to be the forward one, that is, orbiting in the same direction as the shaft rotation. The orbit size is sometimes enlarged when bending deformation of the rotating flexible shaft is added. In contrast with the unbalance vibration explained in Chap. 3, the whirl frequencies of the unstable vibrations are generally lower than the shaft rotating frequency, that is, subsynchronous, close to the natural frequency of the rotor-bearing system. These self-excited vibrations break out when the rotor-bearing systems exceed stability limits. When a linear vibration analysis is applied to the rotor-bearing system, the characteristic equation is derived in the form of an algebraic polynomial equation of the sixth degree from the equations of motion. When the Routh–Hurwitz criterion is applied to the characteristic equation, the stability limit of shaft speed can be obtained in the form of mathematical expression consisting of the rotordynamic coefficients of bearing oil film and the bending stiffness variable of shaft. The stability limit can be also found by means of eigenvalue analysis applied to the characteristic equation. In other words, when all the real parts of the eigenvalues have negative values for the given operating condition, the rotor-bearing system can remain stable. When at least one real part has a positive value, the system becomes unstable, starting oil whirl or oil whip. The effects of operating conditions and journal bearing configurations are demonstrated, and various effective countermeasures are derived. The flow-excited vibrations of a rotor by non-contact seals are characterized by dependence on load, because, with increase in load, the working fluid increases in pressure and flow rate, strengthening the destabilizing force (cross-coupled stiffness force) of the fluid flow. The magnitude of the destabilizing force is strongly dependent on the swirl velocity of the fluid in the direction of shaft rotation at the seal inlet. Consequently, one of the effective countermeasures against seal flow-excited vibration is to support the rotor by anisotropic bearings having oil film forces that

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_4

59

60

4 Stability of a Rotor in Plain Bearings

give rise to elliptical whirl orbits. This is because the backward whirl component of the elliptical whirl orbit reduces the effect of the destabilizing seal force. Keywords Self-excited vibration · Oil whip · Oil whirl · Cross-coupled stiffness coefficients · Stability limit · Stability chart · Flow-excited vibration · Non-contact seal · Swirl velocity · Forward and backward whirl orbits · Anisotropic oil film stiffness

4.1 Cause and Phenomena of Oil Whip 4.1.1 Cause of Oil Whip Oil film bearings can decrease unbalance vibration amplitudes of a rotor to a satisfactorily low level, as shown in Chap. 3. However, in the case of a high-speed rotor or a lightweight rotor, the journal eccentricity ratio is often small enough for a destabilizing oil film force component to become large and cause self-excited vibration. This unstable vibration of a rotor is called “oil whip.” The unique name originates from the experimental findings [29] reported for the first time in 1925. When a violent whirling of journal bearing supported rotor started in a test rig, oil supply to the bearings was accidentally stopped. Immediately, the vibration subsided although it started again after oil supply resumed. Consequently, the vibration was assumed to take place due to some unknown action of the “oil” film at that time. Furthermore, violent whirling of an elastic rotor in oil film bearings appeared as if the bending rotor hit the bearing surface violently in a “whip” motion. In contrast to forced vibration caused by oscillating external forces, self-excited vibrations of rotor are caused by some factors inherent in the mechanical properties of rotating machinery. While a rotor in oil film bearings is running in the stable operation region, on the stable side of the stable–unstable boundary, small perturbations around the equilibrium position always subside and vanish with time, because the damping force in the system works effectively. On the other hand, when the rotor exceeds the boundary with the operating condition being varied and falls into the unstable region of operation, small perturbations around equilibrium grow with time, often resulting in violent whirling of the rotor, because the destabilizing force is larger than the damping force, which increases the small perturbation. Then, the increased perturbation increases the destabilizing force. Thus, the spiral feedback process becomes self-excited. Once self-excited vibration takes hold, it is very difficult to control or predict accurately its vibration amplitude. Violent whirling of rotor in plain bearings may result in damage to the bearings and/or the rotor itself. Consequently, it is of paramount importance to design oil film bearings for rotating machinery so that oil whip never happens as far as the operating conditions remain within predicted maxima.

4.1 Cause and Phenomena of Oil Whip

61

Extensive research work on oil whip has been conducted theoretically and experimentally, and oil whip was soon confirmed to be a self-excited vibration with quite different features from that of any kind of forced vibration, such as unbalance excitation of a rotor explained in Chap. 3. The cause of oil whip has been confirmed to be a destabilizing effect of the cross-coupled stiffness coefficients of the oil film formed in a journal bearing. These coefficients are known to increase in magnitude, generally with increasing Sommerfeld number. Consequently, high-speed and/or lightweight rotors in plain bearings are susceptible to oil whip. Instead of circular bearings that are most susceptible to oil whip, multi-pad fixedbore bearings or tilting pad bearings, described in Chap. 2, are widely used in highspeed rotating machinery to shift the onset of oil whip toward higher shaft speeds.

4.1.2 Oil Whip Phenomena Figure 4.1 illustrates some cases of oil whip. The horizontal abscissa of the bottom , where figures and the middle figures represents the dimensionless shaft speed, /ωn √ ωn is the first bending critical speed of the rotor-bearing system, defined by k/m. The vertical ordinate of the bottom figure represents vibration amplitude, while in the middle figure, it is the dimensionless frequency of vibration, ω/ωn . The top figure shows the waveforms of the vibration corresponding to each coordinate A to K shown in the bottom figure, and the abscissa represents time with each revolution of shaft marked. In the case of the light rotor, Fig. 4.1a shows that unstable vibration starts at a shaft speed (A) lower than the first bending critical speed (B), with the oil film stiffness being considered. The frequency is slightly lower than 0.5 , for example, 0.48 , increasing with shaft speed . The corresponding waveform in the top figure resembles synchronous unbalance response vibration, because the amplitude of unstable vibration component is not so large at this shaft speed. The unbalance vibration is predominant at the critical speed (B), and the corresponding waveform is nearly synchronous with the shaft pulse signal. At the shaft speed  of about 2ωn (D), the vibration frequency approaches ωn , the first bending critical speed, and the vibration amplitude becomes very large. With further increase in , the vibration keeps the large amplitude with the vibration frequency ωn unchanged. The subsynchronous vibration with a frequency lower than  is proof of self-excited vibration, resulting from a negative damping ratio of the rotor-bearing system. In the case of the slightly heavily loaded rotor, shown in Fig. 4.1b, the waveforms of synchronous vibration are observed at the shaft speeds (E) and (F). Self-excited vibration starts at the speed (G), which is higher than the onset speed (A) for Fig. 4.1a and even higher than the critical speed (E). With increasing , a similar excursion of vibration amplitude and frequency is experienced.

4 Stability of a Rotor in Plain Bearings

vibration waveform

62 D

H

C

G

B

F

A

E

J

frequency ω /ω n

one revolution pulse

I one rev.

Ω

1

· 0.5 ω =· 0

vibration amplitude

K

1

ω =ω n

Ω

2

3

0

1

2 H

3

0

F 2

Ω /ω n

(1) lightly loaded

3

0

1

1

2

3 K

I

E C

1

pulse

Ω

1

B

0

one rev.

1

D

A

pulse

Ω

G

J 2

Ω /ω n

3

(2) slightly heavily loaded

0

1

2

Ω /ω n

3

(3) heavily loaded

Fig. 4.1 Oil whip phenomena

In the case of the heavily loaded rotor shown in Fig. 4.1(3), the synchronous vibration is predominant even when the shaft speed, , is much higher than 2ωn (J). Then, violent unstable vibration of frequency ωn starts abruptly at the shaft speed (K). Once it happens, with decreasing shaft speed, the unstable vibration maintains its large amplitude until the shaft speed reduces to about 2ωn . This is called the inertia effect of oil whip and is observed often in the case of a heavily loaded rotor. “Oil whirl” is sometimes defined as the unstable vibration of small amplitude observed at shaft speeds lower than 2ωn shown in Fig. 4.1(1) and (2). Violent unstable vibration of a rotor generates large alternate bending stress in the rotor, which is very dangerous for shaft integrity. It may make the rotating parts rub with stationary parts, resulting in damage to bearings and seals. Consequently, once oil whip is observed as taking place in actual rotating machinery, immediate deceleration or power off is conducted.

4.2 Stability Chart of Circular Bearing Based on Linear Vibration …

(a)

(b) m

63

S

m

{ xR , yR }

ks ks B

S

B 2kij

{ x, y} 2cij

Fig. 4.2 Jeffcott rotor supported in journal bearings

4.2 Stability Chart of Circular Bearing Based on Linear Vibration Analysis 4.2.1 Coordinates and Equation of Motion As violent oil whip is triggered by oil whirl of small vibration amplitude, a linear vibration analysis [30, 31] can be applied to discriminate the stability of small vibration of rotor in the rotor-bearing system. This section applies the method to a rotor in circular bearings. Figure 4.2a shows a schematic of a rotor-bearing system to be analyzed. The symmetrical rotor, termed a “Jeffcott rotor,” consists of a massless shaft (bending stiffness k s ) and a single mass m at the mid-span of the shaft. The shaft is supported in two identical circular bearings at both ends. Because the system is assumed to be perfectly symmetrical, it can be represented by the dynamic system shown in Fig. 4.2b. Consequently, each of the two identical oil film bearings has the same values of linear stiffness coefficients k ij (i, j  x, y) and linear viscous damping coefficients cij (i, j  x, y) that vary with Sommerfeld number. Because of that, oil film forces are represented by 2k ij and 2cij in the figure. Figure 4.3 shows the orthogonal coordinates in the shaft rotating plane with the origin O at bearing center. P is the journal center position at steady-state equilibrium determined by Sommerfeld number on the journal center locus. The journal is assumed to rotate counterclockwise at the angular speed . With the origin P, the vertical and horizontal coordinates x and y are defined as shown in the figure. Q is the mass position at the steady-state equilibrium, slightly displaced downwardly (in the direction of gravity). With the origin Q, the vertical and horizontal coordinates x R and yR are defined as shown in the figure. When the system is perturbed, the journal center is assumed to move from P (0, 0) to B (x, y) with the velocities x˙ and y˙ at an instant in time, while the mass moves from Q (0, 0) to S (x R , yR ) with the velocities x˙ R and y˙ R . Each displacement and velocity are assumed to be small enough to regard the system as being linear. The dynamic oil film force components f x and f y exerting on the journal located at B are given in Eq. (2.26) and can be expressed as follows:

64

4 Stability of a Rotor in Plain Bearings O 2fx y

P

2fy

B (x, y)

steady-state Q P journal center Q mass center

yR S (xR, yR) perturbed

x xR

B journal center S mass center

Fig. 4.3 Coordinates in shaft rotating plane

f x  k x x x + k x y y + cx x x˙ + cx y y˙ f y  k yy y + k yx x + c yy y˙ + c yx x˙

 (4.1)

where k ij (i, j  x, y) and cij (i, j  x, y) are linear stiffness coefficients and linear damping coefficients, respectively. Each equation of motion in the x or y direction is combined in a matrix-vector form as follows (R1_Sect. 3.1.3): 

m 0  m 0

   0 0 x¨ R + x¨ 0 0    0 y¨ R 0 + y¨ 0 0

       ⎫ −ks x˙ R xR 0 ks 0 ⎪ + +  0⎪ ⎬ x˙ −k k +2k x x x 2cx x 2k y+2cx y y˙    s s    x y  ⎪ −ks y˙ R yR 0 ks 0 ⎭ + +  0⎪ y˙ −ks ks +2k yy y 2c yy 2k yx x+2c yx x˙ (4.2)

The stability of the system is determined based on the roots of the following characteristic equation: 2 Ms + Cs + K  0

(4.3)

where ⎡

m ⎢0 M ⎢ ⎣0 0

0 0 0 0

0 0 m 0

⎤ 0 0⎥ ⎥, 0⎦ 0

⎡

−ks ks ⎢ −ks ks + 2k x x K ⎢ ⎣ 0 0 0 2k yx

⎤ ⎡ 0 0 0 0 ⎢ 0 2cx x 0 2k x y ⎥ ⎥, C  ⎢ ⎣0 0 ks −ks ⎦ −ks ks + 2k yy 0 2c yx

0 0 0 0

⎤ 0 2cx y ⎥ ⎥ 0 ⎦ 2c yy

4.2 Stability Chart of Circular Bearing Based on Linear Vibration …

65

4.2.2 Characteristic Equation and Stability Criterion Some symbols are defined as follows: g  gravity acceleration, C  mean radial clearance of bearing, √ ωs  ks /m  natural frequency of rotor of the first bending mode with simple support assumed at bearing positions, α ≡ mg/(ks C)  shaft elasticity index (α  0, when the bending stiffness k s is infinitely large), C C K i j  (mg/2) ki j , Ci j  (mg/2) ci j are eight dimensionless oil film coefficients. Equation (4.3) divided by k s can be expressed with these variables shown above, and then, the characteristic equation is derived as follows: 2 2 s /ωs + 1 −1 0 0   −1 + C s/) 0 α K + C s/ 1 + α(K xx xx xy xy 0 2 2 /ω + 1 −1 0 0 s s     −1 1 + α K yy + C yy s/ 0 α K yx + C yx s/

(4.4)

The expansion of the determinant gives an algebraic equation of the sixth degree in s: C0 s 6 + C1 s 5 + C2 s 4 + C3 s 3 + C4 s 2 + C5 s + C6  0

(4.5)

Each coefficient is expressed as follows:

⎫ 2 2 2gα α α α 2 ⎪ C 0 ≡ 2 A3 , C 1 ≡ A3 + α A2 + α A4 + 1 ⎪ A1 + A5 , C 2 ≡ ⎪ 2 ⎪   ⎪  C ⎪ ⎪ ⎬   2 g g 2α g 1 C3 ≡ A1 + A5 , C4 ≡ (A4 + 2α A2 ) + 2 2 A3  C  C ⎪ C  ⎪ ⎪ ⎪ 2 2 ⎪ ⎪ g g ⎪ ⎭ C 5 ≡ 2 A1 , C 6 ≡ 2 A2 C  C (4.6) where the Ai (i  1–5) coefficients are expressed with dimensionless stiffness coefficients and dimensionless viscous damping coefficients as follows: A1  K x x C yy − K x y C yx − K yx C x y + K yy C x x , A2  K x x K yy − K x y K yx A3  C x x C yy − C x y C yx , A4  K x x + K yy , A5  C x x + C yy

 (4.7)

Equation (4.5) is solved for six complex roots si  ai + jqi (i: 1–6, j: imaginary unit) where the imaginary part q corresponds to a damped natural frequency and the sign of the real part ai represents the stability (through exponential decay or growth of vibrations) of the rotor-bearing system. The system lies on the boundary of the

66

4 Stability of a Rotor in Plain Bearings

stable and unstable operation regions for ai  0. Consequently, the system is stable for ai < 0 (stable root), while it is unstable for ai > 0 (unstable root). Thus, the system is stable when every complex root is a stable root, while it is unstable if one or more unstable roots with positive real parts are obtained. As a result, all complex roots of Eq. (4.5) must be obtained to check the sign of each real part and in turn determine the stability of the system. However, no analytical solutions are available for the complex roots of an algebraic equation of the sixth degree. Although nowadays various computer programs are available to obtain these roots numerically, Routh and Hurwitz proposed different methods independently in the nineteenth century to determine whether general characteristic equations have unstable roots or not, without solving the equations. Both stability criteria are now found to be completely the same mathematically. Here, the Routh–Hurwitz criterion is explained. Let the coefficients of Eq. (4.5) be arranged in a 6 × 6 matrix form shown in Eq. (4.8). ⎤ ⎡ C1 C3 C5 0 0 0 ⎢C C C C 0 0 ⎥ ⎥ ⎢ 0 2 4 6 ⎥ ⎢ ⎢ 0 C1 C3 C5 0 0 ⎥ (4.8) ⎥ ⎢ ⎢ 0 C0 C2 C4 C6 0 ⎥ ⎥ ⎢ ⎣ 0 0 C1 C3 C5 0 ⎦ 0 0 C0 C2 C4 C6 Note that C 1 , C 2 , … C 6 are arranged on the diagonal in that order. The Routh–Hurwitz criterion prescribes that the necessary and sufficient condition for every ai to be negative is that every principal minor determinant is positive. Consequently, if this condition is not satisfied, the rotor-bearing system is unstable and oil whip takes place. When the criteria are applied to Eq. (4.8), the stability criterion of the system can be given as follows [30, 31]: ( A2 + A2 A25 − A1 A4 A5 )( A5 + α A1 ) 1 > 1 2 ν A1 A3 A25

(4.9a)

√ where ν ≡ / g/C. When each oil film coefficient in Eq. (2.28) derived by means of the infinitely short bearing approximation is calculated for journal eccentricity ratio ε0 at a given static equilibrium position, each Ai in Eq. (4.7) can be determined. Then, f 1 (ε0 , α), the right-hand side of Inequality (4.9a) can be calculated for a value of shaft elasticity index α given. Finally, it is concluded that the rotor-bearing system can remain stable as long as Inequality (4.9a) is satisfied; that is, the reciprocal of squared dimensionless shaft speed, 1/ν 2 , remains larger than f 1 (ε0 , α) obtained. In other words, the dimensionless stability threshold shaft speed ν c can be obtained from Inequality (4.9a) as follows:

4.2 Stability Chart of Circular Bearing Based on Linear Vibration …

 νc 

( A21

+

A2 A25

A1 A3 A25 − A1 A4 A5 )( A5 + α A1 )

67

(4.9b)

√ where νc  c / g/C, c  dimensional stability threshold shaft speed. The so-called WFR (whirl frequency ratio) corresponds to the dimensionless threshold shaft speed for a rigid rotor, obtained from Eq. (4.9b) when α  0.

4.2.3 Stability Chart Figure 4.4a [32] shows the stability chart corresponding to Inequality (4.9a). The vertical coordinate is the journal eccentricity ratio ε0 at a static equilibrium position, while the horizontal coordinate, 1/ν 2 , is the left-hand side term of Inequality (4.9a). When various values of ε0 are substituted into f 1 (ε0 , α) for each value of α, each of the six solid curves is obtained, corresponding to each stability boundary. The rotor remains stable in the region on the right-hand side of each boundary. The lower the shaft elasticity index α, the larger the stable operation region is. The stability criterion, Inequality (4.9a), was presented in 1959, when modern computers were not available. Consequently, the stability boundary had to be determined only by means of mechanical calculator and had to be shown on the ε0 and 1/ν 2 plane. Each of the three dashed curves shows the relationship of ε0 and 1/ν 2 at static equilibrium obtained from Eq. (2.19) for each of the three different values of the parameter λ(L/D)2 . Here, the dimensionless bearing modulus, λ, is defined as follows: μ λ≡ pm

 2 √  R g/C , νc  c / g/C C 2π

where pm ≡ mg/(2DL)  specific bearing load defined by load per bearing divided by the product of bearing diameter D and bearing width L. As a result, the Sommerfeld number S can be rewritten as λν. Then, Eq. (2.19) can be written as λ(L/D)2 ν  f 2 (ε0 ). When various values of ε0 are substituted into f 2 (ε0 ) for a fixed value of λ(L/D)2 , corresponding values of 1/ν 2 can be determined. Thus, each of the three dashed curves can be drawn. With increasing shaft speed, the coordinate point (ε0 , 1/ν 2 ) moves on each curve from right to left. When a curve intersects a stability boundary, oil whip occurs. Figure 4.4b shows another stability chart with a horizontal coordinate different from that of (a), that is, dimensionless shaft speed defined by p  /ωs . Then, the dimensionless shaft speed ν can be rewritten as follows:   √ ν ≡ / g/C ≡ (/ωs )(ωs / g/C) ≡ p/ α.

68

4 Stability of a Rotor in Plain Bearings 0 eccentricity ratio ε 0 [−]

unstable

α= 1 α= 2.5

0.2 0.4

0.5

( DL )

2

ε 0 decreasing with

stable 0.8

increasing rotor speed ν or p

λ ( L ) = 0.02 D 2

1.0

1

α= 1/16

0.4

stability boundary

2

1

ε0 0 0.2

α = 15

α = 10 L λ(D ) = 0.2

0.6

0

S

L 2 λ(D ) =2

α =5

5

3 4 1/ ν 2 (a) plot on 1/ν 2 – ε 0 plane

α = 1/2 1/5

2

∞

5 1 2

Q (D L

0.4

) = μWLD 2

2

( CR ) ( DL ) ω2 π 2

s

unstable

0.1 0.6

4 stable

0.01 0.001

0.8 40

Q (D L

1.0 0

1

2

2

) = 400

3

dimensionless rotor speed

4

5

p= Ω ωs

(b) plot on p – ε 0 plane

Fig. 4.4 Stability chart of Jeffcott rotor supported in cylindrical bearings

When√Eq. (4.9b) is expressed as ν  f 3 (ε0 , α), it can be rewritten √ as p  f 3 (ε0 , α) α. When various values of ε0 are substituted into f 3 (ε0 , α) α for each value of α, each of the seven solid curves is obtained, corresponding to each stability boundary on the plane defined by ε0 and p. The rotor remains stable in the region on the left-hand side of each boundary. Each of the four dashed curves shows the relationship of ε0 and p at static equilibrium obtained from Eq. (2.19) for each of the four different values of the parameter Q(D/L)2 where Q ≡ p/S. Then, Eq. (2.19), S(L/D)2  f 2 (ε0 ) can be rewritten as p  f 2 (ε0 )Q(D/L)2 . Thus, the four dashed curves can be drawn for four different values of Q(D/L)2 as shown in Fig. 4.4b. With increasing shaft speed, the coordinates (ε0 , p) of the operating state point move on each curve from left to right. When a curve intersects a stability boundary, oil whip occurs. It is obviously difficult to understand the oil whip stability phenomena intuitively from Fig. 4.4, because the coordinate 1/ν 2 is difficult to perceive. Furthermore, it is not easy to find an oil whip onset shaft speed accurately for a particular rotorbearing system, because an operational curve of the system needs also to be drawn

4.2 Stability Chart of Circular Bearing Based on Linear Vibration …

69

5 (1) cylindrical bearing oil groove 10°× 2 4

L/D = 0.5

ν 3

α=0

νc 2

1/4 1

1

0 0.02

0.05 0.1

0.2

0.5

1

2

5

λ

Fig. 4.5 Stability chart of cylindrical bearing with two grooves

and the intersection with a particular stability boundary curve needs to be determined accurately in the figure. In contrast, Fig. 4.5 gives an alternative style of presenting the oil whip stability chart to solve the problems mentioned above. The vertical coordinate is the dimensionless shaft speed ν (linear scale), while the horizontal coordinate is the dimensionless bearing modulus λ. Both variables have been defined already in this section. Each of the three stability boundary curves shown is determined from Eq. (4.9b) for each of the three values of α (0, 1/4 and 1). Note that Eq. (4.9b) is calculated for ν c by using the oil film coefficients values for cylindrical bearings (L/D  0.5, two oil grooves of 10° arc each) listed in the table [B13, p 14]. Those coefficients are obtained numerically for a finite width bearing, while Fig. 4.4 uses the coefficients given in Eq. (2.28) for a full circular bearing based on the infinitely short bearing approximation. The table gives not only oil film coefficients values, but also the values of Sommerfeld number, S, corresponding to various values of ε0 . While an onset speed ν c (vertical coordinate) is obtained from Eq. (4.9b) for a given value of ε0 , the corresponding value of S is found from the above table. Then, the corresponding horizontal coordinate λ is calculated from the equation of S  λν. The rotor-bearing system remains stable in the region below each boundary curve, but unstable self-excited vibration (oil whirl or oil whip) takes place in the region above each curve. Let λ be 0.2, for example, for given dimensions of bearing and oil viscosity. With increasing shaft speed, the operational state point (0.2, ν) moves vertically upward from the point (0.2, 0) in Fig. 4.5, resulting in the increase in the Sommerfeld number (S  λν), and eventually intersects a stability boundary curve for α  1/4, for example. The vertical ordinate of the intersection gives the dimensionless threshold shaft speed νc ≈ 2. In other words, Fig. 4.5 represents the stability chart with the two coordinates ν and λ, together with the parameter α.

70

4 Stability of a Rotor in Plain Bearings

Thus, Fig. 4.5 makes it much easier and simpler to understand the relationship of oil whip stability boundary and operating condition of bearing and also to determine oil whip onset shaft speed, compared with Fig. 4.4. Engineers can decide reasonable design modifications by studying carefully the chart and choosing skillfully the values of α and λ. Once the system becomes unstable, the vibration amplitudes increase selfexcitedly with time. As shown in Fig. 4.1, oil whirl of small amplitude turns into oil whip of large amplitude when the shaft speed reaches about twice the first bending critical speed ωn in the case of circular bearing. However, the growth of the amplitude cannot be simulated by linear vibration theories, because they are valid only as long as vibration amplitude remains sufficiently small. Hori [30] found from theoretical analysis of circular bearing that small amplitude oil whirl turns into large amplitude oil whip above twice ωn . As described hereinafter, Note 3 (p 82) will outline the analysis.

4.3 Countermeasures Against Oil Whip in a Cylindrical Bearing Figure 4.5 shows obvious countermeasures against oil whip in the case of cylindrical bearing. (1) Shaft elasticity index α The lower α is, the higher the threshold shaft speed is. Consequently, shorter and thicker rotors are more stable than slender ones with regard to small amplitude oil whirl. Furthermore, the former has higher critical speed, ωs , of the first bending mode shifting up the threshold for large amplitude oil whip. (2) Dimensionless bearing modulus λ With increasing λ, the threshold speed ν c decreases regardless of α values, and vice versa. Consequently, operation with lower λ is advisable. This tactic can be translated into the following: i. Specific bearing load pm is increased by decreasing the bearing diameter D and/or bearing width L (downsizing bearings). ii. Bearing clearance C is increased. iii. Lower viscosity oil is used, or supply oil temperature is raised to decrease the temperature-dependent lubricant viscosity μ. With decreasing λ, Sommerfeld number S lowers; thus, journal eccentricity ratio ε increases. Destabilizing cross-coupled stiffness coefficients of an oil film in circular bearings are known to be small in the lower S region (higher ε region). Consequently, operation in the higher ε region is the basic tactic to prevent oil whirl and/or oil whip in circular bearings.

4.4 Stability Charts of Non-circular Bearings and Oil Whip …

71

4.4 Stability Charts of Non-circular Bearings and Oil Whip Countermeasures With increasing rotational speed of machinery, the maximum operating speed often exceeds the oil whip threshold speed, as long as circular bearings are used. In this case, multi-lobe bearings (two-lobe, three-lobe, offset, etc.) shown in Fig. 2.11 are used. Figure 4.6 shows stability charts of two-lobe bearings for each case of preload factor mp  0.5 or 0.75. Preload factor is defined in Eq. (2.31). The stiffness coefficients and the viscous damping coefficients of oil film are listed in the tables [B13, p 41 & 47]. Compared with the stability chart of cylindrical bearing shown in Fig. 4.5, twolobe bearings can attain higher threshold shaft speed (refer to Example 4.1 described hereinafter). Considering the effect of preload factor with the same α, the higher the preload factor, the higher is the threshold speed for stability. While decreasing λ is the only means to raise threshold speed in the case of cylindrical bearing, increasing λ is an alternative in the case of two-lobe bearing. Multi-lobe bearings of higher preload factor are characterized by enabling stable operation equivalent to higher eccentricity ratio operation in a cylindrical bearing.

(3) α=0

13 12

(3) two-lobe bearing

(2) two-lobe bearing

11

oil groove 10°× 2

oil groove 10°× 2

10

L/D = 0.5

L/D = 0.5

9

mp = 0.5

mp = 0.75

8 7 (2) α = 0

6

ν

5 (3) α =1/4

4

νc

(2) α =1/4

3

(3) α =1 (2) α =1

νc 2 1 0 0.02

0.05

0.1

0.2

0.5

λ

Fig. 4.6 Stability chart of two-lobe bearings

1

2

5

72

4 Stability of a Rotor in Plain Bearings

A tilting pad bearing is one of the multi-lobe bearings as shown in Fig. 2.11. Each pad can be set with a preload factor similar to that for a two-lobe or three-lobe bearing. However, different from fixed-bore multi-lobe bearings, each pad can tilt around each pivot on its back surface. Consequently, tilting pad bearings generally generate only direct coefficients of oil film with no cross-coupled stiffness coefficients. Thus, tilting pad bearings are considered to cause no oil whirl and no oil whip because of no destabilizing force generated in the oil film. However, recent reports [33–35] prove that oil whip can take place even with tilting pad bearings, if some conditions are satisfied. Example 4.1 Convert νc ≈ 2.0 into dimensionless threshold speed normalized by the simple support first critical speed ωs , that is, c /ωs in the case of λ  0.2 and α  1/4 in Fig. 4.5 (cylindrical bearing with two oil grooves). Repeat the same task for ν c  2.7 and 3.8 for mp  0.5 and 0.75, respectively, in the case of λ  0.2 and α  1/4 in Fig. 4.6 (two-lobe bearing). Answer √ √ From Eq. (4.9b), νc  c / g/C √  c /( αωs ) Consequently, c /ωs ≡ νc α. For cylindrical bearing with two oil grooves, substituting ν c  2.0 and α  1/4 into the right-hand side of the equation above results in the answer 1.0. For a two-lobe bearing, substituting ν c  2.7 or 3.8 and α  1/4 into the right-hand side of the equation above results in the answers 1.35 for mp  0.5 and 1.9 for mp  0.75.

4.5 Explanation of Mechanism of Oil Whip The previous sections explain that oil whip or oil whirl of rotating machinery supported in oil film bearings is caused by the destabilizing action of cross-coupled stiffness of an oil film; then, the threshold shaft speed can be predicted from stability charts obtained by means of linear vibration theories. Finally, such unstable self-excited vibrations normally take place with increasing shaft speed. This section focuses on the energy balance of the rotor whirling in its orbit, explaining the mechanism of such unstable vibrations. Figure 2.5b defines the downward vertical ordinate, x, and the abscissa, y, of 90° apart in the shaft rotating (counterclockwise) direction at the origin Os (static equilibrium of journal center). When the journal center Oj is now assumed to whirl in a small elliptical orbit around Os , the journal center coordinates are expressed as follows: x  A cos(ωt + ϕx ),

y  B sin(ωt + ϕ y )

(4.10)

where A and B are small amplitudes, ω is whirl frequency, t is time, and ϕ x and ϕ y are phase angles.

4.5 Explanation of Mechanism of Oil Whip

73

As the journal center makes a small-size orbit around the static equilibrium, the oil film changes its shape with time, generating small dynamic oil film force components, f x and f y , besides the static oil film force corresponding to the equilibrium. These two components defined at the equilibrium are represented by four stiffness coefficients and four viscous damping coefficients as shown in Eq. (2.26). While the journal center makes a full orbit, taking the time duration T , the kinetic energy dissipated is expressed as follows:  E





T

f x d x + f y dy 



 f x x˙ + f y y˙ dt

0

 π {ω( A2 cx x + B 2 c yy ) − ABω(cx y + c yx ) sin(ϕx − ϕ y ) − AB(k x y − k yx ) cos(ϕx − ϕ y )}

(4.11)

Figure 2.7 shows the variation of each oil film coefficient of circular bearing with Sommerfeld number, and cxx and cyy are found to always keep positive sign. Consequently, the first term of the right-hand side of Eq. (4.11) contributes to dissipating the kinetic energy, in turn, decreasing the whirl amplitude and giving rise to a stabilizing effect on the system. On the other hand, the sign of the second term changes, depending on the sign of sin(ϕ x− ϕ y ) because cxy and cyx have positive signs. Consequently, the second term contributes to stabilization if the overall sign is positive, and vice versa. However, its contribution is usually small because the magnitudes of cxy and cyx are relatively small compared with cxx and cyy . Finally, the cross-coupled stiffness coefficients, k xy and k yx , in the third term increase the absolute values with Sommerfeld number, approaching asymptotically to the same value. Then, the term (k xy− k yx ) increases its magnitude with Sommerfeld number, always keeping a positive sign. Furthermore, cos(ϕ x −ϕ y ) has a positive sign. Consequently, the third term with the overall negative sign supplies the system with kinetic energy increasing with Sommerfeld number, in turn, increasing the whirl amplitude with time. Thus, the third term works as a destabilizing effect on the system. When the journal whirls around Os in the spinning direction, both the positive k xy force and the negative k yx force work on the journal in line with the direction of the instantaneous journal speed vector, perpendicular to the journal displacement vector, resulting on the journal as a negative damping force. When the journal center is assumed to whirl on a circular orbit, for simplicity, A  B and ϕ x  ϕ y . Then, Eq. (4.11) is expressed as follows:   E  π A2 ω(cx x + c yy ) − (k x y − k yx )

(4.12)

When the term (k xy −k yx ) exceeds ω (cxx + cyy ), the dissipated energy E becomes negative, increasing the kinetic energy of the rotor-bearing system, and eventually, unstable vibrations appear. Consequently, the root cause of oil whirl and oil whip is proven to be the cross-coupled stiffness coefficients of the oil film.

74

4 Stability of a Rotor in Plain Bearings

Note: 1 Stability at high shaft speed Figure 2.7 shows that the cross-coupled stiffness coefficients exceed in value the direct stiffness coefficients in the region of higher Sommerfeld number (higher shaft speed, lower bearing load etc.), that is, in the region of lower journal eccentricity ratio where oil whirl or oil whip appears. Furthermore, the two cross-coupled stiffness coefficients approach asymptotically to the same absolute value with increasing Sommerfeld number, as do the two direct viscous damping coefficients. Consequently, the following approximations are valid at high shaft speeds: cx x ≈ c yy → cd , k x y ≈ −k yx → kc

(4.13)

Then, Eq. (4.12) is expressed as follows: E  2π A2 (ωcd − kc )

(4.14)

which agrees with the explanation in R1_Fig. 7.5. Note: 2 Journal center loci and stability Figure 4.7 shows journal center loci for cylindrical bearing and two-lobe bearings with preload factors of 0.5 and 0.8. Each point C on each of the three loci represents an inflection point, corresponding to the point above and beyond which the cross-coupled stiffness coefficient k yx changes sign from positive to negative. Consequently, if the steady-state journal center position for the maximum operating speed is located below Point C on each of the journal center loci, oil whip or oil whirl should not take place. The figure shows each critical Sommerfeld number value corresponding to each Point C. The Sommerfeld number here is defined by Eq. (2.20) with bearing clearance C for a cylindrical bearing replaced by C b (minimum clearance) for a two-lobe y direction

bearing center 0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 (1) cylindrical (2) two-lobe mp = 0.5

0.2

(3) two-lobe mp = 0.8 S1 = 0.393 C

x direction

0.4 S1 = 0.173 (1) C 0.6

0.8

1

Fig. 4.7 Journal center loci

(2) S1 = 0.455 C

(3)

4.5 Explanation of Mechanism of Oil Whip

75

bearing. Then, the critical Sommerfeld number for a cylindrical bearing is a minimum while that for a two-lobe bearing with preload factor 0.8 is a maximum. A two-lobe bearing with preload factor 0.5 lies between these two cases. Consequently, the threshold for a two-lobe bearing is shifted toward a higher shaft speed than that of a cylindrical bearing. With regard to the effect of preload factor on the threshold, the higher the preload factor, the higher the threshold speed for stability.

4.6 Countermeasures for Actual Journal Bearings and Points of Attention As discussed so far, oil whirl or oil whip, characterized by subsynchronous vibration frequencies, is a self-excited vibration (unstable vibration) of a rotor caused by the destabilizing action of the cross-coupled stiffness coefficients of a bearing oil film. Once it takes place, it is practically impossible to predict the maximum whirl amplitude or to control the amplitude. Furthermore, if some other destabilizing factors exist in the system, the instability worsens. Consequently, rotor-bearing systems should be designed so that unstable vibrations never occur in operation. This self-excited vibration mechanism is quite different from forced vibration such as unbalance vibration of a rotor that is normally acceptable unless the peak amplitude of the unbalance response is excessive at critical speeds. Consequently, when subsynchronous vibration of rotor is detected in actual machinery, oil whip or oil whirl should be suspected, and then, its cause must be uprooted. In selecting countermeasures in actual rotating machinery, it should be noted that real rotor-bearing systems are not always represented by the simple theoretical models mentioned so far. Furthermore, when a bearing design needs to be modified to prevent self-excited vibrations, attention should be paid to the possibility that the design modification may result in impairing other bearing performance. A tradeoff design concept should always accompany anti-vibration design of bearings for rotating machinery. (1) Different from a Jeffcott rotor model, actual rotors are normally asymmetric in the axial direction. Bearing size and bearing load are not always identical at both ends of rotor. Instead of simple theoretical models, the FEM or transfer matrix method can be applied to analyze these complicated rotor-bearing systems in detail. For example, R1_Chap. 12 introduces such a software “MyROT.” (2) Higher shaft bending stiffness by increasing shaft diameter and/or decreasing shaft length contributes to increasing threshold shaft speed of oil whirl or oil whip, as mentioned before. On the other hand, the increase in rotor diameter naturally leads to a heavier rotor and upsizing of the machinery. Consequently, compared with small-size, lightweight rotors, various costs and man-hours would be increased in purchasing, manufacturing, transporting, assembling, and installing

76

4 Stability of a Rotor in Plain Bearings

the systems. In the case of fluid machinery, for a fixed outer size, the crosssectional area for passing fluid needs to be decreased by increasing shaft diameter, resulting in lower performance. This is another trade-off design problem. (3) Each of the countermeasures against oil whirl or oil whip for cylindrical bearings always results in an increase in journal eccentricity ratio. Consequently, attention should be paid so that the minimum oil film thickness in operation should not be too thin. Excessively small oil films result in various tribological problems for bearings. When bearing width is decreased to increase specific bearing load, the lower limit shaft speed to maintain a hydrodynamic oil film increases, resulting in higher risks of meltdown or wiping of soft bearing metals. Smaller bearing surface area decreases the viscous damping effect of oil films against rotor vibration. Higher journal eccentricity ratio corresponds to higher stiffness coefficients of a bearing oil film, reducing journal movement in bearing oil film. Then, viscous damping of an oil film cannot work on a journal effectively, resulting in the increase in the peak amplitude of unbalance vibration at critical speeds. When supply oil temperature is increased to improve stability, oxidation degradation of lubricant oil is expedited, resulting in decrease in lubricant performance and lubricant life. (4) Similar notes can be presented for various multi-lobe bearings. Increasing pad preload factors to improve stability normally results in the increase in stiffness coefficients of the oil film. Then, journal movement in the bearing oil film is more suppressed, and insufficient viscous damping of the oil film acts on the journal, which may result in the increase in the peak amplitude of unbalance vibration at critical speeds. (5) In the case of two-lobe bearings, a higher preload factor results in higher anisotropy of stiffness coefficients of the oil film; that is, the vertical coefficient in the static load direction becomes much higher than the horizontal coefficient. Consequently, the critical speeds of unbalance vibration of rotor separate into vertical and horizontal responses, making balancing and vibration amplitude control in operation more complicated. (6) Multi-pad bearings and tilting pad bearings require more components for assembly and more man-hours to manufacture, compared to a simple cylindrical bearing, resulting in higher cost. Consequently, if these bearings are applied to highvolume products without careful consideration, price competitiveness could be lost.

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration Oil whirl or oil whip is caused by the destabilizing effect of the cross-coupled stiffness coefficients of a bearing oil film. Similarly, another self-excited vibration of a rotor

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration

77

is caused by the cross-coupled stiffness coefficients of a fluid film formed in noncontact seal clearance and in impeller clearance. This vibration is called “flow-excited vibration.” When the vibration was reported to take place in high-speed, high-pressure pumps, and compressors after being commissioned in the 1970s, initially, it was mistaken for oil whip because subsynchronous vibration was observed. However, it was found to occur even if rotors were supported in tilting pad journal bearings. Furthermore, the vibration amplitude was found to depend on output load rather than shaft speed. Thus, the destabilizing effect of the cross-coupled stiffness coefficients of a fluid film in seal clearances was identified as the cause of the vibration. Consequently, countermeasures against the new self-excited vibration are to reduce the destabilizing fluid forces in seals and impellers or to increase the damping effect of a bearing oil film. Bearing specifications need to be selected so as to maximize the stabilizing effect of oil film against destabilizing fluid forces. The destabilizing fluid force is expressed by two cross-coupled stiffness coefficients, k xy  k c and k yx  −k c , which is similar to that for the destabilizing bearing oil film. The magnitude of k c increases with increasing pressure and flow rate of the working fluid, resulting in the flow-excited unstable vibration of the rotor. When a rotor supported in tilting pad journal bearings is assumed to make an elliptical whirl orbit (Eq. 4.10) in the seal clearance with the destabilizing fluid force acting on the rotor, the energy balance during one full orbit can be expressed with the first and the third terms of Eq. (4.11). The first term contributes to dissipating the kinetic energy of the system by the direct viscous damping coefficients, and the third term can be regarded as the destabilizing fluid force that increases the kinetic energy. When k xy and k yx increase with pressure and flow rate and eventually exceed the effect of the first term, the system becomes unstable and the flow-excited vibration takes place. Chapter 8 explains various destabilizing fluid forces and how to decrease them. Next, the effect of orbit direction on stability of flow-excited vibration is discussed. As shown in R1_Eq. (7.34), whirling in an elliptical orbit can be decomposed into two components, that is, forward whirling in a circular orbit (radius af ) and backward whirling in another circular orbit (radius ab ). Referring to R1_Eq. (7.39), the equivalent m-k d -cd -k c system can be expressed as follows: (a 2f + ab2 )(m z¨ + kz) + cd (a 2f + ab2 )˙z − jkc (a 2f − ab2 )z  0

(4.15)

The direct viscous damping coefficient, cd , affects both forward whirling and backward whirling, contributing to stabilizing the system. On the other hand, the cross-coupled stiffness of fluid force k c (> 0) working on forward whirling contributes to destabilizing the system, and when it works on backward whirling, it contributes to stabilizing the system. When the whirl is linear, that is, af  ab , k c makes no contribution. Thus, isotropic bearing oil film characteristics expedites instability because it gives a full circular orbit of only forward whirling, while anisotropic bearing support retards instability because it gives an elliptical orbit consisting of

78

4 Stability of a Rotor in Plain Bearings

circular orbit of whirling y

elliptical orbit of whirling

Forward exciting force provides circular orbit with energy effectively, resulting in unstable whirling promptly

y

x

x

Forward exciting force can provide less energy to elliptical orbit resulting from anisotropic support and instability can be delayed

y

x L/D = 0.5 , mp = 0

5

4.60 Kx

x

2 1 0.5 0.2 10

highly anisotropic Kxx and Kyy K yy

0.36 0.05 0.1 0.2 0.5 0.44

5

4.41

2

1.28

1

2

S

Cxx yy

W

Kxy = Kyx = 0

10

C

LOP

dimensionless damping coeff. Cij dimensionless stiffness coeff. Kij

Fig. 4.8 Stabilization by elliptical orbit of whirling

1 0.5 0.2

Cxy = Cyx = 0

0.05 0.1 0.2 0.5 0.44

1

2

S

Fig. 4.9 5-pad tilting pad bearing (LOP)

a backward whirling component. This relationship is explained in detail in many textbooks. For example, Gasch [36] shows Fig. 4.8 in his book. In particular, Fig. 4.9 shows both stiffness coefficients and viscous damping coefficients are highly anisotropic (large difference between x-component and ycomponent) in the case of a five-pad tilting pad bearing with load direction of load on pad (LOP). In the case of the same bearing with load between pads (LBP), anisotropy decreases. On the other hand, Fig. 4.10 shows perfect isotropy in the case of a fourpad tilting pad bearing with LBP. Consequently, a five-pad tilting pad journal bearing with LOP is suitable for stabilization of rotating machinery with high destabilizing fluid forces. Figure 4.11 shows that the damping ratio of a rotor-bearing system (vertical ordinate) decreases with the increase in destabilizing fluid force (abscissa). With the LBP arrangement (low anisotropy), the system damping ratio is higher than that of the LOP arrangement (highly anisotropic), while destabilizing fluid force is small. However, the system damping ratio with the LBP arrangement decreases to

LBP

W y

x L/D = 0.5 , mp = 0

dimensionless damping coeff. Cij dimensionless stiffness coeff. Kij

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration

79

Kxy = Kyx = 0

10 5

Kxx

2.51 2

yy

isotropic Kxx and Kyy

1 0.5 0.2 10

=K

0.05 0.1 0.2 0.5 0.44

5

2

1

S

3.11

C xx =

2

C yy

1 Cxy = Cyx = 0

0.5 0.2

0.05 0.1 0.2 0.5 0.44

1

2

S

Fig. 4.10 4-pad tilting pad bearing (LBP)

damping ratio

damping ratio bearing width LBP low anisotropy

up

increase

bearing clearance up

increase

preload factor

decrease

up

LOP high anisotropy +

stable destabilizing fluid force

− unstable

Fig. 4.11 Behavior of damping ratio

zero rapidly with increasing destabilizing fluid force, resulting in rapid instability. Consequently, the LBP arrangement is suitable for rotating machinery with relatively small destabilizing fluid forces. With the LOP arrangement, the system damping ratio is lower than that of the LBP arrangement, while the destabilizing fluid force is small, but with increasing destabilizing fluid force the system damping ratio decreases slowly, maintaining stability to a greater extent than the LBP arrangement. Consequently, the LOP arrangement is suitable for rotating machinery with large destabilizing fluid forces. As shown in Fig. 4.11 [37], increasing the bearing width increases the system damping ratio, because the oil film area increases, resulting in a higher viscous damping effect. Increasing bearing clearance also increases the system damping ratio, because decreasing the oil film stiffness enhances journal movement in the bearing

80

4 Stability of a Rotor in Plain Bearings 5 (1) cylindrical bearing L/D = 0.5

4

3 – disk rotor C / R = 0.010

3

vc

C / R = 0.003

α = 0.52

2

1.74 5.23

1 C / R = 0.001 0 0.01

0.05 0.1

0.5

1

5

10

50

λ

Fig. 4.12 Stability chart of cylindrical bearing (short bearing approx.)

oil film, resulting in a higher viscous damping effect. On the other hand, increasing the pad preload factor decreases the system damping ratio, because increasing the oil film stiffness suppresses journal movement in the bearing oil film, resulting in a lower viscous damping effect. Selecting the bearing specification against destabilizing fluid force should be based on the knowledge mentioned above. Example 4.2 R1_Sect. 7.6 illustrates experimental results (R1_Fig. 7.19) and calculation results of complex eigenvalues obtained by means of the software MyROT (R1_Fig. 7.21) for a three-disk rotor supported in circular bearings at both ends, as explained in R1_Chap. 12. Figure 4.12 shows the stability chart for the rotor approximated as a Jeffcott rotor. Calculate the threshold shaft speed for the following three cases from this figure. In addition, compare the results obtained with the experimental results and the calculated complex eigenvalue results. Basic variables are prescribed as follows: D  2R  40 mm, L  24 mm (L/D  0.6), m  51.8 kg, ωs  48.7 Hz  306 rad/s Problem 1: C/R  0.001, C  0.02 mm, oil viscosity μ  27 cP  27 × 10−3 Pa s Problem 2: C/R  0.003, C  0.06 mm, oil viscosity μ  28 cP Problem 3: C/R  0.010, C  0.2 mm, oil viscosity μ  45 cP Answer 2  √ Problem 1 ωg  g/C  700 (rad/s)  111.4Hz → α  ωg /ωs  5.23 27 × 10−3 × 24 × 40 × 10−6  1 2 λ 0.001 (51.8 × 9.8/2) c  νc ωg  0.868 × 111.4  96.7 (Hz)

⎫  ⎪ 9.8/(0.02 × 10−3 ) ⎬  11.4 2π ⎪ ⎭

(4.16)

This solution  96.7, measurement  90, complex eigenvalue analysis  93.7 (rps)

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration

Problem 2 ωg 

√

81

 2 g/C  64.3Hz → α  ωg /ωs  1.74

28 × 10−3 × 24 × 40 × 10−6  1 2 λ 0.003 (51.8 × 9.8/2) c  νc ωg  1.35 × 64.3  86.8 (Hz)



⎫ ⎪ 9.8/(0.06 × 10−3 ) ⎬  0.76 2π ⎪ ⎭

(4.17)

This solution  86.8, measurement  87, complex eigenvalue analysis  86.8 (rps) 2  √ Problem 3 ωg  g/C  35.2Hz → α  ωg /ωs  0.52 45 × 10−3 × 24 × 40 × 10−6  1 2 λ 0.010 (51.8 × 9.8/2) c  νc ωg  2.27 × 35.2  80 (Hz)



⎫ ⎪ 9.8/(0.2 × 10−3 ) ⎬  0.06 2π ⎪ ⎭

(4.18)

This solution  80, measurement  92, complex eigenvalue analysis  88.4 (rps) Example 4.3 Destabilizing fluid force acts on a Jeffcott rotor (Fig. 4.2) supported in oil film bearings at both ends (mass m  296 kg, shaft bending stiffness k s  86.8 MN/m, shaft rotational speed N  7620 rpm). Calculate and compare the critical destabilizing fluid force corresponding to flow-excited vibration instability for the following two cases of support bearings: Case 1: five-pad tilting pad journal bearing (LOP) shown in Fig. 4.9 Case 2: four-pad tilting pad journal bearing (LBP) shown in Fig. 4.10 Use the following values: Journal diameter  76.2 mm, clearance ratio C p /R  2.51/1000, effective oil viscosity  10.8 mPa s, pad preload factor mp  0, L/D  0.5. The natural frequency of a simply supported rotor and the shaft elasticity index are evaluated from these given data as follows: ωs  542 rad/s, α  0.35. Answer Calculate the Sommerfeld number S: S

  μN DL R 2 W C

  10.8 × 10−3 × (7620/60) × 0.0762 × (0.0762/2) 1000 2   0.44 (4.19) (296/2) × 9.8 2.5 Figure 4.9 may be used to determine the dimensionless stiffness coefficients and viscous damping coefficients for Case 1. The bearing support anisotropy for this case is found to be very large: K x x  4.60, K yy  0.36, C x x  4.41, C yy  1.28 (dimensionless)

(4.20a)

k x x  70 MN/m, k yy  5.5 MN/m, cx x  84 kNs/m, c yy  24 kNs/m (dimensional) (4.20b)

82

4 Stability of a Rotor in Plain Bearings

Figure 4.10 may be used to determine the dimensionless stiffness coefficients and viscous damping coefficients for Case 2. The bearing support for this case is perfectly isotropic. K x x  K yy  2.51, C x x  C yy  3.11 (dimensionless)

(4.21a)

k x x  k yy  38 MN/m, cx x  c yy  59 kNs/m (dimensional)

(4.21b)

For each of the two cases, the dynamic properties of the rotor-bearing system and the complex eigenvalues calculated from Eq. (4.4) are as follows:

Tilting pad bearing

5-pad, LOP

4-pad, LBP

Simple support natural frequency

542 (rad/s)

542 (rad/s)

Shaft elasticity index

0.35

0.35

Undamped natural frequency ωx (X dir.)

425 (rad/s)

370 (rad/s)

Undamped natural frequency ωy (Y dir.)

119 (rad/s)

370 (rad/s)

First eigenvalue λ1

−69 ± j175

−56 ± j386

Second eigenvalue λ2

−40 ± j437

−56 ± j386

Third eigenvalue λ3

−1267, −1875

−1260, −1260

where the undamped natural frequencies in the x and y directions are calculated as follows:  √ ωx  2k x x ks /(2k x x + ks )/m  ω y  2k yy ks /(2k yy + ks )/m

(4.22)

Next, the effect of the destabilizing fluid force (dimensionless stiffness coefficient K c  K xy  −K yx ) is discussed. Figure 4.13 shows the variation of calculated eigenvalues with K c . The horizontal coordinate represents K c , and the vertical coordinate the damped natural frequency, Im[λ], and damping ratio, −Re[λ]/|λ|. The damping ratio ζ decreases with K c , becoming negative eventually. Then, flow-excited vibration takes place. The critical values of K cmax corresponding to instability are 2.0 for Case 1 and 1.5 for Case 2. Consequently, the rotor-bearing system for Case 2 becomes unstable when K c reaches 75% of the critical value for Case 1. The anisotropic bearing support is better than the isotropic support in suppressing flow-excited vibration of the rotor. Note: 3 Stability of vertical rotor in oil film bearing The static equilibrium of the journal center of a vertical rotor in circular bearing is located at the bearing center (eccentricity ratio  0), because the radial load is zero. When the journal

800

ω3 600

ω2

400 200

ω1

0.0

0.5

1.0

1.5

2.0

Kc = Kxy = −Kyx

0.8 0.6 0.4

ζ2 ζ1

0.2

0.0 −0.1

0

0.5

1.0

83

800

ω3

600

ω2

400

ω1 200 0.0

0.5

1.0

1.5

2.0

ζ3

0.8 0.6 0.4

ζ2

0.2

ζ1

0.0 0 −0.1

2.0

1.5

Kc = Kxy = −Kyx 1.0

ζ3

damping ratio ζ [−]

damping ratio ζ [−]

1.0

damped eigen frequency ω [rad/s]

damped eigen frequency ω [rad/s]

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration

0.5

Kc = Kxy = −Kyx

1.0

1.5

2.0

Kc = Kxy = −Kyx

(a) 5 –pad, LOP (highly anisotropic)

(b) 4 –pad, LBP (isotropic)

Fig. 4.13 Effect of load direction and pad number on stability z = ε Ce j θ Y

θ

z

Fθ

Fε X

Fig. 4.14 Oil film reaction force (ε0  0)

center is displaced slightly from the bearing center, as shown in Fig. 4.14, the small displacement is expressed as follows: z  εCe jθ

(4.23)

The oil film reaction force of Eq. (2.22) applied to a journal can be linearized with the small displacement εC; then, each of the radial and tangential components is expressed as follows: .

πε , Fε  k0 2

.   π ε(−2 θ ) μR L 3 Fθ  k0 k0  [Ns] 4 C2

(4.24)

84

4 Stability of a Rotor in Plain Bearings

Following the discussion about the stability of the rotor supported by the four oil film coefficients of the isotropic bearing in R1_Eq. (B.1) in Appendix B and R1_Eq. (7.3), the equation of motion is expressed as follows:        π k0 d0  [Ns/m] m z¨ + kd z  2 −Fε e jθ + j Fθ e jθ  d0 z − j z 2 C (4.25) where cd → d0 , kc → d0 /2, G  0

(4.26)

Then, the stability is determined by the sign of the coefficient of a on the right-hand side of the following expression:   d0  da − 1− a (4.27) dt 2m 2ω At the forward whirling natural frequency ωf , that is, when ω  ωf (> 0), the system is unstable and self-excited vibration grows when the coefficient is positive for shaft rotational speed  exceeding twice the natural frequency ωf . At the backward whirling natural frequency ωb , that is, when ω  −ωb (< 0), the system is always stable at any shaft speed; hence, unstable vibration never happens. In the case of oil whip of a horizontal shaft in cylindrical bearings, the journal whirls with a large orbit of radius nearly equal to the bearing radial clearance. The center of whirling is reasonably considered to be the bearing center (eccentricity ratio  0). Consequently, the threshold shaft speed  for oil whip of large amplitude can be expressed as twice the forward whirling natural frequency, ωf , from Eq. (4.27). Note: 4 Transition from oil whirl to oil whip Figure 4.1 showed the transition from oil whirl of small amplitude to oil whip of large amplitude in oil film bearing. The threshold for oil whirl of small amplitude is determined by means of stability criteria based on small vibration theories. On the other hand, oil whip of large amplitude exists in the region of shaft speed greater than twice the natural frequency, from Note 3. This transition is simulated as follows: Figure 4.15 shows the calculated variation of whirl amplitude with increasing shaft speed for an unbalanced Jeffcott rotor. The upper figure (a) shows the envelope of whirling of the mass at the mid-span of the rotor (dynamic component only), while the lower figure (b) shows the waveform and its envelope (inclusive of steady-state component) of the journal center whirling, respectively, with increasing shaft speed. Furthermore, the figures of whirl orbits and results of FFT analyses are shown. The lower figure (b) of the envelope variation shows that the journal center moves upwards as a whole with shaft speed. The unbalance resonance vibration is found to take place around the shaft speed p  /ωn  1 ➁, and the FFT analysis shows a synchronous vibration component only.

4.7 Journal Bearing Specification to Suppress Flow-Excited Vibration

85

4 1

disk center loci

1

2

1 −1

1

1

−1

5

0

(a) disk vibration

1 shaft speed p =

−10

2

−10

2

1 2 frequency

×

3

1

0 vibration spectrum [dB]

−1 2

−1

1

−1

−1

/ω s

3

3 oil whirl

0

1

−0.5

1 shaft speed p =

1

0

0.5 −1

0

(b) journal vibration

0 10

4

1

0

−20

3

0.5 −0.5

−0.5

2 unbalance resonance

−10

1

2

−0.5

0

/ω s

0.5

0.5

1

10

−4 vibration spectrum [dB]

disk vibration amp. × C

4 −10

10

0

journal center loci

−4

−1

−1

vertical envelope of journal center vibration × C

10

4

3

2 unbalance resonance

−20 −40 −60 20

3 oil whirl

0 −20 −40

0

1 2 frequency

3 ×

Fig. 4.15 Transition from oil whirl to oil whip

When the shaft speed reaches around p  1.3 ➂, unstable vibration (self-excited vibration) takes place as the stability analysis predicts. The FFT analysis of the journal whirling shows a peak component corresponding to around half the rotational speed  (not exactly half-speed), so it is often called half-speed whirl. The journal orbit size is very large, nearly equal to the bearing clearance. In contrast, the orbit size of the mass is relatively small, about twice that of the journal center. This unstable vibration is called oil whirl corresponding to nearly rigid rotor mode whirling of small amplitude. With further increase in shaft speed, the whirl amplitude of the mass grows large rapidly toward the shaft speed of p  2 ➃. This is oil whip corresponding to rotor bending mode whirling of large amplitude.

86

4 Stability of a Rotor in Plain Bearings

Thus, at first, a small half-speed whirl of rigid mode starts, and its frequency is around /2. With increasing shaft speed, the whirl frequency approaches the natural frequency of the shaft bending mode ωn , and then, the whirl mode shifts from a rigid mode to a bending one, increasing its whirl amplitude rapidly. Finally, at /2  ωn , the whirl amplitude reaches its maximum. Note that the size of the journal orbit is always nearly equal to the bearing clearance, once the system becomes unstable. Reference [R1_Ref. 66] explains in detail the measurements of waveforms and orbits of the mass center, which was introduced in R1_Sect. 7.6. However, the journal whirl orbit has been seldom compared with the mass center whirl orbit. References [38–57] present recent results of self-excited vibrations of various rotor-bearing systems. Example 4.4 Example 4.1 shows the dimensionless stability threshold shaft speed ν c is nearly equal to 2.0 for λ  0.2 and α  1/4 in the case of the cylindrical bearing with two oil grooves shown in Fig. 4.5. Obtain ν c when the oil viscosity is halved with the other conditions unchanged. Furthermore, find ν c for α  0 in the case of these new conditions given above. Answer The dimensionless bearing modulus λ corresponding to the new conditions is 0.1 from its definition: μ λ≡ pm

 2 √ R g/C C 2π

Consequently, the new corresponding threshold speed vc reads 2.5, which is higher by 25% than that of the original case. Furthermore, in the case of α  0, the corresponding stability limit reads 2.75, which is higher by 38% than that of the original case. Example 4.5 Obtain ν c when the bearing clearance C is doubled with the other conditions unchanged given in Example 4.1. Furthermore, find ν c for α  0 in the case of these new conditions given above. Answer The dimensionless bearing modulus λ corresponding to these new conditions is 0.035, resulting in ν c  3.8. Furthermore, in the case of α  0, the corresponding stability limit reads 4.0.

Chapter 5

Vibration of Rolling Element Bearings

Abstract Rolling element bearings are used widely in rotating machinery. A significant variety of rolling element bearing types is available, and they are categorized by the shape of the ball or roller elements; the radial or axial loading direction; the ball type such as deep groove or angular contact; the roller type such as plain or tapered; etc. As most of them are standardized in ISO/JIS, interchangeability is guaranteed. This chapter introduces the dynamic properties of some rolling bearings, but does not make any reference to the static properties given in bearing manufacturers’ catalogues, such as load capacity, sizing, life estimation, and so on. Keywords Ball/roller bearing · Stiffness formulae · Exciting frequencies · Ball pass frequencies · Ringing · Diagnosis · FFT · Amplitude modulation · Envelope · Fault diagnosis · Case studies · Preload · Deterioration

5.1 Stiffness of Rolling Element Bearings [58, 59, 60] In calculating rotating shaft vibration, a rolling element bearing is usually modeled as an elastic spring element. The bearing stiffness (spring constant) is expressed as a nonlinear function of the applied bearing load, derived using Hertzian contact theory. This section introduces the simplified formula [58] of the spring constant derived from calculated contact deformation. The damping effect of a rolling element bearing is normally assumed to be negligibly small (Table 5.1).

5.1.1 Stiffness in the Radial Direction: kr A radial load, F r , is distributed to each rolling element, depending on each element position. When an element lies just under the load vector, the load it carries becomes the maximum, Qmax , in the direction of contact angle α (Fig. 5.1), and its radial component is given by Qmax cosα, where α is the contact angle. Stribeck [59] derived Eq. (5.1) to give the relation between F r and Qmax , with the centrifugal force acting on the rolling element being neglected. N represents the total number of rolling © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_5

87

88

5 Vibration of Rolling Element Bearings

Table 5.1 Rolling element bearings Ball bearing Types of ball layouts

Radial bearing Thrust bearing Miniature bearing (outer diameter< φ 9mm)

(a) Deep groove (b) Angular contact (c) Self-aligning, etc

Roller bearing Types of rollers Radial bearing

(a) Plain roller (b) Tapered roller (c) Spherical roller . etc

Thrust bearing

referring to ISO 15/104 (JIS 1512 Rolling element bearings)

elements (in the case of double-row bearings, N is replaced by 2 N). The expression for the maximum radial load is Q max 

5Fr (N) N cos α

(5.1)

where the coefficient 5 of Eq. (5.1) corresponds to a bearing with typical clearance. The coefficient in the case of no clearance is theoretically 4.37 according to the published literature, such as [58]. The elastic deflection, δ r , in the radial direction for each type of the following bearings, is derived from Hertzian contact theory as follows:

contact angle α

deep groove

α = 0° (nominal) ball retainer

inner ring ball race

outer ring

Fig. 5.1 Nomenclature for a ball bearing

angular 15 ~ 40°

5.1 Stiffness of Rolling Element Bearings

89

• Deep groove bearing and angular contact ball bearing: δr  4.364 × 10−8

2/3

Q max (m) 1/3 B cos α

(5.2a)

• Self-aligning ball bearing: δr  6.986 × 10−8

2/3

Q max (m) B 1/3 cos α

(5.2b)

• Roller bearing (line contact with inner and outer races): δr  3.059 × 10−10

Q 0.9 max (m) cos α

L 0.8

(5.2c)

where B  ball diameter (m) and L  effective roller length (m). When Qmax in each of Eqs. (5.2) is substituted with that in Eq. (5.1), the radial deflection, δ r , can be expressed as a function of the radial load, F r . By partially differentiating F r with respect to δ r , the bearing stiffness in the radial direction for each type of bearing is derived in the following equations: • Ball bearing kr ≡

∂ Fr  Cr × 106 × Fr1/3 B 1/3 N 2/3 cos5/3 α (N/m) ∂δr

(5.3)

where C r  11.76 (for deep groove bearing and angular contact bearing) C r  7.34 (for single-row self-aligning ball bearing. N is doubled in the case of double-row bearings.) • Roller bearing kr ≡

∂ Fr  Cr × 106 × Fr0.1 L 0.8 N 0.9 cos1.9 α (N/m) ∂δr

(5.4)

where C r  853.3 (line contact assumed on both inner and outer races). Figure 5.2 shows these relations between F r and k r . The load dependence of k r for a roller bearing is smaller than that of an equivalent ball bearing. Example 5.1 Obtain the radial stiffness of the following bearings from Eqs. (5.3) to (5.4) in the case of a radial load of 100 (N): (1) Deep groove ball bearing (ball diameter B  12.7 mm, number of balls N  9, contact angle α  12°) (2) Angular contact ball bearing (ball diameter B  6.35 mm, number of balls N  10, contact angle α  30°)

90

5 Vibration of Rolling Element Bearings 10

10

α 1.9

cos

Roller

9

5

0.9

kr [N/m]

10

Deep groove · Angular contact

N 0.8

B

L

5/3 cos α

N

kr [N/m]

2/3 1/3

Radial stiffness

5

8

10 5

Self-aligning balls (per single row) 107 10

10

2

Radial load

10

3

10

4

Fr [N]

Fig. 5.2 Radial stiffness depending on radial load

(3) Cylindrical roller bearing (effective roller length L  9.62 mm, number of rollers N  12, contact angle α  0°). Answers (1) 53 (2) 37 (3) 308 (MN/m) Example 5.2 Figure 5.3a shows a rotor supported in identical ball bearings (angular type) at both ends. Rotating disk: mass m  11 kg, shaft mass ms  9 kg, and shaft stiffness k s  147 × 103 N/m Ball-bearing specification: ball diameter B  6.35 mm, number of balls N  10, and contact angle α  30° Span between bearings l  1.1 m Bearing inner diameter d i  10 mm (1) Obtain the radial stiffness, k b , of the ball bearing from Eq. (5.3) when the radial force applied to the bearing is 100 N. (2) Obtain the natural frequency, f n , of this system with the equivalent mass of the shaft mass being taken into account.

m = 11 kg

(a) m s = 9 kg

ks =

48EI 3 l

= 147 kN/m

(b)

m eq ks kb

Fig. 5.3 A rotor supported by ball bearings and an equivalent dynamical model

kb

5.1 Stiffness of Rolling Element Bearings

91

Answers (1) From Eq. (5.3), kb  11.76 × 106 × 1001/3 × (0.00635)1/3 × 102/3 × cos5/3 30◦  37 MN/m (2) The equivalent system is shown in Fig. 5.3b with Equivalent mass: m eq  m + 0.49m s  11 + 0.49 × 9  15.4 kg (refer to R1_Fig. 2.4c) Equivalent spring constant: keq  1/(1/ks + 1/(2kb )) ≈ ks  147 kN/m Natural frequency: f n  1/(2π ) keq /m eq  15.5 Hz

5.1.2 Thrust Direction (Axial) Stiffness: kz When the load, F z , is applied to a bearing, only in the thrust (axial) direction, the individual load, Q, carried uniformly by each rolling element is given in the following equation, with the total number of rolling elements as N and the contact angle as α: Q

Fz (N) N sin α

(5.5)

Then, the deflection, δ z , in the thrust (axial) direction is given in each of the following equations: • Thrust ball bearing: δz  5.247 × 10−8

Q 2/3 (m) sin α

B 1/3

(5.6a)

• Deep groove bearing and angular contact ball bearing: δz  4.364 × 10−8

Q 2/3 (m) sin α

(5.6b)

Q 2/3 (m) sin α

(5.6c)

Q 0.9 (m) sin α

(5.6d)

B 1/3

• Self-aligning ball bearing: δz  6.987 × 10−8

B 1/3

• Roller bearing: δz  3.059 × 10−10

L 0.8

92

5 Vibration of Rolling Element Bearings 10

10

α 1.9

sin

kz [N/m]

10

Roller

9

Deep groove · Angular contact

5

0.9

N 0.8

L

α 5/3

sin N B

kz [N/m]

2/3 1/3

Thrust stiffness

5

8

10 5

Self-aligning balls (per single row)

107 10

10

2

10

Thrust load

3

10

4

Fz [N]

Fig. 5.4 Thrust stiffness depending on thrust load

where B  ball diameter (m) and L  effective roller length (m) Then, the axial stiffness can be calculated in the same way as the radial stiffness, resulting in Eqs. (5.7) and (5.8). Figure 5.4 shows the variation of the stiffness with thrust load. • Thrust stiffness of ball bearing kz ≡

∂ Fz C z × 106 × Fz1/3 B 1/3 N 2/3 sin5/3 α (N/m) ∂δz

(5.7)

C z  28.6 (thrust ball bearing) C z  34.4 (deep groove bearing and angular contact ball bearing) C z  21.4 (self-aligning ball bearing. The number of balls, N, in a single row shall be applied even in the case of double-row bearings.) • Thrust stiffness of roller bearing kz ≡

∂ Fz C z × 106 × Fz0.1 L 0.8 N 0.9 sin1.9 α (N/m) ∂δz

(5.8)

C z  3633 (line contact assumed on both of inner and outer races) Example 5.3 Obtain the spring constants of the following thrust bearings, in the case of axial load of 100 (N): (1) Thrust ball bearing (ball diameter B  12.3 contact angle (α)  90°) (2) Angular contact ball bearing (ball diameter B  6.35 contact angle (α)  30°)

mm,

number

of

balls

(N)



10,

mm,

number

of

balls

(N)



10,

5.1 Stiffness of Rolling Element Bearings

93

(3) Tapered roller bearing (roller effective length L  6.2 mm, number of rollers (N)  12, contact angle (α)  30°) Answers (1) 142 (2) 43 (3) 247 (MN/m) Example 5.4 Obtain the thrust stiffness of the deep groove ball bearing defined in Example 5.1, in the case of axial load of 100 (N). Answer k z  34.4 × 106 × 1001/3 (0.0127)1/3 92/3 sin5/3 12◦  12 MN/m

5.2 Excitation Frequency of a Rolling Element Bearing [60, B46] 5.2.1 Whirling and Rotational Frequencies of Rolling Elements Figure 5.5 shows an example of the structural specifications for a ball bearing. The inner race is assumed to revolve at the rotational speed of , and the outer race is fixed to the stationary housing. Under this condition, each rolling element rotates (or spins) and at the same time whirls in the same direction of the rotating shaft. Let us obtain the whirl frequency and rotational frequency with the following variables defined:

α

Fig. 5.5 An example of ball-bearing parameters

Outer Race

T

B

Ball

Retainer

T Inner Race

An example of a ball bearing NSK 6207

P

B = Ball diameter = 11.112 mm φ N = Number of balls = 9 α = Contact angle = 12°

P = Pitch diameter = 53.5 mm φ

94

5 Vibration of Rolling Element Bearings

P  pitch circle diameter of bearing B  diameter of rolling element (B/P ≈ 1/4) α  contact angle r 1  radius of inner race  (P − B cos α)/2 r 2  radius of outer race  (P + B cos α)/2 The whirling speed, ν c , of the rolling element is an average of the surface speed of the inner race (r 1 ) and that of the outer race (0) with no slip of rolling element assumed on both race surfaces. Thus, νC 

1  (r1  + r2 0)  (P − B cos α)/2 2 2

(5.9)

With the whirling frequency, c , of the rolling element, the whirling speed, ν c , can be expressed as follows: νC 

P C 2

Substituting Eq. (5.10) into Eq. (5.9) results in the following equation:   B 1 1 − cos α  C  2 P

(5.10)

(5.11)

With the specifications in Fig. 5.5 applied, an approximate value of c is given as follows: C ≡ β   0.3984  ≈ 0.4 

(5.12)

This equation indicates that the rolling element or the retainer (cage) whirls around the shaft at about 40% of the shaft rotational speed  in the same direction. Thus, this speed c is called the fundamental train frequency (FTF):   B 1 1 − cos α  (5.13) FTF C  2 P Next, the ball spin frequency, s , of a rolling element is calculated. As shown in Fig. 5.6, the whirling rolling element covers the distance c r 2 t on the stationary outer race within t seconds, and this distance is equal to the spinning distance s (B/2) t of rolling element under no-slip condition. Then, the following equation is derived: S

B  C r2 2

(5.14)

Thus, the spin frequency, s , called the ball spin frequency (BSF), is given by the following equation:

5.2 Excitation Frequency of a Rolling Element Bearing Fig. 5.6 Spin and whirl motion of a rolling element

95 t =0

Δ

L

t

t=

S

r1

B C×

r2

Δt

   2 P B 2 BSF  S  cos α  ≈ 2.3  ≈ 2  1− 2B P

(5.15)

A rolling element is found to spin at about twice the shaft rotational speed, , in the opposite direction. If there is a scratch on the surface of rolling element, it hits both surfaces of the inner and the outer races. Then, the impacts occur twice per revolution of the rolling element, which generates the exciting frequency of 2 × BSF.

5.2.2 Exciting Frequency Caused by a Traveling Ball When there is a spot scratch on the outer race surface, an impact vibration response is excited every time a rolling element passes the spot. This frequency of this occurrence is called the ball pass frequency of the outer race (BPFO), which is equal to the product of the whirl frequency, c , of each rolling element and the total number of balls N:   B N 1 − cos α  ≈ 0.4N  (5.16) BPFO C N  2 P Next, let us consider a local scratch on the inner race surface. When observed from the rotating inner race on the shaft, a ball whirl is in the backward sense at a frequency given by the following equation:   B 1 C −   − 1 + cos α  ≈ −0.6  (5.17) 2 P Therefore, the frequency of the balls on passing the scratch (ball pass frequency of the inner (BPFI) race) is given by the following equation:

96

5 Vibration of Rolling Element Bearings

Table 5.2 Exciting frequencies of rolling element bearings (fixed inner race) FTF =

Fundamental Train Frequency FTF (caused by retainer/cage defects or improper movement)

BPFO = N

Ball Pass Frequency of Outer race BPFO (caused by rolling elements passing over defective outer races)

fc = 0.39 fr

fr B 1 − cos α P 2

(

)

fo = 3.6 fr

fr B 1+ cos α P 2 r

Ball Spin Frequency BSF (caused by rolling element defects)

B = Ball diameter = 11.112 mm φ N = Number of balls = 9

BPFI ( − C )N 

)

( ) P f BSF = ( ) (1 − ( PB ) cos α ) B 2 BPFI = N

Ball Pass Frequency of Inner race BPFI

(caused by rolling elements passing over defective inner races)

NSK 6207

(

fr B 1 − cos α P 2

α = Contact angle = 12° P = 53.5mm φ

fi = 5.4 fr

2

2

}

fs = 2.3 fr

a typical example

  B N 1 + cos α  ≈ 0.6N  2 P

(5.18)

In summary, Table 5.2 shows the various exciting frequencies generated by a rolling element bearing. All of the frequencies are collectively referred to as pass frequencies. In Table 5.2, the rotational speed of the shaft is expressed as f r (Hz). Example 5.5 Obtain the pass frequencies of the ball bearing in the case of the stationary inner race and the rotating outer race having the rotational speed , as shown in Fig. 5.7. Answer The whirling speed, vc , of a ball is obtained from Eq. (5.9) as     P B 1  P B + cos α  1 + cos α νC  (r1 0 + r2 )  2 2 2 2 4 P Then, each of the pass frequencies is given as follows: Whirling frequency of ball or cage (FTF, fundamental train frequency):

Fig. 5.7 Fixed inner race and rotating outer race

(5.19)

5.2 Excitation Frequency of a Rolling Element Bearing

FTF C 

97

  1 B νC  1 + cos α  ≈ 0.6 P/2 2 P

(5.20)

Ball pass frequency of the inner (BPFI) race:   N B BPFI C N  1 + cos α  ≈ 0.6N  2 P

(5.21)

Ball pass frequency of the outer (BPFO) race:   B N 1 − cos α  ≈ 0.4N  BPFO ( − C )N  2 P

(5.22)

Ball spin frequency (BSF):    2 2 P B 2 BSF S  Cr1  cos α  ≈ 2.3 1− B 2B P

(5.23)

Example 5.6 Obtain the pass frequencies of the ball bearing when the inner and outer races are rotating at I and O , respectively, as shown in Fig. 5.8. Answers The whirling speed, vc , of a ball is obtained from Eq. (5.9) as     B O P B 1 I P νC  (r1  I + r2  O )  − cos α + + cos α 2 2 2 2 2 2 2

(5.24)

Then, each of the pass frequencies is given as follows: Whirling frequency of ball or cage (FTF, fundamental train frequency):     O B I B νC  1 + cos α + 1 − cos α FTF C  P/2 2 P 2 P

(5.25)

Ball pass frequency of the inner (BPFI) race:

Fig. 5.8 Rotating outer race and inner race

O

I

98

5 Vibration of Rolling Element Bearings

B

contact points B

contact points

P X

r2

r1

X (a) Retainer adhered to the outer race

X

P X

(b) Retainer adhered to the inner race

Fig. 5.9 Ball spin frequency in the case of the retainer (cage) adhered to races

BPFI | I − C |N 

  N B | I −  O | 1 + cos α 2 P

(5.26)

Ball pass frequency of the outer (BPFO) race:   N B BPFO | O − C |N  | O −  I | 1 − cos α 2 P

(5.27)

The ball spin frequency (BSF) is given by the following equation, by substituting the relative whirl frequency of the cage (C − O ) observed from the outer race (contact radius r 2 ) into C of Eq. (5.14):    2 2 P B 2 | I −  O | 1 − BSF  S  |C −  O |r2  cos α (5.28) B 2B P Alternatively, the BSF is given by the following equation, by substituting the relative whirl frequency of the cage (C − I ) observed from the inner race (contact radius r 1 ) into C of Eq. (5.23):    2 2 P B 2 | I −  O | 1 − BSF  S  |C −  I |r1  cos α (5.29) B 2B P Example 5.7 Ball spin frequency when the cage is adhered to the bearing inner or outer races Figure 5.9 shows the following two cases: (1) Obtain the BSF with the cage adhered to the outer ring, as shown in Fig. 5.9a. (2) Obtain the BSF with the cage adhered to the inner ring, as shown in Fig. 5.9b. Answers The balls are assumed to slip on the fixed ring surface and roll on the opposite ring surface. (1) The BSF is given by the following equation, by equating the circumferential surface velocity of the spinning ball with that of the inner race:

5.2 Excitation Frequency of a Rolling Element Bearing

S

99

  B B P  r1  →  S  1 − cos α  2 B P

(5.30)

(2) The BSF is given by the following equation, by replacing the whirl frequency, C , of the cage in Eq. (5.14) by shaft rotational angular velocity :   B B P 1 + cos α  (5.31)  S  r2  →  S  2 B P Addendum 1 In Sect. 5.2.2, the pass frequencies are derived with only one local scratch existing. When the race surfaces are wavy with no scratch, vibrations appear with the frequencies of BPFO and/or BPFI and/or those equal to the integer multiple of BPFO and/or BPFI. As a result, the pass frequencies spread over a wide range, typically between 0 and 5 kHz. In practice, vibration waveforms are detected by means of velocity sensors or acceleration sensors installed at bearing boxes, diagnosed using FFT analyzers. One of the examples is described in VB423.

5.2.3 Race Vibration (Ringing) of Rolling Element Bearings When rolling elements pass local scratches on race surfaces, impact vibrations of inner or outer races may be excited. These vibrations are called ringing vibrations. Although a theoretical model of this vibration has not been established, it can be represented by simplified modeling involving bending vibration [61] of a “cylindrical shell,” as shown in Fig. 5.10. The natural vibration modes of a circular ring have a differing number n of diametral nodes. The natural frequency, f n , of the nth mode of vibration is estimated by the following formula: 1 n(n 2 − 1) fn  √ 2π R 2 1 + n 2

+ −

− + n=1

+ − n=2

EI (Hz) ρA

(5.32)

+

+

− +



− +

− −

+

−

−

−

−

+

+

+ n=3

Fig. 5.10 Elastic vibration modes of ring in rolling element bearings

n=4

100

5 Vibration of Rolling Element Bearings

√ where R  radius of ring (m), C  E/ρ  speed of sound (m/s), I  second moment of area (m4 ), and A  cross-sectional area (m2 ). The natural frequency, f n , of the nth mode of vibration of the outer race is given by the following expression, when the outer race made of steel (ρ  7.8 × 103 kg/m3 , E  206 GPa, and C  5140 m/s) is assumed to have a rectangular cross-section (thickness: h): f n  236

h n(n 2 − 1) (Hz) √ R2 1 + n2

(5.33)

Note that this expression has been widely cited. Example 5.8 Ball bearing (NSK 6207) B  11.112 φ mm, P  53.5 φ mm, N  9, α  12°, and R  33.6 mm Estimate the natural frequencies (up to the fourth mode) of the outer ring with its thickness of 4 mm assumed. Answer The natural frequencies are obtained from Eq. (5.33) as follows: f n  236

0.004 n(n 2 − 1)  {2.2, 6.4, 12.2, 66.5} (kHz) √ 0.03362 1 + n 2

(5.34)

5.3 Diagnosis of Rolling Element Bearing Vibrations and Signal Processing 5.3.1 Acceleration Vibration Waveform of Bearing Box For quick and simple diagnosis, one end of a probe is applied to bearing boxes or bearing housings to enable an operator to listen to the noise from the other end of the stick. For more precise diagnosis, accelerometers are fixed firmly on bearing housing surfaces to detect and transmit vibration waveforms emitted from rolling element bearings for signal processing. Figure 5.11 shows typical waveforms detected and with corresponding frequency analysis results for various defects of a ball bearing (NSK 6207) with its specifications listed in Table 5.2. The rotational pulse signal ➀ at the 0.02 s period interval indicates the shaft rotating frequency, f r  50 Hz. The bearing condition is represented by a normal signal ➁, since the vibration amplitude is small. The vibration amplitude increases somewhat due to normal wear during operation and/or poor lubrication after an extended period of operation ➂, but no dominant vibration frequencies are evident. Case ➃ corresponds to the dominant frequency caused by an outer race defect, showing the BPFO and the frequencies of its integer multiples. Case ➄ corresponds to

5.3 Diagnosis of Rolling Element Bearing Vibrations and Signal …

101

features of beating waveforms caused by an inner race defect, showing the amplitudemodulated with the frequency of f r (shaft rotational frequency). The spectral distribution indicates main peaks of n × BPFI and associated peaks with double sideband frequencies, ±f r , i.e., dominant frequencies, n × f i ± fr . The beating is generated by periodic change of the relative distance between the sensor and the sound source 1 of the scratches on the inner race at a rotation period of f − r . Case ➅ corresponds to the beating waveforms caused by a ball surface defect, showing the amplitude-modulated with the frequency of f c (cage whirling frequency). The spectral distribution includes main peaks of n × 2 × BSF (n  1, 2,…) and associated peaks with double sideband frequencies, ±f c , i.e., dominant frequencies n × 2 f s ± f c , since the relative distance between the sensor and the sound source of the scratches on the defective ball surface varies periodically with 1 the cage whirling period of f − c .

2

0.8

Amplitude

2 Normal condition

Vibration

1 Rotational pulse

0

−2

excessive wear

0.08

0 0.02

0.04

0.06

0.08

0 0

0.02

0.04

0.06

2fs ± fc

200

400

600

800

0

200

400

600

800

400 600 270 ± 50

800

180

0 0

200

0

200

0.8

Amplitude 0.02

0.04

0.06

0.08

0

0

−4 0

0.02

0.04 0.06 Time [s]

0.08

400

600

800

200 400 600 Frequency [Hz]

800

230 ± 20

0.8

4

Amplitude

Vibration

0

0.08

Vibration

4 5 Inner race defects and the induced 0 amplitude modulated by rotational frequency −4 0 fi ± fr

6 Ball defects and the induced amplitude modulated by FTF

0

0.8

4

−4

0

0.8

Amplitude

fo

0.06

−4

Vibration

defects

0.04

4

0

4 Outer race

0.02

Amplitude

3 Inadequately lubrication and/or

Vibration

0

0 0

Fig. 5.11 A diagnosis procedure of ball-bearing defects (waveform and FFT results), fr  50, f c  20, f i  270, f o  180, f s  115

102

5 Vibration of Rolling Element Bearings

5.3.2 Amplitude Modulation Beat vibrations shown in Fig. 5.11 ➄ appear under the condition that the BPFI vibration, caused by a scratch on the inner race surface, is dominant. Also, the vibration is amplitude-modulated with the beat frequency (i.e., shaft rotational frequency, f r , in this case). On the other hand, the beat vibrations shown in Fig. 5.11 ➅ appear under the condition that the 2 × BSF excitation caused by the defective ball is dominant and also that the vibration is amplitude-modulated with the beat frequency (i.e., cage whirling speed, f c , in this case). In both cases, because the exciting sources (the scratch or the defective ball) of bearing repeat approaching and leaving the fixed sensor at the timing of frequencies of f r and f c , the magnitude of the signal detected by the sensor repeats by increasing and decreasing at the same frequency, which means the signal is amplitude-modulated. When an acceleration signal, y1 , of a pass frequency, ω, is detected at the bearing box, the following expression is appropriate: y1  a cos ωt

(5.35)

If the signal is amplitude-modulated with the magnitude of and the frequency ωc  2πf c 0 flows in the coil of the left-side electromagnet, the command circuit ensures the current I  I 0 − i(t) > 0 in the coil of the right-side electromagnet. Note that the circuit does not accept the command of I < 0, because there is no repulsive force capability. This is the conventional method to control coil currents, but in some special methods, the bias current I 0 is set to be zero (I 0  0) in order to save electric energy in the steady state. The difference between the two control forces of the magnetic bearing is now examined, with or without the bias current, I 0 , in the coils. Assuming the sinusoidal input i(t)  sin ωt with no bias current, I 0  0, the left-side coil is energized with i(t) > 0, and the right-side coil is energized with i(t) < 0 alternately. Thus, Fig. 6.11 (1) shows the variation of the corresponding specific control force (Px  F x /(DL)) of the magnetic bearing with U in the case of no bias current, I 0 . Also, the Px –U curve is obtained from Fig. 6.9 with U > 0 and U < 0 implemented. It is seen to have a dead zone in the low U region and a nonlinear characteristic overall. Center line

I0 − i

I0 + i

F2

F1 Fb = K δ0

x Fb = F2 − F1

[

≈ 4K

(

where K ka

(I0 - i) 2 ( δ 0 + x)2

( δ 0 - x) 2

I 02 I0 x i− δ 02 δ 03 μ 0 SN 4

2

)

= kai − knx

cos 22.5°

4K I 02 , δ0

Fig. 6.10 Current–force characteristics; linearization by AMB bias current I 0

]

(I0 - i) 2

−

kn

4K

I 02 δ 03

126

6 Vibration in Magnetic Bearing Rotor Systems

0

−2

Saturation

Saturation

Neglecting

Neglecting

Considering

Dead zone

U [AT] Output 1

2 3 ×10

Px [MPa]

−1

1.0 Px [MPa]

F Px = x DL

Saturation Considering

0.2 0 −0.2

Input = ± 1 000 Amplitude [AT]

Px [MPa]

1.0

−1.0

NI0 = 0

Saturation

Neglecting

Neglecting

Considering

Saturation Considering

0.5 U [AT] 1

−1

(1) Bias

Saturation

Input = ± 1 000

×10

3

Output

Px [MPa]

−1.0

Amplitude [AT]

0 −0.5

(2) Bias

NI0 = 800

Fig. 6.11 Linearization with a bias current

On the other hand, if the bias current, I 0 , has enough magnitude to generate the bias magnetic flux density Bg , e.g., Bg  1 T, and is superimposed on the control currents i(t) in both right and left coils, as shown in Fig. 6.11 (2), the ideal model with the magnetic saturation neglected can give the linear Px –U relationship with the dead zone eliminated. In this manner, the input and output relations are practically improved to achieve a linearized force–current characteristic. The following linear relation between the input control current i and the output magnetic bearing control force F b is also shown in Fig. 6.10: Fb ≈ ka i − kn x (ka > 0, kn > 0)

(6.16)

The resultant force, F b , should be totally positive as it works in the opposite direction of the displacement x. However, the second term (−kn x) in the above equation represents a negative force that increases the displacement of rotor by the action of the bias magnetic flux. Therefore, its coefficient, k n , is referred to as the negative spring (negative stiffness). Figure 6.12 shows the block diagram of the Eq. (6.16), where m represents the rotor mass, k n is a minor loop of the negative stiffness, Gr (s) represents the magnetic bearing controller and the dotted arrow line shows the feedback of the displacement signal. Consequently, a feedback control circuit is needed to produce a positive spring force by adding the dotted line with the controller, Gr (s), which should be designed to provide a positive stiffness and positive damping to the entire system.

6.2 ISO Standards Related to Magnetic Bearings

127 Rotor

Gr

-

i

Controller

ka

+

Fb

+

1 m s2

x

kn

Fig. 6.12 Block diagram of AMB rotor system

6.2 ISO Standards Related to Magnetic Bearings [93] An active magnetic bearing is defined as a bearing in which the magnetic attraction generated by the electromagnet is feedback-controlled to magnetically levitate the journal of the bearing. Since the bearing stiffness is very soft with a large bearing clearance, in contrast to rolling bearings or oil film bearings, and also the bearing force needs always to be controlled by means of control circuits, the mechanism of the bearing function is more difficult to understand than that of conventional bearings. Because of this background, the following ISO standards have been established in order to promote the active magnetic bearing business: ISO 14839-1: ISO 14839-2: ISO 14839-3: ISO 14839-4:

Vocabulary of active magnetic bearing: Explanation of the terms used in this field Evaluation of vibration: Vibration standards for active magnetic bearing rotor Evaluation of stability margin: Quality evaluation of stability in control system Technical specifications for magnetic bearing systems.

6.2.1 Terms Related to Active Magnetic Bearings (ISO 14839-1) Some of the terms defined in this standard are introduced as follows: (1) Symbols of active magnetic bearings Figure 6.13 shows the graphical symbols of some active magnetic bearings, together with those of some rolling element bearings. (2) Number of poles Figure 6.14a illustrates that the poles in the normal heteropolar-type active magnetic bearing have the following circumferential sequence of N and S poles: the repetition of N, S, S and N. On the other hand, Fig. 6.14b illustrates the homopolar-type bearing has the same polarity of poles in the circumferential direction. The number of poles is defined as the total number of N and S poles energized. Figure 6.14 shows eight-pole bearing types.

128

6 Vibration in Magnetic Bearing Rotor Systems Deep groove ball bearing

Thrust ball bearing

Radial active magnetic bearing (with sensor)

Axial active magnetic bearing (with sensor)

Fig. 6.13 Graphical symbols for bearings X

Y

S S

NN NN

Y

S S

X

S

S

S

S

8 poles

(a) Heteropolar type

X

S

N

S

N

Y

N

N

N

N

8 poles

(b) Homopolar type

Fig. 6.14 Number of poles of radial AMB

(3) Active magnetic bearing (AMB) Figure 6.15 illustrates the configuration of an active magnetic bearing. When the journal at equilibrium moves downward, the current is applied to the coil of the upper electromagnet, which is energized to pull the rotor upward. When the journal moves upward, the current is applied to the coil of the lower electromagnet (not shown in the figure), it is energized to pull the rotor back to equilibrium. Thus, the spring action of the electromagnets may keep the journal at its neutral position. If the coil is energized ahead of the upward or downward movement of the journal, damping action can be produced. The feedback control force is generated in the following sequence: Vibration displacement (m) → sensor amplifier (V/m) → controller (V/V) → Power amplifier (A/V) → electromagnet (N/A) → electromagnetic force (N). Therefore, because the electromagnetic force (N) is produced in proportion to the vibration displacement (m), the dynamic characteristics (spring and damping functions) of an active magnetic bearing can be evaluated in the same way as for an oil film bearing. (4) Block diagram of AMB rotor system Figure 6.16 shows a block diagram consisting of the dynamic characteristics of rotor and active magnetic bearing in order to calculate the system’s vibration characteristics.

6.2 ISO Standards Related to Magnetic Bearings

129

Electromagnet DC power supply

Power amplifier [A] F [N]

[V]

x [m]

Rotor

Controller

Displacement sensor

[V]

Fig. 6.15 Principle of active magnetic bearing Fd [N] Disturbance force Reference +

–

AMB Controller [V/V]

Power amplifier [A/V]

Ka Electro magnet [N/A]

+ -

Fb [N] AMB force

Sensor [V/m]

AMB actuator

+

Mechanical plant rotor Gp (s)

x [m] Displacement

Kn Negative position stiffness [N/m]

Active Magnetic Bearing Gr (s)

Fig. 6.16 Block diagram of an AMB system

6.2.2 Vibration Evaluation Criteria (ISO 14839-2) Since a displacement sensor is usually mounted in an active magnetic bearing to detect rotor position, the rotor vibrations can be evaluated. No additional displacement sensors are required. Figure 6.17 illustrates a typical example of unbalance vibration amplitude of an AMB rotor with shaft speed. The rotational speeds corresponding to each resonance up to the maximum operational speed are referred to as critical speeds. Since the rotor is generally softly supported at both ends by the AMBs, the translational (parallel) and tilting (conical) rigid body-like modes and their corresponding resonances are evident at lower shaft speeds. This semi-rigid rotor may be accelerated past these two critical speeds up to the critical speed N C3 of the first bending mode resonance. In the case of a flexible rotor specified by the operation at higher rotational speeds beyond N C1 and N C2 , its rated shaft speed is designated as between N C3 and N C4 , the latter corresponding to the second bending mode resonance.

130

6 Vibration in Magnetic Bearing Rotor Systems

NC 3

NC4 Q3 = ω n / Δ ω Δω

A

NC 2 NC1

0

In the case of rigid rotors

N

ωn

N

Rotational speed Ω

In the case of flexible rotors

Fig. 6.17 Unbalance response curve

An auxiliary (or touch-down bearing) bearing is installed to prevent the journal from hitting or rubbing against the electromagnet surfaces directly on the stator side. To this end, the radial clearance of the auxiliary bearing is normally set to be half that of the AMB, and this is called the minimum clearance, C min , which sets the upper limit of the radial displacement of the shaft. This minimum clearance in an actual AMB rotor can be measured by statically moving the shaft up and down, or left to right inside the bearing until it makes contact with the bearing surface, while it levitates without rotation. The average clearance ratio defined as C min /R (R is the shaft radius) is normally around 5/1000, which is larger than that of an oil film bearing, normally around 1/1000. The shaft vibration amplitude is specified in terms of the ratio of the vibration amplitude over the minimum clearance. Figure 6.18 shows graphically the variation of each boundary of the zone limit values listed in Table 6.3 with the minimum clearance. For example, for an AMB rotor with C min  250 μm, allowable zone limits are designated by colors in the lower left part of the figure and in the lower right part for an oil film bearing rotor. The AMBs can continue their operation, even if shaft vibration amplitudes are higher than those in the case of oil film bearings. This is due to the comparatively larger clearance in an AMB.

6.2.3 Evaluation Criteria of Stability Margin (ISO 14839-3) In general, the stability of a feedback control system is evaluated in terms of gain and phase margins, which can be calculated by an open-loop transfer function (sometimes called a loop transfer function). The related theory is described in R1E_Chap. 8. In this standard, a sensitivity function is adopted as a concept covering both of the gain and phase margin evaluations and it is proposed as an index for evaluating the stability margin.

Maximum peak displacement Dmax [zero-to-peak μm]

6.2 ISO Standards Related to Magnetic Bearings

500

400

131 Ratio =

Recommended criteria of zone limits Zone Limit

Displacement Dmax [μm]

A/B

< 0.3 Cmin

B/C

< 0.4 Cmin

C/D

< 0.5 Cmin

Dmax Cmin

= 0.5

0.4 D C 0.3

300 B 200

A AMB In the case of oil-bearings

100

D C B A

In the case of turbo compressors 250

500

750

1 000

Minimum radial clearance Cmin [μm]

Fig. 6.18 Zone limits for vibration criteria Table 6.3 ISO 14839 zone Zone

Details

A

Newly commissioned machines

B

Long-term operation accepted

C

Limited period operation (with remedial action) Shutdown recommended (sufficiently severe, inspection required)

D

Since this sensitivity function is derived from the open-loop characteristics of the system, how to obtain the function is described by the following measurement procedure steps. Firstly, an external excitation signal, E(jω), such as sinusoidal or random in the form is applied to one end of the closed loop of the feedback control system as shown in Fig. 6.19. Then, the open-loop transfer function, Go (jω), the sensitivity function, Gs (jω), and the closed loop transfer function, Gc (jω), are measured with a two-channel FFT analyzer as shown in the figure. Secondly, the transfer functions should be measured not when rotating, but levitating, and if possible, when rotating at the rotor rated speed. With the measurements above, the vibration characteristics, {natural frequency ωn , damping ratio ζ }, are obtained as follows: • When measuring the open-loop transfer function, gain zero dB crossing (called cross-over) frequency → ωn ,

132

6 Vibration in Magnetic Bearing Rotor Systems Fd Sensitivity function

−

V 1 = + 1 E 1 + Go

Gs =

x

G p : Transfer function of the plant rotor

+

G r : AMB controller

V1

V2 E

Open-loop transfer function Go = −

V2 Gs +

V1

Closed loop transfer function Gc =

V2

A B E

A

B

FFT

Go =− E 1 + Go

Excitation signal

Fig. 6.19 Two-channel measurement of transfer functions

Phase margin φm → Damping ratio ζ  1/2 × tan φm • When measuring the sensitivity function, peak frequency → ωn , Peak value Q → Damping ratio ζ  1/(2Q) Example 6.4 In a single degree-of-freedom (1-DOF) system illustrated in the block diagram in Fig. 6.20a and b, the input is expressed as r, and output is expressed as x. With parameters of (a), the measured functions are shown in (c) and (d).

(1) Obtain the open-loop transfer function, Go , closed loop transfer function, Gc , and sensitivity function, Gs . (2) Obtain the natural frequency, ωn , and damping ratio, ζ , of this system shown in (a). (3) Read the vibration characteristics from the open-loop transfer function shown in (c). (4) Read the vibration characteristics from the sensitivity function shown in (d). Answers (1)

(2) ωn 

√

G o (s) 

k + cs ω2 + 2ζ ωn s ω2 + 2ζ ωn s  n 2 , G c (s)  2 n , 2 ms s s + 2ζ ωn s + ωn2

G s (s) 

s2 s 2 + 2ζ ωn s + ωn2

√ k/m  10 rad/s, ζ  c/(2 mk)  0.1

(6.17)

6.2 ISO Standards Related to Magnetic Bearings

133

x

m k

| G o | [dB]

−100 0

−120

| Go |

Go

−160

c −60

r

−140

φm 10

1

100

Phase [deg.]

−80

40

−180 500 1 000

frequency ω [rad/s] (c) Open-loop transfer function Go

m = 1, c = 2, k = 100

(a) Single-DOF system

r + k + cs

-

Controller

1 ms 2 Rotor

(b) Block diagram

x

| Go | , | G s |

10 8

| Go |

6

| Gs |

4 2 0

0

5

10

15

20

frequency ω [rad/s] (d) Gains of Go and Gs

Fig. 6.20 Vibration characteristics of single-DOF system

(3) Gain cross-over frequency ωn  10.2 rad/s and phase margin φ m  12° → ζ  0.11 1  Q → peak value Q  4.8 → (4) G s max  |G s ( jωn )|  2ζ ζ  0.103 at peak frequency ωn  10 rad/s. A typical example of the open-loop transfer function is shown in Fig. 6.21 in the form of a Nyquist plot. This figure includes the plot of Go discussed in Example 6.4. As the frequency ω increases, Go moves along its trajectory in the figure. As long as the critical point A (−1, 0) remains on the left side of the trajectory, the closed loop system is stable. If the critical point is on its right side, the closed loop system is unstable. If the system is unstable, measurement of the transfer function is impossible, and thus the critical point should always be located on the left side. The intersections of the Nyquist plot with the real axis and the unit circle are denoted by the points G and P, respectively. The gain margin is expressed by the ratio of line segment OA to OG (i.e., |OA|/|OG|  1/|OG|). The angle φm  AOP is the phase margin. The shortest distance between the critical point A and the trajectory of Nyquist plot is here designated by |AB| as the stability margin, which is an index that integrates both the gain and the phase stabilities: |AB| = Min|1+G o ( jω)|

(6.18)

134

6 Vibration in Magnetic Bearing Rotor Systems Im Peak sensitivity at zone limits Peak sensitivity Gs max Critical point

(−1, j 0 ) A

G

A/B

Gs max = 9.5 dB = 3

B/C

Gs max = 12 dB = 4

C/D

Gs max = 14 dB = 5

O 1

B

(Fig. 6·20)

φm P

Re

D

m

in

=

|A

B|

Go

(j ω

)

−1

Fig. 6.21 Nyquist plot of the open-loop transfer function

It is considered that the shorter the distance, the smaller the stability margin. As can be seen from the above definition, the sensitivity function is the inverse number of this distance. That is, the maximum value of the sensitivity function, Gs max , corresponds to the shortest distance: Max|G s ( jω)|  1/Min|1 + G o ( jω)|  1/|AB|

(6.19)

The “real” maximum value among the sensitivity function, Gs max , measured at each AMB control axis should be obtained as the index for ISO stability evaluation. This index used for stability margin and its zone limit is decided by the table inset in Fig. 6.21. Example 6.5 Figure 6.22 shows the quasi-modal model (refer to R1_Fig. 4.17) that corresponds only to the translation system of a certain compressor rotor. When this model is shown in the s domain, it is expressed by the following equation (for details, refer to Fig. 4.12 of Ref. [92]):

P(s)X (s)  Bu(s),

y(s)  C X (s)

(6.20)

where u(s)  − Gr (s)y(s) ⎤ −0.04s − 40 m δ s 2 + 0.076s + 44 −0.036s − 4 ⎥ −0.036s − 4 m 1 s 2 + 0.036s + 4 0 ⎦, −0.04s − 40 0 m 3 s 2 + 0.04s + 40



t B≡ 100 , C≡ 100 ⎡

⎢ P(s) ≡ ⎣

⎤ y(s) ⎥ ⎢ X (s) ≡ ⎣ ξ1 (s) ⎦ ξ2 (s) ⎡

6.2 ISO Standards Related to Magnetic Bearings

135 ξ1

u

0.036 m1 = 0.8

Gr (s)

mδ

( ω z3 =

0.1

4

5 , ζ z = 0.01)

( ω z5 = 20, ζ z = 0.01)

40

m3 0.1

y

ξ2

0.04

(Anti resonance mode)

Fig. 6.22 Equivalent model of LP rotor

(1) Obtain the open-loop transfer function. (2) Obtain the sensitivity function and decide the zone limit value. Answers (1) G o (s)  C P(s)−1 BG r (s) → Fig. 6.23 (2) G s (s)  G o (s)/[1 + G o (s)] → Fig. 6.24 → maximum value  7 dB  2.2 → Zone A.

8°

60

ζ m = 0.01

Gain [dB]

40

φm

°

=2

φm

=

32

180 Phase

20

ω1 = 1

0 −20

ω3 = 6 ω5

Gain

−40

Q = 23

−60 0.1

0.5 1

5

Q = 29

0 −180 Phase φ [deg.]

Fig. 6.23 Open-loop transfer function

50 100

10

Frequency ω 10

Gain [dB]

Fig. 6.24 Gain of sensitivity function

7 dB

6 dB

5 0 5 10 0.1

ω3

ω1

0.5 1

5

10

ω5

50 100

Frequency ω N (Rated speed)

136

6 Vibration in Magnetic Bearing Rotor Systems

6.2.4 Case Study (1) Model [94, 95] Let us use Hitachi’s active magnetic bearing-type centrifugal compressor as an example to apply the ISO standards. This model (Fig. 6.25) is a compressor for use in oil refinery plants, and the active magnetic bearings are used in its low-pressure (LP) stage as well as in its high-pressure (HP) stage. Major specifications of these AMBs are as follows: – Compressor output: 4000 kW Rated rotation speed: 10,900 rpm  182 Hz – Diameter of active magnetic bearing journal: φ 150 mm, Clearance at one side: 0.5 mm – Touch-down bearing (ball bearing) clearance at one side: C min  0.23 mm. The eigen modes, for the example of the LP rotor, are shown in Fig. 6.25a. This is a flexible rotor that accelerates beyond the resonance speeds of its first, second, and third modes and is operated at its rated speed just prior to the fourth modal resonance.

LP

Eigen modes

1st mode

2nd mode

AMB 4th mode

AMB

3rd mode

(a) Eigen modes 9 811

2 145 Driver rating 5,300kW Nomal speed 10 900 rpm

NC1,2,3 = 35, 82, 138 rps t LP 1s

d

d 3r

NC1,2,3 = 29, 72, 148 rps 1st HP

3r

Tubine

2nd

Ps / Pd = Suction / Discharge pressure = 0.79/2.03 MPa Rotor mass m = 965 kg, Shaft length L = 2 935

2nd

Ps / Pd = 0.33/0.91 MPa m = 805 kg, L = 2 905

(b) Centrifugal compressor equipped with AMBs

Fig. 6.25 Train of equipped compressor and critical speed modes

6.2 ISO Standards Related to Magnetic Bearings

137

For ease of vibration control, the AMBs and displacement sensors should be located at the ends of the shaft where it is most susceptible to deflection, as determined by these mode shapes. Therefore, the vibrations observed in the AMB rotor may be relatively large, but it can be stated that the rotor is quite normal and healthy. (2) Critical Speed Map Figure 6.26 shows critical speed maps for the AMB-equipped centrifugal compressor and the oil film bearing-mounted centrifugal compressor. It indicates the differences in critical speeds between these two types of bearing. In general, the natural frequency of the AMB rotor is lower than that of the oil film bearing, because the AMB rotor is long and has a small-diameter shaft. The spring stiffness of the bearing is also weaker by a factor of one-tenth (1/10) or less. These two types of rotor have different specifications; nevertheless, the operational speeds of the centrifugal compressor should be the same from the viewpoints of end users. The active magnetic bearings have to be designed for the rotor to pass the first critical speed N C1 and the second critical speed N C2 of the rigid body modes (under-critical). When the specification requires rotor operation at higher speeds, the rotor has to be designed to be a flexible rotor such that it can pass the first bending mode critical speed, N C3 . The rotor is accelerated up to its rated rotational range below N C4 , after passing through N C3 . (3) Evaluation of Vibrations The minimum bearing clearance, C min , in this model is 230 μm. Since this model was commissioned as a new machine to be classified by Zone A in the ISO standard, the vibration shall be 230*0.3  69 μm or less, as defined in Fig. 6.18. Vibration amplitude data from the factory witness tests (Fig. 6.27) and those from the on-site acceptance test (Fig. 6.28) indicate that Zone A is certifiable. Since the factory test is appropriate for field balancing, the measurement data show smaller vibrations than on-site data.

Eigen frequency ×100 [Hz]

10

OIL NC4

AMB

NC3

NC4

A2 1

NC3

0.1 0.01

B2

NC1

Speed range (under critical)

NC2

A1

NC2

Speed range (super critical)

B1

NC1

0.1

1

10

Bearing stiffness × 70 [MN/m]

Fig. 6.26 Typical critical speed map

100

0.1

1

10

100

Bearing stiffness × 70 [MN/m]

1 000

138

6 Vibration in Magnetic Bearing Rotor Systems

Shaft vibration [μm p-p ]

50

NC4 NC1

NC3 NC2

LP HP

0

5 000

10 000

Rotational speed [rpm]

Fig. 6.27 Unbalance response curves measured during machine runs 100

Shaft vibration [μm p-p ]

LP 80

1 60

2 side 1 Y 2

40

3

4

3 side 2 X 4 side 2 Y

20 0

1 side 1 X

5 Axial 5 1.0

4.0

Time [h]

Fig. 6.28 Unbalance response curves measured during machine runs

(4) Evaluation of Stability Margin The open-loop transfer functions, Go , of the HP and LP rotors were measured for each control axis of all channels (X, Y at the right and left AMBs and thrust direction) when the rotor was not running. The measurement while the rotor was running at the rated rotation speed is omitted. An example of an open-loop transfer function, Go (jω), measurement is shown in Fig. 6.29. The gain cross-over frequency is 133 Hz ( 8000 rpm), which is estimated to be the critical speed N C3 in the first bending mode. In the high-frequency region, there are many peaks, which correspond to higher order bending modes. This open-loop transfer function was transformed into the sensitivity function, Gs (jω), using the definition shown in Fig. 6.19. This sensitivity function is shown in Fig. 6.30. The maximum gain in the sensitivity function is about 3.9 dB. This is in Zone A as defined by the ISO criteria in Fig. 6.21.

6.3 Straight Control of an Active Magnetic Bearing Rotor Fig. 6.29 Open-loop transfer function Go Gain [dB]

20

139 65 Hz 381 Hz 521 Hz 689 Hz

10 0 −10 −20

133 Hz

−30 −40 10

100

1 000

Frequency [Hz] 20

Gs Gain [dB]

Fig. 6.30 Sensitivity function

10

3.4 dB

3.3 dB

3.9 dB

0

−10 10

100

1 000

Frequency [Hz]

6.3 Straight Control of an Active Magnetic Bearing Rotor 6.3.1 Controller Transfer Function As shown in Fig. 6.31, the AMB is controlled in the X and Y directions, without contacting the rotor, such that the center of the shaft is kept at the AMB center. The displacement of the shaft was measured, respectively, in the {X, Y} directions, and the electromagnetic force components {ux , uy } were generated, respectively, in each direction via the controller, Gd (s), set for the X and Y directions. The controllers in each direction have the same specifications (i.e., isotropic). This type of Gd (s) is referred to as XY straight control. As a matter of course, it is possible to generate the cross-coupled electromagnetic forces in the Y and X directions in response to displacement in the X and Y directions, respectively, by modifying the electrical circuit networks. This type of control is referred to as XY cross-control as explained in the next section. The levitation and vibration of the rotor basically depend on the straight control. The cross-control may play a role to compensate for any lack of straight control. The rotor is generally supported by isotropic AMBs having specifications that differ between the right- and left-side bearings, as shown in Fig. 6.32a. This type of control is referred to as side-by-side control. The rotor can also be supported as shown in Fig. 6.32b, where the displacements {x 1 , x 2 } are detected, respectively, at the left-side and right-side bearings. These displacements are separated into the translation mode component p (i.e., average of x 1 and x 2 ) and the tilting mode component t (i.e., difference between x 1 and

140

6 Vibration in Magnetic Bearing Rotor Systems x-y directional form

G

d

x

Y

uy

ux

Gd

X x

y

uy

Gd

Gd

y ux

Complex form

X

Y

u

z Gd

= ux + juy

= x + jy

Fig. 6.31 Layout of AMB b

a X1

Y1

Y2

X2

G x1

x2 ux2

ux1 p

x1

=S

t

x2

p Gd

Gd

Gd

1

1

uy1

y1

Y1

x1 X1

Gd ux

2

1

uy2

y2

Y2

x2 X2

Gp

2

Gt up

ut

ux

2

up

ux1 ux2

(a) Side-by-side control

t

=R

ut

(b) Mode control

Fig. 6.32 AMB control method

x 2 ) by applying an S matrix. Then, the controllers, Gp (s) and Gt (s), are prepared, respectively, for translation and tilting displacements, and each control command {up , ut } from the controllers is re-configured with an R matrix, to generate the magnetic force {ux1 , ux2 } to be distributed into the right- and left-side bearings. In order to maintain isotropic support, the configuration from the Y directional vibrations {y1 , y2 } to electromagnetic forces {uy1 , uy2 } has the same specification as that for the X direction. This type of system is referred to as under mode control. The S and R matrices are given typically as follows: ⎡ ⎤ b a p x where S  ⎣ a + b a + b ⎦, R  S t (6.21) S 1 x2 t −1 1

6.3 Straight Control of an Active Magnetic Bearing Rotor

141

where a and b are the distances between the left and right AMBs and the center of gravity, respectively. In this type of control configuration, the system can be expressed in the s domain by the following equation: s

2

m 0 0 mαt2



p p G p (s) 0 + 0 0 G t (s) t t

(6.22)

where m is the rotor mass, I d  mi2 is the transverse moment of inertia, and i is the radius of gyration, and Id ≡ mαt2 (a + b)2 ⇒ αt  i/(a + b). As shown in Eq. (6.22), the translation mode p and the tilting mode t are controlled independently. The control of the right and left bearings is coupled, but since there is no cross-coupling in the XY directions, this type of control can be regarded as a form of straight control.

6.3.2 Dynamic Characteristics of AMB Figure 6.33 shows a single-DOF AMB system. This rigid rotor is expressed by the following equation of motion: m z¨ (t)  u(t)

(6.23)

where −U(s)  Gd (s)z(s) is straight control. Simple examples of the transfer functions Gd (s) (N/m) of the controller are as follows: PID control : G d (s)  K P + K D s +

KI s

(6.24)

PD control : G d (s)  K P + K D s

(6.25)

Gd Gd

u

Gp =

1 ms2

−

Fig. 6.33 Single DOF of AMB levitation system

z

z m

142

6 Vibration in Magnetic Bearing Rotor Systems

Phase lead control : G d (s)  K g

τs + 1 (0 < α < 1) ατ s + 1

(6.26)

In these equations, the proportional constant, K P , is equivalent to a spring constant, and the derivative gain, K D , is equivalent to a viscous damping coefficient. Since the integration gain, K I , is akin to a spring constant being effective only for the static deformation, it contributes to centering the journal of the shaft against gravity and constant loads. Normally, it is designed such that the integration does not affect the vibration control but it becomes effective only in the extremely low-frequency region. In the field, phase lead control is used to achieve desired vibration characteristics (i.e., natural frequencies and damping ratio). Since an actual machine is usually a multi-DOF system, the order of the control plant may become large. However, the system can be regarded as an array of single-DOF systems corresponding to each eigen frequency band and the concept of the phase lead control may be adopted for all eigen modes. Example 6.6 In the system (Fig. 6.33) that has the following phase lead compensator:

G d (s)  K g

s + 1 10 1 and G p (s)  αs + 1 s + 10 ms 2

(6.27)

where m  1 kg, α  0.4, K g  1 N/m (default setting): (1) Obtain the open-loop transfer function, Go (s). (2) Obtain the characteristic roots of this system, and then find the natural frequency, ωn , and damping ratio, ζ . (3) Estimate the natural frequency, ωn , and damping ratio, ζ , from the Bode plot of the open-loop function, Go (jω). (4) From the gain margin, obtain the stability limit, K g , and the natural frequency, ωn , at the stability limit. Answers (1)

G o (s)  G d (s)G p (s)

(2) Numerator of [1 + G o (s)]  mαs 4 + m(1 + 10α)s 3 + 10ms 2 + 10K g s + 10K g  0 → s  −10.3, s  −1.7, s  −0.23 ± j1.15 → ωn  |s|  1.17, ζ  0.196 (3) From the open-loop diagram of Fig. 6.34, the gain cross-over frequency, ωg , is read as ωg ≈ ωn  1.18 Hz. Phase margin, φ m  18° → ζ  0.5 tanφ m  0.16 (almost identical to the solution to (2)).

Fig. 6.34 Open-loop transfer function (Bode plot)

g Gain [dB] , φ Phase [deg.]

6.3 Straight Control of an Active Magnetic Bearing Rotor

143

40

g

Phase margin

20

φ

0

−20

φ − Go ( j ω ) −40 0.1 0.2 0.5 1

Gain margin 2

5

10

Frequency [Hz]

Fig. 6.35 Controller transfer function (Nyquist plot)

1

Gd ( jω )

Im 1

0 1

100

0 1

2 2

3 3.56

Re

5

1 10 rad/s

2

(4) Gain margin, gm  16 dB  6.3 at ωn  3.56 Hz → stable under the stability limit K g  6.3 → if the stability limit K g  6.3 is set, the natural frequency increases to ωn  3.56 Hz, at which the system enters an unstable region in the fourth quadrant shown in Fig. 6.35.

6.3.3 Modeling and Control of an Active Magnetic Bearing Rotor (One AMB) Firstly, let us consider the rotor, the right end of which is simply supported by the ball bearing and the left end of which is supported by the active magnetic bearing as shown in Fig. 6.36. Assuming the AMB portion as the boundary coordinate in this figure, the quasi-modal modeling provides the 3-DOF model equivalent to this rotor as shown in Fig. 6.37. The bearing journal displacement, y, indicates the motion of the rigid body mode, δ, of the shaft, and the displacements, ξ 1 and ξ 2 , generate the vibration of the bending modes, φ 1 and φ 2 , in the shaft, according to the illustrated mode shapes. The critical speed map, which is the result of the natural frequency analysis using the spring constant of the AMB as a parameter, is shown in Fig. 6.38. When the bearing spring constant, k b  106 N/m, is assumed, the critical speed is used as the design specification. The eigen modes, ϕ 1 and ϕ 2 , of the entire system with this specification are also shown in Fig. 6.36.

144

6 Vibration in Magnetic Bearing Rotor Systems ϕ1

y

1 050 φ 50

900

AMB

#3 φ 300

#2 φ 200

#1 ϕ2

60

c b = 10 3 N·s/m

k b = 10 7 N/m ,

Fig. 6.36 Rotor system with an AMB (dimensions in mm) φ1 δ ξ1

ω z1 = 84 Hz Gd (s ) ≈ k b

m t = Total mass = 182.5 kg

mξ 1

m δ = Equivalent mass of δ -mode

mξ 2

m ξ 1 = 12.3kg ( m δ × 26%)

= 47.2kg (mt × 26%)

mξ 0

ω z2 = 167 Hz

y

m ξ 2 = 30.2kg ( m δ × 64%)

ξ2

m ξ 0 = 4.65 kg = ( m δ × 10%)

φ2

Fig. 6.37 Equivalent rotor model with an AMB

f n [ Hz ]

1 000 100 10 1 4 10

10

5

10

6

7

10

8

10

10

9

10

10

k b AMB stiffness [ N/m ]

Fig. 6.38 Critical speed map

Now, the transfer function, Gd (s), of the controller is expressed by the following equation:   τs + 1 6 Ki 1 : G d (s) ⇔ G 1 (s) ≡ 0.8 × 10 (6.28) Case  + s ατ s + 1 where K i (the integral gain)  2, α  0.5, and τ  0.016. Figure 6.39 ➀ is the Bode plot of this function. An example of the resonance curve of the unbalance vibration and the open-loop characteristics is shown in Fig. 6.40. The resonance around N C1 (≈ 30 Hz) is contained at a low level, whereas the peak

6.3 Straight Control of an Active Magnetic Bearing Rotor

145

Gain [dB]

2

g

140 120

60 2 1

100

0

1

80 60 40 0.01

φ 0.1

10

1

100

Phase [deg.]

160

−60 1000

Frequency [Hz]

Fig. 6.39 Comparison of controllers

300 200 100

N C1

0 0

50

100

150

Rotational speed [rps]

200

60 40 20 0 −20

180 g 0

φ

−40 −60

Phase [deg ]

N C2

400

Gain [ d B ]

Amplitude [ μ m ]

500

−180 3 5

10

20

50 100 200

Frequency [ H z ]

Fig. 6.40 Unbalance vibration and open-loop characteristics (Case ➀)

around N C2 (≈ 130 Hz) is very sharp, which indicates insufficient damping. This is because the phase curve of Fig. 6.39 ➀ shows only a small phase lead with the controller at around 130 Hz. In order to improve the phase lead around this point, a phase bump filter (phase bump; see Ref. [96]) is connected to the system in series: 2 : G d (s) ⇔ G 1 (s)G 2 (s) Case 

(6.29)

  α 1 1−  Phase bump filter 1−α (τ s)2 + 2ζ τ s + 1 α  0.66, ζ  0.46, and τ  0.612 × 10−3 s. The corresponding Bode plot of this controller is shown in Fig. 6.39 ➁, which shows a large phase lead around N C2 . This phase lead corresponds with an increase of the system damping ratio, and thus the resonance amplitude at N C2 is reduced as shown in the unbalanced resonance curve of Fig. 6.41. Since the control performance for the entire system can be seen in the open-loop characteristics, the damping performance in the controller ➀ is compared with that in the controller ➁ by reading the phase margin from the right figures of the Bode plots (Figs. 6.40 and 6.41), and the result of the comparison is listed in Table 6.4. The gain cross-over frequency, ωg , corresponds to the critical speed. Controller ➀ has a damping capability sufficient for N C1 . However, when the damping at N C2 is taken into consideration, controller ➁ is better than controller ➀. The equation expressing the relationship, R1_Eq. (8.13), between the phase margins and the dampwhere G 2 (s) ≡

6 Vibration in Magnetic Bearing Rotor Systems 180

60

200

Gain [ d B ]

Amplitude [ μ m ]

300

N C2 100 N C1 0 0

50 100 150 Rotational speed [rps]

40 20 0 −20 −40 −60

200

g 0

φ

Phase [ d eg]

146

−180 3 5

10 20 50 100 200 Frequency [ H z ]

Fig. 6.41 Unbalance vibration and open-loop characteristics (Case ➁) Table 6.4 Comparison of open-loop characteristics N C1 , N C1

N C2

Controller 1

Gain cross-over frequency ω g

26 Hz

27 Hz

Phase lead

φm

19.2°

39.3°

Damping ratio evaluation

ζ = 1/2tan φ m

Q-value evaluation N C2

Controller 2

Gain cross-over frequency ω g Phase lead

φm

Damping ratio evaluation

ζ = 1/2tan φ m

(Ref: Damping ratio and Q-value estimated by complex eigenvalues)

0.17

0.4

Q = 2.9

Q = 1.2

93 Hz

95 Hz

7°

59°

Invalid approximation of flexible modes ζ = 0.06 / Q = 8.3

ζ = 0.8 / Q = 1

ing ratio is suitable for the rigid mode, but with regard to the bending mode, it is at an insufficient level. Based on this approach of adding the phase lead to improve the modal damping ratio, tuning the controller parameters and observing the response curve are used in the field. The literature [96] explains several examples of using the phase lead for tuning filters that is often used.

6.3.4 Modeling and Control of an AMB Rotor (Two AMBs + Symmetrical Rotor) [97–99, VB550] A typical AMB rotor supported by radial bearings at both ends, as shown in Fig. 6.42a, is addressed to discuss the vibration control problem. The rotor has a thrust bearing at the left end and a flat motor at the right end, and the rotor is thus nearly bilaterally (i.e., left–right) symmetrical. The critical speed map of this rotor is shown in Fig. 6.43. The natural frequencies together with their respective eigen mode are shown when the boundary conditions are free-free (AMB stiffness k  0) at the left end and pinned-pinned (AMB stiffness k  ∞) at the right end. In order to set N C1  30 Hz, the stiffness of each AMB is assumed to be approximately k  106 N/m. Under this assumption, the expected

6.3 Straight Control of an Active Magnetic Bearing Rotor thrust

AMB

WC1

WC 2

WC3

147

WC4

WC5

AMB

motor

(a) k = 106 N/m

thrust

k = 106 N/m ϕ1

(b) Parallel group

In-phase

Out-of-phase

ϕ3

ϕ5

ϕ2

Tilting group

ϕ4

Fig. 6.42 Eigen modes (boundary condition is set by AMB stiffness) AMB

free-free

Critical speed [Hz]

1000

pin-pin

NC7 NC6

350 100

NC5

Rated speed

NC4

NC5 = 290 Hz

NC3

NC4 = 170 Hz NC3 = 80 Hz

NC2

NC2 = 50 Hz NC1 = 25 Hz

NC1

20 105

106

107

108

109

AMB stiffness [N/m]

Fig. 6.43 Critical speed map

critical speeds N C1 , N C2 , … are listed in the inset of the figure. The rated rotation speed is set just above the critical speed, N C5 (the third bending mode). The eigen modes for each of the critical speeds are illustrated in Fig. 6.42b. The vibrations of the rotor at the bearings at both ends have the same phase in the modes associated with critical speeds with even numbers and have the opposite phase in modes with odd numbers. This allows the adoption of mode control for the in-phase and out-of-phase mode groups, separately. These modes may be controlled with the system shown in Fig. 6.44. The controller, Grp , influences the in-phase mode group, and the controller, Grt , controls the outof-phase mode group. Strictly speaking, since the rotor is not bilaterally symmetric, these two mode groups are inseparable in theory, but in terms of practical accuracy they can be considered as separable. Figure 6.45 shows the Bode plot of the plant transfer function that represents the system consisting of the translating system (Gpp  x p /up ) and the tilting system (Gpt  x t /ut ). The peak resonance in this plant transfer function indicates the natural frequency, ωf , under the free-free boundary condition. On the other hand, the anti-

148

6 Vibration in Magnetic Bearing Rotor Systems rotor

AMB

sensor

sensor

Digital controller

x2 In-phase

xp xt xp

Separation x1 1 1 1 x1 = 2 −1 1 x2 xt Out-of-phase

Parallel G rp controller vp i2

AMB

Tilting controller

Grt vt

Recomposition v1 1 1 −1 vp = 2 1 1 vt v2

v2

Amplifier

v1

i1 Amplifier

Fig. 6.44 Structure of mode control system

Gain [dB]

−60

ω f3

ω f5

ω f7

ω f4

ω f6

−100 −140 Gpp

−180 10

ω z3

ω z5

Gpt

ω z7

100

1 000

Frequency [Hz] (a) Parallel mode control

10

ω z4

ω z6

100

1 000

Frequency [Hz] (b) Tilting mode control

Fig. 6.45 Transfer functions of mode control system

resonance point (or zero point frequency), ωz , is the natural frequency of the simply supported system (i.e., with pinned-pinned boundary conditions). Note that the bearing journal is equivalent to the nodal portion in the mode shape corresponding to ωz . An example of the controller prepared for such plant control is shown in Fig. 6.46. This is the final result of a variety of tuning operations applied to the system. The upper views in Fig. 6.46 show the Bode plots of the controllers, Grp and Grt , and the lower plots show the open-loop characteristics. The frequencies, ωn1 and ωn2 , are natural frequencies of the in-phase and out-of-phase modes, respectively. For these dominantly rigid body modes, the controller is required to have a large phase lead in the low-frequency range. In Fig. 6.46, the frequencies, ωn3 through ωn7 , are the bending mode natural frequencies. Regardless of controller gain, the phase lead shall be set between 0° and +180° in this frequency range. In this example, a notch filter (NF) is added at the bending mode frequency, ωn7 , in the translating system to greatly delay the phase into the phase lead region. Regarding the bending mode natural frequencies, ωn4 and ωn6 , in the tilting system, a phase shift filter (PSF) is added in order to adjust the

6.3 Straight Control of an Active Magnetic Bearing Rotor

Gain [dB]

180

Gain

Gain

0

0 Phase

Phase

−40 40 20 0 −20 −40

Grp

Grt

NF

−180

PSF

ω n1

ω n3

ω n5 ω n7

Gop

ω n2

ω n4

Phase [deg.]

Gain [dB]

40

149

ω n6

Got

10

100

1 000 10

100

Frequency [Hz] (a) Parallel mode control

1 000

Frequency [Hz] (b) Tilting mode control

Fig. 6.46 Controller transfer functions and open-loop characteristics

Sensivity gain [dB]

30 20

ω n1

10

ω n3

ω n5 [17 dB]

ω n2 [13 dB] ω n4

ω n7

0 −10 −20

Gsp 10

ω n6

Gst 100

Frequency [Hz] (a) Parallel mode control

1 000

10

100

1 000

Frequency [Hz] (b) Tilting mode control

Fig. 6.47 Stability margin evaluation by sensitivity peak gain

phase significantly so that the bending mode natural frequency range can be located in the phase lead region. It may be left to the intuition of field engineers to decide how to implement the phase lead in the bending natural frequency range and what type of filter has to be added. In the digital control scheme, it is possible to try to improve matters through software modifications. Figure 6.47 shows the sensitivity function, Gs , converted from the measured openloop characteristics, Go . The maximum sensitivity taking into account both the translating and the titling systems is set to be 17 dB, which is in Zone D of the ISO criteria (Fig. 6.21). The ratios of the balancing weights for each order of the bending modes are shown in Fig. 6.48, including each mass ratio of the three-plane balance for N C3 , the four-plane balance for N C4 , and the five-plane balance for N C5 . These ratios are decided such that they have an effect on the bending mode of a certain critical speed targeted, but have no effect on any other lower frequency bending modes. Once the ratios have been determined, as shown in the figure, the single-plane balance

150

6 Vibration in Magnetic Bearing Rotor Systems 0.9

ϕ3

−1

(a) 1st bending mode ( NC3 / ϕ3 ) 1

0.80

ϕ4

−0.80

−1

−1

(b) 2nd bending mode ( NC4 / ϕ 4 ) 1

0.52

−1

ϕ5

0.52

−1

(c) 3rd bending mode ( NC5 /ϕ 5 )

Fig. 6.48 Correction weight ratio

technique is applied to this targeted bending mode. After the balancing of this mode is completed, the next critical speed mode should be targeted for balancing, and the same technique is applied successively. This technique is repeated for each bending mode until the rated speed is reached. The principle of this method is explained in “Modal Balancing” (R1_Sect. 5.4). An example of the resonance curve in this balancing test for each mode is shown in Fig. 6.49. The curve, which is achieved after the three-plane balance around the first bending mode, N C3  110 Hz, and the four-plane balance at around the second bending mode, N C4  170 Hz, is extended to the point A of the broken line. Then, the five-plane balance for passing N C5 , which is of the third bending mode around 290 Hz, is shown. In this example, the rated speed of 300 Hz has been reached by applying this final balance of Fig. 6.48c. In this manner, the modal balancing is undertaken successively for the respective modes at each critical speed. Since the minimum clearance, C min , of this machine is 450 μmp-p , the ISO zone criteria (Fig. 6.18) of A (amplitude is 150 μmp-p or less) are satisfied.

6.4 Cross-Control of an Active Magnetic Bearing

151

90 ° 150 NC5 Balance

180 ° After

A

0

Before 280rps 100 150

0°

[μm p-p] NC4

270 °

(a) Nyquist plot

150

Amplitude [μm p-p]

100 NC5, 290rps 50 296rps

NC2 NC5 N Balance C5 Balance

NC4

100

Before A

NC3

50 NC1

C 0

0

50

100

150

200

250

NC5

After 300

Rotational speed [rps] (b) Unbalance response curves

Fig. 6.49 Passing the third bending mode critical speed by cross-stiffness control [VB629]

6.4 Cross-Control of an Active Magnetic Bearing 6.4.1 Cross-Control [100] When AMB reaction forces in the X and Y directions arise crosswise from rotor displacements in the Y and X directions, they are termed as due to cross-control. The direction of displacement and reaction force is defined in Fig. 6.50a. The reaction force toward the X direction is generated mainly in response to the X directional displacement. However, it should be noted that, at the same time, the reaction force toward opposite to the Y direction is optionally generated. The transfer function, Gd (s), is the main straight control in the system, and the transfer function, Gc (s), is the special optional cross-control. As shown in Fig. 6.50b, in response to the input of the {X, Y} directional displacements {x, y}, the output includes the {X, Y} directional reaction forces {ux , uy }. As shown in the figure, note also that the X-Y displacements are cross-coupled to Y-X forces with ± signs. Figure 6.50c shows input and output relations in the complex notation that can express this relation. The imaginary unit “j” before k c represents the cross-coupling. Assuming that the straight control component is PD control (gains k P and k D , respectively) and that the cross-control component is P control (gain K c∗ ), these input and output relations are given by the following equation: u(s)/z(s)  G d (s) + j G c (s)  k P + k D s + j K c∗

(6.30)

Figure 6.51a shows the feedback system in which the AMB control system includes the direct and cross-actions. As shown in this figure, it is possible to adjust the dynamic characteristics of the AMB with the coefficients of the controller (i.e., spring, k P , cross-spring, K c∗ , and damping action, k D ). Figure 6.51b shows the complex form with dynamic characteristics through a complex block diagram.

152

6 Vibration in Magnetic Bearing Rotor Systems

(b) x

G

d

G

Y

c

(a)

G

X x

y

Gc

Gd

c

uy

x-y directional form + ux Gd − Gc

ux Y

y

Gd

X

(c) − ux

+ +

+

uy

Complex form Gd + jGc z u = ux + juy = x + jy

Fig. 6.50 Straight and cross-controls Fig. 6.51 AMB feedback system

− ux uy

=

kP − Kc* Kc* kP

x y

+

kD 0 0 kD

ux

. x . y

X x

Y Ω

uy

−

rotor

y

(a) x - y directional form −

kP + kDs + jKc*

u

rotor

z

(b) z complex form

6.4.2 Whirl Vibration and Stability of Cross-Control Rotor vibration features a trajectory of its whirl motion, not only unidirectional motion. The stability offered by the cross-spring is affected by the orientation of the whirl. In the system subjected to the effects of fluid force in an oil film bearing and seal, the dynamic characteristics of the system are expressed by the following equation, where the spring, k d , and cross-spring, k c , are those defined in Eq. (3.27) and the damping action is expressed by, cd : u(s)/z(s)  kd + cd s − jkc (kc > 0)

(6.31)

When k c >> 0, the stability of the forward whirl is lowered, ultimately inducing self-excited vibration. However, in the active magnetic bearing, the sign as well as magnitude of the cross-spring, K c∗ , is a controllable parameter, and thus by actively varying this, the system can be adjusted to be stable and to exhibit a high damping characteristic. Figure 6.52a shows the forward whirl vibration, which is expressed by the following equation:

6.4 Cross-Control of an Active Magnetic Bearing

153

z  ae jω f t where; ω f > 0

(6.32)

The displacement signals are expressed by: x(t)  a cos ω f t,

y(t)  a sin ω f t

(6.33)

and the velocity signals are expressed by: x(t) ˙  −aω f sin ω f t

∝

−y(t),

y˙ (t)  aω f cos ω f t

∝

x(t)

(6.34)

The velocity feedback that can provide the damping is substituted by cross-control as shown by the following equation: u x  cd x˙ → u x  K c* (−y) u y  c y y˙ → u y  K c* x

 u(s)/z(s)  + j K c*

(6.35)

The cross-spring effect, K c∗ , of the AMB that is capable of damping the forward whirl motion is achieved, as shown by Gc in Fig. 6.50b. This is accomplished by connecting the inverted y displacement signal to the X direction actuator and by connecting the x displacement signal (without inverting it) to the Y direction actuator. Note that this connection is done to generate the opposite effect to the cross-stiffness, k c , of the oil film bearing. On the other hand, the backward whirl vibration shown in Fig. 6.52b is expressed by the following equation: z  ae− jωb t where ωb > 0

(6.36)

The relation between displacement signal and velocity signal is expressed by: x  a cos ωb t ⇒ x˙  −aωb sin ωb t ∝ y y  −a sin ωb t ⇒ y˙  −aωb cos ωb t ∝ −x

(6.37)

Thus, in order to provide the damping effect for the backward whirl motion, the connection with the opposite sign to that in the previous connection is required.

Fig. 6.52 Whirling motion

y

y

ω wb

ω wf

x

x Ω

(a) Forward whirl

Ω

(b) Backward whirl

154

6 Vibration in Magnetic Bearing Rotor Systems

Therefore, it can be understood that the effective tuning of the cross-control for the forward whirl, and that of in the backward whirl, is a compromise. Addendum: Stabilization of Active Magnetic Bearing with Cross-Spring In fluid machinery, self-excited vibrations are often caused by fluid forces such as from oil whip and labyrinth seal whirling, and all give rise to unstable forward whirl vibrations. Let us consider that the fluid force, D (model in Sect. 8.7), is applied to the m–k system as expressed by the following equation: m z¨ + kz + D(˙z − jλ z)  0

(6.38)

where λ is the average fluid speed; e.g., λ is approximately 1/2 in a cylindrical oil film bearing. In order to stabilize the unstable fluid force with the AMB cross-spring, K c∗ , the following equation holds true: m z¨ + kz + D(˙z − jλΩz) + j K c* z  0

(6.39)

The examination of stability with the asymptotic expansion method reveals the following stability condition, which is derived by referring to Eq. (B.5) in R1_Appendix B:   1

K* 1 da 0) vibration is kept stable when the following equation is satisfied:

/ωn < 1/λ ≈ 2

(6.41)

By adding the cross-spring, K c∗ , the stable rotation range can be expanded as in the following equation:  

/ωn < 1 + K c* /(Dωn ) /λ

(6.42)

However, when the cross-spring is excessively strong, the backward whirl stability condition, where ωn in Eq. (6.40) is replaced by −ωn , cannot be satisfied, and the system becomes unstable. The backward stable condition is: 



1

1 da K c* K c* 0) in the system. Example 6.8 Figure 6.55 shows the resonance curve of this system in which the cross-control is applied to the unbalance vibration by using Eq. (6.51). Examine the effectiveness of applying the cross-spring constant, K c∗ (> 0), to reduce the resonance peak amplitude.

Answer Peak amplitude is Q  1/(2ζ )  5 when K c∗  0. Peak amplitude is Q  1/(2ζ + ζ )  3.3 when K c∗  ζ k p . Peak amplitude is Q  1/(2ζ + 2ζ )  2.5 when K c∗ is 2ζ k p .

Fig. 6.55 Effectiveness of N-cross control

6

Amplitude

5 4

×ε ζ = 0.1 γ =0 ε =1

Kc*= 0 Kc*/ kp = ζ

3

Kc*/ kp = 2 ζ

2

OFF

1 0

ON 0.5

1

1.5

Rotational speed p = Ω / ω n

2

158

6 Vibration in Magnetic Bearing Rotor Systems Fx

γ

x

PID-X

x‘

−

y

−

FxY

α

X

xn

Tracking filter

P

Ux

− β

Rotational pulse

+

Ω

yn

Fy

α

β

y‘ PID-Y

+

Uy

Fy-

+

− X

− Y

γ

P

Cos/Sin oscillator a cos ( Ω t + φ ) synchronized with rotational pulse a sin ( Ω t + φ )

β

0

0

0

N-cross

ON

0

0

N-cross + N-cut FF excitation

ON

ON

0

0

0

ON

1

PID

2 3 4

γ

α

Options

No.

Fig. 6.56 Control network including options (PID, N-cross, N-cut, and FF)

[Case Study 1] Options for Control of Unbalance Vibration 1. Control Options As shown in Fig. 6.56, {x, y}-directional rotor vibrations are input to PID controllers connected with the power amplifier command {U x , U y }. When the command {U x , U y } is positive, the magnetic force {F x , F y } from the upper coil is turned on. When it is negative, the opposite side is turned on. This is a general form of the AMB control system with no cross-talk. Now let us prepare a tracking filter that extracts from the rotor vibration {x, y} only the N-(rotational) vibration components {x n , yn }, which are synchronous with the rotational pulse. When the outputs {x n , yn } of this filter are multiplied by α and added transversally (crosswise) to the control command side {U x , U y }, it is called N (or 1X)-cross control. When the outputs are multiplied by β and subtracted directly from the input side {x, y}, it is called N-cut control. With this method applied, if β  1, the N-vibration component does not flow in the PID controller, and as for the unbalance vibration, it becomes equivalent to the free-free response in terms of the dynamics. No bearing reaction force is generated, resulting in quiet rotation. In the ISO standard, this method is referred to as unbalance force rejection control, which is recommended for long-term, economical, and continuous operation. This method has been adopted in a variety of applications. In addition, let us generate the cos/sin waves that are synchronous with the rotational pulse by using a two-phase oscillator. The amplitude, a, and phase difference,

6.4 Cross-Control of an Active Magnetic Bearing

159

φ, of the {acos( t + φ)} and {asin( t + φ)} signals are variable. These signals are multiplied by γ and added to the control command signals, {U x , U y }. This is equivalent to direct excitation to the rotor synchronous vibration with the rotation and is referred to as feedforward (FF) excitation. FF excitation can be used to cancel the unbalance force by appropriately adjusting the amplitude and phase. It is possible without any influence on the system stability, because there is no feedback. 2. Control Options (In the Case of Compressor) [101] The dotted amplitude curve in Fig. 6.57b shows the amplitude experienced under PID control when the rotor (Fig. 6.57a) is rotating. Two critical speeds in the rigid body (N C1 and N C2 ) modes are located around 3500 rpm, and a fairly sharp peak at around 11,000 rpm corresponds to the critical speed (N C3 ) of the first bending mode. When the N-cross gain is turned on under the resonance condition around the critical speed N C3 , the resonance amplitude is lowered, and when it is turned off, the resonance amplitude returns to its original level. Depending on this ON/OFF switching of N-cross control, the resonance amplitude can be alternately lowered and raised. It is also possible to reduce the unbalance vibration with the FF control, and the experimental data of this rotor are described in R1_Sect. 5.6. The rotor can pass the peak amplitude of its bending critical speed N C3 by changing the gain parameters {α, β, γ } while applying these optional controls. The resonance amplitude and the required control current amplitude at the critical speed are mea3rd bending mode AMB

AMB

50

(a) Cross-section view

μm p-p

50

Amplitude [μm p-p ]

N- cross ON [80%]

NC3

N-cross OFF

N- cross OFF NC1, NC2

Flexible mode Critical speed

Rigid mode N-cross ON

Critical speed

0

2 000

4 000

6 000

8 000

10 000

Rotational speed [rpm] (b) Comparison of with and without N-cross control

Fig. 6.57 Bending mode control of flexible rotor using N-cross control

12 000

160

6 Vibration in Magnetic Bearing Rotor Systems

Control current [A]

20

PID N-cross N-cross & N-cut FF excitation

1 2 3 4

10

4

2

0

1

3 0

10

20

30

40

50

60

Max amplitude [μm p-p]

Fig. 6.58 Comparison of control options

Radius [ mm ]

100 1 2 3 4

Unbalance weight

80 40

AMB

AMB

20 0 0

200 400 600 Rotor length [ mm ]

800

Amplitude [ μm ]

(a) Rotor model 150

Control current [ % ]

60

100

PID N-cross N-cross & N-cut FF excitation

4

1 2

50 3

Bending mode

100

Rigid modes

0

50

0

0 0

2 000 4 000 Rotational speed [ rpm ]

50 100 Amplitude [ % ]

6 000 (c) Comparison of control options

(b) Unbalance response (PID only)

Fig. 6.59 Comparison of control options

sured. Figure 6.58 summarizes these data. In this figure, the nearer the data are to the origin, the lower the power requirements and the lower the amplitude when the rotor passes the critical speed, which are desirable. In this example, ➂ is the best control method among these control options. [Case Study 2] Example of control options (Test Rotor) [102] Figure 6.59b shows the amplitude curve of the unbalance vibrations experienced when the simple AMB-type rotor shown in Fig. 6.59a is rotating. The critical speeds, N C1 and N C2 , for the rigid body modes are around 1500–2000 rpm, and the resonance amplitude around 5000 rpm is the first bending mode critical speed, N C3 . The control options explained previously are applied to this N C3 , and their effectiveness is compared in Fig. 6.59c. In contrast to the previous example, the best option in this case is FF excitation ➃.

6.4 Cross-Control of an Active Magnetic Bearing

161

−20

Motor Ball bearing

Self-excited vibration (73 Hz)

[dB]

Magnetic bearing (Damper)

−80 Water

Unstable Tuning frequency (77.5 Hz) 100

0

Sucrose

200

Cross AMB ON

Drum section

−20

Cross gain : Large ( Ω = 15 000 rpm)

Gap-Sensor

Stabilization

[dB]

Journal bearing

Cross gain : Small ( Ω = 15 000 rpm)

Gap-Sensor

(a) Centrifuge (AMB type)

Tuning frequency (77.5 Hz) (73 Hz) −80 0

Critical speed

Backw ard

Bending mode

20 10

0.6

200

180 ard

Forw

50

Unbalance resonance Flow induced instability 1 2 5 10 20 Rotational speed ×103 [rpm] (b) Instability data (No AMB)

Amplitude [μm p-p]

Frequency [Hz]

100

100 Frequency [Hz]

1 Self-excited vibration (Cross OFF)

160 120

1 / 2 17 000 rpm 13 800 rpm 2 (Cross On)

80

Unstable domain

0 40

0

10 000

20 000

30 000

Rotational speed [rpm] (c) Amplitude curve and FFT displays

Fig. 6.60 Stabilization with cross-control (containing liquid) [VB247]

[Case Study 3] Stabilization of a Centrifuge [103–105] Figure 6.60a shows the configuration of a centrifuge operated continuously in physical and chemical engineering fields. The drive motor is mounted on the upper section, and a rotating drum is suspended in the lower section via the thin rotating shaft. Both ends of motor and of the drum are supported by ball bearings. In this case study, the bearing only on the upper section of the drum is replaced by an AMB, which is required to function as a damper. In the drum, our test mixture of water and sucrose is separated into two layers by centrifugal force. It is well known that a rotor system containing the free surface liquid (two layers of liquid and air) is susceptible to unstable vibration. In this case, since the border surface between two separated layers plays the same role as the free surface liquid, unstable vibrations will happen over a certain speed range. All of the past data of unstable vibrations experienced in similar types of centrifuge are plotted by marks in Fig. 6.60b, where the generated frequency on the ordinate is shown

162

6 Vibration in Magnetic Bearing Rotor Systems

as a function of the rotation speed on the abscissa. Unstable vibrations are observed along the forward natural frequency curve in the conical mode. The amplitude curve, when the rotor in Fig. 6.60a is rotating, is shown in ➀ of Fig. 6.60c. It is shown that self-excited vibration occurs at 13,800 rpm, at which the amplitude shows an excessive increase. The generated frequency, f n , is 73 Hz based on the FFT result. Therefore, the tuning frequency, f t , in the tracking filter is manually set at about f n ≈ 77 Hz in order to extract the self-excited vibration component, and its output is applied in the cross-circuit. As a result, the self-excited vibration component is attenuated and the system is stabilized as shown in the FFT diagram. This f n crosscontrol is maintained until the system goes through this unstable region, and then the cross-control is turned off. The test result is shown in ➁ of the amplitude curves. This type of f n cross-control that is tuned to the generated frequency is an effective countermeasure to cope with self-excited vibration. [Case Study 4] Stabilization of a Centrifugal Compressor [106] In this compressor, the AMB damper is installed inside the sliding bearing at both ends and at the side of the balance drum as shown in Fig. 6.61a. The shaft is longer than the shaft used in the regular design; hence, the system is more susceptible to labyrinth seal whirling instability. Figure 6.61b and c shows the vibration data when the shaft is rotating. When the rotational speed is increased from 13,000 to 14,000 rpm, the amplitude shows significant increase and instability due to the labyrinth whirl. When the f n crosscontrol (tuning frequency f t ≈ whirl frequency) through the AMB is turned on under this condition, the self-excited vibration disappears, and the system becomes stable. [Case Study 5] Aseismic Control [107] (1) Theory: A basic approach to aseismic control with AMBs is shown in Fig. 6.62a

14

80

13

AMB control

Rotational speed

12

AMB Cross-control

(b) Test data of instability and stabilization

cont

rol

ON

com pone nt

OF BC ross -

AM

ON OFF 0 ON 10 0 Time [ min ]

ON

20

1×

Shaft vibration

Wh irl c omp one nt

40

Tim e

60 Shaft vibration [ μm p-p ]

Rotational speed

3

× 10 [rpm]

(a) Centrifugal compressor 15

Frequency (c) Waterfall diagram

Fig. 6.61 Labyrinth seal-induced instability and cross-control stabilization

6.4 Cross-Control of an Active Magnetic Bearing m x

PD k

mα ( t )

Rotor P(t) c

163

0 α (t)

+

PID

−

α (t)

Casing

(a) Foundation excitation of 1-DOF system

gH

+

−

+ P(t)

1 ms 2

x

(b) FF control

Fig. 6.62 Principle of aseismic control

in the form of a single-DOF system. It is assumed that the vibration of the rotor relative to the bearing pedestal is x(t) when the bearing pedestal moves at acceleration α(t). The equation of motion is given by the following equation when the system is also subjected to external force P(t): m x(t) ¨ + c x(t) ˙ + kx(t)  −mα(t) + P(t)

(6.52)

where the damping, c, and spring, k, are controlled by feedback (FB) control of the AMB. If the external force, P(t), is given by the following equation, then the right side of Eq. (6.52) becomes zero, and the excitation caused by acceleration is canceled:

P(t)  mα(t)

(6.53)

This control can be implemented as shown in Fig. 6.62b in which the measured acceleration signal on the bearing pedestal α(t) is multiplied by the proportionality constant, gH , and it is fed forward to the rotor through the AMB. Furthermore, since the frequency bandwidth of a seismic wave is usually low compared with rotor’s eigen frequencies, the above equation can be easily realized by the AMB. (2) Tuning of FF Control Gain: As shown in Fig. 6.63a, the rotor is mounted on a flexible frame. The signal from the displacement sensor on the AMB rotor is used in the PID FB control (natural frequency of rotor shaft system  15 Hz), and on the other hand, the signal from the accelerometer attached on the bearing pedestal is used in the FF control system that generates the excitation force P(t). In the above test bench, the frame is excited by impact excitation. The acceleration, α(t), when the rocking mode vibration (natural frequency  2.7 Hz) occurs on the pedestal, is measured as shown in Fig. 6.63b. The left-side plot in Fig. 6.63b is the vibration, x(t), when the FF control is not applied. Then, as shown in the right-side

6 Vibration in Magnetic Bearing Rotor Systems

X1 Gap sensor Accelerometer X2 Coupling Disk

Gap sensor

Impulse

Motor

Acceleration [m/s 2 ]

164

0.4

0.04G

α (t) Input

Displacement [ μm ]

−0.4 75

AMB AMB Flexible support bench

α (t)

0.04G

0

69μm

6μm

0

−75

fn = 2.7 Hz

X1

X1 Output

0

Time [s]

0.8 0

(1) PD

Time [s] (2) PD + FF

0.8

(b) Tuning of gain g H for aseismic control

(a) Test rotor on flexible support bench

Fig. 6.63 Tuning of aseismic control on flexible support bench

Displacement Displacement [ μm ] [ μm ]

9.8 KOBE, NS Acceleration [ m/s 2 ] (a) Seismic wave α (t) α (t) Seismic wave max. 0.75 G −9.8 0 30 Time [s] (b) Without FF control (c) With FF control 300 0 rpm 0 rpm X2 max. 72 μm

max. 287μm

−300 300

6 000 rpm

X2

6 000 rpm

0

Time [ s ] AMB rotor vibration

X2

max. 139μm

max. 256μm

−300

X2

30 0

Time [ s ]

30

AMB rotor vibration

Fig. 6.64 Test data of seismic excitation

plot in Fig. 6.63b, the gain, gH , in the FF control is applied and adjusted to minimize the vibration x(t). Equation (6.53) is tuned by such experimental adjustment as this. (3) Result of Aseismic Test: After this preparation, this rotor is set on a shaker table. The table is subjected to excitation simulating the Kobe NS seismic wave as shown in Fig. 6.64a, having the maximum seismic acceleration 0.75 G. The rotor response to this excitation is shown in Fig. 6.64b and c. The left-side plot shows the data when FF control is not applied, and the right-side plot shows those when the FF control is applied. Comparison of these two plots reveals that the vibration with the FF controlled vibration is more effectively reduced than the vibration without the FF control, in both cases when the rotor is at rest and when it is running at 6,000 rpm.

6.4 Cross-Control of an Active Magnetic Bearing 350

Max. displacement [μm]

Fig. 6.65 Aseismic effect of FF excitation

165

PD PD+FF

300 250

PID

200

X2 y

onl

X1 X2

150 100

X1 FF PID+

50 0

0

2

4

6

8

10

12

Max. acceleration [m/s 2]

The magnitude of the input seismic wave was varied, and the maximum displacement in the response was examined to ascertain its sensitivity to excitation. The effectiveness of aseismic control is summarized in Fig. 6.65. The abscissa axis of the magnitude of the seismic wave was replicated by varying the input gain of each shaking test. It is shown in this figure that the maximum displacement can be almost halved with the implementation of FF control. Note: Takahashi wrote an interesting paper [108] about the review of AMB-related technologies from point of view of Japanese engineers. In the early stages of investigation of AMB technology, many patents were published to obtain business or commercial leadership. Habermann and Brunet of S2M at that time (as of now SKF) were famous as inventors for strong patents, e.g., [109, 110]. These are recommended to be read by every early career engineer beginning study in this area. Reference [111] presented one of the new applications of AMBs; instead of the bearings supporting rotors, they may be used to excite rotors for the purposes of measuring stability and evaluating the operational dynamic state. The Japanese NEDO grant encouraged the research and development for AMB technology, and the corresponding project team was organized worldwide and coordinated by the Japanese leadership over 2002– 2004. The team achieved ISO standardization for the ISO 14839 series. The results of research were presented at the NEDO conference [112] in 2004. Further information is provided by References [113–141]. As shown in the list, many papers were presented at the International Symposium on Magnetic Bearings (ISMB) series, which play a central role in the AMB engineering community.

Chapter 7

Case Studies of Forced Vibration Problems of a Rotor

Abstract The major problem in forced vibration response is resonance, where the frequencies of exciting forces coincide with the natural frequencies. Rotating machinery has different characteristics in resonance problems from general non-rotating structures. Therefore, the problems need be solved with this knowledge in mind, not only at the design stage, but also on-site. In the case of forced vibration, centrifugal forcing due to unbalance is a very typical example; however, various other external forces produced by different mechanisms may also become problematic and be experienced routinely. Examples include mechanically induced forces in gears, cross joints, and pulley belt systems; an electromagnetic force in a motor/generator; and rotating stall and impeller/blade interaction forces, which are induced by the fluid flow. Furthermore, torsional vibration is inevitable for rotating machinery and may become a large problem. In this chapter, these phenomena and appropriate countermeasures to implement are elucidated while referring to previously mentioned examples. Keywords Vibration criteria · Unbalance vibration · Resonance · Balancing · Newkirk effect · Morton effect · Asymmetric rotor · Gear force · Universal joint · Electromagnetic force · Rotating stall · Blade-passing frequency · Rotor–blade interaction · Belt drive system · Torsional vibration damper

7.1 Approaches to Resonance Problems in Rotating Machinery 7.1.1 Natural Frequencies and Damping Ratios Varying with Rotational Speed or Load Gyroscopic moments experienced by rotors and the mechanical properties of oil film bearings are well known to vary with shaft rotational speed. Furthermore, load also exerts a strong influence on bearing characteristics, resulting in the significant variation of natural frequencies and damping ratios. It is therefore necessary to handle the resonance problems by taking account of these changes. © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_7

167

7 Case Studies of Forced Vibration Problems of a Rotor 1

30 23,600

Natural frequency [cpm]

25

Damping ratio [ − ]

× 10

3

168

20 15 m cp

10 4800

=

rpm

19,100

5880

5 0

2 500 0

5

0.8 0.49

0.6

0.28

0.37

0.4

0.15

0.19

0.2 0

10

15

20

Rotational speed [rpm] ( a ) Damped natural frequencies

25 3

× 10

0

5

10

15

20

Rotational speed [rpm]

25 3

× 10

( b ) Damping ratios

Fig. 7.1 Damped natural frequency and damping ratio variations with respect to rotational speed

Figure 7.1 shows an example of the variation of natural frequencies and damping ratios of a turbocharger rotor supported by oil film bearings with rotational speed as the abscissa. Each pair of natural frequencies and corresponding damping ratios are expressed using the same symbol. Figure 7.1a for natural frequency shows that the straight line “cpm  rpm” intersects with the natural frequency curves at 2500, 4800, 5880, 19,100, and 23,600 rpm. These indicate the critical speeds where the frequency of unbalance force coincides with the natural frequency and resonances take place. Figure 7.1b gives the damping ratios at each critical speed, that is, 0.37, 0.49, 0.28, 0.15, and 0.19, respectively.

7.1.2 Constant Speed Machine and Variable Speed Machine Different solutions are needed for resonance problems between a constant speed machine and a variable speed machine. The resonance problem can be avoided by detuning machines that operate with a constant speed; however, resonance may be inevitable for machines operated over a wide speed range. Consequently, for the latter, it is necessary to design for low Q-value (Q-factor  1/(2ς ), where ς is the damping ratio) in order to reduce the vibration sensitivity and amplitudes at the critical speeds.

7.1 Approaches to Resonance Problems in Rotating Machinery

169

7.1.3 Vibration Amplitude at Critical Speed During Acceleration/Deceleration Vibration amplitudes at resonances can be reduced by passing the critical speeds quickly, in contrast with the quasi-steady speed operation. However, it is often the case that the speed control when decelerating is not possible even when it is possible when accelerating. In addition, though the response amplitude depends on the speed change rate, it is difficult to realize a certain reduction of amplitude without a considerably large rate of change. For this reason, it is more practical and safer in design to make vibration sensitivity low by attaining low Q-values.

7.1.4 Bending Vibrations and Torsional Vibrations For whirling vibration in rotor systems, damping from an oil film bearing is effective. Thus in general, a larger damping ratio can be achieved and a low-sensitivity design is possible. On the other hand, for torsional vibration of a shaft, since rotating machines are designed to minimize the mechanical loss in a shaft spinning motion, it is difficult to exert damping influences on torsional vibration. Therefore, if a torsional resonance does occur, the vibration amplitude may become significantly large. Consequently, the first priority should be given to the design objective of avoiding torsional resonance over the entire operating speed range. However, if resonance is inevitable, the vibration amplitude should be assessed by taking account of the selected material, the structural damping of the shaft and couplings, and the aerodynamic damping of impellers. Furthermore, a torsional damper developed for reducing the torsional vibration may also be needed.

7.2 Criteria for Acceptable Vibration Levels of Rotating Machinery This section describes the criteria for acceptable vibration levels to judge whether the measured vibration is acceptably small or unacceptably large. Although such criteria should be applied individually to each machine depending on its feature against vibration, such as having robust or delicate characteristics, ISO defines general guidelines based on past experiences of the criteria applicable to various machines. There are two relevant ISO Standards. ISO 10816: Mechanical vibration—Evaluation of machine vibration by measurements on non-rotating parts [142] prescribes criteria for vibrations at bearings. The other, ISO 7919: Mechanical vibration of non-reciprocating machines—Measurements on rotating shafts and evaluation criteria [143], prescribes criteria for vibrations at a rotating shaft location. The following section describes the origins and the overview of these standards.

170

7 Case Studies of Forced Vibration Problems of a Rotor

7.2.1 Transition of the Vibration Standards

Origin of vibration standard (Bearing vibration) Rathbone chart (Germany)

Amplitude

Figure 7.2 summarizes the historical transition of the standards for mechanical vibrations. The first work for the standardization of vibration criteria was done by Rathbone in Germany in 1939. He presented a diagram prescribing acceptable vibration amplitudes of a bearing housing with shaft rotational speed on the horizontal axis. This chart is now called the “Rathbone chart.” It gives acceptable vibration displacement amplitudes decreasing with the increase in shaft rotational speed, which forms the basis of current standards. The Verein Deutscher Ingenieure (VDI) continued the work and finalized it as VDI 2056 (vibration of a bearing housing) and VDI 2059 (vibration of a rotating shaft). ISO implemented the standardization based on VDI 2056 and decided to employ the vibration velocity as an index value for the vibration severity. VDI 2056 resulted in ISO 2372 (published in 1974, Mechanical vibration of machines with operating speeds from 10 to 200 rev/s—Basis for specifying evaluation standards) and ISO 3945 (large machines, published in 1985). Both standards had been utilized widely and were revised in 1995, resulting in ISO 10816. As for shaft vibration, ISO had no previous standard. Therefore, ISO 7919 was newly finalized and published in 1996 for shaft vibration, paired with ISO 10816. These two ISO standards will be outlined next.

Bearing vibration

Shaft vibration

VDI 2056

VDI 2059

ISO 2372 , 3945 f

API

Amplitude

Amplitude

rpm

rpm ISO 10816

1. General guideline 2. Large steam turbine 5. Hydraulic machine 3. Coupled industrial machine 6. Reciprocating machine 7. Rotodynamic pump 4. Gas turbine

Fig. 7.2 Historical transition of vibration standard

ISO 7919 4. Gas turbine 1. General guideline 2. Large steam turbine 5. Hydraulic machine 3. Coupled industrial machine

7.2 Criteria for Acceptable Vibration Levels of Rotating Machinery

171

7.2.2 ISO 10816: Mechanical Vibration—Evaluation of Machine Vibration by Measurements on Non-rotating Parts [142] ISO 10816 defines standards for acceptable vibration amplitude measured on a bearing housing (stationary part). Its previous version, as a standard of vibration severity, was revised to this version with respect to its applicability to wider variety of machines and their acceptable values. ISO 10816 prescribes the broadband measurement of absolute vibration by means of a seismic pickup (electrodynamic or piezo sensor) fixed to a bearing housing (non-rotating part). Vibrations are evaluated with the effective√value of vibration velocity (root mean square (r.m.s., RMS) value of velocity, 1/ 2 of half amplitude for a sinusoidal waveform), instead of the vibration displacement. This standard is used widely as a quantitative measure of mechanical vibration, particularly being quoted most frequently in the field of condition monitoring of machinery. The criteria have been modified individually in accordance with different sizes and types of machine, resulting in the seven parts shown in Fig. 7.2. In the case of a specified machine not corresponding to any of the six parts (Parts 2–7), Part 1 (general guidelines) can be applied. Table 7.1 shows the evaluation criteria in Annex B of Part 1. Machine classifications are as follows: Class I: Individual parts of engines and machines integrally connected to the complete machine in its normal operating condition. (Production electrical motors of up to 15 kW are typical examples of machines in this category.)

Table 7.1 Typical zone boundary limits R.m.s. vibration velocity mm/s rms

Class I

Class II

Class III

Class IV

0.28 0.45 0.71

A

1.12 1.80

B

2.8 4.5

C

A A B

18 28 45

B

C C

7.1 11.2

A

B

D

C D D D

172

7 Case Studies of Forced Vibration Problems of a Rotor

Class II: Medium-sized machines, (typically, electrical motors of 15–75 kW output) without special foundations, rigidly mounted engines or machines (up to 300 kW) on special foundations. Class III: Large prime movers and other large machines with rotating masses mounted on rigid and heavy foundations that are relatively stiff in the direction of vibration measurements. Class IV: Large prime movers and other large machines with rotating masses mounted on foundations that are relatively soft in direction of vibration measurements (e.g., turbogenerator sets and gas turbines with outputs greater than 10 MW). Evaluation criteria for each of classes are given as the magnitude of measured r.m.s. vibration velocity amplitude divided into the four Zones A, B, C, and D defined as follows: A: The vibration of newly commissioned machines would normally fall within this zone. B: Machines with vibration within this zone are normally considered acceptable for unrestricted long-term operation. C: Machines with vibration within this zone are normally considered unsatisfactory for long-term continuous operation. Generally, the machine may be operated for a limited period until a suitable opportunity arises for remedial action. D: Vibration values within this zone are normally considered to be of sufficient severity to cause damage to the machine. In Table 7.1, the magnitude scale of the borders has been set by dividing each decade into five ranges, logarithmically (101/5 ≈ 1.6 times). Each part shown in Fig. 7.2 describes details of the evaluation zones for specified machines. In the evaluation, if only a constant velocity value is used over a wide range of frequency, displacement will be too large at a lower frequency, while acceleration will be too large at a higher frequency. Therefore, as shown in Fig. 7.3, the evaluation zones are generally defined by introducing a range regulated by a constant displacement (lower than frequency f 2 ) and a range regulated by a constant acceleration (higher than frequency f 3 ). In many cases, settings f 2  10 Hz and f 3  1 kHz are used and the frequency range between f 2 and f 3 is regulated by velocity. This standard provides two criteria. Criterion I is applied to evaluate the steadystate vibration magnitude by using the above zones. Criterion II is applied to evaluate the amount of change in vibration magnitude from a normal value of reference, because a change is observed when some abnormality occurs in the machine. To meet the requirement for Criterion II, the vibration amplitude in the normal status is set as the standard value (baseline) and some (about 25%) of the upper limit value of Zone B is added to the baseline to set an ALARM value, which indicates that the vibration amplitude is large and remedial action may be necessary. This means that an ALARM value often falls into Zone C while it may fall into Zone B when

7.2 Criteria for Acceptable Vibration Levels of Rotating Machinery

173

Vibration velocity [r.m.s.]

Zone D

Zone C Empirically, Alarm value = 2~3 × Baseline Trip value = 10 × Baseline

Zone B

Zone A f1

f2

Frequeny f

f3

f4

Recommended value of ISO 13373 (Condition monitoring and diagnostics-Vibration coondition monitoring) Alarm value = Baseline + upper limit of Zone B × 0.25

Trip value

= will be in Zone B when Baseline is very small = should be lower than upper limit of Zone B × 1.25 = Zone C ~ Zone D = should be lower than upper limit of Zone C × 1.25

Fig. 7.3 General form of evaluation criteria about vibration velocity and setting of Alarm, Trip values

the baseline is small. On the other hand, a TRIP value, which indicates a borderline for an emergency stop operation of the machine, is set differently from an ALARM value. It is a specific value of the machine, and the same value will be selected for the same design of machine. A TRIP value is usually set in the Zones C or D. Note that Table 7.1 and Classes I to IV have been simplified since the revision in 2009. Refer to ISO 10816-1:1995/Amd.1:2009 for further details.

7.2.3 ISO 7919: Mechanical Vibration of Non-reciprocating Machines—Measurements on Rotating Shafts and Evaluation Criteria [143] ISO 7919, which defines acceptable values for shaft vibration (rotating part), is the counterpart of ISO 10816, and the definitions of Zones A to D are identical. This standard consists of multiple parts (currently five) as shown in Fig. 7.2. Measurement of shaft vibrations is made by using non-contact-type or contacttype displacement sensors at a point close to the bearing in two directions perpendicular to each other. For accurate evaluation, the maximum value of peak-to-peak displacement, S(p−p)max , shown in Fig. 7.4, is preferable since it corresponds to the major axis of an ellipse when the two orthogonal waves are sinusoidal at a single frequency. Since the measurement of S(p−p)max is difficult at this time, it is regarded as acceptable to use the greater of the two peak-to-peak amplitudes: SA(p−p) and SB(p−p) .

7 Case Studies of Forced Vibration Problems of a Rotor 450 400 350 300 250

zone D

time

200 150 zone C 100 90 80 70 60 50

SA(p-p)

Transducer A waveform Transducer B waveform

zone B

40 zone A 30

S(

(x, y) mean value

S p-p )m max ax

20

S B(p-p)

Maximum relative shaft displacement S (p-p)

[ μ m]

174

time

S (p-p) is defined as the higher of the two values: SA(p-p) and S B(p-p)

10 1

2 4 6 10 8 3 Rotational speed × 10 [rpm]

20

30

Fig. 7.4 Recommended evaluation values of relative displacement for coupled industrial machines

Figure 7.4 shows evaluation zones for coupled industrial machines (machines having a maximum operating speed in the range 1000–30,000 rpm with fluid film bearings, excluding steam turbine generator sets over 50 MW and hydraulic machinery over 1 MW), which correspond to the machines described in ISO 7919 Part 3. The abscissa axis represents the rotational speed, which is different from the vibration frequency in ISO 10816. The ordinate axis represents the peak-to-peak relative shaft vibration displacement, S(p−p) . In Fig. 7.4, the acceptable amplitude becomes smaller as the rotational speed increases, making a downward-sloping straight √ line. This line has been determined based on experience and is proportional to 1/ N in the log-log plot, where the rotational speed is N. This differs from the line in ISO 10816 where the line is proportional to 1/N when the ordinate axis is rewritten from acceptable velocity to displacement. Although detailed reasons are limited, these relationships have been derived empirically and have a sufficient track record. Note that these criteria for acceptable vibrations of bearing and shaft are only the guidelines recommended by ISO. It is conceivable that the acceptance criteria of a company may be given preference over the ISO standards. Specific criteria for acceptable vibrations are determined mutually after consultation prior to operation between the manufacturer and the end-user of the machine. In such a case, ISO standards can most easily achieve a consensus.

7.3 Case Studies of Unbalance Vibration Problems

175

7.3 Case Studies of Unbalance Vibration Problems Unbalance is the major cause of problems of large vibration amplitudes exceeding the acceptable value given in above-mentioned ISO 10816 (vibration at a bearing) or ISO 7919 (shaft vibration) evaluation criteria. Since the vibrations are caused by centrifugal forces due to rotor unbalance, the vibration frequency coincides with the rotational frequency, which enables an easy diagnosis of the phenomenon by means of frequency analysis. In particular, when the operating rotational speed is close to a critical speed of the rotor system, a resonance may occur and excessive vibrations could be a problem. Solutions to this problem can be classified into two types below: (1) Reduce the unbalance that causes the excessive excitation force by implementing balancing procedures. (2) Modify the mechanical properties of structures such as the rotor, bearings, and support structures in order to make the system less responsive to excitation. When Solution (1) is adopted, field balancing (in-situ balancing, balancing onsite) is the main method that is implemented. According to circumstances, it may be necessary to bring the machine back to the factory for re-balancing using a balancer (low-speed balancer or high-speed balancer). When Solution (2) is adopted, after carrying out detailed measurement and vibration analysis, specific structural design changes shall be decided. In particular, when a resonance is apparent, Solution (2) is more important than Solution (1) because the mass unbalance often changes during a long-term operation. In case the supporting structure causes a resonance, it is common to stiffen the portions of the structure where the rigidity is insufficient after measuring the vibration mode. The following subsections describe some relevant vibration cases in the field.

7.3.1 Balancing of a Gas Turbine Rotor [VB017] The first option to reduce unbalance vibration is balancing. This subsection explains the reasons of an unsuccessful balancing using the modal influence coefficient method, which is a major field balancing technique. As described in R1_Sect. 5.3, the modal influence coefficient balancing method objective is to reduce the vibration vector, where the vibration mode appears dominant, close to zero. The initial vibration vector is V0  θ0 in terms of magnitude V0 and phase θ0 . When a trial weight (w  β, that is, mass  phase) is added, where the phase β is measured from a rotor-fixed point to rotational direction, assuming that the vibration vector changes from V0  θ0 to V1  θ1 , the influence coefficient a  φ can be obtained following vector operation: a φ 

V1  θ1 − V0  θ0 w β

(7.1)

176

7 Case Studies of Forced Vibration Problems of a Rotor

After removing the trial weight, a correction weight wa  βa for balancing can be calculated as follows using the influence coefficient a  φ: V0  θ0 + a  φ × wa  βa  0

→ wa  βa  −

V0 V0  θ0  (θ0 − φ + 180) (7.2)   a φ a

Whether this balancing procedure succeeds or not depends absolutely on the reproducibility of the influence coefficient a  φ. It needs to be recalled that the reproducibility of the influence coefficient, which is the vibration change ratio to weight, is sometimes poor even in a machine whose vibration repeatability is good under the same operational condition. There are many causes for this problem. If the reproducibility is insufficient, it will naturally be difficult to balance. Figure 7.5 shows the structure of a gas turbine whose centrifugal compressor stage and two turbine stages are connected by Curvic couplings via a tie bolt. Figure 7.6 shows the transition of the vibration vector in a Nyquist polar plot during rotor spin up of the test operation. After passing the first and second modes with small amplitude, the vibration amplitude increased almost beyond the acceptable limit toward the third critical speed. Since the measured vibration was synchronous with rotation, unbalance vibration was assumed as the cause, the influence coefficients were measured by adding trial weights to the second stage turbine in order to implement balancing to the third mode.

Curvic coupling Tightening force

Curvic coupling Shrink fit

Tie-bolt

Nut Ball bearing Impeller

Turbine

Fig. 7.5 Gas turbine shaft model and curvic coupling [VB017] 3rd

Fig. 7.6 Nyquist polar plot of shaft vibration

90°

80 m

1st

B 180°

0° A 2nd 270°

7.3 Case Studies of Unbalance Vibration Problems

177

Table 7.2 Trial weights and measured influence coefficients Influence coefficient [μm/g]

No

Added weight [g]

(1)

0.3 ∠ 120°

7.6 ∠ −272°

(2)

0.5 ∠ 120°

9.2 ∠ −104°

(3)

1.5 ∠ 48°

38.0 ∠ −124°

(4)

0.5 ∠ 196°

8.0 ∠ −107°

(5)

1.5 ∠ 214°

42.0 ∠ −118°

Note : influence coefficient = vibration change between A-B/added weight

Fig. 7.7 Changes of vibration vector between rotational speeds A and B

90° (5) (1) (2)

180°

0 (3)

20

40 μm

0°

(4)

270°

Table 7.2 shows the influence coefficients calculated by Eq. (7.1) for the vibration vector changes corresponding to the rotational speeds between A and B (third mode), which were measured several times by adding different trial weights. From this table, it is found that the influence coefficient has almost no reproducibility, and balancing is unfeasible. Figure 7.7 shows changes of vibration vector between speeds A and B. This figure shows that the vibration changes in either of two ways irrespective of added weights. Subsequent research of this rotor revealed that the rotational speed at which the vibration level sharply increased was shifted upward as the tie bolt tension was increased. In addition, the natural frequency of first bending mode of the tie bolt was found by calculation to be close to the rotational speed in question. From these findings, it was presumed that this excessive vibration was due to the following phenomena: The tie bolt bends near the resonance frequency and comes into contact with the rotor inner side in two different ways. Correspondingly, the tie bolt generates very large unbalance and the two types of vibration change shown in Fig. 7.7 occurred irrespective of the initial unbalance. Based on this understanding, a loosely fitted bush was inserted into the annular clearance between the inside of the impeller and the tie bolt in order to restrict the bending of the tie bolt. Then the vibration growth between A and B was greatly reduced, enabling the rotor to operate successfully up to the rated speed without problem.

178

7 Case Studies of Forced Vibration Problems of a Rotor

7.3.2 Structural Resonance Problems of a Vertical Pump

Transmission / Magnet coupling Motor

Stiffening flange of motor frame Thickening and shortening length of the motor frame Shortening connecting shaft Stiffening flange

Stand

1 400

1 680

1 510

This subsection introduces the case of a countermeasure against unbalance vibrations by modifying the structure. In general, for adequately designed rotor-bearing systems, the Q-value at a critical speed is relatively small because of the damping effect of oil film bearings and it rarely causes a problem. On the other hand, resonances related to the casing structure that supports the rotor-bearing system often cause problems because damping implementation is not easy, leading to a high Q-value. This subsection introduces an example of a tall vertical motor–pump system [144]. Figure 7.8 shows a cross-sectional view of the motor–pump system. This pump has a very complicated integrated structure consisting of a motor, a variable transmission, a magnet coupling, and a multistage pump. Because of the tall structure with the heavy upper part, the natural frequency of the bending motion above the base flange was in the operational speed range, resulting in excessive vibration. As a countermeasure, the stiffness of each part of the structure was increased as shown on the right side of Fig. 7.8 to increase the resonance frequency sufficiently above the maximum operational speed as shown in Fig. 7.9. The vibration amplitude was then reduced successfully. Moderate increases of stiffness did not produce a sufficient effect, and all parts with insufficient stiffness had required further stiffening.

Changing to high stiffness motor stand

Original

Multistage pump

4 360

Stiffening pump base flange

Discharge rate : 504 m3/h Discharge head : 213 m Motor power : 425 kW Operational speed : 500 ~ 980 rpm Countermeasures of natural frequency detuning

Fig. 7.8 Vibration countermeasures for a vertical pump and its cross-sectional view before and after them

Vibration amplitude at the motor [mm]

7.3 Case Studies of Unbalance Vibration Problems 100

179

After the countermeasures Operating range

10−1 Original 10−2 10−3 10−4 10−5

0

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 Rotational speed ×103 [rpm]

Fig. 7.9 Improvement in unbalance vibration of vertical pump

In this case, the stiffness was increased as a countermeasure, but another option, such as decreasing the natural frequency below the operational speed, is also possible by adding mass or decreasing stiffness if the structural strength allows.

7.3.3 Vibration Caused by Low Stiffness of a Fan Framework [VB058] Figure 7.10 illustrates a case of excessive unbalance vibration of a fan installed on a foundation framework. Judging from the vibration mode measured by two vibration sensors (one was fixed to check the relative phase, and the other was moved for each measuring point) at the constant rotational speed, the whole structure was found to experience a rocking motion from the root of the foundation without bending deformation. This indicated that the rigidity of root portion was too low.

Fig. 7.10 Vibration mode of a fan [VB058]

Vibration mode

Casing

Foundation

180

7 Case Studies of Forced Vibration Problems of a Rotor

Consequently, the root of foundation was stiffened as shown in Fig. 7.11. Figure 7.12 shows the vibration modes before and after the stiffening remedy. As a result, the inclination deflection at the foundation root was drastically reduced, decreasing the response to less than one third at the bearing housing, changing the vibration mode to one involving the elastic mode of the machine casing. The reinforcement of weak portions having insufficient stiffness improves the vibration characteristics. The weak portion can be found at a position where the shape of vibration mode bends sharply with a high curvature. Figures 7.13 and 7.14 show the change of the vibration amplitude at the bearing housing before and after the countermeasures. f = 40.6 Hz Casing

Foundation Reinforced section

Fig. 7.11 Vibration mode after stiffening remedy Fig. 7.12 Effect of stiffening remedy

Strong (after stiffening)

Concrete foundation

50 Amplitude [μm]

Fig. 7.13 Response before stiffening remedy

Weak (before stiffening)

1 422

40

1 638 rpm

30 20 Rated speed 1 800 rpm

10 0 0

500 1 000 1 500 Rotational speed [rpm]

2 000

7.3 Case Studies of Unbalance Vibration Problems Fig. 7.14 Response after stiffening remedy

181

Amplitude [μm]

50 40 30

Rated speed 1 800 rpm

20 10 0 0

500 1 000 1 500 Rotational speed [rpm]

2 000

7.3.4 Thermal Bending Vibration Unbalance vibration induced by heat is known as thermal bending vibration. An example is a phenomenon called the “bowed rotor.” After a rotor is heated during operation, then stopped, and left for a certain period of time, this phenomenon may arise when the operation is restarted. The start-up is impossible due to excessive unbalance vibration, which is caused by the rotor being bent like a “bow” due to temperature difference between the upper and lower sides of a horizontally installed rotor arising from upward hot-air convection. This phenomenon can be prevented by means of a “turning operation” in order to cool down the rotor surface uniformly during stoppage periods.

Shaft vibration a [ μ m p-p ]

100 75 50

2 minutes later

End of trajectory 4 minutes later

25 1 minute later 0

16

20

24

4

8

12

16

3 minutes later

20

Time [h] 90°

180°

7 65 4 98 3 10 a θ 2 18 11 1 24 23 17 12 22 21 [h] 20 16 13 15 14

Starting to keep rotational speed at 13 500 rpm

5 minutes later

4 minutes later

14 500 rpm 0°

( b ) Morton effect [VB675]

270° ( a ) Newkirk effect [VB010]

Fig. 7.15 Examples of thermal bending vibration (Newkirk effect and Morton effect)

182

7 Case Studies of Forced Vibration Problems of a Rotor

There is another example where a shaft is heated locally by rubbing with a stator (non-rotational part) at small clearance portions such as a seal or oil thrower location, resulting in rotor bending. This phenomenon is known as contact thermal bending vibration, or the Newkirk effect [145, 146–151]. As already described in R1_Chap. 11: “Stability problem of rotor system,” this phenomenon is classified as unstable vibration the amplitude of which increases with time. This vibration is often observed as a globally stable one when the unstable vibration has reached a limit cycle, where the vibration is characterized by a cyclic change in amplitude of the rotational frequency component. The periods of such cyclic phenomena are known to vary widely from less than one second to several days. Figure 7.15a shows a case over a period of about 20 h. When the vibration of the synchronous rotational component is expressed vectorially in terms of amplitude |a| and phase θ in a Nyquist plot, a locus will be drawn like repeated circles around a specific value of vibration. This cyclic phenomenon occurs because of the phase difference between the direction of unbalance due to thermal bending and the direction of the response as shown in Fig. 7.16. For example, the response phase may be delayed by 90° from that of the unbalance at the critical speed. Therefore, the contact position on the rotor surface gradually shifts with time and eventually moves around the whole circumference of the rotor. When the phase of the response caused by thermal bending coincides with that of response before contact, the vibration is strengthened, and when their phases differ by 180°, the vibration is weakened. This results in cyclic changes in vibration. The cycle and the stability of the vibration depend on the factors such as heat input and dissipation conditions, the heat capacity of the rotor, and the relative relationship between the rotational and the critical speeds. R1_Sect. 11.3 includes an analytical example of this phenomenon, but the analysis is not effective enough to enable quantitative predictions. On the other hand, the countermeasures are very simple. After finding the rubbing mark on the rotor surface, the clearance between the rotor and the stator at the marked position is widened to prevent contact and rubbing from recurring. This will solve the problem in most cases. It is reported that such a phenomenon also occurs due to whirling motion of a shaft in an oil film bearing, which is called the Morton effect [152–154]. This phenomenon occurs due to a similar hot spot area generated on the surface of the journal whirling in the bearing. Figure 7.15b illustrates a case when the rotor speed was lowered from

Contact and heating portion (Hot spot) Generation of thermal bending

Fig. 7.16 Mechanism of contact thermal bending vibration

Vibration developing direction (Not coincident with unbalance direction) Unbalance direction

7.3 Case Studies of Unbalance Vibration Problems

183

14,500 rpm and kept at 13,500 rpm on a high-speed balancer, the spiral vibration occurred and the fluctuation required 12 min to fade away and stabilize. The cause of this case was the use of directed lubrication-type bearings. The countermeasure was conducted by changing the bearings to a conventional flooded lubrication type, and this solved the problem. Note: In this field, references [155–158] provide further information and details on the problem. The reader is referred to these to experience the various ideas and principles.

7.4 Case Studies of Forced Vibration in an Asymmetric Rotor 7.4.1 Secondary Critical Speed of an Asymmetric Rotor Since a rotor is usually axisymmetric in design, the anisotropy in bending stiffness is generally small. However, a synchronous motor rotor having a coil, such as two-pole machine in particular, has considerable asymmetry in bending stiffness. A keyway for installing an impeller also causes asymmetric bending stiffness. The asymmetry results in various phenomena of rotor vibration. It is known that the vibrations of asymmetric rotors can be classified into following categories of phenomena as described in R1_Sect. 11.2. Each phenomenon occurs under the conditions below, where  is the rotational speed and ωn is the nth critical speed: (1) Resonance phenomena at the secondary critical speed:   ωn /2 (2) Unstable vibration at each critical speed:  ≈ ωn (3) Unstable vibration between two critical speeds:   (ωi + ω j )/2 Unbalance vibrations of an asymmetric rotor supported horizontally are characterized by the strong harmonic vibrations of twice the rotational frequency, and the resonance with second harmonic occurs at half the critical speed. In the case of a horizontal rotor with a crack, its bending stiffness varies depending on the circumferential position of the crack because the crack opens and closes during rotation; consequently, harmonic vibrations of twice the rotational frequency arise, and resonance occurs at half the critical speed. These phenomena are represented by the secondary critical speed listed in (1) above. We explain this case first. Also, it is well known that in the phenomenon under (2), an asymmetric rotor changes the circular Nyquist plot of each mode to an ellipse when the rotor passes critical speeds. The ellipticity of a Nyquist plot depends on the asymmetry of the rotor. This will be explained in the next subsection. First, Fig. 7.17 illustrates a fan rotor with asymmetric bending stiffness due to the influence of a keyway for installation of a fan disk. Figure 7.18 gives a FFT analysis of the rotor vibration. The second harmonic vibration component of rotational frequency, 2 fr , had grown extremely large at the rotational frequency 12.5 Hz, and

184

7 Case Studies of Forced Vibration Problems of a Rotor

Fig. 7.17 Fan rotor having asymmetric stiffness [VB065] Fig. 7.18 Rotational component and 2nd harmonic (FFT analysis) Amplitude 0

5

10 15 Frequency [Hz]

20

25

Before the countermeasure Trip Amplitude

Fig. 7.19 Overall vibration amplitudes before and after the countermeasure

2 fr

Rotational speed fr 12.5Hz

After the countermeasure

0

600

700

800

Rotational speed [rpm]

the speed increase was not possible toward the rated speed of 800 rpm (13.3 Hz). The rotor was subcritical and of a non-complex design. Unfortunately, the critical speed was almost coincident with twice the rated speed. Although the asymmetric characteristic should have been corrected as a countermeasure, this approach was not employed because it would have required remanufacture of the impeller and shaft. By reasoning that the rotor had a sufficient rigidity, the shaft was machined to a smaller diameter in order to lower the critical speed by 15%. As a result, the rotor could pass the secondary critical speed at a lower speed with smaller amplitude, reaching the rated speed easily as shown in Fig. 7.19.

7.4 Case Studies of Forced Vibration in an Asymmetric Rotor

185

7.4.2 Elliptical Deformation of Circular Nyquist Plots Due to Rotor Asymmetry When an asymmetric rotor passes a critical speed, the actual damping decreases and the Nyquist plot (Refer to R1_Sect. 5.2) sometimes becomes an extreme ellipse. In the worst case, the rotor cannot pass the critical speed and the amplitude may be excessive [159, 160]. Figure 7.20 shows an example measured on an actual machine. Figure 7.21 explains the reason why such an ellipse is induced. Consider a flatshaped shaft and the forces in a rotating coordinate system o − x y synchronized with the shaft, where the x-axis is taken to represent the direction of higher bending stiffness (rigid direction) and the y-axis is taken to represent the directionof lower bending stiffness (flexible direction). Then, the whirling amplitude is r  x 2 + y 2 . In the radial direction, the inertia force, mr 2 , and the restoring force, kr, balances each other, and in the circumferential direction, the damping force, cr , acts in the direction opposite to the whirling velocity. In this case, the bending stiffness of the shaft (k x : x direction, k y : y direction) is asymmetric and k x > k y , the direction of the true restoring force vector, k x x + k y y, will shift from the direction of the whirling center to a direction of higher stiffness. The true restoring force vector can be resolved into two forces: the apparent restoring force, kr, in the direction of the whirling center and the force, F, in the direction perpendicular to that direction. When F faces the direction of the whirling velocity and exceeds the damping force, cr , acting in the direction opposite to the velocity, unbalance vibration will become large and unstable. As can be seen from Fig. 7.21a, when the rotor deflection direction is in the second and fourth quadrants of the rotation coordinate system, vibration will be stable since F faces in the same direction as the damping force cr . When the deflection is in the first and third quadrants, F acts to cancel the damping force and makes vibration unstable. Thus, the amplitude in the Nyquist plot becomes smaller

90°

78

2nd 79

77 76 75

180°

70 140 139

138

20

80 135 136 130 25

1st

137

0°

[Hz] Rotational speed

3rd

270°

Fig. 7.20 An example of elliptic Nyquist plot (modal circle) of rotor vibration [VB678]

186

7 Case Studies of Forced Vibration Problems of a Rotor

(a)

mrΩ2

Rotation

crΩ

F

kr

Flat shaped shaft

mrΩ2

F

Flexible direction Nyquist plot y crΩ

kr

o o

x Rigid direction

x Rigid direction kr

kr

crΩ

(b)

mrΩ2

kxx + kyy F Flexible direction y

kxx + kyy

F crΩ

mrΩ

2

Damping lowering in 1st and 3rd quadrants

Fig. 7.21 Illustration of elliptic Nyquist plot due to asymmetric stiffness of a rotor

in the second and fourth quadrants and larger in the first and third quadrants. It makes the circle an ellipse, its major axis extending in the first and third quadrants as shown in Fig. 7.21b. For a countermeasure, it is important to eliminate asymmetry of the bending stiffness of the rotor. Adding damping, c, that results in cr  > F as a symptomatic therapy has a similar effect. Equation (7.3) shows the relationship for three parameters, namely the asymmetry of modal stiffness: μ, calculated from two natural frequencies in the rigid and the flexible directions of the shaft, the effective asymmetry: κ, defined as a number μ divided by the damping ratio ς and the ellipticity: A, defined by the ratio of major/minor axis lengths of the ellipse: A2 − 1 μ 1 ωx2 − ω2y  , , κ  μ≡ 2 ωx2 + ω2y ς A2 + 1

 A

1+κ 1−κ

(7.3)

Thus, for example, in order to make the ellipticity A < 1.5 of the Nyquist plot, κ < 0.385 is necessary and the relationship between the required mode asymmetry, μ, and the damping ratio, ς , is obtained from the second expression stated in Eq. (7.3).

0. 01

1.0 Stable 0.5

0.0 0.005

2 ∞ 5.0 2 .0 5 3.0 ζ = 0 3 2.5 .0 ζ= 0 2.0 4 0 ζ = .0 1.5 ζ = 0.05 1.2 ζ = 0.1 1.0 0.015 0.02 2 1 ωx − ω y2 μ= 2 ω x2 + ω y2

ζ=

This case

0

5

.01

0 ζ=

0.01

Modal stiffness asymmetry

0.0

1+ κ 1− κ

ζ

Ellipticity A =

=

0.0

Unstable

ζ=

( Stable range κ < 1 )

κ=

μ ζ

05

1.5

Effective asymmetry

Fig. 7.22 Relationship between ellipticity and asymmetry

7.4 Case Studies of Forced Vibration in an Asymmetric Rotor

187

In the case of Fig. 7.20, three critical speeds were passed. Although a circle is maintained at the first critical speed, the asymmetric nature is strong at the second and third critical speeds. The ellipticity is about 3.0 at the third critical speed. Because the measured damping ratio was ς  0.006, the effective asymmetry of the third mode was κ  0.8 (unstable if κ > 1), and the mode asymmetry was μ  0.005 from the relationship of Fig. 7.22. From Fig. 7.20, the 0° direction of the rotor is the rigid direction and the 90° direction is the flexible direction in the third mode. This rotor had difficulty in passing the third critical speed and failed several times at the early phase of the development. The same phenomena were repeated even after reducing the unbalance to a very small level by means of balancing. The asymmetry was found to be the cause of this problem after noticing that the Nyquist plot of the rotor at the critical speed was an excessively deformed ellipse as shown in Fig. 7.20. As this was due to strong anisotropy in the bending stiffness of the shaft components, quality control was initiated to achieve the uniformity in bending stiffness. This phenomenon did not occur after the implementation of this countermeasure. Note: Ellipticity, A, and effective asymmetry, κ If gyroscopic effects and the gravity are ignored as described in R1_Sect. 11.2, by using the relations Z  (x + j y)e j  t and Z¯  (x − j y)e− j  t to express motions, the equation of motion for the asymmetric rotor can be represented as follows:   Z¨ + 2 ζ ωn Z˙ + ωn2 Z + 2μ e j2  t Z¯  ε 2 e j ( t+θ )

(7.4)

where θ is the phase of unbalance in the rotation direction from the axis of rigid direction (x-axis). As seen in Fig. 7.21, the phase of the major axis of the ellipse is advanced by 45° in the direction of rotation from the x-axis. To express this in a delay simple form, we assume a new coordinate system (o − x  y  ) with a phase  of  ◦    ◦  j45◦  − j45  j  t−45 45°. That results in Z  Z e , thus Z  Z e  x + jy e . By substituting this into Eq. (7.4), the following relationship is obtained:      ◦ −2 + 2 j ζ ωn  + ωn2 x  + j y  + 2 jμ ωn2 x  − j y   ε 2 e j(θ+45 )

(7.5)

Then dividing the both sides by ωn2 and using a non-dimensional rotational speed, p  /ωn , the following equation is obtained: 

1 − p 2 −2(ζ p − μ) 2(ζ p + μ) 1 − p 2



x y



εp

2

cos(θ + 45◦ ) sin(θ + 45◦ )

 (7.6)

Figure 7.23 shows a response by using θ as the parameter. When θ  135◦ , we obtain the following formula:

188

7 Case Studies of Forced Vibration Problems of a Rotor

Fig. 7.23 Elliptic Nyquist plot with respect to unbalance phase θ

y' 20 θ = 135°

y

x 10

θ = 180°

θ = 90°

x' 10

−10 θ = 225° θ = 270°

θ = 45° θ = 0°

−10

ζ = 0.035 μ = 0.01

θ = 315°

−20

  1 − p 2 x  − 2(ζ p − μ)y   −ε p 2   2(ζ p + μ)x  + 1 − p 2 y   0

(p≈1)

(7.7)

An operation (the second equation × x  − the first equation × y  ) is carried out for Eq. (7.7), and p ≈ 1 is set because the rotational speed is considered to be close to the natural frequency. Then, we obtain the following equation:  2  2 2(ζ − μ) y  + 2(ζ + μ) x   ε y 

(7.8)

By introducing an effective asymmetry, κ  μ/ζ , and converting the equation to the standard ellipse equation, we obtain the following formula: 

ε y − 4ζ (1 − κ)  2 ε 4ζ (1 − κ) 

2 +

x 2 2

1 ε 1+κ / 4ζ (1 − κ) 1−κ

(7.9)

Thus, we obtain the equation of an ellipse that is off-centered and elongated in the y  direction. The ellipticity, A, (major axis/minor axis) is found to be:  A

1+κ 1−κ

(7.10)

Note: As for the cracked rotor system, many studies have been reported. Helpful information and discussion are provided by references [161–164].

7.5 Vibrations Induced by Gears

189

7.5 Vibrations Induced by Gears In meshed gears, meshing conditions between teeth surfaces vary with time. Specifically, cyclic changes of meshing stiffness occur during every meshing interaction. The exciting force at the gear mesh frequency ( Nz), i.e., the rotational frequency, N (Hz), multiplied by the number of gear teeth, z, is generated in the direction of the meshing line of action. In addition, vibrations at an integer multiple of the rotational frequency will be generated due to the errors of the gears (deviations from the involute curve including deformation of the teeth resulting from load). The equations of motion for two shafts with meshed gears are expressed as follows: I1 θ¨1 + C1 θ˙ 1 + K 1 θ1  r G1 R (t) I2 θ¨2 + C2 θ˙ 2 + K 2 θ2  −r G2 R(t) R(t)  −k(t)x − c(t)x˙ + W + k(t)e(t) where x  r G1 θ1 − r G2 θ2

(7.11) (7.12)

Here, constants Ii , Ci , K i (i  1, 2) represent the moment of inertia, torsional damping coefficient, and torsional stiffness coefficient of each gear shaft, respectively, while R(t) denotes a force generated from meshing. R(t) acts in the direction of the line of action, which is the external common tangent of the gear’s base circles in Fig. 7.24, and is a resultant force arising from the mesh stiffness, k(t), and mesh damping, c(t), as well as static gear load, W , and reaction force, k(t)e(t), from the gear error, e(t), as in Eq. (7.12). Figure 7.25 shows the changes in mesh stiffness k(t) in the case of spur gears. In spur gears, meshing between one pair of teeth and two pairs of teeth occurs, alternately, during the meshing period, TG  1/(N z). In Eq. (7.11), by setting Ci , K i equal to 0, retaining only the inertia terms of the gear shafts, Ii , and assuming that the shaft motion is as a rigid body, we obtain the following equation by converting the coordinate x  r G1 θ1 − r G2 θ2 : m x¨ + c(t)x˙ + k(t)x  W + k(t)e(t)

Radius of base circle : rG1

c(t) Base circle e(t)

k(t) a

Pitch circle θ 1 Tip circle

Fig. 7.24 Meshing of gears

b rG2

Line of action (plane)

θ2

(7.13)

7 Case Studies of Forced Vibration Problems of a Rotor Meshing stiffness k (t)

190 Fig. 7.25 Meshing stiffness between gears

2 pairs of teeth meshing

TG

1 pair of teeth meshing Time t

Helix angle β

Face width B px a (Meshing start line) pt

Length of pass of contact L

b (Meshing end line) Plane of contact

Transverse contact ratio L Length of pass of contact εα = pt Base pitch

(

)

Overlap contact ratio Face width B εβ = px pitch of Face width direction

(

)

Total contact ratio εα + ε β

Fig. 7.26 Contact ratio of helical gear

where m

I1 I2 2 + I2 r G1

(7.14)

2 I1r G2

0.20 δ

0.15 0.10

1 pair teeth meshing ϕ

0.05

Δ

δ

δ

δ

Δ

Δ

Δ

ϕ ϕ ϕ 2 pairs teeth meshing 0 -0.4-0.2 0 0.2 0.4 -0.4-0.2 0 0.2 0.4 -0.6-0.4-0.2 0 0.2 0.4 0.6 -0.6 -0.4-0.2 0 0.2 0.4 0.6 x x x x Ptn Ptn Ptn Ptn Ptn Ptn Ptn Ptn ( b ) β = 14° ( c ) β = 20° ( d ) β = 30° ( a ) Spur gear ε α = 1.48 ε α = 1.48 ε α = 1.48 ε α = 1.48 ε β = 0.44 ε β = 0.66 eβ = 1.10

1.0 0.8 0.6 0.4 0.2 0

ϕ : Load shearing ratio

Levels of deformation δ , Δ × 10−7 m /(N / m n)

Though Eq. (7.13) is nonlinear and is difficult to solve, the gear shaft system can be regarded as a parametrically excited system where the spring constant and the viscous damping coefficient oscillate with the meshing frequency, Nz. In order to reduce vibration caused by the oscillating meshing stiffness, it is recommended to increase the value of the total contact ratio εα + εβ [for example, 165], where the transverse contact ratio εα and overlap contact ratio εβ (εβ  0 for spur gears) are defined in Fig. 7.26. In particular, it is reported that the vibration of the Nz component is greatly reduced in helical gears with an overlap ratio of εβ > 1.

Fig. 7.27 Teeth deformations of parallel axis gears having different overlap contact ratios (comparisons with spur gear, helical gear β  14°, 20°, 30°, Umezawa (1974)

7.5 Vibrations Induced by Gears

191

Figure 7.27 [166] shows the study results on meshing deformation of spur and helical gears when the overlap contact ratio, εβ , varies from 0 to 1.1 (β from 0° to 30°) while keeping the transverse contact ratio as εα  1.48. This figure indicates the deformation level, δ, for a pair of meshing teeth, the rotation delay, , of a driven gear (integrated deflection of teeth) per unit load normalized by the module, mn , and the load sharing ratio, ϕ, of one tooth (1.0 for one-tooth pair meshing and 0.0 for no meshing). From these figures, it can be seen that the greater the overlap contact ratio, εβ , the smaller the changes of the mesh stiffness (proportional to inverse of deflection ), which reduces vibration at the meshing frequency.

Accelerometer

660Hz (Nz) (2Nz) 990 1 320 970

Microphone

Fin Servomotor

(3Nz) 1 980

Microphone (in operation noise spectrum)

Frequency [Hz] Vibration spectrum

Gearbox

Noise spectrum

Case 1 Case of meshing frequency noise Figure 7.28 shows a case of a meshing frequency noise problem found in the driving device of a machine tool. The frequencies of the major vibration components are 660 Hz, corresponding to Nz, and harmonics of two to three times this frequency. A hammering test implemented at the end of the motor revealed that the first natural frequency of the motor shaft was within 660–680 Hz in addition to the mode of 780 Hz where the motor end vibration was dominant. This case is an example of a rotor resonance at the meshing frequency. The most important countermeasure is to avoid the resonance. Although changing the gear mesh frequency was one possible solution, the first natural frequencies were detuned significantly from Nz by adding a preload spring to the servomotor bearing, and this noise problem was then solved. In addition, the external force in Eq. (7.13) includes a term of forced displacement excitation due to a gear error, e(t). Since the gear error includes a large undulation such as run-out error (eccentricity of the pitch circle) and pitch error in addition to minute variations caused by each tooth profile deviation, the gear errors can be expanded in a Fourier series, by using the angular velocity   2πN as the base frequency:

780 Hz 680 670

1 340 1 630 1 140

Measured natural frequencies by accelerometer (hitting servomotor end)

Frequency [Hz]

Fig. 7.28 Gear meshing frequency noise in servomotor drive system of machine tool [VB402]

192

7 Case Studies of Forced Vibration Problems of a Rotor ∞

e(t) 

a0  + (ak cos k  t + bk sin k  t) 2 k1

(7.15)

Thus, in a gear-shaft system, the vibration response shows the meshing frequency (Nz) and the components of integer multiple of the shaft rotational frequency, where the Fourier components depend on the characteristics of gear error. To sum up, the following countermeasures may be taken against gear vibrations: (1) To replace the spur gear by a helical gear to achieve a large contact ratio (in particular, overlap contact ratio) for the case of mesh frequency attributed to the mesh stiffness variation. (2) To reduce the relevant gear error for the case of characteristic vibrations attributed to low gear accuracy. Case 2 Case of a geared compressor Figure 7.29 shows a case of a geared compressor [167]. It was found that the vibration of a pinion shaft increased sharply at a specific rotational speed, and the dominant frequency was just six times the wheel shaft rotation frequency, N G , as shown in Fig. 7.30. It was also revealed that the pinion shaft was at resonance with this frequency. The possible cause of the frequency, 6N G , was the run-out error (pitch circle eccentricity) of the wheel gear. Figures 7.31 and 7.32 show the measured results. As a countermeasure, the wheel gear was reground to improve the accuracy, and at the same time, the bearing characteristics of the pinion rotor were modified to shift the natural frequency. In this way, the pinion shaft resonance was separated from 6N G .

Pinion shaft

2nd stage

Wheel shaft

Fig. 7.29 Geared compressor [VB120]

Bearing

1st stage

7.5 Vibrations Induced by Gears

193

Fig. 7.30 Measured vibration frequencies

Pinion shaft vibration [dB]

NG : Wheel shaft rotation speed 0

6NG N : Pinion shaft rotation speed

−30

−60

0

1 Non-dim. frequency

2

a 360-degree roll −10 PCDφ

10 μm

Time

Synthesized waveform

6NG + 3NG + 1NG

Fig. 7.31 Pitch circle shape of wheel gear Before re-grinding After re-grinding

Component amplitude

4.0

3.0

1.0

0 1 2 3 4 5 6 7 8 9 10 Order of Fourier component

Fig. 7.32 Wheel gear re-grinding result

Note: Recently, a finite element analysis approach has been used to study the dynamic properties of various gears with the consideration for meshing nonlinearity [168–170].

194

7 Case Studies of Forced Vibration Problems of a Rotor

7.6 Vibration Generated by a Cross Joint A cross joint is also called a universal joint, Hooke’s joint, or Cardan joint. With regard to a cross joint, it is well known that vibrations of twice per rotation (2N) and its harmonics with torque fluctuations arise depending on the magnitude of a joint angle (angle of misalignment). Figure 7.33 shows a case of cross joints used in the drive shaft for an automobile, and we will explain their basic characteristics. Figure 7.34 [171] shows the structure of a cross joint. The driving side of the cross pin is referred to as the  pin and the driven side as the  pin, and the joint angle between the driving and the driven shafts is defined as α. The coordinate system defines the torque direction of the driving shaft as z 1 , the torque direction of the driven shaft as z, the y-axis in the direction perpendicular to the A − z 1 z plane, the x-axis in the direction perpendicular to the y- and z-axes, and the x1 -axis in the direction perpendicular to the y- and z 1 -axes. Thus, the angle between the x1 -axis and the x-axis is α. Rotation angles of the driving and the driven shaft  and  are defined from the point where the  pin and the  pin coincide with the x1 - and y-axes, respectively. In addition, unit vectors in the directions of x-, y- and z-coordinate axes are defined − → − → − → as i , j and k , which are expressed with symbol “→” as a vector notation.

Cross joint A

No.1 joint angle between shafts : α 1

No.2 joint angle between shafts : α 2

No.3 joint angle between shafts : α 3

Section A detail

Fig. 7.33 Cross joint used for automobile [VB197] y

y Θ

Ψ

z

x1

· Ψ=Ω

P

x A

α

x1

· Θ Q

Θ pin

α

A

MΘ

Ψ pin

Drive shaft

T

Driven shaft z1

Driven yoke

Drive yoke (a) Drive side

(b) Driven side

Fig. 7.34 Basic structure of cross joint (relation of drive shaft and driven shaft, Ota and Kato (1984)

7.6 Vibration Generated by a Cross Joint

195

Then, as a geometric relationship, defining one end of the  pin and one end of → → the  pin as P and Q, respectively, unit vectors − p () and − q () are considered in −→ −→ − → − → the A P and AQ directions. From Fig. 7.34, there is relationship i 1  cos α i + − → − → sin α k when the unit vector in the x1 direction is defined as i 1 , and the following relationships are obtained: − → − → − → − → − → − → p ()  cos  i 1 + sin  j  cos α cos  i + sin  j + sin α cos  k − → − → − → q ()  − sin  i + cos  j (7.16) → → As − p () and − q () are orthogonal, the inner product is 0. This is expressed as follows: − → → p () • − q ()  − cos α cos  sin  + cos  sin   0

(7.17)

Using Eq. (7.17) and the relationship sin2  + cos2   1, we obtain a geometric relationship: tan  

cos α cos  sin  tan  , cos   √ , sin   √ cos α H () H ()

(7.18)

where H ()  1 − sin2 α cos2 

(7.19)

When assuming that there is no friction between the cross pin and the yoke, torque −→ from the driving shaft will be transmitted without components in the cross pin A P direction while torque from the cross pin will be transmitted to the driven shaft −→ without components in the AQ direction. Thus, only the torque perpendicular to the cross pin plane PAQ will be transmitted through the cross joint. Since this torque is → → in the direction of outer product of − p () and − q (), assuming the vector of this − → torque is M a and the magnitude is Ma , we obtain the following equation:   − → − → − →   i j k   − →   − → − → M a  Ma p () × q ()  Ma  cos α cos  sin  sin α cos       − sin  cos  0  − → − →  Ma − sin α cos  cos  i − sin α cos  sin  j − → + (cos α cos  cos  + sin  sin ) k (7.20) − → Since the torque of the driven shaft is the z direction component of M a and the − → torque of the driving shaft is the z 1 direction component of M a , the torques of the

196

7 Case Studies of Forced Vibration Problems of a Rotor

driven and the driving shafts are defined as M and M , which can be expressed as − → − → − → the inner product of M a and the unit vectors k and k1 , respectively. Because the − → − → − → − → unit vector k1 in the z 1 direction is given by k1  − sin α i + cos α k , we obtain the following equations by using Eq. (7.18) and (7.19) into Eq. (7.20) to eliminate :  − → − → M  M a • k  Ma H () cos α − → − → − →  − → − → M  M a • k1  M a • − sin α i + cos α k  Ma √ H ()

(7.21) (7.22)

Therefore, by eliminating Ma in the above equations, transmission of torque between the driving shaft and the driven shaft can be expressed by: M 

H () 1 − sin2 α cos2  M  M cos α cos α

(7.23)

The orthogonality relation in Eq. (7.17) remains valid after differentiation with respect to time. By using Eqs. (7.18) and (7.19), after some work we obtain the relation:    1  cos α d − → → ˙  ˙ − H ()  ˙ − M  ˙ 0  p () • − q ()  √ M  dt Ma H () (7.24) Thus, the following equation expresses the relationship between the rotational ˙ (driving shaft), and,  ˙ (driven shaft), and the torque, M (driving shaft), speed,  and, M (driven shaft): ˙  

cos α ˙ , H ()

˙  M  ˙ M 

(7.25)

−→ Then, from Eq. (7.20), in the driven yoke in Fig. 7.34, the moment in the AQ −→ direction and the moment in the T direction, which is perpendicular to AQ and bends the shaft, can be represented as follows: − → → q ()  0 Mq  M a • −  π − →  − → − − → − → →  M a • − cos  i − sin  j  Ma sin α cos  Mq⊥  M a • q  + 2 (7.26) Thus, from Eqs. (7.18) and (7.22), the relationships with regard to the bending moments, Mx and M y in the x and y directions of the static coordinate, hold as

7.6 Vibration Generated by a Cross Joint

197

1−sin2α cos2 Ψ MΨ cos α

MΨ Intput shaft

α

Ψ

Torque ratio

MΘ =

MΘ / MΨ

1.2

Output shaft

joint angle (crossing angle) α = 30° α = 20°

1.1

α = 10° α = 0°

1.0 0.9 0.8

MΘ

0

90 180 270 360 Rotational angle of input shaft Ψ [deg]

Fig. 7.35 Torque variation caused by the joint angle between input and output shafts No.3 joint angle : α 3 [deg] 1 : Size represents an acceleration magnitude

No.2 joint angle : α 2 [deg] −1

0

1

2

3

−1

Fig. 7.36 Vibration amplitude with respect to two joint angles

M sin α (1 + cos 2t) 2 M tan α sin 2t M y  Mq⊥ sin   M tan α cos  sin   2

Mx  Mq⊥ cos   M sin α cos2  

(7.27) (7.28)

Equations (7.27) and (7.28) are functions of the rotation angle  of the driving ˙   to be constant, they reveal that bending moments are axis, and when setting  fluctuating in their 2  component in the x and y directions. From the expressions in Fig. 7.35, we have now explained how the component of twice the torque fluctuation is generated. Figure 7.36 shows the test results of generation of vibration at different joint angles in an actual case of an automobile transmission. The greater the joint angle, the greater the vibrations. In this case, we solved the problem by making the output axis line as close to a straight line as possible. In addition, it is found that in a three-axis system having two joint angles, vibrations can be reduced by setting the first and third axes parallel and setting the installation phase angles of two cross joints equal to each other [172]. Additionally, as cross joints have characteristics intrinsically similar to an asymmetric rotor, research is being carried out on the cases of unstable vibration of a parametric excitation [173, 174].

198

7 Case Studies of Forced Vibration Problems of a Rotor

7.7 Case Studies of Other Forced Vibration 7.7.1 Electromagnetic Vibration Electric motors or generators may also give rise to various types of vibrations. As a characteristic of such electromagnetically induced vibrations, we can find the vibration is attributable to them if the problem ceases as soon as the power is turned off. Electric motors and generators are classified into AC and DC machines. AC machines are used frequently and are subclassified into synchronous machines and induction machines. A synchronous machine rotor rotates in synchronization with the magnetic field and does not tend to cause special vibration except at the rotational component and its harmonics. In these machines, vibrations occur during a start-up, where the machine is operated as an induction machine, or in a hunting phenomenon that is apt to occur at the initiation of load operation after the synchronization with the grid. Large hunting will lead to a loss of synchronization (step-out); therefore, measures to suppress hunting amplitude are necessary. The most widely used machines in industrial applications are induction machines. Figure 7.37 shows examples of bipolar (two-pole) and quadrupolar (four-pole) machines. In a bipolar machine, the magnetic field rotates once per cycle of power supply frequency, f , while it rotates 1/2 per cycle for a quadrupolar machine, and 2/P per cycle for a P-pole machine. Thus, the rotation frequency, f 0 , of the magnetic field is as follows: f 0  2 f /P

(7.29)

The rotor rotates at the rotation frequency, fr , being driven by the rotating magnetic field with slip:

(1) Rotor

S

(1) Rotor

(2) Rotor bar

S

(2) Rotor bar

N

N

(3) Stator winding (4) Rotating magnetic field

N

(a) Bipolar machine

(3) Stator winding

S

(4) Rotating magnetic field

(b) Quadrupolar machine

Fig. 7.37 Rotating magnetic field in squirrel-cage induction motor

7.7 Case Studies of Other Forced Vibration

199 S 2f /P

Ω N

N

S

Fig. 7.38 Static eccentricity

fr  (1 − s) f 0

(7.30)

where s  ( f 0 − fr )/ f 0 is the slip (ratio) and s f 0 is the slip frequency. The following sections introduce some typical vibrations in such induction machines. [1] Static eccentricity of rotor and unbalance of stator circuit As Fig. 7.38 shows, when the rotor center is positioned statically with an eccentricity from the stator center and rotates without vibration, unbalanced magnetic attractive force will vary with a frequency with which a rotating magnetic pole passes the narrow gap. Thus, the excitation frequency will be frequency of the rotation magnetic field f 0 × the pole number P, which becomes twice the power supply frequency (2f ). When there exists an unbalance in the primary circuit (stator-side coil) due to a short circuit or other reasons, the magnetic attractive force will be strong in a specific direction just as in the case of the static eccentricity; hence, the frequency of 2f will be generated in the excitation force. [2] Shaft bending (dynamic eccentricity) Let us now consider a condition where the rotor support center coincides with the stator center and the rotor whirls in the gap all around the circumference with a certain amount of eccentricity, for example, due to a bent rotor. When supposing that the amount of eccentricity is constant, the rotor is fixed in a rotation coordinate as Fig. 7.39. In this coordinate, the magnetic flux density rotates at the frequency of f 0 − fr  2s f /P and is subject to amplitude modulation of 2s f /P × P  2s f as is the case of static eccentricity. In addition, magnetic flux density is inversely proportional to the magnetic gap and the minimum magnetic gap itself rotates at the rotation frequency in the static coordinate system. When the changing rate of a magnetic gap is defined as A and the rate of the modulation of magnetic flux density as B, since the magnetic attractive force F is proportional to the square of the magnetic flux density, by supposing both A and B are considerably smaller than 1, the force can be expressed approximately as follows:

200

7 Case Studies of Forced Vibration Problems of a Rotor

Fig. 7.39 Shaft bending

S 2sf /P

N

N

S

1 + B cos ωs t 2 1 + A cos ωr t    ≈ 1 − 2 A cos ωr t + 3A2 cos2 ωr t + · · · 1 + 2B cos ωs t + B 2 cos2 ωs t 

F∝

 1 − 2 A cos ωr t + 2B cos ωs t + 3A2 cos2 ωr t − 4 AB cos ωr t cos ωs t − 6A2 B cos2 ωr t cos ωs t − 2 AB 2 cos ωr t cos2 ωs t + · · ·

(7.31)

where ωr  2π fr , ωs  2π (2s f ). Thus, as the product-to-sum formula of the trigonometric function indicates above, the major generated vibration frequencies will be as follows: 2s f

fr

fr ± 2s f 2 fr 2 fr ± 2s f · · ·

(7.32)

Frequencies most observed in this phenomenon are the rotational frequency, f r , and its upper and lower sideband components with twice the slip frequency [175, 176], fr ± 2s f . Twice the rotation frequency, 2 fr , and its upper and lower sideband components, 2 fr ±2s f , are also observed. [3] Defects in rotor bar (broken bar) and unbalance of rotor circuit If parts of rotor bars of a squirrel-cage-type rotor break or electrical insulating performance deteriorates locally, impedance unbalance will occur in the secondary circuit (rotor-side circuit). In such a case, the force acting on the rotor will not be even around the circumference and consequently the force in the translational direction arises as shown in Fig. 7.40.

Fig. 7.40 Force generation by defective rotor bar

Torque large Ω Torque small

Translational force

7.7 Case Studies of Other Forced Vibration

201

In a rotor-fixed coordinate system, the position of a defective bar is fixed, and in this state, the force is modulated by the rotation magnetic field at the slip frequency. Thus, the vibration force is basically the same as the previously described dynamic eccentricity and is subject to amplitude modulation of a frequency of 2sf . Therefore, around the rotation frequency, f r , the upper and lower sidebands, fr ± 2s f , will arise [177]. [4] Slip beat sound If impedance unbalance, such as due to shaft bending or a defective rotor bar, exists in a secondary circuit (rotor side), the rotation magnetic field with a frequency of f 0 − fr  s f 0 in the forward direction in the rotor-fixed coordinate system becomes non-axisymmetric (elliptical). This indicates that a magnetic field of an inverse rotation (backward) of − s f 0 has occurred in the rotor. The frequency of this magnetic field of inverse rotation is − s f 0 + fr  ( 1 − 2s) f 0 when viewed from a static coordinate frame. Therefore, the generated fluctuation of the magnetic field intensity is derived by the pole number/2, which results in a frequency of ( 1 − 2s) f , and this will then interfere with the power supply frequency, f , causing a noise signal with a beat frequency of 2sf . This phenomenon is called “slip beat sound” [178], whose period 1/(2s f ) becomes shorter as load increases because the slip, s, becomes larger. [5] Motor Current Signature Analysis (MCSA) Although the frequencies mentioned previously have been observed in terms of vibration, this section assesses them through electric current observation. It is a common practice for the electric current from one phase of a stator to be measured using a clip-on type CT (current transformer, current probe) for a health diagnosis of electric motors. Research has been conducted on the phenomena caused by secondary circuit unbalance such as a broken rotor bar or a rotor eccentricity, and this technique has been established as a diagnostic method [179]. Figure 7.41 shows an example of spectrum measured in a case of a secondary circuit unbalance. It is known for electric current that spectra are generated at the power supply frequency, f (not a rotation frequency f r ), and at its upper and lower sideband components, f ± 2s f . Also, the amplitude ratio (dB) of the sidebands to the power supply frequency is used as an indicator of the soundness. The sidebands are generated at ± 2s f apart from the main frequency, which also result in vibration. In addition, a slot passing frequency or other acoustic vibrations in the range of kHz are generated in electromagnetic machinery. However, if frequency ranges of these phenomena are limited to the range relevant to the rotordynamics, characteristic frequencies can be summarized as in Table 7.3 [180, 181].

202

7 Case Studies of Forced Vibration Problems of a Rotor log Current [A]

f − 2sf

f

f + 2sf [ Guideline of healthiness for bipolar machine ]

ratio dB

−30 dB or more : more than one bar or short-circuit ring is broken −45 dB or less : good

f + 4sf

f − 4sf

Frequency [Hz]

Fig. 7.41 Typical measured spectrum in MCSA Table 7.3 Vibration frequencies occurring chiefly in motor and generator (in ease of induction machine) Generation frequency 2f

Description

Causes

Twice the power supply frequency f

fr

f r ± 2s f

Rotation frequency f r and its harmonics Sum /difference of f r and double slip frequency 2sf

2f r

2f r ± 2s f

Sum /difference of 2f r and 2sf

2s f

Static eccentricity of rotor Primary circuit (stator side) voltage and coil unbalance Dynamic eccentricity (Rotational whirl with an eccentricity) Secondary circuit (rotor side)

· Coil unbalance

Twice(Double) Slip Frequency " Slip beat sound "

(wound type)

· Broken bar f

f ± 2s f

Sum /difference of power frequency and double slip frequency (current)

(squirrel-cage type) s : slip rate

Case Example When a blower was started up using a three-phase induction motor (two-pole machine) as shown in Fig. 7.42, noise that beats with one or two times per second was generated during rated speed operation. Figure 7.43 shows the measurement results of the shaft vibration when increasing rotational speed. As the rotational speed increased, the amplitude of an abnormal vibration at a frequency 58 Hz grew to about seven times that of the rotation frequency 59.2 Hz. The abnormal frequency was slightly lower than the rotation frequency. In this example, we speculated that the phenomenon was caused by electromagnetic force from the motor, because the phenomenon disappeared immediately after shutting off the power.

Fig. 7.42 Three-phase induction motor [VB102]

3-phase wound-rotor induction motor ( 220 kW, Bipolar machine, 60 Hz ) Flexible Turbo blower coupling Slip ring

Common bed Rubber vibration isolator

7.7 Case Studies of Other Forced Vibration

203

Fig. 7.43 Abnormal vibration of electric motor

58 Hz , 50μm p-p Abnormal vibration component near rotation frequency [rpm] fV 3 550

Rotation component ( 59.2Hz ) fr

3 480 3 430 3 320 3 300 3 240 3 180 3 120 3 080 3 000 2 940 2 760 2 610 2 520 2 280 2 160 2 040 1 920 1 800 1 740 1 640 1 510 1 470 1 300 1 100 0

Fig. 7.44 Beat frequency change according to motor load

20

40 60 80 Frequency [Hz]

100

Frequency difference between vibration and power supply f0 − fV

2.5

Frequency [Hz]

2.0 Beat frequency 2s f 0 = f 0 − f V 1.5 2 × Slip frequency = 2s f 0

1.0

0.5

0

Slip frequency s f0 = f0 − fr 0

50 100 Motor load [%]

150

By taking the motor load as the abscissa, the measured slip frequency, s f 0  f 0 − fr , beat frequency, and f 0 − f V : the difference between f 0 and the abnormal vibration frequency, f V , are summarized in Fig. 7.44. From the figure, the beat frequency is almost the same as 2s f 0 ( 2s f for two poles), which can be regarded as the “slip-beat sound” phenomenon mentioned above that occurs in the presence of unbalance in the secondary circuit (rotor side). The motor manufacturer carried out the countermeasures. The inner circumference of the stator was machined to improve roundness. Also, to improve rotor concentricity to the stator, the installation play of the oil film bearings was reduced and the bearing clearance was reduced from 105 to 90 μm. As a result, the abnormal vibration was reduced to below 1/2 of the rotational component and the beat sound problem was drastically alleviated.

204

7 Case Studies of Forced Vibration Problems of a Rotor

7.7.2 Rotating Stall In a compressor, before entering the low-flow surge region, separated flow (stall) occurs at some blades on a part of the whole circumference, where the fluid excites the rotor at a specific frequency. In an axial flow compressor, the stalled fluid cells rotate, lagging behind the rotor rotation, which is called “Rotating Stall” [182, 183]. Although this is an unstable phenomenon of fluid flow, we classify it into a forced vibration from the viewpoint of a rotor. Figure 7.45 shows a 12-stage axial flow compressor in which a rotating stall occurred, and Fig. 7.46 shows a case of a rotating stall in this compressor. During starting-up, a large vibration of frequency 0.436N occurred within the range of Inducer Intake casing Labyrinth seal Intake side bearing box Stator vane

Diffuser

Discharge casing Labyrinth seal

Rotor blade Discharge side bearing box Rotor

Journal bearing

Thrust bearing Intake

Distance between bearings

Discharge

Journal bearing

6 100

Pressure fluctuation frequency caused by rotating stall Frequency of bearing box vibration Amplitude of bearing box vibration

20 15 10

ll sta 1 N ting a rot 3 000 ell 2 c 72 N 8 . 0 2 500 2 000

ll ll 1 ce ting sta rota N 6 0.43

1 500 1 000

Frequency [cpm]

Vibration amplitude [μm]

Fig. 7.45 Multi-stage axial compressor [VB304]

500

5 0

0 0

500

1 000 1 500 2 000 2 500 3 000 3 500 Rotational speed N [rpm]

Fig. 7.46 Vibration frequency, amplitude and pressure fluctuation frequency due to rotating stall

7.7 Case Studies of Other Forced Vibration Pressure sensor (1)

205 (1)

Stall region

Stall regions rotor

rotor 1T

Tacho pulse

1T

Tacho pulse

Pressure sensor (2)

(2) 1.15T

2.3T

Pressure measured by (1)

(1)

Pressure measured by (2)

(2)

(a) In case of one stall cell

(b) In case of two stall cells

Fig. 7.47 Manner of the rotation of stall cell

N  2750 rpm to 2900 rpm. Measurement of the pressure fluctuation revealed that pressure fluctuation of 0.872N was rotating in a wide range of rotational speed and a pressure fluctuation of 0.436N was rotating in this problematic range of rotational speed. In addition, we also confirmed that the value 0.436N was close to the first natural frequency of the rotor, which caused the large response. The characteristic of this phenomenon is that the pressure fluctuation (stall cell) rotates as Fig. 7.47 illustrates, which requires two pressure sensors to confirm the phenomenon. There may be one or multiple stall cells. The number of stall cells, n, can be calculated from the cross-correlation function (time lag can be measured) between pressure fluctuations measured by two sensors installed on the casing periphery apart from each other by θ (°) and the pulsation period, S (s), at a sensor by using the formula n

360◦ τmax θ◦ S

(7.33)

where τmax is the peak time in the cross-correlation function (time lag) between two pressure waveforms. In this case, the number of stall cells in the problematic speed range was one as shown in Fig. 7.46, while two cells existed in other cases. It was also found that, for the case of two cells, since these were 180° in phase apart from each other, the forces were balanced out and the vibration was prevented. It is commonly known that, in axial flow compressors, the frequency of rotating stall from 0.4N to 0.45N occurs. The mechanism in the case of centrifugal machines is more complicated, and a stall cell is known to rotate at a variety of frequencies. Table 7.4 summarizes these events. Note that the values in the table are for the case of the number of stall cells, n  1. If two or more stall cells exist, forced frequencies can be obtained by multiplying the values in the table by the number of stall cells n. The cause of this phenomenon is that the flow rate is too low for the passage area and may require extensive design modifications in some cases [184]. In the above case, this machine was a variable stator vane type and the stator vane angle was changed as a countermeasure. As an easy-to-use approach, ensuring the required flow rate by installing a bypass or recirculation flow path is often effective [185]. When rotating stall occurs at the diffuser without the vanes of a centrifugal machine,

206

7 Case Studies of Forced Vibration Problems of a Rotor

Table 7.4 Frequency and generation area of rotating stall (N: rotation frequency) Type of machine

Generation area

Axial compressor

Rotor blade

Centrifugal fan

Impeller

Generation frequency 0.4 ~ 0.45N 0.7 ~ 0.75N

Impeller Centrifugal compressor /pump

Diffuser (without vanes)

0.06 ~ 0.15N

Diffuser (with vanes)

0.25 ~ 0.33N

countermeasures, such as simply narrowing the diffuser flow path or narrowing the path by installing vanes, are usually effective.

7.7.3 Rotor Blade Fluid turbomachines generate vibrations with frequency Nz resulting from rotation frequency N multiplied by the number of rotor blades, z. This frequency, Nz, is called the blade-passing frequency (BPF), which acts as an excitation force together with its harmonics. This phenomenon is particularly apparent in pumps and other machines whose fluid density is high, in which the fluid force is large. Care must be taken since it often causes resonance when its frequency agrees with the rotor natural frequency. Figure 7.48 shows the structure of a typical volute pump. Pressure at the outer periphery of the impeller is not uniform, and the blades will be subject to large pressure fluctuation when they pass a tongue part in particular [186]. For this reason, the rotor will be excited by the frequency Nz. Figure 7.48a shows a structure of single volute pump that is subject to not only BPF vibrations, but also to a large static load (radial thrust) generated in the radial direction. Another type with reduced such radial force is a double volute pump in Fig. 7.48b, which has a symmetric layout of tongues being 180° apart from each

(Spiral) Single volute casing

Tongue

Double volute casing

Tongue

Impeller Impeller

(a) Single volute pump

Fig. 7.48 Scroll casing structures of volute pump

Tongue

(b) Double volute pump

7.7 Case Studies of Other Forced Vibration

207

Discharge rate 1375m3/h, Discharge pressure 1300kgf/cm2, Motor power 14200kW, Rated speed 5740rpm

Fig. 7.50 Nz vibration component before and after the countermeasure of fluid force balancing

Nz component of bearing [G] vibrational acceleration

Fig. 7.49 Case example of boiler feed pump (double volute) [VB083] 5

Before the countermeasure

4 3 2

After the countermeasure of fluid force balancing

1 0

1 000

2 000 3 000 4 000 Rotational speed [rpm]

5 000

6 000

other, and a diffuser pump (there are two types: with vanes and without vanes), for example. In these pumps, though the amount of dynamic load due to the BPF is reduced below that of the single volute type, Nz vibrations will still persist to a lesser extent. Figures 7.49 and 7.50 show the experience of a large double volute pump [187]. This pump showed a sudden increase in vibration acceleration at the casing around the rated speed. We conducted measurements and found that it was a vibration of the Nz component (number of blades z  5). In order to reduce excitation force that acts on the rotor, we changed the installation angle of the impeller to balance these fluid forces. Figure 7.51 shows the principle. In this double volute pump, the tongue parts are arranged 180° to each other, and the first mode will be excited in Case 1, where the impeller excitation takes place in the same phase. In Case 2, the impeller installation phases are opposite, and excitation takes place in the reverse phase, where the first mode will respond less and the second mode will be excited instead. This machine in question was in Case 2. We carried out countermeasures to make the machine ranked as Case 1. As a result, the machine was able to attain rated operation.

208

7 Case Studies of Forced Vibration Problems of a Rotor Case1 1st mode 0°

0° 2nd mode

0°

180°

Vibration amplitude

Case2 Operating range Case1

Case2

Rotational speed

Fig. 7.51 Principle of balancing of fluid force

7.7.4 Interaction Between the Rotor Blade and the Stator Vane As described above, vibrations at the rotation frequency, N, multiplied by number of rotor blades, Z r ( N Z r ), are often problematic for rotor vibrations. For blade vibrations, a rotor blade is subject to pressure fluctuations resulting from the rotation frequency, N, multiplied by number of stator vanes, Z g ( N Z g ), and a resonance problem of the rotor blades will occur when the natural frequencies of the blades are close to N Z g . For a bladed disk (impeller), since the vibrations of the blades and the disk are coupled, the resultant phenomenon is a little more complicated. Analysis of the natural frequency of the bladed disk reveals that there exist groups of a considerable number of similar natural frequencies, whose mode shapes are almost the same for each blade, but different in shapes for disk. These eigen modes have a different number of nodal diameters of the disk. In the case of a disk having Z r rotor blades, there are INT(Z r ÷ 2) + 1 natural frequencies (eigen modes) with different nodal diameter including a mode of zero (0) nodal diameter, where the INT function means rounding down after the decimal point to the nearest integer. Resonance does not necessarily occur at any of these natural frequencies, but resonances occur only in specific modes because of the interaction between the number of rotor blades, Z r , and the number of stator vanes, Z g . We studied the vibration analysis of a bladed disk in detail in R1_Sect. 10.2 and confirmed that when the number of disk nodal diameters is κ, the bladed disk interacts with vanes and responds only in the modes that have the next relationship [188–190] :

7.7 Case Studies of Other Forced Vibration

209

n Z g ± κ  h Zr

(7.34)

where n and h are any positive integers. And at this time, the resonance frequency of the bladed disk, ωκ , with κ nodal diameters holds as follows where  is the rotational frequency: n Z g   ωκ

(7.35)

Figure 7.52 shows a case of an interaction problem between rotor blades and stator vanes of a centrifugal compressor, which experienced cracking and the loss of an impellor short blade. In the configuration shown, there were 9 long impeller blades, 9 short blades, 11 stator mounted inlet guide vanes (IGV), and 19 diffuser vanes. Since Z r  9 (considering long and short blades as a set) and the number of IGV vanes Z g  11, when n  1, h  1, κ  2 (two nodal diameters, see Fig. 7.53), the condition of Eq. (7.34) is satisfied as 1 × 11 − 2  1 × 9. It was also made clear by an analysis that the blade vibration mode and the locations of the maximum stress were close to the damaged case (the crack locations). As for the diffuser vanes whose number was Z g  19, there was no resonance condition over the operation range. Therefore, it was revealed that resonance occurred when rotation frequency was   ωκ /11, where ωκ is the natural frequencies of this mode. As a countermeasure, it is necessary to detune natural frequencies of the bladed disk from the operation frequency. However, it was difficult to completely detune

Diffuser vane

Loss of short blade Crack generation Inlet guide vane (IGV)

Impeller blade

Crack generation

Fig. 7.52 Interaction between the impeller blade and the stator vane [VB547] Fig. 7.53 Two nodal diameters’ mode causing the problem

+ + -

210

7 Case Studies of Forced Vibration Problems of a Rotor

them since this machine was a variable rotational speed type. It was decided to avoid using IGV’s for flow control, and hence, they were removed, then to operate using rotational speed control only.

7.7.5 Four-Cycle Engine In a four-cycle engine, combustion takes place in each cylinder once every two revolutions and causes a vibration. Thus, the fundamental frequency is 1/2 of the rotational frequency N, ( N /2), and harmonics also occur at integer multiples (in particular the number of combustion cylinders) of the fundamental frequency. Note that in two-cycle engines, combustion takes place once per revolution in each of the cylinders, and the fundamental frequency will be the rotational component. Figure 7.54 shows a compressor driven by a diesel engine, which is set on a common bed supported by rubber vibration isolators. With increase of the rotational speed, the vibration at the end of the common bed reached about 1-mm peak–peak (Fig. 7.55). Measurement by a hammering test could not find natural frequencies that could be problematic within the rotational speed range and it could not be determined whether the rotation component caused the vibration. Figure 7.56 shows the measured frequencies when the engine speed was 20 and 40 rps, respectively. The frequency of both vibrations was the rotation frequency × 2. Since the engine was a four-cylinder type, the main frequency was 4 (number of cylinders) ×N /2 and integer multiples of this frequency occurred.

Compressor

Diesel engine Common bed Rubber vibration isolator

Fig. 7.54 Diesel engine-driven compressor [VB037]

Fig. 7.55 Vertical vibration of common bed

#2

Engine side

#3 Compressor side

#4

#1

7.7 Case Studies of Other Forced Vibration Fig. 7.56 Measured vibration at the common bed

211 Frequency 40.5 Hz

80.5 Hz (900 μm)

1200 rpm (20 rps)

2400 rpm (40 rps)

We decided to increase stiffness of the bed first because the main cause was its insufficient stiffness and then decreased the operational speed a little within a range that would not adversely affect the compressor performance, and these measures solved the problem.

7.7.6 Belt Drive Machines There is often a problem in belt drive machines where the vibration with a period of one rotation of the belt and its harmonics becomes large. This phenomenon can be estimated because the frequency is proportional to the rotational speed. By defining the rotational speed as N (Hz) and the belt with length, L rotates around the pulley with diameter D, and the generated fundamental frequency and harmonics, f (Hz), can be expressed by the formula: f n

πD N (n  1, 2, 3 . . .) L

(7.36)

It may be the case that the connection joint of the belt could have adverse consequences, but these problems are common, even with V-belts whose connection joint is very smooth and hardly visible on the surface. Figure 7.57 shows a Roots blower (three lobes) using four V-belts, where vibration of about 60-μm 0-peak at the common bed near the blower was measured during rated operation (rotational speed 29.4 Hz). The major vibration component was 33.8 Hz. Many other components were also measured, which were organized as shown in Table 7.5 into two vibration types: vibration at integer multiples of the

Blower

Motor

Vibration amplitude of common bed [μm]

60 50

33.8

40 30 20

16.9

10

41.9

0 0

Common bed

Fig. 7.57 Vibrations occurred in belt drive machine [VB828]

29.4 178.1

100 200 Frequency [Hz]

300

212

7 Case Studies of Forced Vibration Problems of a Rotor

Table 7.5 Measured frequencies and phenomena Measured frequency [ Hz ] 1

33.8

Amplitude [ Pm ]

Phenomena

43.3

Belt × 4 Belt × 2

2

16.9

9.5

3

178.1

2.9

Rotational speed × 6

4

29.4

2.8

Rotational speed

5

41.9

1.7

Belt × 5

6

8.4

1.2

Belt

7

25.0

0.9

Belt × 3

8

356.3

0.9

Rotational speed × 12

9

152.5

0.8

Belt × 18

10

89.4

0.6

Rotational speed × 3

T = T0 + ΔT cos νt V

( a ) Experimental system

1.0ν 2 0.5ν 1

Amplitude

Vibration frequency ( × ω n )

belt rotation frequency; and those of integer multiples of the Roots blower rotational speed. Also, the major vibration frequency of 33.8 Hz was just four times the belt rotation frequency, which is close to the natural frequency of the common bed. To solve this, we increased the stiffness of the bed to raise the natural frequency as a detuning countermeasure. As a result, the vibration amplitude reduced by a factor of 1/4 and the problem was fixed. Belt drive machines produce unstable vibrations as in a parametric excitation system, when the forced frequency of the belt tension, which is induced from the connection joint mentioned above or the eccentricity of the pulley, becomes close to an integer multiple of the belt natural frequency of transverse vibration. In particular, a large vibration of the belt will occur when the tension fluctuation frequency agrees with the natural frequency or twice the natural frequency. When the belt tension fluctuates as cos ν t shown in Fig. 7.58a, the equation of motion for transverse vibration of the belt can be expressed by the Mathieu equation (7.37):

0

0

1 2 ν : Excitation frequency ( × ω n )

( b ) Time history response and Campbell diagram

Fig. 7.58 Belt drive device and parametric excitation experiment

7.7 Case Studies of Other Forced Vibration

213

x¨ + ωn2 (1 − h cos ν t) x  0

(7.37)

Figure 7.58b shows a case when ν is increased up to three times of the belt natural frequency of transverse vibration, ωn . From this case, in addition to the first resonance at ν ≈ ωn , a larger vibration with an amplitude several times as large as the first resonance is observed in the vicinity of ν ≈ 2ωn . The latter case is an unstable vibration due to a parametric excitation that occurs when the coefficients of the vibration system (e.g., mass and stiffness) cyclically change. Care should be taken about the excessive unstable vibration in this case. Furthermore, there are some cases for reducing these belt vibrations by using active belt tension control [VB407].

7.7.7 Reducing Torsional Vibration As described in Sect. 7.1, since torsional vibration is characterized by low damping, vibration amplitudes may become large if resonance occurs. Various dampers have been developed to prevent torsional vibration. It is often the case that a vibrational torque fluctuates with the amplitude of a significant proportion of the rated torque in reciprocating machines, such as a four-cycle engine mentioned previously. Therefore, countermeasures against the torsional vibration are indispensable for such machines. Figure 7.59 shows a case involving severe damage where the generator shaft was broken by torsional vibration in an emergency diesel generator set. Detailed research revealed that the excitation torque had a frequency of six times the rotation frequency, which resonated with the lowest order torsional natural frequency at the time of start-up, and became excessive up to several multiples above the rated static torque. The countermeasure employed was the installation of a hood damper (silicon damper) as shown in Fig. 7.60, in which a ring-shaped movable mass is enclosed and the ring gap is filled with silicon oil. The damper was installed at the end of the diesel engine shaft opposite to the generator. The effect of the damper was confirmed by simulations before its installation. After this countermeasure, the machine has been operating without any significant vibration. In addition to such a torsional viscous damper, a torsional dynamic damper (vibration absorber) is often used and effective in many cases. Figure 7.61a shows a simplified torsional vibrating system before attaching the damper; Fig. 7.61b shows a Shaft breakage

Crank shaft (6 cylinders, 4 cycle)

Generator Flywheel

Fig. 7.59 Shaft breakage occurred in a diesel engine generator system [VB190]

214

7 Case Studies of Forced Vibration Problems of a Rotor Outer ring Silicon oil Inner ring (Movable mass)

Fig. 7.60 Silicon damper used for countermeasure n2 =

I M0 cos nΩt Ω

(Cylinder)

k Ω

2a 3r

r

a Ω

a

r

a l n2 =

a l

ϕ

m (a) Torsional vibrating system

(b) Dynamic damper (type1)

m1/2

m2

a 1 + ( m1 / 2m2 ) n2 = r 1 + ( 3m1 / 8m2 ) (c ) Dynamic damper (type2)

Fig. 7.61 Examples of torsional dynamic damper

simple pendulum type torsional damper; and Fig. 7.61c shows two types of damper where inner cylindrical surfaces of the disk are utilized. There are many types of torsional damper. In all of these cases, natural angular frequency of the pendulum under the rotational field of angular speed  can be expressed by the next formula [191]: ω0  n  (Case of a simple pendulum: n 

 a/l)

(7.38)

At this frequency, vibration of the main system behaves as an anti-resonance point against the excitation. Thus, vibration at the frequency of n times the rotational speed, , will be eliminated effectively [192–194]. Let us simulate the effect of the damper using a torsional vibration model of the piston rotary machine driven by the electric motor shown in Fig. 7.62. The dynamic damper is installed on the disk as shown in Fig. 7.61b. The displacement of torsional vibration of the disk is expressed as θ while the total angle of rotation of the disk is expressed as    t + θ . By defining the deflection angle of the centrifugal pendulum as ϕ, the translation displacement of the mass m of dynamic damper in the plane of rotation can be expressed as follows: x  a cos  + l cos( + ϕ),

y  a sin  + l sin( + ϕ)

7.7 Case Studies of Other Forced Vibration

215

Piston Ip Torsional vibration θ

Ω

kθ

Motor a

Deflection l angle ϕ m

Centrifugal pendulum

Fig. 7.62 Torsional vibration system of piston rotary machine and dynamic damper

Thus, kinetic energy T and potential energy V can be expressed by the following equations after defining the shaft torsional stiffness as kθ : Ip 2 ˙ +  2 Ip 2 ˙ +   2 kθ V  θ2 2 T 

 m 2 x˙ + y˙ 2 2     m  2 2 2 ˙ +l  ˙ + ϕ˙ 2 + 2al  ˙  ˙ + ϕ˙ cos ϕ a  2 (7.39)

By setting the Lagrangian, L  T − V , treating θ and ϕ as independent variables and substituting into Lagrange’s equations: ∂L d ∂L d ∂L ∂L −  F(t), − 0 (7.40) dt ∂ θ˙ ∂θ dt ∂ ϕ˙ ∂ϕ In these equations, θ and ϕ are considered to be small and terms of second order or ˙   + θ˙ ≈ , above are neglected. By approximating cos ϕ ≈ 1, sin ϕ ≈ ϕ, and  we obtain the linear equations of motion: I p θ¨ + kθ θ + m(l + a)2 θ¨ + ml(l + a)ϕ¨  F(t) ≡ T0 f (t) ml(l + a)θ¨ + ml 2 ϕ¨ + mla 2 ϕ  0

(7.41)

The excitation torque, F(t), from the piston is expressed with the product of the constant T0 and normalized time variation, f (t), which is approximated as a rectangular wave. This wave becomes positive and negative in one rotation. Therefore, f (t) can be expanded as a Fourier series using the rotation angular speed  as the fundamental frequency:

1 1 4 sin  t + sin 3 t + sin 5 t + · · · (7.42) f (t)  sign(sin  t)  π 3 5

7 Case Studies of Forced Vibration Problems of a Rotor 12

1X

θ

Vibration θ δ

(a)

7X 5X 3X

216

p

p

7X 5X

0.2

Vibration θ δ

12

1X

θ

4 0

7X 5X 3X

(b)

(Operating range)

8

3X

0.4 0.6 0.8 Rotational speed p = Ω / ω θ

1.0

1.2

1.0

1.2

(Operating range)

8 5X 4 0

0.2

3X Resonance disappearing 0.4

0.6

0.8

Rotational speed p = Ω / ω θ Parameter

ζ θ = 0.05, ζ d = 0.01, hd = 0.001, n = 3

Fig. 7.63 Simulation results of torsional dynamic damper (for a case of n  3)

After the preparation above, we carried out simulations with consideration of damping ratio, ζθ , for the major torsional vibration system and damping ratio, ζd , for the pendulum system, by rewriting Eq. (7.41): θ¨ + 2ζθ ωθ θ˙ + ωθ2 θ + h d ϕ¨  ωθ2 δ f (t)   1 + n 2 θ¨ + ϕ¨ + 2ζd n  ϕ˙ + (n )2 ϕ  0

(7.43)

√ where ωθ  kθ /Ia , Ia  I p + m (l + a)2 , h d  ml(l + a)/Ia , δ  T0 /kθ , Figure 7.63 shows the results of this simulation, obtained by solving Eq. (7.43) numerically. Figure 7.63a: Assuming h d  0 (i.e., the vibration absorber is inactive), a piston excitation force as a rectangular wave defined in Eq. (7.42) was input. Resonances with harmonic excitation components at p  /ωθ  · · · 1/7, 1/5, 1/3 can be observed before the resonance, p  /ωθ  1.0. Figure 7.63b: In order to suppress the 3X torsional vibration resonance, we adjusted the length of the pendulum so that n  3.0 is satisfied and added a mass equivalent to h d  0.001. As seen in the response curve, the 3X resonance at p  /ωθ  1/3 has been suppressed; hence, we can confirm the beneficial effect of the dynamic damper. As observed above, the natural frequency of a centrifugal dynamic damper is proportional to the rotational speed, , and it is possible to eliminate resonance with a specific order of rotation over the entire operation speed; however, many resonance points exist. It is also possible to eliminate other resonance points simultaneously, such as 5X and 7X, by using multiple dynamic dampers whose pendulum lengths are set as n  5.0 and 7.0.

Chapter 8

Case Studies of Self-excited Vibration of Rotor Stability Problems

Abstract Frequencies associated with self-excited vibrations are, in most cases, the natural frequencies of the system. The natural frequencies are not proportional to the rotational speed. Therefore, self-excited vibration is a form of non-synchronous vibration. In rotor systems, the lowest natural frequency is often below the rated rotational speed and it may become unstable, hence self-excited vibration is also known as sub-synchronous vibration. This chapter describes various case studies of self-excited vibrations, which are inherent in rotating machinery, as for a journal bearing, seal, centrifugal impeller, and blade for an axial flow machine. Also, the phenomena of internal friction, fluid trapped in a rotor, and rotor contacting with a stator, may produce strong self-excited vibration. While illustrating these unstable phenomena, the cause or mechanism of the instabilities and appropriate solutions are discussed by citing the v_BASE data. In addition, squeeze film dampers, which are used to stabilize the system by adding a damping effect, are explained. Keywords Oil whip · Oil whirl · Labyrinth seal · Rotor blade/impeller force (Thomas force) · Wachel’s formula · Hysteresis whip (Internal damping) · Fluid containing rotor · Dry friction whip · Dry friction whirl · Bently–Muszynska model · Squeeze film damper

8.1 Approaches to Self-excited Vibration Problems in Rotating Machinery 8.1.1 How to Identify Self-excited Vibrations and Apply Countermeasures It is easy to distinguish self-excited vibration from other abnormal vibrations by conducting a frequency analysis of vibration with knowledge of the natural frequencies of the system obtained from an excitation test or hammer test. This is because the frequency of self-excited vibration is typically equal to one of the natural frequencies of the system. However, it is not easy to find the causes of the vibration even if the phenomenon is identified as a self-excited vibration. In the countermeasures against © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_8

217

218

8 Case Studies of Self-excited Vibration of Rotor Stability …

the self-excited vibrations, a top priority is to disable the generation mechanism of the vibrations. In order to identify the causes, we start by varying the parameters of relevance and evaluate their effects on the vibration. An enormous amount of effort may be necessary for a cessation of the mechanism. It is possible, however, to prevent the vibrations even if the real causes are uncertain. Since self-excited vibration arises when the damping ratio becomes negative and the system becomes unstable by a decrease of damping, the vibration can be avoided by adding a damper, for example, to compensate for the decrease. Although this is a symptomatic therapy, this countermeasure is frequently implemented. The occurrence of self-excited vibrations depends on whether or not the system becomes unstable, and the vibration can be prevented by adding a small amount of damping to change the border of the stability limit. To consider the causes and the mechanisms of self-excited vibrations, actual cases in the past are very informative. For this reason, let us study self-excited vibration case examples here. Figure 8.1 summarizes problematic vibrations observed in rotating machines whose frequency is lower than the rotational speed. Although the vibrational frequencies correspond to a low-order natural frequency, they tend to be regarded as unstable vibrations (self-excited vibrations), the self-excited vibrations are grouped into (1) in Fig. 8.1. The vibrations are often referred to generically by the machine type, irrespective of the mechanism, for example, Steam whirl for steam turbines, Gas whirl for centrifugal compressors, and Hydro whirl for pumps, etc. The generic terms are used in the figure. Phenomena caused by unstable flow, for example, surging and rotating stall, are classified into self-excited vibration from the viewpoint of fluid dynamics. However,

Example phenomena Oil whip Steam whirl (1) Self-excited vibration (Unstable vibration) 0

(Unstable Vibration at partial control feeding into control stage stage flow)

Forced vibration

Electric disturbance Surging, Reverse flow

Karman vortex, etc.

Fig. 8.1 Classification of sub-synchronous vibration

Swirl breaker, etc.

Bearing improvement

(Stiffness, inertia asymmetry)

Fluid disturbance

Removal or reduction of cause

Adding damper

Removal or reduction of cause

(Various excitations) (Low flow rate)

(Resonance)

Removal or reduction of cause

Restriction of low flow rate operation

8.1 Approaches to Self-excited Vibration Problems …

219

from the viewpoint of rotor dynamics, these unstable flows are classified as external disturbances, the vibrations are categorized into forced vibration, corresponding with (2) and (3) in Fig. 8.1. These are similar to self-excited vibrations but should be treated as forced vibrations for the solution. In order to discriminate between self-excited vibrations and forced vibrations in an actual machine, it is important to conduct an excitation test to measure the damping ratio with changes in operating parameters. This enables an affirmation about stability. Forced vibrations of (2) and (3) are less dangerous than the unstable vibrations of (1) and are not so problematic if the vibration amplitude is small.

8.1.2 Examples of Destabilizing Force A feature of the mechanism that induces unstable vibration in a rotor system is caused by the changes of the characteristics depending on rotational speed or loads as described in Chap. 7 on forced vibrations. Such vibrations occur, in general, as illustrated in Fig. 8.2. The damping ratio is gradually decreased from the initial value associated with bearings and damper elements due to the rotational speed or various load-dependent destabilizing forces. The damping ratio thus becomes zero then negative. This operating condition is the stability limit and self-excited vibration will develop. Such destabilizing forces arise from a variety of mechanisms, which will be explained in the following sections. (1) Features of destabilizing force acting on a rotor system A damping force is proportional to the velocity and acts in the “braking” direction opposing the vibrational velocity. In a rotor system, a force may also occur in the direction of the whirl velocity, i.e., in the “accelerating” direction that coincides with the tangent to the whirl. Figure 8.3 shows that the force (1) acts as a stabilizing force in the braking direction, and the force (2) acts as a destabilizing force in the accelerating direction, which promotes whirling. It does not matter whether a tangent force is proportional to the velocity or proportional to the displacement. A force that applies in the same direction as the whirl velocity has a great and negative influence on the stability.

0.05

Damping ratio

Fig. 8.2 Decline of damping ratio by operating condition

0

Damping decline due to various unstable effects

50

Load [%]

100

Initiation of unstable vibration (Stability limit)

220

8 Case Studies of Self-excited Vibration of Rotor Stability … y

Whirl direction

Force Δ Fy

(2) x

Δx Displacement

(1)

Rotor

Fig. 8.3 Cross-coupling force causing unstable vibration

The forces that act perpendicular to the displacement are termed as cross-coupling spring forces since they act in the y-direction for a displacement in the x-direction and vice versa. In most cases, except for friction forces caused by a rotor–stator contact, these are forward whirl forces that act in the same direction as the rotor rotation. For this reason, the unstable vibrations in a rotor system are mostly forward whirl vibrations. (2) Tendencies of various destabilizing forces Figure 8.4 shows schematic images of characteristics of various destabilizing forces, with rotational speed as the abscissa axis and the damping ratio as the ordinate axis. (a) In the case of internal damping (internal friction, hysteresis whip) or a fluidcontaining rotor, the destabilizing force will arise after the rotational speed has passed over the critical speed f 1 . Although such a force acts as a stabilizing force below the critical speed, it acts as a destabilizing force above the critical speed and increases as the rotational speed increases. The system becomes unstable when the destabilizing (negative) damping exceeds the initial sum of the (positive) damping values, which is referred to as base damping here. By defining the base damping ratio, ζ0 , internal damping ratio, ζi , natural frequency, f 1 ,and rotational speed, fr , the actual effective damping ratio, ζ , can be represented by the linear approximation of the internal damping mechanism. The stability limit speed, fr , can be calculated by setting ζ  0:     fr ζ0 → fr  f 1 1 + (8.1) ζ  ζ0 + ζi 1 − f1 ζi When the base damping is ζ0  0, self-excited vibrations occur immediately after the rotational speed passes the critical speed f 1 . (b) Unstable vibrations known as oil whip in a cylindrical oil film bearing are caused by a destabilizing force that acts when the rotational speed exceeds twice the natural frequency. Since a multi-lobe bearing for improving stability

Stabilizing action

Less stable seal Improved seal fr 2f1

Base damping 0'

Less stable bearing Improved bearing fr f1

2f1

Base damping

Rotational speed Unstable

( c ) Labyrinth seal

Rotational speed

Unstable

0'

( b ) Cylindrical /multi-lobe oil film bearing

Stabilizing action

( a ) Internal damping (internal friction)

f1

221

0

Destabilizing action

Unstable

Base damping

0

Unstable Destabilizing action

Rotational speed

f1

0'

Stabilizing action

Unstable Destabilizing action

0

fr

0

Destabilizing action

Stabilizing action

8.1 Approaches to Self-excited Vibration Problems …

fr f1

2f1

Base damping

Rotational speed

Unstable

0'

depending on load, etc. ( d ) Rotor blade, impeller

Fig. 8.4 Images of various types of unstable force

has a smaller destabilizing force as shown by the dashed line, the stability is improved on the high-speed side, and the stability limit speed increases. (c) Labyrinth seals used for turbines and compressors also bring out destabilizing forces. Depending on the swirl of gas flow at the seal entrance, the magnitude or the trend of the destabilizing forces will change. Implementing countermeasures such as prevention of swirls and the introduction of an opposite swirl to the rotational direction will shift the curve upward as shown by the dashed line and consequently the vibration becomes stable. On the contrary, if swirl is increased, self-excited whirl will occur readily in the swirling direction, even when the rotor does not rotate. (d) Destabilizing forces of a rotor blade or an impeller will change, depending not only on rotational speed but also on load and flow rate. They may even be destabilizing forces in any operating condition. The forces depend on the design and are complicated in many cases; further research is required in the future.

222

8 Case Studies of Self-excited Vibration of Rotor Stability …

8.2 Self-excited Vibration Problems Caused by Oil Film Bearings or Seals 8.2.1 Self-excited Vibration Caused by Oil Film Bearing (Oil Whip) The dominant case of self-excited vibration in the past was oil whip in oil film bearings, described in Chap. 4. Figure 8.5 summarizes the condition under which oil whip occurs. In a rotor supported by cylindrical oil film bearings, when the rotor speed is increased, self-excited vibration of “oil whirl” (also referred to as “half speed whirl”) with a relatively small amplitude is observed initially at a frequency of slightly less than half the rotational speed N ( f ≈ N /2). The oil whirl will turn into a large amplitude of self-excited vibration called “oil whip” when the frequency of the oil whirl coincides with the natural frequency, f 1 ( f ≈ f 1 ), at the rotational speed of N ≥ 2 f 1 . Hysteresis phenomena (history-dependent characteristics) are sometimes observed in oil whip, where large amplitude of vibration persists even if the speed is lowered considerably below the speed at which the oil whip vibration was initiated. The oil whip mechanism is described in Chap. 4, which shows that the energy balance (mechanical work done by the oil film) for one whirl cycle of the journal can be calculated by Eq. (4.12), supposing a forward whirl having a circular orbit. The following formula can be derived after the conversion into a non-dimensional expression using Eq. (2.27):         W ∗ E E Fx d x + Fy dy  π A2 ω cx x + c yy − k x y − k yx ≡ π A2 Cp    ω C x x + C yy − K x y − K yx (8.2) E∗  

(1)+(2)

f = f1

f1

N

f≈ 2

(1)

0

(2)

(Natural frequency component) (2) Self-excited vibration (1) Unbalance vibration

2 f1 Rotational speed N

f1

Amplitude (solid line)

N f=

Fig. 8.5 Frequencies of oil whirl and oil whip

Measured frequency f (dashed line)

where the oil film coefficients in lowercase letters are dimensional and those in uppercase letters are non-dimensional. In this equation, energy will be dissipated

R

R

R

R

R

2-lobe bearing

3-lobe bearing

Tilting pad bearing (a) Bearing types

Dissipation energy E* ( in case of ω = 0.45Ω )

8.2 Self-excited Vibration Problems Caused by Oil Film Bearings …

223

1.0 0.8

L/D = 0.5

0.6

Tilting pad mp = 0.5 3-lobe bearing mp = 0.75

0.4

2-lobe bearing mp = 0.5 Cylindrical bearing with 2 axial grooves

0.2 0 −0.2

0

0.2

0.4

0.6

0.8

1.0

Sommerfeld number Sb (b) Dissipation energy

Fig. 8.6 Stability-improved bearings and comparison of dissipation energy

when E ∗ > 0, but energy will be supplied from the oil film when E ∗ < 0 and unstable vibrations will occur. It can be said, therefore, that the difference of the crosscoupling stiffness term between the x- and y-directions, K x y − K yx , acts in the same manner as negative damping and becomes a cause of forward self-excited vibration (oil whip). For a cylindrical bearing in particular, the relation of K x y ≈ −K yx > 0 is satisfied at a high rotational speed, and consequently self-excited vibration will occur inevitably. In order to prevent oil whip, multi-lobe bearings, including: 2-lobe, 3-lobe, and 4lobe bearings, shown in Fig. 8.6a have been developed. In these bearings, the speed of the stability limit has been improved considerably and operation at a higher rotational speed is possible. However, the cross-coupling terms of the stiffness coefficients, K x y and K yx , which are the cause of oil whip, are not zero for the multi-lobe bearings. A bearing for which the cross-coupling terms are zero is the tilting pad bearing, which has enabled high-speed rotation without any self-excited vibrations. Figure 8.6b compares the dissipation energy of different bearings when the ratio of vibration frequency/rotational speed is ω/  0.45, plotted against Sommerfeld number, Sb (proportional to the rotational speed, Eq. (2.20)), based on the radial clearance of the assembled bearing, Cb , in the horizontal axis, and m p as the preload factor. Selfexcited vibration occurs when E ∗ becomes negative, but in a tilting pad bearing of course, E ∗ never becomes negative. By calculating numerically dynamic characteristics of these bearings in advance or by using the oil film bearing characteristics databook [B13], and performing a complex eigenvalue analysis of the system, a stable rotor-bearing system can be designed in which self-excited vibration will not occur over the whole operating condition range. Currently, it is not difficult to design the system to be free from oil whip and oil whirl.

224

8 Case Studies of Self-excited Vibration of Rotor Stability …

However, exceptional trouble occurs if prior consideration is insufficient or unusual operation is implemented. Examples are introduced in the following sections. Case 1 Vertical pump motor Considering a vertical rotor supported by oil film bearings, since the radial load is zero, whirl may occur readily around the static point of zero eccentricity, which is irrespective of the rotational speed. Thus, the dynamic characteristics of the bearing based on the Sommerfeld number cannot be applied, and the instability is very strong. Figure 8.7 shows the example of a vertical pump motor [195], which was exposed to working liquid and the bearings were lubricated with this liquid. Fortunately, a slight radial load F acts on the impeller and dynamic characteristics of the upper and lower bearings can be determined based on the Sommerfeld number. A complex eigenvalue analysis was carried out to examine the stability. Favorable results were obtained for the 3-lobe offset (staggered, eccentric) bearing and were compared with the case of a 4-lobe bearing. Figure 8.8 shows the test results. While the vibration of half-speed whirl was about 1.5 times the amplitude of the rotational speed component that occurred with the 4lobe bearing, the 3-lobe offset bearing was able to offer stable performance, with only the synchronous rotational speed component showing in the vibration response. Bearing like effect at seal gap

F

Impeller (a) 4-lobe bearing

Upper bearing Motor rotor

G

Magnetic pull by motor Lower bearing

(b) Offset 3-lobe bearing

Fig. 8.7 Rotor and bearings of a vertical pump motor [VB079]

Amplitude

0.5 0.4 0.3

4-lobe bearing

0.2

Offset 3-lobe bearing

0.1 0

0

500

1 000

Amplitude

Vibration amplitude [mm p-p ]

0.6

1/2 rotation whirl component Rotational component

1 / 2N N

Frequency

Rotational component

1 500

Rotational speed [rpm]

Fig. 8.8 Test run results of vertical pump motor

N

Frequency

8.2 Self-excited Vibration Problems Caused by Oil Film Bearings …

225

Case 2 Instability of a tilting pad bearing supported rotor Concerning a tilting pad bearing, which is believed not to cause oil whip, if the bearing clearance is widened too much, the pad natural frequency in the tilting direction may decrease. Deviations may then arise from the usual characteristics listed in the databook and self-excited vibration may occur [e.g., 33]. The following provides an example. Figure 8.9 shows the case of a rotor supported by tilting pad bearings. When the rotational speed was increasing in a high-speed balancer, a self-excited vibration (oil whip) at a large amplitude of 120 µm 0-peak (88.5 Hz) occurred suddenly at 13,200 rpm (220 Hz). High-speed balancing was not scheduled originally, but became necessary urgently and auxiliary tilting pad bearings in the balancer were employed, because they were similar in size to the original bearing. The clearance ratio of the bearing (bearing radial clearance divided by journal radius) was 0.2% in the original design, but that of the bearings of the balancer was 0.5%. In addition to this, the preload factor of the bearing was negligibly small so that m p  0. The suspected cause of the self-excited vibration was the tilting pads themselves because the bearing clearance was too wide and the preload factor was too small. Therefore, the fluid constraint forces on the pads became so far reduced that the pads gave rise to self-excited vibration. The countermeasure implemented was to reduce the bearing clearance as well as to increase the preload factor in order to be closer to the original design. Specifically, shims were temporarily put on the outside of the bearing pads as shown in the downside of Fig. 8.9a. As a result, the rotational speed could be increased up to the rated speed of 18,850 rpm as shown in Fig. 8.9c.

Shim (100 mm) Pad fixing pin

μmP 40.0

1N

5N

20.0

2N 3N

3N

Carrier ring Pad

1N

13,200 rpm (220Hz)

μmP 30.0

4N

Pad fixing pin

18,850rpm (314Hz) 29 μm 39.7Hz 4 μm

Rotational speed

0 10 Pad

2N

=φ

4N

D

Rotational speed

88.5Hz 120 μm

5N

15.0

Carrier ring 0

( a ) 5-pad bearing and countermeasure against self-excited vibration

200 400 600 [Hz] ( b ) Before the countermeasure

0

200 400 600 [Hz] ( c ) After the countermeasure

Fig. 8.9 Self-excited vibration occurred in a machine using tilting pad bearings [VB608]

226

8 Case Studies of Self-excited Vibration of Rotor Stability …

8.2.2 Self-excited Vibration Caused by Seal (Clearance Flow) Varieties of seals are used in fluid turbomachines. These seals also generate damping and destabilizing forces caused by clearance flows. In particular, a labyrinth seal requires detailed consideration since it has little damping, and therefore even a small destabilizing force may be a problem. High-pressure pumps and high-pressure compressors whose discharge pressure exceeds 10 MPa process fluid of very high energy density and produce a very high destabilizing force when compared to low-pressure machines. During the development of ultra-high pressure compressors in recent years, many self-excited vibration problems have been reported. Unstable vibrations caused by fluid forces produced around the impeller, between the shroud and the casing, or in the seal, for example, have been termed collectively as high-pressure fluid-induced vibration. Research conducted thereafter has revealed that considerable components of these vibrations are caused by labyrinth seals. Destabilizing forces developed in labyrinth seals are greatly influenced by a swirl flow inside the seals and it is known, for example, that increasing pre-swirl at the seal inlet will produce self-excited vibration whirling in the swirling direction even when the rotational speed is zero [196–199]. In addition, swirl flow around the impeller tends to be in the same direction as the rotation, thus causing a forward whirl of self-excited vibration similar to oil whip. The dynamic characteristics of a labyrinth seal are expressed as in the case of oil film bearings, using stiffness and viscous damping coefficients. Assuming an infinitesimally small vibration around the center of the seal clearance, the seal reaction force can be expressed with four dynamic characteristics, namely kd , kc , cd , cc , as shown in Eq. (8.3), which may be calculated by bulk flow theory or, in recent years by using CFD software:   

 k k x x˙ x x˙ F c c kd kc cd cc − x  xx xy + xx xy  + Fy k yx k yy c yx c yy −kc kd −cc cd y y˙ y y˙ (8.3) As is the case in oil whip, the dissipation energy in this phenomenon can also be calculated from Eq. (8.2): E  2π A2 (ω cd − kc ) It is evident from this equation that the cross-stiffness coefficient, kc , is the cause of instability. Labyrinth seals are frequently used in centrifugal compressors as a non-contact design, as shown in Fig. 8.10 [199]. The vibration stability is evaluated by the same complex eigenvalue analysis as for the case of oil film bearings by taking the seal dynamic characteristics of Eq. (8.3) for each seal into account. For seals, however, the prediction accuracy of dynamic characteristics is not as high as in the case of oil film

8.2 Self-excited Vibration Problems Caused by Oil Film Bearings …

Impeller eye

227

Balance drum

Shaft seal

Destabilization

Fig. 8.10 Various labyrinth seals in a compressor (Kirk (1990))

kxy

Steam

High pressure side

kxy −ω cxx

Control stage seal

Inlet swirl velocity circumferential verocity

Stabilization

1

High pressure tip seal

2

3

4

5

− ω cxx Control stage seal

High pressure tip seal

Low pressure side Nozzle root seal High pressure dummy seal

High pressure dummy seal, nozzle root seal

Fig. 8.11 Example of destabilizing effects due to inlet swirl velocity in steam turbine

bearings, and it is not possible yet to evaluate stability quantitatively with sufficient accuracy. Prediction of dynamic characteristics is still difficult, particularly in the cases of the narrower seal clearance, higher rotational speed, and higher differential pressure. Figure 8.11 shows an example of a labyrinth seal for a steam turbine, where the inlet swirl velocity is non-dimensionalized by the circumferential velocity along the horizontal axis and the investigation results of k x y − ωcx x (the same as kc − ωcd in Eq. (8.3)) at the relevant seals are plotted. It can be seen from the figure that a stabilizing effect is expected at points where the dimensionless swirl velocity is smaller than 1.0, whereas a destabilizing effect becomes evident when the velocity is larger than 1.0. Other pump seals and damper seals have similar characteristics as labyrinth seals. Since these fluid seals have a relatively larger spring coefficient, kd , and a larger damping coefficient, cd , compared with a labyrinth seal, in addition to the destabilizing coefficient, kc , these coefficients should be taken into consideration when examining the stability.

228

8 Case Studies of Self-excited Vibration of Rotor Stability … (left /right two cases) Shunt Reverse jet flow hole

Leakage flow Fin

Rotational direction

Shunt holes

Rotational direction Fin

Leakage flow

Swirl breaker

Fig. 8.12 Swirl breaker and shunt hole (Dresser Rand Company)

Note: The characteristics of different types of non-contact seals are described in various papers, for example, Iwatsubo [200], Childs [201, 202], and further helpful information is provided by references [203–207]. Principal countermeasures against this problem are to reduce swirl flows in the seal. Therefore, it is important, firstly, to install the fins (teeth) of the labyrinth seal on the stator side not on the rotor side. It is also effective to adopt partition plates that connect neighboring fins in the seal with each other or to use a honeycomb seal that is not likely to develop swirls, instead of an ordinary labyrinth configuration. A swirl velocity at the seal inlet from the outer circumference of an impeller leakage will become much faster than the shaft circumferential speed, because the radius of the seal is smaller than that of the impeller and the angular momentum of flow is conserved. Therefore, these measures to eliminate the inlet swirl (pre-swirl) are often employed, and there are many example cases [208–210]. Figure 8.12 shows actual examples of swirl cancellers that employ a frequently used swirl breaker (swirl brakes) and shunt holes (several holes around the entire circumference) used in a 50 MPa reinjection compressor for oil drilling as pre-swirl preventive measures [209]. The former restricts pre-swirl flow in the shaft rotation direction with the radial fins installed directly in front of the seal inlet. The latter reduces the swirl flow by imposing reverse injection flow from the discharge pressure region into the seal. This compressor was also equipped with honeycomb seals and damper bearings that have a squeeze film damper on the outer circumference of a tilting pad bearing. Case 3 Steam turbine for cryogenic power generation The rotor has labyrinth fins installed on the rotor side, and considerable portion of the rotor is covered with labyrinth seals (see Fig. 8.13). As a result, self-excited vibrations occurred at 38 Hz corresponding to the first natural frequency as shown in Fig. 8.14. As a countermeasure, we installed radial swirl breakers at the inlets of each labyrinth seal and widened the bearing clearance by 0.1% to increase the damping effect. Also, the static load direction of the bearing was shifted from load-betweenpad (LBP) to load-on-pad (LOP) to increase anisotropy of the bearing stiffness in order to reduce the effects of the destabilizing force (see Sect. 4.7). These countermeasures completely eliminated self-excited vibrations.

8.3 Self-excited Vibration Due to Fluid Force of an Impeller

229

Flow direction

Flow direction Labyrinth seal

Labyrinth seal

Thrust bearing

Shaft coupling side

Tilting pad bearing (5-pad)

High pressure turbine

Low pressure turbine

Tilting pad bearing (5-pad)

Shaft vibration amplitude [mp-p]

Fig. 8.13 Steam turbine for cryogenic power generation [VB004] 62.5 50

38Hz 40.3 mp-p

O.A. 42.4 mp-p

25

4 545 CPM (rotational component) 7.8 mp-p 0

40

80 120 Frequency [Hz]

160

200

Fig. 8.14 Shaft vibration at the neighborhood of bearing

8.3 Self-excited Vibration Due to Fluid Force of an Impeller 8.3.1 Self-excited Vibration Caused by Turbine Blades (Axial Flow) An axial flow turbine disk develops cross-coupling force during eccentric whirling. In this state, the bladed disk experiences a larger force at the side with narrower tip clearance than at the side with wider clearance, which results in a tangential force. Figure 8.15 shows the mechanism to create such a destabilizing force. This figure indicates that a bladed disk displacement, ε, in the radial direction causes a variation in leakage loss at the blade tip, and an unbalance torque force is produced in the direction perpendicular to the disk displacement. It is understood that such unbalance torque force induces forward unstable vibrations for turbines, while it induces backward unstable vibrations for compressors, fans, propellers, or other machines whose forces are in the direction opposite to that of turbine [211, 212].

230

8 Case Studies of Self-excited Vibration of Rotor Stability … Stator

Unbalance torque force

Tip clearance

Rotor

2 0







Fig. 8.15 Destabilizing force caused by torque unbalance

This effect is sometimes referred to as excitation due to steam, but it is generally called the Thomas force [213] or Alford force [214] after the advocates of the mechanism. Thomas defined a cross-coupling spring for a steam turbine disk with the formula: kx y  β

Ttotal D×L

(8.4)

where Ttotal  torque generated on a turbine disk, D  turbine disk diameter, L  turbine blade length, and β  calibration coefficient against experimental values. With respect to the calibration coefficient, β > 0 for a turbine, but β < 0 for an axial compressor where the fluid force is opposite to the rotation direction and results in backward self-excited vibrations. Specifically, β values should be obtained from experiments or CFD calculations around the blade disk by varying the tip clearance. Case 4 Turbine pump In this case, a turbine cross-coupling force was calculated using CFD evaluations from the torque, T1 (when the tip clearance is narrower by the displacement ε), and the torque, T0 (ordinary average tip clearance), of a turbine blade array as shown in Fig. 8.16 [215]: 

T0  T1 − T0 1+ Ft  d F cos θ  cos θ cos θ (8.5) R T0 Ft kx y  (8.6) ε As a result, β  0.56 was obtained from the relationship between the calculated value of k x y  8.1 × 105 N/m and Eq. (8.4). By employing a model which includes seal cross-coupling forces and internal damping, a complex eigenvalue analysis was conducted as shown in Fig. 8.17, with and without the Thomas force. In the case with the Thomas force, the damping ratio increases in 1B (the first backward whirling mode) and 2B and higher order backward whirling modes, without exceptions. In contrast, forward whirling, 1F (the first forward whirling mode), in particular, shows a considerably reduced damping

8.3 Self-excited Vibration Due to Fluid Force of an Impeller dF 

Ft



dF

T1 R

dFcos

231

(

T0 T1 − T0 cos  R 1+ T0

)

T0 R

R  



0

Fig. 8.16 Method for estimation of cross-coupling force in a turbine disk Fig. 8.17 Effect of Thomas force in turbine pump Damping ratio [ % ]

14

Without Thomas force With Thomas force

12 10 8 6 4 2 0 1B

1F

2B 2F Mode number

3B

3F

ratio close to the unstable state. This model faithfully reproduced the status of the actual machine. Countermeasures for this case were to use shunt holes (two holes around the entire circumference) in order to introduce pre-swirl in the direction opposite to the rotation at the seal inlet. As a result, the stability was improved and the problem was solved.

8.3.2 Self-excited Vibration Caused by Centrifugal Impellers (1) Centrifugal compressor Impellers of a centrifugal compressor also produce destabilizing forces during high speed and high load operation. Wachel [216] obtained the destabilizing force generated in each impeller stage in the form of a cross-coupling spring, k x y , for a multistage centrifugal compressor as:   16.0 × M × P ρ D (N/m) (8.7) kx y  D × h × fr ρS where M  gas molecular weight (28.97, or approx. 30 for air), P  rated output per an impeller stage (kW), f r  rotational speed (rps), D  impeller outer diameter

232

8 Case Studies of Self-excited Vibration of Rotor Stability …

(OD) (m), h  width at impeller OD or diffuser width (m), ρ D  gas density at the impeller outlet (kg/m3 ), and ρ S  gas density at the impeller inlet (kg/m3 ). The approximate formula of Eq. (8.7) was adopted in the API 617 standard (2002) [217] for compressors and has been used for the screening of instability of centrifugal compressors. A logarithmic decrement of the system is calculated by a complex eigenvalue analysis, introducing the cross-coupling stiffness k x y above for each impeller stage. Consideration of this item is regarded as a mandatory screening for Level-1 stability analysis for API-compliant compressors. This approximation formula is experimentally based, which contains other destabilizing forces such as from labyrinth seals attached to the impeller, in addition to the impeller force itself. If the logarithmic decrement in the Level-1 calculation becomes negative (i.e., if instability risk is a concern), more detailed consideration on stability is required in a Level-2 stability analysis [e.g., 218]. Wachel’s formula, although it has many actual achievements, can be regarded as an approximation formula for design and therefore the prediction accuracy is insufficient. This field requires continued research in the future. Case 5 Model compressor for experiments This section introduces a case in which unstable vibrations due to destabilizing forces from impellers were reviewed at the laboratory level [219]. By employing a shaft system composed of a lightweight impeller with almost no associated damping, unstable vibrations were reproduced even in low-power operation. Figure 8.18 summarizes the experiments, which are compared with Wachel’s formula by taking valve opening as a parameter for discharge flow control. The results show that as the rotational speed rises, the stability decreases. Finally, an unstable state (damping ratio: ζ < 0) is evident. Figure 8.19 shows the time evolution of how unstable vibrations developed. The figure demonstrates that the forward whirl (direction from the x-axis positive to the y-axis positive) grows exponentially.

Fig. 8.18 Damping ratio of the compressor rotor with respect to rotational speed

0.01

Discharge valve full open 3/4 open 2/4 open 1/4 open Wachel's formula

0.008

Damping ratio ζ

0.006 0.004 0.002 0

0

5

10

15

20

−0.002 −0.004 −0.006 Rotational speed [ rps ]

25

30

8.3 Self-excited Vibration Due to Fluid Force of an Impeller 2/4 open Y [mm]

Amplitude [mm]

Time history waveform 0.3 0.2 0.1 0 −0.1 −0.2 −0.3

233 (0.15, 0.15)

2/4 open 0

0

1

2

3

4 5 Time [s]

6

7

X [mm]

8 (−0.15, −0.15)

Fig. 8.19 Time history and whirl orbit of unstable shaft vibration resulting from compressor impeller force

(2) Pump It is known that a pump impeller develops self-excited vibration because of the interference with the diffuser at the impeller outlet. Case 6 Vibration due to impeller destabilizing force in a vertical pump [144] Vibration of this case has a characteristic that the instability occurs in low-flow conditions. Figure 8.20 shows a vertical pump which experienced self-excited vibration. A flow-dependent unstable vibration of the first mode occurred as shown in Fig. 8.21a and the pump became inoperable due to the self-excited vibration at or below 80– 90% of the rated flow rate. The detailed mechanism of such phenomena has not been

910

Pump (upper)

1 700

Pump (lower)

1 155

2nd mode 35 Hz

Motor

1st mode 22 Hz

Discharge rate : 40 m3 / h

Discharge head : 160 m

Motor power : 45 kW

Rated speed : 3 540 rpm

Fig. 8.20 Vertical pump and eigen modes

Before impeller cut (2C / D = 1.0%)

100 80 60 40 20 0

0

20

40 60 80 100 120 140 Flow rate [%]

Vibration [ mp-p ] displacement

8 Case Studies of Self-excited Vibration of Rotor Stability … Vibration [ mp-p ] displacement

234

After impeller cut (2C / D = 4.6%)

100 80 60 40 20 0

0

20

(a) Before improvement

40 60 80 100 120 140 Flow rate [%]

(b) After improvement

Fig. 8.21 Occurrence of flow rate-dependent self-excited vibration 150

D

Fluid temperature [°C]

C

100

Unstable area

Stable area

50

0

Impeller Diffuser

Vibration extinction (at flow increase) Vibration generation (at flow decrease) 0

20

40

60

80

100

120

......... 2C = 1.0% D ......... 2C = 4.6% D

140

Flow rate Q [%]

Fig. 8.22 Stability limit condition of flow rate and operating fluid temperature

clarified yet and should be studied in the future. However, it is known empirically that the vibration reduces by widening the clearance, C, between the impeller outer circumference (diameter D) and the diffuser inner circumference. For this reason, the solution was to widen the diametrical clearance ratio 2C/D from 1% to 4.6%. As a result, the stability threshold dropped close to 60% of rated flow rate as shown in Figs. 8.21b and 8.22. The unstable area is shifted outside the operating range and operation became possible.

8.4 Self-excited Vibration Due to Internal Damping (Hysteresis Whip) There is a specific terminology in the field of rotor dynamics: internal damping and external damping. The word “internal” here means that the force is working in relative motion, and thus it can be described in a rotational coordinate system fixed in a rotor. For example, internal damping may be generated from frictional force when a shrink fit between a shaft and impeller becomes loose. It may also be generated when a

8.4 Self-excited Vibration Due to Internal Damping (Hysteresis Whip)

235

shaft made of a viscoelastic material is subjected to bending vibrations [220]. The word “external” means the force is working in the non-rotating parts, such as from ordinary bearings and dampers which can be described in a stationary coordinate system. The destabilizing effect from internal damping stated in R1_Sect. 11.1 can be summarized for a one-degree-of-freedom system that becomes unstable at the rotational speed above the stability limit, , in Eq. (8.8). This is higher than the critical speed, ωn , where ce is the external damping and cr is the internal damping:   (1 + ce /cr )ω n

(8.8)

In Eq. (8.8), when the internal damping, cr , is sufficiently larger than the external damping, ce , self-excited vibrations will occur immediately after the rotational speed passes the critical speed, ωn , by the reasoning that ce /cr ≈ 0. Case 7 Gear coupling Figure 8.23 shows self-excited vibrations experienced with a rotor having a gear coupling [221]. Figure 8.23a is a Campbell diagram that shows an increase in the rotational speed to the rated speed in a no-load condition when passing through a critical speed. In contrast, Fig. 8.23b is a Campbell diagram for gradually increased load to the rated load at the rated rotational speed. In Fig. 8.23a, during the increase of rotational speed, the vibration at the natural frequency (self-excited vibration) occurs immediately after passing a critical speed. In Fig. 8.23b, when load is increased, the larger the load, the higher the frequency of the self-excited vibration, and resonance occurs when the frequency coincides with the rated rotational speed. In addition, when the load increases further and the natural frequency exceeds the rotational speed, the self-excited vibration disappears. Such self-excited vibration occurs due to the internal friction damping when lubrication of the gear coupling tooth surfaces is insufficient. Furthermore, when tooth surfaces of a gear coupling are not crowned (crowning is a round-off treatment of

Frequency [Hz]

400

Rotational component (Forced vibration)

Gear (spline)

300 200

80 mp-p

100 0

Diaphragm

Natural frequency component (self-excited vibration) 3

18 6 9 12 15 0 3 Rotational speed × 10 [rpm] (a) At speed ascending

200 400 600 800 Load [kW] (b) At load increasing

(Before)

(After)

(c) Countermeasure by changing coupling

Fig. 8.23 Self-excited vibration generated by a gear coupling [VB030]

236

8 Case Studies of Self-excited Vibration of Rotor Stability …

the tooth surface in the face width direction so as not to constrain displacements in a shaft bending direction), the bending stiffness of the coupling will vary depending on the torque. Hence, the natural frequency is greatly affected by the load as shown. Figure 8.23c shows countermeasures where a gear coupling was replaced by a diaphragm coupling, which is free of internal damping and less dependent on torque. This eliminated the self-excited vibration completely. Case 8 Vertical pump Figure 8.24 shows a vertical pump, which had a long intermediate shaft of length 7.5 m [144]. When increasing the rotational speed beyond the critical speed, severe vibration occurred suddenly, and then the shaft warped like a bow and fractured. The cause of the vibration was the use of bearings that have almost no damping, together with the employment of laminated leaf spring couplings (multi-flexible disk couplings) having internal friction, even though the machine was designed to operate above the critical speed.

1 600

7 490

2 200

Countermeasures were implemented by adding a bearing supported by a rubber bush to the intermediate shaft that increased external damping, ce , and enabled adjustment of the natural frequency, ωn , by changing the support stiffness. These measures were effective in both adding base damping and increasing the critical speed ωn in Eq. (8.8). These achieved the prevention of unstable vibration and a highly safe operation. Thereafter, this machine has operated safely for many years.

Motor

Rigid coupling Intermediate shaft

Multi-flexing disk coupling

Adding rubber supported bearing

Pump

Original configuration

Improved content

Fig. 8.24 Self-excited vibration problem of vertical pump due to internal damping and remedial improvement [VB078]

8.5 Fluid-Containing Rotor

237

8.5 Fluid-Containing Rotor It is known that a flange and other hollow rotating shafts may cause severe vibrations when oil or other liquid becomes trapped internally. It seems at present that this phenomenon may be classified approximately into two modes: pure self-excited vibrations and forced vibrations caused by unstable fluid flow. The following section introduces them in Cases 9 and 10, respectively. Case 9 Pure self-excited vibration For a hollow rotating shaft that contains liquid as shown in Fig. 8.25 [222], self-excited vibration of a first natural frequency occurs after the shaft has passed the first critical speed, as shown in Fig. 8.26. According to the theory on an oscillating boundary layer by Kaneko [223], when the displacement of the hollow shaft is x  εx cos ωt, the fluid force acting on a container is eventually decomposed into two components: one proportional to the shaft acceleration and the other proportional to the shaft velocity, where the phase shifts by 90° and becomes a sine component. The force can be expressed as f x  ρπa 2 hεx ω2 ( f 1 cos ωt + f 2 sin ωt)  ρπa 2 h(− f 1 x¨ − f 2 ω x) ˙

(8.9)

630

Flexible coupling Slip ring Bearing

400 Hollow shaft

Pulley Belt Motor

 200

Bearing

Leaf spring

Leaf spring

Fluid

Fig. 8.25 Test apparatus for self-excited vibration of fluid contained hollow shaft

5

Amplitude 7.5 Frequency Unstable region 0 400

500 600 700 Rotational speed (a) Shaft vibration

5.0

800 900 [rpm]

1.0

Wave height [ mmp-p]

10.0

6

Wave 0.5 height Pressure

4 2

Unstable region

0 500

600

700

Rotational speed

800

0 900

[rpm]

(b) Wave height and pressure fluctuation

Fig. 8.26 Self-excited vibration occurring in fluid contained hollow shaft [VB136]

Pressure fluctuation [kPa p-p]

12.5

Frequency [Hz]

Amplitude [mmp-p]

10

238

8 Case Studies of Self-excited Vibration of Rotor Stability …

where ρ is the density of the fluid, a is the inner diameter of the hollow shaft, and h is the length of the hollow shaft. Assuming the mass, damping, and the spring constant of the hollow shaft without fluid contained to be m, c, and k, respectively, the equation of motion of the hollow shaft when it contains fluid can be expressed from Eq. (8.9) as. x¨ + 2ς ω x˙ + ω2 x  0

(8.10)

where ζ is the converted damping ratio and k ω02 ,  m + ρ πa 2 h f 1 1 + ρ π a 2 h f 1 /m   c + ρ πa 2 hω f 2 ρ π a2 h f2 ω ω ζ     ζ0 + 2m ω 0 ω0 2 m + ρπa 2 h f 1 k

ω2 

(8.11)

√ √ Note that ω0  k/m, ζ0  c/(2 mk) are the natural frequency and the damping ratio, respectively, of the hollow shaft that does not contain any liquid. Figure 8.27 shows an example of a converted damping ratio, ζ , with respect to the rotational speed, which is non-dimensionalized by the natural frequency of a hollow shaft without liquid. The initial damping without fluid, c, is assumed to be 0 and the rotor mass increment ratio at the fully contained condition to be ρ πa 2 h/m  0.1. The dimensionless level of contained liquid, R  b/a (radius at free surface b/container radius a), is 0.8 and the Ekman number at the critical   speed is E 0  1×10−3 (a ratio of viscous force to Coriolis force: E 0  ν/ a 2  , where ν is the dynamic coefficient of viscosity and  is the rotation angular velocity). From the figure, the additional damping from the liquid is effective when the non-dimensional rotational speed is below 1.0, while instability begins when it exceeds 1.0 and will be at its most severe when the speed is approximately 1.3 in this case. The phenomenon in which instability occurs when the rotational speed exceeds the first critical speed is similar to the self-excited vibration caused by internal damping described previously. There was a case in which only several cubic centimeters of oil entered the gap of a shaft-coupling flange and caused this type of self-excited vibration. This occurs

0.04

Converted damping ratio 

Fig. 8.27 Converted damping ratio of fluid containing hollow shaft

0.03 0.02 0.01 0 −0.01

 a2 h m = 0.1 R = 0.8 E0 = 1×10-3

0 00.2 .2 00.4 .4 00.6 .6 0.8 0.8 11.0 .0 1.2 1.2 1.4 1.6 1.8 2.0

−0.02 −0.03 −0.04

Non-dimensional rotational speed ( / 0 )

8.5 Fluid-Containing Rotor

239

because the liquid force develops in the whirling direction, and therefore the following countermeasures will be effective: (1) Prevention of oil from entering as the first priority. (2) A drain hole to discharge entered liquid by centrifugal force. (3) Installing a partition plate to prevent oil from moving in the circumferential direction. These measures, however, are not applicable to a centrifugal separator or other liquid-containing machines. Since self-excited vibrations of this type will not occur at a speed below the first critical speed, the following alternatives should be considered: (4) To design a rotor to operate below the critical speed. (5) If it is difficult to take the above measures, external damping (addition of a damper) can suppress the vibration amplitude to a non-problematic level by narrowing occurrence area of the self-excited vibration, though it is hard to completely eliminate vibrations. (6) There was an actual case where active magnetic bearings were used to suppress the self-excited vibrations [VB247]. Case 10 Phenomenon similar to forced vibration Another example is the case where the frequency is not equal to the natural frequency of the rotor, but proportional to the rotational speed as shown in Fig. 8.28 [225]. Since the liquid contained in a rotor has a free surface, a vibration at the following frequency (viewed from the rotational coordinate system), equivalent to the sloshing frequency in the field of centrifugal force, will occur:  f sk 

1  2π

  πh π r 2 χ tan h χ L L

(8.12)

 kL 2 , for the (s, k)-th combined mode number of vibration, b r  intermediate radius of liquid depth,   shaft rotation angular velocity, h  fluid depth, L  average liquid circumference length (2πr ), b  longitudinal length, s  number of nodes in the circumferential direction (0, 2, 4, …), k  number of nodes in the longitudinal direction (0, 1, 2, …). Equation (8.12) shows that the sloshing natural frequency in the centrifugal force field is proportional to the square root of the centrifugal acceleration, r 2 , so it is proportional to the rotation angular velocity, . Figure 8.28 is an example which shows that the vibrational frequency is proportional to the rotational speed. Figure 8.28a shows the amplitude and the frequency of the vibrations of the (2, 1)-th mode during deceleration, whose node indices are s  2 in the circumferential direction (i.e., the eccentric mode) and k  1 in the longitudinal direction. This indicates that the amplitude of the vibrations becomes larger by resonance when the frequency is close to the shaft natural frequency. Figure 8.28b shows the sloshing pressure fluctuation where χ 

s2 +



8 Case Studies of Self-excited Vibration of Rotor Stability … 4 100 3

10

Calculation Experiment

Natural frequency of shaft

5

0

(0,

1)

Frequency [Hz]

Deceleration −0.05Hz/s

1 0 15

Pressure fluctuation Calculation Experiment Shaft vibration Calculation Experiment

80

2

R sy otati nc on hro no us

Vibration [Hz] frequency

Shaft vibration [mmp-p] amplitude

240

60

(2, 0)

40

)

(2, 1

) (2, 0

20

(2, 10 20 Rotational speed

30 [rps]

40

( a ) Frequency and amplitude of shaft vibration

0

) (2, 1 ) (0, 1

0

10

1)

(0, 1)

20 30 40 Fluid depth [mm]

50

( b ) Frequency and fluid depth (at rotation 60Hz)

Fig. 8.28 Case of vibration frequency being proportional to rotational speed [VB369]

(symbol ◯) measured by a pressure sensor fixed inside the rotor. By subtracting this pressure fluctuation from the rotational speed (60 Hz), the obtained frequency, converted to the static coordinate system, agrees with the whirling frequency of the shaft (symbol ). This indicates that the sloshing natural frequency is estimated almost exactly by Eq. (8.12). This phenomenon can be regarded as a forced vibration similar to the case of a rotating stall caused by unstable internal fluid flow. This case sometimes becomes a problem for aircraft engine rotors consisting a hollow shaft. In addition, since the resonance will be a problem when the sloshing frequency and the natural frequency of the rotor are close to each other, it is necessary to install a partition plate to detune the sloshing natural frequency from the rotor speed.

8.6 Friction Whip Most of the cases described so far relate to self-excited vibrations where a forward whirl becomes unstable. The self-excited vibration presented in this section is a backward whirl. Figure 8.29 shows the mechanism in which a friction force occurs in the direction ω opposite to the rotation  when a rotor makes contact with a stator. That is, supposing the contact force from the stator as Fn , and the friction coefficient μ, the friction force, μ Fn , being proportional to contact force acts in the direction opposite to the rotation, . Thus, when μFn is greater than the damping force, Cr ω, which acts in the direction opposite to the whirl velocity, the backward whirl vibration mode will be unstable and the self-excited vibration, called (dry) friction whip [226–229], will occur.

8.6 Friction Whip

241  Fn

Fn

 Cr

r 

  Fn  Fn

Rotor angular speed Whirling angular speed Contact force in normal direction Contact force in tangential direction (friction force) Cr Damping force Occurrence condition  Fn > Cr of friction whip

Fig. 8.29 Mechanism of friction whip

Rotation angle c ( r+c)

r

A

2π

One circumferential length of rotor surface 2π r A

Whirl angle 2π r ( r+c) In case of rotor rolling on its entire circumference Whirl frequency : f Rotation frequency : f r

r f = c fr

Fig. 8.30 Whirling in the condition of rolling contact without slip (friction whirl)

In addition, another case was also reported in which a rotor came into rolling contact with a stator surface with slight slipping. Figure 8.30 shows the geometric relationship without slip. This vibration is called (dry) friction whirl and is also backward whirl. When the entire rotor circumference has rolled on the stator surface without slipping, the whirl angle in this case will be smaller than 2π , but it is understandable from the figure that the sum of the whirling angle and the rotation angle is 2π . Thus, there is a relationship between the whirling frequency, f , and the rotation frequency, fr :

242

8 Case Studies of Self-excited Vibration of Rotor Stability …

f 

r fr c

(8.13)

where r is the rotor radius at the contacting point, c is the average radial clearance between the rotor and the stator. This relation is also obtained from the condition of zero slipping velocity at the contact point, v  2π (r fr − c f )  0. Since usually the relationship of r c holds, the frequency of rolling contact whirl will be considerably higher than that of the rotational speed. Case 11 Friction whip Figure 8.31 shows a case for an experimental device, which was driven by an electric motor with a limited torque characteristic [230, 231]. The figure summarizes a phenomenon in terms of its whirling locus, vibration amplitude ˙ and the whirling velocity, ψ, ˙ under the in the x-direction, the rotational speed, φ, condition where the radial clearance between the stator is 0.5 mm. A sufficiently unbalanced rotor was accelerated, then the rotational speed passed the critical speed of 20 rps, and was maintained at 21 rps. During start-up, the amplitude gradually increased as the speed increased. Whirling along the entire circumference of a clearance circle (slightly triangular) of c 0.5 mm occurred in the interval of 0.6 s to 1.45 s. During this phenomenon, the rotational speed was 21 rps, the whirling frequency was also 21 rps, and the unbalance vibration of forward whirl synchronized with rotation, were observed. Then, during the interval from 1.45 s to 1.9 s, while the whirling amplitude remained almost the same, the locus became reciprocating in shape in the clearance

1mm

Rotating direction Clearance circle

1mm

·  [rps]

·  [rps]

x, y [mm]

(a) Whirling locus

Whirling direction

1.0 (b) 0.5 0 −0.5 −1.0 30 25 (c) 20 15 10 5 0 50 (d) 0

8 rps

−95 rps

−50 −100 0

Fig. 8.31 Case of friction whip

1

2 3 Time [s]

4

5

8.6 Friction Whip

243

˙ were small, the whirling circle. Although the changes of the rotational speed, φ, ˙ developed to −20 rps (the minus sign means the direction was oppovelocity, ψ, site to the rotation). This indicates that a self-excited vibration (friction whip) of a backward mode whose natural frequency was −20 Hz occurred. The reciprocating locus inside the clearance circle is generated because the backward whirl of the friction whip overlaps the forward whirl of unbalance vibration (whirling will become a straight line if these amplitudes are equal). ˙ gradually increased in The state after 2.5 s shows that the whirling velocity, ψ, the backward direction and reached −95 rps. The whirling direction at this time was purely opposite to the rotation and its amplitude was large (about double of the clearance circle) with accompanying deformations of the stator. In addition, the ˙ dropped to 8 rps, whose assumed cause was an insufficient torque rotational speed, φ, from the driving motor. Considering  thefact that the frequency ratio of the whirling ˙ φ˙   95/8 ≈ 11.9, and the rotor radius, r , is velocity to the rotational speed is ψ/ 6 mm, resulting in r/c ≈ 12, it is then concluded that the rotor is in rolling contact with the stator surface. The most important countermeasure against friction whip is to implement sufficient balancing so that the rotor will not make contact with the stator. It is also important to make the structure such that it does not allow any contact. In the cases of contact-type seals or stoppers that are naturally in contact, and touch-down bearings in active magnetic bearing, which operate for safety when the rotor dropped onto the stator due to power outage, the structure must be designed so that the friction force on the contact or sliding parts will be minimized. It is also important to design the damping ratio to be high by adding external damping.

8.7 Bently–Muszynska Model for Clearance Flow It is known that the frequency of oil whirl (half-speed whirl) due to the effect of fluid forces in oil film bearings is slightly less than half of the rotational speed, . For these self-excited vibrations, including oil whirl and oil whip induced by clearance flow in oil film bearings or seals, a dynamic model [232, 233] has been developed. This model considers fluid force that acts on a rotational coordinate system synchronized with whirling frequency of angular velocity λ  as seen in Fig. 8.32. The fluid force is represented by including the added mass, M f : −FR ≡ −( Fr + j Ft )  [ K 0 + ψ1 (|zr |)]zr + [ D + ψ2 ( |zr |)]˙zr + M f z¨r

(8.14)

 where zr ≡ xr + j yr , |zr | ≡ xr2 + yr2 , ψ1 ( |zr |), and ψ2 ( |zr |) are nonlinear terms of the spring stiffness, K 0 , and the damping, D, which are dependent on the amplitude and increase near the inner bearing surface as shown in Fig. 8.33b. If the journal radius is R, the average velocity of ambient fluid is assumed as λ R, where λ is called average circumferential

244

8 Case Studies of Self-excited Vibration of Rotor Stability …

Fig. 8.32 Rotational co-ordinate O-x r yr and fluid force

y

yr Ft

xr

Fr

Ω t O Ω

x

R

Fig. 8.33 Eccentricity ε and λ, ψ1 , ψ2

O = O (| zr |)

O

0.5

0

H

1.0

( a ) Average circumferential velocity ratio O

\1(| zr |) , \2(| zr |)

Average fluid velocity ΩR

0

1.0

H

( b ) Nonlinear function \1 , \2

velocity ratio whose value is close to 0.5 when the eccentricity is zero and approaches zero as the eccentricity increases, as shown in Fig. 8.33a. Equation (8.14) can be represented by the matrix form instead of complex notation: 

K 0 + ψ1 (|zr | ) 0 F xr − r  Ft yr 0 K 0 + ψ1 ( |zr |)

  Mf 0 x¨r 0 D + ψ2 (|zr | ) x˙r + + y˙r y¨r 0 Mf 0 D + ψ2 (|zr | )

(8.15)

Thus, the model is subject to the isotropic nonlinear spring and damping, and added mass in the λ  rotational coordinate system. This model is known as the Bently–Muszynska model, named after the advocates. The meaning of this model is more comprehensive if the dynamic characteristics are expressed in the stationary coordinate o - x y notation. A coordinate transformation from the λ  rotational coordinate system to the stationary coordinate system by letting Fz  FR e jλt , z  zr e jλ t will result in:   −Fz  − Fx + j Fy

   [K 0 + ψ1 (|z|)]z + [D + ψ2 (|z|)](˙z − jλz) + M f z¨ − 2 jλ˙z − λ2 2 z (8.16)

8.7 Bently–Muszynska Model for Clearance Flow

245

Fig. 8.34 1-mass system model

Ds Ds

K 2

: O

F

K 2

M

Fluid force Fx 0 DO : − F = − Fy = −DO : 0

{ }[

]{ } [ ]{ } x D 0 ẋ y + 0 D ẏ

Rewriting the above equation in x and y by using z ≡ x + j y, and omitting nonlinear terms for simplicity, we obtain:

 Dλ K 0 − M f λ2 2 x F − x  Fy −Dλ K 0 − M f λ2 2 y 

  x¨ Mf 0 D 2M f λ x˙ + + D −2M f λ 0 Mf y¨ y˙

(8.17)

The coordinate conversion shows the centrifugal forces in the diagonal term of the stiffness matrix and Coriolis force in the off-diagonal term of the damping matrix in a skew-symmetric manner, which are proportional to M f . Also, the product of the fluid damping, D, and the angular velocity, λ, appear in the off-diagonal terms of the stiffness matrix in a skew-symmetric manner. In particular, Dλ provides the cross-coupling stiffness, which may cause self-excited vibrations and the effect increases as the rotational speed, , rises. Firstly, let us consider the effects of the cross-coupling stiffness in the 1-mass system model shown in Fig. 8.34. The disk center displacements, x and y, are treated as z  x + j y. The cross-coupling force, ignoring nonlinearity, is D λ(y − j x)  − j Dλz in the complex notation. By denoting the initial damping as Ds , the external force as P( t), the equation of motion obtained is M z¨ + (Ds + D)˙z + K z − j Dλz  P(t)

(8.18)

The characteristic equation may be obtained by substituting z  Aest and by setting it to be zero: Ms 2 + (Ds + D)s + K − j Dλ  0

(8.19)

Since this is a quadratic equation with complex coefficients, the characteristic roots can be calculated as: Ds + D ± s− 2M



Ds + D 2M

2 −

K − j Dλ M

(8.20)

246

8 Case Studies of Self-excited Vibration of Rotor Stability …

Let us consider the approximate solution in the case of Ds  0. If the fluid force, D, is small, assuming D 2 ≈ 0 and Dλ/K > M 2 in amplitude. From Fig. 8.38, it is found that the real part of the eigenvalue Z1 becomes positive  (unstable) when the rotational speed is above K 1 M1 /λ  4360 rpm and the natural frequency is just about λ   0.42 × 4360/60 × 2π ≈ 190 rad/s. This means that oil whirl has occurred. 400 300 200 1Ω 100 0.42Ω

0 −100

Unstable

ZI1 ZI2 ZI3 ZI4

−200 −300 −400 0

( a ) Real part

2 000 4 000 6 000 Rotational speed [rpm]

8 000

( b ) Imaginary part

M1 = 0.68 kg, M2 = 0.013 kg, Mf = 0.005 kg, Ds = 2.63 N·s/m , D = 2150 N·s/m , K0 = 0 N/m, K1 = 2.5 × 104 N/m, K2 = 2.5 × 104 N/m, λ = 0.42, here, ZR4 = – 1.18 × 104 rad/s is out of plot range

30 20

Unstable

10 0 −10

ZR1 ZR2 ZR3

−20 −30 −40 −50 0

2 000 4 000 6 000 Rotational speed [rpm] ( a ) Real part

8 000

Damped natural frequency [rad/s]

Real part of eigenvalues [rad/s]

Fig. 8.38 Complex eigenvalues of 2-mass system model and oil whirl–Case 1 400

0.42Ω

300 200 1Ω 100 0

Unstable

−100

ZI1 ZI2 ZI3 ZI4

−200 −300 −400 0

2 000 4 000 6 000 Rotational speed [rpm]

8 000

( b ) Imaginary part

M1 = 0.68 kg, M2 = 0.013 kg, Mf = 0.005 kg, Ds = 2.63 N·s/m , D = 2150 N·s/m , K0 = 0 N/m, K1 = 1 × 104 N/m, K2 = 4 × 104 N/m, λ = 0.42, here, ZR4 = – 1.19 × 105 rad/s is out of plot range

Fig. 8.39 Complex eigenvalues of 2-mass system model and oil whirl–Case 2

250

8 Case Studies of Self-excited Vibration of Rotor Stability …

 The figure also indicates that the natural frequency approaches asymptotically to (K 1 + K 2 )/M1  271 rad/s (natural frequency for simple support condition), and it is recognized that the phenomenon is shifting from oil whirl to oil whip. Furthermore, in order to confirm the rotational speed at the onset of the oil whirl indicated by Eq. (8.29), the springs are changed in the ratio of K 1 to K 2 from 1:1 (Fig. 8.38) to 1:4 (Fig. 8.39) keeping the total unchanged. This condition means that the location of mass point M 1 becomes closer to M 2 because the total stiffness K 1 + K 2 is unchanged. Figure 8.39  shows the results that the rotational speed at the onset of the oil whirl results in K 1 /M1 /λ  2757 rpm and the oil whirl begins at a lower rotational speed than in the case of Fig. 8.38. These results have also been confirmed experimentally. Figure 8.40 [234] shows an example of the experiment data where the location of mass M1 is moved closer to M2 . Next, the cessation phenomenon of the oil whip can be also calculated by this model. As Fig. 8.41 shows that the model becomes stable at a certain rotational speed, ces , when Ds > 0, whereas it remains unstable when Ds > 0. This rotational speed at cessation of the oil whip is calculated as follows. By substituting the characteristic  root of oil whip s  jωz  j (K 1 + K 2 )/M1 into Eq. (8.26) and assuming M f  M2  0:     Ds jωz −K 2 2    −K 2 j D(ωz − λ) + K 0 + K 2   −Ds Dωz (ωz − λ) − K 2 + j Ds ωz (K 0 + K 2 )  0

(8.30)

The imaginary part is comparatively small, and the real part is set to be zero:   K 22 ωz 1+  0 → ces  λ Ds Dωz2   K 22 M 1 K1 + K2 1+  λ M Ds D(K 1 + K 2 )

(8.31)

Whip

3

W

hi

rl

3

Rotational speed  × 10 [rpm]

Whip

Rotational speed  × 10 [rpm]

Ds Dωz (ωz − λ) +

K 22

Frequency

3

× 10 [1/min]

rl

hi

W

Frequency

3

× 10 [1/min]

Fig. 8.40 Oil whirl onset frequency with respect to changing shaft stiffness ratio of K 1 and K 2 (Muszynska [234], experiment)

Ds = 0 N·s/m Ds = 2.63 N·s/m Ds = 13.15 N·s/m

20

(=10815)  ces

10 0

251

1400

30

0

 st

10 000

(=29600)  ces

20 000

−10

30 000

Damped natural freq. [ rad/s ]

Real part of eigenvalue[ rad/s ]

8.7 Bently–Muszynska Model for Clearance Flow

800

( a ) Real part

0.42

600 400 200 0

−200

rotational speed [ rpm ]

Ds = 0 N·s/m Ds = 2.63 N·s/m Ds = 13.15 N·s/m

1200

1000

0

10 000

20 000

30 000

rotational speed [ rpm ] ( b ) Imaginary part

M1 = 0.68 kg, M2 = 0.013 kg, Mf = 0.005 kg, D = 2 150 N·s/m, K0 = 0 N/m, K1 = 1×104 N/m, K2 = 4×104 N/m, = 0.42, same as Fig. 8·39 except for D s, showing ZR1,ZI1,ZI2 only

Fig. 8.41 Instablility onset angular speed st and cessation angular speed ces

The results are shown in Fig. 8.41. The numerical value of ces in the figure where the cessation of the unstable vibration occurs is calculated from Eq. (8.31). It shows that the larger the value of Ds , the lower the cessation speed ces becomes. Other case studies have also been reported, and the complicated phenomena can be explained easily by employing the Bently–Muszynska model. Readers are encouraged to refer both authors’ books, [B37, B41], for further details. Later, we apply their idea to more general rotor systems by combining with the mode synthesis modeling technique. The stability limit is then summarized, and this allows us to predict the unstable onset speed and resultant unstable vibration frequency. The details are given in Sect. 11.2 entitled “Simplified prediction of the stability limit for an oil film bearing supported rotor.”

8.8 Stabilization with Squeeze Film Damper In order to prevent self-excited vibrations, it is effective to add a damper to increase the base damping. In the case of oil film bearings, it is difficult to increase the damping only, the design usually accompanies changes of the spring coefficients. Also rolling bearings have almost no damping and therefore need to be equipped with a damping mechanism. A typical solution for these issues is to add a squeeze film damper to the circumference of the bearing. As Fig. 8.42 shows, a squeeze film damper is installed outside the bearing and can add a necessary amount of damping independently, without changing the tribological characteristics of the bearing. This section introduces some practical applications of a squeeze film damper and offers the points to note in design.

252 Fig. 8.42 Schematic view of a squeeze film damper (in case with centering spring)

8 Case Studies of Self-excited Vibration of Rotor Stability … Oil feeding

Centering spring

Squeeze film Ball bearing

Shaft

End seal Circumferential oil feed groove

8.8.1 Differences with and Without a Centering Spring A squeeze film damper may be said to be an oil film bearing whose journal does not rotate. Since floating effects due to rotation do not exist, a static equilibrium position will correspond with the eccentricity of 1.0 (the journal is in contact with the damper wall) unless a centering spring is employed. There are two types of dampers in the structure: dampers with and without a centering spring (see Fig. 8.42). A centering spring is used to equalize the oil film thickness of the damper circumferentially and to make the eccentricity ratio of the journal close to zero. As the dynamic characteristics of the damper oil film are a function of the eccentricity ratio, and when the ratio is approximately 1.0, nonlinearity is extremely large as will be described later. If a damper is used under the condition of low eccentricity ratio by using centering spring, nonlinearity is low and a design that uses linearized characteristics about zero eccentricity becomes possible. Although models without a centering spring need an anti-rolling pin to the damper journal, they have a simple structure and are free from fatigue problems. Also, no installation space is required for a centering spring. However, when the vibration amplitude is small, the eccentricity of the journal of this model is close to 1.0 and nonlinearity of dynamic characteristics is high, which makes the optimum design more difficult.

8.8.2 Feeding/Draining of Oil and End Seals Oil feed holes or annular oil feed grooves are frequently used for supplying oil into a damper (Fig. 8.43). Feed holes are used to supply oil into the damper land directly, but it is easy to deduce that a single hole is difficult to provide sufficient oil when taking into account the oil leakage from the damper end. Therefore, a number of holes are always implemented. A circumferential oil feed groove may be used to feed oil sufficiently and uniformly over the entire circumference of the damper. Though this system is frequently employed, care must be taken because the groove

8.8 Stabilization with Squeeze Film Damper

253

makes the damper effective width narrower and has great influence on the dynamic characteristics of the damper. As for oil draining, many dampers adopt a leakage method from the damper end section. Some dampers employ end seals while others do not. These seals aim to reduce the necessary amount of oil supply to the damper and to increase the pressure inside the damper, in order to prevent oil film rupture and performance degradation due to a contamination of external air when the journal vibration amplitude is large. Figure 8.44 shows regular end seals including end face seals, piston rings, and Orings. Since oil does not drain through O-ring seals, implementation of an oil drain hole will be necessary. This drain hole should be designed to have an adequate diameter in order to maintain the pressure inside the damper. It should also be positioned to allow air that entered during assembly to escape easily without remaining in the damper (e.g., on an upper side).

8.8.3 Dynamic Characteristics of a Simple Model of a Squeeze Film Damper The dynamic characteristics of squeeze film dampers have been studied so far using the Reynolds equation (classical lubrication theory) assuming fluid flow in a thin layer between two cylinders. According to the short bearing width approximation theory, the Reynolds equation is   ∂h ∂ h3 ∂ p  12 ∂z μ ∂z ∂t

Fig. 8.43 Different types of oil feeding methods

(8.32)

Oil feed

(a) Feed hole type

Oil feed

(b) Circumferential feed groove

Fig. 8.44 Types of damper end seal End face seal

Piston ring

O ring

254

8 Case Studies of Self-excited Vibration of Rotor Stability …

Fig. 8.45 Coordinate system in squeeze film damper

y



R z

c

F



(x, y) F

x

L 

where p is the oil film pressure, μ is the oil viscosity coefficient, z is the coordinate in the damper axial direction, and h is the oil film thickness, which has the following relationship from Fig. 8.45:  2 c ≈ c(1 + ε cos ψ) (8.33) h  c − x cos θ − y sin θ + o R where c is the average radial clearance, and ε and φ are the eccentricity ratio and the attitude angle of the damper journal, respectively. Integrating Eq. (8.32) in the z direction yields  6μc  p  3 ε˙ cos ψ + εφ˙ sin ψ z(z − L) + p0 (8.34) h where the integration constant p0 is the oil supply pressure to the damper. The integration of Eq. (8.34) with respect to z and ψ results in the following oil film reaction forces for the eccentricity direction and the direction perpendicular to the eccentricity:   μR L 3  μR L 3  Fε (t)  − 2 ε˙ A2 + εφ˙ A1 , Fφ (t)  − 2 ε˙ A1 + εφ˙ A3 (8.35) c c where ψ2 A1  ψ1

sin ψ cos ψ (1 + ε cos ψ)

ψ2 dψ, 3

A2  ψ1

cos2 ψ (1 + ε cos ψ)

ψ2 dψ, 3

A3  ψ1

sin2 ψ (1 + ε cos ψ)3

dψ (8.36)

The pressure integration range [ψ1 , ψ2 ] of Eq. (8.36) depends on the oil film existence area. It is known that such an area depends on the oil supply pressure and amplitude of the journal vibration, etc. Generally, the oil film existence area is in a range from half of the circumference (π ) to the entire circumference (2π ). The section below describes the dynamic characteristics of these two extreme approximate simple condition models, followed by the general expression for damper models. (1) 2π film model When the oil supply pressure, p0 , is so high that the oil film pressure in Eq. (8.34) will not drop below the saturated vapor pressure (or atmospheric pressure when

8.8 Stabilization with Squeeze Film Damper

255

contaminated with air), it may be assumed that there exists no oil film rupture for the entire circumference. In this case, we obtain A 1  0 from the integration range ψ1  0 and ψ2  2π , and Eq. (8.35) becomes Fε (t)  −

π μR L 3 1 + 2ε2  5/2 ε˙ , c2 1 − ε2

Fφ (t)  −

1 π μR L 3  3/2 εφ˙ c2 1 − ε2

(8.37)

As can be seen from the denominator, the viscous damping coefficients are infinite at ε  1, the nonlinearity is found to be extremely large near this condition. (2) π film model When the oil supply pressure, p0 , is sufficiently low compared to the first term in Eq. (8.34), the oil film cavitates or ingests air bubbles, thereby generating no pressure in an area of the ruptured film over approximately half of the damper circumference. The pressure develops on the remaining half circumference where the oil film exists. In this case, the region of positive pressure will change with time according to the journal velocity in Eq. (8.34), and the following inequality will define the range for pressure integration: ε˙ cos ψ + εφ˙ sin ψ ≤ 0

(8.38)

Therefore, defining ψ1 as tan ψ1  −

ε˙ εφ˙

(8.39)

ψ1 must be selected so that the inequality ε˙ cos ψ3 + εφ˙ sin ψ3 ≤ 0 is satisfied, where ψ3  ψ1 + π/2 Thus, Eq. (8.36) can be integrated by setting ψ2  ψ1 +π , the integration constants become ψ2  2 α 1 + 2ε cos ψ sin α + 2ε cos α 2  2ε2 (1 + ε cos ψ)2 ψ1 2(1 − ε)2 α1  α2  2  1 1 A2   5/2 2ε + 1 α − 4ε sin α + sin 2α 2 2 1 − ε2 α1  α2 1 1 A3   3/2 α − sin α 2 2 2 1−ε α1 

A1 

where α is the value defined by the Sommerfeld’s substitution [235]:

(8.40)

256

8 Case Studies of Self-excited Vibration of Rotor Stability …

1 + ε cos ψ 

1 − ε2 1 − ε cos α

(8.41)

In order to make α as a single-valued function of ψ in the transformation, the value is defined so that α1 < α2 is satisfied when ψ1 < ψ2 . As a special case, when the journal whirls concentrically about the damper center ˙ the integration range will be [π, 2π ] according to Eq. (8.38) and, (˙ε √ ωb ωθ 2 r

(9.34)

Example 9.6 Stability of   ωθ −ωb at intersection C: Assume that the vibration solution in Eq. (9.30) is for backward motion, and evaluate stability in a similar manner as described above.

Answer Always stable (It is the same as that at intersection A). Example 9.7 Stability of   ωθ + ωb at intersection D: Assume the vibration solution in Eq. (9.30) is for the backward motion and evaluate stability in a similar manner as described above.

Answer It is the same as the intersection B. The stable condition is the same as Eq. (9.34). Now, we can conclude the discussions in this section as indicated in Fig. 9.31:

294

9 Torsional Vibration and Related Coupled Vibration

  |ωb −ωθ | at intersections A and C → Always stable.   ωb + ωθ at intersections B and D → Eq. (9.34) for stable detuning (σ ) range. (3) Approximate analysis of stabilization with external damping Here, the stability is analyzed about the condition of the rotational speed,   ωb + ωθ , indicated by intersections B and D in Fig. 9.31, with the inclusion of a damping ratios, ζ b , for bending vibration and, ζ θ , for torsional vibration. Adding damping ratios to Eq. (9.15), we obtain the following equations of motion in the inertial coordinate system:  z¨ + 2ζb ωb z˙ + ωb2 z  ε ( + θ˙ )2 − j θ¨ e jφ θ¨ + 2ζθ ωθ θ˙ + ωθ2 θ  εωb2 /r 2 Im(ze− jφ )

(9.35)

In the rotating coordinate system, the equations above are rewritten and expressed in a similar manner to Eq. (9.17) as follows:  z¨r + 2 j˙zr − 2 zr + 2ζb ωb (˙zr + jzr ) + ωb2 zr  ε ( + θ˙ )2 − j θ¨ e jθ e jδ θ¨ + 2ζθ ωθ θ˙ + ωθ2 θ  εωb2 /r 2 Im (zr e− jθ e− jδ )

(9.36)

As well as process from Eq. (9.19) to the derivation of Eq. (9.20), the first-order approximation for small ε yields:  z¨r + 2 j˙zr − 2 zr + 2ζb ωb (˙zr + jzr ) + ωb2 zr  ε j2 θ + 2θ˙ − j θ¨ e jδ θ¨z + 2ζθ ωθ θ˙z + ωθ2 θz  εωb2 /r 2 zr e− jδ θ  Im(θz ) (9.37) Hence, dealing with stabilization at intersection B by adding external damping, we introduce the solution assumed by Eq. (9.30) in (9.37) and the following approximation in relation to small damping ratios, ζ b and ζ θ , is obtained: 2ζb ωb (˙zr + jzr ) ≈ 2ζb ωb (− jωθ + j)Ae− jωθ t ≈ 2 jζb ωb2 Ae− jωθ t 2ζb ωb θ˙ ≈ 2ζb ωb (− jωθ )Be− jωθ t ≈ −2 jζb ωb2 Be− jωθ t These mathematical developments finally provide the following characteristic equation, modified from Eq. (9.32):   ε   2 jω (s + jσ ) + 2 jζ ω2 − ωb2 e jδ   b b b 0  2 (9.38) ε 2 − jδ   − 2 ωb e −2 jωθ s − 2 jζθ ωθ2   r

9.4 Bending–Torsional Coupled Vibration

295

When stability conditions are derived under the worst condition (detuning σ  0), the quantity of external damping required for stabilization is determined. Consequently, stability conditions follow from the following characteristic equations at σ  0:   ε   2 jω (s + ζ ω ) − ωb2 e jδ   b b b 0  2 (9.39) ε 2 − jδ   −2 jωθ (s + ζθ ωθ )   − 2 ωb e r   1  ε 2 ωb3 ∴ (s + ζb ωb )(s + ζθ ωθ ) − 0 (9.40) 8 r ωθ Since Eq. (9.40) is a second-order equation for s, stability requires positive coefficients in the first-order and zero-order (constant) terms. The external damping required for stabilization is    1  ε  ωb (9.41) ζb ζθ > √ ωθ 8 r Consequently, the following comments for the improvement of stability apply: √ (a) Increase damping with respect to the geometrical mean, ζb ζθ , instead of the arithmetic mean, (ζb + ζθ )/2. (b) Balance the rotor well to reduce the eccentricity, ε. (c) The stability is independent of unbalance phase, δ. (4) Calculation of complex eigenvalues The stability condition obtained by the detuning of Eq. (9.34) and the stabilization of Eq. (9.41) are based upon approximate analysis. Hence, we verify the accuracy of this approximation in comparison with the exact analysis by assuming specific numerical parameters and by solving for complex eigenvalues. Substitute the complex displacement, zr  xr + j yr , of the rotating coordinate system into Eq. (9.37), and set the unbalance phase, δ  0. We thus obtain the following linear differential equations concerning real displacements, x r and yr : x¨r − 2 y˙r − 2 xr + ωb2 xr + 2ζb ωb x˙r − 2ζb ωb yr  ε(2θ˙ ) y¨r + 2x˙r − 2 yr + ωb2 yr + 2ζb ωb y˙r + 2ζb ωb xr  ε(2 θ − θ¨ ) θ¨ + 2ζθ ωθ θ˙ + ωθ2 θ  εωb2 /r 2 yr

(9.42)

Letting s denote the characteristic root, it is given by:   2   s + 2ζb ωb s + ωb2 − 2 −2s − 2ζb ωb  −2εs   2 2 2 2 2  s + 2ζb ωb s + ωb −  −ε( − s )   0 2s + 2ζb ωb    s 2 + 2ζθ ωθ s + ωθ2  0 −εωb2 /r 2

(9.43)

The exact complex eigenvalues are given by determining the characteristic root, s.

296

9 Torsional Vibration and Related Coupled Vibration

Referring to Fig. 9.31, the following types of system parameters are prepared with a coupling effect ε/r  1/1000. The rotating speed, c , of the center of a possible instability zone and stability detuning factor h  σ /ωb are as follows: ➀ About intersection B: ωb  1, ωθ  0.4, c  1.4 According to Eq. (9.34), instability region predicted → h ≡ |σ |/ωb < 1.1/1000

(9.44)

➁ About intersection D: ωb  1, ωθ  1.4, c  2.4 Instability region predicted → h ≡ |σ |/ωb < 0.6/1000

(9.45)

Using Eq. (9.43), we calculate the complex eigenvalues of the undamped system with external damping ζ b  ζ θ  0. Figure 9.32 shows the real and imaginary parts of the complex eigenvalues in relation to the rotational speed, , near the intersection. Outside the shaded area of this figure, detuning is sufficient and the solution is a pure imaginary root (i.e., practically stable). Inside the shaded area of this figure, complex roots exist, having positive real part. This means that the region is not stable. The boundary of the shaded area indicates the detuning of a limit of stability. The widths of the unstable speed regions are also shown in the figure: half-width  1.1 × 10−3 for (a) and half-width  0.6 × 10−3 for (b) in Fig. 9.32, which agree well with predictions of the stability detuning in Eq. (9.34). Re

Re

unstable region

1 ×10−3 2×1.1×10−3

2×0.6×10

−3

×10−3

2

unstable Ω

0

Ω

0 stable

−1 1

1.398

1.399

−2 Im

0 1.4

−1

2

Forward natural frequency

1

Ω

1.401

2.398

1.402

Backward natural frequency

Ω ≈ 1.4 B around ω ( ω b = 1, ω θ = 0.4 )

(a) ω b > ω θ

Fig. 9.32 Stable region by frequency detuning

Im

2.399

0 2.4

Ω

2.401

2.402

−1 −2 Ω ≈ 2.4 D around ω ( ω b = 1, ω θ = 1.4 )

(b) ω b < ω θ

9.4 Bending–Torsional Coupled Vibration

297

(5) Stability seen in time history waveforms Using the equation of motion, Eq. (9.35), in the inertial coordinate system, we may simulate numerically free vibration in a time history response analysis through variables: a. Shaft vibration waveform x(t) as viewed from the inertial coordinate system and b. Shaft vibration waveforms x r (t), yr (t) as viewed from the rotating coordinate system. Calculation parameters: intersection B, ωb  1, ωθ  0.4, c  1.4 Required damping for stabilization → ζa > 0.88/1000 (See Figs. 9.31a and 9.32a.) Firstly, Fig. 9.33a shows unbalance excited bending vibrations of an undamped system, assuming the operation at a constant rotational speed of   ωb + ωθ  1 + 0.4  1.4 [rad/s] and no coupling with any torsional vibration. Waveform x(t) shows unbalance vibration having a constant amplitude. When viewed in the rotating coordinate system, waveforms x r (t) and yr (t) are constant. Next, Fig. 9.33b shows the results by assuming that the rotor is rotating at the same constant rotational speed, but is coupled with torsional vibration (ε/r  1/1000). From this figure, we can see that the amplitude of unbalanced vibration waveform, x(t), fluctuates and the system becomes unstable. When viewed in the rotating coordinate system, the amplitudes of waveforms x r (t) and yr (t) increase slowly due to the entry of backward whirl components at angular frequency ωθ −0.4 rad/s (period  15.7 s). Then, we recognize the occurrence of unstable backward whirl motion. From Fig. 9.33c, we can see that waveforms x r (t) and yr (t)√begin to ζb ζ θ  be attenuated and stabilized when external damping ζa  (a) Uncoupled

(b) Coupled / r = 1 / 1 000

x (t)

3

(c) Stabilization

x (t)

3

3

0

0

0

−3

−3

−3

560 570 580 590 600 yr

1 0 −1 −2 −3 1 0 −1 −2 −3

xr 560 570 580 590 600 yr xr 0

200

400

Unstable around the intersection B

600 [s]

960 970 980 990 1 000 yr

1 0 −1 −2 −3 1 0 −1 −2 −3

xr 960 970 980 990 1 000 yr xr 0

ζb =ζ t = 0

x (t)

500

1 000 [s]

960 970 980 990 1 000

1 0 −1 −2 −3 1 0 −1 −2 −3

yr xr 960 970 980 990 1 000 yr xr 0

500

ζ b = 0.01, ζ t = 0.000 2 required for stabilization

Fig. 9.33 Time history response and stability simulation (ωb  1, ωθ  0.4,   0.4)

1 000 [s]

298

9 Torsional Vibration and Related Coupled Vibration

1.4/1000 (ζb  0.01, ζθ  0.0002) is applied to the system. In this state, waveform x(t) returns to vibration of the unbalance excited component only. Case Study: Bending–Torsional Coupled Resonance [264] As shown in Fig. 9.34a, the rotor consists of a rotating shaft having a large disk at each end. A ball bearing rigidly supports the shaft at the right-hand end; an active magnetic bearing (AMB) “softly” supports it at the left-hand end. Through the AMB, the spring constant is controllable and the superposition of excitation is also possible. Figure (a) also shows the bending and torsional eigen modes corresponding to natural frequencies, f b  18 Hz and f θ  55.3 Hz, respectively. In this state, we conducted an experiment to check whether the coupled resonance of a torsional vibration occurs by applying vertical bending excitation to the journal via the AMB at a specific rotating speed. For torsional vibration measurement, a 360 tooth ring was installed on the outer circumference of the right-hand disk for generating a frequency modulated pulse Photo-interrupter output Disk

10 V

Active magnetic bearing φφ30 Disk

−10 1 169 mm Bending eigen mode φ b

sparse dense sparse dense 0

40 ms

Frequency modulation F/V converter 2 Demodulated wave V

18.0 Hz

high 55.3 Hz

−2

Torsional eigen mode φθ (a) Rotor system

18.0 Hz 37.25 Hz

MAG V

0.1

55.3 Hz

18.0 Hz Induced torsional vibration 55.3 Hz

0 Frequency [Hz]

TIME

40 ms

100

(c) fr = fθq − fb = around 37.3 Hz

Induced torsional vibration 18.03 Hz

73.0 Hz 55.3 Hz

0

100

37.3 Hz

0

0.1 MAG V

Induced torsional vibration

0

high low

(b) Torsional vibration measurement

100

0.1 18.0 Hz

MAG V

MAG V

0.1

low

Induced torsional vibration 55.3 Hz

0

73.3 Hz

Frequency [Hz]

100

(d) fr = fθq + fb = around 73.3 Hz

Fig. 9.34 Case studies of bending and torsional coupled vibrations (input: bending excitation) [VB765]

Input: Bending excitation ν = 18 Hz 0.08 37.19 rps 73.18 rps 0.04 0.00 0

20 40 60 Rotational speed [rps]

80

(a) Bending excitation to torsional response

299 Bending amplitude [μm]

Torsional amplitude [°]

9.4 Bending–Torsional Coupled Vibration

Input: Torque (Torsional) excitation ν = 55.17 Hz 0.4 73.17 rps 37.17 rps 0.2 0.0

40 60 Rotational speed [rps]

80

(b) Torsional excitation to bending respone

Fig. 9.35 Amplitude curves of bending and torsional coupled vibrations

wave, which is input to a F/V converter. The F/V converter outputs change in rotational speed by superposing the wave of torsional vibration velocity, as shown in figure (b). Experimental results are shown in figure (c) near the rotational speed f r  f θ− f b  37.3 Hz. We see the upper graph taken at a speed a little off-resonance speed and the lower graph at just on the coupled resonance speed. The bending resonance component (18 Hz) appears in both the upper and lower graphs, but the torsional resonance component (55.3 Hz) only appears in the lower graph. This lower graph shows the bending (18 Hz)–torsional (55.3 Hz)-coupled resonance at the specified speed 37.3 Hz. As shown in figure (d) near the rotational speed f r  f θ + f b  73.3 Hz, we see another example, similar to (c), where the upper and lower graphs correspond to off-resonance and coupled on-resonance, respectively. The lower graphs verify the bending (18 Hz)–torsional (55.3 Hz)-coupled resonance, induced by bending excitation through the AMB. Figure 9.35a shows the amplitudes of torsional vibration measured while applying bending excitation (ν  18 Hz) to the rotor end via the AMB so as to maintain the state of the bending resonance condition, while sweeping through the rotational speed. Clearly, the coupled resonances are also induced on the torsional response at the specified speeds near f r  f θ −f b  37 Hz and f r  f θ + f b  73 Hz. This resulting torsional resonance is induced by the coupling effect with bending excitation. Figure 9.35b shows the result of observing bending vibration when applying a torque excitation continuously at the torsional resonance frequency (ν  55.17 Hz) through a motor/inverter. We can confirm the resulting bending resonance induced by the coupling effect with torsional torque excitation at the specified rotational speed, mentioned above. Note: There are many kinds of coupled rotor lateral and torsional vibrations generated by a variety of mechanisms, for example, unbalance, geared couplings, electric motors, and electric inverters. The following references provide interesting reading: [265] (Ishida, 2009), [266] (Pennacchi, 2010), [267] (Carden, 2014) and [268] (Francis, 2015).

Chapter 10

Signal Processing for Rotor Vibration Diagnosis

Abstract In the vibration diagnosis of rotating machines, the first attention is the evaluation of the unbalance vibration, i.e., the rotational synchronous point of vibration, noted as “1X” or “1N” vibration. A vector monitor (R1_Fig. 5.8) is a measuring instrument including a tracking filter that extracts only the 1X vibration component from the actual vibration waveform. In this chapter, the principle and the application of this vector monitor are explained. In the following, attention is given to the transformation from the waveforms in the time domain to amplitude (spectrum) in the frequency domain by using a FFT analyzer. We learn about the theory of signal processing related to FFT analysis, which is capable of providing numerous displays for easy understanding and effective diagnosis for vibration troubleshooting. Keywords Signal processing · PLL (Phase-locked loop) · Vector monitor · MATLAB · Unbalance forward/backward orbit · CFT · DFT (FFT) · Aliasing error · Fourier transform · Frequency response analysis (FRA) · Zooming · Full spectrum

10.1 Vector Monitor (Balance Analyzer) 10.1.1 What is a Vector Monitor? As shown in Fig. 10.1, the rotor vibration is measured by a displacement transducer together with the rotational pulse signals from a rotating machine (a). The measured vibration waveform (b) contains a 1X vibration component synchronized with the rotation and, additionally, various other frequency components such as ultraharmonics (termed an nX vibration) and low-frequency components. Extracting only the rotation-synchronous component from the waveforms (b) by a 1X filter, we have a waveform that resembles that indicated by the broken line in figure (c). An analyzing instrument that extracts the vibration amplitude, a, and phase difference, φ, between the vibration peak and the pulse signal is called a “vector monitor” or “balance analyzer.” It is one of the most convenient instruments

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_10

301

302

10 Signal Processing for Rotor Vibration Diagnosis

(a)

y3 Displacement transducer

y2 x2

Rotational pulse y1

(b) x3

(c) x1

Rotor Bearings

1X Filter

2a

Ω

ϕ Rotational pulse

Fig. 10.1 Vibration measurement in a rotor system

for rotor balancing work. The principle of function implemented in the vector monitor is explained in the following sections.

10.1.2 PLL (Phase-Locked Loop) A PLL is a common circuit that outputs a periodic function that is synchronous with the input pulse signal. In rotor dynamics, a PLL produces a sine wave or cosine wave that is synchronous with a rotational pulse signal. Understanding the feedback control theory for synchronization embedded in the PLL circuit requires an extensive knowledge of electronics [269, 270], which is beyond the scope of this book, so here we set forth the input–output relations using several software systems. As shown in Fig. 10.2, when a rotation-synchronous pulse signal ➀ is input, a synchronized sawtooth waveform ➁ with amplitude ±2 V is created on the output side via signal processing software for the PLL. The peak–peak values of the sawtooth wave correspond to the angle θ of ±180°. By referring to the sine and cosine function tables, a trigonometric function is obtained: π pulse → sawtooth wave ν(t) → conversion to angle θ (t)  ν(t)  2 cos[θ (t)] → cos t → sin / cos function tables → trigonometric functions sin[θ (t)] → sin t (10.1) As a result, a cosine wave ➂ and a sine wave ➃ synchronized with the rotational pulse are generated as the output of the PLL, as shown in the figure. Examples of the software for executing Eq. (10.1) are given by MATLAB/SIMULINK shown in Appendix Fig. D.1a, b and with the C language in Appendix Fig. D.2.

10.1 Vector Monitor (Balance Analyzer)

1

Rotational pulse

303

4 0

= + 180°

2 2

Sawtooth shock pulse

0 -2 1

3

Cosine

= – 180°

0 -1 1

4

Sine

0 t

-1 0

0.01

0.02

Fig. 10.2 PLL function and cos/sin generator

10.1.3 Bode Diagram and Nyquist Diagram Assuming that cosine and sine waves synchronized with the pulse have been prepared, let us use them to derive the 1X filter, also termed a tracking filter with the rotational speed . The waveform x(t) is measured as the sum of 1X component and other various components n(t): x(t)  ac cos t + as sin t + n(t)

(10.2)

where n(t)  a0 + a2 cos 2t + a3 cos 3t + · · · b2 cos 2t + b3 cos 3t + · · · Note that n(t) represents non-synchronous components corresponding to various frequencies except for the rotational frequency and is regarded here as noise. According to the computational outline as shown in Fig. 10.3 [271], the product of the cosine and sine waves with the input measured waveform is: x(t) × cos t  ac (1 + cos 2t)/2 + as /2 sin 2t + n(t) cos t x(t) × sin t  as (1− cos 2t)/2 + ac /2 sin 2t + n(t) sin t

(10.3)

The synchronous vibration amplitudes, ac and as , are here observed as DC values. In order to extract these DC values, the waveform of Eq. (10.3) is passed through a low-pass filter (LPF; with a time constant of 1 s or so) that is equivalent to a timeaveraging operation. As a result, only the first-term constants ac and as remain as the output signals. Since the second term with the product of noise n(t) is AC, they are then attenuated greatly after passing through the LPF. Hence: ac  2 × LPF[x(t) cos t], as  2 × LPF[x(t) sin t]

(10.4)

304

10 Signal Processing for Rotor Vibration Diagnosis Nyquist plot (Polar plot) as

Output : rotational speed Ω

Input : rotational pulse vibration wave x(t)

amplitude a =

a c2 + a s2

Bode plot a

ϕ

real part of polar plot a c imaginary part of polar plot a s

ac

x(t) 2

2

Ω

ac

LPF

a=

tan ϕ =

as

LPF

a c2 + a s2 as ac

cos Ω t

sin Ω t

LPF for time averaging = PLL Vector monitor

Rotational pulse

τ

a

ϕ

1 τs + 1

1 2π

0.5

Fig. 10.3 Vector monitor (balance analyzer)

The cosine and sine amplitudes, ac and as , are, thus, derived as shown in the figure. If both ac and as are plotted on a XY chart, termed a Nyquist diagram (polar plot), because the phase lag appears every time the rotor passes resonance speeds, this diagram describes a circle in the counterclockwise direction as shown in the figure. Also, note that the synchronous vibration is the sum of these two components: ac cos t + as sin t ⇔ a cos(t − ϕ), Hence, the amplitude a and the phase lag ϕ become: a

 ac2 + as2 ϕ  tan−1 (as /ac )

(10.5)

Plotting this amplitude a and phase lag ϕ versus the rotational speed  will give a Bode diagram. Figure 10.4 illustrates implementation of functions of the vector monitor with the MATLAB [272] software. A rotational pulse signal and a vibration waveform are input from the left side, and various amplitudes of the corresponding 1X vibration, phases, and a rotational speed are output to the right side. The computation of the rotational speed (t) is depicted in the bottom section of Fig. 10.4. In the first step, the input pulse signal is reshaped to ideal pulse signal having 1/0 V levels, that is 1 V only in one-sampling period and otherwise 0 V. This reshaped pulse signal is input to the First LPF system. The response is evaluated continuously and uniformly like the array of a damped pulse signal. Its time average value, obtained by passing through the first LPF with a large time constant and a multiplying gain (here, 5000) indicates the rotational speed (t) [Hz].

10.2 Signal Processing for Unbalance Vibrations Pulse

PLL

cos

305

1 τs + 1

2

ac

ac(t)

as

as(t)

sin x(t)

1 τs + 1

2

LPF for time average τ = 1/(2π ×0.5)

ac2 + as2 a(t)

Unit Delay Pulse shaping 1 in1 z out in2 in out if (in>0, 1, 0) if (in1>0, 0, in2)

Output (D/A)

Input (A/D)

LPF for time average τ = 1/(2π ×0.5)

time average Ω(t) 1 τs + 1 LPF τ = 1/(2π ×100)

1 τs + 1

5000

LPF τ = 1/(2π ×0.5)

Fig. 10.4 Block diagram of a tracking filter

10.2 Signal Processing for Unbalance Vibrations 10.2.1 Waveform and Orbit of Shaft Center of Rotation for Unbalance Vibration As shown in Fig. 10.5a, the acquisition of a rotational pulse signal is indispensable for vibration diagnosis of rotating machinery. A pulse mark is made by a small slot on the rotor surface. It gives a rotational pulse at each revolution when it faces a magnetic or optical pulse sensor installed at the stationary side. A displacement sensor is placed in the X direction—and if possible another one in the Y direction—to detect the rotor position. In Fig. 10.5b, a shaft center of rotation orbit shows a circular whirl motion, because the bearing support dynamic coefficients are isotropic in the X and Y directions (e.g., ball bearings and magnetic bearings). In this case, the X and Y directional unbalance vibration amplitudes will be the same, but the phase of the X directional unbalance vibration waveform is advanced on than that of the Y direction by the difference of 90°, i.e., the circular and forward-oriented orbit. On the other hand, in Fig. 10.5c, the shaft center orbit shows an elliptical whirl motion caused by anisotropic coefficients of the bearing support dynamics in the X and Y directions (e.g., oil film bearings). In this case, a difference emerges in the X and Y direction unbalance vibration amplitudes. At the same time, the phase

306

10 Signal Processing for Rotor Vibration Diagnosis ϕ

ϕy

ϕx

90° t

Pulse x(t)

y(t) U

Y

y(t) Y

Xr θ Ωt Rotor

(a) Rotor

Whirling orbit

x(t)

y(t) Yr

Y Rotor

Pulse X x(t)

t Pulse

Xr θ Os Ω t X

(b) Circular orbit

θ Os Ω t Xr Rotor

X

Whirling orbit

(c) Elliptic orbit

Fig. 10.5 Waveform and whirl orbit of unbalance vibration

difference shifts away from 90°. In the case of an elliptical orbit, the forward-oriented whirl motion is usually seen so that the peak of the Y direction waveform appears after the peak of the X direction waveform. Note that the opposite scenario of a backward-oriented orbit is rare, but potentially possible.

10.2.2 Vibration Measurement As shown in Fig. 10.1, shaft vibration is monitored by displacement sensors in the X and Y directions placed at several key locations along the rotating shaft system, for example, near the bearing portions. A multi-channel system monitoring bearing vibrations may be installed for important plant machines such as a turbine generator set. The rotational pulse signal and the X and Y directional shaft vibrations are detected for field balancing. Usually, the amplitudes of the X and Y directional vibrations will be approximately equal and the shaft center of the rotor will describe a circular whirl orbit due to isotropic bearing supports. In such cases, therefore, either the X direction or the Y direction vibration is used for balancing. The methods for finding balancing weights are already treated in Chap. R1_5. However, in cases where the bearing support coefficients are strongly anisotropic, the rotor shaft center will describe an elliptical whirl orbit and the vibration amplitudes of the X and Y direction will be different. In such situations, it will be more effective to carry out balancing using data on both the X and Y directional vibrations. We will introduce this balancing method for cases of strong elliptical orbits.

10.2 Signal Processing for Unbalance Vibrations

307

10.2.3 Block Diagram of Unbalance Vibration The equation of motion for unbalance vibration is expressed by Eq. (3.17) featuring the anisotropic characteristics of bearing dynamics. The solution of rotor vibration is assumed in the manner of Eq. (3.19): Af  forward complex amplitude Ab  backward complex amplitude Then these amplitudes can be found from 

ms 2 − jGs + k f + c f s kb + cb s ms 2 + jGs + k¯ f + c¯ f s k¯b + c¯b s



 s j

Af Ab



 

 U 2 0 (10.6)

This can be rewritten as:   2 A f  G −1 11 ( j) U  − G 12 ( j)Ab Ab  −G −1 22 ( j)G 21 ( j)A f

(10.7)

where G 11 (s)  ms 2 − jGs + k f + c f s G 12 (s)  kb + cb s G 21 (s)  k¯b + c¯b s G 22 (s)  ms 2 + jGs + k¯ f + c¯ f s As shown in the block diagram of Fig. 10.6, since the unbalance is equal to a forward exciting force, the forward vibration component Af is directly excited by the unbalance. In the case of isotropic supports, k b  cb  0, so G12 (s)  G21 (s)  0, and vibration will appear in the upper block forward   path only, with the circular orbit having the radius of the forward amplitude  A f . Thus, the circular whirl could be utilized to carry out the balancing calculation by selecting either the X direction or the Y direction vibration signals. In the case of anisotropic support, the forward vibration component is generated directly by the unbalance force, and it acts as the input of the backward system.

Fig. 10.6 Block diagram of unbalance vibration (anisotropic support)

Af : Forward UΩ

2

Unbalance

−1 G11 ( j Ω )

+ +

−1

G12 ( j Ω )

G22 ( j Ω ) −Ab Ab : Backward

G21 ( j Ω )

308

10 Signal Processing for Rotor Vibration Diagnosis

The backward vibration component is excited by this input and it is connected to the forward system in a closed loop. Both components having the forward Af and backward Ab amplitudes are coupled in an elliptical whirl orbit. However, from the viewpoint of the balancing calculation, if the Af becomes zero, the Ab automatically becomes zero and a well-balanced state will be achieved. It is said that a new balancing method focusing only on the forward vibration component is sufficient, as is the case of isotropic support. A little innovation is required in order to put this new method into practice, as explained below.

10.2.4 Extraction of Forward Unbalance Vibration Component In the case of isotropic bearing supports, the shaft center orbit will describe a circular locus such as shown in Fig. 10.7a. The Y direction vibration will be retarded by 90° compared to the X direction. The rotor shaft center position, the complex amplitude, and the corresponding Nyquist plot [see Fig. 10.3, detailed in Eq. (10.10)] are definitely expressed by the following relationships: Complex amplitude Nyquist plot Orbit and waveform j(t−ϕ) − jϕ A  ae A  ae jϕ z(t)  ae − jϕ ⇔ A x  ae ⇔ A x  ae jϕ (10.8) x(t)  a cos(t − ϕ) ◦ − j(ϕ+90◦ ) y(t)  a sin(t − ϕ) A y  ae A y  ae j(ϕ+90 )  a cos(t − 90◦ − ϕ)  − j Ax  j Ax

Y

Y

X

(a) Isotropic support Fig. 10.7 Unbalance whirl orbit

X

=

(b) Anisotropic support

ab

Ωt − ϕ f X

ius

Ωt − ϕ

+

rad

ra di us

rad

Y

af

Y

ius

a

Thus, the Nyquist plot of x(t) rotated by 90° counterclockwise is same as the Nyquist plot of y(t). On the other hand, in the case of anisotropic bearing support (for example k x  k y ), an elliptical whirl orbit will be described, as shown in figure (b), and the orbit is the sum of a forward circular whirl and a backward circular whirl, as illustrated in figure (c). In the same way, we can write the relationships as follows:

X Ωt − ϕ b

(c) Forward + Backward

10.2 Signal Processing for Unbalance Vibrations

309

Orbit and waveform Nyquist plot Complex amplitude A f  a f e− jϕ f , Ab  ab e− jϕb z(t)  a f e j (t−ϕ f ) + ab e− j(t−ϕb ) ⇔ ⇔ A x  ax e− jϕx A x  ax e jϕx x(t)  ax cos(t − ϕx ) − jϕ y A y  ay e A y  a y e jϕ y y(t)  a y sin t − ϕ y (10.9) In this case, there is no obvious relationship concerning the phase difference so as to maintain 90° between the shaft center vibration x(t) and y(t), and the two are independent. Methods for extracting the forward vibration component Af from actually measured vibration x(t) and y(t) are considered below. According to the literature [273], two such methods are recommended: called here Methods A and B. Method A: Method Using Two Vector Monitors (Fig. 10.8) Since a vector monitor analyzes the amplitude ac  a cos ϕ of the cosine component and the amplitude as  a sin ϕ as of the sine component, the corresponding complex form ac + jas is obtained. Therefore, the signal processing of this vector monitor gives A as follows: Inputting vibration waveform and pulse → Outputting vector monitor analysis → ac + jas  ae jϕ  A

(10.10)

It is, thus, concluded that the vector monitor finds the conjugate form of the complex amplitude A, i.e., the amplitude a and phase lag ϕ > 0. The complex amplitude expressed in terms of x and y and the complex amplitude expressed in terms of z  x + jy are linked in a relation by Eq. (3.31). It can be rewritten for obtaining the forward vibration amplitude as follows: Af 

    Ax − j A y 1   1    Re A x + Im A y + j Im A x − Re A y 2 2 2

X vibration Re Balance analyzer Ax = ax e jϕ x

Im

+ +

1 2

(10.11)

Real [ Af ] Im Af

Pulse Re Re Balance analyzer Ay = ay e jϕ y Y vibration

Fig. 10.8 In the case of using two balance analyzers

Im

+ −

1 2 Imaginary [ Af ]

310

10 Signal Processing for Rotor Vibration Diagnosis X vibration

+ −

Pulse

1 2

Real [ Af ] Balance analyzer Af Imaginary [ Af ]

Im

Phase shift by 90°

Af Re

Y vibration

Fig. 10.9 In the case of using phase shift

In order to implement the above equation, signal processing is recommended as shown in Fig. 10.8. x(t) signal is passed through one vector monitor, y(t) signal is also passed through the other vector monitor, and a pulse signal is passed through both, so as to derive A x and A y . The results of applying cross-computation to the real and imaginary parts of each of these are to the real and imaginary parts of the forward component A f . Method B: Method Using Phase Shift (Fig. 10.9) Rewriting the above equation Eq. (10.11), we have: 

 A x − (− j A y ) Af  2

(10.12)

and the corresponding signal processing as shown in Fig. 10.9. It shows that it will suffice to retard the Y direction vibration signal by 90°, subtract it from the X direction vibration signal, and multiply it by 1/2. In this way, we define the result as a new signal of the forward vibration waveform. Thus, we will input this new signal to the vector monitors, and finally, the output will be ac and as on x–y axis of the Nyquist plot.

10.2.5 Balancing to Decrease the Forward Unbalance Vibration Amplitude Whether Method A or Method B is employed, it will be possible to extract only the forward vibration component from the unbalance vibration of an elliptical whirl orbit. The Nyquist plot of the forward vibration component is applied to the modal balancing method described in R1_Chap. 5. In the case of strongly anisotropic oil film bearings, this new modal balancing focusing on the forward component can avoid deciding whether to measure the X or the Y direction sensors. Thus, the calculation will be simple and effective.

10.2 Signal Processing for Unbalance Vibrations

311

(a) 1

3

2

4

Wt 5

6

8 (A) Before BAL 6 (B) After BAL 4

(A) ay

ay

2

(B)

ax 0

ax

20 40 Rotational speed [ rps ] 4

Trial weight

Ay

60

Amplitude af , ab [ μm ]

Amplitude ax , ay [ μm ]

8

9

10

(c)

(b)

0

7

8 (A) Before BAL 6 (B) After BAL 4

af

2 0

(B)

30

25 25

4

2 −46°

Initial run (Before )

60

2

−2

C

ab

20 40 Rotational speed [ rps ]

0

Af

−8

af

ab

C B

(A)

25 B

25 −4

A 30 rps

−47° 30

Trial weight −4

30 A Initial run (Before )

Fig. 10.10 Case studies using the phase shift (BAL  Balancing)

Example 10.1 Figure 10.10 shows an example of balancing of a three-disk rotor

(R1_Figs. 12.1 and 12.3). Figure (a) shows the rotor system, supposing initial unbalance masses of 1g 0◦ at node ➂ and 1g 90◦ at node ➆. The resonance curves before balancing are shown concerning the X and Y direction amplitudes in figure (b) and the forward and backward amplitudes in figure (c). The lower graphs provide a comparison of the Nyquist plot before balancing and of the trial run attaching trial weight W t  1g − 90◦ at node ➄. (1) Determine the correction weight Wc if a Y directional vibration sensor is employed. (2) Determine the correction weight Wc if a forward vibration sensor is employed. (3) The resonance curves after balancing are provided for each case. Verify the adequacy of the balanced condition. Answer (refer to Example 5.3 in R1) (1) In the Nyquist plot of figure (b), estimating the influence coefficient at 30 rps: Hi ≡ |AC|/|AB| BAC  1.43 − 46◦ → Wc  Hi × Wt  Hi × 1g − 90◦  1.43g − 136◦

312

10 Signal Processing for Rotor Vibration Diagnosis

(2) In the Nyquist plot of figure (c), estimating the influence coefficient at 30 rps: Hi ≡ |AC|/|AB| BAC  1.36 − 47◦ → Wc  Hi × Wt  Hi × 1g − 90◦  1.36g − 137◦ (3) In both cases, the correction weight has the opposite phase of the initial unbalance, and so these values are adequate.

10.3 Fourier Series Expansion [274, 275] According to Fourier series expansion (FSE), it is considered that “any periodic function of a period T can be expressed by the sum of the trigonometric function series made up of the frequencies nω (ω  2π /T [rad/s]).” Later, in the digital age, this idea achieved completion in the form of Fourier transform theory, which takes as its objects not only periodic functions but any functions, and today is applied in various devices of science and engineering fields. First of all, let us understand the FSE, which is the essence of analog signal processing.

10.3.1 Example: Estimation of an Inherent Cosine/Sine Wave’s Amplitude Consider a waveform consisting of the superposition of several harmonic waves of the frequency nω having n times of the fundamental frequency ω. Let us take examples of three functions: one of them is a DC component, f 0 (t), and other two harmonic components, f 1 (t) and f 6 (t): DC component: f 0 (t)  2

(10.13a)

Fundamental frequency component: f 1 (t)  4 cos ωt 6th order harmonic component: f 6 (t)  2 cos(6ωt − 60◦ )  cos 6ωt +

(10.13b) √ 3 sin 6ωt (10.13c)

We define a waveform including these three components, denoted by w(t): w(t)  f 0 (t) + f 1 (t) + f 6 (t)

(10.14)

When the fundamental frequency ω  2π rad/s  1 Hz, the various components f i (t) are shown in (a) of Fig. 10.11, and the superposed waveform w(t) is shown in (b)

10.3 Fourier Series Expansion

313 f6(t) = 2sin(6ω t−60°)

(a)

4 f0(t) 2 f1(t) 0 f6(t) −2 −4

0.4

0.2

f0(t) = 2

0.6

0.8

t 1 [s]

f1(t) = 4cos ω t Fourier Series Expansion

6 4 w(t) (b)= Σ f (t) 20 i −2 −4

0.4

0.2

0.6

0.8

t 1 [s]

Fig. 10.11 Estimation of involved cosine and sine functions

of the figure. Although the superposed wave w(t) looks partially as complicated, the global view is easily visible to note the periodic function of the fundamental period T  1 s. According to FSE, we can inversely estimate each component consisting of superimposed waveforms as shown in (a) from looking over the waveform in (b). The method for doing so is simple to multiply by the trigonometric function (cos/sin) of the target frequency to be estimated, then derive the time average and double it, as already explained in Eq. (10.4). In fact, you can estimate the amplitude of the fundamental frequency ω components as: 2 T 2 T

T

2 w(t) cos ωtdt  T

T (4 cos ωt + · · ·) cos ωtdt  4

0

0

T

T

w(t) sin ωtdt  0

2 T

(4 cos ωt + · · ·) sin ωtdt  0

(10.15)

0

Also, in a similar way, the√ amplitude of the components of the sixth-order harmonic, 6ω, are found to be 1 and 3. The time average of the superimposed waveform w(t) corresponds to the DC component: 1 T

T 0

1 w(t)dt  T

T {2 + 4 sin ωt + · · ·}dt  2 0

(10.16)

314

10 Signal Processing for Rotor Vibration Diagnosis

Thus, it can be said that the amplitudes of the fundamental frequency ω  2π/T rad/s, and of the nth multiple frequency nω, including the DC component, can be inversely estimated by FSE. Example 1a Proof that√the amplitudes of the sixth-order harmonic component 6ω in Eq. (10.14) are 1 and 3. Answer

2 T

T

2 w(t) cos 6ωtdt  T

0

2 T

T (· · · + cos 6ωt + · · ·) cos 6ωtdt  1 0

T w(t) sin 6ωtdt  0

2 T

T 

···

√

 √ 3 sin 6ωt + · · · sin 6ωtdt  3

0

10.3.2 Principles of Fourier Series Expansion For any periodic function w(t), w(t)  w(t + T )

(10.17)

where ω  2π /T rad/s  1/T Hz  the fundamental frequency. By elaborating the example just given, the reverse estimation of a superposed function f (t) is possible to approximate various harmonic function components of n-times the fundamental frequency ω. The approximate equation is generally as follows: w(t) ≈ f (t) ≡ f 0 + c1 cos ωt + c2 cos 2ωt + c3 cos 3ωt + · · · + s1 sin ωt + s2 sin 2ωt + s3 sin 3ωt + · · ·

(10.18)

2 T 2 T 1 T where cn  w(t)dt 0 w(t) cos nωtdt, sn  0 w(t) sin nωtdt, f 0  T T T 0 and f 0 , cn and sn are called Fourier coefficients. In the case of non-continuous or complicated periodic waveforms, a perfect match would be obtained if the harmonic component waves in the above equation were summed in an infinite number. However, in actual practice, the series of the summation is truncated at a certain number n and the approximation accuracy is thus correlated with this truncated number n. Incidentally, if we combine two coefficients of cosine and sine functions in Eq. (10.18) and rewrite them in terms of the amplitudes and phases, we gain the following expression: f (t)  f 0 + a1 cos(ωt − α1 ) + a2 cos(2ωt − α2 ) + a3 cos(3ωt − α3 ) + · · · (10.19)

10.3 Fourier Series Expansion

315

 where an  cn2 + sn2  amplitude and αn  tan−1 (sn /cn )  phase. Further, simplifying the description by rewriting the above equation with the exponential function gives: ∞      Re f¯n e jnωt (10.20) f (t)  f 0 + Re a1 e j(ωt−α1 ) + a2 e j(2ωt−α2 ) + · · ·  f 0 + n1

2 T f (t)e jnωt dt  complex amplitude. where f n ≡ an e jα  cn + jsn  T 0 Defining amplitudes as a Fourier (amplitude) spectrum fse: fse 



 f 0 a1 a2 a3 · · ·  f 0 | f 1 | | f 2 | | f 3 | · · ·

(10.21)

This fse is displayed in a bar graph with the order numbers n placed along the horizontal axis. Example 10.2 Carry out Fourier series expansion on the sawtooth waveform

w(t)  2 + 20t (−0.5 ≤ t ≤ 0.5), according to the steps shown in Fig. 10.12. (1) Find  w(t) ≈ f (t) ≡ 2 + 20

 sin 2ωt sin 3ωt sin 4ωt sin ωt − + − + ··· π 2π 3π 4π (10.22)

where ω  2π × 1 Hz. (2) Show the Fourier (amplitude) spectrum distribution as a bar graph. (3) Verify that an improvement in the accuracy results from increase of the order numbers n. Answer (1) Fourier (amplitude) spectrum fse of the figure (a):   20 20 20 20 20 20 20 fse  2 ··· ··· π 2π 3π 4π 5π 6π 15π

(10.23)

(2) The spectrum distribution is shown in figure (b) with the truncation n  15. (3) The approximate waveforms are shown in Figure (c). These subscript numbers in f 1 , f 2 , f 15 are the truncated order numbers. When the wave is reconstructed by Eq. (10.22) including higher orders, it becomes closer to the input waveform of the sawtooth and the accuracy is improved.

316

10 Signal Processing for Rotor Vibration Diagnosis w (t) (a) input waveform w(t) w (t) = 2 + 20 (t−Round[t]) −1

−0.5

0.5

1

t [s]

f2 (t) T = 1s (c) Comparison with the order of FSE: by 1st, 2nd and 15th

8 fse =

6

fse

f1 (t)

−4 −8 fundamental period

FSE: Fourier Series Expansion

(b) Fourier amplitude spectrum

f15 (t)

12 8 4

{

4 20 10 2, π , π , · · · , 3π

{

4 2 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Order

f (t) = 2 +

20 sin 2 π t π

–

10 sin 4 π t π

+ ···+

4 sin 30 π t 3π

Fig. 10.12 Fourier trigonometric series of a sawtooth shock pulse

10.4 Discrete Fourier Transformation [276] The discrete Fourier transformation (DFT) is the digital processing version of the Fourier series expansion. The algorithm for the high-speed DFT is the fast Fourier transformation (FFT). The use of FFT computation chips has spread widely and they are installed in various fields of measuring instrument, where they form the nucleus of the signal processing technology.

10.4.1 Principles of the Discrete Fourier Transformation Suppose the discrete data, “dat”, by taking N samples obtained in a single period, 0 to T , of an original waveform w(t), is:  dat  w0 w1 w2 w3 · · · wk  w(kT /N ) · · · w N −1

(10.24)

The period is called the fundamental period. In this case, the Fourier coefficient set defined by Eq. (10.20) for analog integration is replaced by the following equation specified for digital calculation:

10.4 Discrete Fourier Transformation

317

N −1 1  wk  average value, DC N k0   N −1 2  2nπ k  complex amplitude Fn  wk Exp j N k0 N

f0 

an  |Fn | actual amplitude

(10.25)

Thus, the Fourier coefficients are calculated in this way from discrete data, and this is termed the discrete Fourier transformation (DFT). In this case, there are only N discrete data in the time domain that are transferred to the complex amplitude F n in the frequency domain, stored in dft corresponding to n (  0 to N − 1) times the fundamental frequency, 1/T . Hence, the spectrum dft is a sequence of N samples of complex amplitudes: dft 



 f 0 a1 a2 a3 · · · a N −1  f 0 |F1 | |F2 | |F3 | · · · |FN −1 |

(10.26)

Reconstruction of the wave is according to the following equation, equivalent to Eq. (10.20): f (t)  f 0 +

N −1 

  Re F¯n e jnωt

(10.27)

n1

Example 10.3 Figure 10.13 shows the results of carrying out the DFT with the number

of samplings N  16 on the sawtooth wave w(t) of Example 10.2, according to each step below. Scrutinize the processing content of each step. reconstructed wave w (t) w (t) = 2 + 20 (t−Round[t])

10

−10

DFT: Discrete Fourier Transformation 8

fse

(

White bar chart by Fourier Series Expansion

Spectrum

6

)

t

0

(a) Input waveform and sampling w(t)

(b) Fourier spectrum (amplitude) distribution (order:0-15th)

Input wave

0

0.2

0.4

0.6

0.8

1 [s]

(c) Approximation by DFT Mirror phenomenon caused by DFT Display zone dft

4 2

fse

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Order Order of the Nyquist frequency

Fig. 10.13 Discrete Fourier transformation of a sawtooth shock pulse

318

10 Signal Processing for Rotor Vibration Diagnosis

(1) Take samples of the sawtooth wave w(t) as in Fig. 10.13a. (2) Show the DFT (amplitude) spectrum distribution as a bar graph as in Fig. 10.13b. (3) Reconstruct the wave from the DFT results, and compare it with the original waveform as Fig. 10.13c. Answer (1) The number of samplings is N  16 within the period T  1 s, so the sampling frequency  16 Hz. (2) A mirror phenomenon (termed “aliasing”) arises, in which there is left-right symmetry about order N/2  8. (3) The reconstructed waveforms pass unfailingly through the sample points, but have small amounts of undulation between the sample points.

10.4.2 Mirror (Aliasing) Phenomenon of Complex Amplitudes How to calculate the actual value of the complex amplitude as defined by Eq. (10.25) is illustrated in Fig. 10.14, taking the sampling number N  16. Unit circles are drawn in a complex plane, and their circumferences are divided into N  16 equal parts. In the case where, for example, n  1, the calculation is to find the sum for F 1 of the product between each complex value on the unit circle and each sampled value, repeatedly for N times, starting from 0°, as shown in the figure. In the case where

6

5

4

3

2

2 1 w0

7 Im

n=1

8 10 Re

n=0

.

w15

.

.

w0 w1 .

8

10

11

12 12

9 n=15

8

14 15 w0 1

7 6

5

2 4

3

Fig. 10.14 Cause of DFT mirror phenomenon

6 11 w0

3

Im

5 14

13

13

1

n=3

14 11

12

13

15

9

7

14

9

9

4 4

15

15

3 8

n=13

10 w1 3 10 5 w0 11

13 2

7

12

1

6

Re

n=8 5.

..

.4 ..

w0 2

10.4 Discrete Fourier Transformation

319

n  3, one will find the sum for F 3 of the product between the complex value selected on the unit circle by three steps and the sampled value, repeatedly for N times, as shown in the figure. In this way, when n  15 one will extract the unit circle value for every 15th division, i.e., moving counterclockwise at one by one, as shown in the figure. The complex value for moving counterclockwise is the conjugate value for moving clockwise. Hence, F 15 will be the conjugate value for F 1 . With the same reasoning, F 13 is the conjugate value for F 3 , as shown in the figure. In this manner, it is found that the forward half and the backward half, divided by the border N/2  8, are in a conjugate relation: F1  F¯15 F2  F¯14 · · ·

F7  F¯9

(10.28)

Since the absolute value of both sides is equal, they will appear left-right symmetrical about N/2  8 in a bar graph of the DFT amplitude spectra, whatever the input is. This is called the mirror phenomenon as shown in Fig. 10.13b, i.e., one of the features of the DFT.

10.4.3 Sampling Values Figure 10.15 illustrates the situation when N  16 samplings are taken in a period of 1 s (fundamental frequency  1 Hz) for a cosine wave. Given alongside is an illustration of sampling of cosine waves having specified frequencies, 15 Hz and 17 Hz. In these cases, the sampling timings coincide and the sample values are the same. In other words, this means that in the set of sampled value, it is impossible to make a distinction among 1 Hz, 15 Hz, 17 Hz …. and so on. This problem regarding duplication of sampled values arises also in the case of other order numbers. Figure 10.16 illustrates the phenomenon of this duplication of the sampled values concerning with other order numbers n. It is seen that ultimately no 15 Hz

17 Hz

15 Hz

1 0

17 Hz

1 Hz 0.2

0.4

0.6

−1 1 Hz

Fig. 10.15 Sampling and quantizing ( f s  16 Hz)

0.8

1 [s] 0.8

320

10 Signal Processing for Rotor Vibration Diagnosis

n=0

n=1

n=2

n=7

Average

n = 15 n = 17 . . . .

n = 14

n=9

n = N / 2 =8

n = 18 . . . .

n = 31 . . . .

n = 32 . . . .

n = N = 16 . . . .

Fig. 10.16 Duplication of sampling data

distinction can be made among the waves with order number N/2  8 and above. This means that measures will be needed to prevent input of high-frequency components with order number N/2  8 and above.

10.4.4 Aliasing Error In frequency analysis via the DFT, 1/2 of the sampling frequency is termed the Nyquist frequency f q ( f s /2). The entry of frequency components higher than the Nyquist frequency must be filtered from the input signal. If these high-frequency components happen to be input, any component of the Nyquist frequency or above is made to crossover to the low-frequency side by means of aliasing. This is referred to as “aliasing error.” The general theory of aliasing error is set forth in Fig. 10.17. The horizontal axis represents the input frequency, and the Nyquist frequency bands are considered as set of standing panels numbered by 1, 2, 3 …. We observe the display of the first panel as the result of the DFT. However, all the panels are transparent, and so if a high-frequency wave is input in the fifth panel, it is observed in the first panel by means of aliasing, as shown by dotted lines in the figure. Suppose a low-pass filter (LPF) having the Nyquist frequency f q as its cut-off frequency. Then any components of N/2 or higher orders will appear as aliasing errors inside frequency range from 0 to the Nyquist frequency, as shown in the schematic in Fig. 10.18. If the cut-off frequency of the LPF is set by a value smaller than the Nyquist frequency, this error is getting smaller. Such a filter is termed an “anti-aliasing filter.” Generally, the setting will be recommended as follows: • Sampling frequency f s , Nyquist frequency f q  f s /2, maximum display frequency f max  f s /2.56 → LPF cut-off frequency f c  f max • Hence, in Example 10.3 above, f c  f max  f s /2.56  16/2.56  6.25 > the 6-th order, which is set by f max as shown by the display zone in Fig. 10.13b. Example 10.4 In Fig. 10.13b, the pink spectrum (fse) represents the accurate values

produced by the Fourier series expansion, while the black spectrum (dft) represents those produced by the discrete Fourier transformation. For n  0–8, the dft values ought to coincide with those of the fse. Why do they not do so?

10.4 Discrete Fourier Transformation

321

Answer Because the sawtooth wave contains an infinite number of high-frequency components beyond the Nyquist frequency, those of f q  8 Hz and above appear on the low-frequency side as aliasing errors. Example 10.5 Figure 10.19 shows the DFT (N  16) results of the wave of

Eq. (10.29), w(t)  3 +

15 

  ai cos iωt − (i − 1)30◦

(10.29)

i1

where ai  10/i (i  1 · · · 15, i  13), a13  6, ω  2π rad/s = 1 Hz. Scrutinize the processing content of each of the following steps: (1) The FSE of the waveform w(t) produces an amplitude spectrum (real values) → figure (a). (2) Sampling of the waveform w(t) → the circles (“●”) in figure (b). (3) Display all the elements of the DFT (amplitude) spectrum of w(t) → figure (c). (4) Sampling of an input wave v(t) passed through an ideal LPF with setting of the cut-off frequency at 16/2.56  6.25 > the sixth-order harmonic (6 Hz), as an anti-aliasing filter → figure (d):

Spectrum

fma

1 x

2

= fs 2.5 6

fq =

fs/

3

4

2

0

1

0.5 Observed

5

2 1.5

Input of high frequency wave

reflected

2.5 Fre que × f ncy s

Fig. 10.17 Aliasing error and Fusuma (Japanese Partition) Fig. 10.18 Cut-off frequency of anti-aliasing filter

Mirror phenomenon Gain

Display zone

Frequency

0 fc = fs / 2.56 = fmax (Cut-off frequency)

fq = fs / 2 (Nyquist frequency)

10 Signal Processing for Rotor Vibration Diagnosis

(a) fse: Fourier Series Expansion

12 10 8 6 4 2

Spectrum

322

fse

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Order 40

(b) Input wave and sampling

w(t)

20

t

0 −20

0

0.4

0.2

dft: Discrete Fourier Transformation

0.6

0.8

1

Display zone Mirror phenomenon

12 10 8 6 4 2

Spectrum

(c) Fourier spectrum (amplitude)

[s]

dft

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Order 40

(d) Filtered input wave and sampling

v(t)

20

t

0 −20 dft: Discrete Fourier Transformation Spectrum

(e) Fourier spectrum (amplitude)

12 10 8 6 4 2

0

0.2

0.4

[s]

0.6

0.8

1

Display zone Mirror phenomenon dft

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Order fse

Fig. 10.19 Effectiveness of anti-aliasing filter

v(t)  3 +

6 

  ai cos iωt − (i − 1)30◦

(10.30)

i1

(5) Display all the elements of the DFT (amplitude) spectrum (dft) of v(t) → figure (e). Answer (1) The real values fse of the FSE spectrum (exact values) are:

10.4 Discrete Fourier Transformation

323

 fse  3 10 5 10/3 5/2 2 5/3 · · · (2) Sampling of a wave containing all components above the Nyquist frequency f q  f s /2  8 Hz and a 13 Hz harmonic is included, (3) Aliasing error has occurred. For example, the third-order component is the sum of the original third-order component and the aliasing of the 13th order component: √ ◦ ◦ ◦ 10/3e j60 + 6e j0  (5/3 + 6) + j5 3/3  7.67 + j2.89  8.2e j21 This means that the third-order component in the bar graph indicates 8.2, which is greatly different from the true value of 10/3  3.33. (4) The waveform v(t) passes through an anti-aliasing filter, to remove all components above the Nyquist frequency and excluding the 13 Hz harmonic, which completely removes harmonic components of the seventh order and above. (5) Because the high-frequency components have been filtered out, no aliasing to the low-frequency side is present. Therefore, the dft display area, ranging up to the sixth order, coincides exactly with the fse (exact values) as shown in the bar graph (e).

10.4.5 Suppression of Noise Noise is an ever-present problem with signals. An effective countermeasure against noise is to time average the signal. If one does this, the average of the noise should converge to 0 and only the inherent signal components should remain. Generally, the signal S will be proportional√to the iteration number n for averaging, and the noise N will be proportional to 1/ n. Hence, √ the SN ratio (SNR  Signal-to-Noise Ratio  S/N) will improve by a factor of n. Assuming the sampling with a number of samples N  16 in the fundamental period T  1 s (the fundamental frequency f  1 Hz), a signal of a 2 Hz sine wave as the original signal including the noise n(t) is considered: w(t)  sin(4π t) + n(t)

(10.31)

Figure 10.20a shows the signal, w(t), with noise present. The noise contains various frequency components, but excludes frequency components over than f s /2.56  16/2.56  6.25 Hz as determined by the anti-aliasing filter. So far as one observes the waveform, no 2 Hz wave is recognized in it. Figure (b) shows the spectrum obtained from carrying out a DFT on sampled data in the initial 1 s for the time window. Thus, after the first DFT, it is difficult to observe the dominant 2 Hz vibration. Accordingly, a countermeasure for the noise involves time averaging by the successive application of the DFT for each 1 s time window of the waveform, as shown

324

10 Signal Processing for Rotor Vibration Diagnosis

(a)

(b)

2

8s

0 −2

1st DFT

1

0

Index of DFT

1

2

3

5

4

6

7

0

8

1

2 3 4 5 Frequency [Hz]

6

(c) Average by 2nd DFT

1

0

0

1

2 3 4 5 6 Frequency [Hz]

Average by 4th DFT

1

0

0

1

2 3 4 5 Frequency [Hz]

6

Average by 8th DFT

1

0

0

1

2 3 4 5 6 Frequency [Hz]

Fig. 10.20 Effectiveness of averaging ( f s  16 Hz, T  1 s)

(a)

(b)

(c)

2

2

2

0

0

0

−2

−2 0

2

4

6

8 [s]

Window length : 2 times

1

0

0

2

4

6

8 [s]

Window length : 4 times

1

0 0

1

2 3 4 5 Frequency [Hz]

6

−2 0

2

4

6

8 [s]

Window length : 8 times

1

0 0

1 2 3 4 5 Frequency [Hz]

6

0

1

2 3 4 5 Frequency [Hz]

6

Fig. 10.21 Effectiveness of window length (total sampling) time ( f s  16 Hz, T  1–8 s)

by the index of the DFT in (a). As shown in figure (c), the average of each result for the first two DFTs, the first four DFTs and the first eight DFTs succeeds in raising the bar of the 2 Hz component highlighted. Although the signal is hidden in noise at first, after the time averaging, it becomes the most dominant frequency, which demonstrates that the technique is very effective. Another countermeasure against noise is to increase the number of samples at the same sampling frequency. This will mean a longer time window, so that the fundamental frequencies get smaller and the frequency resolution becomes finer. Figure 10.21 clarifies the DFT under such conditions. The time window shows that waveforms are progressively lengthened by factors of 1, 2, 4, and 8, together with the corresponding DFT results, as shown in figures (a), (b), and (c). It is seen how, by lengthening the time window and increasing the number of samples, the noise is progressively counteracted, and this 2 Hz component is then highlighted.

10.4 Discrete Fourier Transformation

325

Increasing the number of samples and making the frequency resolution finer is very effective against noise. With this method, however, an extended amount of memory is required and the calculation time will lengthen by (N/2)Log2 N depending to the number of samples N, so that in actual practice it is often difficult to implement.

10.5 Practicalities of FFT Analysis [276, 60] Figure 10.22 shows the outlook of one of the most popular items of equipment, the CF-350/360 FFT Analyzer made by ONOSOKKI, Japan. As can be seen from the screen illustrated here, this FFT analyzer is able to display waveforms in the time domain and Fourier spectra in the frequency domain. The vertical and horizontal axes in the display include the following definitions:

10.5.1 Basic Specifications (A) Fourier spectrum F n  cn + jsn where cn  real part, sn  imaginary part (B) Power spectrum Spectrum of (half) amplitude display V  |F n | (Half) amplitude spectrum of logarithm display dBV  20log10 |F n | Spectrum of peak-peak amplitude display V pp  2 |F n | √ Spectrum of root-mean-square value display Vr  |Fn |/ 2 (C) Time domain (see Fig. 10.23, dat data of Eq. (10.24)) Span of time window  data sampling time  fundamental period T [s]

Fig. 10.22 FFT analyzer (ONO SOKKI CF-350)

326

10 Signal Processing for Rotor Vibration Diagnosis 5

Fig. 10.23 Wave in time domain (N  1024)

Ts

0.1

0.2

0

t [s]

−5 −10

Sampling period Ts T/N = 9.77 ms

Window 10 0 −10 0

T = 10 s

Amplitude

Mirror phenomenon 8 0.1 Hz

4 0

10

Δ f Frequency resolution

20

30

Frequency [Hz]

40

fq = fs / 2

fmax

= 51.2

fs = 102.4 [Hz]

Fig. 10.24 FFT spectrum analysis (N  1024, L  400)

Fundamental frequency f  1/T [Hz] Fundamental angular frequency ω  2π f [rad/s] Number of sampling data N (normally N (e.g. N  1024/2048) Sampling period Ts  T /N [s] Sampling frequency f s  1/Ts [Hz] (D) Frequency domain (see Fig. 10.24, dft data of Eq. (10.26)) This display shows a result of a DFT. The horizontal axis represents discrete values of the frequency, which are multiples of the fundamental frequency. The dft data are complex valued (amplitude  phase), and thus, the vertical axis of this graph is the absolute value of the dft, i.e. amplitude. Frequency resolution (i.e., resolvable frequency) f  fundamental frequency f  1/T [Hz] Left end  DC component Right end  Nf [Hz] ≈ fs  sampling frequency Middle fq  fs /2 [Hz]  Nyquist frequency (aliasing frequency) Frequency display: Maximum number of lines L  N/2.56 (e.g. L  400/800) Maximum display frequency f max  L f  fq /2.56 [Hz] Anti-aliasing filter: e.g.   Butterworth filter ( 1/ 1 + (ω/ωc )2n ) of the n-th order Cut-off frequency of anti-aliasing filter  f max [Hz]

10.5 Practicalities of FFT Analysis

327

10.5.2 Signal Processing Procedure The term “FFT” was originally a definition for a mathematical algorithm that performs the discrete Fourier transformation (DFT) at high speed. Today, FFT computational chips are installed in all DFT instruments and the acronym DFT has almost been replaced by FFT. The following processing is implemented on a FFT analyzer as shown in Fig. 10.25. (1) LPF (low-pass filter): to exclude frequency components over the maximum display frequency f max . (2) A/D conversion: normally N  1024/2048 samples are taken. (3) Window function: normally, the Hanning window is applied. Details are given in the following section. (4) DFT: Discrete Fourier Transformation is implemented at high speed using the FFT algorithm. (5) Display: the waveform and the Fourier (amplitude) spectral distribution are shown.

10.5.3 DFT of Synchronous and Asynchronous Waveforms In our previous discussion on the Fourier series expansion (FSE) and the discrete Fourier Transformation (DFT), the input signals were assumed implicitly to be periodic functions with the period T . The waveforms were limited to being synchronous1 with the time window.

Waveform

Display of FFT analyzer 0.1

−0.1

0.2

0.4

0.6

Waveform t [s]

LPF

0.8

fc = fmax

0

T

Spectrum

−40 −50 −60 −70 −80

f [Hz] 100 200

0

A–D

300 400 500

fmax

DFT by FFT

Window functions ·Rectangular ·Hanning ·Flat top

Fig. 10.25 Signal processing of FFT analyzer

1 Hitherto

there is the expression “periodic function of period T”, but now “periodic” is changed to “synchronous with the time window having the span T ”.

328

10 Signal Processing for Rotor Vibration Diagnosis

However, when one measures any input waveform and takes N samples of it to carries out the DFT, there is no guarantee that the input signal is ideally window synchronous. The inconvenience that may arise in association with the spectra that appear, when the input signal is asynchronous, are described below. The following discussion requires a general knowledge of the Fourier transformation from the time domain to the frequency domain, which can deal with any synchronous or asynchronous waveforms in time domain. We refer as necessary to Appendix C, which provides an introduction to Fourier transforms. Suppose here that the DFT is to be applied with time window of 1 s span (fundamental frequency  1 Hz  frequency resolution) and the number of samplings N  1024. Assume the following synchronous function w1 (t): w1 (t)  4 + 8 sin 10ωt + 4 sin 20ωt + 10 sin 30ωt (ω  2π × 1)

(10.32)

and the following asynchronous function w2 (t): w2 (t)  4 + 8 sin 9.6ωt + 4 sin 20.4ωt + 10 sin 30.5ωt

(10.33)

Hence, the synchronous function will contain frequency components that are 10 (10 Hz), 20 (20 Hz) and 30 (30 Hz) times the fundamental frequency. The asynchronous function has three components. Each of them is deviated from the integer multiples of the fundamental frequency by approximately 0.5 Hz. Each block of Fig. 10.26 contains waveforms at the upper part and the corresponding amplitude spectra obtained by applying the DFT of the lower part. Note that the amplitude spectra displays are emphasized only over the frequency zone of 0–40 Hz. These blocks are seen as a 2 × 3 matrix. The first row corresponds with a synchronous function at the upper side (a, b, c) and an asynchronous function at the second row (d, e, f). Three columns include the usage of each window function; rectangular, Hanning and flat top at (a, d), (b, e) and (c, f), respectively. In the case of the synchronous function, w1 (t), as shown in the first row, the peaks of the amplitude spectra appear at each frequency of 0, 10, 20 and 30 Hz with amplitude of 4, 8, 4 and 10, respectively. These peaks of the DFT are completely coincident with the input amplitudes and frequencies. The usage of the rectangular window (i.e., no window function) provides the peak only, but completely zero elsewhere. The other two window functions provide the main peak and sideband peaks. On the other hand, the second row of the figure shows the DFT result of the asynchronous function, w2 (t). Both waveforms, w1 (t) and w2 (t), look similar in the time domain, but they differ in the frequency domain. The spectral amplitude corresponding to the asynchronous w2 (t) are a little smaller at 10, 20 and 30 Hz than those of the synchronous w1 (t), and, moreover, the spectra at the other frequencies rise up above the zero level. This is termed the leakage error due to spectral side lobes. In figures (a) and (d), each waveform is directly input to the DFT. This case is called “rectangular windowing”, i.e., no window function. Three peaks of amplitude

10.5 Practicalities of FFT Analysis Win -dow

329

Rectangular

Hanning

20

w1 Periodic functions

t

0 −20 0.0 10

0

[s] 0.2

0.4

0.6

0.8

1.0

20

10 20 30 Frequency f [Hz]

20

t

0 −20 0.0 10

0 0

40

w2 Non-periodic functions

0

0

0.2

0.4

0.6

0.8

10 20 30 Frequency f [Hz]

1.0

[s]

0

40

t

0 −20 0.0 10

[s] 0.2

0.4

0.6

0.8

10 20 30 Frequency f [Hz]

1.0

40

0.4

0.6

0.8

10 20 30 Frequency f [Hz]

1.0

40

(f) 20

20

t

0 −20 0.0 10

0

0.2

0

(e) t

−20 0.0 10

[s]

0

(d) 20 0

Flat top

(c)

(b)

(a)

0

[s] 0.2

0.4

0.6

0.8

10 20 30 Frequency f [Hz]

w1 (t) = 4+8sin10 ω t + 4sin20 ω t + 10sin30 ω t

1.0

40

t

0 −20 0.0 10

0

0

[s] 0.2

0.4

0.6

0.8

10 20 30 Frequency f [Hz]

1.0

40

w2 (t) = 4+8sin9.6ω t + 4sin20.4 ω t + 10sin30.5 ω t

Note: T = 1 s , fmax = 400 Hz , L = 400 These are detailed displays ranging from 0 to 40 Hz

Fig. 10.26 Differences by applying window functions

spectrum (d) are smaller than (a) down to approximately 60%, because of not being window-synchronous. A “trick” for making an asynchronous function resemble a window-synchronous function is to narrow the waveform at both ends of the window, as modifying them into “zero start” and “zero end” in time domain. For this purpose, one applies the “Hanning window”, which is drawn by the enveloping curves in the waveforms in (b) and (e). The original waveform multiplied by the window function is then processed by the DFT, and the corresponding spectral curve is obtained as shown in the lower parts of (b) and (e). In the case of asynchronous functions, the peak amplitudes of spectra are smaller than the exact values, down to approximately 80%, moreover, spectral peaks appear below and over the actual frequencies of 9.6, 20.4, and 30.5 Hz. Compared with the rectangular windows, the side-lobe error is drastically reduced by multiplying by a Hanning window, and, accordingly, FFT analyzers usually apply a Hanning window as standard. Next, the flat top window is introduced, which is indicated by the envelope lines in figures (c) and (f). It is well known as a function for compensating the decreases of peak amplitude of spectra even in the case of asynchronous functions. This window function looks like the Hanning window in time domain, but narrowing at both ends is more pronounced. The corresponding spectral peaks are so spread and it is hard to identify the input frequency components precisely in the frequency domain. However, the peak values indicate the exact amplitudes of the original waveforms

330

10 Signal Processing for Rotor Vibration Diagnosis

whether the input is synchronous or asynchronous. Therefore, flat top windows are recommended for accurate identification of amplitude.

10.5.4 Fourier Transforms In the case of a synchronous function, the peaks of the DFT spectra coincide completely with the actual amplitudes of the input waveform. However, in the case of an asynchronous function, it was found that they are somewhat lower than the actual amplitudes. For understanding what the peak values of the DFT spectra are in general, including the case of an asynchronous waveform, we need Fourier transform theory that treats any given input signal under the assumption that the time window → ∞, or in other words the frequency resolution → 0. The input waveform is multiplied by a window function, and then it is subjected to the DFT processing. Thus, it is known that the peak value of spectral amplitude corresponding to a certain input frequency can be estimated by the Fourier transform of the window functions themselves, as set forth in Appendix C. Figure 10.27 shows the result of the same waveform multiplied by each the window functions at the upper figures and the corresponding Fourier transform curves at the lower figures. This waveform is here set as window-synchronous, as recognized in figure (a). The fundamental frequency of the time window period T is 1/T [Hz], which is the frequency resolution f , as represented in the x-axis of the spectral graph. The DFT spectral curves are displayed by linear and dB scales of the y-axis. The main lobe curve is most spread by the flat top (c) as seen in linear display. Numerous small side-lobe spectra are visible as seen in dB display of the Hanning (b). Here, the reference level is the amplitude  1  0 dB, and so the vertical axes signify the ratio of decrease compared with the exact amplitude. The procedures to read these graphs are as follows. The center value, f  0, is equivalent to the certain input frequency, f in , at the horizontal axis for the FFT and the abscissa axis means f in ± n × f , the frequency difference between the input frequency and the closest multiple resolution frequency. The height of the spectral curves is the amplitude in the vertical axis. The amplitude is less than unity, due to viewing the amplitude corresponding to the frequency away from the center frequency. For example, suppose the frequency resolution, f  1 Hz, and a 30 Hz sine wave as a synchronous is the input function. Then the spectral center in Fig. 10.27 will be interpreted as 30 Hz, while the horizontal axis is the frequency difference from 30 Hz at 1 Hz resolution pitch each to the left and to the right. This case study is illustrated in Fig. 10.28c, where spectra extending with 1 Hz resolution from 26 Hz at the left end to 34 Hz at the right are displayed, being marked with ●, ◯ and ▲ symbols corresponding to the various window functions. Some remarks are given below regarding figure (c):

10.5 Practicalities of FFT Analysis

331 T

T

1

1 t 0.5 [s]

0

−0.5

T 1 t 0.5 [s] −0.5

0

−0.5 −1

−1

(a) Rectangular

−1

(c) Flat top

(b) Hanning −6

Spectrum [dB] −4 −2 0

2

Spectrum (Linear)

(b)

0.5

6

−20

(a)

(c)

4 (b)

(c)

1

t 0.5 [s]

0

(a) −40

−6

−4

−2

0

2

4 6 1 [Hz] Frequency × T

−60 Frequency ×

1 [Hz] T

Fig. 10.27 Window functions and Fourier transformation of windows

(a)

(b)

(c)

Periodic (Synchronized)

1

0

6

7

8

Non-periodic (Non-synchronized)

(d)

9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34

(e)

(f)

1

0 6

7

8

9.6 Hz

9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 26 27 28 29 30 31 32 33 34 Frequency [Hz] 20.4 Hz 30.5 Hz

Fig. 10.28 Spectra of periodic and non-periodic functions

(1) Since the center value of the spectral curves is 1  0 dB, regardless each of the synchronous functions, the original waveform’s amplitude will be indicated as unchanged by the DFT in any case of window function. (2) With regard to the rectangular window ●, all the spectra other than at the spectrum center value  1 are 0, and so no side lobes are present. However, with the Hanning window ◯ or the flat top window ▲, several spectra stand up adjacent to the center spectral value.

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10 Signal Processing for Rotor Vibration Diagnosis

In the case of asynchronous input frequency, for example, considering the input sine wave of an asynchronous frequency of 30.5 Hz, the figure (c) is thus shifted by 0.5 Hz to the right, as shown in figure (f). Three window functions are marked with symbols ●, ◯ and ▲ at the same frequency in the horizontal axis of the figure. Some remarks are given below regarding figure (f): (3) With the rectangular window ●, there are two predominant spectra, of the same height 0.64  −3.9 dB, and so the DFT peak values that are displayed are considerably smaller than the actual amplitude. The side lobes are also prominent. (4) With the Hanning window ◯, two predominant spectra values indicate 0.85  −1.42 dB, meaning that 85% value of the actual amplitude is displayed at 30 and 31 Hz in the x-axis. The side lobes are suppressed. (5) With the flat top window ▲, the spectral peak of the main lobe is spread at both sides, and the height of this peak is 0.99  −0.11 dB, neglecting the decrease of amplitude by the DFT, even in the case of non-synchronization. This window function maintains the amplitude of the input waveform at almost 1:1 for any center frequency. The same Fig. 10.28 also illustrates the spectral amplitude curves of other cases obtained in a similar manner; the synchronous input frequencies of 10 Hz and 20 Hz in (a) and (b), the asynchronous input frequency of 9.6 Hz in (d), the leftward shifted (a) by 0.4 Hz, and 20.4 Hz in (e), and the right shifted (b) by 0.4, respectively. The various graphs in this figure correspond with all the spectral peaks as shown in the previous Fig. 10.26. Example 10.6 In Fig. 10.29, FFT results for analyzing the input wave of 20.25 Hz

(asynchronous) sine wave, with fundamental period, T  2 s (fundamental frequency, f  0.5 Hz), number of samplings, N  1024, and 1 V amplitude of the input signal. Corresponding spectra to each window function are shown in Figs. (1), (2), and (3); those in the right-hand column are enlarged graphs around 20 Hz. Estimate the peak spectral values for each of (1), (2), and (3) when the input signal is changed to a 20.2 Hz (asynchronous) sin wave. Answer Around 20 Hz, synchronous frequencies of 19, 19.5, 20, 20.5, 21…Hz are in the abscissa axis of the FFT graph. The input frequency of 20.2 Hz is different from the synchronous frequency of 20 Hz by 0.2 Hz, that is 0.2/0.5  40% of the resolution 0.5 Hz → The spectrum graph of Fig. 10.28e is applied to this case of 40% offset → (1) 0.75 V  −2.5 dB, (2) 0.89 V  −1.0 dB, (3) 0.99 V  −0.05 dB at 20 Hz.

10.5.5 Resolution In the DFT analysis for any input waveform, including asynchronous functions, we have often to distinguish the difference between two frequencies, f 1 and f 2 (Hz). Here we discuss the resolution that is needed for this distinction.

10.5 Practicalities of FFT Analysis

Spectrum [dB] Spectrum [dB]

150

200

fluctuated

−60 −80 0

50

100

150

−40 −60 −80 0

50

100

150

−40 −60 −80

22

24 f [Hz]

−40 (1) Rectangular window

−60 −80 18

0

20

22

−20

24 f [Hz]

(2) Hanning window

−40 −60 −80

200

(3) −0.11 dB (0.99)

−20

20

−20

200

(2) −1.42 dB (0.85)

−20

18

0

Spectrum [dB]

−40

0

(3)

100

(1) −3.9 dB (0.64)

−20

0

(2)

50

Spectrum [dB]

(1)

0

18

0

Spectrum [dB]

Spectrum [dB]

0

333

20

−20

22

24 f [Hz] (3) Flat top window

−40 −60 −80

(a) Result of FFT

(b) Detailed around 20Hz

Fig. 10.29 Vibration analysis by ONO SOKKI/CF350

(a) 1 1

Resolution 3 2

f1

(b)

4 f2

1

0.7 0.64

1 f1

Resolution

2

4 f2

3

0.7 BW BW

0 −1

WF = 1 0

1

2

×Δf

0 −1

WF = 1.5 0 1

2

3

×Δf

Fig. 10.30 Resolution depending upon windows (BW  bandwidth)

The spectrum for a window function on vibration of f 1 is drawn as ➀ in Fig. 10.30. The Fourier transform is equivalent to an array of band-pass filters with a window factor (WF), that is the same concept as the filter bandwidth (70% gain). As shown in Appendix Table B.2, WF  1 for the rectangular window, WF  1.5 for the Hanning window and WF  3.8 for the flat top window. At the right side of curve ➀, there is an adjacent virtual spectrum ➁ having the same gain, which does not actually exist as shown in the dotted curves. Then, instead of curve ➁, we can see the empty area at ➂. At the right side of curve ➂, the spectrum of frequency f 2 appears at curve ➃.

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10 Signal Processing for Rotor Vibration Diagnosis

If the frequency f 1 is ➀ and the frequency f 2 is ➃, we can distinguish between these two frequency components. This is termed the resolution and is determined by the following equation: Resolution ≡ | f 1 − f 2 |  2 × WF ×  f

(10.34)

Example 10.7 When analyzing motor vibrations caused by electromagnetic excita-

tion, we want to perform a FFT analysis to distinguish the difference between 2f 0  120 Hz, i.e., two times the line frequency f 0 , and 2f r  119 Hz, i.e., two times the rotational speed f r . We use a Hanning window and set a maximum frequency f max  500 Hz. Find (1) the number of lines L and (2) the time window T . Answer (1) Substitute f 1  120 Hz, f 2  119 Hz, WF  1.5 and f  f max /L  500/L into Eq. (10.34). Then, 1  2 × 1.5 × 500/L  1500/L → L  1600→ Select number of samplings N  212  4096. (2) T  1/f  1600/500  3.2 s. Example 10.8 In the same situation as in the previous Example 10.7, but with the

window function changed to a rectangular window, what is (1) the number of lines L, (2) the time window T , and (3) the data collection time with 25% overlap and averaging of 10 collections? Answer (1) 1  2 × 1 × 500/L  1000/L → L  1600 (number of samplings N  4096). (2) T  1/f  1600/500  3.2 s. (3) 3.2 + 0.75 × 3.2 × 9  24.8 s.

10.5.6 Overall (OA) The Overall is the power that synthesizes all of the frequency spectra. It coincides with the time domain power (mean square). Hence, assuming that an anti-aliasing filter works ideally and the input wave is approximated with the number of display lines L without the aliasing error, it follows that Overall OA 

√

2 × rms

rms  root mean square ≡ square root of the mean square of a value

(10.35)

10.5 Practicalities of FFT Analysis

 ≡

N −1 1  2 w N k0 k

1/2

⎡ 1 ⎣ T

335

T

⎤1/2 w(t) dt ⎦ 2

 

f 02

|F1 |2 +|F2 |2 · · · |FL |2 + 2

1/2

0

(10.36) If the DC component, f 0  0, then:

1/2 OA  |F1 |2 +|F2 |2 · · · |FL |2

(10.37)

Let us try this calculation for the previous case of Fig. 10.28, with the resolution of 1 Hz and 400 lines. Figure (a) is an input wave of a single frequency (10 Hz) with amplitude a that is synchronous. We read the following values from the graph and the OA values are then evaluated: Rectangular window: OA  a √ Hanning window: OA  (12 + 0.52 × 2)1/2 a  1.5a √ Flat top window: OA  (12 + 0.9562 × 2 + 0.6452 × 2 + 0.1942 × 2)1/2 a  3.8a Though the actual overall amplitude is the same amplitude a of the input function, these OA values are over-estimated by the square root of the window factor, WF, due to the influence of the window function. The definition of Eq. (10.36) is inconvenient, because of the disagreement due to over-estimation. In order to avoid this inconvenience, actual FFT analyzers define rms and OA in the following modified forms: 1

rms  √ WF



|F1 |2 +|F2 |2 + · · · + |FL |2 + 2 √ OA  2 × rms

f 02

1/2 (10.38) (10.39)

where WF  1 for a rectangular window, WF  1.5 for a Hanning window and WF  3.8 for a flat top window. Example 10.9 Using the spectral graphs in Fig. 10.28, estimate the OAs for 9.6 Hz [in graph (d)] and 20.4 Hz [in graph (e)], which are asynchronous with the window and verify that they agree with the actual amplitude a.

Answer Since both 9.6 and 20.4 Hz are asynchronous by 0.4 Hz, they have the same power. So we read one from both graphs. We read several values of comparably high spectral distribution for 9.6 Hz, and calculate the OA values as follows: Rectangular window: OA  (0.762√ + 0.52 + 0.222 + 0.192 + 0.132 + 0.122 )1/2 a/ 1  a √ Hanning window: OA  (0.92 + 0.792 + 0.232 + 0.122 )1/2 a/ 1.5 √a Flat top window: OA  (12 + 12 + 0.882 + 0.822 + 0.452 + 0.352 )1/2 a/ 3.8  a

336

10 Signal Processing for Rotor Vibration Diagnosis 5 Voltage [V]

Fig. 10.31 FFT analysis

0 −5

Input wave 1 + 2 sin ( 2 π × f t ) 0

2

Time [s]

5 Voltage [V]

DC component ?

0

? OA

? f component

0

200 Frequency [Hz]

Note: Regardless of whether a synchronous or asynchronous function is under consideration, this example concludes that the square-sum of the spectra of a 1 V sine wave coincides with the WF window factor: |F1 |2 +|F2 |2 · · · |FL |2  WF Example 10.10 Figure 10.31 shows a waveform and spectrum on a FFT analyzer. The time window is 2 s, the maximum display frequency is 200 Hz, and the number of lines is 400. The window functions used are rectangular and Hanning windows. The input waveform  1 + 2sin2π ft (f  10 Hz for synchronous or f  10.25 Hz for asynchronous). Each value of the DC component at f  0 Hz, a peak near f  10 Hz, and an OA value ( symbol at the right edge) can be seen in the spectral graph. Fill in the blanks of spectrum values in Table 10.1, indicating the results of FFT analysis.  √ 22  2.46, ➅  2, Answer ➀  ➄  1, ➁  1.28, ➂  ➃  ➇  ➈  2 1 + 2 ➆  1.7 Example 10.11 Figure 10.32 shows the results of the FFT (f max  500 Hz, number of lines 400, Hanning window). Estimate the frequency and amplitude, a, of the actually inputted sine wave with a single frequency. Also, what is the value of the OA?

Table 10.1 FFT analysis Input frequencies 10Hz 10.25Hz

Rectangular window DC component

Component around 10Hz

1

2

1

2

Hanning window OA

DC component

Component around 10Hz

OA

3

5

6

8

4

1

7

9

Displacement [μm]

10.5 Practicalities of FFT Analysis

337

6

4.04

4 1.51

2 0.04 0

?

5.15

0

1.25

0.48

0.15 57.5

60 Frequency [Hz]

0.09

62.5

0.03 65 500

Fig. 10.32 Spectrum T  0.8 s, f max  500 Hz, L  400

Answer Approximately 60.4 Hz. Excluding the DC component,

√ a  (1.512 + 5.152 + 4.042 + 0.482 · · ·)1/2 / 1.5  5.5 Including the DC component, 1/2 1/2   √ √ a2 5.52 2  22 + 2  6.18 OA  f 02 + 2 2 1/2  1 22  2. where the DC component, f 0  √ 22 + 2 1.5

Note on how to find the frequency: Plot {5.15, 4.04}/a  {0.94, 0.748}  {a1 , a2 } in Appendix Fig. C.6, and find the most suitable frequency difference on the horizontal axis. It will be 0.32. Then, 0.32 × (500/400)  0.4 → 60 + 0.4  60.4 Hz. Another Answer Plot the ratio of (the secondly highest amplitude/the highest amplitude)  4.04/5.15  0.784  −2.1 dB in Appendix Fig. C.7, and find the most suitable frequency difference  0.32 on the horizontal axis. It yields 60.4 Hz as mentioned above.

10.5.7 Application of FFT Analyzers [277, 278] The theory of spectral analysis by means of the discrete Fourier transformation is applied to many and various fields; mechanical vibration, audio signal processing, CG image processing, medical technology, computational finance, and so on. FFT analyzers that utilize this theory are indispensable instruments for the field of mechanical vibration problems. (A) Impulse test FFT analyzers are very convenient for impulse testing in the mechanical vibration field, as shown in the right-hand part of Fig. 10.33. The engineer hits lightly a part of a machine structure with an impulse hammer and detects the resulting vibration by

338

10 Signal Processing for Rotor Vibration Diagnosis

(a)

(b)

Acceleration sensor

Waveform (1)

Impulse test Window time T

Spectrum (amplitude) (2)

34 Hz A structure (test specimen)

5 Hz Frequency resolution Δ f = 0.125

Charge amplifier

fmax FFT

Fig. 10.33 Impulse test using FFT analyzer

an acceleration sensor attached to the structure. This impulse acceleration waveform is passed through a FFT analyzer to conduct the spectral analysis. An example of such analysis is shown in the left-hand part of the figure. The input waveform (1) is measured by means of “zero start” and “zero end.” The wave is seen not to be a single frequency—rather, it is somewhat fluctuating because of the superposition consisting of several frequency components. These frequency components can be made clear from the peaks in the amplitude spectrum (2). Some of these peak frequencies indicate natural frequencies of the machine structure. A machine’s natural frequency is the frequency at which it will resonate when excited by external sources of its engine, etc., and is often regarded as the machine’s “weak point” frequency. Avoidance of such resonance is one of the key features of machine design that can be verified by using a FFT analyzer. (B) Vibration diagnosis A FFT analyzer may be useful for on-site measurement of vibration problems. As shown schematically in Fig. 10.34, it can be used to measure the vibration of various points on a machine and to analyze the frequencies. This enables one to estimate problematic frequencies that are related to mechanical malfunctions. According to diagnosis tables indicating the relationship between resulting frequencies and vibration troubles, we investigate the cause using the results from the FFT analyzer.

10.5 Practicalities of FFT Analysis

339 Gear

Bearings

Coupling

Vibration level

Motor

Frequency Low freq. band

Middle freq. band

High freq. band

· Unbalance

· Gear meshing

· Rolling bearing vib.

· Misalignment

· Higher order resonance etc.

· Cavitation excited vib. etc.

· Self-excited vib. etc.

Fig. 10.34 Vibration frequency band depending upon malfunctions f (t)

Input wave

y (t)

4

Transient response

Steady state response

60

2 0

t

t

0

−2 −4

−60

G (s)

G (s = j n ω )

Fig. 10.35 Response under input of periodic functions

(C) Frequency Response Analysis (FRA) [96] Consider a steady-state response, which is given by a particular solution in mathematics when the harmonic excitation acts as the input of a linear system, having a transfer function G(s). As shown in Fig. 10.35, the general responses of the system include a somewhat transient period just after the input starts and periodically steady-state responses after a settling time. Below we describe the frequency response analysis (FRA) method, which is able to find the output steady-state response in a simple manner. The input is a synchronous function of the fundamental frequency ω. As understood from Eq. (10.19),

340

10 Signal Processing for Rotor Vibration Diagnosis

the synchronous function, f (t), is replaceable with the sum of harmonic functions cos(nωt), based upon a Fourier series expansion: f (t)  f 0 +



Fn (t) ≡ f 0 +



n

f n cos(nωt − αn )

(10.40)

n

Thus, as shown by the waveforms on the left side in Fig. 10.36, the sum of the input DC component, the fundamental frequency, ω, and higher-order frequencies, nω, components can be considered as being input independently to the system. The right side figure of Fig. 10.36 shows the steady-state response, y(t), of this linear system, as expressed by the sum of the respective frequency responses: y(t)  a0 +



an cos(nωt − αn + θn )  a0 +

n



Re[an e jθn e j(nωt−αn ) ]

(10.41)

n

Since the complex amplitude of a steady-state response for the input of sin(ωt) or cos(ωt) function is generally equivalent to G(jω), the input–output relation in the present case can be expressed by the following equation: a0  G( j0) f 0 , a1 e jθ1  G( jω) f 1 , · · · , an e jθn  G( jnω) f n , · · · (10.42) If the steady-state response is focused on amplitude only and the phase is ignored, the input–output relation becomes even simpler. As shown in Fig. 10.37, the input amplitude spectrum, f , is: f ≡



f0 f1 f2 · · · fn · · ·



(10.43)

The output response amplitude spectrum, a, is determined by the following equation:

1 n=1

n=2

0

6

7

8

9

t

10

f (t)

0

t

6

7

8

9

10

f (t)

0

t

−10 y (t)

G(s)

6

7

8

9

10 [s]

(a) Input wave Fig. 10.36 FRA: frequency response analysis

s = jω 3

(b)

t 6

7

8

9

10

6

7

8

9

10

60 0

t

−60 y (t)

G(s)

10 0

s = jω 2

3

−3

G(s) s = jω 1

−1 2

−2

n=3

y (t)

f (t)

20 t

0 −20

6

7

8

9

10 [s]

(c) Steady state output wave

10.5 Practicalities of FFT Analysis 4

341

30

×

2 1 0

Gain

3

1 2 3 4 5 Frequency [Hz]

(a) Input spectrum

20

=

10 0

1 2 3 4 Frequency [Hz]

100 80 60 40 20

5

(b) G( j n ω ) : Transfer function

?

0

1 2 3 4 5 Frequency [Hz]

(c) Output spectrum

Fig. 10.37 Calculation of FRA

 a  G( j0) f 0 , |G( jω)| f 1 , |G( j2ω)| f 2 , · · · , |G( jnω)| f n , · · ·

(10.44)

The steady-state response analysis, called frequency response analysis (FRA), is carried out in the frequency domain as follows: “input spectrum × transfer function  response spectrum.” Also, when the input and output response signals have been obtained in the time domain, it will be possible to convert each into the frequency domain and identify the transfer function by ratio: “transfer function  output response spectrum/input spectrum.” Example 10.12 The input waveform as shown in the left-hand side f (t) in Fig. 10.35 consists of the corresponding frequency components of Fig. 10.36a and corresponds to the spectral graph of Fig. 10.37a. The transfer function, G(s), is shown in the center of Fig. 10.37b. Find the resonance peak value at 2 Hz of the output spectrum?

Answer 62 at 2 Hz.

10.6 Zooming of a FFT Analyzer In motor vibration diagnosis, denoting the power line frequency as f [Hz] and the rotational speed by f r [rps], where f r is slightly lower than f due to slipping, we frequently observe the power supply induced vibration of 2f and the rotation-induced vibration of 2f r . These frequencies must be precisely distinguished for troubleshooting activities. For a detailed analysis with such conditions, we need the zoom function of the FFT, as described below.

10.6.1 FFT of the Original Signal While maintaining theoretical generality, let us consider the example of a waveform mixing three frequency components, which are very close to each other, as follows:

342

10 Signal Processing for Rotor Vibration Diagnosis

Fig. 10.38 FFT analysis

10

Amplitude

(a)

Zooming

0 110

120

125

130

10

Amplitude

(b)

115

0 0

100

200 300 Frequency [Hz]

w(t)  a1 cos(2π f 1 t + θ1 ) + a2 cos(2π f 2 t + θ2 ) + a3 cos(2π f 3 t + θ3 )

400

(10.45)

where a1  2, f 1  117 Hz, a2  10, f 2  118.5 Hz, a3  9, f 3  120 Hz, θ1  θ2  θ3  0. Figure 10.38a shows a result of a FFT analysis for this input waveform, with the following parameters: the time window T  1 s, frequency resolution f  1 Hz, number of samplings N  1024, frequency display band is 0–400 Hz and the number of lines L  400. A set of peaks is observed in the neighborhood of 120 Hz in figure (b). This portion was zoomed in geometrically. However, the details of the various frequency components are still unclear, although several peaks can be seen a little below 120 Hz. It will be understood that with these FFT parameters, it is difficult to distinguish the frequency of twice the power frequency  2f  f 3  120 Hz from the frequency of twice the rotational speed vibration  2f r  f 2  118.5 Hz, because they are too similar to each other. For figure (b), the FFT zoom function is evidently to display the frequency range of 110–130 with L  400.

10.6.2 Frequency Shifting and LPF The principle of the zoom function is frequency shifting by multiplication of harmonic waves. This involves generating sine and cosine waves having the zoom center frequency, f c , which is set by f c  f 3  120 Hz in this case, and multiplying them by the original signal w(t): w(t) × cos(2π f c t) 

1 1 ai cos[2π( f c + f i )t + θi ] + ai cos[2π( f c − f i )t − θi ] 2 2 3

3

i1

i1

10.6 Zooming of a FFT Analyzer

343

Fig. 10.39 FFT analysis

2 10V 118.5 Hz 1 9V 120 Hz

3 2V 117 Hz 0

10

LPF

10

237 238.5 240

2 5V 1.5 Hz 1 4.5V 0 Hz

3 1V 3 Hz

280

400 [Hz]

(a) Original signal

540 [Hz]

(b) Frequency shift

w(t) × sin(2π f c t) 

1 1 ai sin[2π( f c + f i )t + θi ] + ai sin[2π( f c − f i )t − θi ] 2 2 3

3

i1

i1

(10.46) These multiplications shift each frequency component: {f c + f i }  120 + {117,118.5, 120}  {237,238.5, 240}  approximately 240 Hz {f c − f i }  120 − {117,118.5, 120}  {3, 1.5, 0} Hz  very low frequencies Figure 10.39 illustrates how this shifting is done. After the shifting, since the area to be focused on is the low-frequency area around 0 Hz (shaded portion), the area around 240 Hz that is unnecessary in this case is eliminated by a low-pass filter. FFT processing of the waveform involving the shaded portion provides the zooming spectral display. This sequential processing for zooming is illustrated in Fig. 10.40. The waveform of Figure (a) is the original input signal, seen as beating due to mixing of three close frequencies f i (i  1–3). The results of multiplying the input by sine and cosine waves of the center frequency are shown in Figure (b), where the beating wave comes from the mixing of shifted frequency components of {f c − f i }. The high-frequency components are removed by an LPF (cut-off frequency  10 Hz) as shown in figure (c). After the LPF, we see only the low-frequency {f c − f i }  {3, 1.5, 0} Hz components that have passed through the LPF. This is expressed by the following equations: wc (t) ≡ LPF[w(t) × cos(2π f c t)] 

1 ai cos[2π ( f c − f i )t − θi ] 2 i1

ws (t) ≡ LPF[w(t) × sin(2π f c t)] 

1 ai sin[2π ( f c − f i )t − θi ] 2 i1

3

3

(10.47)

344

10 Signal Processing for Rotor Vibration Diagnosis 20

(a) Original w (t)

0 −20

0

0.5

×

1

1.5

× cos 2 π fc t

× sin 2 π fc t

20

(b) Multiplication 10

0

0

0.2

0.4 0.6 Time [s]

2 Time [s]

0.8

1

10 0 −10

0.6 0

0.2

0.4 0.6 Time [s]

0.8

1

10

(c) After LPF wc (t)

0

4 0 −4 0

5 Time [s]

10

0

20

5 Time [s]

10

20

Aliasing error

LPF

10 0

0 1 2 3 4 5 6 7 8 9 10 Frequency [Hz]

Amplitude

(d) DFT

Amplitude

FFT 10 0

0 1 2 3 4 5 6 7 8 9 10 Frequency [Hz]

Fig. 10.40 Wave processed by frequency shift and its FFT results

In this way, though the amplitude has been reduced by half, the frequency axis has been shifted as being the difference from 120 Hz. Looking at the filtered waveform of figure (c), the high-frequency components of figure (b) are eliminated and only the low-frequency components appear around 0 through 3 Hz. This elimination of highfrequency components is essential for the prevention of an aliasing error. Otherwise, we would see aliasing spectra around 7 Hz through to 10 Hz as enclosed by the yellow area in figure (d).

10.6.3 DFT and Half-Spectrum Displays Rewrite Eq. (10.47) of low-frequency components by using complex amplitude: 1   − jθi 2π( fc − fi )t  Re ai e e 2 i1 3

wc (t) 

3 3  1   − jθi 2π( fc − fi )t  −1   Im ai e e Re jai e− jθi e2π( fc − fi )t  ws (t)  2 i1 2 i1

10.6 Zooming of a FFT Analyzer

345

The DFT processing gives the result in the following equations: 0 Hz 1.5 Hz 3 Hz dftc ≡ DFT[wc (t)]  {a3 cos θ3 · · · a2 e jθ2 · · · a1 e jθ1 · · ·}/2 dfts ≡ DFT[ws (t)]  {−a3 sin θ3 · · · ja2 e jθ2 · · · ja1 e jθ1 · · ·}/2

(10.48)

With the parameters of the time window span, T  20 s, frequency resolution, f  1/20 Hz, number of samplings, N  512, frequency display band, 0 − f max =10 Hz, number of lines, L  200, the above equations then give the following values (amplitude @ frequency), as confirmed in Fig. 10.40d:  dftc  9@0 Hz · · · [email protected] Hz · · · 2@3 Hz · · · /2  dfts  0@0 Hz · · · 10 [email protected] Hz · · · 2 j@3 Hz · · · /2

(10.49)

As seen here, the spectra are all plotted on the right side, according to their differences from the center frequency 120 Hz. Also, it is reiterated that, as indicated by the portion surrounded by broken lines in the right side of figure, if DFT (d) is carried out without the LPF (c), an aliasing error occurs around 8 Hz.

10.6.4 Spectrum Zoom Displays In this display, however, it is impossible to know whether the frequencies are higher or lower compared with the center frequency (120 Hz). Here, we denote the spectra on the higher side as dftH and those on the lower side as dftL and define them by means of the following equations: dft H  dftc + jdfts dft L  dftc − jdfts

(10.50)

This conversion gives the following values:  dft H  4.5@120 Hz · · · [email protected] Hz · · · 0@123 Hz · · ·  dft L  4.5@120 Hz · · · [email protected] Hz · · · 2@117 Hz · · ·

(10.51)

Gathering the DFTs of both enables a zoom display of the spectra. The result is:  {dft L } + {dft H }  · · · 2@117 Hz · · · [email protected] Hz · · · 9@120 Hz

(10.52)

This display is shown in Fig. 10.41. The spectra are plotted on both sides, according to their differences from the center frequency 120 Hz.

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10 Signal Processing for Rotor Vibration Diagnosis

Fig. 10.41 Zooming of FFT

Amplitude

10

5

0 110

w (t)

×

wc

LPF

DFT

+

120 125 Frequency [Hz]

130

Higher comp. Zoom

+

cos ωc t

fc − fmax

sin ω c t ×

115

+ ws

LPF

DFT

j

−

Conjugate

Lower comp. Zoom

−

fc

+

fc + fmax

Fig. 10.42 Signal processing for zooming

Example 10.13 Regarding the zoom function described in this section, summarize the signal processing for determining the amplitude spectra by a block diagram.

Answer Figure 10.42.

10.7 Full Spectrum 10.7.1 Concept By setting two displacement sensors in X and Y directions as shown in Fig. 10.43a, we focus on the orbit of the shaft center of rotation. This method is convenient to carry out diagnosis by separating forward whirl components that are the same direction as the rotor spin direction, , and backward whirl components that are the opposite spin direction. When FFT analysis is carried out using a signal of a single sensor, e.g., the X directional vibration only, the dominant vibration frequencies will be obtained, that is the speed of whirling motion. However, the direction of the whirling motion is not obvious. Then, by observing the vibration in both the X and Y directions, the whirling speed and direction are analyzed, i.e., enabling the use of the full spectrum. The full spectrum makes it possible to diagnose phenomena, such as forward whirl

10.7 Full Spectrum

347 y

Y Dir.

X Dir.

Ω

y Ω

Ω

x

0 +10 Hz

−30 Hz

(a) XY Displacement

x

Spectrum

x

Pulse

y

f

(b) Full spectrum

(c) Orbit

Fig. 10.43 Full spectrum and whirling motions

motion featuring oil whip and backward whirling components caused by rubbing, more precisely. Forward whirl frequencies are displayed as positive values and backward whirl vibration frequencies as negative values. For example, when spectra are displayed as in Fig. 10.43b, the +10 Hz peak spectrum will be read as indicating the occurrence of 10 Hz forward whirl motion. At the same time, we can see the −30 Hz peak spectrum, finding that simultaneously a 30 Hz backward whirl motion is contained in the orbit of the shaft center as shown in figure (c). Even when the actual orbit has a complicated path such as shown in the figure, it will be possible to make it clear by using the full spectral display having ± frequency axis. Below we describe the signal processing to obtain this full spectrum.

10.7.2 One-Channel FFT To begin with, let us review analysis using one-channel FFT analyzer (1-channel FFT). When an input waveform is expressed in the form of complex amplitude, the resultant FFT values are the conjugate values of the complex amplitudes, as shown in following examples: Input waveform → Complex amplitude form ] a cos ωt  Re[ae jωt   j ωt−90

→

FFT values a

◦

]  Re[− jbe jωt ] b sin ωt  Re[be a cos ωt + b sin ωt  Re[(a − jb)e jωt ]

jb a + jb

Initially, sampling values of the input waveform are put into N memories of the 1-ch FFT analyzer, and after the DFT calculations, they are all replaced with FFT values. Usually, the magnitudes |a + jb| of the memory contents are taken and displayed as the spectral amplitudes. Each of memory consists of a real part and an imaginary part (a, b).

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10 Signal Processing for Rotor Vibration Diagnosis

10.7.3 Two FFTs for Full Spectrum How to produce a full spectrum using two 1-channel FFT analyzers is now introduced. As shown below, for a given frequency ω, FFT values corresponding to the X and Y directional vibration signals x(t) and y(t) can be written as: x(t)  a cos ωt + b sin ωt → 1-ch FFT → FFT value  a + jb

(10.53)

y(t)  c cos ωt + d sin ωt → 1-ch FFT → FFT value  c + jd

(10.54)

Letting the x-y plane be a complex plane, orbit is: z(t)  x(t) + j y(t)

(10.55),)

the orbit of the present case: z(t)  a cos ωt + b sin ωt + j(c cos ωt + d sin ωt)  (a + jc) cos ωt + (b + jd) sin ωt 1 1  (a + jc)(e jωt + e− jωt ) + (b + jd)(e jωt − e− jωt ) 2 2j 1 1  [a − jb + j(c − jd)]e jωt + [a + jb + j(c + jd)]e− jωt (10.56) 2 2 Thus, the complex amplitude is derived mathematically from the above equation. Hence: Frequency + ω → forward complex amplitude  [a − jb + j(c − jd)]/2 (10.57) Frequency − ω → backward complex amplitude  [a + jb + j(c + jd)]/2 (10.58) x(t) y(t) X Dir.

FFT

a+jb

1/2

FFT

c+jd

j/2

+

Backward

+

complex amplitude

Spectrum

Y Dir.

1/2 Orbit

Fig. 10.44 Signal processing for a full spectrum

j/2

−300 Hz + −

Conjugate

+200 Hz f

Forward complex amplitude

10.7 Full Spectrum

349

The FFT calculations for these Eqs. (10.57) and (10.58) are realized by using two 1-ch FFT analyzers as described in the block diagram in Fig. 10.44. The rearrangement of each FFT calculation result of the two x(t) and y(t) waveforms provides two complex amplitude values corresponding to the positive and negative frequency areas of a full spectrum.

10.7.4 Example of Vibration Diagnosis Using a Full Spectrum The book written by Bently [B37] gives a good example of vibration diagnosis using a full spectrum, as referred below. Figure 10.45 illustrates case studies of rubbing vibration. The results of a 1-ch FFT for a X directional waveform and those for a Y directional waveform are shown. It is seen that 1X and 1/2X vibrations are predominant. A full spectrum is also shown in the figure, as obtained by rearranging results of these X and Y directional FFT calculations. The full spectral graph indicates two main orbits: positive (higher height) and negative (lower height) frequency spectra of 1X vibration, i.e., forward elliptical orbits, and they generally have similar magnitudes of positive and negative frequency spectra of 1/2X vibration, i.e., a sharp elliptical orbit is seen almost as a one-directional straight line, indicating some strong rubbing effects. In fact, actual orbits can be inferred by the combination of the both orbits, as shown in figure ➀. In figure ➁, the forward 1X vibration spectrum is higher compared with other components in spectra, so that the forward 1X orbit is dominant and the state seems normal, showing no evidence of strong rub. Figure 10.46 illustrates a case of oil whip. As shown in the 1-channel FFT, vibration of approximately 1/2X predominates, closely resembling the 1/2X rubbing vibration mentioned above. However, looking at the full spectrum for this case, it can be seen that the problematic approximately 1/2X vibration is a forward whirl motion. Also, similar orbits are shifted little by little without tracing the same orbital locus. Hence, it can be identified as vibration due to oil whip. As seen in these case studies, a full spectrum can provide information regarding forward/backward whirl orbits, which is very effective for diagnostic identification.

0

4

20

40 Time [ms]

60

X

2 0 -2 -4

0

20

4 2 0 -2 -4

10

40 Time [ms]

60

Y

X

-4 -2 0 2 4 4410 rpm

Y

5 0

X

-5 -10

1 mil/div

2580 rpm

Amplitude [mil p-p]

-4

Amplitude [mil p-p]

2

0 -2

10 8 6 4 2 0 10 8 6 4 2 0

Amplitude [mil p-p]

Weak rubbing (turbine)

2

Orbit Displacement [mil]

1

Y

Orbit Displacement [mil]

Displacement [mil]

Strong rubbing

4

10 8 Full spectrum +1X 6 +1/2X 4 -1X -1/2X 2 0 -6 -4 -2 0 2 4 6 Backward Forward Whirling frequency [kcpm]

Amplitude [mil p-p]

10 Signal Processing for Rotor Vibration Diagnosis Displacement [mil]

350

10

1/2X Y-FFT 1X

0

1 2 3 4 5 Frequency [kcpm]

1X

X-FFT

1/2X 0

1 2 3 4 5 Frequency [kcpm]

6

+1X

Full spectrum 5

6

-1/2X +1/2X -2X -1X

+2X

+3/2X -3/2X 0 -6 -4 -2 0 2 4 6 Backward Forward Whirling frequency [kcpm]

Fig. 10.45 Vibration diagnosis of rubbing Fig. 10.46 Vibration diagnosis of oil whip

20 10 0 −10 −20

Y

20

0.49X

15 X

Y-FFT 1X

10 5

−20 −10 0 10 20

0

5 070 rpm 20 Fluid-induced instability 15 10 5

0

1 2 3 4 5 Frequency [kcpm]

6

X and Y Full spectrum +0.49X +1X

−0.49X −1X 0 −6 −4 −2 0 2 4 6 Forward Backward Whirling frequency [kcpm]

10.7 Full Spectrum Ω

351 Ω

Forward

ω

ω

Ω

Ω

Ω

Backward

ω

ω

Forward

ω

Backward

Fig. 10.47 Whirling orbits +1X

+1X −1X

−1X R

F

R

F

−1X −1X

+1X

−1X

R

F

R

+1X F

R

+1X F

Fig. 10.48 Full spectrum (F: Forward, B: Backward)

Example 10.14 Corresponding to the whirl orbits shown in Fig. 10.47, draw the full

spectrum. Answer Figure 10.48. Note: In our studies of rotor orbit analysis, multi-frequency analysis and forward/backward analysis, we referred to the following references [275–285]. We can see further information of signal processing related to rotor dynamics in the references [286–290].

Chapter 11

Our Latest Topics Relating to Simplified Modeling of Rotating Systems

Abstract This chapter is divided into three parts, which include our latest areas of interest. (1) Guyan reduction or mode synthesis: With respect to vibration analysis for actual machines, many types of codes, 3D-FEM and 1D-FEM (e.g., MyROT introduced in R1_Chap. 12), are available for engineering design based upon multiDOF (degree of freedom) modeling. However, case studies of vibration problems show the benefits for understanding the vibration mechanisms and their practical and potential solutions using simplified models. In this book, Guyan reduction and the mode synthesis technique are strongly recommended for this purpose. (2) System instability in oil film bearing rotors: This model order reduction (MOR) is applied to a flexible rotor supported by oil film bearings to analyze system instability, often referred to as oil whip, oil whirl, and casing whirl. Stability analysis is generally performed by using complex eigenvalue codes in the 1D-FEM modeling. However, this chapter yields an alternative way without complex eigenvalue calculation, i.e., how to predict the limiting speed for stability and the resultant unstable frequency. (3) Blade modeling coupled with shaft vibration: The MOR by the mode synthesis technique is applied to a flexible blading system to analyze the vibration coupled with the rotating shafting system. Uncoupled eigen frequencies and modes may be analyzed initially for the rotating blading system only by a 3D-FEM code, based upon the rotating coordinates. As the result, the corresponding 2-DOF models for each eigen mode of blading are proposed by two methods depending upon the number of the nodal diameter, κ  0 and κ  1. The former is coupled with torsional and axial shaft vibration models and the latter with bending (lateral) shaft vibration models, respectively. These blade reduced models are combined into shaft vibration data by a 1D-FEM code for the evaluation of the coupling effect. The effectiveness of these procedures is confirmed by numerical examples. Keywords Model order reduction (MOR) · Guyan reduction · Mode synthesis · Eigen frequency · Oil whirl/whip · λ model · Blade–shaft coupled vibration · 1D-FEM · 3D-FEM

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_11

353

354

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

11.1 Practical Techniques Recommended for Model Order Reduction (MOR) In R1_Chap. 4, we introduced an overview of several modeling techniques for mechanical vibration systems, with Guyan reduction and mode synthesis models being recommended for vibration problems in practical engineering fields. Here, we again discuss these modeling techniques as applied to a 5-DOF mass-spring (m–k) system and other case studies for evaluating the relative advantages and disadvantages of both methods.

11.1.1 Procedures for Guyan Reduction For the 5-DOF mass-spring system shown in Fig. 11.1(1), we use a reduced 3-DOF system consisting three masses colored pink. These selected masses are called master coordinates (equivalently called the “A-set” in Nastran), while the other masses are called slave coordinates. Here, of the five full-order coordinates noted by {x 1 , x 2 , x 3 , x 4 , x 5 }, three coordinates {x 1 , x 3 , x 5 } are designated as master, while the other two {x 2 , x 4 } are as slave. According to this master-and-slave assignment, the mass and stiffness matrices should be rearranged. To do so systematically, we define a coordinate transformation matrix from {x 1 , x 2 , x 3 , x 4 , x 5 } to {x 1 , x 3 , x 5 } and {x 2 , x 4 }, noted by T B , as follows: ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ x1 10000 x1 x1 ⎢x ⎥ ⎢ x ⎥ ⎢ 0 0 0 1 0 ⎥⎢ x ⎥ ⎥⎢ 3 ⎥ ⎢ 3⎥ ⎢ 2⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (11.1) ⎢ x 3 ⎥ ≡ ⎢ 0 1 0 0 0 ⎥⎢ x 5 ⎥ ≡ T B ⎢ x 5 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ x2 ⎦ ⎣ x 4 ⎦ ⎣ 0 0 0 0 1 ⎦⎣ x 2 ⎦ 00100 x5 x4 x4

Fig. 11.1 5-DOF and Guyan’s reduction by δG  [δ1 , δ2 , δ3 ]

x

x

1

1

(1)

x

2

1

1

1

x

3

1 1

x

4

1 1

1

2

2 0.5

1 0.5

1 2 1

(2)

G

3

0.5

0.5

1

5

11.1 Practical Techniques Recommended for Model Order Reduction (MOR)

355

The stiffness matrix, K, is rearranged automatically by a congruence transformation (i.e., TBt K T B ) as follows: ⎡ x1 x2 x3 x4 x5 ⎤ 3 −1 0 0 0 ⎢ −1 2 −1 0 0 ⎥ ⎢ ⎥ → K  ⎢ ⎥ ⎢ 0 −1 2 −1 0 ⎥ ⎢ ⎥ ⎣ 0 0 −1 2 −1 ⎦ 0 0 0 −1 3 ⎡ x1 x3 x5 x2 x4 ⎤ 3 0 0 −1 0 ⎢ 0 2 0 −1 −1 ⎥ ⎥ TBt K TB  ⎢ ⎢ ⎥ ⎢ 0 0 3 0 −1 ⎥ ⎢ ⎥ ⎣ −1 −1 0 2 0 ⎦ 0 −1 −1 0 2

≡

x x x x x 1 3 5 2 4 K 11 K 12 K 21 K 22

(11.2)

While the static state of restricting all master coordinates to zero indicates a nondeformed system, we release one of these master coordinates and move it with unity displacement. At the same time, the slave coordinates are then forcibly moved by certain displacements. We alternate this unity-moving point for all master coordinates one by one, and then calculate the forced displacement of slave coordinates at each step. To input the unity displacement of {x 1 , x3 , x 5 }, the resulting deflection, δ G , as shown in Fig. 11.1(2), of slave points {x 2 , x 4 } can then be calculated as follows: ⎤ ⎡ ⎤ ⎡ ⎤   x1 x1 x1 x x [K 21 ]⎣ x3 ⎦ + [K 22 ] 2  0 → 2  −[K 22 ]−1 [K 21 ]⎣ x3 ⎦  [δG ]⎣ x3 ⎦ x4 x4 x5 x5 x5 (11.3) ⎡ ⎤ ⎡ ⎤  x1 x1 x [K 11 ]⎣ x3 ⎦ + [K 12 ] 2 ≡ K eq ⎣ x3 ⎦ → K eq  [K 11 ] − [K 12 ][K 22 ]−1 [K 21 ] x4 x5 x5 ⎡

Note that the force needed to generate the unity displacement refers to the equivalent stiffness denoted by K eq as shown above. The coordinate transformation, T G , for the scaled-down form of the global coordinates of the 5-DOF system {x 1 , x 2 , x 3 , x 4 , x 5 } to the master coordinates of the 3-DOF {x 1 , x 3 , x 5 } is defined as follows: ⎡ ⎤ ⎡ ⎤ 1 0 0 x1 ⎤ ⎡ ⎤ ⎡ ⎢ 0.5 0.5 0 ⎥⎡ x ⎤ ⎢x ⎥  x1 x1 ⎢ ⎥ 1 ⎢ 2⎥ E3 ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ (11.4) x3 ≡ TG ⎣ x3 ⎦  ⎢ 0 1 0 ⎥⎣ x3 ⎦ ⎢ x 3 ⎥  TB ⎢ ⎥ ⎢ ⎥ δG ⎣ 0 0.5 0.5 ⎦ x5 ⎣ x4 ⎦ x5 x5 x5 0 0 1

356

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

where E 3 is the three-dimensional identity matrix. The original mass and stiffness matrices (M and K) are also scaled down by the congruence transformation to obtain the reduced mass and stiffness matrices, M G and K G . Therefore, the equation of motion of a scaled-down 3-DOF system (called the Guyan reduced model) is written as follows: MG X¨ G + K G X G  0

(11.5)

where ⎡

⎡ ⎤ ⎤ 1.25 sym. x1 ⎦, X G ≡ ⎣ x3 ⎦ MG ≡ TGt M TG  ⎣ 0.25 1.5 x5 0 0.25 1.25 ⎡ ⎤ 2.5 sym. ⎦  K eq K G ≡ TGt K TG  ⎣ −0.5 1 0 −0.5 2.5 Thus, natural frequency, ω g , and eigenvector, φg , are calculated from the following eigenvalue problem: ωg2 MG φg  K G φg

(11.6)

In the case of Fig. 11.1, these solutions of the obtained 3-DOF reduced model are given as:

 eigenvalues ωg2  0.42 2 2.72

 Compared with the exact values, ωe2  0.38 1.38 2.62 3.62 4 , of the 5-DOF system, the relative errors of ε in each eigenvalue are evaluated as follows: 2

ωg − ωe2



 ε≡  0.099 0.45 0.04  −20 −7 −28 dB (11.7) ωe2 The error of the first mode is very low as indicated by ε  0.099  −20 dB, compared with the second mode. Though the 10% error for the first mode would be permissible in a practical sense, the accuracy is insufficient for computational codes. Generally speaking, an accuracy less than 1%  −40 dB may be necessary at least. In order to improve accuracy, we must increase the number of master coordinates.

11.1.2 Mode Synthesis Technique As shown in Fig. 11.2, we designate {x 1 , x 5 } of the left and right masses as the boundary coordinates (equivalently, the “A-set” in Nastran), and more important

11.1 Practical Techniques Recommended for Model Order Reduction (MOR) Fig. 11.2 5-DOF and mode synthesis modeling [δ1 , δ2 , φz ]

x

x

1

1

(1)

x

2

1

1

1

357 x

3

1

x

4

1

5

1 1

1

2

2 1

0.75

0.5

1 (2) 2

0.25

0.5 (3)

0.5

0.25

q

0.75 1

0.707

0.5

z

master coordinates such as bearing potions and other points as the inner system (equivalently, the “q-set” in Nastran). The A-set produces the deflection modes (δ q ) described in the previous section, and the q-set is expressed by modal modes (φz ) that are eigenvectors of the inner system. The transformation matrix (TB ) used to separate the boundary coordinates {x 1 , x 5 } and others is defined as follows: ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ x1 10000 x1 x1 ⎢x ⎥ ⎢ x ⎥ ⎢ 0 0 1 0 0 ⎥⎢ x ⎥ ⎥⎢ 5 ⎥ ⎢ 5⎥ ⎢ 2⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ (11.8) ⎢ x 3 ⎥ ≡ ⎢ 0 0 0 1 0 ⎥⎢ x 2 ⎥ ≡ T B ⎢ x 2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ x3 ⎦ ⎣ x 4 ⎦ ⎣ 0 0 0 0 1 ⎦⎣ x 3 ⎦ 01000 x5 x4 x4 The stiffness matrix is then rewritten as follows: ⎡ x1 x2 x3 x4 x5 ⎤ 3 −1 0 0 0 ⎢ −1 2 −1 0 0 ⎥ ⎥ → K  ⎢ ⎢ ⎥ ⎢ 0 −1 2 −1 0 ⎥ ⎢ ⎥ ⎣ 0 0 −1 2 −1 ⎦ 0 0 0 −1 3 ⎡ x1 x5 x2 x3 x4 ⎤ 3 0 −1 0 0 x x x x x ⎢ 0 3 0 0 −1 ⎥  1 5 2 3 4 ⎢ ⎥ t TB K TB  ⎢ ⎥ ≡ K 11 K 12 ⎢ −1 0 2 −1 0 ⎥ ⎢ ⎥ K 21 K 22 ⎣ 0 0 −1 2 −1 ⎦ 0 −1 0 −1 2

(11.9)

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11 Our Latest Topics Relating to Simplified Modeling of Rotating …

As described in the previous section, the static deflection mode for Guyan reduction is obtained here by inputting the forced displacements at {x 1 , x 5 } and outputting the induced deflections at {x 2 , x 3 , x 4 }. However, this answer is predicted easily by the straight lines δ q as shown in Fig. 11.2(2). Deflection modes, δ q , and equivalent stiffness, K eq , are calculated as follows: ⎡ ⎤ ⎤  3/4 1/4    x2 ⎣ x3 ⎦  −[K 22 ]−1 [K 21 ] x1  ⎣ 1/2 1/2 ⎦ x1 ≡ δq x1 x5 x5 x5 1/4 3/4 x4  2.25 −0.25 K eq  [K 11 ] − [K 12 ][K 22 ]−1 [K 21 ]  −0.25 2.25 ⎡

(11.10)

Next, we treat the eigenvalue problem for the inner system consisting mass matrix M 22  Diagonal [m2 , m3 , m4 ] and stiffness matrix K 22 in Eq. (11.9), renamed as M z and K z as follows: x x x x x x ⎡2 3 4⎤ ⎡ 2 3 4 ⎤ 100 2 −1 0 Mz  ⎣ , Kz  ⎣  K 22 0 1 0⎦ −1 2 −1 ⎦ 001 0 −1 2

(11.11)

As shown in Fig. 11.2(3), this is the 5-DOF m–k system subject to the restriction of boundary displacement  0 at both ends. The eigenvalue problem is defined for the inner system as follows: ωz2 Mz φz  K z φz

(11.12)

Only the first eigen mode (φz1 ) corresponding to natural frequency ω z1 is employed here for mode synthesis modeling:  t 2 ωz1  0.586 and φz1  0.5 0.707 0.5 Using the normal coordinate η 1 of the eigen mode φz1 , the transformation matrix T q , from the 5-DOF system to the 3-DOF mode synthesis model, is defined by the sum of two deflection modes (δ q ) and one eigen mode φz1 as follows: ⎡ ⎤ ⎡ ⎤ 1 0 0 x1 ⎡ ⎤ ⎢ ⎡ ⎤ ⎥⎡ ⎤ ⎢x ⎥  x1 x1 ⎢ 3/4 1/4 0.5 ⎥ x1 ⎢ 2⎥ 0 E ⎥ ⎢ ⎥ 2 ⎣ x5 ⎦ ≡ Tq ⎣ x5 ⎦  ⎢ ⎢ 1/2 1/2 0.707 ⎥⎣ x5 ⎦ (11.13) ⎢ x 3 ⎥  TB ⎢ ⎥ ⎢ ⎥ δq φz1 ⎣ 1/4 3/4 0.5 ⎦ η1 ⎣ x4 ⎦ η1 η1 x5 0 1 0

11.1 Practical Techniques Recommended for Model Order Reduction (MOR)

359

The original mass and stiffness matrices (M and K) are also scaled down by the congruence transformation to obtain the reduced mass and stiffness matrices, M q and K q . Therefore, the equation of motion of the reduced 3-DOF system (called the mode synthesis model) is written as follows: Mq X¨ q + K q X q  0

(11.14)

where ⎡

⎡ ⎤ ⎤ 1.875 sym. x1 ⎦, X q ≡ ⎣ x5 ⎦ Mq ≡ Tqt M Tq  ⎣ 0.625 1.875 η1 0.854 0.854 1 ⎡ ⎤  2.25 sym. K eq 0 t ⎣ ⎦ K q ≡ Tq K Tq  −0.25 2.25  2 0 ωz1 0 0 0.586 The following eigenvalue problem yields natural frequencies as follows: ωq2 Mq φq  K q φq

(11.15)

 The solution yields the eigenvalues ωq2  0.382 2 2.94 . The relative errors compared with the exact values of ωe2 are thus given as follows: 2

ωq − ωe2



 ε  0.0004 0.45 0.12  −68 −7 −18 dB (11.16) 2 ωe Although the first mode accuracy of the Guyan model is insufficient as shown in Eq. (11.7), it is much improved in the case of the mode synthesis model, as verified by comparing error 0.099  −20 dB in Eq. (11.7) and error 0.0004  −68 dB in Eq. (11.16). However, the relative errors for the second and third modes remain too high, as is also the case for the Guyan model. The accuracy is generally considered as being maintained depending on the number of coordinates in the q-set. Thus, this case only guarantees the first mode. In reference to the N + 2 balancing described in R1_Sect. 5.5.3, if we want perfect balancing of the N-th critical speed for flexible modes, N + 2 correction planes should be prepared. This is called the “N + 2” balancing rule. The number of flexible modes could be equal to the number of coordinates in the q-set (i.e., number of flexible eigen modes of the inner system) as shown in Fig. 11.2(3). As the dimension of the mode synthesis model is 3 DOF in Eq. (11.14), the N + 2 rule states that the number of accurate modes is 3 − 2  1. Thus, it is well known that “the degree of the model – 2” is the limitation of the accurate mode number, which is reconfirmed in the following section.

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11 Our Latest Topics Relating to Simplified Modeling of Rotating …

11.1.3 Accuracy Comparison Between Guyan and Mode Synthesis Models Next, we consider the 11-DOF m–k system shown in Fig. 11.3, which is our target model in discussing the accuracy of both reduction models. (1) Guyan Model Prior to the calculation of the Guyan modeling, we must decide which points are to be assigned as master coordinates. As shown in Fig. 11.3, when you select the three red points as master coordinates, the 3-DOF Guyan model is obtained. As shown in Fig. 11.4, we see other examples of the selection of 5/7/9-DOF models. We selected each order of model and performed the corresponding eigenvalue calculation. The obtained frequencies (ωg2 ) were compared with the exact values   (ωe2 ) in the manner shown in Eq. (11.7). The relative error of ωg2 − ωe2 /ωe2 are summarized for each model order as shown by the red circles in Fig. 11.5. Note that lower values are better because the vertical scales are logarithmic (in dB). (2) Mode Synthesis Model Before entering the calculation of mode synthesis, we must declare what are the A-set and q-set coordinates. As shown in Fig. 11.6, we assign two boundary points (marked by red circles) at both ends (i.e., A-set). The corresponding deflection modes are two straight lines, as well as those in Fig. 11.2(1). Generally, each boundary point has varying parameters depending on the rotational speed or frequency, such as for oil film bearings and magnetic bearings, respectively. The other points are for the inner system, which provides the eigen mode of the shafting system subject to pinned-pinned restriction at the designated boundary points. In this case, if you select the first pinned-pinned eigen mode coordinate as the q-set, as shown in (1) of the figure, you obtain the 3-DOF reduced model, which

Fig. 11.3 Mass-spring system

x 1

x

1

1

1

x

2

1

1

x

6

1

1

11

1

1

1

1

Fig. 11.4 Assignment of master masses for Guyan method

x

10

Node

3-DOF 5-DOF 7-DOF 9-DOF

1

2

3

4

5

6

7

8

9

10

11

11.1 Practical Techniques Recommended for Model Order Reduction (MOR) 3 4

2

3-DOF

– 80

Error [dB]

2

4

– 20 – 40

0 Error [dB]

MS

– 60

0

Model order 10 8

Guyan

– 20 – 40

6

6

Model order 10 7 8

yan

Gu

– 60 – 80

MS

2

5 6

4

Model order 10 8

Guyan

– 20 – 40

MS

– 60

5-DOF

– 80 2

0 Error [dB]

Error [dB]

0

– 20

6

4

Model order 8 9 10

Guyan

– 40 – 60

MS

– 80 – 100

7-DOF

361

9-DOF

Fig. 11.5 Accuracy and modeling order Fig. 11.6 Mode synthesis modes

1

Node

(1)

1

2

3

4

5

6

7

8

9 10 11

1

3-DOF (2) 5-DOF

(3) 7-DOF

(4) 9-DOF

arises from the 2 boundary points of the A-set plus one eigen mode of the q-set. By including two eigen modes as shown in (2) of the figure, the 5-DOF reduced model is obtained completely. The figure also shows eigen modes in (3) for the 7-DOF and in (4) for 9-DOF reduced models obtained in the same way. We executed eigenvalue analysis for each DOF model. Expressing these calculated values as ωq2 , we evaluated the relative errors compared with the exact values. The   errors defined by the ratio ωq2 − ωe2 /ωe2 are summarized by the blue circles in Fig. 11.5.

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11 Our Latest Topics Relating to Simplified Modeling of Rotating …

Fig. 11.7 Shaft supported by two bearings and model order reduction

10 500L

xL

xR

 , A , L steel (1) kb

(2)

1 ms= 0.31 kg Kb= 49.4 1

1

2

3

4

5

System

kb

6

Guyan

2

φ1

Mode synthesis

(3) φ2

φ3

(3) Comparison of Accuracy Figure 11.5 can be summarized as follows: (a) As seen in the figure, the mode synthesis model is better than the Guyan model by approximately 20 dB or more for lower orders of eigen modes. Thus, it can be said that the mode synthesis model provides maximum accuracy using the least number of degrees of freedom. (b) The figure shows that by increasing the number of DOFs for both types of modeling, the accuracy is improved. The accuracy of that with model order 2 from the right side is very poor for each DOF. If you have the answer for the n-DOF model, the calculation results for n-th and (n − 1)-th eigen frequencies are generally very poor. In other words, if you search for the correct answers using the n-th eigenvalues, you have to prepare the (n + 2)-th DOF model by including two marginal DOFs. (c) In the case of Guyan modeling, we must increase the number of master coordinates by as much as possible, while compromising the increase in calculation time and size.

11.1.4 Discrete Modeling for Continuous Medium (1) Exact Solution We consider a uniform shaft supported by springs at both ends, as shown in Fig. 11.7(1). The continuous shafting system has an infinite number of eigen frequencies. Here, x denotes the longitudinal position from the left, and u denotes lateral vibrational displacement. The general solution is given by the following form: u(x)  A sin βx + B cos βx + C sinh βx + D cosh βx

(11.17)

11.1 Practical Techniques Recommended for Model Order Reduction (MOR)

363

ρA 2 ω , ρA  line density, EI  bending stiffness, ωn  natural frewhere β 4  EI n quency. The coefficients A, B, C, and D are integral constants determined by the boundary conditions. Here, we introduce non-dimensional longitudinal position by using ξ  x/L and rewrite the above equation as follows: u(ξ)  A sin λξ + B cos λξ + C sinh λξ + D cosh λξ

(11.18)

 λ2 E I . L 2 ρA Regarding the boundary conditions at both ends, we consider the shear force Q determined by the spring force of k b u and zero moment as follows:

where λ  β L, contributing to the natural frequency formula of ωn 

d 3u E I d 3u d 2u E I d 2u   −k u and M  E I   0 at ξ  0 b dx3 L 3 dξ 3 dx2 L 2 dξ 2 E I d 2u E I d 3u (11.19) Q  3 3  kb u and M  2 2  0 at ξ  1 L dξ L dξ

Q  EI

As these boundary conditions are applied to the solution of Eq. (11.18), the corresponding characteristic equation is expressed as a function with coefficient λ for the natural frequency calculation and K b  non-dimensional bearing stiffness as follows: |D(λ, K b )|  0

(11.20)

⎤ 0 −1 0 1 ⎢ −λ3 EI Kb λ3 Kb ⎥ ⎥ D⎢ ⎣ − sin λ − cos λ sinh λ cosh λ ⎦, kb  K b L 3 D42 D43 D44 D41 ⎡

where D41  −λ3 cos λ − K b sin λ, D42  λ3 sin λ − K b cos λ, D43  λ3 cosh λ − K b sinh λ and D44  λ3 sinh λ − K b cosh λ. When K b  49.4 is given as the non-dimensional stiffness, the answer for λ is obtained, and the natural frequencies are then known as stated after Eq. (11.22). When we draw the function of sign [D(λ)] as shown in Fig. 11.8, the roots of the characteristic equation can be solved graphically by checking where the unknown variable λ crosses the abscissa axis. The array of answers is continued infinitely since it corresponds with a uniform shaft.

 λe  2.68 4.03 5.52 8.06 11.1 · · · The answer corresponds to natural frequencies f n as follows:

(11.21)

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11 Our Latest Topics Relating to Simplified Modeling of Rotating … 4.03

8.06

14.2

sign ( |D| )

1 0.5 2 -0.5

2.68

4

6

8

5.52

10

12

14

16

11.1

18

20



17.3

-1

Fig. 11.8 Eigen solutions for a uniform shaft supported by spring

β 2 fn  2π



λ2 EI  ρA 2πL 2

 EI ρA

(11.22)

The exact values of natural frequencies fne are

 f ne  58.6 132.3 248.1 529.7 999.3 · · · Hz

(11.23)

(2) Approximate Solution by Guyan Model For the continuous system in Fig. 11.7(1) and (2), we select the master coordinates (colored pink) and put the node number on them for discretization (called 1D-FEM) as shown in R1_Fig. 4.2. For this example, if a n-DOF is selected, we divide this shafting into (n − 1) segments. According to the discretization, we formulate the mass matrix M and stiffness matrix K. Eigen frequencies are obtained by solving the eigenvalue problem of M −1 K. In the case of the 5-DOF model, the calculated natural frequencies are given as follows:

 f nG  58.9 125.2 194.3 352 633 Hz

(11.24)

(3) Approximate Solution of Mode Synthesis Model For the continuous system in Fig. 11.7(1) and (3), the A-set of coordinates denotes the spring portions located at both ends, while the inner coordinates denote the other part of the shaft. The A-set provides the static defection modes i.e., two straight lines for the two A-set portions, as shown by the blue straight lines in (3) of the figure:

δ1  (L − x)/L  1 − ξ

δ2  x/L  ξ



(11.25)

Eigen modes of the inner system under the pinned-pinned conditions are shown by the red sinusoidal waves in (3) of the figure. These theoretical formulae are given in R1_Table 3.1(5): φk  sin λk ξ

 where λk  π 2π 3π · · ·

(11.26)

11.1 Practical Techniques Recommended for Model Order Reduction (MOR)

365

The combination of these deflection modes and eigen modes defines the mode synthesis matrix T q as follows:   Tq  δ1 δ2 φ1 φ2 φ3 · · ·

(11.27)

The congruence transformation by the mode synthesis matrix is applied to the uniform shaft, which then gives the equation of motion of the reduced model as follows: Mq X¨ q + K q X q  0

(11.28)

⎤ 2 1 sym. ⎥ ⎢ 1 2 ⎥ ⎢ ⎥ ⎢ 6/π 6/π 3 ⎥ 1 t ρAL ⎢ ⎥ ⎢ where Mq ≡ ρAL Tq Tq dξ  ⎥ ⎢ 3/π −3/π 0 3 6 ⎢ 0 ⎥ ⎥ ⎢ 2/π 2/π 0 0 3 ⎦ ⎣ .. .. .. .. .. . . . . . . . . t    X q ≡ x L x R η1 η2 η3 · · · , K q ≡ Diagonal kb kb k1∗ k2∗ k3∗ · · · ⎡

where

ki∗

  2 ρAL (iπ)2 E I (iπ)4 E I   2 L2 ρA 2 L3

The corresponding eigenvalue problem ωq2 Mq φq  K q φq

(11.29)

gives the natural frequency and eigenvector. For example, if you want a 5-DOF model, remove the elements of the first five rows and columns for the reduced mass and spring matrix. The eigen frequencies of the model are calculated for the 5-DOF model as follows:

 f nq  58.6 132.5 248.5 545 1030 Hz

(11.30)

(4) Comparison of Accuracy The approximate eigen frequencies of the Guyan model are shown in Eq. (11.24), and those of the mode synthesis model are shown in Eq. (11.30). These approximate solutions are compared with the exact values of Eq. (11.23). The result of the error evaluation is shown in Fig. 11.9, in the upper part for the mode synthesis model and the lower part for the Guyan model. The x-axis denotes the number of eigen solutions of each n-th modeling order, and the y-axis denotes relative error on a linear scale. Note that the error in the upper chart is zoomed to be about ten times larger than that in the lower chart. This figure brings out the following points:

366

11 Our Latest Topics Relating to Simplified Modeling of Rotating … Model order 0.04

Model order

0.04

Model order

0.04

Model order

Error

0.04

0 1 2 3 0

0 1 2 3 4 5 0

0 1 2 3 4 5 6 7 0

0 1 2 3 4 5 6 7 8 9 0

Error

Guyan

Mode synthesis

0.06

-0.4

-0.4

-0.4 3-DOF

5-DOF

-0.4 7-DOF

9-DOF

Fig. 11.9 Relative error of natural frequency calculation (Guyan model vs. mode synthesis)

(a) In every case, we recognize that the mode synthesis technique can yield more accurate solution than the Guyan model by a factor of about ten. (b) In the Guyan model, the accuracy is acceptable for the lower order modes, but not for the higher order modes. Eigen frequencies of the Guyan model are generally slightly lower than the exact values, due to the appearance of mostly negative values in the error direction. (c) In the case of the mode synthesis model, the accuracy of the n-th order model is maintained within 1 to (n − 2)th orders. It is not necessarily true that the solutions of the last two eigen modes are sufficiently accurate. Eigen frequencies of the mode synthesis model are generally slightly higher than the exact values, as indicated by the positive direction on the error axis.

11.1.5 Mode Separation In Figs. 11.6(3-DOF) and 11.7(3), we show two deflection modes in the form of straight lines for the inner system. There are two angled straight lines: If the left end is the top, then the right end is zero and vice versa. Then it is called “side by side” modeling. Instead of these two angled straight lines, we can introduce the parallel and tilting straight lines as shown in Fig. 11.10(2). The modeling by these parallel and tilting mode separations is called the “mode” separation method. The coordinate transformation is then defined as follows: ⎤ ⎡ 1 xL ⎢ xR ⎥ ⎢ 1 ⎢ ⎥ ⎢ ⎢ η1 ⎥ ⎢ 0 ⎢ ⎥⎢ ⎢ η ⎥ ⎢0 ⎣ 2⎦ ⎣ .. .. . . ⎡

−1 1 0 0 .. .

0 0 1 0 .. .

0 0 0 1 .. .

⎤⎡ ⎤ xp ··· ⎢ ⎥ · · · ⎥⎢ xt ⎥ ⎥ ⎢ ⎥ ···⎥ ⎥⎢ η1 ⎥ ⇔ X q  Tm X m ⎢ ⎥ ···⎥ ⎦⎣ η2 ⎦ .. .. . .

(11.31)

11.1 Practical Techniques Recommended for Model Order Reduction (MOR) I10 u500 L U , A , L steel

xL

(1) system

xR

kb

kb

Gp

1

367

Gt

(2)

1

I2

I1

(3)

I3

I6

I3

(a) Parallel mode group

I4

(b) Tilting mode group

Fig. 11.10 Mode separation into parallel and tilting groups

The congruence transformation of this matrix transforms the equation of motion in (11.28) as follows: Mm X¨ m + K m X m  0

(11.32)

where ⎡

⎤ 1 0 2/π 0 · · · ⎢ 0 1/3 0 −1/π · · · ⎥ ⎢ ⎥ ⎢ ⎥ Mm  Tmt Mq Tm  ρAL ⎢ 2/π 0 1/2 0 · · · ⎥ ⎢ 0 −1/π 0 1/2 · · · ⎥ ⎣ ⎦ .. .. .. .. . . . . . . .   t K m  Tm K q Tm  Diagonal 2kb 2kb k1∗ k2∗ · · · According to the noted alignment of zero elements inside the M m matrix, all elements can be separated into two groups: the parallel group of odd numbers 1, 3, 5… and the tilting group of even numbers 2, 4, 6… with respect to the diagonal element number of the M m matrix. Then two sets of decoupled equations of motions are obtained as follows: M p X¨ p + K p X p  0 Mt X¨ t + K t X t  0 where

(11.33)

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11 Our Latest Topics Relating to Simplified Modeling of Rotating …

⎡

⎤ 1 sym. ⎢ 2/π 1/2 ⎥   ⎢ ⎥ M p  ρAL ⎢ 2/(3π) 0 1/2 ⎥, K p  Diag. 2kb k1∗ k3∗ · · · ⎣ ⎦ .. .. .. . . . . . . ⎡ ⎤ 1/3 sym. ⎢ −1/π 1/2 ⎥   ⎢ ⎥ Mt  ρAL ⎢ −1/(2π) 0 1/2 ⎥, K t  Diag. 2kb k2∗ k4∗ · · · ⎣ ⎦ .. .. .. . . . . . . The former is parallel motion. Bearing journal movements might be seen in the same phase at both ends in the case of an oil film bearing equipped rotor. The latter is called tilting motion, where both bearing journals might move out of phase. When you consider the levitation of the uniform shaft by active magnetic bearings located at both ends as shown in Fig. 6.32a, it called “side-by-side control.” The controllers for both ends of x L and x R are independent of each other. In “mode control” as shown in Fig. 6.32b, since two displacement signals are measured at both ends, the parallel and tilting displacements are known indirectly by a matrix calculation involving the sum and difference. The controller for the parallel motion works for the vibration of an odd number of eigen modes. The tilting controller regulates the vibration of an even number of eigen modes. Each controller shares vibration control for all eigen modes in their own group; the respective controller duty is thus reduced by half compared with side-by-side control. The mode controller based on the latter modeling is recommended due to its superiority from the standpoint of industrial vibration control technology. Example 11.1 The shaft-bearing system shown in Fig. 11.10(1) yields the mode synthesis model of Eq. (11.33) and the natural frequency of Eq. (11.30). In this case, we add a disk (mass  m a ≡ ρAL × μa  ρAL × 1.0, where μa  disk mass ratio compared to shaft mass) at the middle point on the shaft, as shown in Fig. 11.11(1). The bearing stiffness is set by kb ≡ K b E I /L 3  40 × 103 N/m, where K b  49.4 is the non-dimensional stiffness. Obtain the mass and the stiffness matrices for a 4-DOF reduced model of this new rotor system and calculate the natural frequencies. Note that the mode shapes for the mode synthesis transformation, shown in Fig. 11.10(3), are invariant even when fitting disks on the shaft.

Answer For parallel modes, the mass matrix M pa and stiffness matrix K pa are:  M pa  M p + m a

11 and K pa  K p 11

For tilting modes, the mass matrix M ta and stiffness matrix K ta are:

11.1 Practical Techniques Recommended for Model Order Reduction (MOR) Fig. 11.11 Shafting systems with disks

369

10 500L xR

xL (1)

ma

kb

ms= 0.31kg Kb= 49.4

xL (2)

kb

kb

xR

ma

ma

ma

kb

ma = 0.31kg

 Mta  Mt + m a

00  Mt and K ta  K t 00

Eigen frequencies are {36.6, 222} Hz for parallel modes and {132.5, 545} Hz for tilting modes. Note the latter is the same as Eq. (11.30) in the case of no attached disk. Hints For parallel and tilting systems, transformation matrices for mode synthesis are defined by p and t with respect to only these disk locations of ξ  {1/4, 2/4, 3/4} as follows: ⎡

⎤ ⎡ ⎡ ⎤ ⎡ ⎤ ⎤  p1 (1/4) 1 0.707 t1 (1/4) −0.5 1  p ≡ ⎣  p1 (1/2) ⎦  ⎣ 1 1 ⎦ and t ≡ ⎣ t1 (1/2) ⎦  ⎣ 0 0 ⎦  p1 (3/4) t1 (3/4) 1 0.707 0.5 −1 

where φ p1 (ξ)  δ p  1 φ1  sin πξ and 

φt1 (ξ)  δt  2(ξ − 1/2) φ2  sin 2πξ . Then, in addition to M p and M t of Eq. (11.33) for the uniform shaft system, mass matrices of the mode synthesis model, M p and M t , include attached disk components and they are calculated as follows:  M pa  M p + φtp1 m a Diagonal 0 1 0 φ p1 and   Mta  Mt + φtt1 m a Diagonal 0 1 0 φt1 Since there is only a middle disk, the additional mass matrix is defined by the inclusion of Diagonal [0, 1, 0]. Example 11.2 In the previous example, if three disks are attached on the shaft in

the same interval, as shown in Fig. 11.11(2), calculate the mass and the stiffness matrices, and the natural frequencies for a reduced 4-DOF system. Answer For parallel modes, the mass matrix M pa and stiffness matrix K pa are:

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11 Our Latest Topics Relating to Simplified Modeling of Rotating …

 M pa  M p + m a

3 2.41 and K pa  K p 2.41 2

For tilting modes, the mass matrix M ta and stiffness matrix K ta are:  Mta  Mt + m a

1/2 −1 and K ta  K t −1 2

Eigen frequencies are {27, 204} Hz for parallel modes and {76, 408} Hz for tilting modes. Hints: Since the additional mass matrix is defined  by setting Diagonal [1, 1, 1],   3 2.41 M pa  tp1 m a Diagonal 1 1 1  p1  m a and 2.41 2    1/2 −1 Mta  tt1 m a Diagonal 1 1 1 t1  m a −1 2

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported Rotor Unstable vibration caused by oil film bearings is called oil whip/oil whirl, and the characteristics of such vibration are shown in Figs. 4.1/4.2/4.15. This section presents methods to predict quickly and approximately the stability limit, without executing complex eigenvalue analysis.

11.2.1 Single-Degree-of-Freedom System and Simplified Stability Criterion [B37, 293] Let us consider the parallel mode of a rigid rotor with bilateral symmetry as shown in Fig. 11.12. The dynamic characteristics of the oil film bearings are defined by the Bently-Muszynska (B/M) model in Sect. 8.7, expressing oil film bearing dynamic characteristics in the Laplace domain as BRG  kd + cd (s − jλ) based on only the forward whirl component. Therefore, the isotropic equation of motion can be expressed, with total mass mt of the rigid rotor, as follows: m t z¨ + kd z + cd (˙z − jλz)  0

(11.34)

which can be simplified by dividing by the mass mt to z¨ + 2ζd ωd (˙z − jλz) + ωd2 z  0

(11.35)

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

371 z

Fig. 11.12 Rigid rotor supported by oil film bearings

mt mt kd

cd s - jO :

:

Oil film brg.

Oil film brg. (a) Rigid rotor

(b) Single-DOF system

where ωd2  kd /m t , called here the “rigid mode oil frequency”, and 2ζd ωd  cd /m t . The characteristic equation of Eq. (11.35) is s 2 + 2ζd ωd (s − jλ) + ωd2  0

(11.36)

Figure 8.35 shows a calculation example of the characteristic roots in Eq. (11.36). The root should be s ≈ jω √d , as is well known when D is smaller than 10,000 Ns/m (equivalently ζd  D/(2 M K )  1), and s ≈ jλ as can be approximated when D is greater than 10,000 Ns/m. The latter corresponds mainly to many rotor systems supported by oil film bearings. This section offers a method for predicting more accurate eigen solutions and stability criteria. (1) For ζ d > 1, substituting the approximate solution s  jλ+ into Eq. (11.36), considering large ζ d and small , and ignoring infinitesimal quantities higher than first order, results in the following: ωd2 − (λ)2 ωd2 + ( jλ + )2 + 2ζd ωd   0 →   − → ωd2 − (λ)2 + 2ζd ωd   0 2ζd ωd

(11.38)

In both cases, the real parts () of these eigenvalues determine the stability and a zero real part (  0) indicates the stability limit by λ  ωd . Therefore, we would observe the appearance of the unstable vibration with the frequency, ω d , when the rotational speed reaches the onset speed,   ωd /λ. For a rigid rotor system supported by oil film bearings, the natural frequency (ω d ) may be normalized to unity and the characteristic roots (s) were then calculated at various damping ratios (ζ d ). Figure 11.13a shows the real and imaginary parts

372

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 1.4

1.5 Im

 d =0.02

1.2

Im [s]

1.0 0.8 0.6

 d =0.5

0.5

1

0.4 2

0.2 0.0

Re [s]

0.2 0.1 - 0.1 - 0.2 - 0.3

Stability limit (Unstable onset) 

1.0

2.0

3.0

1.0 Stability limit

0.5

d =1 d =2

 d =0.02

Re 1.0

2.0 Rotational speed 

2

d =0.02

3.0 

- 0.8

- 0.6 - 0.4

- 0.2

1

0.5 (a) Re [s] and Im [s]

(b) Root loci of s

Fig. 11.13 Eigen frequency and stability (1-dof system with m  1, kd  ωd2 and c  2ζd ωd )

of the characteristic roots, where the rotational speed is at the stability limit when Re[s] changes from negative to positive. Stable roots ◯ and unstable roots × on the imaginary part, Im[s] (damped natural frequency), show that the stability limit exists at λ  ωd , where self-excited vibration commences. Figure 11.13b shows the root locus. It shows the separation between s ≈ jωd +  and s ≈ jλ +  roots, and indicates that ζ d  1 is the boundary damping ratio between them. Since the stability limit exists when the real part equals zero, the position (0, ω d ) is always at the stability limit for any case of this single-DOF system.

11.2.2 Two-DOF System and Undamped Natural Frequency [294] For the flexible rotor system (total mass mt and shaft stiffness k z ) with bilateral symmetry, as shown in Fig. 11.14a, the mode synthesis model consisting the parallel mode δ  1 (a straight and flat line) and flexural eigen mode φ (the first mode shape being subject to pinned-pinned boundary condition at both ends, having the natural frequency of shaft ω z ) can be reduced to the following 2-DOF system with the dynamic characteristcs, noted BRG, at oil film bearings:

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …



mc mt mc m∗ → 1



  B RG 0 z1 z¨ 1 + 0 η¨2 0 k ∗ → ωz2 η2

373

(11.39)

where the total mass m t  δ t Mδ  1t M1, cross-mass m c  1t Mφ, modal mass m ∗  φt Mφ → 1 and modal stiffness k ∗  m ∗ ωz2 → ωz2 are calculated with the mass matrix (M), shaft stiffness matrix (K), parallel mode (δ) and eigen mode (φ) for the n-dimensional original system. The cross-mass, mc , exists because the bending displacement η 2 is a relative quantity. The eigen mode φ is here set as normalized by the modal mass, m* . These modeling techniques are already stated in Example 11.1 in the previous section. Defining the absolute coordinate of the center of the rotor at the middle of the shaft, ξ 2 , and the coordinate transformation in R1_Eq. (4.20) as η2  a(ξ2 − z 1 ), where a  mc /m* the quasi-modal model, i.e., the equivalent two-DOF system, can be obtained as follows: 

m1 0 0 m2



  z¨ 1 B RG + k2 −k2 z 1 + 0 −k2 k2 ξ2 ξ¨2

(11.40)

where m 2  a 2 m ∗ is the equivalent mass, i.e., being effective at the middle of the shaft, m 1  m t − m 2 is the equivalent mass of the journal, and k2  m 2 ωz2 is the equivalent shaft stiffness. Figure 11.14b shows the schematic of Eq. (11.40). When normalizing to m*  1 in Eq. (11.39), the cross-mass is the square root of the equivalent mass at the middle of the shaft because m 2  a 2  m 2c in the above equation. In the mode synthesis model in Eq. (11.39), normalizing displacements z and η by masses mt and m* results in the following:

Fig. 11.14 Flexible rotor supported by oil film bearings

2

mt ks

m2



 2 /a

k2 Oil film brg.

Oil film brg.

z1

m1 kd

2

z1

cd  s - j 

 =1

{  , z } (a) Mode synthesis

(b) Two-DOF system

374

11 Our Latest Topics Relating to Simplified Modeling of Rotating …



z1 η2





 √ 0 1/ m t √ z  0 1/ m ∗ η

(11.41)

which can be coordinate-converted and rewritten as follows: 

1 μc μc 1

   2 z z¨ ωd + 2ζd ωd (s − jλ) 0 0 + 0 ωz2 η η¨

(11.42)

where μ2c  m 2c /(m t /m ∗ )  (m t /m ∗ )2 m ∗ /m t  m 2 /m t < 1 shows the mass ratio. The characteristic equation of Eq. (11.42) is   2  s + 2ζd ωd (s − jλ) + ω 2 μc s 2  d 0   μc s 2 s 2 + ωz2 

(11.43)

Therefore, the undamped natural frequencies s  jωn can be determined as follows: D20

   −ω 2 + ω 2 −μc ω 2  2





n n d   ω − ω 2 ω 2 − ω 2 − μc ω 2 2   n d n z n −μc ωn2 −ωn2 + ωz2  ≡ y1 − y2  0

(11.44)

where y1 ≡ (ωn2 − ωd2 )(ωn2 − ωz2 ) y2 ≡ μ2c ωn4 . The upper diagram in Fig. 11.15a shows graphical solutions. The undamped natural frequencies are at the intersections between the curves of y1 and y2 . (a)-(1) in the figure corresponds to a stiff shaft with a heavy rotor; (a)-(2) corresponds to a flexible rotor with a light rotor—the category covering most of the industrial pumps and compressor rotors experienced by the authors. The intersections, denoted by ω12 and ω22 , between both curves yield the first and second natural frequencies (ω 1 and ω 2 ). In both cases, the lowest natural frequency (ω 1 ) is smaller than both natural frequencies for the “rigid mode oil frequency” (ω d ) and the shaft bending frequency (ω z ). The second natural frequency (ω 2 ) is higher than both frequencies.

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

(2)  z <  d

(1)  d <  z (a)

y =  c2 n 4

y1= ( n2  d 2 ) (  n2   z2 )

d2 =1

y1

1 0.76

 1 0.87

mode 



mode z

2



3

4

z  1.4

 2 2.27 

whirl 



whip

=5.16  n2 5



 

1 0.76  1 0.87



 22

y1

 12

6



y2

 z2 =1  d2 =2

 22

 12

(b)

2

y2

 z2 =2

375

2

3

 z 1.0



4

=5.16  n2 5

6

 2 2.27







whirl 





whip

Fig. 11.15 Stability criteria for λ and ωz modes (ζd  large, λ  0.49 and μ2c  0.5)

11.2.3 Simplified Stability Criterion for Two-Degree-of-Freedom System Let us discuss approximate solutions and stability in the 2-DOF system as well as the single-DOF mentioned previously. (1) For ζ d > 1: When the oil film damping is overdamped, Eq. (11.43) can be approximated by the following equation when ζ d >> 1: 2

s + 2ζd ωd (s − jλ) + ωd2 (s 2 + ωz2 ) − (μc s 2 )2 ≈ 2ζd ωd (s − jλ)(s 2 + ωz2 )  0 where the fundamental solutions are s  jλ and s  jωz . (a)

2.5

2.5

Frequencies

d = 2.0 0.02 1.5

d



s1

1

2

 d = 0.02 3

4

5

2.5

1.5

d

s2



s1

z 1

0.0

0

whirl 1

 d = 0.02

2

3

4

5

2.5 2 z

1.5

s2

d

1.5

whip whirl

0

1

2 d

s2



s1

z

1.0

1

0.5

2.0



s1

1.0

0.0

2

0.5

whirl 0

2.0

1.0

1

0.5

d 2.0 =2 Frequencies

s2

1.0

0.0

(b)

2 z

d = 2

2 3 4 Rotational speed 

1

0.5 5

0.0

0

(1) Case of  d =1 and  z =1.4

Fig. 11.16 Frequency and stability (λ  0.49 and μ2c  0.5)

whirl 1

whip

2 3 Rotational speed 

d = 2

4

(2) Case of  z =1 and d =1.4

5

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

377

The characteristic equation can be obtained by letting the approximate solution for one of the fundamental solutions λ of rotor rigid motion be s  jλ + , so that   2  ω + ( jλ + )2 + 2ζd ωd  μc ( jλ + )2  0  d  μc ( jλ + )2 ( jλ + )2 + ω 2 

(11.48)

z

Considering small quantities,  and large quantity, ζ d , and referring to Appendix Eq. (F.7), manipulation of Eq. (11.48) yields the following stability criteria:    2 ω1 − (λ)2 ω22 − (λ)2   Re[]  − 2ζd ωd ωz2 − (λ)2

(11.49)

Therefore, the positive/negative switching point of  in Eq. (11.49), which expresses the stability boundary, is the intersection between λ and each natural frequency (ω 1 , ω 2 , or ω z ). It is plotted as the λ mode in Fig. 11.15b. Instability occurs when λ exceeds the first natural frequency, ω 1 , and stability returns when it exceeds the shaft natural frequency, ω z , but instability occurs again when it exceeds the second natural frequency, ω 2 . The characteristic equation can be obtained by letting the approximate solution for another fundamental solution of shaft bending motion be s  jωz + :   2  ω + ( jωz + )2 + 2ζd ωd ( jωz +  − jλ) μc ( jωz + )2  0  d  μc ( jωz + )2 ωz2 + ( jωz + )2  Solving with the small quantities, , and the large quantity, ζ d , yields:   2  ω − ω 2 + 2(ζd ωd + jωz ) + 2 jζd ωd (ωz − λ) −μc ω 2 + 2 μc j ωz   z z   d   2 jωz  −μc ωz2 + 2 μc j ωz   2   ω − ω 2 + 2 jζd ωd (ωz − λ) −μc ω 2 + 2 μc j ωz   z z 0   d  2 jωz  −μc ωz2 + 2 μc j ωz  then the following approximation is obtained in a similar way for Eq. 11.49: Re[]  −

μ2c ωz3 4ζd ωd (ωz − λ)

(11.50)

Therefore, the stability limit is at the intersection between λ and shaft bending frequency, ω z . It is plotted as the ω z mode in Fig. 11.15b. Instability occurs when λ multiplied by the rotational speed, λ, exceeds the shaft bending frequency, ω z , and it continues afterwards. Figure 11.16b shows the loci of the imaginary parts (damped natural frequency) of the exact complex eigenvalues to verify the approximate estimation mentioned above by comparison with the exact complex eigen frequency calculation for ζ d 

378

11 Our Latest Topics Relating to Simplified Modeling of Rotating … Im  2 = 2.27 2.0 s2

1.5

 2 = 2.27

d =1  z = 1.4 d =2

2.0

z s2

0.0

0.5

(a) Case of  d =1 and  z =1.4

1.0  1 = 0.87

s1

0.5

0.5 Re

Re - 0.5

 d = 1.4  z =1 d =2

1.5

z

1.0  1 = 0.87 s1

Im

-0.5

0.0

0.5

(b) Case of  z =1 and d =1.4

Fig. 11.17 Root loci (λ  0.49 and μ2c  0.5)

2. These loci of damped natural frequency have two lines: One is an upward-sloping line for the λ mode (oil whirl); the other is a horizontal line for the ω z mode (oil whip). Strictly speaking, the sloping solution is close to λ, but slightly different from it. The beginning of instability is at the intersection between the straight line of λ and the first natural frequency (ω 1 ), and the unstable root λ represents whirl phenomena. Afterwards, this upward-sloping line λ of oil whirl motion approaches the shaft bending frequency (ω z ) of oil whip motion. Finally, it shifts as to be saturated and transitions to the full oil whip phenomenon. On the left side of the figure, (1) shows an example when oil whip occurs after long-term oil whirl; on the right side, (2) shows immediate oil whip. Both cases correspond to the examples shown on the right and left sides of Fig. 8.40. Figure 11.17 shows the root locus for large oil film damping at ζ d  2. For every locus, the frequency at the stability limit, which occurs just on the imaginary axis, is the first natural frequency, ω 1 . Also, it can be seen that the self-excited vibration frequency approaches the shaft bending frequency, ω z , on the imaginary axis when the rotational speed becomes high.

11.2.4 Calculation Example for a Rotor System with Three Disks [295] Figure 11.18 represents a rotor model, which is shown in R1_Fig. 12.1 with the shaft dimensions in detail. The corresponding eigenvalue analysis is shown in R1_Fig. 7.21 (complex eigenvalues), Fig. 7.22 (root locus), and Fig. 7.23 (response curves). Let us estimate the stability limit for case studies using the methods stated in the previous section.

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

379

1062 200

200

 240

 40

200

200

Sliding bearing

Sliding bearing 40

Fig. 11.18 Shafting system with three disks (dimensions in mm) Table 11.1 Bearing dynamic properties (× 104 N/m, × 104 Ns/m) C/R

Ω rps

kxx

kxy

kyy

cxx

kyx

cxy

cyy

147

6

147

6

1.8

6.0

1.8

0.5

0.5

0.5

0.001

90

0.003

90

669

2,188

1,042

− 1,572

7.2

0.010

90

539

533

262

− 25

1.5

1,622 41,680 3,230 − 41,438

cyx

Table 11.2 Bearing dynamic properties (× 104 N/m, × 104 Ns/m) C/R

kd

kc

cd

ωd Hz

ζd

kc λ= c d Ω

ω1 Hz

kb

cb

0.001

4,852

83,118

294

154

29.3

0.5

45.6

1,626

12.0

0.003

1,711

3,760

13.2

91.4

2.2

0.503

42.7

720

3.79

38.6

579

1.41

0.010

801

558

2.0

62.5

0.49

0.493

Although the dynamic properties of journal bearings depend on the rotational speed as shown in Examples 2.1 (C/R  0.003), 2.2 (C/R  0.001) and 2.3 (C/R  0.01), let us adopt the properties for   90 rps for simplicity, which is assumed to be available for the entire speed range. Table 11.1 summarizes the dynamic properties of these bearings. The bearing characteristics in the XY form are rewritten in complex form for the application to our methods, as shown in Table 11.2. In Fig. 11.19, we have the equivalent 2-DOF model of the rotor of Fig. 11.18 as obtained by the quasi-modal technique stated at R1_Fig. 4.14. This 2-DOF model has the total mass, m t  51.8 kg, the shaft bending frequency, ωz  47.3 Hz, and the coefficient of coupled mass, μc  0.9 of Eq. (11.42). The model shows that the equivalent mass at the center of the shaft is 81% and the mass of the journal bearings is 19%, which is obtained from μ2c  0.81  81%. Figure 11.19 also includes the system and dynamic properties of the bearings for simplified analyses. Table 11.2 includes the first undamped natural frequencies, ω 1  {45.6, 42.7, 38.6} Hz, calculated by the forward system of Eq. (11.44) for Fig. 11.19. Therefore, we may draw a graphical evaluation of the stability limit, as shown in Fig. 11.20. The

380

11 Our Latest Topics Relating to Simplified Modeling of Rotating … m t  51.8kg 81%

z1

kd

19%

 z  47.3Hz

cd s- j  2 0.003

C/R

1 0.001

kd cd

48×10

2.94×10

 d

0.5

0.13×10 0.503

29.3

2.2

6

17×10 6

3 0.01

6

8×10 6

6

0.02×10

6

0.493 0.49

Fig. 11.19 Two-DOF model 50

 z  47.3 Hz

A Estimation ( Isotropic Brg )

C/R=0.003 2

40  1  45.6 Hz

Frequency [Hz]

C/R=0.001 1

 1  42.7 Hz  1  38.6 Hz

30

C/R  1 1

1

C/R=0.010 3

03

.0 20

2

10

.0 30

3

78.3 84.8 20

40

60

Rotational speed

3

[Hz]

1 01 .0 =0

R C/

0

2

B Exact complex eigenvalue k and c at = 90rps ( Anisotropic Brg )

20

10

1

0.5 0.503 0.493 45.6 42.7 38.6 91.1 84.8 78.3

80 [rps]

rps 91.1 100

C/R  1e 1e e

2 3 1 45.6 42.4 33.3 91.1 82.3 83.6 0.5 0.51 0.4 [Hz]

Fig. 11.20 Stability limits (forward whirl bearing force only)

flat lines correspond with the ω 1 and the slope lines with λ. These intersections between the lines give the stability limit. The onset speed, where the instability would commence, is 1  ω 1 /λ  {91.1, 84.8, 78.3} rps. These values are summarized in the inset table placed at the right side of the figure. Because the shaft bending frequency, ω z , is lower compared to the rigid mode oil film frequency, ω d , the stability limit is determined mainly by the lower frequency at ω1 ≈ ωz and 1 ≈ ωz /λ. If we assume that bearing properties obtained at   90 rps (Table 11.1) are constant for the entire speed range (no rotational speed dependency), the corresponding eigenvalue calculation gives answers about the resulting unstable frequencies ω 1e  {45.6, 42.4, 33.3} Hz at the onset speeds 1e  {91.1, 82.3, 83.6} rps for C/R  {0.001, 0.003, 0.010}, respectively. Note that the λ coefficients are thus observed as λe  ω 1e /1e  {0.5, 0.51, 0.4}. The summarization is listed in the inset table to the right side of the figure. (estimation based upon forward whirl only, Comparing the inset tables isotropic bearing) and (exact eigenvalue calculation including also backward whirl,

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

381

anisotropic bearing), the estimated values almost coincide with the exact values for practical cases such as C/R  0.001 and C/R  0.003. However, in the case of C/R  0.010 giving rise to large anisotropy of the bearing dynamic properties, the difference between the estimated and exact values cannot be ignored. It is still necessary to recalculate the estimation by taking backward components |k b | and |cb | into account. The recovery of the accuracy is stated in Sect. 11.2.7.

11.2.5 Example of Numerical Calculation for Muszynska’s Rotor System [293] Figure 11.21 shows the model of the rotor, taken from Ref. B41_Fig. 4.2.1. The left side is assumed to be supported by a rolling element bearing. The right side is a boundary coordinate (A-set in Nastran®) and supported by a journal bearing. The rigid mode, δ, when the boundary coordinate undergoes a forced displacement, is indicated by the inclined line, and the rigid mode mass, mδ , is obtained by the sum of one-third of the shaft mass, ms . Other masses include the journal mass, mj , and the disk mass, md . Therefore, the left figure has a relatively large mass, mδ  0. 95 kg, and the right figure has a relatively small mass, mδ  0.27 kg. The shaft bending frequency, ω z  61 Hz, of an eigen mode, φ, is considered to be the same in both cases because both ends are simply supported. The dynamic characteristics of the bearings are estimated with reference to Muszynska’s book and papers. The reduced two-mass model corresponding to each shaft-disk-bearing system, (1) and (2), are also shown in the same figure. Then we know the following frequencies: for (1) having the “rigid mode oil frequency, ω d  28.3 Hz”, we know the first and z1

md = 0.68

ms

m= m s / 3+m j +md = 0.95kg

ms = 0.22kg

z1

mj

m j= 0.196 m= m s / 3+m j = 0.27kg

kd = 30,000N/m( d  28.3Hz)

kd = 30,000N/m( d  53.1Hz)

cd = 3,000Ns/m d = 8.9

cd = 3,000Ns/m d = 16.6

 = 0.44

= 0.44

z1

kz 1.3%

kd 4%

98.7%

 z 61Hz

cd  s -j 

(1) Right-side Disk

z1

kz

kd

96%

 z 61Hz

cd  s -j 

(2) Left-side Disk

Fig. 11.21 Rotor system with movable disk supported by oil film bearing at the right

382

11 Our Latest Topics Relating to Simplified Modeling of Rotating … (2) Left-side disk

(1) Right-side disk

3

0

2 4 6 8 3 Rotational speed × 10 [rpm]

114 rps (6,866 rpm)

1

64rps (3,859 rpm) 10

0 0

2 4 6 8 3 Rotational speed × 10 [rpm]

10

10

8  d = 28 Hz (1,680 rpm) 6 64 rps 4 (3,859 rpm)

 z = 61 (3,660)

2  1 =28.3 Hz (1,698 rpm)

 2 = 62 (3,698)

0

1 = 51 Hz (3,064 rpm)

2

1

8 6 4

2 3 4 Frequency ×10 3 [rpm]

5

 d = 53.1 Hz (3,186 rpm) 1 = 51 Hz (3,064 rpm)

2 0

 z = 61 (3,660)

114 rps (6,866 rpm)

0

 2 = 65 (3,912)

3 2 4 5 Frequency ×10 3 [rpm]

1

Whip

Whip rl

Rotational speed  × 103 [rpm]

(c)

3

10

0

d = 53.1 Hz (3,186 rpm)

hi

3

Rotational speed  × 10 [rpm]

 1 = 28 Hz ( 1.680 rpm)

1

 z = 61 Hz (3,660 rpm)

4

d = 28.3 Hz (1,698 rpm)

2

 2 = 65 Hz (3,912 rpm)

5

 z = 61 Hz (3,660 rpm)

4

0

(b)

 2 = 62 Hz (3,698 rpm)

5

W

d

Frequency ×103 [rpm]

(a)

l

hir

W

Frequency ×10 3 [1/min]

Frequency ×10 3 [1/min]

Fig. 11.22 Approximate stability evaluation and experimental data

second undamped natural frequencies of ω 1  28 Hz and ω 2  62 Hz, respectively; and for (2) having the “rigid mode oil film frequency, ω d  53.1 Hz”, the first and second undamped natural frequencies of ω 1  51 Hz and ω 2  65 Hz, respectively, are known. These frequencies are shown by horizontal flat lines in Fig. 11.22a or vertical straight lines in Fig. 11.22b. Between these figures, the X and Y axes are interchanged. The inclined line corresponds with λ. Then, the intersection between the lines of λ and ω 1 reveals the onset of instability. Let us review the simplified stability criteria of the rotor system in Fig. 11.22a and b. On the left-side disk (a)-(2) and (b)-(2), instability occurs at the rotational speed

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

383

of   51/0.44  114 rps at the intersection of ω1  λ, soon after which oil whip occurs. The instability frequency coincides with the shaft bending frequency, ω z  61 Hz. On the right-side disk (a)-(1) and (b)-(1), because the lowest natural frequency is ω 1  28 Hz, the stability limit is given by ω1  λ and the oil whirl instability occurs at the rotational speed,   28/0.44  64 rps. After that, since the whirl frequency λ is much lower than the whip frequency ω z , the λ frequency of oil whirl is continued. As the rotational speed increases further, the instability frequency, λ, converges at the shaft bending frequency ωz  61 Hz and the state of the oil whirl transitions to oil whip. In figure (c), the corresponding experimental results are shown. The right- and left-side figures show that the state changes from whirl to whip, which coincides well with the aforementioned calculation results obtained by simplified estimation.

11.2.6 Example of Numerical Calculation for Casing Whirl As well as oil whirl/whip, casing whirl vibration may become unstable if the casing (or housing) system is supported by low rigidity on the basement. The casing contains a rotor system equipped with oil film bearings, as shown in Fig. 11.23a. The casing whirl phenomenon happens when the natural frequency of the casing rocking motion is below that of any internal shaft-bearing system. Thus, √ since the lowest natural frequency is determined by the casing frequency ωc ≡ kc /(m c + m t ), the stability limit could be due to the coincidence of ω c and λ. Let us consider a casing-rotor-bearing system with three masses supported by flexible casing pedestals as shown in Fig. 11.23b. This 3-DOF system is used for calculations. The mass of the casing is mc  500 kg (i.e., ten times larger than that of the rotor system) and the natural frequency of the casing is ω c  35 Hz (< shaft bending frequency ω z  47.3 Hz < “rigid mode oil frequency” ω d  154 Hz).

zc

r

z2

b

z1

mt = 51.8 kg mt= 51.8 kg

m c = 500 kg

19%

81% m2

c  0

kd

kz shaft

m1

 z  47.3 Hz

cd  s - j  

kc

Brg.

c  35 Hz mc = 500 kg

 c 0.9

 cr 0.28

 c b 0.31

(a) Coupling system

Fig. 11.23 Rotor-casing coupled system

(b) 3-DOF model (C/R=0.001)

384

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

In the calculation model shown in Fig. 11.23b, the absolute displacement is denoted by zc , the relative displacement of the bearing journal by η b , and the relative displacement from the journal of the bearing at the center of the shaft by η r . They are normalized by the mass matrix in a similar way to that of Eq. (11.41). Then the equations of motion with only springs can be expressed as follows, referring to Eq. (G.5) as detailed in Appendix G: ⎡

⎤ ⎡ 2 ⎤⎡ ⎤ ⎤⎡ μcb μcr ωc 0 0 zc z¨ c ⎣ 1 μc ⎦⎣ η¨b1 ⎦ + ⎣ ωd2 0 ⎦⎣ ηb1 ⎦  0 sym. 1 η¨r 1 ηr 1 sym. ωz2 1

(11.51)

√ √ √ √ where ηb1 ≡ ηb m t ≡ (z 1 − z c ) m t and ηr 1 ≡ ηr m 2 ≡ (z 2 − z 1 ) m 2 . The corresponding characteristic equation with dampers is   2  s + 2ζc ωc s + ωc2 μcb s 2 μcr s 2    s 2 + 2ζc ωd (s − jλ) + ωd2 μc s 2   0   sym. s 2 + ωz2 

(11.52)

Figure 11.24 shows the results of eigenvalue and stability limit analyses from the characteristics in Eq. (11.52). The undamped natural frequencies {33.5, 49.6, 370} Hz are obtained for ζ c  ζ d  0. The first undamped natural frequency, ω 1  33.5 Hz, is just under the casing frequency, ω c  35 Hz, and the intersection between ω 1 and λ gives the stability limit. It can therefore be estimated that the instability onset speed is onset  ω 1 /λ  67 rps. The results of complex eigenvalue calculations are also shown. The damped eigen frequencies, Im[si ], are expressed with respect to the rotational speed, where the stability roots are designated by ◯ and instability roots by ×. The switching point from ◯ to × is the exact stability limit. It coincides well with the aforementioned estimation. On the left side, for a casing damping ratio ζ c  0, because instability occurs at   67 rps with λ  ω 1 and the frequency remains at ω 1 afterwards, it is obvious that it is due to casing whip, which continues with increasing rotational speed. On the right side, for a casing damping ratio ζ c  0.002, instability occurs at the same speed of λ1 , but soon disappears and the system returns to the stable state at   85 rps.

11.2.7 Anisotropic Properties of Oil Film Bearing Dynamic Force (1) Single-DOF System As stated in Eq. (3.17), the dynamic coefficients of an oil film bearing are expressed by the forward components, {k f cf }, and the backward components, {K b C b }. Though the isotropic system is defined only by the former, the anisotropic system is expressed by both components. Thus, inserting the backward components into Eq. (11.34),

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

50 40

Im [ s1] ω 2 = 49.6 Hz

0

] [s 3

99 rps

67 rps

Im

20

sc

Im [s c ]

ω 1 = 33.5 Hz

10

60 80 40 Rotational speed [ rps ]

20

40 60 80 Rotational speed [ rps ]

100

0.4 Re[ sc ]

0.2

0.2

unstable

Re [ s ]

99 rps

67 rps

100

0.4

20

40

60

80

100

unstable 20

40

60

80

100

-0.2

-0.2 stable -0.4

s3

ω 2 = 49.6 Hz Im [ sc] ω c = 35 Hz

Im [ sc ]

ωc = 35 Hz

30 20

Im [ s1 ]

Re [ s ]

Im[s] = Frequency [ Hz ]

60

385

stable s1

-0.6

-0.4

s1

-0.6 (a) No casing damping  c = 0

(b) Small casing damping  c = 0.002

Fig. 11.24 Instability of casing whirl and the stabilization

the equation of motion of a single-DOF anisotropic system may be written in the following form: m z¨ + kd z + cd (˙z − jλz) + K b z¯ + Cb z˙¯  0

(11.53)

The characteristic equation is derived by using Eq. (3.30)   2   s + ω 2 + 2ζd ωd (s − jλ) kb + cb s d   D2   2 2 ¯kb + c¯b s s + ωd + 2ζd ωd (s + jλ) 

(11.54)

where, ωd2  kd /m, 2ζd ωd  cd /m, kb  K b /m and cb  Cb /m. This equation is simplified as follows:

2  



D2  s 2 + ωd2 + (2ζd ωd )2 s 2 + (λ)2 + 2(2ζd ωd ) s 2 + ωd2 s − (kb + cb s) k¯b + c¯b s  D2a + D2b

2  

D2a ≡ s 2 + ωd2 + (2ζd ωd )2 s 2 + (λ)2 − kb k¯b + cb c¯b s 2



D2b ≡ 2(2ζd ωd ) s 2 + ωd2 s − kb c¯b + k¯b cb s (11.55) Let s  jω1a denote an unstable eigen frequency, ω1a , as observed by unstable vibration in the anisotropic system. The stability limit is given by D2 (s)  0, that

386

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

is D2a (s)  0 for the real part and D2b (s)  0 for the imaginary part. The latter requires D2b (s  jω1a )  0: s 2 + ωd2 

Re[kb c¯b ] 2ζd ωd

→

2 ω1a  ωd2 −

Re[kb c¯b ] 2ζd ωd

(11.56)

In this manner, we find the change of the unstable natural frequency; ω 1 of the isotropic system to ω1a of the anisotropic system:  ω1a  ωd 1 − δa

(11.57)

Re[kb c¯b ] . 2ζd ωd3 If δ a > 0, the eigen frequency is decreased and vice versa. On the other hand, the real part, D2a (s  jω1a )  0, requires:

where δa 

s 2 + (λ)2 

kb k¯b cb c¯b s 2 Re[kb c¯b ]2 + − (2ζd ωd )2 (2ζd ωd )2 (2ζd ωd )4

(11.58)

Then the stability limit speed, λ, is modified by the anisotropic effect as follows:  λ  ω1a 1 + δk − δc − δkc where δk 

(11.59)

kb k¯b cb c¯b Re2 [kb c¯b ] , δ  and δ  . c kc 2 2 ω1a ω1a (2ζd ωd )2 (2ζd ωd )2 (2ζd ωd )4

(2) Two-DOF System The equation of motion is expressed as follows, adding backward properties of oil film bearing dynamic force into Eq. (11.39): 

mt mc mc m∗



    z¨ 1 BRG_F 0 z1 BRG_B 0 z¯ 1 + + 0 η¨2 0 k ∗ η2 0 0 η¯2

(11.60)

where BRG_F  kd + cd (s − jλ) and BRG_B  K b + Cb s. The above equation may be rewritten by using a non-dimensional mass matrix: 

1 μc μc 1



    brg_F 0 z 11 brg_B 0 z¯ 11 z¨ 11 + + 0 η¨21 0 ωz2 η21 0 0 η¯21

where brg_F  (B RG_F)/m t  ωd2 + 2ζd ωd (s − jλ), brg_B  (B RG_B)/m t  kb + cb s, ωz2  k ∗ /m ∗ , √ √ z 11  m t z 1 and η21  m ∗ η2 .

(11.61)

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

387

The corresponding eigenvalue equation becomes:  2   s + 2ζd ωd (s − jλ) + ωd2 μc s 2 kb + cb s 0    s 2 + ωz2 0 0  μc s 2  2 2  ¯ 0 s + 2ζd ωd (s + jλ) + ωd μc s 2  kb + c¯b s   0 0 μc s 2 s 2 + ωz2   2     s + 2ζd ωd (s − jλ) + ω 2 μc s 2  s 2 + 2ζd ωd (s + jλ) + ω 2 μc s 2  d d      μc s 2 s 2 + ω22  μc s 2 s 2 + ωz2 

2

− (kb + cb s) k¯b + c¯b s s 2 + ωz2  0 (11.62) When the natural frequency of unstable vibration at the stability limit speed is denoted by s  jω1a , the above equation is divided into two parts: For the real part:

2  



2 D 2F0 + (2ζd ωd )2 s 2 + ωz2 s 2 + (λ)2 − kb k¯b + cb c¯b s 2 s 2 + ωz2

(11.63)

For the imaginary part:



2 2(2ζd ωd )s s 2 + ωz2 D F0 − kb c¯b + k¯b cb s s 2 + ωz2  0

(11.64)

  2  s + ω 2 μc s 2  d   0 gives the first undamped natural frequency, ω1 , where D F0   μc s 2 s 2 + ωz2  for the isotropic system. The above Eq. (11.64) may be used to determine the unstable natural frequency, denoted by s  jω1a , for the anisotropic system:

Re[kb c¯b ] 2 s + ωz2 → 2ζd ωd

2



2

2 2 s + ωd s + ωz2  μc s 2 + δa ωd2 s 2 + ωz2 D F0 

(11.65)

This first natural frequency, ω 1a , is explained graphically in Fig. 11.25. In the figure, the intersection between curves of y1 and y2 gives the unstable natural frequency, ω 1 , for the isotropic system having a forward whirl component only. When considering the forward and backward whirl components, the unstable natural frequency, ω 1a , is defined by the intersection between curves of y1 and y3 . This frequency, ω 1a , is slightly less than the frequency, ω 1 , because ω1a < ω1 . Then, the stability limit speed, λ, is redefined from Eq. (11.63) as follows: kb k¯b + cb c¯b s 2 D 2F0 −

2 (2ζd ωd )2 (2ζd ωd )2 s 2 + ωz2 kb k¯b + cb c¯b s 2 Re2 [kb c¯b ]  − (2ζd ωd )2 (2ζd ωd )4

s 2 + (λ)2 

(11.66)

388

11 Our Latest Topics Relating to Simplified Modeling of Rotating … y1= ( n2  d 2 )(  n2   z2 )

y

y =  c2 n 4

y3   a ( z2  n2 ) d 2  y2

 z2d2

2

 a z2 d2 1 a2

12

d2

(new) (old)

 n2 2

z

 2a2

22

Fig. 11.25 Natural frequency revised by backward properties

Inserting s  jω1a into the above equation, the stability limit speed, λ, is calculated as   cb c¯b Re2 [kb c¯b ] kb k¯b 2 2 (λ)  ω1a 1 + 2 − − 2 ω1a (2ζd ωd )2 ω1a (2ζd ωd )4 (2ζd ωd )2 2  ω1a (1 + δk − δc − δkc )

(11.67)

(3) Case Study on Accuracy Compensation for Sect. 11.2.4 [295] Stability limit analysis is given in Fig. 11.20 by treating only the forward whirl component, assuming an isotropic system. However, the accuracy of the onset speed is not enough for the case of C/R  0.010 having strong anisotropy in the bearing dynamic properties. We again discuss the accuracy by treating backward whirl components featured in the anisotropic system based upon the new Eqs. (11.65) and (11.67). We show the process in Table 11.3, step by step, corresponding to the calculation of the new equations for each case of C/R  {0.001, 0.003, 0.010}. Nos. 1 and 2 are copies form the inset table of Fig. 11.20. Prior to solving Eq. (11.65) graphically as shown in Fig. 11.25, we reviewed the non-dimensional coefficient, δ a , as shown in Table 11.3_No. 3, which shows the comparably large effect in the change from ω 1 to ω 1a for the case of C/R  0.010 and no significant changes for the other clearance cases. In fact, the solution of Eq. (11.65) gave unstable eigen frequencies ω 1a  {45.6, 42.4, 33.3} Hz in No. 4 for the anisotropic system, instead of ω 1 Hz in No. 1 for the isotropic system. Next we calculate the onset speed of the instability as shown in Nos. 5–9. The ratio δ k indicates stability increase due to the anisotropic magnitudes of bearing stiffness. The ratio δ c indicates stability decreases due to the anisotropic magnitudes of bearing damping. The ratio δ kc shows the stability influence generated by the product of anisotropic magnitudes of bearing stiffness and damping. The sum of these ratios is shown in Row 8. Owing to strong anisotropic stiffness for the case of C/R  0.010, the ratio δ k  +1.9 become comparably large thus increasing the stability limit. As shown in No. 9, we see the onset speeds, 1a  {91.0, 82.0, 83.5} rps, instead of the original values 1 in No. 2. Consequently, the lambda coefficients

11.2 Simplified Prediction of the Stability Limit for an Oil Film Bearing Supported …

389

Table 11.3 Parameter values influenced by backward components Note: F = only Forward whirl bearing force and FB = both of Forward and Backward whirl bearing forces C/R No.

0.001

0.003

0.010

λ = kc ( cd Ω )

0.5

0.503

0.493

1

F

ω 1 Hz

45.6

42.7

38.6

2

F

Ω1 = ω 1 λ

91

84.8

78.3

3

FB

δa

0.002

0.08

0.49

4

FB

ω 1a Hz

45.6

42.4

33.3

5

FB

δk

0.0004

0.042

1.9

6

FB

−δ c

− 0.002

− 0.08

− 0.5

7

FB

− δ kc

0

− 0.0014

− 0.87

8

FB

δ k − δ c − δ kc

− 0.001

− 0.042

0.53

9

FB

Ω1a_onset rps

91.0

82.0

83.5

10

FB

λ a = ω 1a / Ω 1a

0.50

0.51

0.40

are improved in estimation to λa  ω 1a /1a  {0.50, 0.51, 0.40} in No. 10. As indicated in Table 11.3, for each case of C/R  0.001 and 0.003, these ratios for the accuracy recovery are very small and the anisotropic effect can be considered as negligible. However, as we see for δ k  1.9, the system stability of C/R  0.010 is drastically recovered by the anisotropic stiffness and the stable zone can be extended. Therefore, the stability chart of Fig. 11.20, the forward component only of the system, is redrawn in Fig. 11.26, which takes account of the additional backward component, as stated in Table 11.3. We see the resulting unstable frequencies ω 1a  {45.6, 42.4, 33.3} Hz at the onset speeds 1a  {91, 82, 83.5} rps for C/R  {0.001, 0.003, 0.010}, respectively. The λ coefficients are thus calculated as λa  ω 1a /1a  {0.5, 0.51, 0.4}. These values are summarized in the inset table in the figure. Since the significant revision was achieved for C/R  0.010, all of these estimation values agree with exact values of the inset table , colored yellow at the bottom-right corner. The proposed estimation procedure is therefore acceptable. For further information, more exact eigenvalue analysis of R1_Fig. 7.21 was recalculated by assuming that the oil film bearing dynamic characteristics vary with the rotational speed. The eigen solution with rotational speed dependency yielded the following values: ω1e  {47.0, 43.5, 33.5} Hz@1e  {93.7, 86.8, 88.4} rps, where λe  ω1e /1e  {0.5, 0.5, 0.38}. These estimated values are thus validated by the experimental data reported in Ref. [295] or as seen in R1_Fig. 7.19:

390

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 50

C/R = 0.001 1 0.003 2

45.6 Hz

Frequency [Hz]

C Estimation ( Anisotropic Brg )

47.3 Hz

40

42.4 Hz

30

33.3 Hz

2 3 1 C/R  1a 45.6 42.4 33.3 1a 91 82 83.5 a 0.50 0.51 0.40 [Hz]

3

.00 2 0

0.010 3

1

.00 10

20

B Exact complex eigenvalue k and c at = 90rps ( Anisotropic Brg )

010 =0.

/R 3C 10

0

20

2 82

40 60 Rotational speed [rps]

1 3 83.5 91 rps 80

100

C/R  1e 1e e

2 3 1 45.6 42.4 33.3 91.1 82.3 83.6 0.5 0.51 0.4 [Hz]

Fig. 11.26 Stability limit revised by backward properties

ω1exp  {48.1, 43.1, 35.4} Hz at 1exp  {90.3, 87, 88} rps, where λexp  ω1exp /1exp  {0.53, 0.5, 0.4}.

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems [296] Blade-and-shaft coupled vibration is categorized into two groups as shown in R1_Table 10.4: (1) Torsional and axial vibrations of shaft coupling with nodal diameter κ  0 blade vibration. (2) Bending vibration of shaft coupling with nodal diameter κ  1 blade vibration. As mentioned in Sect. 9.3, conventional analysis of blade-and-shaft coupling vibration excludes the coupling of shaft axial vibration with blade vibration in Case (1), so coupling analysis is incomplete. Thus, a systematic and accurate method of analysis is proposed to integrate (1) and (2) above in a global manner.

11.3.1 Coupling Mass of Mode Synthesis Model Modeling by the mode synthesis method is described in R1_Chap. 4. Let us check it with the example below.

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems x0

x1 0.643

0.766

k0

391  1 (1 =1.14)

x2

0.542 2

1 1

m0

2

3

 2 ( 1 = 2.05) – 0.484

Fig. 11.27 2-DOF model with boundary basement x0

Example 11.3 Find the mode synthesis model corresponding to a system having three

masses and springs as shown in Fig. 11.27, where x 0 is the boundary coordinate (master coordinate), and x 1 and x 2 are internal coordinates (slave coordinates). Answer The equation of motion is expressed as M X¨ + K X  F

(11.68)

⎡ ⎤ ⎡ ⎤ k0 + 4 −1 −3 x0   Here, M  Diag. m 0 1 2 , K  ⎣ −1 3 −2 ⎦, X  ⎣ x1 ⎦ x2 −3 −2 5 When statically offsetting the boundary coordinate by unity (x 0  1), the forced deformation mode, δ, is defined by the shape of the static deflection of internal coordinates, i.e., {x 1  1, x 2  1}. Thus, it is the rigid body mode δ  [x 0 , x 1 , x 2 ]t  [1, 1, 1]t  1. Also, when setting the boundary coordinate to zero (x 0  0), eigen modes of the internal system are obtained as φ1 and φ2 as

11.27. The 

shown in Fig. corresponding first and second natural frequencies are ω1 ω2  1.145 2.05 . The coordinate transformation matrix used to obtain the mode synthesis model is defined as: ⎡

⎤ ⎡ ⎤⎡ ⎤  x0 1 0 0 x0   ⎣ x1 ⎦  ⎣ 1 0.643 0.766 ⎦⎣ η1 ⎦ ≡ δ [φ] x0 η x2 η2 1 0.542 −0.454

(11.69)

Here, eigen modes [φ] are normalized with respect to the mass matrix M, which means that φit Mi φi  1 for i  1, 2 The equation above yields the mode synthesis model as follows: 

m δ Mct Mc E



  x0 x¨0 k 0 + 0  2 0 η¨ η 0 ωn

(11.70)

392

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 

 1tF

m c 1 = 1.726 m c1

k0

1



m0 + 3 m c2 t

2

1

x0

 F

1

 12= 1.314

 22 = 4.186

mc 2 = – 0.142

 F t

Fig. 11.28 Mode synthesis model

t Here,  2  m δ  δ Mδ  m 0 + 3 (total mass of the global system), ωn  Diag.[1.314 4.186] (eigenvalues of the internal system) and

 Mc  [φ]t Mδ  [φ]t M1 ≡

m c1 m c2



 

1.726 −0.142



(coupling mass between the internal and boundary coordinates). The mass matrix described in Eq. (11.70) is not diagonal, so the corresponding m (mass)–k (spring) system cannot be drawn precisely. Figure 11.28 is one of our proposed schematic drawings of Eq. (11.70). The corresponding m–k system can be drawn precisely, excluding the off-diagonal elements, mc1  1.726 and mc2  − 0.142. To interpret off-diagonal elements visually, the system was originally drawn with a coupling mass, mc , connecting the physical coordinate, x 0 , for the boundary point and modal coordinate η of the eigen modes of internal vibration. This coupling mass, mc , located between the boundary coordinate and internal modal coordinate, is called the modal participation factor in the general code Nastran®, where it is defined as a mass coupling effect between the A-set and q-set.

11.3.2 Modeling of Mode Synthesis Method with 3D Finite Element Method (FEM) Assuming that vibration analysis of a general structure shown in Fig. 11.29(2) is conducted via FEM, set the origin as the boundary node (or A-set for Nastran®) [297]. Here, three translational and three rotational forced displacement modes against the origin should be considered, and they must correspond with a translational parallel static mode, δ p (forced displacement  1), and rotational linear mode, δ t , around the origin (slope  1 rad). These static modes are straight lines. Thus, the physical characteristics of the target vibration system are derived in terms of the rigid body form:

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems 2

1

y

393

T2

δt

R2

1

1

fixed 2

1

Eigen modes of inner system

T1

x

R1 Rigid modes

δp

R3

T3 z

Fig. 11.29 Vibration analysis using 3D-FEM

δ tp Mδ p  total mass δtt Mδt  moments of inertia around 3 axes through the origin

(11.71) (11.72)

As eigen modes of the internal system represent cases where the origin is fixed, it is drawn schematically as eigen mode φ of a cantilever beam shown on the left in Fig. 11.29(1). The modal coordinates form the q-set, and the number of eigenvectors is designated when conducting analysis with Nastran®. Then the coupling mass can be expressed by the inner product involving the mass matrix M between the forced displacement modes (δ p and δ t ) and the eigen modes (φ), as follows: δ tp Mφ for translational parallel linear mode δ p , 3 lines for x yz-directions (11.73) δtt Mφ for rotational linear mode δt around origin, 3 lines around x yz-axes (11.74) Thus, six coupling masses are obtained per eigen mode. Let us check the coupling mass with the example below. Example 11.4 The rotational shaft system as shown in Fig. 11.30 has three nodes (0, 1, 2) and 1-m pitch. The in-plane vibrational displacement at each node {translating deflection, tilting angle}  {xi θi } is taken into account. Here, node 0 at the left end is the boundary coordinate {x0 θ0 } for a bearing (master node), while nodes 1 and 2 are internal coordinates {x1 θ1 x2 θ2 } (slave nodes). Note that only the masses of the disks, 4 and 1, are considered, while their moments of inertia are ignored. The shaft is massless. Figure 11.30(2) represents the eigen mode, φ, when the left end is constrained. Find the values of rigid mass, coupling mass, and the mode synthesis model under those conditions.

Answer t  Physical coordinate X ≡ x0 θ0 x1 θ1 x2 θ2   Mass matrix M  Diag. 0 0 4 0 1 0 As shown in Fig. 11.30(1), the parallel mode, δ p , corresponds with the mode when the left end is offset by the displacement {x0  1 θ0  0}, and the tilting mode, δ t ,

394

11 Our Latest Topics Relating to Simplified Modeling of Rotating … l

x0

1

l

x1

x2

θ0 =1

x0

2

δt

φ2 ( ω 2 = 2.21)

δp

x0 =1 1 θ1

0 free T

x0 θ0

x1 θ1

m=4

z

2 θ2

x2 θ2

m=1

0 fixed

x0 =0 θ 0 =0

x1 φ 1 ( ω 1 = 0.51)

x2

1 θ1

2 θ2

Id 1 = 0

x1 θ1

Id 2 = 0

z x2 θ2

Fig. 11.30 Shaft with 2 disks and mode synthesis modes

is the mode when offset by the displacement of {x0  0 θ0  1}. Thus, the rigid mode matrix is determined as  t   101010 δ G ≡ δ p δt  011121

(11.75)

The elastic eigen mode matrix, , of the internal system is expressed with the proper values shown in Fig. 11.30(2) as follows: 

 ≡ φ1 φ2





0 0 0.28 0.475 0.828 0.548  0 0 −0.414 0.0628 0.561 1.431

t (11.76)

The transformation of coordinates from the original physical coordinate, X, to the coordinates of the mode synthesis model means that the sum of the boundary coordinates {x0 θ0 } and modal coordinates {η1 η2 } of the internal coordinates is determined as follows:    t  X  δG  x0 θ0 η1 η2 ≡ δG  wt

(11.77)

The congruence transformation of the above matrix yields the mass matrix of the mode synthesis model as  t   δG MδG δGt M δGt M δG   Mq  t 1 t MδG 

(11.78)

The matrix element (1, 1) above is rigid body mass, and element (2, 1) is coupling mass between the boundary and internal eigen mode. Element (2, 2) indicates that the eigen mode is normalized with respect to the mass matrix. Then, the equation of motion for the mode synthesis model is determined as follows: ¨ + Kq w  0 Mq w  where K q  Diag. 0

0 ω12  0.26

  t ω22  4.88 , w  x0 θ0 η1 η2

(11.79)

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

395

Table 11.4 Rigid and coupling masses Translating T: δ p

Tilting R: δ t

Translating T: δ p

δ p Mδ p = 5

δpt Mδt = 6

Tilting R: δ t

δtt Mδp = 6

δ tt M δt = 8

t

Eigen mode φ 1

mc11 = φ 1 M δ p = 1.949

mc12 = φ 1t M δ t = 2.777

Eigen mode φ 2

mc21 = φ 2t M δ p = – 1.096

mc22 = φ 2t M δ t = − 0.534

t

⎤ sym. δ tp Mδ p ⎥ ⎢ δ t Mδ p δ t Mδt t t ⎥ Mq  ⎢ t t ⎣ φ Mδ p φ Mδt 1 0 ⎦ 1 1 φt2 Mδ p φt2 Mδt 0 1 ⎡

Table 11.4 lists the specific values of the mass matrix of the mode synthesis model (M q ). Examples are shown for verification: Mq (2, 1)  δt (3) × 4 × δ p (3) + δt (5) × 1 × δ p (5)  6 Mq (3, 1)  φ1 (3) × 4 × δ p (3) + φ1 (5) × 1 × δ p (5)  1.946 Mq (4, 2)  φ2 (3) × 4 × δt (3) + φ2 (5) × 1 × δt (5)  −0.534.

11.3.3 Mode Synthesis Model of Blade The shapes of turbine blades are usually complicated as shown in R1_Fig. 1.1 and R1_Fig. 10.12. Here, the method of analyzing coupling vibration is demonstrated with a simplified blade system shown in Fig. 11.31. The boss that connects with the shaft is the boundary node, so the boundary boss is in the A-set and the blades are in the q-set of Nastran®. Boundary displacement at the boss node has 6 degrees of freedom (dof). Bending vibration has 4-dof, including translational displacements x 0 and y0 , for the X and Y directions, and tilting angles, θx0 and θy0 , around the Y and X axes, respectively. Assuming that the boss is fully constrained, vibration analysis for the global blade system provides eigen modes and eigen frequencies for each nodal diameter κ  0, 1, 2 …, as being calculated sequentially. However, the blade vibration for nodal diameters κ  0 and 1 is possible only when coupled with the shaft vibration. Also, note that nodal diameter κ  0 eigen mode couples with axial vibration, a0 , and torsional vibration, ϕ0 , of the shaft, and the nodal diameter κ  1 eigen mode couples with shaft bending vibration, as mentioned previously. Coupling mass is thus automatically calculated by a 3D-FEM program as shown in Table 11.5. Each row contains information on nodal diameter, mode number,

396

11 Our Latest Topics Relating to Simplified Modeling of Rotating … Y

Y

R1

X

R2 θ x0

θy 0

T2

ϕ0

1

Y

1

Z

X R3

T1

X

Z

Ry

Rx

1

1

T3

T3 (2) side view

(1) Impeller

(3) side view

Fig. 11.31 Definition of boss movement Table 11.5 Coupling masses for mode synthesis model of blade Nodal Diameter

Mode No.

κ

FREQ.

T3

R3

T1

R2

T2

− R1

Hz

a0

ϕ0

x0

θx0

y0

θy0

0

1: φ 0

f0

b1

b2

0

0

0

0

1

2: φ 1x

f1

0

0

a1

a2

a3

a4

1

3: φ 1y

f1

0

0

a3

± a1

± a2

a4

±

±

eigenvector, and eigen frequency. Eigen mode and eigen frequency of nodal diameter κ  0 are defined as φ0 and f 0 (Hz), respectively. There are also two eigen modes (φ1x and φ1y of nodal diameter κ  1) related to an identical frequency f 1 (Hz), where the eigen modes, φ1x or φ1y , have similar shapes, but the dominant vibrating portions of the mode shape are different in the X or Y directions, respectively. The columns represent each of the 6-dof of the boss. T1, T2, and T3 in the figure represent rigid parallel modes generated by forced unit displacements of {x 0  1, y0  1, a0  1} in the X, Y, and Z directions individually; likewise, R1, R2, and R3 represent rigid tilting modes according to forced unit angles of {θx0  1, θy0  1, ϕ0  1}. From the viewpoint of the rotation around the X axis, θy is negative as defined in R1_Fig.6.4b, prepared for shafting rotor dynamics. But the R1 mode of Fig.11.31(2) for blade dynamics is in the positive direction. Hence they have opposing definitions. Therefore, R2 and R1 directional tilting modes are applied to assess the coupling effect with the shaft vibrations of θx and θy . The independent calculation result of blade vibration itself yields the following coupling masses: {b1 b2 }  {φt0 M T3 φt0 M R3 }

(11.80)

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

{a1 a2 a3 a4 }  {φt1 M T1 φt1 M R2 φt1 M T2

397

− φt1 M R1 }

(11.81)

Here, Eq. (11.80) corresponds to the eigen mode, φ0 , of nodal diameter κ  0, and Eq. (11.81) corresponds to the eigen mode, φ1 , of nodal diameter κ  1. Once we obtain a1 , a2 , a3 , a4 for the φ0x mode, the next row for the φ0y mode is automatically determined as −a3 , −a4 , a1 , a2 or a3 , a4 , −a1 , −a2 . Non-coupling elements are zero in value. The coupling of shaft torsional and axial vibration under the modal coordinate η 0 of κ  0 blade vibration can be expressed as follows: ⎡

⎤⎡ ⎤ ⎤⎡ ⎤ ⎡ m 0 b1 00 0 a0 a¨ 0 ⎣ 0 I p b2 ⎦⎣ ϕ¨ 0 ⎦ + ⎣ 0 0 0 ⎦⎣ ϕ0 ⎦  0 b1 b2 1 η¨0 η0 0 0 ω02

(11.82)

Likewise, the coupling of the shaft bending and modal coordinates, η x and η y , with κ  1 blade vibration yields the equations of motion: ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎡ 0000 0 0 x0 x¨0 m 0 0 0 a1 −a3 ⎢ 0 I 0 0 a −a ⎥⎢ θ¨ ⎥ ⎢ 0 0 0 0 0 0 ⎥⎢ θ ⎥ d 2 4 ⎥⎢ 0x ⎥ ⎢ ⎥⎢ 0x ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎢ 0 0 m 0 a3 a1 ⎥⎢ y¨0 ⎥ ⎢ 0 0 0 0 0 0 ⎥⎢ y0 ⎥ (11.83) ⎥⎢ ⎥0 ⎢ ⎥⎢ ¨ ⎥ + ⎢ ⎢ 0 0 0 Id a4 a2 ⎥⎢ θ0y ⎥ ⎢ 0 0 0 0 0 0 ⎥⎢ θ0y ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥⎢ ⎣ a1 a2 a3 a4 1 0 ⎦⎣ η¨x ⎦ ⎣ 0 0 0 0 ω12 0 ⎦⎣ ηx ⎦ η¨ y ηy −a3 −a4 a1 a2 0 1 0 0 0 0 0 ω12 Equation (11.83) can be simplified by introducing complex displacements, z 0  x0 + j y0 , θ0  θ0x + jθ0y , η1  ηx + jη y , a13  a1 + ja3 and a24  a2 + ja4 as follows: ⎡

⎤⎡ ⎤ ⎡ ⎤⎡ ⎤ m 0 a13 z¨ 0 00 0 z0 ⎣ 0 Id a24 ⎦⎣ θ¨0 ⎦ + ⎣ 0 0 0 ⎦⎣ θ0 ⎦  0 a¯ 13 a¯ 24 1 η¨1 η1 0 0 ω12

(11.84)

The equation above is a degenerate model of the mode synthesis method corresponding to the non-rotating coordinate with   0. It can be modified to the equation below for the rotating coordinates with angular velocity  by considering that the gyroscopic factor of a bladed disk is equivalent to that of a thin disk, namely γ  2, as stated in R1_Sect. 6.3.2: Mq z¨ − jG q z˙ + K q z  0  t where z  z 0 θ0 η1

(11.85)

398

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

⎡

⎡ ⎡ ⎤ ⎤ ⎤ m 0 a13 0 0 0 00 0 ⎦ Mq  ⎣ 0 Id a24 ⎦, G q  ⎣ 0 I p 2a24 ⎦ and K q  ⎣ 0 0 0 2 2 a¯ 13 a¯ 24 1 0 2a¯ 24 2 0 0 ω1 −  Note that how to obtain Eq. (11.85) is explained mathematically in Appendix H. Example 11.5 Figure 11.32 shows a bladed disk subject to vibration analysis (single

blade mass ρAL  2090/8  261 kg, blade length L  1 m, boss radius ratio a  1/3) conducted with 3D-FEM, and Table 11.6 lists the calculation results. Verify the coupling masses shown in the table. Note that the data written in Table 11.6 are evaluated as follows: total mass m  N ρAL  2090 kg, polar moment of inertia  (1+a)L  x 2 dx  N ρAL I p  N ρA

1

(a L + ξ L)2 dξ

, and  m L 2 × 0.772  1614 kg × m2 transverse moment of inertia Id  I p /2  807 kg × m2 . aL

0

Answer Basic mathematical formulae are prepared for the single blade shown in Fig. 11.33. In the figure, the blade is modeled as a cantilever beam, and the first-order mode shown in R1_Table 3.1➅ is applied for the single blade eigen mode:

side view

3D

Fig. 11.32 Bladed disk (stagger angle  0°) Table 11.6 Coupling masses for mode synthesis model of bladed disk (stagger angle  0°) Nodal Diameter

Mode No.

κ

FREQ.

T3

R3

T1

R2

T2

− R1

fn Hz

a0

ϕ0

x0

θx0

y0

θy0

0

1: φ 0

107

b1 = 0

b2 = 37.9

0

0

0

0

1

2: φ 1x

107

0

0

0

0

25.3

0

1

3: φ 1y

107

0

0

− 25.3

0

0

0

Note: m = 2 090 kg, transverse Id = 807 kgm2 and polar Ip = 2×Id for rigid global blade

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

 φ1 

399

 sinh λξ − sin λξ √ cosh λξ − cos λξ − / 0.1082 cosh λ − cos ξ sinh λ − sin ξ

(11.86)

Here, λ  1.875. 1 Modal displacement is normalized by 0 φ21 (ξ)dξ  1. It is illustrated graphically as the φ1 curve. The translational parallel mode with displacement  1 is denoted by T, and the tilting mode with 1 rad rotation of the axis through the origin by R. Then, the coupling mass calculation is expressed as follows: 1

1 φ1 Tdξ 

φ1 dξ  0.783

0

0

1

1 φ1 Rdξ 

0

φ1 (a + ξ)dξ  0.569 + 0.783a → 0.83 if a  0.33

(11.87)

0

With the preparation above, solve the problem with the condition a  0.33, number of blades N  8, and stagger angle  0°. Vibration is only in the in-plane circumferential direction, so the eigen mode for the nodal diameter κ  0 gives the mass matrix, M q , of Eq. (11.82) in the following equivalent form. Then, it should be rewritten to replace the (3, 3) element by unity as follows: ⎡

⎡ ⎤ ⎤ 0 0 m 0 0 Mq  ⎣ I p 0.83N ρAL 2 ⎦ ⇒ Mq  ⎣ 0 I p 37.9 ⎦, sym. N ρAL 0 37.9 1 m

√ √ √ √ because b1  0, b2  ρAL 2 × 0.83 × N /( ρAL N )  ρAL L × 0.83 × N  37.9. For the normalization to be unity at only M q (3, 3), we applied similar matrix manipulation as shown in Eq. (11.41) or Eq. (G.4). Next, for the eigen mode with nodal diameter κ  1, the mass matrix, M q , of Eq. (11.84) is given, and it is then modified as to set the (3, 3) element to unity for the normalization as follows: Fig. 11.33 Blade eigen mode φ1 and rigid modes T and R

R

φ1

2.0 d.

1.0 1.0 0

1ra

T 0.5

Y aL Z

L

1.0

ξ X

400

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

⎡ ⎢ ⎢ Mq  ⎢ ⎣

m

N ρAL 2 0 N ρAL 2

0 j0.783

0

Id N ρAL − j0.783 0 2

⎤

⎡ ⎥ ⎥ ⎥ ⇒ Mq  ⎣ ⎦

⎤ m 0 j25.3 0 Id 0 ⎦ − j25.3 0 1

because a1  0, a2  0,     a3  ρAL × 0.783 × N /2/( ρAL N /2)  ρAL × 0.783 × N /2  25.3 and a4  0. Example 11.6 Figure 11.34 shows the bladed disk subject to vibration analysis (stagger angle  90°) conducted with 3D-FEM, and Table 11.7 lists the calculation results. Verify the coupling masses shown in this table.

Answer In the case of a  0.33, number of blades N  8, and stagger angle  90°, vibration is only in the out-of-phane direction, so the eigen mode for the nodal diameter κ  0 could similarly give the mass matrix of Eq. (11.82) in the following manner:

3D

side view

Fig. 11.34 Blade disk (stagger angle  90°) Table 11.7 Coupling masses for mode synthesis model of bladed disk (stagger angle  90°) Nodal Diameter

Mode No.

κ

FREQ.

T3

R3

T1

R2

T2

− R1

fn Hz

a0

ϕ0

x0

θx0

y0

θy0

0

1: φ 0

107

35.8

0

0

0

0

0

1

2: φ 1x

107

0

0

0

26.8

0

0

1

3: φ 1y

107

0

0

0

0

0

26.8

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

⎡

401

⎡ ⎤ ⎤ 0 0.783N ρAL m 0 35.8 ⎦ ⇒ Mq  ⎣ 0 I p 0 ⎦ Mq  ⎣ 0 Ip sym. N ρAL 35.8 0 1 m

√ √ because b1  ρAL × 0.783 × N  35.8 and b2  0. For eigen modes with the nodal diameter κ  1, the mass matrix of Eq. (11.84) is calculated similarly: ⎡ ⎤ m 0 0 ⎡ ⎤ m 0 0 ⎢ ⎥ N ρAL ⎢ ⎥ Id 0.83L Mq  ⎢ ⇒ Mq  ⎣ 0 Id 26.8 ⎦ 2 ⎥ ⎣ ⎦ N ρAL 0 26.8 1 sym. 2 √ √ because a1  0, a2  ρAL L × 0.83 × N /2  26.8, a3  0, and a4  0. Example 11.7 Figure 11.35 shows the bladed subject to vibration analysis (stagger

angle  30°) conducted with 3D-FEM, and Table 11.8 lists the calculation results. Verify the coupling masses shown in this table. Answer In the case of a  0.33, number of blades N  8, and stagger angle  30°, vibration is in both the cos30°  0.866 to in-plane circumferential direction and the sin30°  0.5 to out-of-plane direction, so the eigen mode for the nodal diameter κ  0 yields: 3D

side view

30°

Fig. 11.35 Blade disk (stagger angle  30°) Table 11.8 Coupling masses for mode synthesis model of bladed disk (stagger angle  30°) Nodal Diameter

Mode No.

κ

FREQ.

T3

R3

T1

R2

T2

− R1

fn Hz

a0

ϕ0

x0

θx0

y0

θy0

107

17.9

32.8

0

0

0

0

− 13.4

21.9

0

0

13.4

0

1: φ 0

1

2: φ 1x

107

0

0

0

1

3: φ 1y

107

0

0

21.9

0

402

11 Our Latest Topics Relating to Simplified Modeling of Rotating …

b1  b2 

 

√

N sin 30◦  17.9 √ ρAL L × 0.83 × N cos 30◦  32.8. ρAL × 0.783 ×

The eigen mode for the nodal diameter κ  1 yields:   a1  0, a2  ρAL L × 0.83 × N /2 sin 30◦  13.4,   a3  ρAL × 0.783 × N /2 cos 30◦  21.9, a4  0. Example 11.8 Coupled vibrations between shaft (torsion/axial) and blade (κ  0)

Let us demonstrate vibration analysis of blade (κ  0) and shaft torsional and axial vibration coupling [298]. The target blade–shaft system has two turbine-model bladed disks on both ends as shown in Fig. 11.36. Table 11.9 lists the coupling mass coefficients of the blade itself as obtained by 3D-FEM calculation. Figure 11.37 shows a schematic 1D-FEM model for calculation of a shafting system assuming rigid blades located on both sides, the shaft has an eigen frequency of 108 Hz for the torsional mode, and 1605 Hz for the axial mode. The following Eqs.

Shaft only with rigid blades

Blade only flexible blades

torsional mode ωt =108 Hz axial mode ωa =1,605 Hz

m = 2.77 k g -3 Id = 41.4×10 kg ∙ m2 I p = 2×Id

shaft system ≈ φ 35φ ×580 L mass = 9.19 kg

Node : 246,464

ωb = 77.2 Hz @ 600 rpm for nodal diameter κ =0

Elements : 187,136

Fig. 11.36 Test error for interaction of shaft–blade coupling vibration Table 11.9 Coupling masses for mode synthesis model of blade Nodal Diameter

Mode No.

κ

FREQ.

T3

R3

T1

R2

T2

− R1

fn Hz

a0

ϕ0

x0

θx0

y0

θy0

0

0

0

− 0.0207

0.125

0.0189

0.137

0

1: φ 0

77.2

0.768

0.0987

0

1

2: φ 1x

144

0

0

0.0189

1

3: φ 1y

144

0

0

0.0207 − 0. 125

0.137

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

403

m = 2.77 kg -3

Blade

Boss @15

60 40 20 0

200

I d = 41.4×10 kg ∙ m 2 I p = 2×Id ωb = 77.2Hz @ 600rpm for nodal diameter κ =0

400

600

Bearing @node 7

Blade

Boss @78

800 Bearing @node 84

Fig. 11.37 Computational model for shaft independent system

(11.88) and (11.89), of the 2-dof equivalent shafting model, are obtained by applying the Guyan reduction technique according to the designation of both bosses as master coordinates; here, torsional displacement at the blade boss and axial displacement are defined as a and ϕ, with suffixes of station numbers, 15 and 78, for the left and right blading, respectively. The 1D-FEM model provides a reduced order model for axial vibration as follows: Ma a¨ + K a a  0

(11.88)

  1 −1 6.76 0.7 a15 8 . where Ma  ,a  , K a  3.04 × 10 a78 −1 1 0.7 6.58 It also gives the reduced order model for torsional vibration as 

Mt ϕ¨ + K t ϕ  0  where Mt 

(11.89)

  1 −1 84.7 0.1132 ϕ15 . ,ϕ × 10−3 , K t  19.7 × 10−3 ϕ78 −1 1 0.113 84.6

(1) Find eigen frequencies corresponding to the eigen modes of the blade–shaft coupling (both torsional and axial). Refer to Fig. 11.38. (2) Find eigen frequencies corresponding to the eigen modes of blade–shaft torsional coupling (disregarding axial coupling). (3) Compare and discuss the split ratio  (ASO−PSI)/PSI in cases (1) and (2) above.

404

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 1 (T) only

Relative Response

0

2 (A+T) coupled 2 73.7Hz

1 73.4 PSI mode

2 86.1Hz

1 82.1 0

ASO mode 1 121

PSO mode

2 121 Hz

0

(1) Shaft-blade coupled modes

Shaft Only

Coupled System

Blade Only

108

73.7

86.1

121

77.2 50

100 Frequency (Hz)

150

(2) Split of eigen frequencies

Fig. 11.38 Eigen modes and natural frequencies

Answer (1) Appropriate coordinates for both blades are defined as η 1 and η 2 , and the equation of motion for the blade–shaft coupling system is determined as follows: ⎡

⎤⎡ ⎤ ⎤⎡ ⎤ ⎡ Ma 0 Mc1 a a¨ Ka 0 0 ⎣ 0 Mt Mc2 ⎦⎣ ϕ¨ ⎦ + ⎣ 0 K t 0 ⎦⎣ ϕ ⎦  0   Mc1 Mc2 1 0 0 ωb2 η η¨

(11.90)

  η 1 0  2 2 1 0 ,η 1 where 1  , ωb  (2π × 77.2) η2 01 01   10 −3 1 0 Mc1  0.768 , Mc2  98.7 × 10 01 01 

The answer is {PSI  73.7, ASO  86.1, PSO  121} Hz, which corresponds with the modes shown in the figure. (2) {PSI  73.4, ASO  82.1, PSO  121} Hz are calculated by assigning no axial vibration a  0 to the equation above; it is the eigenvalue analysis without the influence of the axial vibration. (3) The split ratio is 17% in the case (1) and 12% in the case (2) [298]; moreover, the ASO eigen frequency changes significantly. These results suggest that the coupling with shaft axial vibration should be considered in order to accurately grasp the coupled eigen frequencies. ISO22266-1:2009 [250] currently recommends vibration analysis be undertaken with blade-and-shaft torsional vibration coupling, but the results suggest that such an approach would be inadequate.

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems (Bearing span) Ball bearing (Blade span)

405

Ball bearing 13 Kbrg = 10 [ N/m ]

Fig. 11.39 Test rotor system

Example 11.9 Coupled vibrations between shaft (bending) and blade (κ  1)

Another coupling effect may be evidenced between shaft bending vibration and blade vibration with κ  1 of nodal diameter. Figures 11.36 and 11.37 show the target blade–shaft system, which is supported by ball bearings in the test equipment shown in Fig. 11.39. (A) Shafting Only: Assuming that the blades are rigid, the equation of motion for the uncoupled shafting system is written as follows, using Z (complex displacement vector), M (mass including rigid blades), K (stiffness), and G (gyroscopic) matrices: M Z¨ − jG Z˙ + K Z  0

(11.91)

At standstill with   0, two eigen frequencies, ω 1  96.5 and ω 2  282 Hz, and their eigen modes, φ1 and φ2 , are shown in Fig. 11.40, called the first and second bending modes, respectively. We read the shafting deflection and angle (δ, θ) of the eigen mode shapes at the blading station as shown in the figure, which could be applied to modal analysis afterward to create a reduced order model. During the rotation, eigen frequencies behave as shown in Fig. 11.41. They split in two ways for forward (upward curve) and backward (downward curve) whirling motions. if we apply the modal transformation by using the modal matrix   In addition,  ≡ φ1 φ2 , obtained at 0 rps, the following equation of motion for the modal t  coordinates, σ ≡ σ1 σ2 , is obtained: M ∗ σ¨ − jG ∗ σ˙ + K ∗ σ  0

(11.92)

  ∗ 0.506 0 m1 0  where M ∗ ≡ t M ≡ , ki∗  m i∗ ωi2 (i  1, 2), 0 m ∗2 0 0.276  0.301 −0.029 G ∗ ≡ t G  and −0.029 0.282

406

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 0.2

Blade @15

0.1 0

Blade @78

δ = – 0.098 θ = – 1.175 200

δ = 0.072 θ = – 1.39

400

– 0.1

600

800

φ 2 ( ω 2 = 282 Hz )

δ = – 0.07 θ = 1.35

– 0.2

δ = – 0.109 θ = – 1.32

– 0.3

φ 1 ( ω 1 = 96.5 Hz )

Fig. 11.40 First and second eigen modes only for shafting (0 rps) 200 e 0 = inertial frame e r = rotational frame

Eigen frequency [ Hz ]

180 160 141

+ ωb +

Blading

Ω on e 0

Blade frequency

– ωb + Ω on e 0

120

Forward

96.5

100

145.4

± ωb on e r

140

106 Backward Shafting

80 0

5

10 15 20 Rotational speed [rps]

88 25

30

Fig. 11.41 Uncoupled eigen frequency map



k∗ 0 K ≡  K ≡ 1 ∗ 0 k2 ∗

t





0 186 × 103  0 866 × 103



(B) Blading Only; The blading structure is considered to be flexible as detailed in Figs. 11.42 and 11.43. The vibration of the cyclic structure may be analyzed efficiently by 3D-FEM codes, and the first eigen frequency, ω b , of single nodal diameter (κ  1) behaves as shown by the green line of Fig. 11.41. It varies from ω b  141 Hz at   0 rps to ω b  145.4 Hz at   30 rps. Note that this is the value observed on the rotational coordinate, er . It is rewritten in term of the two values, ±ωb + , measured on the inertial coordinate e0 . At the same time, we know the coupling mass parameters, a1 to a4 , as shown in Table 11.9. (C) Coupled Blade-Shaft; As shown in Fig. 11.37, the left and right blades are connected with the shaft at nodal stations 15 and 78, respectively. The displacements (deflection and angle; δ 15 , θ15 , δ 78 , and θ78 ) of the first and second eigen modes are listed in Fig. 11.40. When the complex displacement vectors

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

407

Fig. 11.42 Test disk with continuous coupled blades

Fig. 11.43 Blade–disk mesh model Cover

Cyclic symmetry

Nodes : 2007 Elements : 1334

Cyclic symmetry Disk

of other nodal points are included in the variable Z 0 , the equation of motion of the global system for blade–shaft coupled system is written in the following 1D-FEM manner: M7 Z¨ 7 − jG 7 Z˙ 7 + K 7 Z 7  0

(11.93)

t  t  where Z 7 ≡ Z η1 η2 ≡ Z 0 δ15 θ15 δ78 θ78 η1 η2 ηi (i  1, 2): normal coordinates for the first eigen modes of left and right blades.

408

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 200 173 Eigen frequency [Hz]

+ 196

F

180

+ 165 –161

B

160 133

F

1D-FEM 3D-FEM

140 120 89.5

100

– 106 + 104

Backward

80 60

B Forward

0

5

– 74.8 10 15 20 Rotational speed [rps]

25

30

Fig. 11.44 Coupled eigen frequencies including comparison with 3D-FEM

⎡

0 a13 a24 0 0 1 0

⎢ ⎢ ⎢ [M] ⎢ ⎢ M7  ⎢ ⎢ ⎢ ⎢ ⎣ 0 a¯ 13 a¯ 24 0 0 0 0 0 a¯ 13 a¯ 24 ⎡ 0 ⎢ 0 ⎢ ⎢ 2a24 [G] ⎢ ⎢ G7  ⎢ 0 ⎢ ⎢ 0 ⎢ ⎣ 0 0 2a¯ 24 0 0 2 000 0 2a¯ 24 0

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥   ⎥ a13 ⎥, K 7  Diagonal [K ] wb2 wb2 , ⎥ a24 ⎥ ⎥ 0 ⎦ 1 ⎤ 0 ⎧ 2 wb ≡ ωb2 − 2 ⎪ ⎥ ⎪ 0 ⎥ ⎪ ⎪ ⎪ ⎪ 0 ⎥ ⎨ a13 ≡ a1 + ja3 ⎥ ⎥  0.0189 − j0.0207 0 ⎥ and ⎪ ⎥ ⎪ ⎪ 2a24 ⎥ a24  a2 + ja4 ⎪ ⎪ ⎥ ⎪ ⎩ 0 ⎦  0.137 + j0.125 2

The coupled vibration may be analyzed using Eq. (11.93) by the 1D-FEM code, e.g., MyROT introduced in R1_Chap. 12. Eigenvalue analysis of the above equation provides the relation between the eigen frequency and rotational speed, as shown in Fig. 11.44. The solid lines denote forward and dotted lines backward whirl motions. The coupled eigen frequency points are very different from uncoupled lines in Fig. 11.41. That is the reason why we need to recalculate the eigen frequency of the coupled system at the final design stage. Eigen frequencies of the coupled system are also calculated by 3D-FEM and the results are plotted by small white-painted blue circles. The comparison between the 1D- and 3D-FEMs agrees so well that the effectiveness of the reduced order modeling for blading is demonstrated. In the following step, we consider how to create a lower order model, which could be beneficial for an accurate approximation of eigen solutions. To implement

11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

409

it, we will define a transformation matrix, Tq , for model order reduction by mode synthesis. If we assume that there is no A-set and only two q-sets for the shafting, only the modal reduction is then taken into account in the modal matrix  stated above (section (A): Shafting Only). The matrix Tq is thus formulated as ⎡

φ10 ⎢ −0.109 ⎡ ⎤ ⎢ ⎢ −1.32 φ1 φ 2 0 0 ⎢ ⎢ Tq  ⎣ 0 0 1 0 ⎦  ⎢ −0.07 ⎢ ⎢ 1.35 0 0 01 ⎢ ⎣ 0 0

φ20 −0.098 −1.175 0.072 −1.39 0 0

0 0 0 0 0 1 0

⎤ 0 0⎥ ⎥ 0⎥ ⎥ ⎥ 0⎥ ⎥ 0⎥ ⎥ 0⎦ 1

for ⎡

⎤ Z0 ⎢δ ⎥ ⎡ ⎤ 15 ⎥ ⎡ ⎤ ⎢ σ1 ⎢ ⎥ Z ⎢ θ15 ⎥ ⎢ σ2 ⎥ ⎢ ⎥ ⎥ Z 7  ⎣ η1 ⎦ ≡ ⎢ δ78 ⎥  Tq ⎢ ⎣ η1 ⎦ ⎢ ⎥ ⎢ θ75 ⎥ η2 ⎢ ⎥ η2 ⎣ η1 ⎦ η2

(11.94)

The congruence transformation by Eq. (11.94) for Eq. (11.93) yields the following equation of motion for mode synthesis mode of the 4-DOF system: Mq q¨ − jG q q˙ + K q q  0

(11.95)

 t where q ≡ σ1 σ2 η1 η2 , E 2  2-dimensional unit matrix,  ∗  ∗ M Mcq G G cq , G q ≡ Tqt G 7 Tq  ¯ t and Mq  Tqt M7 Tq  ¯ t Mcq E 2 G cq 2E 2  ∗ 0 K t K q ≡ Tq K 7 Tq  0 ωb2 − 2 E 2 (1) M*, G*, and K* are already known by Eq. (1.92). What are the coupling terms of M cq and Gcq ? (2) Transition values of eigen frequency before and after coupling Eq. (11.93) are drawn in Fig. 11.45. What are the eigen frequencies of Eq. (11.95) for the 4-DOF system when   0? Note that the first three eigen mode shapes for the coupled system are indicated in Fig. 11.46. In the first coupled mode, the shaft vibrates mainly in the first bending mode with small amplitude of blading in the out-of-phase. In the second coupled mode, the blade vibrates mainly in-phase with small amplitude of the

11 Our Latest Topics Relating to Simplified Modeling of Rotating … 200

1D-FEM 3D-FEM

150 133 100

96.5

50

ωb =141 89.5

Shaft only

Eigen frequency [Hz]

173

Blade-Shaft coupling

Blade only

410

Fig. 11.45 Transition behavior pre-/post-coupling (  0 rps) f1 = 89.5 Hz

f2 = 133 Hz

shaft

shaft

2nd bent

1st bent

Blade : out-of-phase

Blade : in-phase

f3 = 173 Hz

shaft

1st bent

Blade : out-of-phase

Fig. 11.46 Mode shapes of blade–shaft coupled system

shaft second bending mode. The third coupled mode indicates that both shaft and blade vibrations are of comparable importance in the first shaft bending mode and the blading reverse phase (out-of-phase) mode. (3) The transition values before and after coupling Eq. (11.93) at 30 rps are drawn in Fig. 11.47. What are the eigen frequencies of Eq. (11.95), if   30 rps? Answer (1) From the corner parts of M q and Gq for the mode synthesis reduction:  Mcq  Tqt M7 Tq 

−0.109a13 − 1.32a24 −0.07a13 + 1.35a24 −0.098a13 − 1.175a24 0.072a13 − 1.39a24



11.3 Vibration Analysis of Blade-and-Shaft Coupled Systems

Eigen frequency [ Hz ]

200

196 165

106 Hz

411 175 Hz

104

100

Note ωb = 145.4 Hz 0

1D-FEM 3D-FEM

– 74.8 – 100

– 106

– 88

– 115

– 161 – 200

Shaft only

coupled

Blade only

Fig. 11.47 Transition behavior pre-/post-coupling (  30 rps)

   G cq  Tqt G 7 Tq 

−2.64a24 −2.35a24

 −0.183 − j0.163 0.183 + j0.170 −0.163 − j0.145 −0.189 − j0.175    −0.362 − j0.330 0.37 + j0.337 2.7a24  −0.322 − j0.294 −0.380 − j0.347 −2.78a24

  (2) Solving ωn2 Mq − K q   0 with ω b  141 Hz gives eigen frequencies for  

 0: ωn  89.5 133 174 389 Hz   (3) Solving  Mq s 2 − jG q s + K q   0 with ω b  145.4 Hz gives eigen frequencies for   30 rps:

 s  j 104 167 197 396 Hz and

 s  j −74.7 −108 −162 −385 Hz Note that these approximate solutions of (2) and (3) above agree well with the values calculated by the 1D-FEM (large red circles) and 3D-FEM (small whitepainted blue circles), as seen in Figs. 11.45 and 11.47, respectively.

Chapter 12

Exercises of ISO Certification Examination for Vibration Experts

Abstract According to ISO 18436-2, entitled “Condition monitoring and diagnostics of machines—Requirements for training and certification of personnel—Part 2: Vibration condition monitoring and diagnostics,” the introduction of the certification examination for vibration experts is mentioned. The certification level is divided into four categories: 1—for beginners; 2—for elemental experts (apprentices); 3—for advanced experts; and 4—for super-experts. This chapter, including 100 questions, is prepared for ambitious experts challenging categories 3, 4, and higher. Every question requires knowledge strongly related to basic engineering mathematics, practical signal processing, vibrational dynamics theory, standard modal analysis, model order reduction, and so forth. It excludes knowledge gained purely from experience. If you are students, it is a good opportunity to know how to apply undergraduate knowledge to the field. If you are vibration consultants, apply these simple and understandable modeling techniques in troubleshooting for your customers. Based on the experience of one of the authors as a JSME (Japan) examiner, he found that questions of the examination are divided approximately into two groups. The first group includes questions related to knowledge of field experience in the maintenance service, for example, permissible vibration levels decided by API and/or ISO standards, how to measure and report vibration data, and so on. The second group involves questions concerning mathematics and dynamics, for example, rotor balancing, the FFT algorithm, modal analysis, applying complex numbers. Many examinees may have difficulty to recall and manage the theoretical background for questions if they graduated some time ago. A total of 100 questions have been selected, specifically to enrich an expert’s understanding of the mathematics and dynamics for them. The following advises the reader on how to use this chapter: (1) Questions 1–30: The first 30 are multiple choice questions with five options. It is in the same style as the certification exam; hence, examinees can check their level. The actual exam contains 100 questions in a 4–5 h period. Therefore, it is recommended that you complete these 30 questions within 60 min. Finally, check your answers with the correct answers in Table Q30, located in Sect. 12.1. (2) Questions 31–100: The aim of these questions is to sharpen examinees’ thinking, so the questions may be slightly beyond their knowledge. On average, the difficulty of these questions is almost equivalent to vibration category level 3 © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0_12

413

414

12 Exercises of ISO Certification Examination for Vibration Experts

and 4; some may be near category 4 and higher. Do not be afraid to try these questions to test your knowledge. From Question 31 onwards, the content is divided into two ways: Questions stated in Sect. 12.2 with answers and hints associated with each question in Sect. 12.3. Please try these exercises, which challenge readers to become world-leading experts in the rotor dynamic vibration field. Keywords ISO 18436-2/3 · Expert certification · Vibration monitoring · Analysis and diagnosis · Rotor dynamic-related knowledge · FFT-related knowledge · System modeling · Practical balancing · Vibration characteristics evaluation Note (1) ISO 18436-2 states the requirement for vibration experts as follows: Vibration analysis using measurements to monitor condition and diagnose faults in machinery has become a key activity in predictive maintenance programs for most industries. Those in the manufacturing industry who have diligently and consistently applied these techniques have experienced a return on investment far exceeding their expectations. However, the effectiveness of these programs depends on the capabilities of individuals who perform the measurements and analyze the data. ISO 18436 defines the requirements against which personnel in the machine condition monitoring and diagnostics technologies associated with vibration analysis are to be certified and the methods of testing such personnel. Conformity assessment for certification in vibration analysis will be performed by a body accredited to the requirements of ISO 18436-3. (2) At the time of writing this book, the Vibration Institute (VI) and the Mobius Institute in the USA, the Japan Society of Mechanical Engineers (JSME) in Japan, the Korean Society for Noise and Vibration Engineering (KSNVE) in Korea, and the British Institute of NDT (BINDT) in the UK are conducting the certification exam.

12.1 The First 30 Questions and Multiple Choice Answers Question 1 Determine the correct combination (1, 2, 3, 4, or 5) of (a), (b), and (c) to complete the conversion of vibration magnitudes as shown in Fig. Q1.

12.1 The First 30 Questions and Multiple Choice Answers

peak-peak

2

415

÷2

overall peak

peak (c)

rms

(a)

(b)

Fig. Q1 Conversion of vibration magnitudes

Question 2 Figure Q2 shows the key phase pulse signal and rotor vibration waveform. What is the rotational speed (rpm)? 6000 3000 1980 1500 60

Displacement [µm]

1. 2. 3. 4. 5.

40 30 20 10

rotational pulse 0.01

0.02 vibration waveform

0

Fig. Q2

0.03

0.04

sec

416

12 Exercises of ISO Certification Examination for Vibration Experts

Question 3 What is the peak-to-peak amplitude (μm) in Fig. Q2? 1. 2. 3. 4. 5.

36 μm pp 18 μm pp 15 μm pp 10 μm pp 8 μm pp

Question 4 What is the phase lag angle between the pulse and the vibration peak in Fig. Q2? 1. 2. 3. 4. 5.

0° 30° 60° 120° 360°

Question 5 A vibration vector, which consists of the amplitude from Q. 3 and the phase angle from Q. 4, is plotted in Fig. Q5. What is the correct point of the vibration vector? 1. 2. 3. 4. 5.

point A point B point C point D point E 90° C B

D

Phase lag A

180° E

10

0° 20 m

270°

Fig. Q5

Question 6 −→ For field balancing, we measured an initial vibration vector noted OA in Fig. Q6(1). We put a trial weight of 5 g on the rotor as shown in Fig. Q6(2) and then measured −→ a trial vibration vector noted OB in Fig. Q6(1). What is the amplitude and phase lag −→ angle of the initial vibration vector OA?

12.1 The First 30 Questions and Multiple Choice Answers

1. 2. 3. 4. 5.

417

60 μm 60° 80 μm 90° 80 μm 210° 139 μm 240° 139 μm 270° 90° B Trial run Phase lag 180°

O

A Initial run

0° 100 Pm

5g

(rotation)

120°

key phase

270° ( 1 ) Vibration vector on polar plot

( 2 ) Cross-section of rotor

Fig. Q6

Question 7 −→ What is the amplitude and phase lag angle of the trial vibration vector OB in Fig. Q6(1)? 1. 2. 3. 4. 5.

60 μm 60° 80 μm 90° 80 μm 210° 139 μm 240° 139 μm 270°

Question 8 What is the angle of  OAB in Fig. Q6(1)? 1. 2. 3. 4. 5.

0° 30° 60° 90° 120°

Question 9 −→ What is the effective vector AB, i.e., the amplitude  phase lag angle, as shown in Fig. Q6(1)? 1. 60 μm 60° 2. 80 μm 90° 3. 80 μm 210°

418

12 Exercises of ISO Certification Examination for Vibration Experts

4. 139 μm 60° 5. 139 μm 90° Question 10 What is the correction weight for balancing after removing the trial weight? 1. 2. 3. 4. 5.

0.5 g 3g 5g 10 g 15 g

Question 11 Where should we add the correction weight in Fig. Q11? 1. 2. 3. 4. 5.

point A point B point C point D point E C B

(rotation)

A

D 120°

E

key phase

Fig. Q11

Question 12 A waveform in the time domain and the FFT spectrum in the frequency domain are shown in the upper and lower plots of Fig. Q12, respectively. The sampling rate of the waveform is 1024 points in a time interval of 4 s. What is the sampling frequency? 1. 2. 3. 4. 5.

256 Hz 512 Hz 1024 Hz 2048 Hz 4096 Hz

12.1 The First 30 Questions and Multiple Choice Answers

419

50

REAL mV x1 – 50 0 MAG dBV

4s

– 40 – 60 – 80 0 X : 68.5 Hz

Fig. Q12

100 Hz Y : – 45.44 dBV ( 0 dBV = 1V )

Question 13 What is the frequency resolution in Fig. Q12? 1. 2. 3. 4. 5.

0.01 Hz 0.05 Hz 0.125 Hz 0.25 Hz 0.5 Hz

Question 14 What is the line number on the FFT diagram when the frequency span is 0–100 Hz in Fig. Q12? 1. 2. 3. 4. 5.

200 400 800 1024 2024

Question 15 What are the dominant frequency ( ) Hz and the corresponding vibration amplitude ) dBV  ( ) mV in Fig. Q12? (

1. 2. 3. 4. 5.

68 Hz 10 Hz 68 Hz 10 Hz 68 Hz

-45 dBV -58 dBV -58 dBV -45 dBV -45 dBV

2.2 mV 5.6 mV 2.2 mV 5.6 mV 5.6 mV

420

12 Exercises of ISO Certification Examination for Vibration Experts

Question 16 What is the overall value indicated on the right side of the FFT diagram as designated by the symbol  in Fig. Q12?

1. 2. 3. 4. 5.

-40 dBV -45 dBV -45 dBV -58 dBV -58 dBV

2.0 mV 2.2 mV 5.6 mV 2.2 mV 5.6 mV

Question 17 In the case of Fig. Q12, suppose we changed the measuring time from 4 to 8 s. What would be the maximum frequency Hz on the frequency axis of the FFT display? 1. 2. 3. 4. 5.

25 Hz 50 Hz 100 Hz 200 Hz 500 Hz

Magnitude [V]

Question 18 We measured a vibration waveform by an impulse test as shown in Fig. Q18a. Select the best corresponding spectrum from the FFT displays shown below. 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 0.0

0.2

0.4 0.6 0.8 Time [sec]

Fig. Q18(a) Waveform

1.0

12.1 The First 30 Questions and Multiple Choice Answers

Magnitude [dBV]

(3)

(2)

0 10 20 30 40 50 60 70 80 90 100 Frequency [Hz]

10 0 -10 -20 -30 -40 -50

Magnitude [dBV]

10 0 -10 -20 -30 -40 -50

(4)

0 10 20 30 40 50 60 70 80 90 100 Frequency [Hz]

Magnitude [dBV]

(5)

10 0 -10 -20 -30 -40 -50

Magnitude [dBV]

Magnitude [dBV]

(1)

10 0 -10 -20 -30 -40 -50

10 0 -10 -20 -30 -40 -50

421

0 10 20 30 40 50 60 70 80 90 100 Frequency [Hz]

0 10 20 30 40 50 60 70 80 90 100 Frequency [Hz]

0 10 20 30 40 50 60 70 80 90 100 Frequency [Hz]

Fig. Q18(b) Corresponding FFT

Question 19 Table Q19 shows the monthly trend of vibration velocity of a machine. What is the most suitable graphical form from 1, 2, 3, 4, or 5 for displaying the trend? Table Q19 Monthly vibration level Measurement month April May June July August September October November December

Alerm Vib. velocity level (mm/s) (mm/s) 19 30 23 30 18 30 19 30 30 22 19 30 27 30 30 26 27 30

422

12 Exercises of ISO Certification Examination for Vibration Experts

Vib. velocity [mm/s]

(1)

4

5

6

7 8 9 10 11 12 Month

Alarm level

Month 12 11 10 9 8 7 6 5 4

( 3 ) Vibration velocity [mm/s] 19

27 26 Fig. 019(1) 27

0 (4)

Alarm level

10 20 30 Vibration velocity [mm/s]

Vibration velocity [mm/s] 30 4 Alarm level 12 5 20 10

11

6

0

7

10 9

8 Month

40

19

22

Month 4 23 5 6 7 18 8 9 10 19 11 12

( 5 ) 35 Vib. velocity [mm/s]

(2)

35 30 25 20 15 10 5 0

Alarm level

30 25 20 15 10 5 0 4

6

8 Month

10

12

Fig. Q19 Trend data diagrams

Question 20 What action is required after reading the trend data of Fig. Q19? 1. Though an increase of vibration level was recognized, we do not need to be concerned about the machine condition because the present level is still less than the alarm level. 2. Around October, a large increase of vibration level is recognized, but we do not need to be concerned about the machine condition because no further increase of the vibration level occurred in the following months. 3. Around October, a large increase of vibration level is recognized. We need to be concerned about the machine condition but not stop the machine because the present level is still less than the alarm level. 4. Around October, a large increase of vibration level was recognized. We recommend operators to stop the machine immediately. 5. The present level is still less than the alarm level, so we do not need to do anything for machine maintenance.

12.1 The First 30 Questions and Multiple Choice Answers

423

Question 21 Figure Q21 shows the section of a rotor when it crosses the x-axis on a whirling orbit. If a bolt is removed from the rotor when passing the critical speed, what is the correct drawing of the rotor cross-section at unbalance vibration due to the loss of a bolt? Note that  is the rotational spin direction in these figures. 1. 2. 3. 4. 5.

Fig. A Fig. B Fig. C Fig. D none of these figures Y Whirl orbit

Y

bolt removed

bolt removed X

X Cross-section of rotor A Y

bolt removed

B Y

X

X

C

D

bolt removed

Fig. Q21 Drawings of rotor sections at unbalance vibration due to the bolt removed

Question 22 Figure Q22(1) shows vibration waveforms measured at four different locations along the shaft, denoted by x i (i  1–4), of a machine during a harmonic excitation test. We can guess the mode shape of the machine by reading these vibration data. Select the correct mode shape in Fig. Q22(2).

12 Exercises of ISO Certification Examination for Vibration Experts

x3 [µm]

x2 [µm]

x1 [µm]

424 2 0 -2 4 2 0 -2 -4 4 2 0 -2 -4

0.1

0.2

0.3

0.4

t [s]

x1

x2

1

x4 [µm]

x4

1. 0.1

0.2

0.3

0.4

t [s] 1

-2 2

2

-3 3

2. 2 0.1

0.2

0.3

0.4

t [s]

3.

1 -2

-3 3

6 4 2 0 -2 -4 -6

x3 2

2

2

4.

0.1

0.2

0.3

0.4

2

1

t [s]

1

2

5. -3

Fig. Q22(2) Mode shapes

Fig. Q22(1) Waveforms

Question 23 What are the natural frequency (Hz), damping ratio, and Q-value by reading from the damped vibration waveform as shown in Fig. Q23?

1. 2. 3. 4. 5.

Natural freq.(Hz)

Damping ratio Q-value

0.5 0.5 1.0 2.0 2.0

0.01 0.05 0.1 0.05 0.1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6

100 10 10 20 5

1.0

time [s]

Fig. Q23 Damped vibration

Question 24 Figure Q24(1) shows a rotor model consisting of a shaft and a disk, denoted by ms and m for each mass, respectively. The mass eccentricity of the disk is ε  10 μm.

12.1 The First 30 Questions and Multiple Choice Answers

425

What is the mass eccentricity ε* of the rotor considering both weights of the shaft ms and the disk m? 1. 2. 3. 4. 5.

4 μm 8 μm 10 μm 20 μm 40 μm l = 750 mm mε

lb = 500 mm

θ

m 40kg

ms = 30kg

2 j t

e

mε ms 3

x

m

2 j t

e

x

A kb

cb

( h = lb / l )

kb h 2

cb h2

x sensor

Disc Vib. [μm]

( 1 ) Rotor

( 2 ) Equivalent 1-dof model

Apeak

ε* 0

Rotational speed

[rps]

c ( 3 ) Resonance curve

Fig. Q24 Unbalance vibration of an overhung rotor

Question 25 Figure Q23 shows a damped vibration waveform measured from an impulse test of the rotor shown in Fig. Q24(1). When we rotate this rotor, what is the peak amplitude Apeak of unbalance vibration of Fig.Q24(3)?

Apeak 1. 2. 3. 4. 5.

4 8 10 20 40

m m m m m

Question 26 In Fig. Q26, figure (1) shows a damped vibration waveform measured from an impulse test of a rotating shaft and figure (2) indicate the relationship of enveloped amplitudes bi and bi+1 measured at each time interval of one second. What is the damping ratio of the system?

426

1. 2. 3. 4. 5.

12 Exercises of ISO Certification Examination for Vibration Experts

0.12 0.09 0.06 0.003 0.001 50 b0 =46 20 0 – 20 – 50

detailed b1 =26 b2 =14 b3=8 b4 =4 b5 =2 t [s]

fn [Hz]

30 0

1.0

1.1

[s] 1.2

– 30

1 4 5 [s] 2 3 zoomed ( a ) Waveform part

( b ) Zoomed

Fig. Q26(1) Vibration waveform of lightly damping system

50 40 30 bi 20 10 0

0

5

10 15 20 25 30 b i+1

Fig. Q26(2) Plot of amplitude data  b1 b5 

Question 27 A motor (mass 200 kg) is coupled to an overhung fan (unknown mass) through the massless shaft as shown in Fig. Q27. The weight of the fan makes the static deflection of 100 μm with the elastic shaft as shown in the figure. What is the natural frequency of the fan lateral vibration? 1. 2. 3. 4. 5.

5 Hz 25 Hz 50 Hz 75 Hz 100 Hz

m 1 =200 kg

ms 0 kg 100μm m2 = ? kg

(a) before fixing a fan

(b) with fan

Fig. Q27 Fan

12.1 The First 30 Questions and Multiple Choice Answers

427

Question 28 We measured the amplitude A0  46 μm during an initial run. As shown in Fig. on periphery of the rotor Q28(1), we attached a trial weight of 3 g at , , and in three trial runs and measured the amplitudes A1  63 μm, A2  54 μm, and A3  29 μm during each trial run, respectively. What are the correction weight and the phase for balancing?

2

54  m

0°

12 63  m

1



12

0°

A0  46  m

3

29  m

Fig. Q28(1) Data for balancing

Question 29 Figure Q29(1) shows a rotor model consisting of a shaft (the mass is unknown.) and a disk (the mass is 10 kg). When we added a 3 kg mass at the disk as shown in Fig. Q29(2), the natural frequency decreased from 100 to 90 Hz. What is the shaft mass, ms , approximately? 1. 2. 3. 4. 5.

0.5 kg 1 kg 5 kg 10 kg 20 kg

k=

3EI l3 ms ?

10kg m

k=

3EI l3 ms ? l

l 100Hz ( 1 ) Original

90Hz ( 2 ) Mass added

Fig. Q29

10kg ma = 3kg (Added) m

428

12 Exercises of ISO Certification Examination for Vibration Experts

k =

GJ l

d = 500

Question 30 Figure Q30 shows a steel shaft–disk system and its specification. What are the polar moment of inertia I p , torsional shaft stiffness kθ , and natural frequency f n ? Note that the mass density ρ  7800 kg/m3 and modulus of transverse elasticity G  80.4 × 109 Pa of the steel, then use related formulas J  π d 4 /32 and I p  md 2 /8.

Ip

Ips 10 500mm ds l t = 10mm Fig. Q30

Q1–Q30_Answers

Table Q30 Answers for the first 30 Questions Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 3

2

1

3

3

3

2

2

4

2

3

1

4

2

5

Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 3

2

4

5

3

4

1

5

2

5

4

3

5

4

1

Q26_Hint We know the natural frequency 6/0.2  30 Hz by the zoomed diagram of Fig. Q26(1) and the slope of the ratio m  bi /bi+1  tan(60°) of a straight line shown in Fig. Q26(2). Hence 1 1 ln m  0.003 30 2π     k mg/δ g 9.8     313rad/s  50 Hz. Q27_Hint ωn  m m δ 100 × 10−6 ζ 

12.1 The First 30 Questions and Multiple Choice Answers

429

Q28_Hint Ans.: 3 × 2.3  6.9 g  − 140◦ . See Fig. Q28(2). Im 54  m

2 0°

1

12

0°

A0  46  m

B

m

Re

Wc 3



 m

m

63  m

#1  

12

#2

29  m

A

A

140°

(a) Rotor section

P Wc C

 

m

#3

(b)

Fig. Q28(2) 4-run method (in case of changing test run sequence)

Q29_Hint Since the mass is inversely proportional to square of natural frequency, the following equation is obtained: 1002 m + m s /4 + m a  , 902 m + m s /4 Introducing m  10 and m a  3 gives m s  11 kg. Q30_Hint Concerning the thin disk placed at the right end, the mass m  15.4 kg connected to the gives the polar moment of inertia I p  md 2 /8  0.48 kgm2 . Since it torsional stiffness kθ  G J/ls  157 Nm/rad, the answer is: f n  kθ /I p /(2π )  2.8 Hz.

12.2 Additional 70 Questions Question 31 rms The root-mean-square of a signal indicates: 1. 2. 3. 4. 5.

a peak value of the signal a scalar value of the signal power an average value of the signal information on the peak of the signal information on the signal strength

Question 32 A/D converter A signal includes noise with the level 240 mV rms when a FFT resolution is 8 Hz. What resolution is needed to reduce the noise level to 60 mV rms? 1. 0.5 Hz 2. 1 Hz

430

12 Exercises of ISO Certification Examination for Vibration Experts

3. 2 Hz 4. 4 Hz 5. 16 Hz Question 33 A/D converter The range of a 12-bit A/D converter is set to ±20 V, and the signal peak is 7 V of an input waveform. What is the lowest voltage that can be detected? 1. 2. 3. 4. 5.

9.77 mV 6.84 mV 4.88 mV 3.42 mV 2.44 mV

Question 34 A/D converter If a signed 12-bit AD converter is replaced with a signed 16-bit AD converter, how many dB are improved for the dynamic range? 1. 2. 3. 4. 5.

4 dB 12 dB 16 dB 24 dB 28 dB

Question 35 Two-dof model In Fig. Q35, figure (a) is a rotor system having a disk (the mass, m  41.2 kg) at the mid-span of the steel shaft (of mass, ms  10 kg), supported by a spring k b and a damper cb in parallel, identically at both ends. Figure Q35b shows its reduced two-dof model. The further simplified 1.5-dof model is shown in Fig. Q35c, called a Jeffcott rotor. When the rotor system is simply supported at both ends as shown in Fig. Q35e, its natural frequency is f z  42.5 Hz. When the shaft is assumed rigid and is supported by the spring at both ends as shown in Fig. Q35d, its natural frequency is f d  54.5 Hz. (1) What are the shaft stiffness, k s , and the bearing stiffness, k d ? (2) Estimate the first undamped natural frequency, f 1 , of the system shown in Fig. Q35b. (3) An optimal design chart of R1_Fig. 8.28 for lowering the response severity applies to the Jeffcott rotor of Fig. Q35c. For this Jeffcott rotor, find approximately the optimum damping factor cd  2cb and predict the lowest obtainable response severity Af peak by referring to the chart.

12.2 Additional 70 Questions

431 ms 2

800 mm

m

m=41.2 kg m cb

shaft k s cb m s =10 kg

kb

kb

kd =2kb

ms 2

m

m+ms ks

kd

cd =2cb

(b) 2-dof model

( ) Rotor system ms 2

m1 ms =m+ 2 ks m m2 = s 2

cd

(c) 1.5-dof model

ms m

kd

fd =54.5Hz fz=42.5Hz (e) Simple supported model

(d) 1-dof model

Fig. Q35 System modeling for rotor-bearing systems

Question 36 Hilbert transform Figure Q36(1) shows three curves of time-varying amplitude a(t) of the first three eigen modes in torsional damped vibrations generated by dump test for turbine generator set. These curves are obtained using the Hilbert transform. Eigen frequencies are identified by 16.8 Hz for the first mode, 31.1 Hz for the third mode, and the 34.9 Hz for fourth mode. Evaluate the damping ratio ζ n of each eigen mode.

Amplitude [dB]

0

1st 16.8Hz

20 3rd 31.1 Hz

40 60

4th 34.9H z 0

Time

[s]

31.25

Fig. Q36(1) Torsional vibration

Question 37 Damping ratio measurement As shown in Fig. Q37(1) and Table Q37(1), we measured a damped vibration waveform and read peak amplitude data at each cycle, i.e., data  {x 0 , x 1 , x 2 , x 3 … x n }. What is the damping ratio of this system?

432

12 Exercises of ISO Certification Examination for Vibration Experts

Table. Q37(1) 10 2 x1 xn : Reading values x2 x3 x 1 4 x5 x 6 x7 x8 x9 x10 1 2 3 4 5 6 sec 1 2

Fig. Q37(1) Reading at each cycle

n 1 2 3 4 5 6 7 8 9 10

xn 21.5 16.5 11.5 8.5 6.5 4.5 4.5 2.5 2.5 1.5

Question 38 Damping ratio measurement As shown in Fig. Q38(1) and Tabel Q38, we measured a damped vibration waveform and read peak values at each half-cycle as denoted by amplitude data  {y0 , y1 , y2 , y3 … yn }. What is the damping ratio of this system? 10 2 y1 1 0 1 2

yn : Reading values y3

y2

Table. Q38 Reading values n 1 2 3 4 5 6

y5

1 y4 1.5 2 y6

2.5 3sec

Fig. Q38(1) Reading at every half cycle

yn 14.5 8.5 4.5 2.5 1.5 0.5

Question 39 Measurement for Id As shown in Fig. Q39a, we hung a shaft, of mass, m1  4 kg and transverse moment of inertia, I 1 , by massless chords at both ends. After an impulse test, the shaft oscillates in tilting motion, horizontally around the gravity center G, and the measured the period of the natural frequency was T 1  0.1 s. Next, we fixed identical disks at each shaft end, as shown in Fig. Q39b. Then, we measured the period of the natural frequency as the period T 2  0.3 s. Hence, the added mass is ma  2 kg, and the distance between the gravity center G and the shaft end is l2  1 m. What is the transverse moment of inertia I 2 of the shaft with the two masses shown in Fig. Q39b?

h

2a

kθ

m 2 Ι2

G G m1 Ι1 (a) Shaft m1 = 4 kg

l2

l2 ma = 2 kg ma = 2 kg (b) Shaft with disks m 2 = 8 kg

Fig. Q39 Measuring I1 and I2

12.2 Additional 70 Questions

433

Question 40 Eigen frequency prediction Figure Q40(1) shows a geared pump driven by a turbine, where I i is the polar moment of inertia of each disk part, N i denotes the rotating speed of the input/output gear shafts, and the torsional stiffness is k i for each shafting part. Select the estimation of the first natural frequency ωn in this torsional system when letting the speed ratio n  N 2 /N 1  2.   1 1 8 1 2. ωn ≈   0.9 1. ωn ≈ 1 1/4 + 1/1 10   1 2 3. ωn ≈  0.32 4. ωn ≈  0.42 1 + 10  10 5  0.35 5. ωn ≈ 1 + 40 

Rotational speed N1 = 1

(1)

Pump Ι1 = 1

Ι21 = 0.1 k1 = 1 k2 = 1 Ι22 = 0.1 Gear

N2 = 2

Turbine Ι3 = 10

Fig. Q40(1) Torsional model of a 2-shaft system

Question 41 Prediction of natural frequencies of a torsional train system Figure Q41(1) shows a geared train consisting of a boiler feed pump (BFP), a turbine, and a booster pump (BP), where the polar moment of inertia is I i for each disk part, the rotating speed is N i for the input/output gear shafts, and the torsional stiffness is k i for each shafting part. (1) Letting the speed ratio n  N 2 /N 1  3.3, torsional angles of the BP shaft are normalized by that of the turbine pump shaft as shown in Fig. Q41(2). What are I *1 [ kg m2 ], k *1 [Nm/rad] and I *2 [kg m2 ]? (2) Figure Q41(2) indicates the first eigen mode. Using this mode shape, estimate the first natural frequency ω1 [Hz] of this torsional system.

434

12 Exercises of ISO Certification Examination for Vibration Experts I21 = 5.1

1.0

Ii = [ kg × m2]

N1 = 1

ki [ N m / rad ]

k1 BP

0.81 BP

k2 = 9.64×104

I1 = 1.1

k1* I1*

N2 = 3.3

k1 = 49×104

k2

I22 = 0.53

1 System

Gear

k3

Turbine k2

I2*

I3

BFP k3

I4

2 1st Eigen mode

Turbine

BFP

I3 = 340

I4 = 13

k3 = 270×104

Fig. Q41 Torsional train system

Question 42 Vibration transmissibility As shown in Fig. Q42(1), a peak vibration velocity of 3.05 mm/s at 29.55 Hz was measured in the horizontal direction at the base of a vertical pump, denoted by Y for absolute displacement. An inaccessible drive motor is directly attached to the pump, denoted by X for absolute displacement. An impact test on the pump revealed a natural frequency of 27.5 Hz and a damping ratio of ζ  0.03. Estimate the vibration velocity level of the motor. X m = Motor Motor Pump casing k

Y

c

Y=3.05 mm/s @ 29.55Hz

Floor

Floor

Impeller

Fig. Q42(1) Vertical pump

Question 43 Modal analysis (a) Modal parameters of two-dof system Approximate eigen modes of a two-dof system are given as shown in Fig. Q43. In the case of m 1  m 2  m and k1  k2  k, the corresponding modal parameters are summarized in Table Q43. Fill out blank parts of the table.

12.2 Additional 70 Questions

435

Table Q43 Modal parameters 1st mode

2nd mode

13 × m

(1?) × m

Modal stiffness

(2?) × k

34 × k

Natural frequency

(3?) × √k/m

Modal mass

m1

3

x1

(4?) × √k/m

-2

k1 m2

x2

2

3

k2

Fig. Q43a Eigen modes of 2-dof system ( m1=m2=m, k1=k2=k )

(b) Mode shape of deflection mode Confirm that Fig. Q43b shows the deflection mode δ due to gravity, and find the first natural frequency by replacing the first eigen mode by it.

m1

x1

3

mg k1 m2

x2

2

mg k2

Fig. Q43b Deflection modes of 2-dof system ( m1=m2=m, k1=k2=k )

Question 44 (a) Measurement of equivalent mass We attach small masses m to node m1 of Fig. Q43a and then measure changes of the natural frequency. We summarize these measurement data as shown in Fig. Q44, where the natural frequency decreases as the added mass increases. Estimate the equivalent mass meq of this first mode with respect to nodal point m1 .

12 Exercises of ISO Certification Examination for Vibration Experts natural frequency / k / m

436

0.64 0.63 0.62 0.61 0.60 0.59 0.58 0.57 0.02

0.04

0.06

0.08

0.10

mass ratio = Δm / m

Fig. Q44 Equivalent mass measurement

(b) Modal mass and equivalent mass

 t If we replace the first eigen vector φ1  [3 1.85]t by φ1a  1 0.62 normalized by unity the displacement of the mass m1 , what is the first-order modal mass? Confirm that it agrees with the equivalent mass meq .

Question 45 Equivalent mass measurement A fan is driven by an electric motor as shown in Fig. Q45(1). The fan vibrates in the first bending mode indicated by the dashed curve. When we add small masses to the fan, the natural frequency decreases as shown in Fig. Q45(2). (1) What is the equivalent mass with respect to the fan portion? (2) Assuming ε  10 μm as the eccentricity of the fan, what is the modal eccentricity ε* ? (3) What is the frequency resolution when measuring FFT data of Fig. Q45(2)?

60kg 200kg

40kg

Added mass

Fig. Q45(1) Added mass

Frequency [Hz]

37.5

37.0 = 0.27 m 37.4 36.5

0

1 Added mass (kg)

2

Fig. Q45(2) Equivalent mass measurement

12.2 Additional 70 Questions

437

f cos ωt k2

k1

m2

m1 x1 ω 1=

c2

x2

ω2 k k1 m c2 , ζ= , R= 2 , ω2 = 2 , ν = ω m2 m1 1 m1 2m2ω1

Amplitude A1 ( = X1 / Xs t )

Question 46 Tuning of a dynamic damper In Fig. Q46(1) and (2), we can see a two-DOF system connecting the main system of m1 -k 1 with dynamic damper m2 -k 2 -c2 . The graph includes four response curves induced by harmonic exciting force, where the horizontal axis indicates the non√ dimensional exciting frequency λ  ω/ω1 , hence defined by ω1 ≡ k1 /m 1 , and the vertical axis the√amplitude of the main system under the condition of R  m 2 /m 1  0.05 and ω2 ≡ k2 /m 2  ω1 . These four curves correspond to values of the damping ratio ζ , defined by ζ  c2 /(2ω1 m 2 ). Select an appropriate combination from 1 to 5 for the relationship between the damping ratios and the curves.

20 18 C 16 A D 14 B B 12 C 10 D 8 A 6 4 2 0 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 Exciting frequency λ ( = ω/ω1 )

( 1 ) Dynamic damper

( 2 ) Bode plot

Fig. Q46(1)(2)

438

12 Exercises of ISO Certification Examination for Vibration Experts

Question 47 Response analysis in the frequency domain An m-k-c system is subject to pulse excitation forcing as shown in Fig. Q47(1). When 11 times the pulse frequency is equal to the natural frequency, estimate the amplitude of the 11th resonance component according to the following steps: (1) Referring to Table Q48 about Fourier series expansion (FSE), the input pulse waveform in the time domain is transformed by the FSE as shown in Fig. Q47(2). What is the magnitude of 11th component? (2) The frequency response of the m-k-c system is given in Fig. Q47(3). What is the Q-value? (3) The response of the system against the input pulse is shown in the FSE of Fig. Q47(4). What is the amplitude of the 11th resonance component?

F(t)

F0

(1)

(2) 1 ×F0 0.8 F(t) 0.6 0.4 0.2 0 0 5 (3) 10 ×δst

π ω

2π T= ω Waveforms

t

8 6 4 2 0

G( jnω)

0

10

15 G(s)

(4) 1 ×δst 0.8 Response Xp(t) an 0.6 0.4 0.2 0 0 5 10 15 frequency order n

5 10 15 frequency order n

Fig. Q47 Waveform and FRA

Question 48 Nonlinear excitation In a torsional shafting system consisting of a motor, two pistons, and a fan, the pistons generate torque pulsation acting on the torsional system, as shown in Fig. Q48. Referring to Table Q48 about Fourier series expansion (FSE), the input pulse waveform in the time domain includes 1X (synchronized with the rotational frequency), 2X (2 times the rotational frequency), 3X, 4X, 5X, and others. If we assume 100% as the magnitude of the 1X component, evaluate the relative magnitudes of the other components; 1X  100%, 2X  ?, 3X  ?, 4X  ?, 5X  ?

12.2 Additional 70 Questions

439 Piston

fan Torsional vib. k

M

(1) Tortional shafting system Torque pulsation 180° 360°

0

angle shaft

(2) Excitation torque

Fig. Q48 Torsional vibration excited by rectangular torque

Table Q48 Fourier series g1 (θ )  Original wave

1 0

(1)

f1 (θ )  Approximation of Fourier series

1

2

3

5× π

4

1 for

0 θ α

f 3 (θ )  2 f1 (θ , π )  1 

,

2

3

4

5× π 0.8

g 3 (θ )   1 1

π

for

α  θ  2π

1

1

2

3

4

0

5×π

0

0

1

2

3

4

5

6

7

8

7

8

1

1

1 g 7 (θ ) 

α π 1

(sin θ  3 sin 3θ  5 sin 5θ      )

4

1 0

0 1

g 3 (θ )  1

(2)

1.6

1

α 

Spectrum

α  π 1

2

3

4

5× π

1

θ π

1

for

0  θ  2π

f 7 (θ )  2 f 6 (θ , 2 )  1  

2



(sin θ  21 sin 2θ  13 sin 3θ      )

0

0

1

2

3

4

5

6

Question 49 Campbell diagram In Q48, we spin the rotor system and measure the torsional vibration. The vibrational amplitude is displayed in the Campbell diagram of Fig. Q49(1). The circles correspond to resonances for each order, nX. Assuming that the Q-value is invariant through the rotation, which is the most appropriate diagram, where ωθ  torsional natural frequency and   rotational speed?

12 Exercises of ISO Certification Examination for Vibration Experts 5X 4X

3X

1X

ωθ

Rotational speed

Frequency

3

5X 4X

3X

5X 4X

2X

1X

Rotational speed

2X

5X 4X

P =  /ωθ

3X

2X

1X

ωθ

P =  /ωθ

3X

3X

ωθ

4

1X

Rotational speed

5X 4X

P =  /ωθ

ωθ

5

Frequency

2

2X

Frequency

Frequency

1

Frequency

440

Rotational speed

P =  /ωθ

2X

1X

ωθ

Rotational speed

P =  /ωθ

Fig. Q49(1) Campbell diagram and vibration magnitude

Question 50 A single mass and flexible shaft system under parallel motion We consider the design of a rotor-bearing system as shown in Fig. Q50(1) of a Jeffcott rotor and Fig. Q50(2) of its equivalent model, also called a 1.5-dof system. The rotor has a thin disk placed at the center of a massless shaft, which is supported by bearings at the both ends. The thin disk is specified by the mass m  3 kg and the polar moment of inertia I p  0.1 kg mm2 . To achieve the design target for the first critical speed of f 1  20 Hz or above and the damping ratio of ζ 1  0.2 or above, the following steps are made. m = 3kg, IP = 0.1 kg˜mm m=3 kg

cb

kb

Fig. Q50( 1 ) Jeffcott rotor

cb

kd

48EI l3 kd = kb cd = cb

ks =

ks

ks kb

m

cd

Fig. Q50(2) 1.5-dof of translation model

12.2 Additional 70 Questions

441

Q(1) Instead of this mechanical system, we consider an equivalent mechatronic system as shown in the block diagram of Fig. Q50(3). A mass is actively magnetically levitated by an actuator characterized by a controller transfer function of G r (s)  k g (τ s + 1)/(ατ s + 1). Determine the controller parameters of k g α τ to achieve the design target from the viewpoints of mechatronic engineers. u

Gr(s)

u

GP (s) =

x

m

m= 3 kg

r +

Gr(s)

x

1 ms2

Fig. Q50(3) Magnetic levitation system : G r (s) = k g

τ s +1 α τ s +1

Q(2) The open-loop transfer function is defined by G o  G r G p in Fig. Q50(3). Confirm that curves of the gain |G o | and the phase  − G o   G r are drawn in Fig. Q50(4) and they satisfy the design target.

Gain [dB] , Phase [deg]

60

| Go |

Gr

40

m

20 0 -20

fg 1

5

10

50

100

Frequency[Hz]

Fig. Q50(4) Open-loop transfer function for parallel motion

Q(3) Convert these controller parameters obtained from Q(1) to the mechanical parameters of Fig. Q50(2) and (1). Q(4) An example of an impulse response is shown in Fig. Q50(5). Estimate the natural frequency, f 1 , and the damping ratio, ζ 1. a1 2 1

Step

a3

0 -1 -2 0.0

a2

a5

a4 Impulse 0.1

0.2 0.3 Time [s]

0.4

0.5

Fig. Q50(5) Responses (Time histories)

442

12 Exercises of ISO Certification Examination for Vibration Experts

Q(5) An example of a frequency response analysis (FRA) undertaken for this system is shown in Fig.Q50(6). Evaluate the natural frequencies f 1 and Q-value for first eigen mode. Confirm that the Q-value agrees with ζ 1 of Q(4) . 1.4

f2=?

2nd

1.2 Amplitude

1.0 0.8

1st

f1

0.6

=?

Q1

0.4

Q2

=?

=?

0.2 0

20

40

60

80

100

Exciting frequency [Hz]

Fig. Q50(6) Q-value measurement in FRA

Question 51 A single mass and flexible shaft system under tilting motion As stated in R1_Fig. 4.14, the mode synthesis technique and quasi-modal modeling are applied to a Jeffcott rotor, and the corresponding 1.5-dof reduced model is then completed by two types of motion/mode as shown in Fig. Q51(1) and (2). Each motion provides the respective 1.5-dof models of Fig. Q51(1) for parallel motion and Fig. Q51(2) for tilting motion. Since the system parameters of Fig. Q51(1) have been selected in the previous Q50, then all parameters of Fig. Q51(2) are automatically determined using similar deductions.



1

m=3 ks cd

kd

(a) Parallel modes

(b) 1.5-dof

Fig. Q51(1) Parallel modes and 1.5-dof reduced model

Id



2

2

k kd

(a) Tilting modes

cd

 L  k    ks  2  

12 EI L

(b) 1.5-dof

Fig. Q51(2) Tilting modes and 1.5-dof reduced model

12.2 Additional 70 Questions

443

Q(1) With regard to the system specified by the transverse moment of inertia I d and the bearing span L  0.5 m in Fig. Q51(2),what are the parameters of the shaft stiffness kθ and bearing dynamics kdθ + cdθ s? Q(2) Predict the second natural frequency f 2 and the damping ratio ζ 2 of Fig. Q51(2). Q(3) Confirm that this ζ 2 agrees with the Q-value measured for the second mode of Fig. Q50(6). Question 52 Two-dof flexible system with single controller Instead of Q50 for rigid body mass control, we consider the vibration control for a flexible system having a spring between two masses as shown in Fig. Q52(1). The total mass mt  3 kg is divided into two separate parts, m1  1.8 kg (60%) and m2  1.2 kg (40%). Both masses are connected by a spring constant, k, where the antiresonance√frequency ωz , called the zero frequency, exists at a very high frequency of ωz  k/m 2 ( 200 Hz). It is found that the zero response of m1 occurs, even when exciting it at the frequency of ωz . This system is helpful for us to understand the spill-over instability induced by the control of flexible mechanical systems. Q(1) Verify the plant transfer function Gp (seen in the same figure (b)) from the input control force u to the output displacement x 1 of the target mass m1 . What is the pole frequency ωp ? Gr

u

x1

m1 m1 = 1.8 kg

y=x-x 2

k

2

m2 = 1.2 kg

Gr

1

x2 m2 (a) 2-dof

2

G p=

 s   z   1  

x1

2

 s     1 m t s 2      p    

Total mass mt = m1 + m2 (b) Block diagram of a flexible system

Fig. Q52(1) 2-dof system with single input, single output control

Q(2) To attain the design targets, including the first natural frequency f 1 ≥ 20 Hz and the damping ratio ζ1 ≥ 0.2 for the first eigen mode, we use the same controller of Q50. Instead of Fig. Q50(4), we measure the open-loop transfer function as shown in Fig. Q52(2). Read the cross-over frequency and phase margin and verify that they agree well with the design target for the first eigen mode.

444

12 Exercises of ISO Certification Examination for Vibration Experts

Gain [dB] , Phase [deg]

60

Im

40

fg

20 0 | Go |

-20

fg

-40 -60

Gr 1

5 10

50 100

500 1000

Frequency[Hz]

Fig. Q52(2) Go(s) and

G r(s)

Q(3) The closed loop transfer function Gc can be defined by G c  G o /(1 + G o ), and it provides the step response and the impulse response as measured by sensor x 1 and the relative displacement sensor y2  x 2 − x 1 , as shown in Fig. Q52(3). Read the damping ratio of the first eigen mode from the impulse response ➀, and verify the agreement with ζ 1 of Q(2). 4

Step

1

2

0

0

y2

x1

2

-1 0.1

a1

0.2 0.3 0.4 Time [s] 1 Sensor x1

0.5

-4

a2 Impulse a3 a4 a

5

-2

Impulse

-2 0.0

a0

Step 0

1

2 3 4 Time [s] 2 Sensor y2

5

Fig. Q52(3) Responses (Time histories)

Q(4) Read the damping ratio of the second eigen mode from the impulse response ➁. Question 53 Modal control for torsional vibrations Modal control is offered for a typical torsional flexible system consisting of m1 -k-m2 as shown in Fig. Q53(1). The displacements, x 1 and x 2 , are actually measured at the both ends and may be transformed to the parallel motion x p and the tilting motion x t by the following matrix T m : 

x1 x2



 

1 h1 1 −h 2



xp xt



 ≡ Tm

xp xt

where h 1  m 2 /m t and h 2  m 1 /m t

(1)

The control forces, up and ut , are defined independently by each controller, Grp and Grt , respectively. These forces on the modal coordinates are converted to the actual 1 t forces u1 and u2 on the physical coordinate by B− m , where Bm = T m . The controller Grp works for the parallel motion, i.e., the average motion of the both bodies for driving control. The controller Grt plays the role of maintaining the relative displacement (i.e., the spring force) for loading control.

12.2 Additional 70 Questions

445 u1

m1 = 1.8 kg

1

x1

m1

rp 1

2

Tm

k x2

m2 = 1.2 kg

m2

m t = m1 + m2

Grp

xp xt

Grt

 xp   x1     Tm    x2   xt 

up 1

ut

1 h1  Tm    1  h2  m m h1  2 h 2  1 mt mt

Bm

rt h 1  Bm1   2   h1  1

1  h 2 h1  Tm     1 1 

Bm  Tmt u2

Fig. Q53(1) 2-dof system and mode control layout

Q(1) Find eigen frequencies and eigenvectors for the m1 -k-m2 system, as shown by φ 1 and φ 2 in Fig. Q53(1),  and note that this transformation matrix T m is the modal matrix, i.e., Tm ≡ φ1 φ2 . Q(2) Verify the diagonalization by the following forms of the plant transfer function Gp from the modal control forces, up and ut , to the modal displacements, x p and x t : 

xp xt



 ≡ Gp

up ut





G pp 0  0 G pt



up ut

(2)

Note that each mode is independent and without cross-coupling. Find the diagonal elements, Gpp and Gpt .   Q(3) The same controller having k gp  48 × 103 τ p  0.012 α p  0.458 of Q50 is used for the parallel controller Grp to provide the target specification satisfying the natural frequency ≥ 20 Hz and the damping ratio mode controller Grt is set by the modified parameters ≥ 0.2. The tilting k gt  48 × 103 τt  0.0012 αt  0.458 to attain a high damping effect (i.e., significant phase lead) in the high frequency domain, with the maximum phase lead around 200 Hz. Then, the open-loop transfer function is defined as follows, as shown in the Bode plots of Q53(2):

      G o ≡ Diag. G op G ot  Diag. G r p / m t s 2 G r t / m eq s 2 + ω2p where G r p (s) ≡ k gp

τps + 1 τt s + 1 m1m2 , G r t (s) ≡ k gt , m eq  αpτps + 1 αt τt s + 1 mt

(3)

Estimate the natural frequency and damping ratio of each mode from these plots.

12 Exercises of ISO Certification Examination for Vibration Experts 60 40 20 0 -20 -40 -60

| G op | I m

Grp

fg 1

5 10 50 100 Frequency[Hz]

1000

Gain [dB] , Phase [deg]

Gain [dB] , Phase [deg]

446

60 40 20 0 -20 -40 -60

Im

I m* = 21º

Grt

Im

Im* Re k = mωp2 m1 m2 fg | G ot | m = m m = 0.72 1+ 2 ωp = 2 π ×258 Hz k = 1.8×106 5 10 1000 50100 ω Grt ( j p ) = (71.6 + j27.7) ×103 Frequency[Hz] = 76.8 ×103∠21º (b) Tilting mode G rt

1

(a) Parallel mode

Fig. Q53(2) Open-loop Bode plots for mode control

Q(4) The closed loop transfer function Gc is defined by G c  (1 + G o )−1 G o , which provides the step and impulse responses ➀ of x p as shown in Fig. Q53(3). Read the natural frequency and the damping ratio from the damped waveform of x p , and confirm the agreement with the design target. Q(5) From the waveform x t of the impulse response ➃ in Fig. Q53(3), read the damping ratio ζ 2 and confirm the agreement with prediction for the tilting mode. Input rp : Parallel mode 1

2 Step

Sensor xp

1 0

0.1

-1

0.2 0.3 Impulse

Sensor xt

2

1 0.4

0.5

0

Not coupled 0.1

0.2

0.3

0.4

0.5

-1

-2 6 4 2 0 -2 -4 -6

Input rt : Tilting mode 2

-2 3

Not coupled 0.1

0.2

0.3

0.4

6 4 2 0 0.5 -2 -4 -6

4 Step

0.1

0.2

0.3

0.4

0.5

Impulse

Fig. Q53(3) Responses of mode control

Question 54 Reduced two-dof modeling A rotor shafting (mass, ms  0.31 kg) having a disk mass (md  0.37 kg) at the right end is shown in Fig. Q54(1), being subjected to the pinned boundary condition at the left end and supported by the right-side bearing specified by a spring k b and a damper cb . The shaft vibration is measured at the right end by the y1 sensor. Assuming a certain system unbalance, an example of an unbalance vibration amplitude curve is given in Fig. Q54(2) as measured by the y1 sensor.

12.2 Additional 70 Questions

447 cb

φ 10 × 500L steel

md

ms = ρAl, mδ =ms / 3, kδ = 0 φ1

(b) 1

y1 δ

0

Amplitude

(a)

kb 60 50 40 30 20 10

f1 = ? Q= ?

0

l

1 Fig. Q54( 1 ) Rotor system (left pinned)

20 40 60 80 Frequency [Hz]

100

Fig. Q54(2) Unbalance response

Q(1) Read the first critical speed, f 1 , and the Q-value from the amplitude curve. Q(2) Referring to R1_Fig. 4.11b, the shafting can be reduced to a two-dof model by applying two steps: firstly, the mode synthesis is undertaken in terms of the δmode (k eq  0) and φ-mode (natural frequency f z  79 Hz); then, the quasi-modal technique is applied. Consequently, we obtain a reduced model as shown in Fig. Q54(3), consisting of m 1  0.39m δ + m d and m 2  0.61m δ , where m δ  m s /3. Find these values, m1 and m2 . a2

φ1

a2 – a1 δ

a1

m2 61%

a1

ξ fz = ks 2 79 Hz m1 39% md

kb

y1

cb

Fig. Q54(3) 2-dof reduced model

Q(3) Determine the amplitude, a1 and a2 , of the first eigen mode, by assuming a1  1. Q(4) Determine the modal mass, m ∗1  m 1 a12 + m 2 a22 , of the first eigen mode. Q(5) Determine the bearing stiffness k b to satisfy the modal relationship of the first eigen mode; m ∗1  k1∗ ω12 , where the modal stiffness is k1∗  ks (a1 − a2 )2 + kb a22 and the first natural angular frequency ω1  2π f 1 measured in Q(1). Q(6) Determine the bearing damping coefficient cb to satisfy the modal relationship of the first eigen mode: 2ζ1 ω1  c1∗ /m ∗1 , where the modal damping is c1∗  cb a12 and the first modal damping ratio ζ1  1/(2Q), determined in Q(1). Question 54a Reduced two-DOF modeling Instead of the pinned boundary at the left side as shown in Fig. Q54(1), the rotor system is subjected to the clamped boundary condition as seen in Fig. Q54a(1). Assuming a certain unbalance, the amplitude curve obtained is shown in Fig. Q54a(2).

12 Exercises of ISO Certification Examination for Vibration Experts cb (a)

φ 10 × 500L steel

ms = ρAl, mδ =ms / 4, kδ = (b) 1

φ1

kb 50

md

3EI l3

Amplitude

448

y1

δ

0

30 20 10 0

l

1

f1 = ? Q= ?

40

Fig. Q54a( 1 ) Rotor system (left fixed)

20

40 60 80 100 120 Frequency [Hz]

Fig. Q54a(2) Unbalance response

Q(1) Determine the critical speed, f 1 , and the Q-value from the amplitude curve. Q(2) Referring to the φ 1 mode of R1_Fig. 4.12, the shafting can be reduced to a two-DOF model by applying two steps: the mode synthesis using the δ-mode (shaft stiffness kδ  3E I /l 3  2424 N/m) and φ-mode (natural frequency f z  123 Hz) and then the quasi-modal technique. Consequently, we obtain a reduced model as in Fig.Q54a(3), consisting of m 1  0.48m δ + m d , m 2  0.52m δ , where m δ  m s × 33/140. Find the values, m1 and m2 , and the equivalent shaft stiffness, kδ . a2 φ1

a2 – a1 δ

m2 52%

a1

fz = 123 Hz

ks

ξ2

m1 48% md

y1

a1 kδ

kb

cb

Fig. Q54a(3) 2-dof reduced model

Q(3) Determine the amplitudes, a1 and a2 , of the first eigen mode, by assuming a1  1. Q(4) Determine the modal mass, m ∗1  m 1 a12 + m 2 a22 m ∗ , of the first eigen mode. Q(5) Determine the bearing stiffness k b to satisfy the modal relationship of the first eigen mode; m ∗1  k1∗ ω12 , where the modal stiffness is k1∗  ks (a1 − a2 )2 +(kδ + kb )a22 and the first natural angular frequency ω1  2π f 1 . Q(6) Determine the bearing damping coefficient cb to satisfy the modal relationship of the first eigen mode: 2ζ1 ω1  c1∗ /m ∗1 , where the modal damping is c1∗  cb a12 and the first modal damping ratio ζ1  1/(2Q). Question 55 Field balancing using the Nyquist plot An unbalance vibration response of a rotor is shown by the “initial run” of the Nyquist plot of Fig. Q55(1), with non-dimensional rotational speed, p  /ωn , where   rotational speed and ωn  natural frequency. When the trial weight W t  1.5 g < −

12.2 Additional 70 Questions

449

130° is attached to the rotor, the corresponding Nyquist plot is noted by “Trial run,” as labeled in the same figure. What is the correction weight W c ? 6 Im Wt =1.5 230q Trial run

4 2

1.0 -6 -4 95

p

. =0

C 0.9 0 B

4

-2

Re 6

p = 0.95 A

1.1 -4 Before (Initial run) -6

1.0

Fig. Q55(1)

Question 55a Retry of field balancing Since the rotor was well-balanced in Q55, it has been operated over a long period until the unbalance has increased again, as shown in Fig. Q55a(1). Re-balance the rotor using the same influence vector of Q55. What is the correction weight W c for the re-balancing? 6 Im 4

1.1

2 Re

C

1.0 -6

-4

D P = 0.95

-2 0.9

0

2

4

6

-4 -6

Fig. Q55a(1)

Question 56 Field balancing by observing waveforms An unbalance vibration of a rotor is shown in the waveform (A) of Fig. Q56(1), measured at a certain speed near the critical speed. When the trial weight W t  0.5 g 90° is attached to the rotor, the waveform is measured as shown in the waveform (B). Q(1) Read the complex amplitudes of (A) and (B). Q(2) What is the correction weight W c ?

450

12 Exercises of ISO Certification Examination for Vibration Experts 10

Trial run Wt = 0.5g q (B)

5

–5 Initial run (A)

–10

Fig. Q56(1)

Question 56a Retry of field balancing In Q56, the rotor was well-balanced. After this balancing, the rotor was operated over a long period until the unbalance became established again, as shown by the curve (D) in Fig. Q56a(1). Re-balance of the rotor is required by using the same effective vector of Q56. Q(1) Read the complex amplitude of (D). Q(2) What is the correction weight W c for this re-balance? 10 5

-5 -10

Initial run (D)

Fig. Q56a(1)

Question 57 Balancing using three trial runs We measured the amplitude A0  46 μm during the initial run. As shown in Fig. Q57, we attached a trial weight of 3 g at ➀, ➁, and ➂ on the periphery of the rotor in three trial runs and measured the amplitudes A1  63 μm, A2  49 μm, and A3  37 μm at each trial run. What are the correction weight and phase for balancing? 2

°

49 m

5 13

63 m 1

90



°

A0  46  m

3

37 m

Fig. Q57(1) Example of calculation

Question 58 An unbalanced rotor is whirling close to the critical speed as shown in Fig. Q58. We see the mass center of gravity G and a notch for the rotational pulse signal. At the

12.2 Additional 70 Questions

451

moment when the rotor was passing the X-direction, we took the photograph of the rotor section. Which photograph is correct?

Ω

Y

Y

Y Orbit Unbalance

Ω G

X Rotational pulse Cross-section A of rotor Y G

X

Ω X G C

B Y G Ω

X G Ω D

X

E

Fig. Q58 Whirl motions and unbalance G

close to sensor

Amplitude

away from sensor

Question 59 Sensor layouts and Bode plot A rotor, having the critical speed of 3000 rpm, is rotating at 300 rpm. A vibration waveform is shown in Fig. Q59, which also indicates the measurement system with the layout of the displacement sensor and pulse signal. Note that both are separated by the angle of 135°. The output of the displacement sensor is negative as the rotor moves closer to the sensor and positive vice versa. Which is the optimum position among {A, B, C, D, E, F. G. H} when placing the correction weight for balancing? Vibration Pulse

Vibration time C D Pulse

B Ω

E

A H G

F

Fig. Q59 Layout of pulse sensor and vibration sensor

Question 60 Sensor layouts and Bode plot In Q59, without the correction weight, we increase the rotational speed and pass the critical speed.

452

12 Exercises of ISO Certification Examination for Vibration Experts

Amplitude

4 3 2 4 Rotation by 180°

1

Phase lag [deg]

0 360

180

3 2 1

90

5

0

0.0

Re 3 –4 Rotation by 90°

–2

2

( 1 ) Bode plot

4

–2

2 Rotation by 45°

0.5 1.0 1.5 2.0 2.5 3.0 Rotational speed p = Ω ⁄ ωn

5 Rotation by 270°

2

4

270

4 Im

1

–4 ( 2 ) Nyquist plot

Fig. Q60(1)(2)

Q(1) The corresponding Bode plot is drawn in Fig. Q60. Which is the appropriate curve in Fig Q60(1) to show the phase lag? Q(2) Which is the corresponding Nyquist plot in Fig. Q60(2)? Question 61 Balancing for first bending mode In Fig. Q61(1), a rotor passed two critical speeds of Nc1 (rigid parallel mode) and Nc2 (rigid tilting mode) under well-balanced conditions. When we passed the Nc3 critical speed (first free-free bending mode shown in the figure), we needed 230 g · mm as balancing weights near the bearing portions. These values are measured by FF (feedforward) excitation using an AMB, as stated in relation to Fig. 6.56. We intend to replace these two balancing weights by three balancing weights of Wc1, Wc2, and Wc3, with the well-balanced state of rigid modes (Nc1 and Nc2) being invariant. What are these three correction weights? Correction planes 230 g⋅mm

AMB

Wc1

Wc2

Wc3

230 g⋅mm

AMB

3rd eigen mode (a) FF excitation to cancel unbalance

Fig. Q61(1) Balancing by FF excitation of AMB

Question 62 Balancing for first bending mode In Q62, we change the location of correction planes by a set of {W c4, W c2, W c5} as shown in Fig. Q62(1). In the similar way, we want to replace these two balancing weights of 230 g · mm by W c4, W c2, and W c5. What are these three correction weights?

12.2 Additional 70 Questions

453

Correction planes: Wc4, Wc2 and Wc5 230 g⋅mm 230 g⋅mm Wc2 Wc5

Wc4

AMB

AMB

3rd eigen mode (a) FF excitation to cancel unbalance

Fig. Q62(1) Balancing by FF excitation of AMB

Question 63 Rigid rotor balance for a compressor A compressor for air conditioners consists of a rotating shaft, two oil film bearings, a twin-rotary type piston, and an electric motor as shown in Fig. Q63(1). Hence, we regard the unbalances U/−U of the two pistons as moment unbalance 20U [20 mm × U] and apply two-plane balancing using U 2 and U 3 . What are the correction weights U 2 and U 3 for rigid mode balance to compensate for the unbalance moment 20U?

#1 Brg. L #2 Brg. 20 20 20 M LU = 20mm U 3 4 5 1 2 20 30 30 20 U M

S1

Motor 6 100 60

16

M = Dynamic Unbalance of Twin Rotary Piston

U kb

cb

kb

cb U2

U3 7

kb=10 N/m

3

cb=10 Ns/m

Fig. Q63(1) 2-plane balancing for compressor (lengths in mm)

Question 64 N + 2 balance of a twin-rotary compressor Instead of two-plane balancing, three-plane balancing using U 1 , U 2 , and U 3 , as shown in Fig. Q64(1). These three planes are set to achieve rigid mode balance plus bending mode balance, called N + 2 plane balancing. Hence, N  1 for a bending mode φ 1 , as shown in Fig. Q64(2). What are the balancing weights U 1 , U 2 , and U 3 to completely compensate for the unbalance moment M  20 U?

454

12 Exercises of ISO Certification Examination for Vibration Experts M = Dynamic Unbalance of Twin Rotary Piston

U1

U kb

cb

kb

S1

Motor 6 100 60

16

#1 Brg. L #2 Brg. 20 20 20 M LU = 20mm U 3 4 5 1 2 20 30 30 20 U M cb U2

kb=107 N/m

U3 cb=103 Ns/m

Mode shapes

Fig. Q64(1) 3-plane balancing for compressor (lengths in mm)

Rigid mode (δ1 & δ2) 3.7 3 3 Bending mode ( 1) a2 =1/60 δ2 2 1.33 1.33 1 1 0.5 1 0.16 0.49 0 −0.33 −0.19 a1 = −1/60 −0.33 −1 #1 #2 δ1 −2 −2 Brg. Brg. 0 200mm 50 100 150

Fig. Q64(2) Modes for N+2 plane balancing

Question 65 A motor-fan system is operated at 3000 rpm, and the vibration waveform of the fan is observed, arriving at the limit of 50 μm, as shown in Fig. Q65(1) and (2), respectively. To reduce the unbalance vibration, balancing is required. An attempt was made to refit the fan on its axis by rotating it by 180°, and the vibration waveform was measured as shown in Fig. Q65(3). The mark F  0° on the fan is upward in the former and downward in the latter responses. Answer the following questions. Q(1) Read the complex amplitudes (amplitude phase lag) of waveforms (2) and (3). Q(2) If we rotate the fan by a certain angle and refit it, how many degrees are recommended for the best balancing level? Q(3) What vibration level is expected by this best refitting? Pulse sensor

A

Vibration sensor Impeller

Coupling

Shaft

Motor

Fan

Fig. Q65(1) A motor-fan system

12.2 Additional 70 Questions

455 Y F=0°

Phase lag θ amp [ μm]

100 80 40 0

-80 -100

270°

90°

-40

 0

5

10

15 20 time [msec]

25

X

180°

30

Fig. Q65(2) Original impeller layout

Y

Phase lead

0° 180°

40 0

90°

270°

X

-40

-80 -100

0

5

10

15 20 time [msec]

25

30

F=0°

amp [ μm]

100 80

Fig. Q65(3) Reversed impeller layout

Question 66 A steam turbine rotor, having a total mass 3600 kg and a rated speed N  4950 r/min, is put on a balancing machine as shown in Fig. Q66(1). We see the bearing span is 18 m, and the center of gravity location from the left is 8 m. Since balancing quality is required by G2.5, the permissible residual eccentricity ε is 4.8 μm as seen in Fig. Q66(2). In this situation, what is the mass [kg] loading on each bearing and the permissible unbalance [g · mm] at each bearing, A and B?

1. 2. 3. 4. 5.

Bearing A Mass 1500 [kg] 1600 2000 1600 2000

Bearing B Unbalance Mass 17.3 [kg × mm] 1500 [kg] 7.7 2000 7.7 1600 9.6 2600 9.6 1600

Unbalance 17.3 [kg × mm] 9.6 9.6 7.7 7.7

456

12 Exercises of ISO Certification Examination for Vibration Experts m= 3600kg

G A

B 8m

10m

G=

G 100

40

ε = 4.8

G= 16

G 40

1 6 G=

G 16

.3

0.5

G 6.3

2.5

G= 0.4

1000

G 1.0 G=

100

G 2.5

1.0

0.05

G=

N = 4 950

0.1

10000

Gyroscopes Spindles and drives of high-precision systems

0

G 250

Compressors Computer drives Electric motors and generators (above 950 r/min) Gas turbines and steam turbines Machine-tool drives Textile machines

10

G=

5

Audio and video drives Grinding machine drives

0

25

G=

G 630

Agricultural machinery Crankshaft drives (rigidly mounted) Crushing machines Drive shafts (cardan shafts, propeller shafts)

0

10

Aircraft gas turbines Centrifuges Electric motors and generators up to 950 r/min Fans Gears Machine-tools Process plant machines Pumps Turbo-chargers Water turbines

63

G 1600 50

Crankshaft drives, inherently unbalanced, rigidly mounted

G=

100

Complete reciprocating engines for cars, trucks and locomotives

500 00

Permissible residual eccentricity ε

1000

16 G=

[ μm ]

5000

Cars (wheels, drive shafts) Crankshaft drives (elastically mounted)

Crankshaft drives for large slow marine diesel engines

10000

Crankshaft drives, inherently unbalanced, elastically mounted

Fig. Q66(1) Mass center and bearings

G 0.4

100000.

Service speed N [r/min]

Fig. Q66(2) ISO 1940 - 1 : 2003(E) : Balance quality requirements of rigid rotors in a constant (rigid) state

Question 67 Center spring supports for mass center A compressor bedplate having a total mass 1400 kg is supported softly by rubber springs for the vibration isolation of 50% as shown in Fig. Q67(1). An excitation force is generated by unbalance due to the eccentricity 10 μm of the compressor rotor of mass, 300 kg, at rotational speed 1800 rpm. Answer the following questions using transmissibility curves as shown in Fig. Q67(2): Q(1) What is the excitation force F e and the transmitted force F? Q(2) If the damping ratio is assumed by ζ  0.3, what is the natural frequency of this system? Q(3) What is the stiffness k of the rubber spring?

12.2 Additional 70 Questions

457 G a1 = 1m

a2 = 3m m = 1400kg

Ω

I = m i2 (i = 2m) Fe

k

c F

Fig. Q67(1) Vibration isolation

TR = F / Fe

2.0

ζ=0

1.5 1.0

ζ = 0.3

0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 f / fn

Fig. Q67(2) Transmissibility (TR)

Left and Right Spring Supports For designing the actual support system, the rubber spring k is considered to be divided into two rubber springs placed at the left and right sides, as shown in Fig. Q67(3), as only the translation vibration of the bed is generated, but no tilting vibration. Q(4) What are values of spring constants k 1 and k 2 ? m = 1400kg

Ω

c1

I = m i2 (i = 2m)

k1

k2

c2

Fig. Q67(3) Supported by 2 springs

Question 68 FFT specification As shown in Fig. Q68, we can see the waveform in the time domain and the corresponding FFT spectra in the frequency domain, when inputting a sinusoidal wave of a single frequency 20.5 Hz. Determine the:

458

12 Exercises of ISO Certification Examination for Vibration Experts

Q(1) Fundamental frequency (frequency resolution) [Hz] Q(2) Number of lines Q(3) Sampling number Q(4) Sampling time [μs] Q(5) Sampling frequency [kHz] Q(6) Amplitude of waveform [V] Q(7) Window function Q(8) Dominant frequency [Hz] Q(9) Amplitude of dominant component [dB]  [V] Q(10) Overall spectrum magnitude

1.0

0

Time [s]

MAG [V]

[V]

1.0

1.0

1.0 0.8 0.6 0.4 0.2 0.0

OA

0

( a ) waveform

100 200 300 400 Frequency [Hz]

0 20 40 60

200

300

400 OA

MAG [V]

MAG [dBV]

( c ) FFT (linear scale) 100

100 200 300 400 Frequency [Hz]

0

1.0 0.8 0.6 0.4 0.2 0.0 16

20 22 24 18 Frequency [Hz]

( d ) FFT (zoomed)

( b ) FFT (log scale)

Fig. Q68 FFT analysis (20.5Hz)

Question 68a In Q68, we used a rectangular window for a single frequency of 20.5 Hz. Now, we change it to a Hanning window. Q(1) Which FFT analysis shows this change in following five figures? Q(2) What is the single frequency of the inputting wave and window used for each figure? (b)

1.0

1.0 0.8 0.6 0.4 0.2 0.0

0

Time[s]

100 200 300 Frequency [Hz]

1.0 OA

400 MAG [V]

1.0 0.8 0.6 0.4 0.2 16

ｰ1.0 0 MAG [V]

[V] MAG [V]

ｰ1.0 0

MAG [V]

1.0

[V]

(a)

18 20 22 Frequency [Hz]

24

Time[s]

1.0 OA

0

100 200 300 Frequency [Hz]

400

16

18 20 22 Frequency [Hz]

24

1.0 0.8 0.6 0.4 0.2 0.0 1.0 0.8 0.6 0.4 0.2

Fig. Q68(a)(b)

12.2 Additional 70 Questions

0

ｰ1.0 0

400

1.0 0.8 0.6 0.4 0.2 16

20 22 18 Frequency [Hz]

24

1.0 0.8 0.6 0.4 0.2 0.0

1.0 OA

Time[s]

ｰ1.0 0

MAG [V]

[V] 100 200 300 Frequency [Hz]

1.0

0

100 200 300 Frequency [Hz]

400

1.0 0.8 0.6 0.4 0.2

MAG [V]

1.0 0.8 0.6 0.4 0.2 0.0

1.0 OA

Time[s]

MAG [V]

ｰ1.0 0

(e )

1.0

[V]

(d)

1.0

MAG [V]

MAG [V]

MAG [V]

[V]

(c)

459

20 22 18 Frequency [Hz]

16

Time[s]

1.0 OA

0

100 200 300 Frequency [Hz]

400

16

18 20 22 Frequency [Hz]

24

1.0 0.8 0.6 0.4 0.2 0.0 1.0 0.8 0.6 0.4 0.2

24

Fig. Q68(c)(d)(e)

Question 69 A rotor supported by cylindrical oil film bearings exhibits a typical amplitude curve, as shown in Fig. Q69, including unbalance excited vibration resonance at the critical speed ωn . Also, a small instability of oil whirl is generated at the rotational speed, , less than 2 × ωn , and a strong instability of oil whip occurs at the rotational speed around  ≈ 2ωn . The corresponding calculation with respect to the complex eigenvalue λ  α + jq is to be performed with the various rotational speeds. Which is the correct curve for the real part of the complex eigenvalue of the unstable mode?

Amplitude

Mode shapes Oil whip

Oil whirl

0

1

2

3

Ω / ωn

Re [ λ ]

positive 0

1 2 3 4

negative

5

Fig. Q69 Oil whirl and oil whip

Question 70 The fan is rotating at 1740 rpm, driven by an induction motor specified at 1800 rpm. If we identify separately both frequencies using a flat top window and f max  500 Hz, what is required for the line number? Question 71 An electric motor, specified by two poles and 2000 HP, is rotating at 3570 rpm, which

460

12 Exercises of ISO Certification Examination for Vibration Experts

is close to the line frequency 60 Hz. Using a Hanning window, what are recommended for the number of lines, L n , and the maximum frequency f max ? Question 72 Note that the roll-off (cut-off) frequency of an anti-aliasing filter is defined √ mainly by the frequency span, f max . If a Butterworth filter of the gain G(ω)  1/ 1 + ω2n , having a high roll-off with the slope of 120 dB/decade  36 dB/octave (n  6), were to be used for anti-aliasing in a FFT application, what is the magnitude of the aliasing spectrum at the roll-off frequency? Question 73 In Fig. Q73, the time waveform includes a DC component and sine waves of 10, 20, and 30 Hz in the window time of 8 s with sampling number 1024. In the corresponding FFT spectrum using the rectangular window, what are the rms and the overall value?

Volts

4 2 0 -2 -4

Waveform 0

5

1

2 Time [s]

3

8

FFT

Volts

4 3 2 1 0

0

10

20 30 Frequency [Hz]

40

50

Fig. Q73 Frequency spectrum

Question 74 We measure a sinusoidal waveform of 1 V @ 200 Hz in the time and frequency domains as shown in Fig. Q74; hence, the window time T  0.2 s, sampling number N  1024, and a rectangular window is employed. If the signal frequency is changed from 200 to 202.5 Hz, what are the spectral value at 200 Hz and the overall values?

12.2 Additional 70 Questions

461

OA

0

200Hz

1.0001V

2kHz

0.2 s

0

Fig. Q74 FFT analysis

Question 75 From Q74, we now change the window function to a Hanning window. What are the peak spectrum and the overall values for the input frequency of 202.5 Hz ? Question 76 Figure Q76 shows the results of a FFT (f max  500 Hz, line number L n  400, Hanning window) for a sinusoidal waveform having a single frequency.

Velocity [mm/s]

Q(1) What are the frequency and the amplitude of the input waveform? Q(2) What is the overall value? 12 8 4 0

PWR SP 10.30 8.08 0.96 0.06 0.18

3.02 0.30 0.08 0.04

96 97 98 99 100 101 102 103 104 Frequency [Hz]

Fig. Q76 Power Spectrum

Question 77 Figure Q77(1) shows a motor-driven pump system. The vibration data are acquired on the motor, specified by 18.5 kW, two poles and the line frequency f 0  50 Hz, for a suspected air-gap problem. Figure Q77(2) shows a typical example of zoomed FFT data. What are the slip ratio s and the rotating speed f r (rps)?

462

12 Exercises of ISO Certification Examination for Vibration Experts

Fig. Q77(1) Motor-driven pump system

Range : 31.6 ( 3 Hz − 1 kHz ) LIN

0020 / 0020

2.46 E−1

mm/s

7 x2

100 Hz 99.56 Hz

0 93.750 ZMx16 Hz 99.562Hz O.A Freq.Span 200Hz 3200Line

− +

106.250 O.A 5.51E−2 mm/s 1.87E−1 mm/s Hanning

Fig. Q77(2) Abnormal vibration

Question 78 Instead of the zoomed display shown in the previous Question 77, we wish to reset the FFT parameters in order to display these two peaks in a full frequency span ranging from 0 to f max . Q(1) What is the required resolution f ∗ to distinguish these two peaks using a Hanning window? Q(2) If we set f max  200 Hz, what number of lines, L n , is recommended, then what is the frequency resolution (fundamental frequency) f ? Q(3) What is the window time T ? Q(4) What are the sampling number N and the sampling frequency f s ? Q(5) What is the data acquisition time T 16 for this FFT setup if sixteen averages with 25% overlap are employed? Q(6) If we change f max into f max  300 Hz, what are corresponding to the number of lines, L n , the frequency resolution (fundamental frequency), f , and the data acquisition time, T 16 ? Question 79 Figure Q79 shows the vibration diagnosis using a FFT spectrum for induction motors

12.2 Additional 70 Questions

463

having some fault, hence rotational speed, f r , line frequency, f , and slip ratio, s. Which figures are associated with each of the following faults? Faults: (A) (B) (C) (D)

air-gap (static eccentricity) eccentric rotor (dynamic eccentricity) broken rotor bar stator short circuits

(1)

A:MAG

BFP MOTOR 28+15

RMS:10

2f

Vel. [ in/s ]

.25

0 BW:4.7743Hz

START:0Hz

(2)

Vel. [ in/s ]

.1 A:MAG

RMS:3

fr

2sf

Vel. [ in/s ]

.5

(4)

near f r

2f

SPAN CENTER BW:75mHz 20Hz 120Hz

near 2 f r

B:STORED

SPAN 20Hz

RMS:4

fr

Vel. [ in/s ] 0

2fr

2sf

BW:75mHz

CENTER 60Hz

(3)

RMS:3

A:MAG

x = 57.8 Hz y = .016 IPS rms

0

.2

STOP:500Hz

2sf

2sf x = 60.425 Hz y = .0095 IPSrms

CENTER:59.5Hz

BW:37.5Hz

SPAN:10Hz RMS:10

A:MAG

4f

6f

2f

0 START:0Hz

BW:7.6388Hz

STOP:800Hz

Fig. Q79 Fault diagnosis for electric motor

Question 80 Concerning the amplitudes and the frequencies of the beating waveform shown in Fig.Q80(1), find the two harmonic vibration components, i.e., {? Hz, ? V} and {? Hz, ? V}.

464

12 Exercises of ISO Certification Examination for Vibration Experts 1.0

[V]

0.5 0.0 0.5 1.0 0

0.5

1 Time [s]

1.5

2

Fig. Q80(1) Beating waveform

Question 81 Concerning the amplitudes and the frequencies of the undulating waveform shown in Fig. Q81(1), find two harmonic vibration components, i.e., {? Hz, ? V} and {? Hz, ? V}. 1.0

[V]

0.5 0.0 0.5 1.0 0

0.5

1 Time

1.5 [s]

2

Fig. Q81(1) Wavy waveform

[V]

Question 82 Concerning the amplitudes and the frequencies of the AM (amplitude-modulated) waveform shown in Fig. Q82(1), find the three spectral peaks, i.e., {? Hz, ? V}, {? Hz, ? V} and {? Hz, ? V}. 3.0 2.0 1.0 0.0 1.0 2.0 3.0 0.0

0.5

1 Time

1.5

2

[s]

Fig. Q82(1) AM waveform

Question 83 Run-out Figure Q83 includes Nyquist plots (1)–(5) and Bode plots (a)–(e) of five cases of unbalance vibrations with run-out. Find the five pairings of both plots.

465

1.1 -5

2 0

(b) 0.5 0.8 0.9

-5

4

p = 1.0

5

(2)

6

5

1.3 1.1

0

6

– 180

4 2 0

(c) 0.5

-5

0.8 0.9

p = 1.0

5

p = 1.0 -5

0.5 0.8

0.9

(d)

1.3

0

6

– 180

4 2

5

0.9

p = 1.0

0.8 0.5

-5

1.3

-5

180 0

6

– 180

4 2 0

1.1 5

(e)

1 2 3 Frequency p = Ω / ω n 180

10 Amplitude a

5

3

8

-5

(5)

1 2 Frequency p = Ω / ω n

10 Amplitude a

5

180

0

1.1

3

8

-5

(4)

1 2 Frequency p = Ω / ω n

10 Amplitude a

5 1.3

180

8

-5

1.1

3

10

p = 1.0

(3)

1 2 Frequency p = Ω / ω n

Phase φ

0.9

1.3

0 – 180

Phase φ

5

8

Phase φ

-5

0.8

Amplitude a

0.5

Amplitude a

(1)

180

10

8

0

6

– 180

Phase φ

(a)

5

Phase φ

12.2 Additional 70 Questions

4 2 0

1 2 3 Frequency p = Ω / ω n

Fig. Q83 Nyquist and Bode plots

Question 84 Digital Filter Data1  {x0 x1 x2 · · ·} are collected by an A/D convertor (sampling frequency f AD  4 × 1028 Hz  4096 Hz, T AD  1/f AD [s]) for data acquisition.

466

12 Exercises of ISO Certification Examination for Vibration Experts

Q(1) In order to analyze by the FFT (f max  400 Hz, number of lines L  400), we must filter Data1 by a certain cut-off frequency f c for an anti-aliasing. What is the frequency f c ? Q(2) If we use a first-order LPF for the transfer function, G 1 (s)  1/(1 + τ s), for the anti-aliasing filter, the digitization of the filter is obtained using the bilinear transformation:   2 a0 + a1 z −1 1 − z −1 ≡ × G 1 (z)  G 1 s  T AD 1 + z −1 1 − b1 z −1 Data2  {y0 y1 y2 · · ·} are produced by the following recurrence relation of G1 (z) with the cutoff frequency f c : y0  x0 and yk  b1 yk−1 + a0 xk + a1 xk−1 What are the constant values of b1 , a0 , and a1 ? Q(3) If we assume Data1  {1,2,3….}, what are Data2  {y1 y2 y3 · · ·}? Q(4) In order to match the sampling frequency, f AD , of the A/D converter to the sampling frequency, f s , of the FFT analysis, we must decimate Data2 by the interval of N d . What is the value of N d , where f AD = N d × f s ?

Gain

Question 85 note: Q72 concerning Butterworth filter The Butterworth filter (here denoted√ by BWF) is often used as an anti-aliasing filter, which is defined by: |G n ( jω)|  1/ 1 + ω2n , as shown in Fig. Q85(1): 1.0

[dB] 10

0.8

0

0.6



0.2 0 0

n=1

3 4 5

0.4

0.5

1.5 2.0 f / fmax ( a ) Linear Y-axis

2



2

1.0

n=1

3 4 5

 2.5

3.0

 0.1

1.0 f / fmax ( b ) Logarithmic Y-axis

10

Fig. Q85(1) Transfer functions of Butterworth filters

1.0

1.0 Analog

Gain

0.8 Nd = 4

0.6 0.4

Nd = 4

0.6

Analog

0.4

?

0.2

0.2 0.0

0.8

100

200 300 Frequency [Hz]

( a ) 1st Butterworth

400

0.0

? 100

200 300 Frequency [Hz]

400

( b ) 3rd Butterworth

Fig. Q85(2) FFT analyses for input frequencies {100, 200, 300, 400,x}Hz

12.2 Additional 70 Questions

467

We prepare a test signal including five harmonic functions with the same amplitude of 1 V and frequencies of {100 Hz, 200 Hz, 300 Hz, 400 Hz, unknown x Hz}. The test signal is input to the FFT analyzer set with f max  400 Hz and number of lines L  400. The FFT results are shown in Fig. Q85(2) (a) and (b), corresponding to two cases of first BWF (n  1) and third BWF (n  3) for anti-aliasing. Q(1) In these cases, we can see peaks at 100, 200, 300, and 400 Hz corresponding to the input frequencies, but the peak at 324 Hz does not correspond to any input frequency. What is the unknown x Hz that is likely to be generating it? Q(2) What is the magnitude of the spectrum at 324 Hz in each case? Question 86 ISO 13373-1 introduces two types of condition monitoring system (CMS) as shown in Figs. 1.6 and 1.7, called online CMS and portable/offline CMS. Condition monitoring systems may take many forms. A decision to select the appropriate CMS depends upon a number of factors, such as in the following table. Answer which is more acceptable for a CMS by filling H  high or L  low in residual spaces (1)–(10) of the table. Table Q86 Features of CMS No

Machines

Portable CMS

On-line CMS

1

Criticality of the machine operation

(1)

(2)

2

Cost of catastrophic failure

(3)

(4)

3

Cost of machine

Low

High

4

Speed rate of progress of the failure mode

(5)

(6)

5

Accessibility of repair/maintenance (e.g. in nuclear plants or other remote locations)

Not difficult

Difficult

6

Accessibility of the appreciate measuement position

Easy

7

Quality of the measurement/diagnostic system

(7)

8

Operational modes of the machine

—

—

9

Safety

—

—

10

Environmental impact

(9)

(10)

Hard (8)

Question 87 The online monitoring system shown in Fig. 1.6 is applied to the following machines. Which is a proper statement? 1. 2. 3. 4. 5.

Machines to be replaced in a short exchange period. Oil-supply pump for plain bearings. Machines giving criticality of operation. Machines showing a slow rate of progress of the failure mode. Machines being distantly placed from the central monitoring room.

468

12 Exercises of ISO Certification Examination for Vibration Experts

Question 88 When monitoring x- and y-directional vibrations and a pulse wave (around 3600 rpm) of a rotor as shown in Fig. Q88, we observed 1X vibrations in figure (1). After a while, they changed into the larger vibrations in Fig. Q88(2). The following diagnosis is reported, and you are required to fill out the brackets: In Fig. Q88(1), the unbalance vibration is dominant and the state is quite normal. In Fig. Q88(2), the non-synchronous vibration appears and the system falls into an abnormal state. The abnormal vibrations of (A) Hz presents (B) whirl motion so that the occurrence of (C) vibration is guessed. Then the machine is commanded to (D) operation. y

sensor

Rotor x sensor pulse sensor Y vibration

Y

X vibration

X

pulse (1)

(2)

Fig. Q88 Before and after the instability

Question 89 Figure Q89 shows orbits of a rotor, which is rotating in an anticlockwise sense. In Fig. Q89(1), when we measure the rotor position on the X-axis, we see three peaks at the instants indicated by the yellow vertical lines. Then, we know the existence of combinations of forward 1X vibration and forward 3X vibration, that is, 1X and 3X in a full FFT. In a similar way, determine the full FFT of orbital whirl motion that shows how many multiples of the rotational speed with sign of forward (+) and backward (−) apply for each Fig. Q89(2–8).

12.2 Additional 70 Questions

469

(1)

(3)

(5)

(7)

(2)

(4)

(6)

(8)

Fig. Q89 Orbits and full FFT

Question 90 In a pulley belt drive system as shown in Fig. 7.57, a pulley of 0.3 m diameter rotates at 6000 rpm driven by a V-belt with 2.0 m perimeter length. What is the excitation frequency to the pulley rotor caused by the influence of the passing of the joint seam of the belt? Question 91 A pump rotor has 18 blades for the impeller (Z r  18) and the stator has 24 vanes for the diffuser (Z g  24) as shown in Fig. Q91(1). Q(1) What is the excitation frequency for the rotor, which is referred to as the blade passing frequency (BPF), at speed 3000 rpm? Q(2) What is the excitation frequency acting upon the rotor caused by stator vanes at the rotor speed 3000 rpm? Q(3) In the case of the operation at variable rotating speeds, what is the lowest number of the nodal diameter modes of the disk where the resonance will occur? 24 defuser blades

18 impeller blades

Fig .Q91(1) Blade passing frequency

470

12 Exercises of ISO Certification Examination for Vibration Experts

Question 92 As shown in Fig. Q92(1), a flexible rotor is supported by a ball bearing and an oil film bearing, and the vibration is observed. Oil whip instability occurred at the rotational speed of 68 rps as shown in Fig. Q92(2). The FFT analysis at the onset of this instability occurrence is shown in Fig. Q92(3). Which is the spectrum of oil whip frequency in options {➀, ➁, ➂, ➃, ➄}? Flexible Rotor

Ball bearing

Motor

130

18

70

Oil bearing

625

Fig. Q92(1) Test rotor

Overall amp. [m]

160 Oil whip 120 A

80

1st critical speed

40 0

0

10

20 30 40 50 Rotational speed [rps]

60

70

Fig. Q92(2) Amplitude

1

2

3

4

5

PWR SP [dB]

20

65.5rps

40 60 80

0

10

20

30 40 50 60 70 Excitation frequency [Hz]

80

90

100

Fig. Q92(3) FFT

Question 93 Figure Q93 shows the various shutdown records of a compressor installed in a plant, including overall vibration, denoted by OA, synchronous vibration amplitude, denoted by 1X, and rotational speed, denoted in RPM ranging from the rated

12.2 Additional 70 Questions

471

speed down to standstill. Concerning each figure, the diagnosis is stated as follows. Select the pairing of the statement and the figure. Diagnosis statements:

1x Time

RPM Motor power off Rotation stops B Trip O.A. 1x

Time RPM Motor power off Rotation stops

Vibration amplitude

O.A.

Vibration amplitude

Vibration amplitude

A Trip

Vibration amplitude

Noise in the measurement system Unbalance occurrence due to blade loss Oil whip instability Electric motor induced vibration Unbalance occurrence due to thermal bow Vibration amplitude

(1) (2) (3) (4) (5)

C Trip O.A.

1x

Time RPM Motor power off Rotation stops D

O.A.

Trip 1x Time RPM Motor power off Rotation stops E Trip O.A.

1x

Time RPM Rotation Motor power off stops

Fig. Q93 Vibrations recorded at shutdown

Question 94 Ball bearing vibration As shown in Fig. Q94, a FFT spectrum of ball bearing vibration shows dominant frequencies at 77.3 Hz and at multiple frequencies of the rotational speed, f r  20 Hz. What is the type of fault in the ball bearing specified by the number of balls, 10, and the contact angle of 0°?

472

12 Exercises of ISO Certification Examination for Vibration Experts Range : ACCe EXP 0021 / 0032 1.58 E0

3.16

( 10 Hz  20 kHz )

fo

1 x2

0  

0.000 ZMx1 77.3 Hz O.A

Freq.Span 500Hz

Hz

500.000 O.A 2.88E0

3200Line

Hanning

Fig. Q94 FFT of ball bearing vibration

Question 95 Ball bearing vibration As shown in Fig. Q95, a FFT spectrum of ball bearing vibration shows dominant frequencies at 7.7, 82.406 Hz and sideband frequencies of the rotational speed, fr  20 Hz. What is the type of fault in the ball bearing specified by the number of balls, 10, and the contact angle of 0°? Range : ENVELOPE LIN 0008 / 0008

3.16

( 10 Hz  20 kHz )

7.90 E1

2 fb 2 fb fc

2 x2

0

0.000 ZMx1 82.406 Hz O.A Freq.Span 100Hz 



Hz





3200Line

2fb fc

100.000 O.A 1.20E2 3.26E0 Hanning

Fig. Q95 FFT of ball bearing vibration

Question 96 Ball bearing vibration As shown in Fig. Q96, a FFT spectrum of ball bearing vibration shows dominant frequencies at 391.2 Hz, 469.3 Hz, 312.8 Hz, 234.6 Hz and so on. The rotational speed is f r  20.2 Hz. What is the type of fault in the ball bearing (NSK #1302) specified by the number of balls, 10, and the coefficient β  FTF/ fr  0.386?

12.2 Additional 70 Questions

473 Range : 3.16 ( 10 Hz  20 kHz ) ACC EXP 0032 / 0032 1.97 E1

391.2 Hz 312.8 Hz

469.3 Hz

234.6 Hz m/s 2

4 x2

0

0.000

ZMx1

Hz

1000.00

O.A Freq.Span 1kHz

O.A

0.00E0 m/s 2 6.09E1 m/s 2 Hanning

DC

 

3200Line

Fig. Q96 FFT of ball bearing vibration

Question 97 Ball bearing vibration As shown in Fig. Q97, a FFT spectrum of ball bearing vibration, dominant frequencies are observed 357.1 Hz and others at the rotational speed f r  49.84 Hz. What is the type of fault in the ball bearing specified by a number of balls, 12, and the contact angle of 0°? Range : 10 ( 3 Hz  20 kHz ) EXP 0005 / 0032 1.25 E0

49.84 Hz

357.1 Hz 407.0 Hz

3 x2

0

0.000 ZMx1 357.1 Hz O.A Freq.Span 500Hz

307.3 Hz

Hz

 

3200Line

456.8 Hz

500.000 O.A 8.07E1 2.52E0 Hanning

Fig. Q97 FFT of ball bearing vibration

Question 98 In R1_Example 5.6, we completed balancing at two rotational speeds of 48 Hz and 68 Hz near the first and second critical speeds as shown in R1_Figs. 5.22 and 5.23, respectively. Then, we had already attached correction weights as follows: {1, 1, 1} × 3.5 g −55° at {#1, #2, #3 } for the first Nc1 balancing and {2, 0, −1} × 6 g −180° at {#1, #2, #3 } for the second Nc2 balancing. Afterward, the rotor was spun up to the third critical speed, Nc3, as shown by curves noted by “Run 4, Before” in Fig. Q98(1). We intend to achieve the Nc3 balance according to the following procedures:

474

12 Exercises of ISO Certification Examination for Vibration Experts

[ μm ]

200

Nc3

160

Amplitude

S1

(Before) Run 4

120 80 Nc2

40

S2

(After) Run 6

S1

Nc1

0

100 speed

50

S 2 200

150 [ Hz ]

Fig. Q98(1) Comparison of before and after 3rd mode balance

Q(1) R1_Fig. 5.20 shows the balancing planes of #1, #2, and #3, and eigen modes of Nc1, Nc2, and Nc3. Prior to the third mode balancing, we have to decide the ratio of the pair of weights for the Nc3 mode balance so as not to influence the already well-balanced states of the Nc1 and Nc2 modes. Prove that one of the ratios is {− 1.8, 2.4, −1} at {#1, #2, #3 }. Q(2) We put a trial weight {−1.8, 2.4, −1} × 2 g −60° at {#1, #2, #3} and the corresponding Nyquist plot as drawn as “Run 5, Trial” in Fig. Q98(2). Obtain the correction weight for Nc3 balance by comparing Run 4 and Run 5. The correct answer would give the Bode plot denoted by “Run 6, After,” having small amplitudes over the entire operational speed range. Q(3) We attached, three correction weights for Nc1, Nc2 in the past, then additional weights for Nc3. Summarize these paired correction weights as single masses to be attached at the three respective correction planes. B

[ μm ] 170 Hz Run 5

Trial S1

160 Hz – 200

– 150

– 100

50

C

– 50

50 [ μm ]

? A

Run 4

S1

– 50

170 Hz 160 Hz

Fig. Q98(2) Polar plot for 3rd mode balance

Question 99 Figure Q99(1) shows a rotor of mass m1  52 kg, polar moment of inertia I p1  0.31 kg m2 , and transverse moment of inertia I d1  2.32 kg m2 around the mass

12.2 Additional 70 Questions

475

center, i.e., the center of the shaft, which is located at x g1  531 mm from the left end. We add a thin disk, having the mass ma  13.5 kg and I pa  0.1 kg m2 , at the right end, as shown in Fig. Q99(2). Concerning the total system of Fig. Q99(2): Q(1) What is the total mass m2 ? Q(2) What is the total polar moment of inertia I p2 ? Q(3) Where is the mass center, x g2 ? Q(4) What is the transverse moment of inertia around the mass center I d2 ? 1062 400

40

400

oil brg.

240

oil brg. G Brg.

Brg.

Fig. Q99(1) Rotor (m1=52 kg,

Ip1= 0.31 kg⋅m2, Id1= 2.32 kg⋅m2

@ G, lengths in mm)

1062 400

40

400

oil brg.

240

oil brg. G Brg.

Brg. xg2

Fig. Q99(2) Rotor with a thin disk (ma= 13.5kg, Ipa= 0.1kg⋅m2, lengths in mm)

Question 100 Figure Q100 shows a vertical pump rotor system consisting of the shaft (shaft mass, ms  62 kg, length L  1 m, diameter φ100, Young’s modulus E  206 × 109 N/m2 ) and impeller (impeller mass, mp  154 kg, I dp  48 kg m2 ) located at the bottom. The impeller is subject to a spring k 1  0.5 × 106 N/m and damper c1  4.2 × 103 N s/m for the translation motion and a rotational spring k 2  0.2 × 106 N m/rad and moment damper c2  0.42 × 103 N m s/rad for the tilting motion. Q(1) What is the natural frequency of shaft–impeller system without springs and dampers? Q(2) What are the natural frequency and the damping ratio of shaft–impeller system with spring k 1 and damper c1 ? Q(3) What are the natural frequency and the damping ratio of shaft–impeller system with springs (k 1 and k 2 ) and dampers (c1 and c2 ) ? Q(4) When we consider the cross-stiffness effect kc  jc2 λ2  (note, k c  k xy  − k yx ) for tilting motion, we replace c2 z˙  c2 sz by c2 z˙ − jkc z  c2 (s − jλ2 )z, using the impeller displacement of z and the differential operator of s, as shown in the left side of Fig. Q100. Note that the notation of λ is detailed in Fig. 8.33 and Fig. 11.12.

476

12 Exercises of ISO Certification Examination for Vibration Experts

In the case of λ2  0.48, what are the onset speed [rps] and the resultant frequency [Hz] of the instability?

 c2( s – j λ ) c

c

k

k

Fig. Q100 Vertical pump system

12.3 Answers and Hints for the Additional Questions 31–100 Q31_Ans. 2 Q32_Ans. 1 Q32_Hint: If the√frequency resolution is reduced by 1/2  0.5, the noise level is decreased by 1/2  0.7  −3dB. The reduction from 240 to 60 mV is 20Log(60/240)  −12 dB  −3 dB × 4, which requires the frequency resolution of 8/(1/2)4  8/16  0.5 Hz. Q33_Ans. 1 Q33_Hint: Since one bit is excluded for the sign, the lowest detectable voltage is 20 V/(211 − 1)  20/2047  0.00977. Q34_Ans. 4 Q34_Hint: The dynamic range is 211 − 1  2047  66.2 dB for the 12-bit and 215 − 1  32,767  90.3 dB for the 16-bit. The difference yields the improvement of the dynamic range, i.e., 90.3 − 66.2  24 dB. Q35_Ans. (1) k s  3.3 × 106 and k d  6×106 N/m, (2) f 1  34 Hz, (3) cd  32.8 × 103 N s/m and Afpeak  7 Q35_Hint: (1) Using m 1  m + m s /2  41.2 + 5  46.2 kg and m 2  m s /2  5 kg, ks  (m + m s /2)(2π f z )2  3.3 × 106 N/m, kd  (m + m s )(2π f d )2  6 × 106 N/m. (2) R1_Eq. (3.41) gives the following eigenvalue equation:

12.3 Answers and Hints for the Additional Questions 31–100

477

−1    1 m1 0 ks −ks (kd + ks )m 1 /(kd ks ) m 2 /kd φ  φ, φ  −ks ks + kd m 1 /kd m 2 /kd 0 m2 ωn2 The obtained eigenvalues ω2n  {45.4 × 103 , 1.88 × 106 } indicate the first eigen frequency f 1  34 Hz from ω1  2π f 1 . Q35_A different solution: (2) According to R1_Eq. (3.43), the first eigen frequency is thus approximated as follows: 1 m1 m2 + m1 1 1 1 (kd + ks )m 1 m 2 ≈ +  + → 2  2+ 2 2 kd ks kd ks kd fz ω1 f1 fd 1 1  + → f 1  33.5 Hz 42.52 54.52 (3) By referring to R1_Fig. 8.28, we see ζ d  0. 94 and Afpeak  7 on  1.8 on the abscissa. Then, the ordinate √ √ for σ  kd /ks  6/3.3 cd  2ζd kd (m + m s )  2 × 0.94 6 × 106 × 51.2  32, 800 [N s/m] is required. The response sensing for harmonic excitation at the center disk is calculated as follows:  −1    m 1 s 2 + ks −ks 1 y(s)  1 0 for s  j2π f. −ks m 2 s 2 + ks + kd + cd s 0 The Bode plot of the response amplitude is drawn by three curves in Fig. Q35f for the optimal cd and two other values assuming the mis-tuned. In fact, we can confirm that the optimal cd gives the lowest peak amplitude compared with miss-tuned cases. Note that the predicted amplitude, Afpeak  7 mentioned above¸ is the value to be designed for the Jeffcott rotor of Fig. Q35c and the actual maximum amplitude  5.5 observed in Fig. Q35f belongs to the system of Fig. Q35b. 10

0.5 × cd 2 × cd

6

Af

peak

/ δs

8

4 2 0

Optimal cd 10

20

30

Exciting frequency

40

50

[ Hz ]

Fig. Q35(f) Bode Plot

Q36_Ans. ζ 1  −0.00016, ζ 3  −0.00039, ζ 4  −0.00016

60

478

12 Exercises of ISO Certification Examination for Vibration Experts

Amplitude [dB]

Q36_Hint: Referring to Example 9.4, the modal damping ratio is obtained by ζ  m d B /54/ f n : As seen in the first mode of Fig. Q36(2), the parameters  {m d B  (13.2 − 8.8)/29.7  0.148, f 1  16.8} give ζ1  1.6 × 10−4 . For the third and the fourth modes, each damping ratio can be obtained in a similar way. 0 8.8dB 1st 10 20 30.8dB 30 3rd 40 50 43.7dB 4th 60 0 Time

13.2dB

29.7s 50.4dB 52.7dB

[s]

31.25

Fig. Q36(2) Evaluation

Q37_Ans. ζ  0.047 Q37_Hint: We plotted the data on a X–Y graph in the relation of x i (Y -axis) and x i+1 (X-axis) as seen in Fig. Q37(2). These plotted points are approximated by a straight line through the origin, and the slope of the line is read as mx  tan53°  1.34. Then, 1 ln(m x )  0.047 for one cycle of data. ζ  2π cle

25

:1

15

xn

xn

cy

20

10 5

mx 5

10

15 20 xn+1

25

Fig. Q37(2) Data plot xn

Q38_Ans. ζ  0.23 Q38_Hint: We plotted amplitude data on a X–Y graph in the relation of yi (Y -axis) and yi+1 (X-axis) on Y -axis and yn+1 on X-axis as shown in Fig. Q38(2). These plotted points are approximated by a straight line through the origin, and the slope of the 2 ln h y  0.23 by the line is determined by hy  tan64°  2.054. Then, ζ  2π conversion from “half” to “one” cycle of data.

yn

15

ycl f-c hal yn :

20

e

25

hy = tan64º= 2.054 = ln ( hy2 ) = 2ln ( hy ) = 1.44

10 hy

5 5

10

15 20 yn+1

25

Fig. Q38(2) Data plot yn

12.3 Answers and Hints for the Additional Questions 31–100

479

Q39_Ans. I 2  4.23 kg m2  Q39_Hint: From the formula, ωn  mga 2 / h/I , for calculating the natural frequency, when hanging a rotor shown in Fig. Q39b, the relationship between figures (a) and (b) is as follows: I1 (2π/T1 )2 /m 1  I2 (2π/T2 )2 /m 2 . By inserting I2  I1 + 2m a l22 , m 1  4 and m 2  m 1 + 2m a  8 kg into the above equation, we can obtain I 1 and I 2 as follows: I1 /0.12 /4  I2 /0.32 /8 → I1  0.235 → I2  4.23. Q40_Ans. 2 Q40_Hint: Letting the speed ratio n  2, the torsional twist angles of the turbine shaft ∗  I22 n 2  0.4, k2∗  k2 n 2  4, are normalized by the pump shaft as follows: I22 ∗ 2 and I3  I3 n  40. The equivalent single shaft system is shown in figures (2) and then (3). Assuming the first mode shape of “out of plane,” the corresponding natural frequency is estimated by means of the two-DOF system ignoring I *2 .  −1 keq  1/k1 + 1/k2∗  0.8   √   → ωn  keq I1 + I3∗ / I1 I3∗ ≈ keq /I1  0.8  0.9. Q40_A different solution: Referring to Sect. 9.1.2, the matrices for the four-DOF system of figure (1) are: ⎡

I1 ⎢0 M4  ⎢ ⎣0 0

0 I21 0 0

0 0 I22 0

⎤ ⎡ 0 k1 ⎢ −k1 0⎥ ⎥ K4  ⎢ ⎣ 0 1⎦ 0 I2

−k1 k1 0 0

0 0 k2 −k2

⎤ 0 0 ⎥ ⎥. −k2 ⎦ k2

Define the following transformation matrix from this four-DOF system to a threeDOF one of Fig. (3): ⎡

1 ⎢0 T1  ⎢ ⎣0 0

0 1 n2 0

⎤ 0 0⎥ ⎥. 0⎦ n2

The matrices for the three-DOF system of Fig. (3) are then obtained: ⎡

⎡ ⎤ ⎤ 1 0 0 1 −1 0 M3  T1t M4 T1  ⎣ 0 0.5 0 ⎦ K 3  ⎣ −1 5 −4 ⎦. 0 0 40 0 −4 4

480

12 Exercises of ISO Certification Examination for Vibration Experts

The solution of the eigenvalue problem with system matrix M3−1 K 3 provides eigen frequencies  {0, 0.898, 3.21} and the corresponding eigenvectors  {[1, 1, 1]t , [36.5, 7.06, −1]t , [11, −102, 1]t } (2)

(3)

N1 Pump

Ι21 = 0.1 k1 = 1 k =4

Ι1 = 1

Ι22 = 0.4

N2

N1 Pump

Turbine

k1 = 1

Ι1 = 1

k =4

Turbine

Ι 2 = 0.5

Ι 3 = 40

Ι 3 = 40

Fig. Q40(2)(3) Torsional system modified to a single equivalent shaft

Q41_Ans. (1) I1∗  I1 /n 2  1.1/3.32  0.101 kg m2 , k1∗  k1 /n 2  49 × 104 /3.32  45 × 103 Nm/rad and I2∗  I21 /n 2 + I22  5.1/3.32 + 0.53  1.0 kg m2 (2) ω1  47 Hz Q41_Hint: We measure the deflection of the first eigen mode by a scale, i.e., {1, 0.81, 0,0} at { I *1 , I *2 , I 3 , I 4 }portions. Referring to R1_Sect. 3.2.2, we calculate the modal mass m* and the modal stiffness k * as follows: m ∗  I1∗ × 12 + I2∗ × 0.812 + I3∗ × 02 + +I4∗ × 02  0.74 and k ∗  k1∗ × (1 − 0.81)2 + k2 × 0.812 + k3 × 02  64.9 × 103 Then, the natural frequency is ω1 

√

k ∗ /m ∗  292 rad/s  47 Hz.

Q42_Ans. 18.2 mm/s Q42_Hint: The equation of motion of the mass M of the motor/pump is written as a following single-DOF model: M X¨ + c( X˙ − Y˙ ) + k(X − Y )  0 The vibration transmissibility is defined by the transfer function T (s)  X(s)/Y (s), and curves of the frequency response TR  |T (jω)| are drawn in Fig. Q42 (2): 2ζ ωn s + ωn2 cs + k  Ms 2 + cs + k s 2 + 2ζ ωn s + ωn2       1 + j2ζ p   2ζ ωn jω + ωn2   T R  |T ( jω)|   −ω2 + 2ζ ωn jω + ωn2   1 − p 2 + j2ζ p   1 + (2ζ p)2   , 2 1 − p 2 + (2ζ p)2

T (s) 

12.3 Answers and Hints for the Additional Questions 31–100

481

where p  ω/ωn Referring to p  ω/ωn  29.55/27.5  1.075 and ζ  0.03., TR  X/Y  5.98 is obtained by the above equation or by the point ◯ indicated in Fig. Q42(2). Then, X  5.98 × Y  5.98 × 3.05  18.2 mm/s. 20 = 0.01

TR

15

0.03 0.05

10

0.1 5 0 0.8

0.9

1.1

1.0

1.2

f / fn

Fig. Q42(2) Transmissibility (TR)

Q43_Ans. (a) (1), (2), (3), (4)  (13, 5, 0.62, 1.62), (b) 0.62 × Q43_Hint:

√

k/m

(a) Letting M and K denote the mass and stiffness matrices, respectively, and φ 1 and φ 2 as the first and second eigen modes:  M

   k1 −k1 m1 0 3 −2 , K  , φ1  and φ2  . −k1 k1 + k2 0 m2 2 3

Note that these eigen mode shapes are approximate. Then, the modal parameters are calculated as follows:

First mode Modal mass m

Second mode

φ1t Mφ1  m 1 × 32 + m 2

Modal stiffness k * √ Natural frequency k ∗ /m ∗

φ1t K φ1  k1 × (3 − 2)2

× 22 + k2

√ 0.62 k/m

 13m

× 22

 5k

φ2t Mφ2  m 1 × 32 + m 2 × 22  13m φ2t K φ2  k1 × (3 + 2)2 + k2 × 32  34k √ 1.62 k/m

Note that the modal stiffness equals to the total potential energy stored in each spring. Additional information about the exact solution obtained by the eigenvalue problem ωn2 Mφ  K φ includes: ⎧ √ ⎨ ω1  0.618 k/m, 3 ⎩ φ1  1.85

for 1st mode and

482

12 Exercises of ISO Certification Examination for Vibration Experts

⎧ √ k/m, ⎨ ω2  1.62  −1.85 ⎩ φ2  3

for 2nd mode

(b) The deflection mode is calculated as follows:      2 1 m1 3 mg 1 −1 m 1 g g Kδ  Mg → δ  K m2 1 1 m2 2 k 1 Since the deflection mode due to gravity is the same as Q43, the same natural fre 1/2  √  0.618 k/m. It is thus found quency is therefore obtained by ( δ t K δ/δ t Mδ that the deflection mode due to the gravity is sufficient for the first eigen frequency estimation. Q44_Ans. (a) meq  1.47m, (b) modal mass meq  1.44m for the eigenvector φ 1a Q44_Hint: (a) The graph gives the slope α  (0.619 − 0.598)/(0.1m)  0.21/m rad/s/kg as the gradient of this straight line and the original natural frequency ω0 at m  0. Thus, R1_Eq. (3.29) gives the equivalent mass meq ; m eq  ω0 /(2α)  0.619/(2 × 0.21)m  1.47m. (b) Since the modal mass is 13 m with respect to the eigenvector φ 1 , the modal t Mφ1a  13m(1/3)2  mass of the eigenvector φ 1a  φ 1 /3 becomes φ1a 1.44m, which agrees with the equivalent mass mentioned above in (a) with respect to the nodal point x 1 . Q45_Ans. (1) meq  69.2 kg, (2) ε*  8.7 μm, (3) f  0.125 Hz Q45_Hint: (1) (2) From the graph, we read α  (37.4-36.86)/2  0.27 as the slope of the straight line and ω0  37.4 Hz. Thus, the equivalent mass m eq  ω0 /(2α)  37.4/(2 × 0.27)  69.2 kg, and the modal eccentricity ε∗  m fan × ε/m eq  10 × 60/69.2  8.67 μm. (3) In this graph, the values of measured frequency are distributed as if like a step down function having the same change of the height at each step, i.e., the frequency resolution (fundamental frequency), of 0.125 Hz. Q46_Ans. 3 Q46_Hint: If the damping ζ is infinite, the two-DOF system is reduced to a one-DOF system and thus the curve has one infinite peak. On the other hand, no damping, ζ  0, gives two infinite peaks, because the two-DOF system has two resonant frequencies. When we set a certain damping ζ , the peaks are suppressed to be under a certain finite level. Too much damping ζ gives a frequency response that approximates to the single-DOF undamped curve. Too little damping ζ gives a frequency response that approximates to the two undamped resonances. Thus, the best tuning of the dynamic damper is obtained between both by the following tuning:

12.3 Answers and Hints for the Additional Questions 31–100

483

(1) The mass ratio, R  m2 /m1 , is first given, being generally less than 0.1. The √ best tuning could promise the achievable reduced amplitude of A1  1 + 2/R for any R. (2) Insert and select the stiffness k 2 to complete an undamped system, as the intersections of P and Q are the same height, i.e., P  Q. The best tuning requires ω2  ω1 /(1 + R) → k2  m 2 ω22 . (3) Insert and select the damping factor c2 , as the curve become maximal at the intersections P and Q with respect  to this damped system. 3R The best tuning requires ζ  → c2  2ζ m 2 ω1 8(1 + R)3

Amplitude A1 ( = X1 / Xs t )

When the best tuning is employed, the response curve of figure (2) is improved as in figure (3). ] =f

20

]=0

Optimum: ] = 0 Q = 0.95 ] = 0.15

15 10

P

5 0 0.4

Q

0.6 0.8 1 1.2 1.4 Exciting frequency λ ( = ω/ω1)

Fig. Q46(3) Optimal damping

Q47_Ans. (1) 0.058 F 0 , (2) Q  10, (3) 0.58 δ st where δ st  F 0 /k Q47_Hint: (1) According to Table Q48(1), F(t) is transformed to the following FSE:  1 2 1 + (sin ωt + sin 3ωt + · · · F(t)  F0 2 π 3  1 1 + sin 11ωt + · · · + sin nωt + · · · 11 n The 11th component becomes F 0 × 2/π /11  0.058F 0 . (3) In the frequency domain, (FSE of input force) × (the Bode plot of the system)  (FSE of output response) is satisfied, i.e., figure (2) × figure (3)  figure (4). For 2 F0  0.058F0 × the frequency of 11ω  ωn only, the spectrum is (1) : f 11 th  11π {(2):Q  10 }  {(3):a11th  0.58δ st }. Q48_Ans. 1X  100%, 2X  0%, 3X  1/3  33%, 4X  0% 5X  1/5  20% Q48_Hint: Referring to Table Q48(1), this waveform f (θ ) is transformed into the following FSE:

484

12 Exercises of ISO Certification Examination for Vibration Experts

f (θ ) 

  1 1 4 sin θ + sin 3θ + sin 5θ · · · π 3 5

Q49_Ans. ➀ Q49_Hint: The magnitude of torsional vibration response is proportional to the magnitude of the input torque, i.e., the answer of Q48. As shown in a simulation of Fig. Q49(2), the radii magnitudes of the circles are proportional to the amplitude peak of the vibration waveform |θ (t)|. 3X

2X

Amplitude

(t)

Frequnecy

5X 4X

1X

0

0

0.5

P =  /ωθ

Rotational speed

P =  /ωθ

Fig. Q49(2) Campbell diagram and vibration magnitude

Q50_Ans. (1) k g  48 kN/m, τ  0.0117 s, α  0.46. kg 1 + τ s 16 × 103 1 + 0.0117s  . (2) G o (s)  G r (s)G p (s)  ms 2 1 + ατ s s2 1 + 0.00539s (3) ks  103 kN/m, kb  43.7 kN/m, cb  514 N · s/m. (4) f 1  25 Hz, ζ1  0.21 → Q 1  1/(2ζ1 )  2.38. (5) Q  2.3 @ f 1  25 Hz for first mode. Q50_Hint: Refer to R1_Sect. 8.3. (1) To achieve the targeted damping ratio ζ 1  0.2, the relationship of R1_Eq. (8.12), ζ  1/2 tan φm , requires the necessary phase lead, φm  tan−1 (2 × 0.2)  22◦ , which is realized by setting the coefficient of α  0.46 according to R1_Eq. (8.48). The definition α  σ/(σ +1) thus gives σ  0.85. Since the targeted first critical speed is f 1  20 Hz. The shaft frequency is 1/4 of R1_Eq. (8.50). given by f s  f 1 /α1/4  24.3 Hz according to f 1 / f√ s α )/ α  0.0117 according to The time constant is calculated by τ  1/(2π f 1 √ τ ω1  1/ α of R1_Eq. (8.51). The total gain is thus assigned approximately to k g  m(2π f 1 )2  48 kN/m. (2) The open-loop transfer function of Fig.Q50(4) proves that the cross-over gain frequency f g is approximately the first critical speed, f 1 ≈ f g  25 Hz, and the phase margin is φm  22◦ , being equal to damping ratio ζ  0.21.

12.3 Answers and Hints for the Additional Questions 31–100

485

(3) The conversion is made by R1_Eq. (8.36) as follows: k g  ks kd /(ks + kd ), α  kd /(ks + kd ) and τ  cd /kd → ks  103 kN/m, kd  87.5 kN/m and cd  1028 N s/m for Fig. Q50(2) → kb  kd /2  43.7 kN/m and cb  cd /2  514 N s/m for Fig. Q50(1) (4) Since the vibration period is read as 0.04 s from the impulse response waveform of Fig. Q50(5), the damped natural frequency is estimated by f 1  1/0.04  25 Hz. We read the peak amplitudes at every half cycle, i.e., a  {1.4, − 0.75,0.34, −0.16, 0.08} and plot their absolute values as the pair of {ai ai+1 } as shown in Fig. Q50(7). The slope of the fitted straight line is 1.93. The damping ratio is thus estimated by ζ1  1/π ln(1.93)  0.21. We therefore confirm that the design target is achieved.

Amplitude a i

1.5 1.0

= fit

3x

1.9

1st Eigenvalue

0.5 0

0.2

0.4 0.6 0.8 Amplitude a i+1

1.0

Fig. Q50(7) Evaluation of damping ratio

(5) The Q-value measurement is performed according to R1_Fig. 2.24. Q51_Ans. (1) I d  0.05 kg m2 , k θ  6.43 kN m/rad, k d θ  5.47 kN m/rad, cd θ  64.2 N m s/rad. (2) f 2  54 Hz, ζ2  0.128 → Q 2  1/(2ζ2 )  3.9 (prediction). (3) Q  3.6 @ 55 Hz for second mode is measured from Fig. Q50(6). Q51_Hint: (1) For the tilting system of the second eigen mode, as shown in Fig. Q51(2), the 1.5-dof model is defined by the following system parameters: I d  I p /2  0.05 kg m2 , kθ  12E I /L  ks L 2 /4  103/16  6.43 kN m/rad, kdθ + cdθ s  (L/2)2 (kd + cd s) → kdθ  87.5/16  5.47 kN m/rad and cdθ  1028/16  64.2 N ms/rad. (2) As well as the parallel motion, the open-loop transfer function of the tilting motion becomes, with the series coupling of springs denoted by “//”: G o (s) 

kθ //(kdθ + cdθ s) 1 kθ (kdθ + cdθ s) ,  × 2 2 Id s Id s kθ + kdθ + cdθ s

486

12 Exercises of ISO Certification Examination for Vibration Experts

Gain [dB] , Phase [deg]

60

| Go |

40

Gr

m

20 0 fg -20

1

5

10

50

100

Frequency[Hz]

Fig. Q51(3) Open-loop response for tilting mode

which provides Fig. Q51(3) for prediction of the zero-crossing frequency and the phase margin. The zero-crossing frequency ωg is observed at about 53 Hz, kθ //kdθ → which is estimated by the second eigen frequency of ω2  Id ω2  54 Hz. f2  2π On the other hand, the phase margin is read as φm  15◦ , which is calculated exactly as follows: φm   − G o   G r  −1   1 kdθ + cdθ s 1    14.4◦ ,  +  kθ kdθ + cdθ s kθ + kdθ + cdθ s where s  jω2 . Therefore, the modal damping ratio ζ2  1/2 tan(14.4◦ )  0.128 → Q  1/(2 × 0.128)  3.9 (3) The prediction Q  3.9 of Q(2) agrees well with the Q-value measured at the second peak, Q  3.6. Q52_Ans. (1) ω p  258 Hz. (2) The cross-over frequency is f 1  24.3 Hz. The phase margin is φm1  23◦ → damping ratio ζ1  0.21 → agrees well with the design target. (3) ζ 1  0.2 @ f g  f 1  24 Hz. (4) ζ 2  0.0003 @ f 2  258 Hz Q52_Hint: (1) The equation of motion is given as follows: 

m1 0 0 m2 M



   x¨1 k −k x1 1 +  u x¨2 x2 −k k 0 X¨ K X  B u

12.3 Answers and Hints for the Additional Questions 31–100

x1  [1 0]X ≡ C X

487

(1)

The plant transfer function is obtained:  2  2 −1 s + ωz2 1 (s/ωz )2 + 1   (2) G p (s)  C Ms + K B ×  2 mt s2 m 1 s 2 s 2 + ω2p s/ω p + 1 Letting the equivalent mass m eq  m 1 m 2 /m t , it follows that  ωp 

 k  m eq



km t  ωz m1m2

mt 200 √  258 Hz, m1 0.6

(3)

τs + 1 where k g  48 × 103 kN/m, α  0.46 and τ  (2) G r (s)  k g ατ s + 1 0.0117 s. G o  G r G p → Fig. Q52(2) → cross-over frequency f g  f 1  24 Hz, phase margin φm  22◦ and damping ratio ζ1  1/2 tan φm  0.2. These values satisfy the design targets of f 1 ≥ 20 Hz and ζ1 ≥ 0.2. (3) Since the impulse waveform of x 1 of Fig. Q50(5) is very similar to ➀ sensor x 1 of Fig. Q52(3), the design target of ζ 1  0.2 is thus achieved for the first eigen mode. (4) For the vibration waveform ➁ sensor of y2 of Fig. Q52(3) for the second eigen mode, we read the envelope of amplitudes at every second, i.e., a  {3.6, 2.21,1.49, 0.92, 0.59, 0.387}, and plot the pair of {ai , ai+1 } as shown in Fig. Q52(4). The slope of the fitted straight line is 1.585. The damping ratio is thus estimated by ζ2  1/(2π × 258) ln(1.585)  0.0003. ai 4 3 2

fit

3x

.58

=1

1 0

0.5 1.0 1.5 2.0 2.5 3.0 a i+1

Fig.Q52(4) Evaluation of damping ratio

Q53_Ans. (1)

ω12  0 ω22  k/m eq

 1 1  m 2 /m t φ2  −m 1 /m t φ1 

where m eq  m 1 m 2 /m t and m t  m 1 + m 2

(4)

488

12 Exercises of ISO Certification Examination for Vibration Experts

(2)

G pp 

1 mt s2

G pt 

1  2 m eq s + ω2p

(5)

(3) For the first mode, the gain cross-over frequency is f 1  25 Hz and the phase margin φm  22◦ (equivalent damping ratio ζ 1  0.20). For the second mode, the gain cross-over frequency f 2  258 Hz, and effective phase margin φm  0.8◦ (equivalent damping ratio ζ 2  0.007). (4) f 1  25 Hz and ζ 1  0.2. (5) f 2  258 Hz and ζ 2  0.0074. Q53_Hint: (1) The equation of motion is given as follows: 

m1 0 0 m2



M

   x¨1 k −k x1 u1 +  x¨2 x2 u2 −k k ¨ X + K X  u

(6)

The eigenvalue problem, ωn2 Mφ  K φ

(7)

gives the eigensolutions of Eq. (4).   Since the modal matrix Tm  φ1 φ2 is orthogonal with respect to the mass and stiffness matrices, the congruence transformation gives the diagonal matrix as follows:  Tmt M Tm 

mt 0 0 m eq



 Tmt K Tm 

0 0 0 m eq ω2p

where ω p  ω2

(8)

(2) The actual displacement vector X is transformed to the modal displacement vector consisting of the parallel displacement x p and tilting displacement x t , specified by the following relationship tuned by h 1  m 2 /m t and h 2  m 1 /m t :  X≡

x1 x2



 

1 h1 1 −h 2



xp xt



 ≡ Tm

xp xt

(9)

Then, the modal model is obtained as follows: 

mt 0 0 m eq



    0 0 x¨ p xp up t u1 +  T ≡ m 0 m eq ω2p x¨t xt u2 ut

The plant transfer function Gp (2 × 2) is thus obtained:

(10)

12.3 Answers and Hints for the Additional Questions 31–100



xp xt



 ≡ Gp

up ut





G pp 0  0 G pt

489



up , ut

(11)

xp , xt

(12)

Since the controller system is described as follows: 

up ut



 ≡ Gr

xp xt



 

Gr p 0 0 Gr t



the open-loop transfer function Go (2 × 2)  Gp Gr is thus expressed by the diagonal matrix of Eq. (3) of Q53. (3) The result of Fig. Q53(2)(a) for the parallel mode is the same as Fig. Q50(4) indicating f g  25 Hz and ζ  0.2. Regarding the tilting mode of Fig. Q53(2)(b), we read the gain cross-over frequency f 2  258 Hz, i.e., f 2 ≈ ω p /(2π ), and the phase margin φm∗  21◦ . As stated for the phase margin of the higher order mode in R1_Fig. 8.19, the effective phase margin must be re-calculated using the modal representation in reference to the triangle placed at the right side in Fig. Q53(2)(b), as follows:    φm2   m eq ω2p + G r t jω p  0.8◦ , and the damping ratio ζ2  1/2 tan(φm2 )  0.007. According to a modal single-DOF system as shown in R1_Fig. 8.18a, the restoring force is the sum of the modal stiffness m eq ω2p and the controller force Grt (s), and the angle of that force thus gives the effective phase margin. (4) Since the damped waveform ➀ (x p due to a r p impulse) is very similar to Fig. Q50(5), ζ 1  0.2 is then expected. (5) We read the envelope of the damped waveform ➃ (x t due to a r t impulse) at every 0.05 s interval and obtain the amplitude data a  {6.5, 3.5, 2.0, 1.1, 0.68, 0.29 }. The average slope of ai /ai+1 is identified by 1.826, as shown in Fig. Q53(4), and the damping ratio ζ 2 is estimated as follows: ζ2 

1 1 ln 1.826  0.0074 ( f 2  258 Hz), which agrees with Q(3). 2π f 2 /20

Amplitude a i

4 3 2

x

26

= fit

1.8

1 0

0.5

1.0

1.5

2.0

Amplitude

2.5

3.0

ai+1

Fig. Q53(4) Evaluation of damping ratio

Q54_Ans.

490

12 Exercises of ISO Certification Examination for Vibration Experts

(1) f 1  65 Hz and Q  13.1, (2) m1  0.411 kg and m2  0.063 kg, (3) a1 : a2  1 : 3.1, (4) m ∗1  1, (5) kb  105 N/m, (6) cb  30 N s/m. Q54_Hint: (2) m1  ms /3 × 0.39 + md  0.41, m2  ms /3 × 0.61  0.063 f z2 ks a2    (3) −m 2 ωn2 a2 + ks (a2 − a1 )  0 → a1 −m 2 ωn2 + ks f z2 − f n2 792  3.1 → a1  1, a2  3.1 2 79 − 652 (4) m ∗1  m 1 a12 + m 2 a22  0.41 × 12 + 0.063 × 3.12  1 (5) k1∗  m ∗1 ω12  ks × (a1 − a2 )2 + kb a12 → 1 × (2π × 65)2  m 2 (2π × 79)2 × 2.12 + kb × 12 → kb  105 (6) cb  2ζ1 ω1 m ∗1 /a12  1/Q × (2π × 65) × 1.0/12  30 Q54a_Ans. (1) f 1  92 Hz and Q  11.2, (2) m1  0.41 kg and m2  0.038 kg (3) a1 : a2  1 : 2.3, (4) m ∗1  0.601, (5) k b  1.6 × 105 N/m, (6) cb  31 N s/m. Q54a_Hint: (2) m1  ms × 33/140 × 0.48 + md  0.41 and m2  ms × 33/140 × 0.52  0.038 f z2 ks a2    (3) −m 2 ωn2 a2 + ks (a2 − a1 )  0 → a1 −m 2 ωn2 + ks f z2 − f n2 1252  2.3 → a1  1, a2  2.3 1232 − 922 ∗ 2 (4) m 1  m 1 a1 + m 2 a22  0.41 × 12 + 0.038 × 2.32  0.6 (5) k1∗  m ∗1 ω12  ks × (a1 − a2 )2 + (kb + kδ )a12 → 1 × (2π × 92)2  m 2 (2π × 123)2 × 1.32 + (kb + kδ ) × 12 → kb  1.6 × 105 (6) cb  2ζ1 ω1 m ∗1 /a12  1/Q × (2π × 92) × 0.6/12  31 Q55_Ans. W c  1.0 g −150°. Q55_Hint: As shown in the Nyquist plot of Fig. Q55(2), we mark A at p  0.95 on the point on the curve of the initial run and B at the same speed on the point of −→ the curve of the trial run. The segment from A to B, i.e., AB, is the influence vector. Therefore, the vector generated by the correction weight should point from A to −→ C; hence, C  the origin, to cancel the initial vector CA. Therefore, the correction weight is calculated as follows: − → |AC| AC  BAC Wc  Wt −→  Wt |AB| AB    1.5 − 130◦ × 0.67 − 20◦  1.0 − 150◦

The balanced state is confirmed by the Bode plot as noted by “After,” as compared with “Before.”

12.3 Answers and Hints for the Additional Questions 31–100

491

6 Im 4

Trial run

2 6 4 A p = 0.95

-4

-6

95

p

. =0

Before (Initial run)

Re

C B

After

-4 Before (Initial run) -6

Fig. Q55(3)

Fig. Q55(2)

Q55a_Ans. Wc  1.2 g  + 90◦ . Q55a_Hint: As shown in Nyquist plot of Fig. Q55a (2), we mark D at p  0.95 on the second initial run. Then, we copy the segment AB and paste it from the point D to E in the same figure as the influence vector. For re-balancing, the correction vector should point from D to C (origin) instead of from D to E. Therefore, the correction weight is calculated as follows: −→ |DC| DC ◦ ◦ ◦ Wc  Wt − →  Wt |DE|  EDC  (1.5 − 130 ) × (0.8 − 140 )  1.2 + 90 DE 6 Im

2 1.0

E

P=

Re

C

-6 -4 -140˚

-2

5 0.9 D 0.9

0

2

4

6

Amplitude

4

1.1

Before (Initial run) After

-4 -6

Fig. Q55a(2)

Fig. Q55a(3)

Q56_Ans. (1) (A)  6.5 −45° (B)  7.5 −15° (2) W c  0.9 g + 180°. Q56_Hint: (1) We read the complex amplitude as 6.5 −45° for (A) and 7.5  −20° for (B). (2) Plot both complex amplitudes on the polar diagram as shown in Fig. Q56(2), −→ including the effective vector AB as the influence vector. The correction vector −→ should point from A to C (origin) to cancel the initial vector CA. Therefore, the correction weight is calculated as follows: Wc  Wt ×  0.9 180◦

|AC|  BAC  (0.5 90◦ ) × (1.8 90◦ ) |AB|

492

12 Exercises of ISO Certification Examination for Vibration Experts

The balancing state is confirmed by the Bode plot noted by “After,” as compared with “Before” in Fig. 56(3). E 12

10 −120°

10

5 -10

-5

5 C

+90°

-5

10 B

Amplitude

D

8

(A)

6 4 2

A

-10

Fig. Q56(2) Related vectors

Fig. Q56(3)

Q56a_Ans. (1) 8  100°, (2) W c  1 g −35°. Q56a_Hint: (1) We read the complex amplitude as 8 100° for (D). (2) Plot the complex amplitude of (D) on polar diagram as shown in Fig. Q56(2) −→ and translate the influence vector AB of Q56 from D to E. The correction vector should point from D to C (origin). Therefore, the correction weight is calculated as follows: −→ |DC| DC Wc  Wt × − →  Wt × |DE|  EDC DE  (0.5 90◦ ) × (2 − 125◦ )  1.0 − 35◦ The correction weight provides the “After” curve of the well-balanced state as shown in Fig. 56a(2). 12 Amplitude

10 8 6 4 2

Before (D) (Initial run) After

Fig. Q56a(2)

Q57_Ans. W c  6.9 g −140°. Q57_Hint: According to R1_Fig. 5.48, we draw four circles as shown in Fig. − → Q57(2)(b) and the intersection, noted by P, gives the effective vector OP. For bal− → −→ ancing, the effective vector OP should be equal to OA to cancel the initial vibration A0 . Therefore, the correction weight is calculated as follows:

12.3 Answers and Hints for the Additional Questions 31–100

493

−→ OA ◦ ◦ ◦ Wc  Wt − →  (3 0 ) × (2.3 − 140 )  6.9 − 140 OP − → Note that the phase of the correction weight agrees with the radial direction OP. Im #2

#1

5 13

63m

°

A0  46  m

135

°

1

49m 2

90 °



3

37m

(a) Rotor section

m 

Wc

m  135°

A A

 140°



B m 135° 20μm

Re

P C  m #3 (b) Correction weight

Fig. Q57(2) 4-run method (in case of changing test run sequence)

Q58_Ans. C. Q58_Hint: The vibration vector (output) is lagging compared with the heavy spot G (unbalance input) at the resonance speed. Therefore, when the rotor whirls in a clockwise direction and just passes at the position of the X-axis direction, the corresponding heavy spot G is oriented toward the negative Y direction, as shown in (C) of Fig. Q58. Q59_Ans. H. Q59_Hint: Figure Q59 shows that the high spot of the vibration waveform is lagging, compared with the pulse signal, by 45°. Then, when the position H rotates to the position A by 45°, the sensor indicates the peak value (high spot) so that the rotor leaves the sensor by the largest displacement. This displacement of D is the unbalance direction (heavy spot), because the rotational speed is very low compared to the critical speed. Thus, it is recommended to set the correction weight at the opposite direction of the heavy spot, i.e., H. Q60_Ans. (1) 2, (2) 2. Q60_Hint: The phase curve starts from 45° between the pulse and the high spot, as shown in curve 2. Q61_Ans. {W c1 W c2 W c3}  {339 Q61 (2).

− 600 261}g × mm as shown in Fig.

494

12 Exercises of ISO Certification Examination for Vibration Experts 261 g⋅mm

339

AMB

AMB 600 (b) Correction weights

Fig. Q61(2) Balancing by FF 3 planes:Wc1,Wc2 and Wc3

Q61_Hint: Referring to deflections of five portions (left AMB, W c1, W c2, W c3, right AMB) of the following three modes as shown in Fig. Q61 (3), the calculation of modal balancing is undertaken as below: Rigid parallel mode balancing for the mode shape  {1, 1, 1, 1, 1} W c1 + W c2 + W c3  0

(1)

Rigid tilting mode balancing for the mode shape  {25, 10, 0, −13, −24} 10 × W c1 + 0 × W c2 − 13 × W c3  0

(2)

The first bending mode balancing for the shape  {10, −5, −12, −7, 6} −5 × W c1 − 12 × W c2 − 7 × W c3  10 × 230 + 6 × 230 From these three equations, we obtain the solution {Wc1 Wc2 Wc3 } {339 − 600 261}g × mm Correction Plane 339 g∙mm 17

230 g∙mm

Wc2

Wc1

(3) 

261 g∙mm Wc3

21 230 g∙mm

Wc4

Wc5

10

-7

-5 30

1st bending mode

-12 10

13 600 g∙mm

AMB

AMB

6

33

Fig. Q61(3) Rotor balancing and bending mode shape

Q62_Ans. {Wc4 Wc2 Wc5 }  {62 − 119 57}g × mm as shown in Fig. (2). Q62_Hint: Referring to deflections at five axial locations (W c4 , left AMB, W c2 , right AMB, W c5 ), the calculation of modal balancing is performed for the following three modes as shown in Fig. Q61(3):

12.3 Answers and Hints for the Additional Questions 31–100

495

Rigid parallel mode balancing for the mode shape  {1, 1, 1, 1, 1} W c4 + W c2 + W c5  0

(1)

Rigid tilting mode balancing for the mode shape  {30, 25, 0, −24, −33} 30 × W c4 + 0 × W c2 − 33 × W c5  0

(2)

The first bending mode balancing for the shape  {17, 10, −12, −6, 21} 17 × W c4 − 12 × W c2 + 21 × W c5  10 × 230 + 6 × 230

(3)

From these three equations, we obtain the solution {W c4 W c2 W c5}  {62 − 119 57}g × mm 230 g⋅mm Wc4

AMB

62

230 g⋅mm AMB

Wc2

Wc5 57

119 g·mm

Fig. Q62(2) Balancing weights at Wc4, Wc2 and Wc5

Q63_Ans. U 2  0.2U and U 3  −0.2 U. Q63_Hint: translating force  0 → U2 + U3  0 Moment around the piston center  0 → 20 × U + 50 × U2 + 150 × U3  0 The solution of these equations gives the answer, as shown in Fig. Q63(2). M = Dynamic Unbalance of Twin Rotary Piston 

#2 Brg. #1 Brg. L 20 20 20

Motor

M LU = 20mm U U M Brg.

U Brg.

S1

0.2U 0.2U

Fig. Q63(2) 2-plane balancing by U2 and U3

Q64_Ans. U1  0.22U, U2  −0.24U, U3  0.02U .

496

12 Exercises of ISO Certification Examination for Vibration Experts

Q64_Hint: Translating force  0 → U1 + U2 + U3  0 Moment around the piston center  0 → 20U − 50U1 + 50U2 + 150U3  0 Bending mode unbalance  0 → 0.16 × U1 + 0.49 × U2 + 3.7U3 + 20U × 0 ( slope)  0, Because the φ 1 mode gives the deflection  −0.19 and the slope  0 at the piston, these equations give the correction weights as shown in Fig. Q64(3). M = Dynamic Unbalance of Twin Rotary Piston Ω

#2 Brg. #1 Brg. L 20 20 20 M = LU = 20mm×U U M 0.22U

Brg.

Motor

S1

0.24U

U Brg.

0.02U

Fig. Q64(3) 3=N+2 plane balancing

Q64_Another solution:  t  t  t Using δ1  1.33 −0.33 −2 , δ2  −0.33 1.33 3 , φ1  0.16 0.49 3 .7 ,   U  − U1 U2 U3 and the slope of the {δ1 δ2 φ1 } mode  {α1  −1/60 α2  +1/60 α3  0} at the piston and M  20U , the threeplane balancing weight of U should be determined by the following equation satisfying each modal balance: δ1t U − Mα1  0, δ2t U − Mα2  0, φ1t U − Mα3  0 Q65_Ans. (1) A  80 μm −30°and B  80 μm +30°, (2) -90°, (3) 29 μm. Q65_Hint: As shown in figures (2) and (3), we can read the complex amplitude A  80 μm −30° and B  80 μm +30°, respectively, and plot these values on the Nyquist chart as shown in figure (4). The mid-point M between A and B indicates the point when assuming that there is no unbalance of the fan itself. Excepting the fan, the unbalance of many other parts is hence assumed and it thus generates the unbalance vibration vector OM. When we rotate the fan for refitting, the resultant vibration vector rotates in the same manner around the point M, such that it rotates on the circle noted by the dashed line. The point C on the dashed circle is the closest to the origin O. Thus, the point F is recommended for finally fixing of the fan, rotated from the upward point F by 90° in the contra-rotational direction. It would be the best option for the fan assembly. This recommendation gives the expectation of the vibration level of the point C, i.e., approximately |OC|  29 μm.

12.3 Answers and Hints for the Additional Questions 31–100 90°

180°

100 μm 0-p

30° ∠− A m μ

80 O C

497

M B

Phase lag 0° 40 μm 0-p

29 μm 270°

Fig. Q65(4) Section for the best fitting

Q66_Ans. 5 Q66_Hint: For bearing A, the mass  3600 × 10/18  2000 kg and U  4.8 × 2000  9.6 kg × mm For bearing B, the mass  3600 × 8/18  1600 kg and U  4.8 × 1600  7.68 kg × mm Q67_Ans. (1) F e  106 N and F  53 N, (2) 15 Hz, (3) k  12.5 × 106 N/m, (4) k 1  9.3 × 106 N/m and k 2  3.1 × 106 N/m. Q67_Hint: (1) F e  300 kg × (2 π 1800/60)2 × 10 × 10−6  106 N and F  0.5 × F e  53 N (see TR  0.5 @ f / f n  2.0 in the transmissibility curve with ζ  0.3) (2) TR  0.5 is obtained by f / f n  2 from the curve of ζ  0.3. Thus, f n  f /2  30/2  15 Hz (3) k  mωn2  1400 × (2π × 15)2  12.5 × 106 N/m (4) Total stiffness and no tilting vibration require constraint equations of k1 +k2  k and a1 k1  a2 k2 , respectively. The solution is thus obtained by k1  3k/4  12.5 × 106 × 3/4  9.3 × 106 N/m and k2  k/4  12.5 × 106 /4  3.1 × 106 N/m Q68_Ans. (1) f  1/(Window time 1 s)  1 Hz, (2) L  f max / f  400/1  400, (3) N  2.56L  1024, (4) t  1 s/1024  977 μs, (5) f s  1/ t  1024 Hz, (6) a  1 V, (7) Rectangular, (8) FFT peak frequencies 20 Hz and 21 Hz, (9) 0.64a V  0.64  −3.9 dBV, (10) 0 dB  1 V. Q68_Hint: Refer to Fig. C.6 for Q(9). Since the input signal frequency 20.5 Hz is just the mid-frequency between two adjacent synchronous frequencies of 20 Hz and 21 Hz, we know the magnitude of the FFT spectrum becomes 0.64 [V] and −3.9 dBV from the left and right figures, respectively, when using a rectangular window. Q68a_Ans. (1) (b) (2) Table Q68a

498

12 Exercises of ISO Certification Examination for Vibration Experts

Table Q68a Frequency and window used (a)

(b)

(c)

(d)

(e)

Frequency [ Hz ]

20.0

20.5

20.0

20.2

20.8

Window

Rectungular

Hanning

Hanning

Hanning

Hanning

Figure

Q68a_Hint: Refer to Example 10.11. For figure (d), we see the magnitude 0.97 @ 20 Hz and 0.65 @ 21 Hz and the ratio of both is 0.65/0.97  0.67  −3.47 dB. The ratio is placed on the Y -axis in Appendix Fig. C.7, and the value, 0.2, is obtained on the X-axis. Then the frequency becomes 20 + 0.2  20.2 Hz. For figure (e), since the same ratio is calculated by 0.68/0.97  0.67  −3.47 dB for 0.68 @ 20 Hz and 0.97 @ 21 Hz, this gives the same offset ratio, 0.2, and then the frequency becomes 21 − 0.2  20.8 Hz. Q69_Ans. ➁ Q69_Hint: The instability happens when the real part of the complex eigenvalues becomes positive. Q70_Ans. 6400 Q70_Hint: The fundamental frequency (frequency resolution), f , is given by (1800 − 1740)/60  0.13 Hz. The corresponding number of Eq. (10.34): f  2 × 3.8 lines, L n , is thus determined as follows: f max 500   3846 < L n  6400

f 0.13 for an available option among the possible numbers of {400, 800, 1600, 3200, 6400, …}. Q71_Ans. L n  1600, f max  200 Hz. Q71_Hint: The fundamental frequency (frequency resolution), f , is given by 60 − 3570/60  0.16 Hz. Since the frequency span is recomEq. (10.34): f  2 × 1.5 mended to be over three times the target frequency, 3 × 60  180 < f max  200. The corresponding number of lines, L n , is 200 f max   1250 < L n  1600 as an available option.

f 0.16 Q72_Ans. 0.067  −23 dB Q72_Hint: The Bode plot of this anti-aliasing filter is described by ➀ in Fig. Q72, as setting the roll-off frequency  1 in non-dimensional form being normalized by the frequency span f max . It also includes the sampling frequency f s  2.56 × f max and the Nyquist frequency  f s /2  1.28 × f max . Referring to Fig. 10.17, the anti-aliasing spectral gain is given by the function G (2.56 − ω) of the Bode plot ➁. Hence, the

12.3 Answers and Hints for the Additional Questions 31–100

499

question asks for the magnitude of the point on the curve ➁, that is, G (2.56 − 1)  G (1.56)  0.067  −23 dB. Therefore, it is said that if we input a harmonic wave having the amplitude 1 V at the frequency of 1.56 × f max , then we can see the aliasing spectrum of 0.067 V at f max . f max

0

-10

1

-20 2

-30

0

Gain [dB]

Gain [dB]

-10

fs / 2=1.28

A

-40

-20

1 2

A

-30 -40

-50 0.1 0.15 0.2 0.3

0.5 0.7 1.0 1.5 2.0 fs =2.56 Frequency X f max

-50 0.0

0.5

1.56 1.0 1.5 2.0 f max fs / 2=1.28 fs =2.56 Frequency X f max

Fig. Q72 Butterworth (n =6)

Q73_Ans. rms  2.56 V, OA  3.75 V. Q73_Hint: Because each component of DC  2 V, 1 V @ 10 Hz, 2 V @ 20 Hz and 1 V @ 30 Hz are determined, 

a 2 + a22 + a32 + · · · rms  √ f 02 + 1 2 WF  12 + 22 + 12 + · · · √ 1  7  2.65 V and  √ 22 + 2 1 √ OA  2 × rm s  3.74 V are given. 1

Q74_Ans. 0.64 V @ 200/205 Hz, OA  1 V Q74_Hint: Since the frequency resolution is f  1/0.2  5 Hz, the input waveform with the single frequency 200 Hz is synchronized with the window time 0.2 s. On the other hand, the input waveform with the single frequency 202.5 Hz is not synchronized, and it is the mid-frequency between 200 and 205 Hz, where both are synchronized, being within the window time 0.2 s. Referring to Fig. C.6, the spectral peak of the mid-frequency is 64% of the input amplitude. The power for the rms and OA is invariant nevertheless with the change of the window function, the window time, etc. Q75_Ans. Peak spectrum: 0.85 V @ 200/205 Hz and OA  1 V. Q75_Hint: Referring to Fig. C.6, the spectral peak of the mid-frequency is 85% of the input amplitude in the case of the Hanning window. The power for the rms and OA is invariant, as stated in Q74. See the test result in Fig. Q75, and confirm the answer.

500

12 Exercises of ISO Certification Examination for Vibration Experts

f = 202.5 Hz Hanning T = 0.2 s  f = 5 Hz

OA

200Hz

0.8488V

2kHz

0.85 0.85

0.17 0

0.17

0.02

0.2 s

0.02

190 195 200 205 210 215 Frequency Hz (1) FFT analysis (Hanning window)

(2) Zoomed around 200Hz

Fig. Q75

Q76_Ans. (1) 11 mm/s @ 99.6 Hz (2) OA  11 mm/s. Q76_Hint: Since the peak ratio  (the second highest peak/the highest peak)  8.08/10.3  0.784, we can see the intersection ● in Fig. C.7, which is the frequency difference ratio. It is 0.32 from the frequency of the highest peak. Then, the input frequency is 100 − 0.32 × 500/400  99.6 Hz. √  2 8.082 + 10.32 + 3.022 + 0.962  11 mm/s  amplitude OA  √ 2 1.5 Q76_Another solution to find the peak According to Fig. C.6, the above frequency difference ratio  0.32 on the x-axis, which gives the gain  0.748 on the y-axis for a Hanning window. We can guess the inputting harmonic wave with the actual amplitude  8.08/0.748  11 mm/s. Q77_Ans. s  0.0044 and f r  49.78 rps. Q77_Hint: In the zoomed FFT, we see twice the rotational frequency and the rotational speed, i.e., 2f 0  100 Hz and 2f r  99.56 Hz. Since 2 f 0 −2 fr  2s f 0  100−99.56  0.44 Hz referring to Eq. (7.32), these answers are s  0.0044 and fr  99.56/2  49.78 rps. Q78_Ans. (1) f ∗  0.146 Hz, (2) L n  1600, f  0.125 Hz, (3) T  8 s, (4) N  4096 → f s  512 Hz, (5) T16  98 s, (6) L n  3600, f  0.0937 Hz, T16  131 s. Q78_Hint: (1) f ∗  | f 1 − f 2 |/(2W F)  |100 − 99.56|/(2 × 1.5)  0.146 Hz f max / f ∗  200/0.146  1363 ≤ L n → L n  400 × 4  1600 (2)

f  f mb /L n  200/1600  0.125 Hz

12.3 Answers and Hints for the Additional Questions 31–100

(3) (4) (5) (6)

501

T  1/ f  1/0.125  8 s N  L n × 2.56  4096  212 → f s  4096/T  512 Hz T16  T + 0.75T × 15  98 s

f ∗  0.146 Hz, 300/0.146  2045 ≤ L n  3200, f  300/3200  0.0937 Hz T  10.67 s, N  8192, f s  768 Hz, T16  13l s

Q78_Note: The frequency span, f max , is recommended to be set according to the following table. Table Q78, referred to in the Book [B30].

Table Q78 Default frequency spans for data analysis by FFT Vibration Shaft Gearbox Rolling element bearings

Frequency span 10 × RPM 3 × GM 10 × BPFI

Memo Revolutions Per Minute Gear Mesh frequency Ball Pass Freq. Inner race

Pumps

3 × VP

Vane Pass frequency

Motor / Generator

3 × 2LF

Line Frequency

Fans Sleeve bearings

3 × BP 10 × RPM

Blade Pass frequency Revolutions Per Minute

In this case, this recommendation gives f max  3 × 2 × LF  3 × 2 × 50  300 Hz for electric motors. The required frequency resolution, f ∗ , is explained in Eq. (10.34). For your reference, the relationship f  f max /L n between the required frequency resolution,

f , and the frequency span, f max , is shown in Fig. Q78, together with the available number of lines, L n , which should be 1/2.56 times the sampling number N  2n . The area under each straight line of the border can be set for our option. In Q78, the intersection ● between f ∗  0.146 and the recommendation of f max  300 Hz belongs to the area of L n  3200, as stated in Q(6). However, since the intersection ◯ between f ∗  0.146 and the recommendation of f max  200 Hz is close and under the straight line L n  1600, it is also available as stated in Q(1)–Q(5).

N= 16, 384

12 Exercises of ISO Certification Examination for Vibration Experts

100

0.00

92

,1

8 N=

96 4,0

20

0

N=

3,

6,4 00

200

Ln =

f max [Hz]

300

n=

400

L

502

00 1,6 L n= 800 L n= L n= 400

0.05

0.10

48

N=2,0

N=1,024 0.146

0.15

0.20

Required frequency resolution ∆ f * [Hz]

Fig. Q78 Relationship of  f * 

f1  f 2 2  WF



f max Ln

Q79_Ans. A—(1), B—(2), C—(3), D—(4). Q79_Hint: Refer to Table 7.3 and the guidelines in Table Q79, included in the book [B30].

Table Q79 Malfunctions due to magnetic force of electric motor Fault

Frequency

Spectrum;Time Waveform/Orbit Shape

Correction/Comment

120Hz plus sidebands,beating 2x with 120Hz

center armature relieving distortion on frame; eliminate and/or any other condition that causes rotot to be off center with stator

(A)

air-gap variation

(C)

broken rotor bars 1x loose rotor

1x and sidebands equal to (number of poles × slip frequency)

replace loose or broken rotor bars

(B)

eccentric rotor

1x

1x, 2x/120Hz beats possible

may cause air-gap unbalance variation

120Hz

2x/120Hz beats

stiffen stator structure

impacting in axial direction

remove source of axial constraint - bearing thrust, coupling

stator flexibility

120 Hz

off magnetic 1x,2x,3x center (D)

stator short circuits

120Hz 120Hz and and harmonics harmonics

replace stator

Q80_Ans. {10 Hz, 1 V} and {11 Hz, 0.2 V} as shown in Fig. Q80(2).

12.3 Answers and Hints for the Additional Questions 31–100

503

2.0 [V]

1.5 1.0 0.5 0

5

10 Frequency

15 [Hz]

20

Fig. Q80(2) FFT

Q81_Ans. {10 Hz, 1 V} and {1 Hz, 0.2 V} as shown in Fig. Q81(2). 2.0

[V]

1.5 1.0 0.5 0

5 10 15 Frequency [Hz]

20

Fig. Q81(2) FFT

Q82_Ans. {10 Hz, 2 V}, {9 Hz, 0.5 V} and {11 Hz, 0.5 V} as shown in Fig. Q82(2). 3.0

[V]

2.0 1.0 0

0

5

10 Frequency

15 [Hz]

20

Fig. Q82(2) FFT

Q83_Ans. (1) (a), (2) (b), (3) (c), (4) (d), (5) (e). Q84_Ans. (1)  f c  f max  400 Hz, (2)b1  0.5305 and a0  a1  0.2348, (3) data2  1 1.23477 1.82885 · · · , (4) N d  4. Q84_Hint: (2) See Table Q84, 2π f c 2−α T AD   0.5305 and a0  a1   0.6136, b1  where α  τ f AD 2+α α  0.2348 2+α (3) y1  1, y2  b1 y1 +a0 x2 +a1 x1  1.23477, y3  b1 y2 +a0 x3 +a1 x2  1.82885 (4) f s  2.56 × f max  1024, Nd  f AD / f s  4096/1024  4

504

12 Exercises of ISO Certification Examination for Vibration Experts 2n Table Q84 Butterworth LPF : Gn( j ) = 1/ 1 + 

1st BWF

3rd BWF

G1( j ) = 1/ 1+  2

Transfer Gain Trans. Func. G(s)

G1( s ) = 1 / ( 1 + s )

Trans. Func. G(z)

G 1( z ) =

G3( j) = 1/ 1+  6 G3( s ) = 1 / ( 1 + 2 s + 2 s2 + s3 )

a0 + a1 z -1 1 - b1 z -1

G3( z ) =

a0 + a1 z -1 + a2 z -2 + a3 z -3 1 - b1 z -1 - b2 z -2 - b3 z -3

Note: The coefficients ai and bi are given as follows in the case of bi-linear transformation for the digitization. Letting fc = the cut-off frequency of LPF and fAD = the sampling frequency of A/D converter give  2  fc / fAD For G1(z), [ a 0 a 1 ] = [   ] / ( 2 +  ) and b1 =

2 - 2 +

For G3(z), [ a 0 a 1 a 2 a 3 ] = [ 1 3 3 1 ]  3/ ( 8 + 8 + 4  2+  3 ) and b1 b2 = b3

24

8

-4

-3

-12 10 8 -8

-3 4

0 -1

1  2 3

/ ( 8 + 8  + 4 2+  3 )

Q85_Ans. (1) x  700 Hz, (2) 0.45 for n  1 and 0.14 for n  3. Q85_Hint: (1) Since the x Hz may generate the aliasing error seen at 324 Hz, we could guess the cause: x  512 + (512 − 324)  1024 − 324  700 Hz. (2) Referring to Fig. Q85(1) for an analog anti-aliasing filter, we see that the gains for f /f max  700/400  1.75 are 0.5 and 0.18 for n  1 and n  3, respectively. These gains are observed as spectral values at 324 Hz as indicated by ◯ in Fig. Q85(2) and (3). In the case of the digital filter, these gains are 0.45 and 0.14 for n  1 and n  3, respectively, as indicated by ● in the figures. These values of the gain appear as the spectral height at 324 Hz, due to reflection of the unfiltered component at 700 Hz. They are called aliasing errors. 324Hz

1.0

512Hz Memo

Analog

Analog

Signal

0.8

0.6

Digital W(t)

Digital

FFT 0

200

400 600 Frequency [Hz]

800

1000

FFT

fc = fmax A/D

LPF

fAD = 4 096 Hz

0.2

N 1 024

Analog fAD = fs fc = fmax = 1 024 Hz

fAD = Nd * fs

0.4

0

A/D

LPF

fs = 2.56 fmax

Alias Analog

Gain

W(t)

fAD for AD

Digital Nd = 4

N 1 024

Digital Decimate Nd = 4

Fig. Q85(3) Anti-aliasing filter for FFT with fmax= 400Hz and Line = 400

12.3 Answers and Hints for the Additional Questions 31–100

505

Q86_Ans. L for (1), (3), (5), (7), and (9). H for (2), (4), (6), (8), and (10). Q86_Hint: See ISO13373-1 for more information. Q87_Ans. 3 Q87_Hint: See ISO13373-1 for more information. Q88_Ans. (A) 25 Hz, (B) forward, (C) oil whip, (D) stop. Q89_Ans. (2) {1, −3}X, (3){1, 4}X, (4) {1, −4}X, (5) {1, 2}X, (6) {1, −2}X, (7) {0.5, 1}X, (8) {−0.5, 1}X. Q89_Hint: (1) Another solution: When we see the orbit on the rotational coordinate, the faster small orbital component is observed as the forward 2X. Since it is superposed on the rotational speed (+1X), it is observed by whirl motion of mixture of 2X + 1X  3X and 1X in the stationary coordinate. Then, the answer is 1X and +3X whirl motions. (2) In the similar way, because of −4X in the rotational coordinate, −4X + X  −3X and 1X whirl motions are observed. (3) 1X and +3X + 1X  +4X are in the stationary coordinate. (4) 1X and −5X + 1X  −4X are in the stationary coordinate. (5) 1X and +1X + 1X  +2X are in the stationary coordinate. (6) 1X and −3X + 1X  −2X are in the stationary coordinate. (7) 1X of the case of (5) is recommended to be replaced by 2X of the case of (7). Then, the full FFT would be {1X, 2X} for (5) → {1X, 2X}/2  {0.5X, 1X} for (7). (8) In the same way, 1X of the case (6) is requested to be replaced by 2X and the full FFT becomes {1X, −2X} for (6) → {1X, −X}/2  {0.5X, −1X} for (7). However, since 1X forward whirl orbit is standard, the full FFT should be rewritten by {−0.5X, 1X}. Q90_Ans. 47.1 Hz and the harmonics. Q90_Hint: The pulley rotation frequency is 100 Hz. Referring to Eq. (7.36), the frequencies are: f n

3.14 × 0.3 πD N n × 100  n × 47.1 (n  1, 2, 3 . . .) (Hz) L 2.0

Q91_Ans. (1) f  900 Hz and the harmonics. (2) f  1200 Hz and the harmonics. (3) Zero nodal diameter mode. Q91_Hint: (1) BPF is f  N × Z r  50 Hz × 18  900 Hz, and the harmonics are the excitation frequencies.

506

12 Exercises of ISO Certification Examination for Vibration Experts

(2) Refer to Eq. (7.35), f  Z g × N  24 × 50  1200 Hz, and the harmonics are the excitation frequencies. (3) The bladed disk has many natural frequencies with similar modes for each blade, but different modes for the disk. In these instances, only a few modes are excited by the interaction between rotating blades and stationary vanes. Referring to the resonance condition, n Z g ± κ  h Z r , as stated in Eq. (7.34), the equation holds when n  3, κ  0, h  4, and 3 × 24 ± 0  4 × 18. Hence, κ  0 is lowest mode.

Natural frequency ωκ

ω9

Ω 48 ×

96×Ω 72× Ω

According to the same approach as R1_Fig. 10.20, the resonance condition is drawn in Fig. Q91(2) from a general viewpoint. The intersections in the figure indicate the resonances: (A) κ  6 for Z g  24 in backward propagation; (B) κ  6 for 2 × Z g  48 in forward propagation; and (C) k  0 for 3 × Z g  72 in a standing wave.

ω6

ω0

= Ω Ω J× 24×

9 27 8 10 7 11 6 12 24 48 96 5 13 4 14 3 15 2 16 1 17 19 72 0 18 K

Rotaional speed Ω

Fig .Q91(2) Campbell diagram

Q92_Ans. ➂ Q92_Hint: The onset frequency of the oil whip instability is approximately half the rotational speed. Q93_Ans. (1) (E), (2) (B), (3) (C), (4) (A), (5) (D). Q93_Hint: (A): When the motor power was switched off, the abnormal vibration amplitude immediately disappeared and reverted to 1X vibration. (B): Since the 1X vibration amplitude changed, probably due to blade loss and continued through the trip-shutdown process. (C): Non-synchronous vibration occurred and it disappeared slowly when the rotational speed began to decrease, probably due to the inertia effect of the oil whip phenomenon. (D): The amplitude of 1X vibration fluctuated slowly, because it may be due to shaft bow caused somehow by a thermal input. (E): Afterward, overall vibration is increased as if the offset has been induced. Q94_Ans. Outer race fault

12.3 Answers and Hints for the Additional Questions 31–100

Q94_Hint: β   77.3 Hz

507

f0 77.3  0.3865, f r  20 Hz, FTF f c  7.7 Hz, BPFO  N fr 10 × 20

Q95_Ans. Ball fault fc 7.7  0.385, FTF f c  7.7 Hz, BPFO  77 Hz, BPFI   Q95_Hint: β  fr 20 123 Hz, BSF f b  41.8, 2f b  83.6 Hz, 2 BSF ∓ FTF  2×41.8 ∓ 7.7  75.8/91.3 Hz Q96_Ans. Outer race default Q96_Hint: In Table 5.2, letting β  0.387, we can rewrite f c  β fr  7.82 Hz for FTF, f 0  β N fr  78.2 Hz for BPFO, f i  (1 − β)N fr  124.4 Hz for BPFI and f b  2β(1 − β)/(1 − 2β) fr  83.5 Hz for BSF. Since the ratios of dominant frequencies/BPFO are {234.6, 312.8, 391.2, 469.3}/78.2  {3, 4, 5, 6} as multiple numbers of BPFO, we therefore know that the disturbances are caused by an outer race defect. Q97_Ans. Inner race fault fi 357.4  0.39, FTF  19 Hz, BPFO   1− Q97_Hint: β  1 − N fr 12 × 49.84 229 Hz, BPFI  357 Hz, BSF  106 Hz, BPFI ± f r  309/406 Hz Q98_Ans. (2) {−1.8, 2.4, −1} × 2.4 g 255° (3) {8.9 170, 8.6 −87, 8.7 −2} where the notation indicates g degree. Q98_Hint: (1) We know eigenvectors, φ 1 and φ 2 , of Nc1 and Nc2 modes from R1_Fig. 5.20:  t  t φ1  0.49 0.78 1 and φ2  0.69 0.09 −1 . t  If the ratio is set by h  h 1 h 2 h 3 , these values are determined so as not to influence responses of the Nc1 and Nc2 modes, and hence, the inner product t between eigenvectors and the ratio should vanish  according to h φ1  0 and t h φ2  0. These equations yield the ratio h  −1.8 2.4 −1 h 3 . (2) In Fig. (2), AC/AB  1.3 − 45◦ determines the correction weight as 2 g −60° × 1.3 −45°  2.6 g 255° (3) {1, 1, 1} × 3.5 g −55 + {2, 0, −1} × 6 g −180° + {−1.8, 2.4, −1} × 2.6 g 255° Q99_Ans. (1) m2  65.5 kg, (2) I p2  0.41 kg m2 , (3) x g2  640 mm, (4) I d2  5.4 kg m2 . Q99_Hint: (1) m2  m1 + ma  52 + 13.5  65.5 kg (2) I p2  I p1 + I pa  0.31 + 0.1  0.41 kg m2 (3) x g2  (x g1 × m1 + 1062 × ma )/m2  (531 × 52 + 1062 × 13.5)/65.5  640 mm

508

12 Exercises of ISO Certification Examination for Vibration Experts

(4) With respect to the transverse moment of inertia around the left end: m2 × x 2g2 + I d2  m1 × x 2g1 + I d1 + ma × 1.0622 + I pa /2 Then, I d2  52 × 0.5312 + 2.32 + 13.5 × 1.0622 + 0.1/2-65.5 × 0.642  5.4 kg m2 Q100_Ans. (1) f 1  21.3 Hz, (2) f 2  23 Hz, ζ2  0.086, (3) f 3  24.4 Hz, ζ3  0.1, (4) f 3 Hz at 277 rps. Q100_Hint: (1) Concerning the shafting, the shaft mass is ms  62 kg and the shaft stiffness is ks  3E I /L 3  3.03 × 106 N/m. Since the single-DOF of impeller–shaft system consists of the mass meq  154 + 62/4  169 kg, we obtain the natural frequency: 1 f1  2π



ks 1  m eq 2π



3.03 × 106  21.3 Hz 169

(2) Since the total stiffness is the sum of the shaft stiffness and the translating spring, k eq  k s + k 1  3.53 × 106 N/m, we obtain: 1 the natural frequency f 2  2π



keq 1  m eq 2π



3.53 × 106  23 Hz and 169

c1 4200 the damping ratio ζ2    0.086.  √ 2 m eq keq 2 3.53 × 106 × 169 (3) When we set the impeller displacement as unity, the bending mode shape gives the slope at the shaft end as θb  3δ/(2L)  3/(2L), referred from the deformation formula of an uniform cantilever beam end. Then, the spring and damping for the tilting motion provide the following modification: the natural frequency  1 keq + θb2 k2 f3  2π m eq  1 3.53 × 106 + 1.52 × 0.2 × 106  24.4 Hz and  2π 169 c1 + θb2 c2 5143 the damping ratio ζ3      2√3.98 × 106 × 169  0.1 2 m eq keq + θ 2j k2 (4) The equation of motion of the single-DOF system for the impeller displacement z:

12.3 Answers and Hints for the Additional Questions 31–100

509

  m eq z¨ + c1 + c2 θb2 z˙ + ks + k1 + k2 θb2 − jc2 θb2 λ z  0. If the λ form, stated in Sec. 8.7, is introduced by letting λeq c2 θb2 λ/ c1 + c2 θb2 , the above equation can be rewritten as



   m eq z¨ + c1 + c2 θb2 s − jλeq  z + ks + k1 + k2 θb2 z  0. Since λ  0.48 gives λeq  0.088, the instability happens at the onset speed st  f 3 /λeq  276 rps with the unstable frequency of f 3  24.4 Hz.

Appendix A

Spring and Damping Coefficients of a Cylindrical Bearing Assuming Infinitely Short Width

The coefficients in Eq. (2.24) are obtained by partially differentiating Eq. (2.22) with respect to the state variables, which results in the following equations:

 k0 X2e0 1 þ e20 k0 p 1 þ 2e20 k0 Xe20 k0 2e0 Fe0 ¼
2 ; k11 ¼
3 ; c11 ¼
2 5=2 ; c12 ¼ 
2 2 2 1  e0 1  e0 1  e20 2 1  e0
 k0 Xp 1 þ 2e20 k0 Xpe0 k0 2e0 k0 p Fh0 ¼
; k21 ¼
; c21 ¼
2 ; c22 ¼ 
  3=2 3=2 5=2 2 1  e0 2 1  e20 4 1  e20 4 1  e20 ðA:1Þ where; k0 ¼ lRL3 =C 2

ðA:2Þ

Let us denote the reaction forces by (−Fe, Fh) in the polar coordinates Oj (e, h), and (Fr, Ft) in the polar coordinates Os (e0, h0) at the static equilibrium point, i.e. the center of vibration, as shown in Fig. 2.5b. The relationship between them is given in the following complex equation:
 Fr þ jFt  ðFe þ jFh ÞejDh  Fe þ jFh þ ðFe0 þ jFh0 ÞjDh þ O D2

ðA:3Þ

By neglecting the second order small term D2 and higher orders, the oil film forces can be rewritten in the following equations by equating the real part and imaginary part of both sides: :

:

Fr ¼ Fe þ Fh0 Dh ¼ Fe0 þ k11 De þ k12 e0 Dh þ c11 D e þ c12 e0 D h  Fe0  fr :

:

Ft ¼ Fh  Fe0 Dh ¼ Fh0 þ k21 De þ k22 e0 Dh þ c21 D e þ c22 e0 D h  Fh0 þ fh ðA:4Þ © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

511

Appendix A: Spring and Damping Coefficients …

512

where k12 

Fh0 k0 Xp Fe0 k0 Xe0 ¼
¼ 
2 3=2 ; k22   2 e0 e 0 1  e20 4 1  e0

ðA:5Þ

The relationships between the polar coordinates and x–y coordinates around the static equilibrium point are given as follows:   
 f cos h0  fx þ jfy ¼ ðfr þ jft Þejh0 , x ¼ fy sin h0

x þ jy ¼ CðDe þ je0 DhÞejh0   De  CT2 e0 Dh

sin h0  cos h0

   x cos h0 , ¼C sin h0 y



fr ft

 sin h0 cos h0





  T1

De e0 Dh



fr ft



ðA:6Þ

ðA:7Þ

Therefore, from Eq. (A.6) and the dynamic component of Eq. (2.25), they can be written as      fx fr k ¼ T1 ¼ T1 11 fy ft k21

k12 k22



  De c þ T1 11 c21 e0 Dh

c12 c22



:

D e: e0 D h

 ðA:8Þ

By applying Eq. (A.7) to Eq. (A.8), the coefficient matrix in Eq. (2.26) can be obtained as follows: 

kxx kyx

kxy kyy



 ¼ T1

k11 k21

 k12 T2t ; k22 C



cxx cyx

cxy cyy



 ¼ T1

c11 c21

 c12 T2t c22 C

ðA:9Þ

Appendix B

Elliptical Coordinates

This Appendix describes how to transform from Eq. (3.7) into Eq. (3.8). To simplify the description, we use a and / in place of a0 and /0, respectively. Using an angle (a0 ! a) between xy and XY axes, both coordinate systems are linked as follows:    cos a x ¼ y sin a

 sin a cos a

   cos a X ¼ Y sin a

 sin a cos a



a cosðXt  a þ /Þ b sinðXt  a þ /Þ

 ðB:1Þ

Letting u = −a + / for simplicity, this equation can be rewritten:    x a ¼ y a  a ¼ a

cos a cosðXt þ uÞ  b sin a sinðXt þ uÞ cos a sinðXt þ uÞ þ b cos a sinðXt þ uÞ



cos a cos u  b sin a sin u

a cos a sin u  b sin a cos u

sin a cos u þ b cos a sin u

a sin a sin u þ b cos a cos u



cos Xt



sin Xt

ðB:2Þ The comparison of the above equation with Eq. (3.3) yields the following equation: 

xR yR

xI yI



 ¼

a cos a cos u  b sin a sin u a sin a cos u þ b cos a sin u

a cos a sin u  b sin a cos u a sin a sin u þ b cos a cos u



ðB:3Þ Then, expressions for xR, xI, yR and yI can be connected with variables a, b, a and u, as follows:

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

513

514

Appendix B: Elliptical Coordinates

x2R þ x2I þ y2R þ y2I ¼ a2 þ b2 x2R þ x2I  y2R  y2I ¼ ða2  b2 Þ cos 2a xR yI  xI yR ¼ ab

ðB:4Þ

2ðxR yR þ xI yI Þ ¼ ða  b Þ sin 2a  2 2 2 xR þ x2I  y2R  y2I þ 4ðxR yR þ xI yI Þ2 ¼ ða2  b2 Þ 2

2

These equations give the relationships shown in Eqs. (3.8) through to (3.11).

Appendix C

Fourier Transformation [291]

The theory of Fourier series expansion stipulates that any periodic function (Period = T) can be approximated by the summation of cos and sin functions having the fundamental frequency (1/T Hz) and their harmonics. Its resolution is the fundamental frequency. On the other hand, as the period T approaches infinity (T ! ∞), the fundamental frequency approaches zero (1/T ! 0), which indicates that, in addition to the periodic function, any waves can be approximated by the superposition of cos and sin functions. That is, when period T approaches infinity, the resolution approaches zero, and any waves can be approximated continuously by cos and sin waves having any appropriate frequency. The above statements form the concept of Fourier transformation theory. FFT analyzers of vibration waveforms have been developed basically for the ideal situation in which periodic functions are assumed. However, in reality, signal analysis with a FFT analyzer has to be applied to waveforms other than periodic functions. The Fourier transformation theory fills the gap between the ideal and actual situations and may be used to interpret the cases of aperiodic functions.

C.1 Definition of Fourier Transformation The Fourier transformation of the waveform f(t) in the time domain is given by the following equation: Z1 f ðtÞe

FðxÞ  1

jxt

ZT dt ¼ lim

T!1

f ðtÞejxt dt

ðC:1Þ

T

The first expression of the integral on the right side is the definition and the actual calculation that is performed by the second integral involving the limit. The second indicates that, in order to filter out waveforms similar to a function ejxt from © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

515

516

Appendix C: Fourier Transformation

the actual waveform, f(t), the inverse function e−jxt is used to multiply the waveform, f(t), and then the averaging operation is used to rectify the DC component of the integration. As mentioned with respect to the approach of the Fourier series expansion in Sect. 10.3, the fundamental concept is based on the combination of two steps, i.e. multiplying by the inverse function and successively rectifying it. The function F(x) after the transformation is defined in the frequency domain. The inverse Fourier transformation is given by the following equation: 1 f ðtÞ ¼ 2p

Z1

1 lim FðxÞe dx ¼ 2p X!1

ZX

jxt

1

FðxÞejxt dx

ðC:2Þ

X

Functions, f(t) in the time domain, and F(x) in the frequency domain, are linked with a bidirectional arrow as shown below, which is referred to as a Fourier transformation pair: f ðtÞ , FðxÞ

ðC:3Þ

Several features of the Fourier transformation pair are listed in Table C.1. Some of the features are explained for the preparation of future applications. (F4) Time axis shifting f ðt  ta Þ , FðxÞejxta

ðC:4aÞ

Proof By letting x = t − ta; Z1

f ðt  ta Þejxt dt¼

1

Z1

Z1

f ðxÞejxðx þ ta Þ dx ¼ ejxta

1

f ðxÞejxx dx ¼ FðxÞejxta

1

(F5) Frequency axis shifting f ðtÞejxa t , Fðx  xa Þ Proof Z1 f ðtÞe 1

jxa t jxt

e

Z1 dt ¼ 1

f ðtÞejðxxa Þt dt ¼ Fðx  xa Þ

ðC:4bÞ

Appendix C: Fourier Transformation

517

Table C.1 Fourier transformation pair f(t) , F(x) f (t )

F (ω )

a1 f1 ( t) + a 2 f 2 ( t)

a1 F1 (ω) + a 2 F2 (ω)

F1

Linearity

F2

Duality

F (t )

2 π f (– ω )

F3

Scaling in time domain

f (at )

1 F (ω / a ) |a|

F4

Translation Modulation (Shift in frequency domain)

F6

Time derivative

F7

Time integral

F8

Frequency derivative

F9

Conjugation

F10

Convolution in time domain

F11

Convolution in frequency domain

F12

Parseval's theorem

f (t ± t 0 )

+ –

F (ω ) e ± j ω t 0 + –

leftward rightward

f (t ) e ± jω 0 t

F (ω ω 0 )

d n f (t ) dtn

( jω ) n F (ω )

t

∫– ∞

F (ω ) jω

f ( t ) dt

( – jt ) n f (t )

d n F (ω ) dω n

f * (t )

F *( – ω )

( f1 * f 2 ( ( t )

F1 (ω) F2 (ω)

f1 ( t) f 2 ( t)

1 ( F F ( (ω) 2π 1 * 2

E=∫

∞

–∞

forward backward

±

F5

(Shift in time domain)

1 2 | f ( t ) | dt = 2π

∫– ∞ | F (ω ) | 2dω = ∫– ∞ | F ( Hz ) | 2 dHz ∞

∞

Note that the notation of convolution is ( f * g )( x ) = ∫– ∞ f ( p ) g ( x − p ) dp = ∫– ∞ f ( x − p ) g ( p) dp ∞

∞

(F11) Product of time functions , Convolution integration of frequency functions f1 ðtÞf2 ðtÞ ,

1 ðF1  F2 ÞðxÞ 2p

ðC:4cÞ

Proof Z1 FðxÞ ¼ 1

f1 ðtÞf2 ðtÞejxt dt ¼

Z1 1

0 @1 2p

Z1 1

1 F1 ðyÞejyt dyAf2 ðtÞejxt dt

518

Appendix C: Fourier Transformation

The following expression is obtained by changing the order of integration of y and t: 1 FðxÞ ¼ 2p 

Z1

0 F1 ðyÞ@

1

1

Z1

jyt jxt

f2 ðtÞe e 1

1 dtA dy ¼ 2p

1 ðF1  F2 ÞðxÞ 2p

Z1 F1 ðyÞF2 ðx  yÞdy 1

C.2 Examples of Fourier Transformations (1) Rectangular wave (Rectangle pulse) whose area is unity  pða; tÞ 

ðjtj\a=2Þ ðjtj [ a=2Þ

1=a 0

ðC:5Þ

As shown in Fig. C.1a, this waveform in the time domain is the rectangular shape whose area is equal to 1. According to the definition, the Fourier transformation is calculated as follows: Z1 Pða; xÞ ¼

pða; tÞejxt dt ¼

1

 a=2 1 ejxt 1 ejxa=2  ejxa=2 sinðxa=2Þ ¼ ¼ jx a jx a=2 a xa=2 ðC:6Þ

p 5

P

a = 0.2

1 0.8

4

a=1

0.2 a=1 t

0 Time [s]

a = 0.5

0.4

a = 0.5

1 −0.5

a = 0.2

0.6

3 2

a = 0 ( Delta function )

1 fa = a [Hz]

0.5

(a) Function in time domain p ( a, t )

f

0 −0.2 −6

−4

0 2 −2 Frequency × fa [Hz]

4

(b) Fourier transformation P ( fa , f )

Fig. C.1 Fourier transformation of rectangular function (area = 1)

6

Appendix C: Fourier Transformation

519

Thus, the Fourier transformation pairs are given by: pða; tÞ , Pða; xÞ ¼

sinðxa=2Þ xa=2

ðC:7aÞ

pða; tÞ , Pðfa ; f Þ ¼

sinðpf =fa Þ pf =fa

ðC:7bÞ

where; x ¼ 2pf ½rad=s; fa ¼ 1=a ½Hz. These pairs are expressed with the angular frequency x (rad/s) in Eq. (C.7a) and the frequency f (Hz) in Eq. (C.7b). For the ease of field application, Eq. (C.7b) is used in the following discussions. The graphical representations of the Fourier transformation pairs are given in Fig. C.1b. They resemble the wavy surface of water, having their peaks at fa = 0, and rippling out in the right and left directions while the peaks of the ripples are gradually damped. It is evident that the larger a is, the wider the ripples are spread. When the width, a, approaches zero, the waveform in the time domain becomes the delta function, whose Fourier transformation is 1, i.e. a flat line. In other words, it can be said that the impulse function d(t) includes every frequency component uniformly in amplitude. As seen in Fig. C.1b, the rectangular wave has the following characteristics in the frequency domain: • The major component is the DC component at zero frequency. • In general, low frequency components are major and the high frequency components are minor. • Any of the fundamental frequency components and their harmonics are not included, because of zeros at frequencies = fa, 2fa, 3fa, …. An example of the approximate reconstruction of the input wave is depicted in Fig. C.2, corresponding to the rectangular wave in Fig. C.1a when a = 1. This reconstruction is calculated by the integration of Eq. (C.2). The accuracy of the Including orders between −10 fa ~ +10 fa

Input function 1 0.8 0.6 0.4 0.2 0 −1

−0.5

0 Time × T [s]

Fig. C.2 Input function and approximate functions

−3 fa ~ +3 fa − fa ~ + fa t 0.5

1

520

Appendix C: Fourier Transformation

approximate reconstruction depends to the order of cos and sin in the case of Fourier series expansion. However, in this case of the Fourier transformation, it is the finite range for the integration instead of the infinite range. As shown by ±10fa (integration range: −10fa to +10fa), the precision of the reconstruction increases as the integration range expands. (2) Rectangular window  w1 ðtÞ 

ðjtj\T=2Þ ðjtj [ T=2Þ

1 0

ðC:8Þ

The time waveform represents the rectangular shape whose height is 1 and width is T, as shown in the upper part of Fig. C.3a. As stated with respect to the discrete Fourier transformation in Sect. 10.5.3, this weighting function is used to filter out the raw waveform of the input between time before −T/2 and after T/2, hence is called a rectangular window. Since the function of Eq. (C.8) is similar to Eq. (C.5), this Fourier transformation is calculated by referring to Eq. (C.7b) and F1 in Table C.1 as follows: w1 ðtÞ ¼ T  pðT; tÞ , T  PðT; xÞ

ðC:9aÞ

w1 ðtÞ , T  Pðf0 ; f Þ  W1 ðf Þ ðf0  1=TÞ W1 ð0Þ ¼ T

ðC:9bÞ

(a) Rectangular or Uniform

(b) Hanning

1 0 0

Gain

0 −1

−1

0.5

−0.5

0

1 0.8 0.6 0.4 0.2 0

1 0.8 0.6 0.4 0.2 0 0

−0.5

0.5

Time × T [s]

Time × T [s]

Gain [dB]

1

0

−1 −0.5

(c) Flat top

1

0

0.5

Time × T [s] 1 0.8 0.6 0.4 0.2 0 0

−20

0 −20

−20

−40

−40

−40

−60 −6 −4 −2 0 2 4 6 Frequency × 1/T [Hz]

−60 −6 −4 −2 0 2 4 6 Frequency × 1/T [Hz]

−60 −6 −4 −2 0 2 4 6 Frequency × 1/T [Hz]

W1 (0) = T

W2 (0) = 0.5T

Fig. C.3 Window function and its Fourier transformation pair

W3 (0) = 0.216T

Appendix C: Fourier Transformation

521

The lower part of Fig. C.3a gives the graphical representations of these pairs. The frequency is multiplied by 1/T such that the maximum gain of the Fourier transformations becomes unity (i.e. 0 dB). (3) Hanning window w2 ðtÞ  w1 ðtÞ½1 þ cos x0 t=b

ðC:10Þ

where the fundamental frequency, f0 ¼ 1=T ½Hz; x0 ¼ 2pf0 ½rad=s; b ¼ 2. This function is the most practical and is used widely for the window of the FFT, namely the Hanning window. The function is shown in the upper part of Fig. C.3b, as to narrow the inputting waveform down to zero at the both ends of the window. By the reduction of this narrowing, any inputting waveform resembles a periodic function with start and end values of zero. It can be regarded as a periodic function whose time width is T. The above Eq. (C.10) can be rewritten as: w2 ðtÞ ¼ 0:5w1 ðtÞ þ 0:25w1 ðtÞejx0 t þ 0:25w1 ðtÞejx0 t

ðC:11Þ

Thus, the Fourier transformation of this equation is given by using F5 in Table C.1: w2 ðtÞ , W2 ðf Þ  0:5W1 ðf Þ þ 0:25W1 ðf  f0 Þ þ 0:25W1 ðf þ f0 Þ W2 ð0Þ ¼ 0:5W1 ð0Þ ¼ 0:5T

ðC:12Þ

The Fourier transformation pairs are shown graphically in the lower part of Fig. C.3b. The gain is adjusted by the factor b = 2 such that W2(0) = 1. The distribution of W2(f) is as significant as W1(f) of the rectangular window, however, the main lobe at around f = 0 expands as wide as ±2f0. (4) Flat top window

w3 ðtÞ  w1 ðtÞ½b0 þ b1 cos x0 t þ b2 cos 2x0 t þ b3 cos 3x0 t þ b4 cos 4x0 t ðC:13Þ where: f0 ¼ 1=T; x0 ¼ 2pf0 b0 ¼ 1=b; b1 ¼ 1:93=b; b2 ¼ 1:29=b; b3 ¼ 0:388=b; b4 ¼ 0:032=b; b ¼ 4:64: As shown in the upper part of Fig. C.3c, this time function narrows more rapidly than the Hanning window. The Fourier transformation pairs of this window are obtained in a similar manner as for Eq. (C.11):

522

Appendix C: Fourier Transformation

w3 ðtÞ , b0 W1 ðf Þ þ

4 X

bk =2½W1 ðf  kf0 Þ þ W1 ðf þ kf0 Þ

k¼1

W3 ð0Þ ¼ b0 W1 ð0Þ þ

4 X

ðC:14Þ bk =2½W1 ðkf0 Þ þ W1 ð þ kf0 Þ ¼ b0 T ¼ 0:216T

k¼1

This graphical representation is shown at the lower part in Fig. C.3c. The gain is adjusted by the factor b = 4.64 such that W3(0) = 1. The flat top window function, as it can be literally implied, features a flat and wide crest of W3(f), and the lobe at around f = 0 expands as wide as ±4f0. (5) Window function  cos wave Hence, the Fourier transformation for a waveform, multiplied by the cos wave having arbitrary frequency xa and the window function wi(t), is discussed: f ðtÞ  wi ðtÞ cos xa t ¼

 wi ðtÞ
jxa t e þ ejxa t 2

ðC:15Þ

The corresponding Fourier transformation pairs are obtained by using F5 in Table C.1 as follows: 1 1 wi ðtÞ cos xa t , Wi ðx  xa Þ þ Wi ðx þ xa Þ 2 2 1 1 wi ðtÞ cos xa t , Wi ðf  fa Þ þ Wi ðf þ fa Þ ðfa ¼ xa =ð2pÞÞ 2 2

ðC:16aÞ ðC:16bÞ

As shown in these equations, the curve of these Fourier pairs is the same as the Fourier pairs Wi(f) of the window itself, whose frequency axis is shifted by fa. Figure C.4 shows examples of the Fourier transformation. In the examples, the span of the time window is T = 1 (f0 = 1 Hz), and the inputting cosine waveform has frequency of fa = 10.6 Hz. As shown in Fig. C.4a, this function f(t) is multiplied by the rectangular window wl(t) and is then subjected to the Fourier transformation. Since the frequency 10.6 Hz of this function, f(t), does not agree with an integer multiple of the fundamental frequency 1 Hz, it may not represent a periodic function. The Fourier transformation pair for Wl(f) of the window function is shown in Fig. C.4b, and the Fourier transformation pair for f(t) is shown in Fig. C.4c. The curve in figure (c) is essentially identical with the curve in figure (b) except that it is shifted from 0 to fa = 10.6 Hz. In the case of the discrete Fourier transformation, the amplitude spectrum corresponding to only the integer multiples of the fundamental frequency f0 = 1 Hz is displayed as shown in Fig. C.4d. These red dots can be copied from those in figure (c). Since the inputting frequency is fa = 10.6 Hz, f = −0.6 Hz and f = 0.4 Hz in figure (b), corresponding to 10 and 11 Hz in figure (c), which are 10 times and 11

Appendix C: Fourier Transformation Rectangular windows

(a)

1

−0.5

(b)

0.5 10.6 Hz

0

−1

(d)

523

Time × T [s]

0

−6

6

8

10 10.6 12

Frequency × 1/T [Hz]

14

0 −4 −2 Frequency

2 4 6 × 1/T [Hz]

1

0 4

0.4

0

(c)

1

− 0.6

1

4

6

8

10

12

14

10.6 Frequency × 1/T [Hz]

Fig. C.4 Uniform window function and Fourier transformation of non-periodic function

times of the fundamental frequency f0 = 1 Hz. It is shown that the spectrum at 11 Hz is larger than that at 10 Hz because the inputting frequency is closer to 11 Hz than to 10 Hz. The form of figure (c) is due to the shift of figure (b). In general, we can state that the spectrum of the aperiodic function, a frequency that is not identical to the integer multiple of the fundamental frequency f0, can be predicted only by observing figure (b), which shows the Fourier pair W2(f) of the window function, regardless of any actual inputting waveform. As noted between figures (b) and (c) with an arrow, we replace the frequency f = 0 by the frequency fa.

C.3 Comparison of Amplitude Spectrum of Window Function As understood in Fig. C.4, the spectrum of the discrete Fourier transformation, including of aperiodic functions, is predictable based on the curve around f = 0 of the Fourier pair of the window function. Therefore performance of the window function can be evaluated with the curves shown in Fig. C.3. The amplitude spectrum of the window function is discussed in the following. An example of the rectangular window function is shown in Fig. C.5. The axis of the ordinate is changed from a linear scale to a dB scale so that the details of low peaks are conspicuous. The most prominent peak at around f = 0 is referred to as a main lobe and a series of lower peaks at the both outsides of the main lobe are referred to as the side lobes. The smallest difference between the peak of the main lobe and that of side lobe (maximum side lobe level) is referred to as roll-off. When an aperiodic function is input, the leakage error creates a spectrum of the side lobes appearing at around the major components of the main lobe, and thus it can be said that the larger the roll-off is the better the window is. In this example, the roll-off is −13 dB from the main lobe to the first side lobe.

524

Appendix C: Fourier Transformation Bandwidth (BW = 0.89) 0

−3 dB = 0.7 Main lobe

Roll-off

[dB]

( Maximum side lobe −10 level ) −13 dB = 0.22 −20

side lobe WF = 1

−30

Equivalent bandwidth

−40 −6

−4

−2 0 2 Frequency × 1/T [Hz]

4

6

Fig. C.5 Main lobe and side lobe

The frequency bandwidth of the point, that is 3 dB lower than the maximum value of the main lobe, is defined as the Bandwidth (BW). In this example, BW = 0.89f0 for the rectangular window. Based on an idea similar to the BW, one defines an equivalent bandwidth (EBW). Its concept is to approximate the shape of main and side lobes in comparison with that of the rectangular window from the viewpoint of the spectral power. The total power (energy) of each window function is calculated as follows, using of F12 (Parseval’s theorem) in Table C.1: Z1 Z1 W1 ðf Þ 2 1 1 Rectangular window : E1 ¼ jw1 ðtÞj2 dt ¼ ¼ f0 W ð0Þ df ¼ T 2 T 1 1

Z1 Hanning window : E2 ¼ 1

1

ðC:17Þ 1 2 Z W2 ðf Þ 1:5 df ¼ 1 ¼ 1:5f0 j2  w2 ðtÞj2 dt ¼ W ð0Þ 2 T T 2 1

ðC:18Þ Z1 Flat-top window : E3 ¼ 1

¼ 3:8f0

W3 ðf Þ 2 12 W ð0Þ df ¼ T 2 3

Z1 j4:64  w3 ðtÞj2 dt ¼ 1

3:77 T ðC:19Þ

The EBW is 1.0f0, which is slightly larger than the BW = 0.89f0 for the rectangular window. The details are shown in Table C.2. The coefficient of the EBW is called the Window Factor (WF), and it is WF = 1 for the rectangular window, WF = 1.5 for the Hanning window, and WF = 3.8 for the flat top window.

Appendix C: Fourier Transformation

525

Table C.2 Characteristics of windows Rectangular Uniform

Hanning

Flat top

narrowest

narrow

wide

0.89 f 0

1.44 f 0

3.73 f 0

1

1.5

3.8

Side lobe

large

middle

small

Roll-off

– 13dB

– 32dB

– 43dB

– 3.9dB

– 1.42dB

–0.1dB

64%

85%

99%

Impulse Test

Frequency identification

Amplitude identification

Windows Main lobe Bandwidth (BW) Equivalent bandwidth (WF)

Gain at the mid (non-periodic) frequency

Recommended applications

In Fig. 10.27, the amplitude spectrum of every window function is shown on the dB scale as well as on the linear scale. This figure can be used to compare performance of one window function with others. However, the gain in this figure is adjusted such that the maximum amplitude spectrum is equal to 1 = 0 dB. • Rectangular window function: When the function is periodic, one peak of the main lobe exists at f = 0 Hz and no side lobe exists. When the function is aperiodic, two amplitudes exist in the main lobe. Their peak values are smaller than 1, making it difficult to identify the amplitude of the original inputting waveform just by reading spectrum peaks. In addition, the relatively large leakage error of the side lobe appears over the wide range. This rectangular window is used in the “Start with zero and end with zero” type “periodic” waveform, such as the impulse response waveform. • Hanning window function: Since the width of the main lobe is twice that of the rectangular window, two additional peaks exist with the main lobe in addition to the peak at f = 0 Hz, even with the synchronous function, but there is no side lobe. However, in the case of the aperiodic function, four amplitude peaks exist with the main lobe, and their peaks are smaller than 1. Because of the rapid decay of the side lobes, the leakage error is small with this window, highlighting the existence of the main component more clearly than in the case of the rectangular window. This window function is most suitable for vibration diagnosis based on the frequency information. • Flat top window function: Since the width of the main lobe is about four times wider than that of the rectangular window, there exists as many as three spectra, whose peaks are approximately 1 for the main lobe, including the peak at f = 0 Hz even with the synchronous function, but there is no side lobe. In the

526

Appendix C: Fourier Transformation (3) Flat top

(3) Flat top 1

a1

a2

a4

(3)

a1 −0.1 dB a4

−1

a3

(2) Hanning

(2)

0.4

−2

a3

Hanning (2)

−3

−3.9 dB (2)

a2

(1) 0.2

(1)

−4

(1)

Rectangular

(1) Rectangular −5

0

(

(3)

−1.42 dB 0.6

Gain [dB]

Gain ( Linear )

0.8

0

0

)

0.5

0.6

1.0

0.32 Example 10.11 Non-periodic frequency × 1/T [Hz]

0

0.5

0.6

1.0

Non-periodic frequency × 1/T [Hz]

Fig. C.6 Fourier transformation per resolution

case of the aperiodic function, the number of these peaks reduces down to two on the main lobe, but their peak values do not change. It is possible with this function to read the amplitude precisely and at any time without the leakage error, which is smaller than the error with the Hanning window. This is the most significant advantage of this function for vibration diagnosis based on amplitude information. Figure C.6 shows to how much the amplitude spectrum decreases relative to the actual amplitude in the case of the aperiodic function. For example, when the input waveform has the amplitude a = 1 and the frequency f = 10.32 Hz with the Hanning window whose resolution is 1 Hz, we can regard that both ends of the left and right sides of this frequency correspond to 10 and 11 Hz, respectively. Hence, the red circles ○ of a1 and a2 placed at 0.32 Hz indicate the actual reading of the dominant spectra; a1 = 0.94 = −0.53 dB at 10 Hz and a2 = 0.75 = −2.5 dB at 11 Hz. When the inputting frequency is hence changed to f = 10.6 Hz, we can also regard the left and right ends of the frequency correspond to 10 and 11 Hz, respectively. The blue circles ● of a1 and a2 are placed at 0.6 Hz in the figure so that the actual readings of the dominant spectra are: a1 = 0.785 = −2.1 dB at 10 Hz; and a2 = 0.9 = −0.9 dB at 11 Hz. As one of the modifications of Fig. C.6, a relationship between the amplitude ratio, defined by (the second highest amplitude/the highest amplitude), and the frequency difference ratio, calculated by (actual inputting a non-periodic frequency − the closest periodic frequency)/(frequency resolution), is presented in Fig. C.7. The former ratio is set in the axis of the ordinate and the latter ratio in the abscissa axis.

Appendix C: Fourier Transformation

527 (3) Flat top

(3) Flat top

0 0.78

0.8

Spectrum ratio [dB]

Spectrum ratio (linear)

1.0

0.6 (2) Hanning

0.4 (1) Rectangular

0.2

0.32 0.1

0.2

0.3

0.4

-2.16dB

-4 (2) Hanning

-6 -8

0.0 0

-2

0.5

Frequency difference × 1/T [Hz]

-10 0

(1) Rectangular

0.32 0.1

0.2

0.3

0.4

0.5

Frequency difference × 1/T [Hz]

Fig. C.7 Spectrum ratio and frequency resolution

For example, when we know the spectrum data: a1 = 0.94 at 10 Hz; and a2 = 0.75 at 11 Hz for the case of a single aperiodic frequency, the amplitude ratio = 0.75/0.94 = 0.78 = −2.16 dB gives the frequency difference ratio = 0.32 as shown in Fig. C.7. Finally, we can identify the inputting frequency = 10 + 0.32  (1/T) = 10.32. In the other case of {a1 = 0.785 at 10 Hz and a2 = 0.9 at 11 Hz}, mentioned previously, the amplitude ratio = 0.785/0.9 = 0.87 = −1.2 dB gives the frequency difference ratio = 0.4 and then the identified inputting frequency = 11−0.4  (1/T) = 10.6 Hz. Advantages and disadvantages in each window function are summarized briefly in Table C.2. In terms of the resolution, the best choice is the rectangular window function. When we place special emphasis on the readability of the amplitude, the best choice is the flat top window function. The Hanning window function has an intermediate characteristic, and this is reason why the Hanning window function is the most frequently used in the field.

Appendix D

PLL Circuit and Synchronized Sinusoidal Wave Generation Circuit

(1) MATLAB/SIMULINK program ! Fig. D.1

(a) Pulse Generator

if (in > 2 or in 0 , 0, 1)

PLL

out_vco

PI(s) = 0.009 937 +

Unit Delay 1 z

reset 0/1

out_vco +

Sawtooth shock pulse

out in

0.050 32 0.010 006 281 4 − 0.009 993 718 4z−1 PI(z−1 ) = s 1 − z−1 Sampling frequency 4KHz

(b) ref_pfc

z−1

PLL

xor1

xor8 NAND a c b c = Not (a AND b)

xor2 z−1 xor3 z−1 z

xor7

+ +

xor4

out_pfc

−1

xor5 out_vco

z−1 z

xor9 xor6

−1

in out if (in > 0, −1, 0)

Fig. D.1 Program for PLL by Matlab (a), Program for PLL/Sawtooth shock pulse by Matlab (b)

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

529

530

Appendix D: PLL Circuit and Synchronized Sinusoidal Wave Generation Circuit

(2) C language program ! Fig. D.2

/*****

PLL → LPF out_pfc=−1(Delay) ; =0 (ln phase) ; = 1 (Lead)

*****/

if(pulse > = 8192) ref_pfc = 0; else ref_pfc = 1; /* 0/1 shaping to pulse */ xor1 = !(ref_pfc && xor8);

xor2 = !(xor1 && xor3);

xor3 = !(xor2 && xor7); xor4 = !(xor5 && xor7); xor5 = !(xor4 && xor6); xor6 = !(out_vco && xor9); xor7 = !(xor1 && xor2 && xor5 && xor6); xor8 = !(xor1 && xor2 && xor7); xor9 = !(xor5 && xor6 && xor7); if((xor8 == xor9)) out_pfc = 0.0; else{if(xor8) out_pfc = 1.0; else out_pfc = −1.0;} /*****

LPF for sampling frequency = 4kHz

*****/

out_lpf=out_lpf + 0.0100062814 * out_pfc−0.0099937184 * pfc_1; pfc_1=out_pfc;

x=x−out_lpf * 0.3;

if( x> 2.0 ) x = −2.0;

/* reset to pll wave x = −2 */

if( x0 ) out_vco=0;

/* reset to out_vco = 0/1 */

else out_vco = 1;

/* define M_PI by 3.141592654 */

teta = ( M_PI*(0.5*x)); /*−2 < x < 2 < = > −π < Θ(rad) < π */ cosv = Cos(teta); sinv = Sin (teta);

/* outputs of cos & sin waves */

Fig. D.2 Program for PLL/cos/sin by C-language

/* xor8 /* 0 /* 0 /* 1 /* 1

xor9 out */ 0 0 */ 1 -1 */ 0 1 */ 1 0 */

Appendix E

Campbell Diagram

This Appendix illustrates our own method on how to draw the Campbell diagram as a 3D-display with respect to rotational speed, the frequency and the amplitude of dominantly resultant vibration. The upper part of Fig. E.1 shows the block diagram of the signal processing, and the lower part of the figure shows the resultant output. Inputs to this diagram are the vibration waveform, w(t), and the pulse signal. The harmonic functions, cosXt and sinXt, are synchronized with the rotational pulse. In the block of the tracking filter, the amplitude, a1, of the rotation-synchronized component 1X is obtained by the multiplication by this synchronized harmonic function, followed by integration executed with a LPF having a large time constant. The magnitude of this amplitude, a1, corresponds to the size of circle on the 1X line. Similar to this operation, the amplitude, an, of the vibration component that is synchronized with n times the rotational speed X is obtained by the multiplication by cosnXt and sinnXt, followed by the integration. These magnitudes, an, are indicated by the size of the circles on the nX line. In this example, up to the 6th order of nX vibration (i.e. 6X) has been analyzed. In general these nX vibrations are the forced vibrations. The residual after subtracting these nX forced vibration components from original input vibration waveform is noted by the fluctuation and it is regarded as free vibration. At a certain interval during the rotation, the waveform data acquisition of this free vibration is undertaken, e.g., for 1024 sampling points, FFT processing is then performed. Selecting pairs of frequency and the magnitude of the noteworthy “peaky” spectra among the FFT spectrums, convert the “peaky” magnitude to the size of the circle, and the plot is of the circle at the coordinate point {X = corresponding rotation speed, Y = the frequency}. The 1X vibration and overall spectrum are referred to the amplitudes of the Bode plot shown in the lower part of Fig. E.1. In this example, there exists a natural frequency at about 23 Hz, which can be obtained with FFT analysis of the fluctuation vibration, and large circles at around 1400 rpm at the intersection with the 1X line indicates the unbalanced responses © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

531

532

Appendix E: Campbell Diagram Rotational pulse

Input vibration wave w(t) fs = 4kHz for sampling Tracking filter

PLL

cn =

cos t sin t

sn =

2 T 2 T

T

∫0w(t) cos n T

∫ w(t) sin n 0

6

wn ( t ) = ∑ cn cos n n=1

t dt

+

−

wn ( t )

t dt t + sn sin n

t

N=1024 FFT

cn2+ sn 2 Synchronous n

Forced vibration

6X 5X

4X

Fluctuation

Free vibration

3X

2X

2a1

1X

Frequency [Hz]

60

40

20

Amplitude

0 1X 8

Overall

4 0

1000 Rotational speed [rpm]

2200

Fig. E.1 Our signal processing for Campbell diagram analysis

around resonance. Even though they are shown with small circles, there is a super-harmonic resonance at the intersection of the 2X line and the natural frequency due to system non-linearity. Through the entire speed range, we see small and large circles as overlapping, letting it be known that the height of 23 Hz indicates one of the natural frequencies. As we can see, the Campbell diagram is suitable for understanding the global nature of the problems. The model rotor [292] of a turbine blade is shown in Fig. E.2. A ball bearing is mounted on the left side of the rotor shaft, and the motor is mounted on the left end of the shaft. There are 8 blades mounted on the rotating disk (boss) on the right side with a stagger angle of 45°. The Active Magnetic Bearing (AMB) on the right end

Appendix E: Campbell Diagram

533 Strain gauges for blade vibration

Synchronous motor

Shaft displ. pick-up

Ball bearing Slip ring

θ 0 = 45°

Boss

AMB

45°blade

(a) Blade layout

(b) Photo of test rig having 45° blade

Fig. E.2 Turbine model rotor for blade-shaft coupled vibration analysis

is used for magnetic levitation as a non-contacting support as well as an exciter. A displacement sensor placed on the AMB at the right end detects the shaft vibration and a strain gauge attached on a blade detects the blade vibration. The vibration measurements produce for the Campbell diagram as shown in Fig. E.3. The lower graph shows shaft vibration measured with the displacement sensor (signal in the stationary coordinate system: e0). The 1X vibration due to unbalance is prominent, but it is also shown that a lower frequency of self-exited vibration Ⓒ exists at the high rotation speed of approximately 30 rps. The upper graph shows the blade vibration measured by the strain gauge (signal in the rotating coordinate system: er). Some resonances are seen at Ⓐ on the 6X line, Ⓑ on the 3X line, and Ⓒ in blade vibrations. It seems that the shaft vibration is coupled with the blade vibration, because the unstable vibration also appears in the blade vibration Ⓒ at the same speed shown in the upper graph. On the other hand, the 6X resonance Ⓐ and the 3X resonance Ⓑ are found only in the blade vibration at the upper graph, there being no coupling with the shaft vibration at the lower graph. These resonances characterize the independent blade vibration.

534

Appendix E: Campbell Diagram

Fig. E.3 Campbell diagram featuring blade-shaft coupled vibration [VB752]

6X 5X

4X

2X

3X

Frequency [Hz] on er

80

|−

60

40 A

1X

C

B

|±

wb 20 ws

45° blade

10 20 Rotational speed

b

| |

s−

|+

00

|

s−

30

40

[rps]

(a) Blade vibration by strain gauge on er

6X 5X

4X

3X

2X

Frequency [Hz] on e0

80 |+

60

s+

|

1X

40

wb 20

| ± ωs |

ws

C

0 0

10

20

Rotational speed

30 [rps]

(b) Shaft vibration by displ. pick-up on e0

|−

b+

|

Appendix F

Details for Obtaining Eqs. (11.46) and (11.49)

A mathematical formula for a determinant is: a11 þ b11 a21 þ b21

a12 þ b12 a11 ¼ a22 þ b22 a21

a12 a11 þ a22 a21

b12 b11 þ b22 b21

a12 b11 þ a22 b21

b12 b22 ðF:1Þ

Equation (11.45) is thus modified as follows, taking account of fd > 1 as follows: © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

535

536

Appendix F: Details for Obtaining Eqs. (11.46) and (11.49)

ðkXÞ2 þ 2jkXD þ x2 þ 2f x D l ðkXÞ2 þ 2jl kXD d d c c d 2 2 2 lc ðkXÞ þ 2jlc kXD ðkXÞ þ 2jkXD þ xz ðkXÞ2 þ x2 lc ðkXÞ2 2jkXD þ 2fd xd D lc ðkXÞ2 d ¼ þ lc ðkXÞ2 2jlc kXD ðkXÞ2 þ x2z ðkXÞ2 þ x2z ðkXÞ2 þ x2 þ 2jl kXD 2jkXD þ 2f x D 2jl kXD d d c c d þ þ l ðkXÞ2 þ 2jl kXD 2jkXD 2jkXD c c h i h i
  D20 ðkXÞ þ 2fd xd D ðkXÞ2 þ x2z þ 2jkXD x2d þ x2z  2 1  l2c ðkXÞ2 ¼ 0 ðF:4Þ Then, the above equation is rewritten as follows: h

ih i x21  ðkXÞ2 x22  ðkXÞ2 þ aD þ jbD ¼ 0

ðF:5Þ

where h i h i
 and a  2fd xd ðkXÞ2 þ x2z ; b  2kX x2d þ x2z  2 1  l2c ðkXÞ2

2 ðkXÞ þ x2d D20 ðkXÞ ¼ lc ðkXÞ2

h ih i lc ðkXÞ2 ¼ x21  ðkXÞ2 x22  ðkXÞ2 ; 2 2 ðkXÞ þ xz



 because the characteristic roots of D20 ðsÞ ¼ s2  x21 s2  x22 are x1 and x2 . Therefore, the approximate solution, D, is determined as h D¼

ih i x21  ðkXÞ2 x22  ðkXÞ2 a 2 þ b2

ða  jbÞ

ðF:6Þ

In fact, since the stability depends upon the real part of the approximate solution, D, the stability index is expressed as follows: h ReðDÞ ¼ 

ih i x21  ðkXÞ2 x22  ðkXÞ2 a 2 þ b2

a¼

h ih i x21  ðkXÞ2 x22  ðkXÞ2 að1 þ b2 =a2 Þ ðF:7Þ

Thus it is found that the stability limit depends upon the numerator and a. The above equation leads to Eq. (11.49), by letting b=0 for simplicity.

Appendix G

Details for Obtaining Eq. (11.51)

With respect to absolute displacements {zc, z1, z2}, of the casing-rotor-bearing system of Fig. 11.23b, the equation motion is written as follows with considering only spring elements: €3 þ K3 X3 ¼ 0 M3 X

ðG:1Þ

where 2 M3  4

mc

0 m1

sym:

3 2 kc þ k d 0 0 5; K3  4 sym: m2

kd kd þ kz

3 2 3 zc 0 kz 5; X3  4 z1 5 kz z2

By employing relative displacements, the following transformation matrix (Tq) is defined: 2

3 2 1 0 zc X 3  4 z1 5  4 1 1 1 1 z2

32 3 2 3 0 zc zc 0 54 gb 5  Tq 4 gb 5  Tq Xq ; 1 gr gr

ðG:2Þ

and it gives the modification of the equation motion: €q þ Kq Xq ¼ 0 Mq X

ðG:3Þ

where 2 Mq  Tqt M3 Tq ¼ 4

mc þ mt sym:

mt mt

3 2 kc m2 0 m2 5; Kq  Tqt K3 Tq ¼ 4 kd m2 sym:

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

3 0 0 5; kz

537

538

Appendix G: Details for Obtaining Eq. (11.51)

mt ¼ m1 þ m2 In order to set the diagonal element of Mq as unity, the following transformation (Tq1) is made: 2

3 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 3 32 0 0 1= mc þ mt zc zc1 zc1 p ffiffiffiffiffi 0 54 gb1 5  Tq1 4 gb1 5  Tq1 Xq1 ; 0 1= mt X q  4 gb 5  4 pffiffiffiffiffiffi 0 0 1= m2 gr gr1 gr1 ðG:4Þ and it gives the modification of the equation motion: €q1 þ Kq1 Xq1 ¼ 0 Mq1 X

ðG:5Þ

where 2 t Mq Tq1 ¼ 4 Mq1  Tq1

1 sym:

lcb 1

3 2 2 xc lcr 0 t lc 5; Kq1  Tq1 Kq Tq1 ¼ 4 x2d 1 sym:

3 0 0 5; x2z

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffi mt m2 m2 where lcb ¼ ,l  ,l  mc þ mt cr mc þ mt c mt x2c ¼ kc =ðmc þ mt Þ; x2d ¼ kd =mt and x2z ¼ kz =m2 When we add damping elements of oil film bearing and casing support, the above equation reaches to Eq. (11.51).

Appendix H

Mode Synthesis Modeling for Rotational Blade Assemblies [299]

H.1 Variables and Definition We discuss here an eight-blading-system (N = 8) as shown in Fig. H.1. (1) Blading and notations The blades are here approximated by assembling single-DOF of m–k systems as shown in Fig. H.1. They are assigned a number beginning from #0, which is the reference blade. A fixed coordinate system {Xri, Yri, Zri} is provided with blade #i at the angle si X r2

vib. displ. { u2 , v2 , w2 } X r3 vib. displ. { u3 , v3 , w3 }

vib. displ. { u1 , v1 , w1 }

m

Y r2

Y r1

m

m

Y r3

Yr

vib. displ. { u0 , v0 , w 0 } Y r0

θ xr

X r4

Ω

m

S ( xr , yr )

vib. displ. Y r4 { u4 , v4 , w4 }

θ yr

Xr

m X0

X r5

m

X r0

vib. displ. { u7 , v7 , w7 }

Y0

Z0

X r1

m

Y r5 vib. displ. { u5, v5, w5}

vib. displ. = vibration displacement

m X r6

Yr6

Y r7 X r7

vib. displ. { u6 , v6 , w6 }

Fig. H.1 Blading system and coordinate axes © Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

539

540

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

from blade #0. Three axes {Xri, Yri, Zri}, are set in the radial direction, the circumferential direction, and axial direction of the shaft respectively. We regard the vibration displacement to each of three directions as {ui, vi, wi}, respectively. We assume the reference coordinate system is fixed on blade #0 as {Xr, Yr, Zr}, and the translating motion of the shaft center is represented on the rotating coordinate system as {xr, yr}, and the tilting motion is represented by {hxr, hyr} as shown in the figure. The three axes noted by {X0, Y0, Z0} denote the inertial coordinate system. (2) Definitions of coordinate transformation matrix An original coordinate system {X1, Y1, Z1} is rotated by h about an axis to a new coordinate system {X2, Y2, Z2} (Fig. H.2). Each axis before and after the rotation has its own unit vectors for three directions, as denoted by en ¼ ½ in jn kn  (n = 1, 2) as a (1  3)-matrix form. Then the rotating coordinate transformation is defined using unit vectors as follows: 2

3 0 sin h 5 ¼ e2 T1 ðhÞ cos h

ðH:1Þ

3  sin h 0 5 ¼ e2 T2 ðhÞ cos h

ðH:2Þ

3 0 0 5 ¼ e 2 T 3 ð hÞ 1

ðH:3Þ

1 0 about X 1 : e1 ¼ e2 4 0 cos h 0  sin h 2

cos h about Y 1 : e1 ¼ e2 4 0 sin h

0 1 0

2

cos h about Z 1 : e1 ¼ e2 4  sin h 0

sin h cos h 0

(3) Uncoupled blading eigen modes As described in Fig. H.3, eight modes /i (i = 0, …, 7) exist when the number of blades is eight, and there are five corresponding eigen frequencies xi (i = 0, …, 4), having the same number (j) for nodal diameters. Then, though the array of the eight eigen modes is /i (i = 0, …, 7), the number of nodal diameters and eigen Fig. H.2 Transformation of coordinate systems

Y1

Z1

Y2

θ2

θ1

Z2

Y1 Y2 X2

θ3 X1 X2

Z1

Y1

X1

Y2

X2

Z2

Z1

X1

Z2

T1 (θ)

T2 (θ)

T3 (θ)

e1 = e 2 T 1( θ 1 )

e1 = e 2 T 2( θ 2 )

e 1 = e 2 T 3( θ 3 )

Fig. H.2 Transformation of coordinate systems

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies ω0 φ0

ω1 φ1

ω2 φ2

541

ω3 φ3 ω4 φ4

k

0

k

1

k

2

k

3

k

7

k

6

k

5 k

φ7 ω7

φ6 ω6

4

φ5 ω5

Fig. H.3 Blading eigen mode and nodal diameter

frequencies correspond to j = {0, 1, 2, 3, 4, 3, 2, 1} and xn = {x0, x1, x2, x3, x4, x3, x2, x1}, respectively. Though the modes of /1 and /7 are the same number of the nodal diameter of j = 1 and the same frequency of x1, but the directions of the nodal diameters are different, in vertical or horizontal directions. The directions of other pairs for /2–/6 and /3–/5 behave in the same way. Note again that only the pair of eigen frequency and modes {x1, /1, /7} is the subject to be considered for the shaft (bending)—blade (j = 1) coupling effect in this section. Both components of in-plane (circumferential direction) and out-of-plane (axial direction) components are seen at the same time in an eigen mode, as drawn by arrows and the notation (±) for up-and -down, respectively. It is thus said that both components are superposed in an eigen mode shape.

H.2 Equation of Motion for an Entire Blading System

(1) Kinetic energy of ani-th blade Focusing on the i-th blade, the equation of motion is discussed for the formulation. The transformation of unit vectors between e0 of the inertia coordinate X0Y0Z0 and ei of the i-th blading coordinate XiYiZi is defined step by step, as shown in Fig. H.4:

e0 ¼ er T3 ðXtÞ for transformation from the inertia to the rotational coordinates,

ðH:4Þ

542

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies vib. displ. { ui , vi , w i }

Xi

P

Y0 Xi

Y0

τi

Yi

S

y

Yi

θxr

θx

l #i

xr

yr

#0

t

S

(1) Translating displ. ( x r , y r )

θ yr

t

θy

X0

x

Xr

τi

Xr

Yr

O

Yr

Xr

(2) Tilting angles ( θ x r , θ y r )

e0

er

e1

e2

ei

X0

Xr

X r1

X r2

Xi

Y0 Z0

T3 t

Yr Zr

X0

T1

θyr

T2

Yr1 Z r1

θ xr

Yr2 Z r2

T3

τi

Yi Zi

(3) Transformation of unit vectors

Fig. H.4 Definitions of coordinate system and unit vectors for blade vibrations


 er ¼ e1 T1 hyr ;

e1 ¼ e2 T2 ðhxr Þ

ðH:5Þ

for tilting motion of the shaft center and e2 ¼ ei T3 ðsi Þ

ðH:6Þ

for transformation for the 2nd to the i-th blading. On the i-th blading coordinate, we set mass displacement by {r + ui, vi, wi}, where writing r + ui as the radial geometrical position plus vibrational displacement of the mass, but no radial (u = 0) vibration is considered. We assume only circumferential (vi) and out-of-plane (wi) vibrations for simplicity. Therefore, the dynamic position of the mass is defined by the following form: 2

3 2 3 2 3 x0 xr r ! 6 7 6 7 6 7 OP  x0 i0 þ y0 j0 þ z0 k0  e0 4 y0 5 ¼ er 4 yr 5 þ ei 4 vi 5 z0 0 wi 02 3 2 31 xr r t 6 7C B6 7  t t t t ¼ e0 T3 ðXtÞ@4 yr 5 þ T3 ðsi ÞT2 ðhxri ÞT1 ðhyri Þ 4 vi 5A 0 wi

ðH:7Þ

The first part is the translational motion and the second part is the titling motion of the blade boss.

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

543

(2) Equation of motion for i-th blading The kinetic energy T is calculated by the following formula:  1
T ¼ m x_ 20 þ y_ 20 þ z_ 20 2

ðH:8Þ

The potential energy V due to the blading deformation is ignored at this stage, but it will be introduced later as a part of the stiffness matrix. The velocity of the absolute displacement {x0, y0, z0} of Eq. (H.7) is introduced into Eq. (H.8), and the following Lagrange equation is applied with respect to vibrational displacement variables {xr, yr, hrx, hry, vi, wi}: d @T @T @V  þ ¼Q dt @ q_ @q @q

ðH:9Þ

  where q ¼ xr yr hxr hyr vi wi , V ¼ 0 (temporarily) and Q ¼ 0. Then, by neglecting second order and higher of these small order variables, we obtain the equation of motion concerning the i-th blade mass: for the shaft translational motion, €i þ XGi V_ i  X2 Mi Vi ¼ 0 ði ¼ 1; 2; 3. . .7Þ Mi V where Vi  ½ xr

ðH:10Þ

v i t

yr

2

1 Mi  m4 0  sin si 2

0 Gi ¼ 2m4 1 cos si

sym: 1 cos si

1 0 sin si

3 5;

1

3  cos si  sin si 5 ¼ Gti 0

and for the shaft tilting motion, € i þ Ki Wi ¼ 0 Ii W where Wi ¼ ½ hxv

hyr

ði ¼ 1; 2; 3. . .7Þ

ðH:11Þ

wi t . 2

r 2 cos2 si 2 4 Ii  m r cos si sin si r cos si

r 2 sin2 si r sin si

sym: 1

3 5

544

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

2

r 2 cos2 si 24 2 Ki ¼ mX r cos si sin si r cos si

sym:

r 2 sin2 si r sin si

3 5

0

(3) Equation of motion for entire blading system Accordingly, for the global blading system, including the 0th to 7th blading, the total equation of motion is rewritten by superposing of Eqs. (H.10) and (H.11) as follows: 2

32 3 2 €xr 0 1ts 1tc 54 €yr 5 þ 2mX4 8 €v 32 3 1c E8 8 0 1ts xr 8 1tc 54 yr 5  mX2 4 0 1s 1c E8 v ¼0

8 0 1s

8 0 8 m4 0 1s 21c

2

4r2 4 m 0 r1c

ðH:12Þ

32 3 2 2 r1tc 4r h€xr r1ts 54 €hyr 5 þ mX2 4 0 E8 r1c € w

0 4r 2 r1s

32 3 1tc x_ r 1ts 54 y_ r 5 08 v_

0 4r 2 r1s

32 3 hxr r1tc r1ts 54 hyr 5 ¼ 0 w 08 ðH:13Þ

where, v ¼ ½ v0

   v7  t , w ¼ ½ w 0

v1

1tc ¼ ½    cos si 1ts ¼ ½    sin si

w1



w7 t

 pffiffiffi pffiffiffi pffiffiffi      ¼ 1 1= 2 0 1= 2    1= 2  pffiffiffi pffiffiffi pffiffiffi      ¼ 1 1= 2 0 1= 2    1= 2

E8 = 8-th dimensional unit matrix, 08 = 8-th dimensional zero matrix. Combining Eqs. (H.12) and (H.13) gives the coupled equation of motion with respect to the entire blading system and the shaft center movement: €20 þ XG20 X_ 20  X2 K20 X20 ¼ 0 M20 X where X20 ¼ ½ xr

yr

hxv

v 2

M20

8 6 0 6 6 1s ¼ m6 6 0 6 4 0 0

hyr 0 8 1c 0 0 0

ðH:14Þ

w t 1ts 1tc E8 0 0 0

0 0 0 4r 2 0 r1c

0 0 0 0 4r 2 r1s

3 0 0 7 7 0 7 7 r1tc 7 7 r1ts 5 E8

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

2

G20

0 68 6 6 1c ¼ 2m6 60 6 40 0 2

K20

8 6 0 6 6 1s ¼ m6 6 0 6 4 0 0

0 8 1c 0 0 0

8 0 1s 0 0 0 1ts 1tc E8 0 0 0

1tc 1ts 08 0 0 0

0 0 0 0 0 0

0 0 0 4r 2 0 r1c

0 0 0 0 0 0

545

3 0 07 7 07 7 07 7 05 08

0 0 0 0 4r 2 r1s

3 0 0 7 7 0 7 7 r1tc 7 7 r1ts 5 08

Note that this derivation is shown only for the mass part corresponding to the kinetic energy. The stiffness part, which would normally be derived from the potential energy, is ignored by setting V ¼ 0 (temporarily) at the beginning of Eq. (H.9). This procedure will then be compensated for, without the process of considering the blading potential energy in Lagrange’s equation, by straightforwardly inserting the modal stiffness at the last stage shown later in Eq. (H.19).

H.3 Mode Synthesis Model for Blading

(1) Coordinate transformation for mode synthesis model As shown in Fig. H.5, the blade is generally set with the stagger angle r. An eigen mode is thus divided into two parts: in- plane (circumferential) component and out-of-plane (axial) component. In this simple example, for a blade eigen mode shape /, the circumferential and axial components are expressed by /cos r = a = /v and /sin r = /b = /w, respectively. As shown in Fig. H.6, we focus on circumferential components of two eigen modes, noted /1v and /7v. In the mode shape of /1v, blades #0 and #4 have reverse phases at the antinode, and the sum of the movements of both masses results in the upward (+Yr) direction, as shown in the left figure (1). Blades #2 and #6 become nodes of one-nodal diameter. On the other hand, in mode shape of /7v, blades #0 Yr

Fig. H.5 Blade setting angle, called stagger angle

φ1

σ

l Zr

Xr

546

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies mode displacement

Fig. H.6 One-nodal diameter mode (in-plane)

mode displacement

(1)

(2) #2

0.7

1.0 #2

0.7

#3

#3

#1

#1

1.0

0.7

1.0 #4

#0

#4

#5

#5

#7

#6

vi

φ 1 v mode

#0

φ 7 η rx

φ 1 η ry

#6

#7

φ 7 v mode

and #4 become nodes of one-nodal diameter, and blades #2 and #6 have reversed phases at the antinode. The sum of both masses acts in the leftward (−Xr) direction, as shown in right figure (2). In this manner, it can be concluded for these eigen modes (j = 1) that the sum of all the blade mass movements are equivalent to the movement of the center of gravity of each blade; /1v mode for the (yr) direction and /7v for the (−xr) direction. In a similar manner, the behavior of vibrations in the out-of-plane direction is considered concerning noted /1w and /7w, as shown in Fig. H.7. Blade #4 moves as if sinking when blade #0 rises in mode /1w. All tilting movements of the blades are, thus, equivalent to the behavior of reverse rotation about the Yr axis, i.e., (−hxr) rotation, as shown in left figure (1). In mode /7w, they are also equivalent to (−hyr) rotation as shown in right figure (2). mode displacement

mode displacement

Yr

(1)

(2)

θ xr θyr

Xr

φ 1 w mode

wi

φ 1 η rx

φ 7 η ry

Fig. H.7 Tilting mode of one-nodal diameter (out-of-plane)

φ 7 w mode

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

547

Therefore, when rewriting the vibration of v and w by modal coordinate ηx and ηy based upon the equivalence mentioned above, the coordinate transformation is defined as follows: 2

3 2 1 xr 6 yr 7 6 0 6 7 6 6 v 7 60 6 76 6 hxr 7 6 0 6 7 6 4 hyr 5 4 0 w 0

0 1 0 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0 0 /7 a 0 0 /1 b

2 3 32 3 xr xr 0 6 yr 7 6 yr 7 0 7 6 7 76 7 6 7 6 7 /1 a 7 76 hxr 7  W6 hxr 7 6 7 6 hyr 7 7 0 76 hyr 7 6 7 4 4 gxr 5 5 5 gxr 0 gyr gyr /7 b

ðH:15Þ

(2) Equation of motion on the rotational coordinate, er The transformation by W provides the following equations of motion described on the rotational coordinate, er, €r  XGr Y_ r  X2 K0r Xr ¼ 0 Mr X Mr Y€r þ XGr X_ r  X2 K0r Yr ¼ 0

ðH:16Þ

where Xr ¼ ½ xr 2

8m ¼ mt Mr ¼ 4 0 4am

hxr

0 4mr ¼ Id 4bmr 2

2

16m 0 0 Gr ¼ 4 0 8am 0 2

8m ¼ mt K0r ¼ 4 0 4am

gxr t ;

 Yr ¼ yr

hyr

gyr

t

3  4am Mr 5 from 4bmr 0 4mða2 þ b2 Þ ¼ m 3  8am 0 5 0 from Gr 0

0 4mr 2 ¼ Id 4bmr

Gr 0

0 Mr

 ¼ Wt M20 W

 ¼ Wt G20 W

3  4am K0r 5 4bmr from 0 4ma2

0 K0r

 ¼ Wt K20 W

The above equations can be rewritten in terms of complex forms: Zr ¼ ½ z r

hr

gr  t

ðH:17Þ

548

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

where zr ¼ xr þ jyr ; hr ¼ hxr þ jhyr ; gr ¼ gxr þ jgyr , which gives the following complex equation of motion: Mr Z€r þ jXGr Z_ r  X2 K0r Zr ¼ 0

ðH:18Þ

Now, we review some elements of these reduced matrices. Mr(1,1) is equal to the total mass mt and Mr(2,2) to the transverse moment of inertia Id, obtained by
 assuming the rigid blading system. Mr(3,3) is the modal mass m ¼ 4m a2 þ b2 of eigen mode /1. The modal stiffness is defined by k  ¼ m x2b , where the blade frequency xb should be identical to the eigen frequency x1 of the first mode. This frequency is variant depending the rotational speed due to centrifugal force and longitudinal force generated in the blading structure. However, we assume the eigen mode shapes are invariant throughout the entire range of rotational speed. Therefore, we directly reset K0r(3,3) = k*, and the above equation of motion is finally replaced by the following form on er: Mr Z€r þ jXGr Z_ r þ Kr Zr ¼ 0 where


Kr ¼ X2 K0r



2

mt X2  4 and ðreset Kr ð3; 3Þ ¼ k Þ ¼ 0 4amX2

ðH:19Þ

0 Id X2 4bmrX2

3 4amX2 4bmrX2 5 m x2b

(3) Equation of motion in the inertial coordinate system, e0 The displacement relationship between the rotational coordinate er and the inertial coordinate e0: Z ¼ Zr ejXt

ðH:20Þ

yields the following equation of motion: M Z€  jXGZ_ þ KZ ¼ 0 in e0 where Z ¼ ½ z 2

h

mt 6 M ¼ Mr ¼ 4 0 4am

ðH:21Þ

g t 0 Id

3 4am 7 4mrb 5;

m 2 0 0 6 2 2 K ¼ K r  X G r  X Mr ¼ 4 0 0

2

0 0 6 G ¼ Gr þ 2Mr ¼ 4 0 2Id ¼ Ip

4mrb

0

0 0

0 k   m X2

3 7 5

0

8mrb

3 0 7 8mrb 5 2m

Appendix H: Mode Synthesis Modeling for Rotational Blade Assemblies

549

Since a blading system is equivalent to a thin disk, the polar moment of inertia (Ip) is twice the transverse moment of inertia (Id), and the gyroscopic factor is thus c = Ip/Id = 2. It is to be understood that twice the 2  2 matrix belonging to the right-lower corner in the mass matrix M agrees with the same part of the gyroscopic matrix G, i.e., 

Id 4mrb

  4mrb 2Id ¼ Ip in M , m 8mrb

 8mrb in G: 2m

ðH:22Þ

Referring to the use of coupling terms {a1, a2, a3, a4} in Eq. (11.81) and Table 11.5, a more general description is proven as follows: 2 6 4

M¼ mt 0

0 Id

a1  ja3

a2  ja4

3

G¼

2

0 a1 þ ja3 7 6 a2 þ ja4 5 , 4 0 m

0

0 Ip

3 0 7 2ða2 þ ja4 Þ 5

2ða2  ja4 Þ

ðH:23Þ

2m

With respect to the natural frequency of uncoupled blading system, it is xb when measuring blade vibration by the strain gauge method (i.e., on er). This uncoupled eigen frequency corresponds with the root of the characteristic equation only for the (3,3) part of Eq. (H.19):
 m s2 þ k ¼ m s2 þ m x2b ¼ m s2 þ x2b ¼ 0

in er

However, if we measure the uncoupled vibration of blading by a gap sensor from a stationary part (i.e., on e0), the eigen frequency will be xb  X. This is obvious from the characteristic equation only for the (3,3) part of Eq. (H.21):

References

The following abbreviations are used in the body text, reference cited and bibliography. R1 R2

VB APC ICORD ISCORMA ISMB VIRM

Vibrations of rotating machinery volume 1. Basic rotordynamics: introduction to practical vibration analysis (2017) Springer Vibrations of rotating machinery volume 2. Advanced rotordynamics: applications, troubleshooting and diagnosis (2019). Springer (The previous book, Vol. 1, and this Vol. 2 are denoted by R1 and R2, respectively) Data of vibration database in The Japan Society of Mechanical Engineers (JSME) American Power Conference International Conference on Rotor Dynamics International Symposium on Stability Control of Rotating Machinery International Symposium on Magnetic Bearings International Conference on Vibrations in Rotating Machinery

R2 References (JSME v_BASE Data Book)1 VB001. VB004. VB010. VB017. VB030. VB037.

Vibration due to oil whip of steam turbine Self-excited vibration caused by labyrinth seal of steam turbine rotor Thermally induced bending vibration in large turbine due to contact with bearing seal Tie bolt resonance in assembled rotor Hysteresis whip in gas turbine connected with gear coupling Baseplate vibration of compressor driven by reciprocating engine

1

https://www.jsme.or.jp/dmc/Links/vbase/data_english.html (Accessed on 31 Mar 2018)

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

551

552

VB038. VB055. VB058. VB065. VB078. VB079. VB083. VB102. VB117. VB120. VB136. VB190. VB197. VB218. VB245. VB247. VB268. VB304. VB369. VB402. VB407. VB423. VB518. VB547. VB550. VB608. VB621. VB629. VB675. VB678. VB752. VB765. VB828 VB958.

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Index

A Acoustic Emission (AE) sensor, 6 Active Magnetic Bearing (AMB), 119, 127, 128, 298, 536 Added mass effect, 256 Aircraft-engine, 2, 240 ALARM (value), 172, 421 Alford force, 230 Aliasing, 317, 318 Aliasing error, 320, 509 Alignment, 367 Aluminum (Al) shaft, 285 American Petroleum Institute (API), 9 Amplitude Modulation (AM), 102, 104, 468 Anisotropy, anisotropic, 36, 40, 54, 76, 77, 305, 381, 384 Anti-aliasing (filter), 105, 320–322, 462, 470, 510 API 617, 10, 232 Aseismic control, 162 A-set, 356, 357, 392 Asymmetric rotor, 15, 183, 185 Asymmetry of modal stiffness, 186 Attitude angle, 21, 26, 254 Average fluid velocity, 244 Axial flow compressor, 204, 230 Axial flow turbine disk, 229 B Backward, 46, 95, 201, 229, 291, 473 Backward complex amplitude, 307 Backward whirl, 45, 77, 153, 201, 229, 240, 291, 297, 306, 346, 347, 380, 387 backward-oriented orbit, 306 Balance analyzer, 301

Balancer (low speed, high speed), 175 Balancing, 9, 76, 149, 175, 413, 418, 457, 478 Ball bearing, 88, 136, 143, 161, 298, 305, 405, 476 Ball Pass Frequency of Inner race (BPFI), 95, 96 Ball Pass Frequency of Outer race (BPFO), 95, 96 Ball Spin Frequency (BSF), 96 Band Pass Filter (BPF), 104 Base circle, 189 Baseline, 172 Bearing alignment, 37 Bearing dynamic (coefficients, properties, characteristics), 370, 379, 381 Bearing pad, 225 Belt drive, 211, 474 Belt rotation frequency, 211 Belt tension control, 213 Bending stiffness, 53, 63, 284, 363 Bending-torsional coupled system, 290 Bending-torsional coupled vibration, 287, 290 Bently-Muszynska (B-M) model, 244 Bent shaft, 106 Bias current, 125 Bipolar (2 pole) machine, 198 Blade and disk (coupled vibration), 208 Blade-and-shaft coupled (system vibration), 390, 402, 538 Blade Pass Frequency (BPF), 206, 474, 506 Blade-shaft-coupled (mode, model, vibration), 277 Blade-shaft-coupled system, 275, 286 Blade-shaft-coupled torsional vibration, 275, 279

© Springer Japan KK, part of Springer Nature 2019 O. Matsushita et al., Vibrations of Rotating Machinery, Mathematics for Industry 17, https://doi.org/10.1007/978-4-431-55453-0

571

572 Block diagram, 128, 305, 441, 443 Bode plot, 142, 304, 445, 469 Boiler Feed Pump (BFP), 207 Bowed rotor, 15, 181 Breakdown Maintenance (BM), 6 Bulk flow theory, 226 Butterworth filter, 471, 509 C Campbell diagram, 16, 212, 274, 283, 439, 535 Cardan joint, 194 Carrier ring, 36 Case studies of vibration troubleshooting, 17 Casing, 1, 2, 111, 163, 187, 274, 382, 434, 542 Casing-rotor-bearing system, 383, 541 Casing whirl, 383 Category, 12, 413 Causal relationship matrix, 14 Cause-and-effect matrix, 15 Centering spring, 252, 256 Center pivot, 32 Centrifugal compressor, 136, 162, 209 Centrifugal (force, field), 14, 87, 161, 205, 239, 245, 552 Centrifugal impeller, 231 Centrifugal pendulum, 214 Centrifuge, 161 Certification system (of vibration diagnostics engineers), 12 Cessation of oil whip, 250 CFD, 226, 230 Characteristic equation, 64, 553 Circumferential oil feed groove, 253 Clearance, 50, 65, 88, 388 Complex amplitude, 48 Complex eigenvalue, 3, 45, 80, 146, 224, 295, 370, 462 Complex eigenvalue analysis, 3, 80, 81, 223, 224, 226, 230, 232, 370, 376 Condition Based Maintenance (CBM), 6, 7 Condition Monitoring System (CMS), 6, 10, 12, 171, 471 Contact angle, 87, 94, 112, 476 Contact-type seal, 243 Control option, 158 Coriolis force, 245 Correction weight, 176 Coupled mode, 279, 404, 409 Coupling, 164, 263, 265, 457 Coupling mass, 390, 402 Cracked rotor, 188 Criterion I,II, 172 Critical speed, 61, 129, 149, 359, 423 Critical speed map, 137, 146

Index Cross-control, 140, 152, 156, 161 Cross-coupled/coupling term, 223 Cross-coupled effect, 50 Cross-coupled stiffness, 59, 72, 74 Cross-coupling (spring) force, 151, 220, 223, 229, 245, 445 Cross joint, 194 Cross pin, 194 Cross-spring, 151, 155 Crowning, 235 Current Transformer (CT), 201 Curvic coupling, 176 Cut-off frequency, 320 4 cycle engine, 210 Cylindrical bearing, 20, 53, 68, 515 Cylindrical bearing with two oil grooves, 33, 72 D Damper Reynolds number, 257 Damper seal, 227 Damping capability, 145 Damping coefficient, 30, 63, 255, 256, 447 Damping ratio, 424, 426, 432 Defects in rotor bar, 200 Degradation, 107, 108 bearing degradation, 107–110 Destabilizing (fluid) force, 60, 79, 219, 226, 229, 231 1D-FEM, 364, 408 3D-FEM, 392, 408 Diagnosis, 13, 175, 201 Diagnostics, 9, 10, 12 Diaphragm coupling, 236 Diesel (engine generator, generator set), 213 Diffuser pump, 206 Diffuser vane, 209 Dimensionless, 30, 51, 53, 65, 66 Dimensionless bearing modulus, 67 Dimensionless major vibration amplitude, 53 Dimensionless shaft speed, 53, 61 Dimensionless stability threshold shaft speed, 66 Direct coefficient, 72 Direct stiffness coefficient, 36, 74 Direct viscous damping coefficient, 74, 77 Discoloration, 110 Discrete Fourier Transformation (DFT), 316, 320, 327, 337, 526 Displacement sensor, 173 Dissipated energy, 73 Dominant frequency, 100, 324, 419 Drain hole, 239, 253 Drive/Driving/Driven shaft, 194

Index Dry screw compressor, 110 Dynamic damper, 213, 437 Dynamic eccentricity, 199, 466 Dynamic oil film reaction force, 28 Dynamic properties of bearing pedestal, 56 Dynamic properties (of oil film), 47, 48, 52 Dynamic stiffness, 47, 50 dimensionless dynamic stiffness, 50 E Effective asymmetry, 186 Effective value of vibration velocity, 171 Eigen mode, 136, 142, 262, 358, 435, 544 Electric motor, 198, 436, 506 Electromagnetic vibration, 198 Elliptical coordinates, 517 Elliptical deformation of circular Nyquist plot, 185 Elliptical (whirl) orbit, 44, 46, 49, 72, 306 Ellipticity, 183, 186 End seal, 252, 253, 257, 258 Energy balance, 72, 77 Engine shaft system, 265 Envelope, 84, 103, 104, 329 Equation of continuity, 23 Equivalent mass (model), 281, 435 Equivalent oil film coefficients, 49, 51 Estimation of the causes, 5 Excessive vibration, 56, 175, 275 Excitation torque, 213, 215, 289, 439 Exciting (excitation) force, 17, 45, 47, 49, 51, 78, 104, 114, 163, 175, 206, 438, 458 Exciting frequency, 95, 274, 437 Exciting source, 102, 189, 307, 437 External damping, 294 External excitation signal, 131 F Fast Fourier Transformation (FFT), 16, 84, 99, 101–105, 107–109, 111, 112, 131, 132, 161, 162, 183, 184, 267, 270–272, 283, 284, 301, 316, 325–327, 329-338, 341–344, 346–350, 413, 414, 418–421, 429, 436, 460–465, 470, 471, 473–478, 502, 505, 508, 510, 519, 525, 535, 536 Feedback (FB) control, 119, 127, 302 Feed Forward (FF) control, 159, 163 FFT (analyzer), 84, 316, 325, 338, 413, 419, 460, 466, 471, 505 Field balancing, 137, 175, 306, 450 Finite difference method, 24 Finite Element Method (FEM), 75, 285 First bending, 61 First critical speed, 42, 61, 237, 239, 475

573 Five-pad tilting pad bearing (LBP, LOP), 54, 55, 78 Flat top window, 329, 462, 525 Flow-excited vibration, 77 Fluid-containing rotor, 237 Fluid force balancing, 207 Forward, 46, 291, 473 Forward complex amplitude, 307 Forward-oriented orbit, 305 Forward whirl, 45, 77, 152, 220, 222, 347, 380, 387, 510 Four-cycle engine, 210 Fourier Series Expansion (FSE), 312, 321, 340, 438, 519 Fourier transform, 156, 330, 333 Fourier transformation, 317, 519 Four-pad tilting pad bearing, 36, 53, 55, 78 Free-free, 146, 280, 453 Frequency domain, 326, 341, 460 Frequency Modulation (FM), 298 Frequency resolution, 324, 330, 342, 460 Frequency Response Analysis (FRA), 339, 340, 442, 439 Frequency Shifting, 342 Friction clutch, 263 Friction whip, 15, 240, 242 Friction whirl, 241 Full spectrum, 347 Fundamental Train Frequency (FTF), 94, 512 G Gain margin, 130, 143 Gas turbine, 2, 10, 170, 175 Gas whirl, 218 Gear, 111, 189, 434 Gearbox, 269 Gear coupling, 235 Geared compressor, 191 Geared shaft system, 267 Gear error, 189 Gear pulse, 265 Gümbel condition, 25 Guyan reduction, 354, 403 Gyroscopic (matrix, term), 43, 52 H Half speed whirl, 222, 243 Hammer test, 217 excitation test, 217, 219, 423 Handheld vibration meter, 8 Hanning window, 327, 329, 460, 525 Helical gear, 190 Helix angle, 190 High-pressure fluid-induced vibration, 226

574

Index

High Pressure (HP) rotor, 2, 111, 136, 282 High-Speed (Dynamic) Balancer (HDB), 175 Hilbert transform, 431 Hollow rotating shaft, 237 Homopolar, 127 Honeycomb seal, 228 Hood damper, 213 Hooke's joint, 194 Hunting, 198 Hydrodynamic lubrication theory, 22 Hydrodynamic pressure distribution, 20 Hydro whirl, 218 Hysteresis whip, 234

ISO 21940-12, 9 ISO 21940-31, 9 ISO 22266, 275, 404 ISO 7919, 9, 11, 170, 173 JIS B 0910, 9, 11, 169, 173 Isotropy, Isotropic, 36, 40, 47, 77, 139, 305

I International Electrotechnical Commission (IEC), 9, 10 Impact vibration (response), 95 Impeller, 15, 176, 183, 206, 209, 221, 227, 229, 234, 396, 474 Impulse induced vibration, 103 Impulse test, 338, 420, 425, 529 Induction machine, 198 Inertia coefficient, 48, 256 Inertia matrix, 50 Infinitely short bearing approximation, 24, 31 Influence coefficient, 175 Infrared thermography, 6 Inlet Guide Vane (IGV), 209 Inner race defect/damage, 109 In-phase (mode group), 147, 280, 409 In-plane, 287, 393, 399, 401, 545, 550 In-situ balancing, 175 Interaction between rotor blade and stator vane, 208 Interference with diffuser, 232 Intermediate shaft, 236 Internal damping, 15, 220, 234 International Organization for Standardization (ISO), 8, 115, 169 Iron core, 115 ISO 10814, 9 JIS B 0911, 9 ISO 10816, 9, 11, 170 JIS B 0906, 9, 11, 169, 171 ISO 13373, 10, 11, 13, 173 ISO 14839, 9, 127 ISO 15, 88 ISO 18436 10, 12, 414 ISO 1940, 9, 458 ISO 2041, 9 JIS B 0153, 9 ISO 20816, 9, 11 ISO 21940-11, 9

K Key pivot, 36 Keyway, 183

J Japanese Industrial Standard (JIS), 88 Jeffcott rotor, 52, 63, 430 Joint angle, 194 Journal center locus, 63 Journal eccentricity ratio, 24, 48, 60

L Labyrinth seal, 16, 162, 221, 226 Lagrange's equation, 179, 183, 215, 439, 547 Laminated leaf spring coupling, 236 L/D, 28 Leakage flux, 122 Linearization of magnetic force, 125 Linear vibration (theory, analysis), 59, 63, 70 Load Between Pads (LBP), 36, 53–55, 78, 228 Load direction, 35, 36 Load on Pad (LOP), 36, 54, 55, 78, 228 Long bearing width theory, 257 Loop transfer function open-loop transfer function, 130, 441 Loose fitting, 114 Low Pressure (LP) rotor, 2, 136, 271, 283 Low-Speed (Dynamic) Balancer (LDB), 176 M Magnetic attraction force, 117, 122 Magnetic bearing, 9, 118, 305, 360 Magnetic circuit, 115 Magnetic flux (density), 115, 117 Magnetic gap, 117, 199 Magnetic pole, 119, 199 Magnetic saturation, 126 Magnetic steel sheet, 119, 123 Magnetomotive force, 115 Magnification factor (Q-value), 42 Main lobe, 525 Main peaks, 101 Mass matrix, 43, 364, 553 Mathieu equation, 212 Mean bearing radial clearance, 24, 27 Mean radial clearance of bearing, 65 Mesh frequency, 189, 506

Index Meshing of gears, 189 Mesh stiffness, 189 Minimum oil film thickness, 21, 37, 76 Mirror phenomenon, 318, 326 Misalignment, 5, 15, 194, 339 Modal balancing, 310 Modal influence coefficient method, 175 Mode control, 140, 147, 368, 445 Model Order Reduction (MOR), 362, 409 Mode separation, 366 Mode synthesis, 286, 354, 391, 447, 543, 549 Module, 190 Morton effect, 182 Motor Current Signature Analysis (MCSA), 201 Multi-lobe bearing, 33, 71, 223 Multi-pad bearing, 34, 76 Multi-pad fixed-bore bearing, 61 MyROT, 75 N N + 2 (plane) balancing, 359, 456 Nastran, 356 Natural frequency, 53, 65, 91, 131, 262, 338, 372, 356, 424 Navier-Stokes equations, 22 N-cross, 158 N-cut, 158 Newkirk effect, 181 Nodal diameter, 208, 474, 545, 550 Noise, 17, 100, 191, 201, 323, 475 Non-contact seal, 228 Non-synchronous vibration, 217, 512 Number of combustion cylinders, 210 NX vibration, 301, 535 Nyquist plot (or diagram), 133, 143, 304, 308, 448, 468 O Offline monitoring, 7, 8 Offset, 34, 71, 107 Offset bearing, 224 Offset pivot, 37 Oil feed hole, 252 Oil film, 515 Oil film bearing, 60, 127, 305, 370, 474 Oil film coefficient, 30, 51, 65 Oil film rupture, 25, 253, 255, 338 air bubble, 255 cavitation, 5, 13, 339 Oil film thickness, 22, 24 Oil groove, 69 Oil whip, 42, 60, 154, 222, 243, 347, 370, 462 Oil whirl, 62, 222, 243, 370, 462

575 Online monitoring, 7 Open-loop characteristics, 131, 144 Open-loop transfer function, 130, 441, 489 Optional control, 158 Outer race fault, 512 Out-of-phase (mode group), 147, 280, 410 Out-of-plane, 400, 545, 550 Overall (OA), 73, 125, 334, 415, 420, 460, 463, 475 Overlap contact ratio, 190 P Pad (arrangement), 32, 35, 36, 72 number of pad, 35, 36 Pad preload factor, 35 Parametric excitation, 212 Partial arc bearing, 49 Partition plate, 228, 239 Pass frequency, 96, 102, 506 Peak (amplitude), 42, 54, 157, 283, 415 Peak-to-Peak (PP) maximum value of peak-to-peak displacement, 173 Peak-(to-) peak (displacement, amplitude), 173, 302, 325, 415, 416 Peak-peak amplitude, 325 Pedestal stiffness, 56 bearing pedestal stiffness, 56, 57 Phase angle of unbalance, 43 Phase Locked-Loop (PLL), 302 Phase margin, 130, 143, 443 2p/p film model, 255, 256 Pinion shaft, 192 Pinned rotor, 53 Piston rotary machine, 214 Pitch error, 191 Pivot, 34, 72 Plain bearing, 60, 472 PLL circuit, 302, 533 Poles, 127, 462 Polor plot, 304 Precise diagnosis, 100 Predictive maintenance, 6, 414 Preload, 34, 71, 113, 191, 225 Preload factor, 32, 55, 71, 74 Pressure distribution, 20 Preventive Maintenance (PM), 6 Primary circuit (stator-side coil), 199, 202 Pulley, 211, 474 Pump, 7, 77, 110, 178, 206, 224, 236, 374, 480 Q Q-set, 357, 393 Quadrupolar (4 pole) machine, 198

576 Q-value (Q-factor), 146, 155, 157, 424, 490 R Radial stiffness, 89 Radial thrust, 206 Rathbone chart, 170 Rectangular window, 328, 502, 524 Resolution, 103, 324, 503, 530 Retainer adhered, 98, 114 Retainer (cage) defect/damage, 109 Reynolds condition, 26, 31 Reynolds equation (classical lubrication theory), 23, 24, 28 Reynolds number, 31, 39 Ringing, 99 Ringing frequency, 104 Ringing vibration, 99 race vibration, 99, 104 Risk Based Maintenance (RBM), 6 Rolling contact (whirl), 241, 243 Rolling (element) bearing, 16, 87, 102, 103, 120, 127, 381 Rolling pivot, 36 Root loci, 378 Root Mean Square (RMS), 171, 325, 334, 415, 429 Roots blower, 211 Rotating magnetic field, 198 Rotating stall, 15, 204 Rotational coordinates, 406, 545 Rotational frequency, 93, 438 Rotor, 1, 383, 536 Rotor-bearing system, 21, 49, 178, 383 Rotor blade, 204, 206, 221 Rotor system, 298, 302, 368 Routh-Hurwitz (criterion), 66 Rub, 114 Rubbing (vibration), 347, 349, 350 Runout error, 191, 468 S Sampling value, 319, 347 Saturation, 124 SC2, 9 SC5, 9 Secondary circuit (rotor-side circuit), 201, 203 Secondary critical speed, 183 Seismic excitation, 164 Seismic pickup, 171 Self-excited vibration, 60, 152, 218, 372 Sensitivity function, 130, 149 Shaft, 42, 87, 118, 261, 305, 362

Index Shaft-blade coupling vibration, 402 Shaft elasticity index, 53, 65 Shifting from oil whirl to oil whip, 250 Short bearing width theory, 257 Shunt hole, 228, 231 Sideband (frequency), 101, 200, 328 Side by side control, 139, 368 Side lobe, 328, 330, 527 Signal to Noise Ratio (S/N ratio), 323 Single shaft system, 262 Slip-beat sound, 201 Slip frequency, 199 double slip frequency, 202 Slip ratio, 199 Sloshing frequency, 239 Sommerfeld condition, 25 Sommerfeld number, 27, 32, 48 modified Sommerfeld number, 32, 33, 39 Sommerfeld's substitution, 256 Specific bearing load, 39 Spectral analysis, 338 Spiral vibration, 183 Spot scratch, 95 Spur gear, 189 Squeeze film damper, 228, 251 Stability, 3, 63, 127, 290, 370 Stability(bending-torsional coupled vibration), 290 Stability chart (criteria), 67, 375, 389 Stability margin, 130 Stability threshold shaft speed, 66 Stagger angle, 401, 549 Static (bearing) load, 26, 28, 30, 36, 76 Static eccentricity, 199, 466 Stator, 1 Steady-state characteristics, 24 Steady-state equilibrium, 26, 63 Steady-state oil film reaction force, 26 Steam turbine, 11, 170, 218, 227, 275, 281 Steam whirl, 218 Step-out, 198 loss of synchronization, 198 Stiffening, 178, 179 Stiffness coefficient, 30, 61 Stiffness matrix, 43, 262, 355, 547 Straight control, 139, 151, 152 Strain gauge, 264, 553 Sub-Committee (SC), 8 Sub-synchronous vibration, 217 Swirl, 218, 221, 226, 231 Swirl breaker (Swirl brakes), 228 Symbol of AMB, 127

Index Synchronous Machine, 198 System damping, 56, 78, 145 T TC108, 8, 13, 115 Technical Committee (TC), 8 Thermal bending vibration, 181 Thomas force, 230 Three-lobe bearing, 72 Thrust stiffness, 92 Tie bolt, 176 Tilting pad (journal) bearing, 32, 34, 61, 223, 225, 229 Time averaging, 304, 323 Time-Based Maintenance (TBM), 6, 7 Time domain, 102, 156, 317, 341, 418 Tip clearance, 229 Tongue part, 206 Tooth profile deviation, 191 Torsional dynamic damper, 213 Torsional model, 262 Torsional viscous damper, 213 Touch-down bearing, 136, 243 Tracking filter, 155, 303, 535 Transfer function, 130, 139, 147, 151, 339 Transfer matrix method, 75 Transverse contact ratio, 190 Trend, 15, 422 Trial weight, 175, 427 TRIP value, 172 Troubleshooting, 4, 17, 108, 301 Turbine, 270, 395 Turbine generator set, 270 Turbocharger, 1, 2, 16 Turbulent lubrication theory, 32 Turning operation, 181 Twice slip frequency, 12, 23, 202 Two-lobe bearing, 35, 54, 71 Two mode group, 147 Two-pole machine, 201 U Unbalance, 130, 270, 305 Unbalance mass, 42, 311 Unbalance vibration (response), 5, 16, 49, 60, 144, 175, 176, 178, 183, 242, 290, 306, 423, 475

577 1N vibration, 301 1X vibration, 301 Universal joint, 194 Unstable vibration, 60, 161, 218, 370, 537 self-excited vibration, 3, 69, 72 V Vector monitor, 301 Verein Deutscher Ingenieure (VDI), 170 Vertical pump, 178, 224, 233, 236, 434, 480 Vibration evaluation vibration criteria, 8, 11, 131, 167, 170 Vibration database (v_BASE) Committee, 17 Vibration evaluation (criteria), 8, 11, 129, 170 Vibration response, 49, 155 Vibration severity, 170 Viscosity, viscosity coefficient, 22–24, 69 Viscous damping coefficient, 30, 63, 142 Volute pump, 206 W Wachel's formula, 232 Waterfall diagram, 162 Water-lubricated bearing, 48 Wheel gear, 192 Whirl amplitude, 42, 73, 84 Whirl frequency ratio, 67 Whirl orbit, 44, 46, 72, 77, 305, 306, 308, 347, 349, 423, 473 See also Lissajous orbit orbit, 41, 44, 46, 50, 53, 59, 60, 77, 84, 86, 233, 306, 308, 310, 349, 351, 423, 510 Window function, 327, 460, 527 Window time, 104, 338, 463 Work done by oil film, 222 Working Group (WG), 8, 115 X X-Y anisotropy, 42 XY cross control, 139 XY Straight control, 139 Z Zone, 103, 172 Zoom, Zoomed, Zooming, 109, 341, 346, 426, 465