A must-read for any practicing engineer or student in this area There is a renaissance that is occurring in chemical and
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Table of contents :
Cover
Half-Title Page
Series Page
Title Page
Copyright Page
Contents
Preface
Acknowledgments
13 Rules of Thumb—Summary
13.0 Introduction
COMPRESSORS, FANS, BLOWERS, AND VACUUM PUMPS
CONVEYORS FOR PARTICULATE SOLIDS
COOLING TOWERS
CRYSTALLIZATION FROM SOLUTION
DISINTEGRATION
TOWERS
TRAY TOWERS
PACKED TOWERS
DRIVERS AND POWER RECOVERY EQUIPMENT
DRYING OF SOLIDS
EVAPORATORS
EXTRACTION, LIQUID–LIQUID
FILTRATION
FLUIDIZATION OF PARTICLES WITH GASES
HEAT EXCHANGERS
INSULATION
MIXING AND AGITATION
PARTICLE SIZE ENLARGEMENT
PIPING
PUMPS
REACTORS
REFRIGERATION
SIZE SEPARATION OF PARTICLES
UTILITIES, COMMON SPECIFICATIONS
VESSELS (DRUMS)
VESSEL (PRESSURE)
VESSELS (STORAGE TANKS)
14 Process Planning, Scheduling, and Flowsheet Design
14.1 Introduction
14.2 Organizational Structure
14.2.1 Process Design Scope
14.3 Role of the Process Design Engineer
14.4 Computer-Aided Flowsheeting
14.5 Flowsheets—Types
14.5.1 Block Diagram
14.5.2 Process Flowsheet or Flow Diagram
14.5.3 Piping Flowsheet or Mechanical Flow Diagram, or Piping and Instrumentation Diagram (P&ID)
14.5.4 Combined Process and Piping Flowsheet or Diagram
14.5.5 Utility Flowsheets or Diagrams (ULDs)
14.5.6 Special Flowsheets or Diagrams
14.5.7 Special or Supplemental Aids Plot Plans
14.6 Flowsheet Presentation
14.7 General Arrangements Guide
14.8 Computer-Aided Flowsheet Design/Drafting
14.9 Flowsheet Symbols
14.10 Line Symbols and Designations
14.11 Materials of Construction for Lines
14.12 Test Pressure for Lines
14.13 Working Schedules
14.14 Information Checklists
14.15 Basic Engineering and Front End Engineering Design (FEED)
References
15 Fluid Flow
15.1 Introduction
15.2 Flow of Fluids in Pipes
15.3 Scope
15.4 Basis
15.5 Incompressible Flow
15.6 Compressible Flow: Vapors and Gases [4]
15.7 Important Pressure Level References
15.8 Factors of “Safety” for Design Basis
15.9 Pipe, Fittings, and Valves
15.10 Pipe
15.11 Total Line Pressure Drop
15.11.1 Relationship Between the Pipe Diameter and Pressure Drop (ΔP)
15.11.2 Economic Balance in Piping and Optimum Pipe Diameter
15.12 Reynolds Number, Re (Sometimes Used NRe)
15.13 Pipe Relative Roughness
15.14 Darcy Friction Factor, f
15.15 Friction Head Loss (Resistance) in Pipe, Fittings, and Connections
15.15.1 Pressure Drop in Straight Pipe: Incompressible Fluid
15.16 Oil System Piping
15.16.1 Density and Specific Gravity
15.16.2 Specific Gravity of Blended Products
15.16.3 Viscosity
15.16.4 Viscosity of Blended Products
15.16.5 Blending Index, H
15.16.6 Vapor Pressure
15.16.7 Velocity
15.16.8 Frictional Pressure Drop, ft of Liquid Head
15.16.9 Hazen–Williams Equation
15.16.10 Transmission Factor
15.16.11 Miller Equation
15.16.12 Shell–MIT Equation
15.17 Pressure Drop in Fittings, Valves, and Connections
15.17.1 Incompressible Fluid
15.17.2 Velocity and Velocity Head
15.17.3 Equivalent Lengths of Fittings
15.17.4 L/D Values in Laminar Region
15.17.5 Validity of K Values
15.17.6 Laminar Flow
15.17.7 Expressing All Pipe Sizes in Terms of One Diameter
15.17.8 Loss Coefficient
15.17.9 Sudden Enlargement or Contraction
15.17.10 For Sudden Contractions
15.17.11 Piping Systems
15.18 Resistance of Valves
15.19 Flow Coefficients for Valves, Cv
15.20 Flow Meters
15.20.1 Process Design of Orifice Meter
15.20.2 Nozzles and Orifices
Conclusion
15.21 Estimation of Pressure Loss Across Control Valves
Liquids, Vapors, and Gases
15.22 The Direct Design of a Control Valve
15.23 Water Hammer
15.24 Friction Pressure Drop for Compressible Fluid Flow
Vapors and Gases
15.24.1 Compressible Fluid Flow in Pipes
15.24.2 Maximum Flow and Pressure Drop
15.24.3 Sonic Conditions Limiting Flow of Gases and Vapors
15.24.4 The Mach Number, Ma
Flow Rate of Compressible Isothermal Flow
Pipeline Pressure Drop (.P)
15.24.5 Critical Pressure Ratio
15.24.6 Adiabatic Flow
15.24.7 The Expansion Factor, Y
15.24.8 Misleading Rules of Thumb for Compressible Fluid Flow
15.24.9 Other Simplified Compressible Flow Methods
15.24.10 Friction Drop for Flow of Vapors, Gases and Steam
15.25 Darcy Rational Relation for Compressible Vapors and Gases
15.26 Velocity of Compressible Fluids in Pipe
15.27 Procedure
15.28 Friction Drop for Compressible Natural Gas in Long Pipe Lines
15.29 Panhandle-A Gas Flow Formula
15.30 Modified Panhandle Flow Formula
15.31 American Gas Association (AGA) Dry Gas Method
15.32 Complex Pipe Systems Handling Natural (or Similar) Gas
15.33 Two-Phase Liquid and Gas Flow in Process Piping
15.33.1 Flow Patterns
15.33.2 Flow Regimes
15.33.3 Pressure Drop
15.33.4 Erosion–Corrosion
15.33.5 Total System Pressure Drop
15.33.6 Pipe Sizing Rules
15.33.7 A Solution for All Two-Phase Problems
15.33.8 Gas–Liquid Two-Phase Vertical Down Flow
15.33.9 Pressure Drop in Vacuum Systems
15.33.10 Low Absolute Pressure Systems for Air
15.33.11 Vacuum for Other Gases and Vapors
15.33.12 Pressure Drop for Flashing Liquids
15.33.13 Sizing Condensate Return Lines
15.34 UniSim Design PIPESYS
15.35 Pipe Line Safety
15.36 Mitigating Pipeline Hazards
15.37 Examples of Safety Design Concerns
15.38 Safety Incidents Related With Pipeworks and Materials of Construction
15.39 Lessons Learned From Piping Designs
15.40 Design of Safer Piping
15.40.1 Best Practices for Process Piping
15.40.2 Designing Liquid Piping
15.40.3 Best Practices for Liquid Piping
16 Pumps
16.1 Pumping of Liquids
16.2 Pump Design Standardization
16.3 Basic Parts of a Centrifugal Pump
16.4 Centrifugal Pump Selection
16.5 Hydraulic Characteristics for Centrifugal Pumps
16.6 Suction Head or Suction Lift, hs
16.7 Discharge Head, hd
16.8 Velocity Head
16.9 Friction
16.10 Net Positive Suction Head (NPSH) and Pump Suction
16.11 General Suction System
16.12 Reductions in NPSHR
16.13 Charting NPSHR Values of Pumps
16.14 Net Positive Suction Head (NPSH)
16.15 NPSH Requirement for Liquids Saturation With Dissolved Gases
16.16 Specific Speed
16.17 Rotative Speed
16.18 Pumping Systems and Performance
16.19 Power Requirements for Pumping Through Process Lines
16.20 Affinity Laws
16.21 Centrifugal Pump Efficiency
16.22 Effects of Viscosity
16.23 Temperature Rise and Minimum Flow
16.24 Centrifugal Pump Specifications
16.25 Number of Pumping Units
16.26 Rotary Pumps
16.27 Reciprocating Pumps
16.28 Pump Selection
16.29 Selection Rules-of-Thumb
16.30 Case Studies
16.31 Pump Cavitations
16.32 Pump Fundamentals
16.33 Operating Philosophy
16.34 Piping
16.35 Troubleshooting Checklist for Centrifugal Pumps
17 Compression Equipment
17.1 Introduction
17.2 General Application Guide
17.3 Specification Guides
17.4 General Considerations for Any Type of Compressor Flow Conditions
17.4.1 Fluid Properties
17.4.2 Compressibility
17.4.3 Corrosive Nature
17.4.4 Moisture
17.4.5 Special Conditions
17.5 Reciprocating Compression
17.6 Suction and Discharge Valves
17.7 Specification Sheet
17.8 Performance Considerations
17.9 Compressor Performance Characteristics
17.10 Hydrogen Use in the Refinery
17.10.1 IsoTherming Technology for Kerosene, Vacuum Gas Oil, and Diesel Hydroprocessing
Nomenclature
Greek Symbols
Subscripts
References
Glossary of Petroleum and Technical Terminology
Appendix D
Appendix E
Index
About the Author
Also of Interest
EULA
Petroleum Refining Design and Applications Handbook Volume 2
Scrivener Publishing 100 Cummings Center, Suite 541J Beverly, MA 01915-6106
Publishers at Scrivener Martin Scrivener ([email protected]) Phillip Carmical ([email protected])
THE COMPANION WEBSITE FOR THIS BOOK, WHICH CONTAINS DOWNLOADABLE PROGRAMS, SPREADSHEETS, AND OTHER MATERIALS RELATING TO THIS VOLUME, CAN BE FOUND AT THE LINK BELOW: http://www.scrivenerpublishing.com/coker_volume_two/ PASSWORD: Refining
Petroleum Refining Design and Applications Handbook Volume 2 • Rules of Thumb • Pumps
• Process Planning, Scheduling and Flowsheet Design • Compressors
A. Kayode Coker
• Process Piping Design • Process safety Incidents
This edition first published 2021 by John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, USA and Scrivener Publishing LLC, 100 Cummings Center, Suite 541J, Beverly, MA 01915, USA © 2021 Scrivener Publishing LLC For more information about Scrivener publications please visit www.scrivenerpublishing.com. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, except as permitted by law. Advice on how to obtain permission to reuse material from this title is available at http://www.wiley.com/go/permissions.
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Cover image: Refining Plant - Photobeps Cover design by Kris Hackerott Set in size of 11pt and Minion Pro by Manila Typesetting Company, Makati, Philippines Printed in the USA 10 9 8 7 6 5 4 3 2 1
In Loving Memory of Phillip Ekundayo Jacobs Medellis Delucia Seeta Onslow The most beautiful spiritual beings ever encountered and a privilege to have known. Wishing both the Almighty Father’s grace and blessing, the most wonderful and beautiful journeys in the Luminous Realm of joyful activities in the beyond. Gratitude to the Elemental and Created beings in higher realms, faithfully fulfilling the Will of the Almighty Father. “God wills that His Laws working in Creation should be quite familiar to man, so that he can adjust himself accordingly, and with their help can complete and fulfill his course through the world more easily and without ignorantly going astray.” Abd-ru-shin (In the Light of Truth) The Laws of Creation The Law of Motion The Law of the Attraction of Homogeneous Species The Law of Gravitation The Law of Reciprocal Action
“What is Truth?” “Only the truth is simple.”
Pilate (John 18, 38) Sebastian Haffner
“Woe to the people to whom the truth is no longer sacred!” Friedrich Christoph Schlosser “Truth does not conform to us, dear son but we have to conform with it.” Matthias Claudius “Nothing will give safety except truth. Nothing will give peace except the serious search for truth.” Blaise Pascal
“Truth is the summit of being; justice is the application of it to affairs.” Ralph Waldo Emerson “The ideals which have lighted my way, and time after time have given me new courage to face life cheerfully, have been Kindness, Beauty and Truth.” Albert Einstein “It irritates people that the truth is so simple.”
Johann Wolfgang von Goethe
“Aglow with the Light of the Divine, I surrender my whole attention to the Presence of Truth that guides my path.” Michael Bernard Beckwith “Truth means the congruence of a concept with its reality.” “Truth is the revealing gloss of reality.”
G.W. Friedrich Hegel Simone Well
“We are the Multi-dimensional Universe becoming aware of Itself. Live in this One Truth – That God is Real As your very Life!” Michael Bernard Beckwith “Truth is a torch, but a tremendous one. That is why we hurry past it, shielding our eyes, even terrified of getting burnt.” Johann Wolfgang von Goethe “Truth is the spirit’s sun.” You will recognise the Truth, and the truth will set you free
Marquis de Vauvenargues John, 8:32
“Truth is the Eternal – Unchangeable! Which never changes in its form, but is as it has been eternally and will ever remain, as it is now. Which can therefore never be subjected to any development either, because it has been perfect from the very beginning. Truth is real, it is ‘being’! Only being is true life. The entire Universe is “supported” by this Truth!” Abd-ru-shin
Truth To honour God in all things and to perform everything solely to the glory of God Abd-ru-shin (In the Light of Truth) Awake! Keep the heart of your thoughts pure, by so doing you will bring peace and be happy. Love thy neighbour, which means honour him as such! Therein lies the adamantine command: You must never consciously harm him, either in his body or in his soul, either in his earthly possessions or in his reputation! He who does not keep this commandment and acts otherwise, serves not God but the darkness, to which he gives himself as a tool! Honour be to God Who only sows Love! Love also in the The Law of the destruction of the darkness! Abd-ru-shin (In the Light of Truth)
Love & Gratitude Crystal Images © Office Masaru Emoto, LLC
Contents Prefacexv Acknowledgementsxvii 13 Rules of Thumb—Summary 13.0 Introduction
1 1
14 Process Planning, Scheduling, and Flowsheet Design 19 14.1 Introduction 19 14.2 Organizational Structure 20 14.2.1 Process Design Scope 21 14.3 Role of the Process Design Engineer 23 14.4 Computer-Aided Flowsheeting 24 14.5 Flowsheets—Types 26 14.5.1 Block Diagram 26 14.5.2 Process Flowsheet or Flow Diagram 26 14.5.3 Piping Flowsheet or Mechanical Flow Diagram, or Piping and Instrumentation Diagram (P&ID)27 14.5.4 Combined Process and Piping Flowsheet or Diagram 32 14.5.5 Utility Flowsheets or Diagrams (ULDs) 32 14.5.6 Special Flowsheets or Diagrams 36 14.5.7 Special or Supplemental Aids 36 14.6 Flowsheet Presentation 36 14.7 General Arrangements Guide 36 14.8 Computer-Aided Flowsheet Design/Drafting 38 14.9 Flowsheet Symbols 40 14.10 Line Symbols and Designations 43 14.11 Materials of Construction for Lines 46 14.12 Test Pressure for Lines 47 14.13 Working Schedules 56 14.14 Information Checklists 61 14.15 Basic Engineering and Front End Engineering Design (FEED) 63 References64 15 Fluid Flow 15.1 Introduction 15.2 Flow of Fluids in Pipes 15.3 Scope 15.4 Basis 15.5 Incompressible Flow 15.6 Compressible Flow: Vapors and Gases 15.7 Important Pressure Level References 15.8 Factors of “Safety” for Design Basis 15.9 Pipe, Fittings, and Valves 15.10 Pipe
65 65 65 70 72 72 73 75 75 75 75 ix
x Contents
15.11 Total Line Pressure Drop 78 15.11.1 Relationship Between the Pipe Diameter and Pressure Drop (ΔP) 80 15.11.2 Economic Balance in Piping and Optimum Pipe Diameter 82 15.12 Reynolds Number, Re (Sometimes Used NRe) 83 15.13 Pipe Relative Roughness 85 15.14 Darcy Friction Factor, f 85 15.15 Friction Head Loss (Resistance) in Pipe, Fittings, and Connections 94 15.15.1 Pressure Drop in Straight Pipe: Incompressible Fluid 94 15.16 Oil System Piping 96 15.16.1 Density and Specific Gravity 97 15.16.2 Specific Gravity of Blended Products 98 15.16.3 Viscosity 98 15.16.4 Viscosity of Blended Products 100 15.16.5 Blending Index, H 101 15.16.6 Vapor Pressure 101 15.16.7 Velocity 101 15.16.8 Frictional Pressure Drop, ft of Liquid Head 104 15.16.9 Hazen–Williams Equation 105 15.16.10 Transmission Factor 107 15.16.11 Miller Equation 112 15.16.12 Shell–MIT Equation 113 15.17 Pressure Drop in Fittings, Valves, and Connections 116 15.17.1 Incompressible Fluid 116 15.17.2 Velocity and Velocity Head 116 15.17.3 Equivalent Lengths of Fittings 117 15.17.4 L/D Values in Laminar Region 120 15.17.5 Validity of K Values 122 15.17.6 Laminar Flow 122 15.17.7 Expressing All Pipe Sizes in Terms of One Diameter 124 15.17.8 Loss Coefficient 128 15.17.9 Sudden Enlargement or Contraction 134 15.17.10 For Sudden Contractions 134 15.17.11 Piping Systems 136 15.18 Resistance of Valves 136 15.19 Flow Coefficients for Valves, Cv 137 15.20 Flow Meters 138 15.20.1 Process Design of Orifice Meter 138 15.20.2 Nozzles and Orifices 142 Conclusion 167 15.21 Estimation of Pressure Loss Across Control Valves 169 15.22 The Direct Design of a Control Valve 173 15.23 Water Hammer 173 15.24 Friction Pressure Drop for Compressible Fluid Flow 175 15.24.1 Compressible Fluid Flow in Pipes 176 15.24.2 Maximum Flow and Pressure Drop 177 15.24.3 Sonic Conditions Limiting Flow of Gases and Vapors 177 15.24.4 The Mach Number, Ma 182 15.24.5 Critical Pressure Ratio 197 15.24.6 Adiabatic Flow 200 15.24.7 The Expansion Factor, Y 201 15.24.8 Misleading Rules of Thumb for Compressible Fluid Flow 203
Contents xi
15.24.9 Other Simplified Compressible Flow Methods 204 15.24.10 Friction Drop for Flow of Vapors, Gases and Steam 205 15.25 Darcy Rational Relation for Compressible Vapors and Gases 213 15.26 Velocity of Compressible Fluids in Pipe 215 15.27 Procedure 228 15.28 Friction Drop for Compressible Natural Gas in Long Pipe Lines 231 15.29 Panhandle-A Gas Flow Formula 235 15.30 Modified Panhandle Flow Formula 237 15.31 American Gas Association (AGA) Dry Gas Method 237 15.32 Complex Pipe Systems Handling Natural (or Similar) Gas 237 15.33 Two-Phase Liquid and Gas Flow in Process Piping 239 15.33.1 Flow Patterns 239 15.33.2 Flow Regimes 242 15.33.3 Pressure Drop 243 15.33.4 Erosion–Corrosion 248 15.33.5 Total System Pressure Drop 250 15.33.6 Pipe Sizing Rules 257 15.33.7 A Solution for All Two-Phase Problems 261 15.33.8 Gas–Liquid Two-Phase Vertical Down Flow 270 15.33.9 Pressure Drop in Vacuum Systems 277 15.33.10 Low Absolute Pressure Systems for Air 279 15.33.11 Vacuum for Other Gases and Vapors 281 15.33.12 Pressure Drop for Flashing Liquids 284 15.33.13 Sizing Condensate Return Lines 286 15.34 UniSim Design PIPESYS 295 15.35 Pipe Line Safety 300 15.36 Mitigating Pipeline Hazards 301 15.37 Examples of Safety Design Concerns 301 15.38 Safety Incidents Related With Pipeworks and Materials of Construction 303 15.39 Lessons Learned From Piping Designs 319 15.40 Design of Safer Piping 320 15.40.1 Best Practices for Process Piping 320 15.40.2 Designing Liquid Piping 321 15.40.3 Best Practices for Liquid Piping 322 Nomenclature 324 Greek Symbols 326 Subscripts 327 References 327 16 Pumps 331 16.1 Pumping of Liquids 331 16.2 Pump Design Standardization 336 16.3 Basic Parts of a Centrifugal Pump 336 16.4 Centrifugal Pump Selection 341 16.5 Hydraulic Characteristics for Centrifugal Pumps 359 16.6 Suction Head or Suction Lift, hs 367 16.7 Discharge Head, hd 369 16.8 Velocity Head 369 16.9 Friction 370 16.10 Net Positive Suction Head (NPSH) and Pump Suction 370 16.11 General Suction System 378
xii Contents
16.12 Reductions in NPSHR 384 16.13 Charting NPSHR Values of Pumps 384 16.14 Net Positive Suction Head (NPSH) 386 16.15 NPSH Requirement for Liquids Saturation With Dissolved Gases 388 16.16 Specific Speed 390 16.17 Rotative Speed 394 16.18 Pumping Systems and Performance 395 16.19 Power Requirements for Pumping Through Process Lines 399 16.20 Affinity Laws 405 16.21 Centrifugal Pump Efficiency 417 16.22 Effects of Viscosity 421 16.23 Temperature Rise and Minimum Flow 436 16.24 Centrifugal Pump Specifications 440 16.25 Number of Pumping Units 441 16.26 Rotary Pumps 448 16.27 Reciprocating Pumps 452 16.28 Pump Selection 456 16.29 Selection Rules-of-Thumb 456 16.30 Case Studies 459 16.31 Pump Cavitations 464 16.32 Pump Fundamentals 474 16.33 Operating Philosophy 475 16.34 Piping 485 16.35 Troubleshooting Checklist for Centrifugal Pumps 485 Nomenclature 493 Subscripts 494 Greek Symbols 495 References 495 17 Compression Equipment 497 17.1 Introduction 497 17.2 General Application Guide 498 17.3 Specification Guides 499 17.4 General Considerations for Any Type of Compressor Flow Conditions 501 17.4.1 Fluid Properties 501 17.4.2 Compressibility 502 17.4.3 Corrosive Nature 502 17.4.4 Moisture 502 17.4.5 Special Conditions 502 17.5 Reciprocating Compression 503 17.6 Suction and Discharge Valves 514 17.7 Specification Sheet 523 17.8 Performance Considerations 524 17.9 Compressor Performance Characteristics 557 17.10 Hydrogen Use in the Refinery 594 17.10.1 IsoTherming Technology for Kerosene, Vacuum Gas Oil, and Diesel Hydroprocessing 595 Nomenclature 829 Greek Symbols 832
Contents xiii
Subscripts 832 References 833 Glossary of Petroleum and Technical Terminology
837
Appendix D
929
Appendix E
1005
Index 1019 About the Author
1025
Preface Petroleum refining is a complex industry that worldwide produces more than $10 billion worth of refined products. Improvements in the design and operation of these facilities can deliver large economic value for refiners. Furthermore, economic, regulatory and environmental concerns impose significant pressure on refiners to provide safe working conditions and at the same time optimize the refining process. Refiners have considered alternative processing units and feedstocks by investing in new technologies. The United States, Europe and countries elsewhere in the world are embarking on full electrification of automobiles within the next couple of decades. Furthermore, the current pandemic of the coronavirus with lock downs in many countries has restricted the movement of people, less use of aviation fuel and motor gasoline. This has resulted in the barrel of crude being sold at $42.0 per barrel presenting problems to oil producers and refiners. The venture of electrification still poses inherent problems of resolving rechargeable batteries and fuel cells and providing charging stations along various highways and routes. Oil and natural-gas will for the foreseeable future form an important part of everyday life. Their availability has changed the whole economy of the world by providing basic needs for mankind in the form of fuel, petrochemicals and feedstocks for fertilizer plants and energy for the power sector. Presently, the world economy runs on oil and natural gas, and the processing of these feedstocks for producing fuels, and value-added products has become an essential activity in modern society. The availability of liquefied natural gas (LNG) has enhanced the environment, and recent development in the technology of natural gas to liquids (GTL) has further improved the availability of fuel to transportation and other sectors. The complex processing of petroleum refining has created a need for environmental, health, and safety management procedures and safe work practices. These
procedures are established to ensure compliance with applicable regulations and standards such as hazard communications (PHA, HAZOP, HAZAN, Inherently Safer Design, MoC, and so on), emissions, Waste Management pollution that includes volatile organic compounds (VOC), carbon monoxide, sulfur oxides (SOx), nitrogen oxides (NOx), particulates, ammonia (NH3), hydrogen sulfide (H2S), and toxic organic compounds) and waste minimization. These pollutants are often discharged as air emissions, wastewater or solid wastes. Furthermore, concern over issues such as the depletion of the ozone layer that results in global warming is increasingly having a significant impact on Earth’s nature and mankind, and carbon dioxide (CO2) is known to be the major culprit of global warming. Other emissions such as H2S, NOx, and SOx from petroleum refining have adversely impacted the environment, and agencies such as Occupational Safety and Health Administration (OSHA), and Environmental Protection Agency (EPA), Health and Safety Executive (U.K. HSE) have imposed limits on the emissions of these compounds upon refiners. Flaring has become more complicated and concerns about its efficiency have been increasing and discussed by experts. The OSHA, EPA and HSE have imposed tighter regulations on both safety and emission control, which have resulted in higher levels of involvement in safety, pollution, emissions and so on. Petroleum refining is one of the important sectors of the world economy, and it’s playing a crucial and pivotal role in industrialization, urbanization, and meeting the basic needs of mankind by supplying energy for industrial and domestic transportation, feedstock for petrochemical products as plastics, polymers, agrochemicals, paints, and so on. Globally, it processes more materials than any other industry, and with a projected increase in population to around 8.1 billion by 2025, increasing demand for fuels, electricity and various consumer products made from the petrochemical route is expected via the petroleum and refining process. xv
xvi Preface
Petroleum Refining Design and Applications Handbook, Volume Two, is a continuation of volume one; comprising of five chapters, a glossary of petroleum and technical terminology, appendices, Excel spreadsheet programs, computer developed programs, UniSim – Design simulation software excises, cases studies and a Conversion Table, interspersed with Process Safety Incidents. Chapter 13 provides the rules of thumb of process equipment and the heuristics for designers, which can be applied by engineers who are substantially familiar with the topics. However, such rules should be of value for approximate design and cost estimation, and should not provide the inexperienced engineer with a perspective, and a foundation where detailed and computer-aided results can be determined; Chapter 14 provides organization structure and design scope and roles of the process design engineer. The functions of these roles are used in various chapters of volumes 2, 3 and 4 of these volume series. Other pertinent functions in this chapter are flowsheets involving a block diagram, process flow (PFD) diagram and process and instrumentation (P & ID) diagram, computer-aided flowsheet design, symbols and basic engineering and front-end engineering design (FEED). Chapter 15 is on fluid flow in process piping, showing the scope, the basis for incompressible and compressible fluids, oil systems piping, pressure drop in process lines, including fittings, resistance of valves, water hammer, two-phase liquid and gas flows in process piping; application of UniSim design PIPESYS, mitigating pipeline hazards, pipeline safety and safety incidents related with pipework and materials of construction and design for
safer piping. This chapter further provides the root causes, findings and recommendations of these incidents in the refinery and chemical plants ensuring that lessons are learned and thus preventing further deaths; Chapter 16 reviews pumping of liquids, centrifugal pump selection, hydraulic characteristics for centrifugal pumps, net positive suction head and requirement for liquid’s saturation with dissolved gases, pump cavitation, affinity laws, centrifugal pump efficiency, rotary pumps, reciprocating pumps, screw pumps, operating philosophy, troubleshooting and checklist for centrifugal pumps, pumps reliability, root causes of pump failures and their impact, cases studies of pump failures in the refinery, their root causes, findings and recommendations. Process safety management involving mechanical integrity and management of change (MOC). Chapter 17 describes compression equipment with specification guides, general application guide, and performance consideration, hydrogen use in the refinery, and UniSim design case studies. The chapter further describes various compressor types, advantages and disadvantages, probably causes and troubleshooting as well as process safety incidents involving compressors’ malfunctions. Furthermore, the chapter describes integrally geared compressors that have wide application in carbon dioxide (CO2) service for enhanced oil recovery (EOR) with an added benefit to the environment, as nearly all of the injected CO2 is permanently sequestrated in the depleted oil fields long after these fields have ceased operation. Appendix D provides construction commissioning start-up checklists of rotary equipment such as pumps, compressors, and other equipment such as blowers, fans and mixers.
Acknowledgments This project is the culmination of five years of research, collating relevant materials from organizations, institutions, companies and publishers, developing Excel spreadsheet programs and computer programs; using Honeywell’s UniSim steady state simulation programs and providing the majority of the drawings in the text. Sincere gratitude to Honeywell Process Solutions for granting permission to incorporate the use of UniSim Design simulation and many other suites of software programs in the book. I express my thanks to Dr. Jamie Barber of Honeywell Process Solutions for his friendship and help over many years of using the UniSim software. To Mr. Ahmed Mutawa formerly of SASREF Co., Saudi Arabia for developing the Conversion Table program for the book. Many organizations, institutions and companies as Gas Processor Suppliers Association (GPSA), USA, Honeywell Process Solutions, Saudi Aramco Shell Refinery Co., (SASREF), Absoft Corporation, USA., American Institute of Chemical Engineers, The Institution of Chemical Engineers, U.K., Chemical Engineering magazine by Access Intelligence, USA., Hydrocarbon Processing magazine have readily given permission for the use of materials and their release for publication. I greatly acknowledge and express my deepest gratitude to these organizations. I have been privileged to have met with Phil Carmical, Publisher at Scrivener Publishing Co.,
some twenty years ago. Phil initiated the well-known Ludwig’s project at the time during his tenure at Gulf Publishing Co., and Elsevier, respectively. His suggestions in collaborating on these important works some seven years ago were timely to the engineering community, as I hope that these works will be greatly beneficial to this community world-wide. I’m deeply grateful to Phil for agreeing to collaborate with me, his suggestions and assistance since. It is my believe upon completing this aspect of the project that the book will save lives in the refinery industry. I also wish to express my thanks to the WileyScrivener team: Kris Hackerott- Graphics Designer, Bryan Aubrey – Copy editor, Myrna Ting – Typesetter and her colleagues. I am truly grateful for your professionalism, assistance and help in the production of this volume.
Finally, Bow down in humility before the Greatness of God, whose Love is never-ending, and who sends us his help at all times. He alone is Life and the Power and the Glory for ever and ever. A. Kayode Coker
xvii
13 Rules of Thumb—Summary
13.0 Introduction An engineering Rule of Thumb is an outright statement regarding suitable sizes or performance of equipment that avoids all requirements for extended calculations. These are safely applied by engineers who are substantially familiar with the topics. However, such rules should be of value for approximate design and cost estimation, and should not provide the inexperienced engineer with perspective and a foundation where detailed and computer-aided results can be determined. Experienced engineers often know where to find information and how to make accurate calculations; they also retain a minimum body of information in mind, which is made up largely of shortcuts and heuristics. The compilation below may fit into such a minimum body of information that boosts to the memory or extension in some instances into less often encountered areas.
COMPRESSORS, FANS, BLOWERS, AND VACUUM PUMPS 1. F ans are used to raise the pressure by about 3% [12 in. (30 cm) water], blowers raise to less than 2.75 barg (40 psig), and compressors to higher pressures, although the blower range is commonly included in the compressor range. 2. For vacuum pumps use the following: Reciprocating piston type down to 133.3 Pa (1 torr) Rotary piston type down to 0.133 Pa (0.001 torr) Two lobe rotary type down to 0.0133 Pa (0.0001 torr) Steam jet ejectors 1 stage down to 13.3 k Pa (100 torr) 3 stage down to 133.3 Pa (1 torr) 5 stage down to 6.7 Pa (0.05 torr) 3. A three-stage ejector needs 100 kg steam/kg air to maintain a pressure of 133.3 Pa (1 torr). 4. In-leakage of air to evacuated equipment depends on the absolute pressure (torr) and the volume of the equipment, V in m3 (ft3), according to W = kV2/3 kg/h (lb/h), with k = 0.98 (0.2) when P > 90 torr, k = 0.39 (0.08) when P is between 0.4 and 2.67 kPa (3 and 20 torr), and k = 0.12 (0.025) at p less than 133.3 Pa (1 torr). 5. Theoretical adiabatic horsepower
A. Kayode Coker. Petroleum Refining Design and Applications Handbook Volume 2, (1–18) © 2021 Scrivener Publishing LLC
1
2 Petroleum Refining Design and Applications Handbook Volume 2 a (SCFM )T1 P2 − 1 THP = 8130a P1
where T1 is inlet temperature in Rankine, R = °F + 460 and a = (k − 1)/k, k = Cp/Cv. Theoretical reversible adiabatic power = mɀ1RT1[({P2/P1}a − 1)]/a, where T1 is inlet temperature, R = Gas Constant, ɀ1 = compressibility factor, m = molar flow rate, a = (k − 1)/k and k = Cp/Cv. Values of °R = 8.314 J/mol K = 1.987 Btu/lb mol R = 0.7302 atm ft3/lb mol° R. 6. Outlet temperature for reversible adiabatic process a
P T2 = T1 2 P1
7. T o compress air from 37.8° C (100°F), k = 1.4, compression ratio = 3, theoretical power required = 62 hp/million ft3/day, outlet temperature 152.2°C (306°F). 8. Exit temperature should not exceed 167–204° C (350–400° F); for diatomic gases (Cp/Cv = 1.4), this corresponds to a compression ratio of about 4. 9. Compression ratio should be about the same in each stage of a multistage unit, ratio = (Pn/P1)1/n, with n stages. 10. Efficiencies of reciprocating compressors: 65% at compression ratio of 1.5, 75% at 2.0, and 80–85% at 3–6. 11. Efficiencies of large centrifugal compressors, 2.83–47.2m3/s (6000–100,000 acfm) at suction, are 76–78%. 12. Rotary compressors have efficiencies of 70%, except liquid liner type which have 50%.
CONVEYORS FOR PARTICULATE SOLIDS 1. S crew conveyors are suited to transport of even sticky and abrasive solids up inclines of 20° or so. They are limited to distances of 3.81 m (150 ft) or so because of shaft torque strength. A 304.8 mm (12 in.) diameter conveyor can handle 28.3–84.95 m3/h (1000–3000 ft3/h), at speeds ranging from 40 to 60 rpm. 2. Belt conveyors are for high capacity and long distances (a mile or more, but only several hundred feet in a plant), up inclines of 30° maximum. A 609.6-mm (24 in.) wide belt can carry 84.95 m3/h (3000 ft3/h) at a speed of 0.508 m/s (100 ft/min), but speeds up to 3.048 m/s (600 ft/min) are suited to some materials. Power consumption is relatively low. 3. Bucket elevators are suited to vertical transport of sticky and abrasive materials. With 508 × 508-mm (20 × 20-in.) buckets, capacity can reach 28.3 m3/h (1000 ft3/h) at a speed of 0.508 m/s (100 ft/min), but speeds up to 1.524 m/s (300 ft/min) are used. 4. Drag-type conveyors (Redler) are suited to short distances in any direction and are completely enclosed. Units range in size from 19.4 × 10−4 to 122.6 × 10−4 m2 (3–19 in.2) and may travel from 0.15 m/s (30 ft/ min) (fly ash) to 1.27 m/s (250 ft/min) (grains). Power requirements are high. 5. Pneumatic conveyors are for high capacity, short distance (122 m (400 ft)) transport simultaneously from several sources to several destinations. Either vacuum or low pressure 0.4–0.8 barg (6–12 psig) is used with a range of air velocities from 10.7 to 36.6 m/s (35–120 ft/s); depending on the material and pressure and air requirements, 0.03–0.2 m3/m3 (1–7 ft3/ft3) of solid is transferred.
COOLING TOWERS 1. W ater in contact with air under adiabatic conditions eventually cools to the wet bulb temperature. 2. In commercial units, 90% of saturation of the air is feasible.
Rules of Thumb—Summary 3 3. Relative cooling tower size is sensitive to the difference between the exit and the wet bulb temperatures:
∆T, °F 5 15 25 Relative volume 2.4 1.0 0.55 4. T ower fill is of a highly open structure so as to minimize pressure drop, which is in standard practice a maximum of 497.6 Pa (2 in. of water). 5. Water circulation rate is 48.9–195.7 L/min m2 (1–4 gpm/ft2) and air rate is 6344–8784 kg/h m2 (1300– 1800 lb/h ft2) or 1.52–2.03 m/s (300–400 ft/min). 6. Chimney-assisted natural draft towers are hyperboloidally shaped because they have greater strength for a given thickness; a tower 76.2 m (250 ft) high has concrete walls 127–152.4 mm (5–6 in.) thick. The enlarge cross section at the top aids in dispersion of exit humid air into the atmosphere. 7. Countercurrent-induced draft towers are the most common in process industries. They are able to cool water within 2°F of the wet bulb. 8. Evaporation losses are 1% of the circulation for every 10°F of cooling range. Windage or drift losses of mechanical draft towers are 0.1–0.3% Blowdown of 2.5–3.0% of the circulation is necessary to prevent excessive salt buildup.
CRYSTALLIZATION FROM SOLUTION 1. C omplete recovery of dissolved solids is obtainable by evaporation, but only to the eutectic composition by chilling. Recovery by melt crystallization also is limited by the eutectic composition. 2. Growth rates and ultimate sizes of crystals are controlled by limiting the extent of supersaturation at any time. 3. The ratio S = C/Csat of prevailing concentration to saturation concentration is kept near the range 1.02–1.05. 4. In crystallization by chilling, the temperature of the solution is kept almost 1–2°F below the saturation temperature at the prevailing concentration. 5. Growth rates of crystals under satisfactory conditions are in the range of 0.1–0.8 mm/h. The growth rates are approximately the same in all directions. 6. Growth rates are influenced greatly by the presence of impurities and of certain specific additives, which vary from case to case.
DISINTEGRATION 1. P ercentages of material greater than 50% of the maximum size are about 50% from rolls, 15% from tumbling mills, and 5% from closed-circuit ball mills. 2. Closed-circuit grinding employs external size classification and return of oversize for regrinding. The rules of pneumatic conveying are applied to the design of air classifiers. Closed circuit is most common with ball and roller mills. 3. Jaw crushers take lumps of several feet in diameter to 102 mm (4 in.). Stroke rates are 100–300/min. The average feed is subjected to 8–10 strokes before it becomes small enough to escape. Gyratory crushers are suited to slabby feeds and makes a more rounded product. 4. Roll crushers are made either smooth or with teeth. A 610-mm (24-in.) toothed roll can accept lumps of 356 mm (14 in.) diameter. Smooth rolls affect reduction ratios up to about 4. Speeds are 50–90 rpm. Capacity is about 25% of the maximum, corresponding to a continuous ribbon of material passing through the rolls. 5. Hammer mills beat the material until it is small enough to pass through the screen at the bottom of the casing. Reduction ratios of 40 are feasible. Large units operate at 900 rpm, smaller ones up to 16,000 rpm. For fibrous materials the screen is provided with cutting edges.
4 Petroleum Refining Design and Applications Handbook Volume 2 6. Rod mills are capable of taking feed as large as 50 mm and reducing it to 300 mesh, but normally the product range is 8–65 mesh. Rods are 25–150 mm in diameter. The ratio of rod length to mill diameter is about 1.5. About 45% of the mill volume is occupied by rods. Rotation is at 50–65% of critical. 7. Ball mills are better suited than rod mills to fine grinding. The charge is of equal weights of 1.5-, 2-, and 3-in. balls for the finest grinding. The volume occupied by the balls is 50% of the mill volume. Rotation speed is 70–80% of critical. Ball mills have a length-to-diameter ratio in the range 1–1.5. Tube mills have a ratio of 4–5 and are capable of very find grinding. Pebble mills have ceramic grinding elements, used when contamination with metal is to be avoided. 8. Roller mills employ cylindrical or tapered surfaces that roll along flatter surfaces and crush nipped particles. Products of 20–200 mesh are made.
TOWERS 1. D istillation usually is the most economical method of separating liquids, superior to extraction, absorption, crystallization, or others. 2. For ideal mixtures, relative volatility is the ratio of vapor pressure, α12= P2/P1. 3. Tower operating pressure is most often determined by the temperature of the available condensing medium, 38–50°C (100–120°F) if cooling water, or by the maximum allowable reboiler temperature, 10.34 barg (150 psig) steam, 186°C (366° F) to avoid chemical decomposition/degradation. 4. Sequencing of columns for separating multicomponent mixtures: a. Perform the easiest separation first, that is, the one least demanding of trays and reflux, and leave the most difficult to the last. b. When neither relative volatility nor feed concentration vary widely, remove the components one by one as overhead products. c. When the adjacent ordered components in the feed vary widely in relative volatility, sequence the splits in the order of decreasing volatility. d. And when the concentrations in the feed vary widely but the relative volatilities do not, remove the components in the order of decreasing concentration in the feed. 5. The economically optimum reflux ratio is about 1.2–1.5 times the minimum reflux ratio Rm. 6. The economically optimum number of theoretical trays is near twice the minimum value Nm. 7. The minimum number of trays is found with the Fenske–Underwood equation:
Nm =
log{[x /(1 − x )]ovhd /[x /(1 − x )]btms } log α
8. M inimum reflux for binary or pseudobinary mixtures is given by the following when separation is essentially complete (xD ≌ 1) and D/F is the ratio of overhead products to feed rate:
Rm D 1 = α −1 F ( R + 1)D α = when feed is at the dew point m F α −1 when feed is at the bubble point
9. A safety factor of 10% of the number of trays calculated by the best means is advisable. 10. Reflux pumps are made at least 10% oversize.
Rules of Thumb—Summary 5 11. The optimum value of the Kremser—Brown absorption factor A = (L/VK) is in the range 1.25–2.0. 12. Reflux drums usually are horizontal, with a liquid holdup of 5 min half-full. A takeoff pot for a second liquid phase, such as water in hydrocarbon systems, is sized for a linear velocity of that phase of 0.15 m/s (0.5 ft/s) minimum diameter of 406.4 mm (16 in.). 13. For towers about 914 mm (3 ft) diameter, add 1219 mm (4 ft) at the top for vapor disengagement and 1829 mm (6 ft) at the bottom for liquid level and reboiler return. 14. Limit the tower height to about 53 m (175 ft) maximum because of wind load and foundation considerations. An additional criterion is that L/D be less than 30 (20 < L/D < 30 often will require special design).
TRAY TOWERS 1. F or reasons of accessibility, tray spacings are made 0.5–0.6 m (20–24 in.). 2. Peak efficiency of trays is at values of the vapor factor Fs = µ (ρv)0.5 in the range of 1.2–1.5 m/s (kg/m3)0.5 [1–1.2 ft/s (lb/ft3)0.5]. This range of Fs establishes the diameter of tower. Roughly, linear velocities are 0.6 m/s (2 ft/s) at moderate pressures and 1.8 m/s (6 ft/s) in vacuum. 3. Pressure drop per tray is of the order of 747 Pa (3 in. water) or 689.5 Pa (0.1 psi). 4. Tray efficiencies for distillation of light hydrocarbons and aqueous solutions are 60–90%; for gas absorption and stripping, 10–20%. 5. Sieve trays have holes of 6–7 mm (0.25–0.50 in.) diameter, hole area being 10% of the active cross section. 6. Valve trays have holes of 38 mm (1.5 in.) diameter, each provided with a liftable cap, with 130–150 caps per square meter (12–14 caps per square feet) of active cross section. Valve trays are usually cheaper than sieve trays. 7. Bubble cap trays are used only when liquid level must be maintained at low turndown ratio; they can be designed for lower pressure drop than either sieve or valve trays. 8. Weir heights are 50 mm (2 in.), weir lengths are about 75% of trays diameter, and liquid rate a maximum of about 1.2 m3/min-m of weir (8 gpm/in. of weir); multi-pass arrangements are used at higher liquid rates.
PACKED TOWERS 1. S tructured and random packings are suitable for packed towers less than 0.9 m (3 ft) when low pressure drop is required. 2. Replacing trays with packing allows greater throughput and separation in existing tower shells. 3. For gas rates of 14.2 m3/min (500 ft3/min), use 25.4-mm (1-in.) packing; for 56.6m3/min (2000 ft3/ min) or more use 50-mm (2-in.) packing. 4. Ratio of tower diameter/packing diameter should be >15/1. 5. Because of deformability, plastic packing is limited to 3–4 m (10–15 ft) and metal packing to 6.0–7.6 m (20–25 ft) unsupported depth. 6. Liquid distributors are required every 5–10 tower diameters with pall rings and at least every 6.5 m (20 ft) for other types of dumped packing. 7. Number of liquid distributions should be >32–55/m2 (3–5/ft2) in towers greater than 0.9 m (3 ft) diameter and more numerous in smaller columns. 8. Packed towers should operate near 70% of the flooding rate (evaluated from Sherwood and Lobo correlation). 9. Height Equivalent to a Theoretical Stage (HETS) for vapor–liquid contacting is 0.4–0.56 m (1.3–1.8 ft) for 25-mm (1-in.) pall rings and 0.76–0.9 m (2.5–3.0 ft) for 50-mm (2-in.) pall rings.
6 Petroleum Refining Design and Applications Handbook Volume 2 10. G eneralized pressure drops
Design pressure drops (cm of H2O/m of packing)
Design pressure drops (in. of H2O/ft of packing)
Absorbers and Regenerators (non-foaming systems)
2.1–3.3
0.25–0.40
Absorbers and Regenerators
0.8–2.1
0.10–0.25
Atmospheric/Pressure Stills and Fractionators
3.3–6.7
0.40–0.80
Vacuum Stills and Fractionators
0.8–3.3
0.10–0.40
Maximum value
8.33
1.0
DRIVERS AND POWER RECOVERY EQUIPMENT 1. E fficiency is greater for larger machines. Motors, 85–95%; steam turbines, 42–78%; gas engines and turbines, 28–38%. 2. For under 74.6 kW (100 hp), electric motors are used almost exclusively. They are made for up to 14,900 kW (20,000 hp). 3. Induction motors are most popular. Synchronous motors are made for speeds as low as 150 rpm and are thus suited, for example, for low-speed reciprocating compressors, but are not made smaller than 50 hp. A variety of enclosures are available, from weather-proof to explosion-proof. 4. Steam turbines are competitive above 76.6 kW (100 hp). They are speed-controllable. They are frequency used as spares in case of power failure. 5. Combustion engines and turbines are restricted to mobile and remote locations. 6. Gas expanders for power recovery may be justified at capacities of several hundred hp; otherwise any pressure reduction in a process is done with throttling valves. 7. The following useful definitions are given:
shaft power =
theoretical power to pump fluid (liquid or gas) efficiency of pump or compressor , ε dr
drive power =
shaft power efficiency of drive, ε dr
Overall efficiency, ε ov = ε sh ⋅ ε dr
DRYING OF SOLIDS 1. D rying times range from a few seconds in spray dryers to 1 h or less in rotary dryers and up to several hours or even several days in tunnel shelf or belt dryers. 2. Continuous tray and belt dryers for granular material of natural size or pelleted to 3–15 mm have drying in the range of 10–200 min. 3. Rotary cylindrical dryers operate with superficial air velocities of 1.52–3.05 m/s (5–10 ft/s), sometimes up to 10.67 m/s (35 ft/s) when the material is coarse. Residence times are 5–90 min. Holdup of solid is 7–8%. An 85% free cross section is taken for design purposes. In countercurrent flow, the exit gas
Rules of Thumb—Summary 7
4.
5.
6.
7.
is 10–20°C above the solid; in parallel flow, the temperature of the exit solid is 100°C. Rotation speeds of about 4 rpm are used, but the product of rpm and diameter in feet is typically between 15 and 25. Drum dryers for pastes and slurries operate with contact times of 3–12 s, and produce flakes 1–3 mm thick with evaporation rates of 15–30 kg/m2-h. Diameters are in the range of 1.5–5.0 ft; and rotation rate is 2–10 rpm. The greatest evaporative capacity is of the order of 1360.7 kg/h (3000 lb/h) in commercial units. Pneumatic conveying dryers normally take particles 1–3 mm diameter but up to 10 mm when the moisture is mostly on the surface. Air velocities are 10–30 m/s. Single-pass residence times are 0.5–3.0 s, but with normal recycling the average residence time is brought up to 60 s. Units in use range from 0.2 m in diameter by 1 m long to 0.3 m in diameter by 38 m long. Air requirement is several SCFM per lb of dry product/h. Fluidized bed dryers work best on particles of a few tenths of a mm in diameter, but particles of up to 4 mm in diameter have been processed. Gas velocities of twice the minimum fluidization velocity are a safe prescription. In continuous operation, drying times of 1–2 min are enough, but batch drying of some pharmaceutical products employs drying times of 2–3 h. Spay dryers: Surface moisture is removed in about 5 s, and most drying is completed in less than 60 s. Parallel flow of air and stock is most common. Atomizing nozzles have openings 3–3.8 mm (0.012–0.15 in.) and operate at pressures of 21–276 bar (300–4000 psi). Atomizing spray wheels rotate at speeds of 20,000 rpm with peripheral speeds of 76.2–183 m/s (250–600 ft/s). With nozzles, the length-todiameter ratio of the dryer is 4–5; with spray wheels, the ratio is 0.5–1.0. For the final design, the experts say, pilot tests in a unit of 2 m diameter should be made.
EVAPORATORS 1. L ong tube vertical evaporators with either natural or forced circulation are most popular. Tubes are 19–63 mm (0.75–24.8 in.) in diameter and 3.66–9.14 m (12–30 ft) long. 2. In forced circulation, linear velocities in the tubes are in the range of 4.57–6.09 m/s (15–20 ft/s). 3. Elevation of boiling point by dissolved solids results in temperature differences of 3–10°F between solution and saturated vapor. 4. When the boiling point rise is appreciable, the economic number of effects in series with forward feed is 4–6. 5. When the boiling point rise is small, minimum cost is obtained with 8–10 effects in series. 6. In backward feed the more concentrated solution is heated with the highest temperature steam so that heating surface is lessened, but the solution must be pumped between stages. 7. The steam economy of an N-stage battery is approximately 0.8 N-lb evaporation/lb of outside steam. 8. Interstage steam pressures can be boosted with steam jet compressors of 20–30% efficiency or with mechanical compressors of 70–75% efficiency.
EXTRACTION, LIQUID–LIQUID 1. Th e dispersed phase should be the one that has the higher volumetric rate, except in equipment subject to back-mixing where it should be the one with the smaller volumetric rate. It should be the phase that wets the material of construction less well. Since the holdup of continuous phase is greater, that phase should be made up of the less expensive or less hazardous material. 2. There are no known commercial applications of reflux to extraction processes, although the theory is favorable. 3. Mixer–settler arrangements are limited to at most five stages. Mixing is accomplished with rotating impellers or circulating pumps. Settlers are designed on the assumption that droplet sizes are about 150 µm in diameter. In open vessels, residence times of 30–60 min or superficial velocities of 0.15–0.46
8 Petroleum Refining Design and Applications Handbook Volume 2
4. 5.
6. 7. 8.
9.
m/min (0.5–1.5 ft/min) are provided in settlers. Extraction-stage efficiencies commonly are taken as 80%. Spray towers as tall as 6–12 m (20–40 ft) cannot be depended on to function as more than a single stage. Packed towers are employed when 5–10 stages suffice. Pall rings 25–38 mm (1–1.5 in.) in size are best. Dispersed-phase loadings should not exceed 10.2m3/min-m2 (25 gal./min-ft2), and HETS of 1.5–3.0 m (5–10 ft) may be realized. The dispersed phase must be redistributed every 1.5–2.1 m (5–7 ft). Packed towers are not satisfactory when the surface tension is more than 10 dyne/cm. Sieve tray towers have holes of only 3–8 mm diameter. Velocities through the holes are kept below 0.24 m/s (0.8 ft/s) to avoid formation of small drops. Re-dispersion of either phase at each tray can be designed for. Tray spacings are 152–600 mm (6–24 in). Tray efficiencies are in the range of 20–30%. Pulse packed and sieve tray towers may operate at frequencies of 90 cycles/min and amplitudes of 6–25 mm. In large-diameter tower, HETS of about 1 m has been observed. Surface tensions as high as 30–40 dyn/cm have no adverse effect. Reciprocating tray towers can have holes of 150 mm (9/16 in.) diameter, 50–60% open area, stroke length 190 mm (0.75 in.), 100–150 strokes/min, and plate spacing normally 50 mm (2 in.) but in the range of 25.0–150 mm (1–6 in.). In a 760-mm (30-in.) diameter tower, HETS is 500–650 mm (20–25 in.) and throughput is 13.7 m3/min-m2 (2000 gal./h-ft2). Power requirements are much less than those of pulsed towers. Rotating disk contractors or other rotary agitated towers realize HETS in the range of 0.1–0.5 m (0.33– 1.64 ft). The especially efficient Kuhni with perforated disks of 40% free cross section has HETS of 0.2 m (0.66 ft) and a capacity of 50 m3 /m2-h (164 ft3/ft2-h).
FILTRATION 1. P rocess are classified by their rate of cake buildup in a laboratory vacuum leaf filter: rapid, 0.1–10.0 cm/s; medium, 0.1–10.0 cm/min; and slow, 0.1–10.0 cm/h. 2. Continuous filtration should not be attempted if 1/8 in. cake thickness cannot be formed in less than 5 min. 3. Rapid filtering is accomplished with belts, top feed drums, or pusher centrifuges. 4. Medium rate filtering is accomplished with vacuum drums or disks or peeler centrifuges. 5. Slow-filtering slurries are handled in pressure filters or sedimenting centrifuges. 6. Clarification with negligible cake buildup is accomplished with cartridges, precoat drums, or sand filters. 7. Laboratory tests are advisable when the filtering surface is expected to be more than a few square meters, when cake washing is critical, when cake drying may be a problem, and when precoating may be needed. 8. For finely ground ores and minerals, rotary drum filtration rates may be 15,000 lb/day-ft2 at 20 rev/h and 18–25 in. Hg vacuum. 9. Coarse solids and crystals may be filtered at rates of 6000 lb/day-ft2 at 20 rev/h and 2–6 in. Hg vacuum.
FLUIDIZATION OF PARTICLES WITH GASES 1. P roperties of particles that are conducive to smooth fluidization include rounded or smooth shape, enough toughness to resist attrition, sizes in the range of 50–500 µm diameter, and a spectrum of sizes with ratio of largest to smallest in the range of 10–25. 2. Cracking catalysts are members of a broad class characterized by diameters of 30–150 µm, density of 1.5 g/ml or so, and appreciable expansion of the bed before fluidization sets in, minimum bubbling velocity greater than minimum fluidizing velocity, and rapid disengagement of bubbles. 3. The other extreme of smoothly fluidizing particles are typified by coarse sand and glass beads, both of which have been the subject of much laboratory investigation. Their sizes are in the range of
Rules of Thumb—Summary 9 1 50–500 µm, densities 1.5–4.0 g/ml, have small bed expansion and about the same magnitudes of minimum bubbling and minimum fluidizing velocities, and they also have rapidly disengaging bubbles. 4. Cohesive particles and large particles of 1 mm or more do not fluidize well and usually are processed in other ways. 5. Rough correlations have been made of minimum fluidization velocity, minimum bubbling velocity, bed expansion, bed level fluctuation, and disengaging height. Experts recommend, however, that any real design be based on pilot-plant work. 6. Practical operations are conducted at two or more multiples of the minimum fluidizing velocity. In reactors, the entrained material is recovered with cyclones and returned to process. In driers, the fine particles dry most quickly so the entrained material need not be recycled.
HEAT EXCHANGERS 1. F or conservative estimate set F = 0.9 for shell and tube exchangers with no phase changes, q = UAF∆Tlm. When ∆T at exchanger ends differ greatly then check F, reconfigure if F is less than 0.85. 2. Take true countercurrent flow in a shell-and-tube exchanger as a basis. 3. Standard tubes are 19.0 mm (3/4 in.) outer diameter (OD), 25.4 mm (1 in.) triangular spacing, 4.9 m (16 ft) long.
A shell of 300 mm (1 ft) diameter accommodates 9.3 m2 (100 ft2); 600 mm (2 ft) diameter accommodates 37.2 m2 (400 ft2); 900 mm (3 ft) diameter accommodates 102 m2 (1100 ft2).
4. 5. 6. 7. 8. 9.
ube side is for corrosive, fouling, scaling, and high-pressure fluids. T Shell side is for viscous and condensing fluids. Pressure drops are 0.1 bar (1.5 psi) for boiling and 0.2–0.62 bar (3–9 psi) for other services. Minimum temperature approach is 10°C (20°F) for fluids and 5°C (10°F) for refrigerants. Cooling water inlet temperature is 30°C (90°F), maximum outlet temperature 49°C (120°F). Heat-transfer coefficients for estimating purposes, W/m2°C (Btu/h-ft2-°F): water to liquid, 850 (150); condensers, 850 (150); liquid to liquid, 280 (50); liquid to gas, 60 (10); gas to gas, 30 (5); and reboiler 1140 (200). Maximum flux in reboiler is 31.5 kW/m2 (10,000 Btu/h-ft2). When phase changes occur, use a zoned analysis with appropriate coefficient for each zone. 10. Double-pipe exchanger is competitive at duties requiring 9.3–18.6 m2 (100–200 ft2). 11. Compact (plate and fin) exchangers have 1150 m2/m3 (350 ft2/ft3), and about four times the heat transfer per cut of shell-and-tube units. 12. Plate and frame exchangers are suited to high sanitation services and are 25–50% cheaper in stainless steel construction than shell-and-tube units. 13. Air coolers: Tubes are 0.75–1.00 in. OD., total finned surface 15–20 ft2/ft2 bare surface, U = 450–570 W/m2°C (80–100 Btu/h-ft2 (bare surface)-°F). Minimum approach temperature = 22°C (40°F). Fan input power = 1.4–3.6 kW/(MJ/h) [2–5 hp/(1000 Btu/h)]. 14. Fired heaters: radiant rate, 37.6 kW/m2 (12,000 Btu/h-ft2), convection rate, 12.5 kW/m2 (4000 Btu/ h-ft2); cold oil tube velocity = 1.8 m/s (6 ft/s); approximately equal heat transfer in the two sections; thermal efficiency, 70–75%; flue gas temperature, 140–195°C (250–350°F) above feed inlet; and stack gas temperature, 345–510°C (650–950°F).
INSULATION 1. U p to 345°C (650°F), 85% magnesia is used. 2. Up to 870–1040°C (1600–1900°F), a mixture of asbestos and diatomaceous earth is used. 3. Ceramic refractories at higher temperatures.
10 Petroleum Refining Design and Applications Handbook Volume 2 4. C ryogenic equipment −130°C (−200°F) employs insulations with fine pores of trapped air, for example, PerliteTM. 5. Optimum thickness varies with temperature: 12.7 mm (0.5 in.) at 95°C (200°F), 25.4 mm (1.0 in.) at 200°C (400°F), 32 mm (1.25 in.) at 315°C (600°F). 6. Under windy conditions, 12.1 km/h (7.5 miles/h), 10–20% greater thickness of insulation is justified.
MIXING AND AGITATION 1. M ild agitation is obtained by circulating the liquid with an impeller at superficial velocities of 30.48– 60.9 mm/s (0.1–0.2 ft/s), and intense agitation at 213.4–304.8 mm/s (0.7–1.0 ft/s). 2. Intensities of agitation with impellers in baffled tanks are measured by power input, hp/1000 gal., and impeller tip speeds: Operation
hp/1000 gal.
Tip speed (ft/min)
Tip speed (m/s)
Blending
0.2–0.5
Homogeneous reaction
0.5–1.5
7.5–10
0.038–0.051
Reaction with heat transfer
1.5–5.0
10–15
0.051–0.076
Liquid–liquid mixtures
5
15–20
0.076–0.10
Liquid–gas mixtures
5–10
15–20
0.076–0.10
Slurries
10
3. P roportions of a stirred tank relative to the diameter D: liquid level = D; turbine impeller diameter = D/3; impeller level above bottom = D/3; impeller blade width = D/15; four vertical baffles with width = D/10. 4. Propellers are made with a maximum of 457.2-mm (18-in.) turbine impellers to 2.74 m (9 ft). 5. Gas bubbles sparged at the bottom of the vessel will result in mild agitation at a superficial gas velocity of 0.0051 m/s (1 ft/min), severe agitation at 0.02 m/s (4 ft/min). 6. Suspension of solids with a settling velocity of 0.009 m/s (0.03 ft/s) is accomplished with either turbine or propeller impellers, but when the settling velocity is above 0.05 m/s (0.15 ft/s) intense agitation with a propeller is needed. 7. Power to drive a mixture of a gas and a liquid can be 25–50% less than the power to drive the liquid alone. 8. In-line blenders are adequate when a second contact time is sufficient, with power inputs of 0.1–0.2 hp/gal.
PARTICLE SIZE ENLARGEMENT 1. Th e chief methods of particle size enlargement are compression into a mold, extrusion through a die followed by cutting or breaking to size, globulation of molten material followed by solidification, agglomeration under tumbling or otherwise agitated conditions with or without binding agents. 2. Rotating drum granulators have length-to-diameter ratios of 2–3, speeds 10–20 rpm, pitch as much as 10°. Size is controlled by speed, residence time, and amount of binder; 2–5 mm diameter is common. 3. Rotary disk granulators produce a more nearly uniform product than drum granulators: fertilizer, 1.5–3.5 mm diameter; iron ore 10–25 mm diameter. 4. Roll compacting and briquetting is done with rolls ranging from 130 mm diameter by 50 mm wide to 910 mm diameter by 550 mm wide. Extrudates are made 1–10 mm thick and are broken down to size for any needed processing, such as feed to tableting machines or to dryers.
Rules of Thumb—Summary 11 5. T ablets are made in rotary compression machines that convert powders and granules into uniform sizes. The usual maximum diameter is about 38.1 mm (1.5 in.), but special sizes up to 101.6 mm (4 in.) diameter are possible. Machines operate at 100 rpm or so and make up to 10,000 tablets/min. 6. Extruders make pellets by forcing powders, pastes, and melts through a die followed by cutting. A 203.2-mm (8-in.) screw has a capacity of 907.2 kg/h (2000 lb/h) of molten plastic and is able to extrude tubing at 0.76–1.52 m/s (150–300 ft/min) and to cut it into sizes as small as washers at 8000/min. Ring pellet extrusion mills have hole diameters of 1.6–32 mm. Production rates are in the range of 30–200 lb/h-hp. 7. Prilling towers convert molten materials into droplets and allow them to solidify in contact with an air stream. Towers as high as 60 m (196.9 ft) are used. Economically the process becomes competitive with other granulation processes when a capacity of 200–400 tons/day is reached. Ammonium nitrate prills, for example, are 1.6–3.5 mm diameter in the 5–95% range. 8. Fluidized bed granulation is conducted in shallow beds 304.8–609.6 mm (12–24 in.) deep at air velocities of 0.1–2.5 m/s or 3–10 times the minimum fluidizing velocity, with evaporation rates of 0.005–1.0 kg/m2s. One product has a size range 0.7–2.4 mm diameter.
PIPING 1. L ine velocities (υ) and pressure drops (∆P): (a) For a liquid pump discharge, υ = (5 + D/3) ft/s and ∆P = 0.45 bar/100 m (2.0 psi/100 ft); (b) For liquid pump suction, υ = (1.3 + D/6) ft/s, ∆P = 0.09 bar/100 m (0.4 psi/100 ft); (c) for steam or gas flow: υ = 20D ft/s and ∆P = 0.113 bar/100m (0.5 psi/ 100 ft), D = diameter of pipe in inches. 2. Gas/steam line velocities = 61 m/s (200 ft/s) and pressure drop = 0.1 bar/100 m (0.5 psi/100 ft). 3. In preliminary estimates set line pressure drops for an equivalent length of 30.5 m (100 ft) of pipe between each of piece of equipment. 4. Control valves require at least 0.69 bar (10 psi) pressure drop for good control. 5. Globe valves are used for gases, control and wherever tight shut-off is required. Gate valves are for most other services. 6. Screwed fittings are used only on sizes 38 mm (1.5 in) or less, flanges or welding used otherwise. 7. Flanges and fittings are rated for 10, 20, 40, 103, 175 bar (150, 300, 600, 900, 1500, or 2500 psig). 8. Approximate schedule number required = 1000 P/S, where P is the internal pressure psig and S is the allowable working stress [about 690 bar (10,000 psi)] for A120 carbon steel at 260°C (500°F). Schedule (Sch.) 40 is most common.
PUMPS 1. P ower for pumping liquids: kW = (1.67) [Flow (m3/min)] [∆P(bar)]/ε[hp = Flow (gpm) ∆P (psi)/ (1, 714)(ε)]. (ε = fractional efficiency). 2. Net positive suction head (NPSH) of a pump must be in excess of a certain number, depending upon the kind of pumps and the conditions, if damage is to be avoided. NPSH = (pressure at the eye of the impeller-vapor pressure)/(ρg). Common range is 1.2–6.1 m (4–20 ft) of liquid. 3. Specific speed Ns = (rpm)(gpm)0.5/(head in ft)0.75. Pump may be damaged if certain limits of Ns are exceeded, and efficiency is best in some ranges. 4. Centrifugal pumps: Single stage for 0.057–18.9 m3/min (15–5000 gpm), 152 m (500 ft) maximum head; multistage for 0.076–41.6 m3/min (20–11,000 gpm), 1675 m (5500 ft) maximum head. Efficiency: 45% at 0.378 m3/min (100 gpm), 70% at 1.89 m3/min (500 gpm), and 80% at 37.8 m3/min (10,000 gpm). 5. Axial pumps for 0.076–378 m3/min (20–100,000 gpm), 12 m (40 ft) head, 65–85% efficiency. 6. Rotary pumps for 0.00378–18.9 m3/min (1–5000 gpm), 15,200 m (50,000 ft) head, 50–80% efficiency. 7. Reciprocating pumps for 0.0378–37.8 m3/min (10–10,000 gpm), 300 km (1,000,000 ft) maximum head. Efficiency: 70% at 7.46 kW (10 hp), 85% at 37.3 kW (50 hp), and 90% at 373 kW (500 hp).
12 Petroleum Refining Design and Applications Handbook Volume 2
REACTORS 1. Th e rate of reaction in every instance must be established in the laboratory, and the residence time or space velocity and product distribution eventually must be found from a pilot plant. 2. Dimensions of catalyst particles are 0.1 mm (0.004 in.) in fluidized beds, 1 mm in slurry beds, and 2–5 mm (0.078–0.197 in.) in fixed beds. 3. The optimum proportions of stirred tank reactors are with liquid level equal to the tank diameter, but at high pressures slimmer proportions are economical. 4. Power input to a homogeneous reaction stirred tank is 0.1–0.3 kw/m3 (0.5–1.5 hp/1000 gal.) but three times this amount when heat is to be transferred. 5. Ideal CSTR (continuous stirred tank reactor) behavior is approached when the mean residence time is 5–10 times the length needed to achieve homogeneity, which is accomplished with 500–2000 revolutions of a properly designed stirrer. 6. Batch reactions are conducted in stirred tanks for small daily production rates or when the reaction times are long or when some condition such as feed rate or temperature must be programed in some way. 7. Relatively slow reactions of liquids and slurries are conducted in continuous stirred tanks. A battery of four or five in series is most economical. 8. Tubular flow reactors are suited to high production rates at short residence times (seconds or minutes) and when substantial heat transfer is needed. Embedded tubes or shell-and-tube constructions then are used. 9. In granular catalyst packed reactors, the residence time distribution is often no better than that of a five-stage CSTR battery. 10. For conversions under about 95% of equilibrium, the performance of a five-stage CSTR battery approaches plug flow. 11. The effect of temperature on chemical reaction rate is to double the rate every 10°C. 12. The rate of reaction in a heterogeneous system is more often controlled by the rate of heat or mass transfer than by the chemical reaction kinetics. 13. The value of a catalyst may be to improve selectivity more than to improve the overall reaction rate.
REFRIGERATION 1. A ton of refrigeration is the removal of 12,700 kJ/h (12,000 Btu/h) of heat. 2. At various temperature levels: −18°C to −10°C (0–50°F), chilled brine and glycol solutions; −45 to −10°C (−50 to −40°F), ammonia, Freon, and butane; −100 to −45°C (−150 to −50°F), ethane or propane. 3. Compression refrigeration with 38°C (100°F) condenser requires kW/tonne (hp/ton) at various temperature levels; 0.93 (1.24) at −7°C (20°F), 1.31 (1.75) at −18°C (0°F); 2.3 (3.1) at −40°C (−40°F); 3.9 (5.2) at −62°C (−80°F). 4. Below −62°C (−80°F), cascades of two or three refrigerants are used. 5. In single-stage compression, the compression ratio is limited to 4. 6. In multistage compression, economy is improved with interstage flashing and recycling, the so-called “economizer operation.” 7. Absorption refrigeration: ammonia to −34°C (−30°F) and lithium bromide to 7°C (45°F) is economical when waste steam is available at 0.9 barg (12 psig).
SIZE SEPARATION OF PARTICLES 1. G rizzlies that are constructed of parallel bars at appropriate spacings are used to remove products larger than 50 mm in diameter. 2. Revolving cylindrical screens rotate at 15–20 rpm and below the critical velocity; they are suitable for wet or dry screening in the range of 10–60 mm.
Rules of Thumb—Summary 13 3. F lat screens are vibrated, shaken, or impacted with bouncing balls. Inclined screens vibrated at 600– 7000 strokes/min and are used for down to 38 µm, although capacity drops off sharply below 200 µm. Reciprocating screens operate in the range of 30–1000 strokes/min and handle sizes to 0.25 mm at the higher speeds. 4. Rotary sifters operate at 500–600 rpm and are suited to a range of 12 mm–50 µm. 5. Air classification is preferred for fine sizes because screens of 150 mesh and finer are fragile and slow. 6. Wet classifiers mostly are used to make two product size ranges, oversize and undersize, with a break commonly in the range between 28 and 200 mesh. A rake classifier operates at about 9 strokes/min when making separation at 200 mesh and 32 strokes/min at 28 mesh. Solids content is not critical, and that of the overflow may be 2–20% or more. 7. Hydrocyclones handle up to 600 ft3/min and can remove particles in the range of 300–5 µm from dilute suspensions. In one case, a 20-in. diameter unit had a capacity of 1000 gpm with a pressure drop of 5 psi and a cutoff between 50 and 150 µm.
UTILITIES, COMMON SPECIFICATIONS 1. S team: 1–2 bar (15–30 psig), 121–135°C (250–275°F); 10 barg (150 psig), 186°C (366°F); 27.6 barg (400 psig), 231°C (448°F); 41.3 barg (600 psig), 252°C (488°F) or with 55–85°C (100–150°F) superheat. 2. Cooling water: For design of cooling tower use, supply at 27–32°C (80–90°F); from cooling tower, return at 45–52°C (115–125°F); return seawater at 43°C (110°F); return tempered water or steam condensate above 52°C (125°F). 3. Cooling air supply at 29–35°C (85–95°F); temperature approach to process, 22°C (40°F). 4. Compressed air at 3.1 (45), 10.3 (150), 20.6 (300), or 30.9 barg (450 psig) levels. 5. Instrument air at 3.1 barg (45 psig), −18°C (0°F) dew point. 6. Fuels: gas of 37,200 kJ/m3 (1000 Btu/SCF) at 0.35–0.69 barg (5–10 psig), or up to 1.73 barg (25 psig) for some types of burners; liquid at 39.8 GJ/m3 (6 million British Thermal unit per barrel). 7. Heat-transfer fluids: petroleum oils below 315°C (600°F) Dowtherms below 400°C (750°F), fused salts below 600°C (1100°F), and direct fire or electricity above 232°C (450°F). 8. Electricity: 0.75–74.7 kW (1–100 hp), 220–550 V; 149–1864 kW (200–2500 hp), 2300–4000 V.
VESSELS (DRUMS) 1. 2. 3. 4. 5.
rums are relatively small vessels to provide surge capacity or separation of entrained phases. D Liquid drums are usually horizontal. Gas/liquid phase separators are usually vertical. Optimum length/diameter = 3, but the range 2.5–5.0 is common. Holdup time is 5 min half-full for reflux drums and gas/liquid separators, 5–10 min for a product feeding another tower. 6. In drums feeding a furnace, 30 min half-full drum is allowed. 7. Knockout drums placed ahead of compressors should hold no less than 10 times the liquid volume passing through per minute. 8. Liquid/liquid separators are designed for a setting velocity of 0.85–1.27 mm/s (2–3 in./min). 9. Gas velocity in gas/liquid separators, υ = k ρL /ρV − 1 m/s (ft/s), with k = 0.11 (0.35) for systems with a mesh deentrainer and k = 0.0305 (0.1) without a mesh deentrainer. 10. Entrainment removal of 99% is attained with 102–305 mm (4–12 in.) mesh pad thickness; 152.5 mm (6 in.) thickness is popular. 11. For vertical pads, the value of the coefficient in step 9 is reduced by a factor of 2/3. 12. Good performance can be expected at velocities of 30–100% of those calculated with the given k; 75% is popular.
14 Petroleum Refining Design and Applications Handbook Volume 2 13. Disengaging spaces of 152–457 mm (6–18 in.) ahead of the pad and 305 mm (12 in.) above the pad are suitable. 14. Cyclone separators can be designed for 95% collection of 5-µm particles, but usually only droplets greater than 50 µm need be removed.
VESSEL (PRESSURE) 1. D esign temperature between −30 and 345°C is 25°C (−20° F and 650°F if 50°F) above maximum operating temperature; higher safety margins are used outside the given temperature range. 2. The design pressure is 10% or 0.69–1.7 bar (10–25 psi) over the maximum operating pressure, whichever is greater. The maximum operating pressure, in turn, is taken as 1.7 bar (25 psi) above the normal operation. 3. Design pressures of vessels operating at 0–0.69 barg (0–10 psig) and 95–540°C (200–1000°F) are 2.76 barg (40 psig). 4. For vacuum operation, design pressures are 1 barg (15 psig) and full vacuum. 5. Minimum wall thickness for rigidity: 6.4 mm (0.25 in.) for 1.07 m (42 in.) diameter and under, 8.1 mm (0.32 in.) for 1.07–1.52 m (42–60 in.) diameter, and 9.7 mm (0.38 in.) for over 1.52 m (60 in.) diameter. 6. Corrosion allowance 8.9 mm (0.35 in.) for known corrosive conditions, 3.8 mm (0.15 in.) for noncorrosive streams, and 1.5 mm (0.06 in.) for steam drums and air receivers. 7. Allowable working stresses are one-fourth the ultimate strength of the material. 8. Maximum allowable stress depends sharply on temperature Temperature (°F)
−20–650
750
850
1000
(°C)
−30–345
400
455
540
Low-alloy steel, SA 203 (psi)
18,759
15,650
9550
2500
(bar)
1290
1,070
686
273
Type 302 stainless (spi)
18,750
18,750
15,950
6250
(bar)
1290
1290
1100
431
VESSELS (STORAGE TANKS) 1. 2. 3. 4.
or less than 3.8 m3 (1000 gal.), use vertical tanks on legs. F For 3.8–38 m3 (1000–10,000 gal.), use horizontal tanks on concrete supports. Beyond 38 m3 (10,000 gal.) use vertical tanks on concrete foundations. Liquids subject to breathing losses may be stored in tanks with floating or expansion roofs for conservation. 5. Freeboard is 15% below 1.9 m3 (500 gal.) and 10% above 1.9 m3 (500 gal.) capacity. 6. A 30-day capacity often is specified for raw materials and products but depends on connecting transportation equipment schedules. 7. Capacities of storage tanks are at least 1.5 times the size of connecting transportation equipment; for instance, 28.4-m3 (7500 gal.) tanker trucks, 130-m3 (34,500 gal.) rail cars, and virtually unlimited barge and tanker capacities.
Source: The above mentioned rules of thumb have been adapted from Walas, S.M., Chemical Process Equipment: Selection and Design, copyright 1988 with permission from Elsevier, all rights reserved.
Rules of Thumb—Summary 15 Physical Properties Heuristics. Units
Liquids
Liquids
Gases
Gases
Gases
Water
Organic material
Steam
Air
Organic material
2.0
1.0
2.0–4.0
Heat capacity
kJ/kg °C
4.2
1.0–2.5
Density
kg/m3
1000
700–1500
Latent heat
kJ/kg
1200–2100
200–1000
Thermal conductivity
W/m °C
0.55–0.70
0.10–0.20
0.025–0.07
0.025–0.05
0.02–0.06
Viscosity
kg/ms
0°C 1.8 × 10−3
Wide Range
10–30 × 10−6
20–50 × 10−6
10–30 × 10−6
10–1000
1.0
0.7
0.7–0.8
1.29 at STP
50°C 5.7 × 10−4 100°C 2.8 × 10−4 200°C 1.4 × 10−4 Prandtl no.
1–15
Source: Turton, R. et al., Analysis, Synthesis, and Design of Chemical Process, Prentice Hall International Series, 2001.
Typical Physical Property Variations with Temperature and Pressure. Liquids
Liquids
Gases
Gases
Property
Temperature
Pressure
Temperature
Pressure
Density
ρl (Tc − T)0.3
Negligible
ρg = MW P/ZRT
ρg = MW P/ZRT
Viscosity
µl = AeB/T
Negligible
Vapor pressure
P* = aeb/(T+c)
–
µgα
T 1.5 (T + 1.47Tb )
–
Significant only for >10 bar –
Note: T is temperature (K), Tc is the critical Temperature (K), Tb is the normal boiling point (K), MV is molecular weight, P is pressure, Z is compressibility, R is the gas constant, and P* is the vapor pressure. Source: Turton, R. et al., Analysis, Synthesis, and Design of Chemical Processes, Prentice Hall International Series, 2001.
Capacities of Process Units in Common Usagea. Process unit
Capacity unit
Maximum value
Minimum value
Comment
Horizontal vessel
Pressure (bar)
400
Vacuum
L/D typically 2–5
Temperature (°C)
400b
−200
Height (m)
10
2
Diameter (m)
2
0.3
L/D
5
2 (Continued)
16 Petroleum Refining Design and Applications Handbook Volume 2 Capacities of Process Units in Common Usagea. (Continued) Process unit
Capacity unit
Maximum value
Minimum value
Comment
Vertical vessel
Pressure (bar)
400
400
L/D typically 2–5
Temperature (°C)
400b
−200
Height (m)
10
2
Diameter (m)
2
0.3
L/D
5
2
Pressure (bar)
400
Vacuum
Normal Limits Diameter
L/D
Temperature (°C)
400b
−200
0.5
3.0–40c
Height (m)
50
2
1.0
2.5–30c
Diameter (m)
4
0.3
2.0
1.6–23c
L/D
30
2
4.0
1.8–13c
Powerd (kW)
250
2500
Square reduction
2
All
D1 D2 NRe
Flow
q
K based on inlet velocity head 4 160 D1 K = 1.2 + − 1 NRe D2 2 D D K = [0.6 + 0.48f D ] 1 1 − 1 D2 D2
Multiply K from Type 1 by θ sin for 45° < θ < 180° 2
Tapered reduction
θ Or 1.6 sin for 0° < θ < 45° 2 3
D1
NRe ≤ 2500
D2 Flow
NRe
NRe > 2500
Thin, sharp orifice
4
NRe ≤ 4000
D2 D1
Flow
NRe
Square expansion
5
All
D2 D1 q
NRe
NRe > 4000
D1
Rounded D2
Re1
Flow
2 4 2 D 4000 D1 D1 K = 2.7 + 1 − − 1 1 D2 NRe D2 D2
D K = 2 1 − 1 D2
4
2 2 D1 K = [1 + 0.8f D ] 1 − D2
If θ >45o, use K from Type 4, otherwise multiply K from Type 4 by
Flow
θ 2.6 sin 2
Tapered expansion
6
2 4 2 D 120 D1 D1 K = 2.7 + 1 − 1 − 1 1 − D2 NRe D2 D2
All
4 50 D1 K = 0.1 + 1 − Re1 D2
Pipe reducer
(Continued)
136 Petroleum Refining Design and Applications Handbook Volume 2 Table 15.12 Excess head loss K correlation for changes in pipe size. (Continued) Type 7
Fitting D1
D2 L
Inlet NRe
K based on inlet velocity head
All
If L/D2 > 5, use Case A and Case F; otherwise multiply K from Case D by:
Flow
0.0936 0.584 + 1.5 (L/D) + 0.225
Thick orifice
8
All
D2
Use the K for Case F
D1 Flow
Pipe reducer
Source: (W.B. Hooper [Chem. Eng. Nov. 7, 1988, pp 89–92]).
Then
v2 h f = K1 1 , ft (m) 2g
(15.146)
Table 15.12 shows how K varies with changes in pipe size.
15.17.11 Piping Systems The K coefficient values for each of the items of pipe, bends, valves, fittings, contractions, enlargements, entrance/ exits into/from vessels are additive as long as they are on the same size (velocity) basis. Thus the resistance equation is applicable to calculate the head or pressure loss through the specific system when the combined K value is used.
L v2 hf = f D 2g
(15.117)
v2 hf = K 2g
(15.118)
or
where K = summation of all K values in a specific system, when all are on the same size (internal flow) basis. See discussion in “Common Denominator” section (Note: The frictional energy loss, or head loss, is additive even if the velocities change).
15.18 Resistance of Valves Figures 15.9a and 15.9b present several typical valves and connections, screwed and flanged, for a variety of sizes or internal diameters. These do not apply for mixtures of suspended solids in liquids; rather specific data for this situation is required (see [2]). Reference [4] presents data for specific valves.
Fluid Flow 137 Valves such as globes and angles generally are designed with changes in flow direction internally, and thereby, exhibit relatively high flow resistances. These same types of valves exhibit even greater resistances when they are throttled down from the “wide open” position for control of flow to a smaller internal flow path. For design purposes, it is usually best to assume a ½ or ¼ open position, rather than wide open. where K1 = refers to coefficient for smaller diameter K2 = refers to coefficient for larger diameter β = ratio of diameters of smaller to larger pipe size θ = angles of convergence or divergence in enlargements or contractions in pipe systems, degrees. From Reference [4], K values for straight-through valves, such as gate and ball (wide open), can also be calculated. These types of valves are not normally used to throttle flow, but are either open or closed. For sudden and gradual (Note: Sub 1 = smaller pipe; Sub 2 = larger pipe)
K2 = K1/β4
(15.147)
For θ ≤ 45°, as enlargements:
K2 = 2.6[(sin θ/2)(1 − β2)2]/β4
(15.148)
For θ ≤ 45°, as contractions
K2 = [0.8(sin θ/2)(1 − β2)]/β4
(15.149)
For higher resistance valves, such as globes and angles, the losses are less than sudden enlargements or contractions situations. For these reduced seat valves the resistance or loss coefficient K, can be calculated as [4]: At θ ≤ 180°, for sudden and gradual enlargements:
K2 = [(1 − β2)2]/β4
(15.150)
At θ ≤ 180°, for gradual contraction:
K2 = [{0.5(sin θ/2)1/2}(1 − β2)]/β4
(15.151)
The use of these equations requires some assumptions or judgment regarding the degree of opening for fluid flow. Even so, this is better than assuming a wide open or full flow condition, which would result in too low a resistance to flow for the design situation.
15.19 Flow Coefficients for Valves, Cv Flow coefficients (not resistance) for valves are generally available from the manufacturer. The Cv coefficient of a valve is defined as the flow of water at 60°F, in gallons per minute, at a pressure drop of one pound per square inch across the valve [4], regardless of whether the valve ultimately will be flowing liquid or gases/vapors in the plant process (Manufacturers give values of Cg or C1, the coefficient for gas flow, for valves flowing gas or vapor). It is expressed as:
Cv = 29.9 d2/(K)1/2
(15.152)
Cv = Q[ρ/(ΔPc)(62.4)]1/2
(15.153)
Q = Cv[ΔPc(62.4/ρ)]1/2
(15.154)
= 7.90 Cv[ΔPc/ρ]1/2
(15.155)
138 Petroleum Refining Design and Applications Handbook Volume 2
ΔP = [Q/Cv]2[ρ/62.4]
(15.156)
where d = internal pipe diameter, in. Cv = flow coefficient for valves; expresses flow rate in gallons per minute of 60°F water with 1.0 psi pressure drop across valve. K = resistance (loss) coefficient Q = flow rate, gpm ΔP = pressure drop across the control valve, psi ρ = fluid density, lb/ft3
15.20 Flow Meters There are many different types of flow meters that are in use in the petroleum and chemical process industries. One type is the traditional differential pressure (DP) type volumetric flow meters. In many applications, the volumetric flow rate is of direct interest to the operators; because of its accuracy, simplicity and relative lower cost, these flow meters are popular in various facilities. By multiplying the flow rate with the actual density, mass flow rate can be obtained. Pitot tube, orifice meter, Venturi meter, and flow nozzle are classified under this category. Among these, orifice meter is by far the most popular in the chemical process industry (CPI). Figure 15.15 shows the schematics of this meter.
15.20.1 Process Design of Orifice Meter The orifice meter is widely used for flow meter in the refinery and petrochemical industries as compared to the Venturi and other flow meter types. Advantages of the orifice meter are: 1. 2. 3. 4.
ixed cost is less F Easy to fabricate and install Occupies less space as compared to the Venturi meter Provides more flexibility. Orifice plate can be easily replaced.
Vena Contracta Taps 1 D and − D (Radius) Taps 2
Flange Taps Corner Taps
1”
25.4 mm −12 D
D Direction of flow Orifice plate
Figure 15.15a Various tap locations for orifice meter.
d
Distance to the mean location of Vena Contracta [Fig. (c)]
D
Fluid Flow 139 0.9 0.8 0.7
β
0.6 0.5 0.4 0.3 0.2 0.1
0.3
0.4
0.5
0.6
0.7
0.8
Figure 15.15b Pipe diameter from the inlet face of the orifice plate. Extracted from the ASME Meter Computation Handbook, 1961.
1
2 Sharp edged orifice
Flow
Vena contracta
zm
Figure 15.15c Orifice meter with vena contracta formation.
Flow
D 1
Figure 15.15d Flow nozzle with differential gauge.
d 2
140 Petroleum Refining Design and Applications Handbook Volume 2 Inlet 1
Throat 2
V1 p1
d
V2
p2
D
Figure 15.15e Venturi meter.
Drive Coil
Flow of Fluid
Tube Oscillation
RTD for Temperature
Electromagnetic Velocity Detector
Figure 15.15f Coriolis mass flow meter (source: Micro Motion, Inc. USA).
Disadvantage Power consumption and hence operating cost of orifice meter is higher than the same of Venturi meter and rotameter. In orifice flow meter, as shown in Figure 15.15c, a square edged or sharp-edged orifice plate is mounted between two flanges at the flanged joint. When fluid flows through the orifice, it forms free flowing jet. This free flowing jet first contracts and then expands. The minimum flow area achieved by the free flowing jet is referred to as the vena contracta. The energy balance equation between two points 1 and 2 (Eq. 15.4) is modified Bernoulli equation for steady flow in a pipe with no pump in the section.
P1 α v 12 g P αv 2 g + + z1 = 2 + 2 + z 2 + e f ρ 2g c g c ρ 2g c g c
(15.157)
Consider points 1 and 2 in Figure 15.15c. At point 1 in the pipe, the fluid flow is undisturbed by the orifice plate. The fluid at this point has a mean velocity v1 and a cross-sectional flow area A1. At point 2 in the pipe, the fluid attains its maximum mean velocity v2 and its smallest cross-sectional flow area A2. This point is known as the vena contracta. It occurs at about one half to two pipe diameters downstream from the orifice plate. The location is a function of the flow rate and the size of the orifice relative the size of the pipe. Let the mean velocity in the orifice be vo and let the diameter and cross sectional area of the orifice be do and Ao, respectively. For the steady state of an incompressible fluid of density ρ between points 1 and 2 in a pipe with no pump and friction, and α = 1. Applying the principle of continuity,
Q = ρ v1A1 = ρ v2A2 = ρ voAo Q=
π 2 π π d1 v 1 = d 22 v 2 = d o2 v o 4 4 4
(15.158) (15.159)
Fluid Flow 141
v1 = v 2
A2 A1
(15.160)
Assuming that ef, the frictional energy loss is negligible (i.e., ef = 0), and Rearranging Eq. 15.157 and substituting Eq. 15.160 gives: 2 v 22 A 2 P1 − P2 g 1− = + (z1 − z 2 ) 2g c A1 ρ gc
(15.161)
Using Eq. 15.159 to substitute for v1 and v2 in Eq. 15.161 gives the velocity vo at the orifice
P − P g 2g c 1 2 + (z1 − z 2 ) gc ρ v2 = 2 A 1 − 2 A1
(15.162)
Or
v2 = where β =
2g{∆h + (z1 − z 2 )} A 2 1 − 2 A1
(15.162a)
do diameter of orifice = d1 Inside diameter of pipe
Provided that location 1 is always the upstream pressure tap and location 2 the downstream tap, Eq. 15.162 is applicable for both upward and downward flow, but note that the sign of (z1–z2) will change. The value of ∆P, and consequently ∆h, will be negative for downward flow if the pressures drop due to flow is smaller than the static pressure difference. For horizontal pipe, z1 = z2, the volumetric flow rate Q, through the orifice from Eq. 15.162 is:
Q = Cd v o Ao P − P 2g c 1 2 πd ρ = Cd 4 [1 − β 4 ] 2 o
(15.163)
πd 2 2g c ρ(P1 − P2 ) G = ρQ = Cd o 4 [1 − β 4 ]
(15.164)
The mass flow rate, G is
where Cd = dimensionless discharge coefficient, which accounts for geometry and friction;
142 Petroleum Refining Design and Applications Handbook Volume 2
1.18
0.725
1.16 d2
d1
Flow
1.14
0.70
1.12
0.675
1.10
0.65
1.08
0.625
1.06
0.60 0.575 0.55
1.04 C=
Cd
0.50 0.45 0.40 0.30 0.20
1.02
1—β4
1.00
Example: The flow coefficient C for a diameter ratio β of 0.60 at a Reynolds number of 20,000 (2 × 104) equals 1.03.
0.98 0.96
Ratio of Nozzle Diameter to Pipe Diameter
d2
d1/d2=β 0.75
C 1.20
d2 2
0.94 0.92
2
4
6 8 104 2 4 6 8 105 2 Re - Reynolds Number based on d2
4
6 8 106
2
Figure 15.16 Flow coefficients “C” for nozzles. C based on the internal diameter of the upstream pipe (source: from Crane [4]).
The flow coefficient C is defined by
C=
Cd 1− β
4
(15.165)
The values of C for orifices and nozzles are shown in Figures 15.16 and 15.17.
15.20.2 Nozzles and Orifices These piping items shown in Figures 15.16 and 15.17 are important pressure drop or head loss items in a system and must be accounted for to obtain the total system pressure loss [4]. For liquids (Note: The ΔP in these equations is NOT a “loss” pressure).
q = C′ A 2g c (144 )(∆P)/ρ = C′ A[2g c h L ]1/ 2
(15.166)
SI units
q = C′ A
2 ∆P = C′ A[2gh L ]1/ 2 ρ
(15.167)
where q = cubic ft/s (m3/s) of fluid at flowing conditions C = flow coefficient for nozzles and orifices
C′ = C d / 1 − β 4 , corrected for velocity of approach
(15.168)
Fluid Flow 143 C 1.3 1.2 d1 β=—=.80 d2 = .75 = .70 = .65 = .60 = .50
1.1 1.0 0.9 0.8 0.7
d2 2
d2
0.6
d1 β=—=.40 d2 = .30 =0 to .20
0.5 0.4 d2
d1
0.3
3 4
6 8 10
20
40 60 80 102
2
4
6 8 103
2
4
6 8 104
Re - Reynolds Number based on d1 Flow
d1/d2=β
0.78
C=
Cd 1—β4
0.74
0.75
0.72
0.725
0.70
0.70
0.68
0.85
0.66
0.60
0.64
0.55 0.50 0.45 0.40 0.30 0.20
0.62 0.60 0.58
2
4
2 6 8 104 2 4 6 8 105 Re - Reynolds Number based on d2
4
Ratio of Orifice Diameter to Pipe Diameter
0.76
6 8 106
Figure 15.17 Flow coefficients “C” for squared-edged orifices (source: Crane [4]).
Note: C = Cd for Figures 15.16 and 15.17, corrected for velocity of approach. Cd = discharge coefficient for nozzles and orifices hL = differential static head or pressure loss across flange taps when C or C values come from Figures 15.16 and 15.17, ft of fluid. Taps are located one diameter upstream and 0.5 diameter down from the device. A = cross-section area of orifice, nozzle, or pipe, ft2 (m2) h = static head loss, ft (m) of fluid flowing ΔP = differential static loss, lbs/in2. (N/m2) of fluid flowing, under conditions of hL above β = ratio of small to large diameter orifices and nozzles and contractions or enlargements in pipes For discharging incompressible fluids to atmosphere, take Cd values from Figures 15.16 or 15.17 if hL or ΔP is taken as upstream head or gauge pressure. For flow of compressible fluids use the net expansion factor Y (see later discussion) [4]:
q = Y C A[2gc(144)(ΔP)/ρ]1/2
(15.169)
144 Petroleum Refining Design and Applications Handbook Volume 2 SI units,
q = YC′ A
2∆P ρ
(15.170)
where Y = net expansion factor for compressible flow through orifices, nozzles, and pipe. The expansion factor Y is a function of: 1. Th e specific heat ratio, k 2. The ratio (β) of orifice or throat diameter to inlet diameter of pipe 3. Ratio of downstream to upstream absolute pressures. Y = 1 for liquids = 1 – [((1 – r)/k)(0.41 + 0.35β4)] for gases C P2 , ratio of downstream to upstream pressure k = p , specific heat ratio P1 Cv C = flow coefficient from Figure 15.16 or 15.17. P = inlet gauge pressure (also see critical flow discussion).
where, r =
Standard Location of Pressure Taps The five locations of pressure taps are: (i) Corner taps: Static holes made in upstream and downstream flange. They are very close to the orifice plate. With corner taps, it is possible to drill both static holes in the orifice plate itself. Then entire orifice meter can be easily inserted in any flanged joint without drilling the holes in pipe or flanges. (ii) Flange taps: Static holes made at a distance of 1 in. (25.4 mm) on upstream and 1 in. (25.4 mm) on downstream side. (iii) Radius taps: Static holes located at a distance one pipe diameter on upstream side and ½ pipe diameter on downstream side. Radius taps are best from practical stand point of view as it gives reasonably good pressure difference means more accurate measurement of flow rate. (iv) Vena contracta taps: Upstream static hole is ½ to 2 times pipe diameter from the plate. Downstream tap is located at the position of minimum pressure. Vena contractra taps give the maximum pressure difference for a given flow rate. But it is not suitable, if the orifice size is changed from time to time. (v) Pipe taps: Static holes are located at 2.5 times pipe diameter upstream side and 8 times pipe diameter on downstream side. This means fluid is flowing normally on both sides without being affected by turbulence, created by the orifice plate. For Re > 30,000. Cd = between 0.595 to 0.62 for vena contracta taps. Cd = between 0.595 to 0.8 for radius taps. Cd = 0.62 for corner taps. Stolz has provided relation between discharge coefficient, Cd, β, and ReD by:
C d = 0.5959 + 0.0312β 2.1 − 0.184β8 + 0.0029 β 2.5 (106 / Re D )0.75
+ 0.09L 1 β 4 (1 − β 4 )−1 − 0.0337L 2 β3
(15.170)
Fluid Flow 145 where ReD = Reynolds number based on internal diameter of pipe D
l l L1 = 1 , L 2 = 2 D do
where β = do/D do = diameter of orifice D = internal pipe diameter l1 = distance of the upstream tapping from the upstream face of all orifice plate, mm l2 = distance of the downstream tapping from the downstream face of the orifice plate, mm. Orifice Problems The classes of problems involving orifices or other obstruction meters that process designers might encounter are similar to the types of problems encountered in pipe flows. These are: 1. U nknown pressure drop. 2. Unknown orifice diameter. 3. Unknown flow rate. Each involves relationship between the same five basic dimensionless variables, namely: Cd, ReD, β, ∆P/P1, and Y, where Cd represents the discharge coefficient of the meter. For liquids, the variables result to four as Y = 1 by definition. The basic orifice equation relates these variables:
π D2 β 2 YC d P1 ρ1 G= 1 − β 4 4
Re D =
1/ 2
1/ 2
P2 2 1 − P1
(15.171)
4G d , β= π Dµ D
(15.172)
and Y = f(β, ΔP/P1) Cd = f(β, ReD)
A. Unknown Pressure Drop The pressure drop is determined for a given fluid flow at a given rate through a given orifice. Given: G, µ, ρ, D, d (β = d/D), P1
Find: ∆P
The procedure is as follows: 1.
Calculate ReD, and β = d/D from Eq. 15.172
2.
Get Cd = Co from Figure 15.18 (Continued)
146 Petroleum Refining Design and Applications Handbook Volume 2 (Continued)
3.
Assume Y = 1 and solve Eq. 15.171 for (∆P)1: 2
4G 1 − β 4 (∆P)1 = πD2 β 2 C o 2ρ1 4.
(15.173)
Using (∆P)1/P1 and β, get Y from Figures 15.16 and 15.17 or
∆P (0.41 + 0.35β 4 ) for radius taps kP 1 ∆P Y = 1− [0.333 + 1.145(β 2 + 0.7β5 + 12β13 )] for pipe taps kP 1 Calculate ΔP = (ΔP)1/Y2 Y = 1−
5. 6.
(15.174) (15.175) (15.176)
Use the value of ∆P from step 5 in step 4 and repeat steps 4–6 until there is no change. 1.00
β=0.8
0.90
0.75 0.70
Discharge coefficient Co
0.80
β=0.5
0.65 0.60
0.70 β=0.2
0.60
0.30
0.40
0.50 0.40
0.30
0.20 1
2
4 6 8 10 2 4 6 8 102 2 4 6 8103 2 4 6 8 104 2 Bore Reynolds number NRed= NRe /β
4 6 8105
D
Figure 15.18 Orifice discharge coefficient for square-edged orifice and flange, corner, or radius type (From Miller, 1983).
B. Unknown Diameter For design purposes, the proper size orifice (d or β) must be determined for a specified (maximum) flow rate of a given fluid in a given pipe with a ∆P device that has a given maximum range. Given: ∆P, P1, ρ, µ, D, G 1.
2.
Find: d (i.e., β)
Solve Eq. 15.171 for β, i.e.,
X β= 1 + X
14
2
8 G , X= ρ1 ∆P πD2 YC o
(15.177)
Assume Y = 1, and Co = 0.61 (Continued)
Fluid Flow 147 3.
Calculate Red = ReD/β, and get Co from Figure 15.18 and Y from Figure 15.16 or 15.17 or Eq. 15.174 or 15.175.
4.
Use the results of step 3 in step 1 and repeat steps 1–4 until there is no change. The required orifice diameter d = β D
C. Unknown Flow Rate For an unknown flow rate, the pressure drop across a given orifice is measured for a fluid with known properties, and the flow rate is to be determined. Given: ∆P, P1, D, d (β = d/D), ρ1, µ
Find: G
1.
Using ∆P/P1 and β, get Y from Eq. 15.174 or 15.175 or Figure 15.16 or 15.17
2.
Assume Co = 0.61
3.
Calculate G from Eq. 15.171
4.
Calculate ReD from Eq. 15.172
5.
Using ReD and β, get Co from Figure 15.18
6.
If Co ≠ 0.61, use the value from step 5 in step 3, and repeat steps 3–6 until there is no change.
Example 15.6: Flow measurement by office meter in a vertical pipe Oil of density 850 kg/m3 flows up a vertical pipe section of diameter 230 mm. A manometer filled with fluid density of 1080 kg/m3 is used to measure the pressure drop across an orifice plate with a throat diameter of 80 mm. Determine the flow rate of oil if the deflection of the manometer fluid is 0.5 m. Assume a discharge coefficient, Cd of 0.65 for the orifice.
2 z2 Orifice
1 z1 h
Flow Datum
148 Petroleum Refining Design and Applications Handbook Volume 2 Solution Applying the Bernoulli equation between points 1 and 2
P1 α v 12 g P αv 2 g + + z1 = 2 + 2 + z 2 ρ 2g c g c ρ 2g c g c
or
P1 − P2 g v2 − v2 = (z 2 − z1 ) + 2 1 ρ 2g c gc
For the steady state of an incompressible fluid of density ρ between points 1 and 2 and α = 1. Applying the principle of continuity,
Q = ρ v1A1 = ρ v2A2 = ρ voAo
v1 = v 2
A2 A1
Substituting for v2 and rearranging gives:
P1 − P2 = ρ
g ρ (z 2 − z1 ) + 2g c gc
A2 1 − A1
where gc = 1 (SI units). For the manometer, the pressure balance is:
P1 – P2 = ρg(z2 – z1) + (ρm – ρ)g h The theoretical velocity through the orifice is:
2 ρv o2 A o 1− = (ρm − ρ)gh 2 A1
vo = where
β=
d o 80 = = 0.348 d1 230
2gh(ρm − ρ) A 2 ρ 1 − o A1
Fluid Flow 149
2gh(ρm − ρ) ρ[1 − β 4 ]
vo =
2 × 9.81 × 0.5 × (1080 − 850) 850[1 − 0.3484 ]
=
= 1.641 m/s.
Note: that this expression does not contain the terms in z. The velocity and therefore flow rate are independent of the orientation of the pipe. The actual flow rate is:
Q = Cd Ao v o π 0.082 × 1.641 4 = 5.3 × 10−3 m3 /s. = 0.65 ×
Example 15.7: Design an orifice for a compressible flow on the following data Name of fluid:
Chlorine gas
Flow rate, G:
2000 Nm3/h
Operating pressure, P:
1.5 atm a
Operating temperature, T:
30°C
Viscosity of Cl2 at 30oC, µ:
0.0145 mPa.s or cP
Inside diameter of pipe, ID:
154.0 mm (6 in. Sch. 40)
Specific heat ratio of Cl2 gas, k:
1.33
Molecular weight, Mw
71
Solution Let β = 0.5
β=
do or d o = 77 mm D
Density of chlorine gas at normal condition at 25°C (298K):
ρ=
m P • M w 101.325(71) = = v RT (8.314 )(298)
= 2.904 kg /m3 Mass flow rate, G = 2000 × 2.904 = 5808 kg/h
150 Petroleum Refining Design and Applications Handbook Volume 2 Reynolds number, ReD is:
kg 4G 4 × (5808/3600) 1 1 h = × × × −3 kg π (ID)µ π × 0.154 × 0.0145 × 10 m s h ms = 919, 910 (Turbulent flow )
Re D =
If corner taps are made, then l1 = l2 = 0 Coefficient of orifice meter can be determined by Stolz’s equation as:
C d = 0.5959 + 0.0312β 2.1 − 0.184β8 + 0.0029 β 2.5 (106 / Re D )0.75 + 0.09L1 β 4 (1 − β 4 )−1 − 0.0337L 2 β3 106 = 0.5959 + (0.0312 × 0.5 ) − (0.184 × 0.5 ) + (0.0029 × 0.5 ) 919910 2.1
8
= 0.6030
Expansion factor Y: Y = 1 for liquids = 1 − [((1 – r) / k)(0.41 + 0.35β4)] for gases For the first trial calculation, let r = 0.8, Y = 0.9362 The mass flow rate G, of chlorine gas:
π 2g c ∆p ρ G = C d Y d o2 4 (1 − β 4 )
Density of chlorine gas at operating condition:
ρ=
151.95 × 71 (8.314 )(303)
= 4.283 kg /m3
5808 π 2 × 1 × ∆p × 4.283 = 0.6030 × 0.9362 × (0.077 )2 3600 4 (1 − 0.54 )
Δp = 41.22 kPa Δp = p1 – p2 or p2 = p1 –Δp = (1.5 × 101.325) – 41.22 p2 = 110.77 kPa
0.75
2.5
r=
=
P2 , ratio of downstream to upstream pressure P1
110.77 = 0.7288 151.99
(15.170)
Fluid Flow 151 25'
T-1 90psig 4" pipe
V-1 80 psig
100 '
3' 10'
10'
5'
20'
6" pipe
5'
3' 20'
25' Pump
2 1
Figure 15.19 Piping layout for Example 15.8.
Y = 1 − [((1 − r )/ k )(0.41 + 0.35β 4 )]
1 − 0.7288 (0.4319) = 1 − 1.355 = 0.9136 ∆p ∝
1 or ∆p1 Y12 = ∆p2 Y22 2 Y
Y ∆p2 = ∆p1 1 Y2
2
0.9362 = 41.22 0.9136
= 43.28 kPa
2
Δp2 = 43.28 kPa = 173.75 in. H2O (4.4 m H2O) For the different variations in the flow rate, Δp can be determined and a differential pressure (DP) transmitter having a range of 4500 mm WC range can be selected. Example 15.8: Calculation of Pressure at Points in System Figure 15.19 shows the layout of a system in which light naphtha is pumped from tank V-1 to tower T-1 by pump P-1. The suction pressure and discharge pressure at the pump, points 1 and 2, respectively, are required. If the pump efficiency is 70%, calculate the hydraulic and brake horsepower for the pump.
152 Petroleum Refining Design and Applications Handbook Volume 2 The following data apply: Fluid: light naphtha, 86.2oAPI at 60°F, Kw = 11.8*. Flow rate 500 gpm measured at 60°F Pumping temperature: 106°F Piping: Schedule 40. Fittings are standard welding-type. Valves are flanged type. *
Specific gravity at 60/60°F = 0.650 Specific gravity at 106/60°F = 0.625 Density at 106°F: (0.625) (62.4) = 39.0 lb/ft3 Viscosity at 106°F = 0.17cSt = (0.17) (0.625) = 0.106cP Flow at 106°F = (500) (0.65/0.625) = 520 gpm Pipe and fittings
Suction side (6 in.)
Discharge side (4 in.)
Straight length of Pipe, ft
10 + 20 + 20 + 25 = 75
25 + 5 + 3 + 10 + 3 + 5 + 100 + 25 = 176
Gate valves
2
2
90° elbows
3
7
Check valves (swing)
–
1
Exit from tank
1
–
Entrance to tank
–
1
Solution Applying the first law of thermodynamics (conservation of energy) between upstream point V and downstream point 1 gives
Pv g v2 P g v + z v + v = 1 + z1 + 1 + e f + w − q ρ gc 2g c ρ g c 2g c
Assuming w = q = 0
(
)
v2 − v2 lb − ft P1 Pv g = + (z 2 − z1 ) + v 1 − e f , f lbm ρ ρ gc 2g c
Applying the changes in pressure head using Table 15.1 gives the following. 1. Pressure head at starting point V-1.
lbf • ft lbm
Pv = 80 + 14.7 = 94.7 psia ρ (Continued)
*Kw is the characterization factor for the naphtha. It is used in evaluating properties of naphtha.
Fluid Flow 153 in 2 lb Pv 94.7 1 = × 144 2 • f2 • ρ 39 ft in lbm ft 3 2. Velocity head term,
349.66
v 2v − v 12 2g c
For 6 in. Schedule 40 pipe, ID = 6.065 in. Large tank, V-1, therefore velocity at V-1 is negligible, vv = 0 Q 6.065 , ID, ft = = 0.5054 ft. A 12 π(ID)2 π(0.5045)2 A= = = 0.2006 ft 2 4 4 v1 =
v1 =
gal 1 520 1 min • 2• • = 5.78 ft./s 0.2006 × 7.48 × 60 min ft gal s ft 3
2 2 2 2 Velocity head, v V − v 1 = 0 − (5.78) ft • 1 • 1 = −0.519 2 2g c 2 × 32.174 s lbm ft lbf s 2
−0.519
Since the velocity head is higher at point 1 than at starting point V-1, therefore the pressure head is decreased g lb ft 3. Static head term (z v − z1 ) , f g c lbm g ≈1.0 gc zv = 3 + 10 + 25 = 38
38
Downstream point 1, is lower than starting point V- 1; therefore, the pressure head is increased
lbf • ft lbm
4. Frictional loss term, ef Reynolds number, Re 50.6 Q ρ Re = dµ =
50.6(520)(39) = 1596180 (6.065)(0.106)
= 1.6 × 106 (Turbulent flow ) Pipe roughness, ε = 0.0018 in. Relative pipe roughness, ε = 0.0018 = 0.000296 d 6.065 From Moody chart (Figure 15.5), fD = 0.0152
L Resistance coefficient due to friction, Kf: K f = f D d Total pipe length at suction side L = 10 + 20 + 20 + 25 = 75ft. (Continued)
154 Petroleum Refining Design and Applications Handbook Volume 2 (Continued) 75 × 12 K f = 0.0152 = 2.26 6.065 Total KTotal = 3.74 K Total
−1.94
lb ft (5.78)2 v 12 = 3.74 = 1.94 f 2g c 2(32.174 ) lbm
Frictional loss term, ef is in the direction of flow, therefore the pressure head is decreased. P1 , pressure head at point 1 = ρ 385.181
lb ft = 349.66 − 0.519 + 38 − 1.94 = 385.181 f lbm P1 =
385.181 × 39 lbf ft lbm ft 2 • 3 • 2 144 lbm ft in
= 104.319 psia (89.62 psig )
Pipe Fittings at the Suction side The loss coefficient K for a straight pipe
L K = fD D
(15.106)
Using Darby’s 3-K method (Eq. 15.139)
K1 K + K i 1 + 0 ,d3 Re Dn ,in .
Kf =
Outside diameter of a 6” pipe Sch. 40, Dn = 6.0 in.
Re = 1.6 × 106 Using Darby’s 3-K Method (from Table 15.11b) Fittings
N
K1
nK1
Ki
nKi
Kd
Kf
Gate valve
2
300
600
0.037
0.074
3.9
0.243
90° Elbows(r/D = 1.5)
3
800
2400
0.071
0.213
4.2
0.737
Exit from tank
1
Kf = (160/Re + K∞)
Total
K∞ = 0.5
0.5 1.48
Fluid Flow 155 Applying the first law of thermodynamics (conservation of energy) between upstream point 2 and downstream point T gives
P2 g v2 P g v + z 2 + 2 = T + z T + T + ef ρ g c 2g c ρ g c 2g c
(
)
v T2 − v 22 P2 PT g lb • ft = + (z T − z 2 ) + + ef , f ρ ρ gc 2g c lbm
5. Pressure head at starting point T-1.
lbf • ft lbm
PT = 90 + 14.7 = 104.7 psia ρ in 2 lb PT 104.7 1 = × 144 2 • f2 • ρ 39 ft in lbm ft 3
386.58
2 2 6. Velocity head term, v T − v 2 2g c
For 4 in. Schedule 40 pipe, ID = 4.026 in. Large tank, T-1, therefore velocity at T-1 is negligible, vT = 0
lbf • ft lbm
Q 4.026 , ID, ft = = 0.3355 ft. A 12 π(ID)2 π(0.3355)2 A= = = 0.0884 ft 2 4 4 v2 =
gal 1 520 1 min • 2• • gal 0.0884 × 7.48 × 60 min ft s ft 3 = 13.106 ft./s
v2 =
Velocity head ,
v 2T − v 22 0 − (13.106)2 ft 2 1 1 = • 2• 2g c 2 × 32.174 s lbm ft lbf s 2 = −2.67
−2.67
Since the velocity head is higher at point 2 than at starting point T-1, therefore the pressure head is decreased. g lb ft 7. Static head term (z T − z 2 ) , f g c lbm g ≈1.0 gc (Continued)
156 Petroleum Refining Design and Applications Handbook Volume 2 (Continued)
zv = 25 + 100 = 125
125
Downstream point 2, is lower than starting point T- 1; therefore, pressure head is increased 8. Frictional loss term, ef Reynolds number, Re Re = =
50.6 Q ρ dµ 50.6(520)(39) = 2404578 (4.026)(0.106)
= 2.4 × 106 (Turbulent flow ) Pipe roughness, ε = 0.0018 in. Relative pipe roughness,
ε 0.0018 = = 0.00045 d 4.026
lbf • ft lbm
From Moody chart (Figure 15.5), fD = 0.0164 Resistance coefficient due to friction, Kf: L K f = fD d Total pipe length at discharge side L = 25 + 5 + 3 + 10 + 3 + 5 + 100 + 25 = 176 ft. 176 × 12 K f = 0.0164 = 8.603 4.026 Total KTotal = 13.419
K Total
lb ft v 22 (13.106)2 = 13.419 = 35.820 f 2g c 2(32.174 ) lbm
35.82
Frictional loss term, ef is counter to the direction of flow, therefore, pressure head is increased P2 , pressure head at point 2 = ρ lb • ft = 386.58 − 2.67 + 125 + 35.82 = 544.73 f lbm P2 =
544.73 × 39 lbf ft lbm ft 2 • 3 • 2 144 lbm ft in
= 147.53 psia (132.83 psig )
544.73
Fluid Flow 157 Pipe Fittings at the discharge side The loss coefficient K for a straight pipe (Eq. 15.105)
L K = fD D
Using Darby’s 3-K method (Eq. 15.139)
K1 K + K i 1 + 0 ,d3 Re Dn ,in .
Kf =
Pipe Fittings at the Discharge side Outside diameter of 4” pipe Sch. 40, Dn = 4.0 in.
Re = 2.4 × 106 Using Darby’s 3-K Method (from Table 15.11b) Fittings
n
K1
nK1
Ki
nKi
Kd
Kf
Gate valve
2
300
600
0.037
0.074
3.9
0.265
90° Elbows (r/D = 1.5)
7
800
5600
0.071
0.497
4.2
1.876
Check valve
1
1500
1500
0.46
0.46
4
1.675
Entrance to tower
1
1.0
Total
4.816
Pressure drop across the pump
ΔPpump = P2 – P1 = 132.83 – 89.62 = 43.21 psi The hydraulic horse power, hp =
Brakehorse power, Bhp
Q ∆Ppump (520)(43.21) = = 13.11 hp 1714 1714
Hydraulic horsepower , hp 13.11 = = 18.7 Bhp Pump efficiency , η 0.7
Excel program (Example 15.8.xlsx gives the solution of Example 15.8). Example 15.9: Case Study Figure 15.20a shows the piping and instrumentation diagram (P&ID) of a crude distillation unit, debutanizer section, and Figure 15.20b further illustrates the debutanizer column C-1007, overhead gas line 8”-P10170-3101C-P to an air cooler condenser E-1031, accumulator V-1008, reflux pump P1017A/B, suction line 8”-P10174-3101C, and discharge line 6”-P10176-3101C, respectively. The operating pressure of the debutanizer C-1007 is 16.59 barg. The condensates from E-1031 are collected in the debutanizer accumulator V-1008 and separated from sour water, which is returned to vessel V-1002. The liquefied petroleum gas (LPG) from the accumulator V-1008 operating at 12.69 barg is refluxed to the top of the debutanizer, C-1007 via a centrifugal pump P1017A/B. Determine the frictional losses (ΔPf ) of both the suction and discharge of pump P-1017 A/B carrying liquefied petroleum gas (LPG) at 1432.2 tons/day from the accumulator vessel V-1008 to the suction of pump P1017 A/B,
3”
LO
SPL 10 FH.1
6”
10 TI 248
V
10 FT 85
10 FRCS 95
8”
6”
8”
D
D
10 PdIC 8
4”
10 HX 75
10 TI 267
1”
10 TW 34
10 TW 33
¾“
FROM 12-LRC-5 ON LP SEPARATOR V-1202 IN HDS UNIT 1200 DRG. T.1.487.144
10 X 113
8”-P10150-3101
6”
4”
3”
3101 3123 4”
8“-P10150-3101
E-1028A
10 TW 30
E-1028B
E-1028C
8“
V
10 TI 266
RATIO
10 TW 29
D
10 FH 80
2”
UL
10 P1 141
DRG. T.1.487.144 VIA. T.1487.793
V
10 TW 32
10 TI 253
10 P1 153
10 HX 71
UC 2”
M
I ST
1 atmosphere)
0.5
Leads to exhaust header 4. Relief valve discharge Relief valve, entry point at silencer
1.5 0.5 vs vs
180 Petroleum Refining Design and Applications Handbook Volume 2 are much different from the velocities below the speed of sound. The ratio of the actual fluid velocity to its speed of sound is called the Mach number [27]. The velocity of sound at 68°F in air is 1126 ft/s. For any gas, the speed of sound is:
v s = kgp′′/ρ , ft /s
(15.184)
where k = ratio of specific heat of gas, at constant pressure to that at constant volume, = Cp/Cv (See Table 15.14) g = 32.2. ft/s2 p = pressure, pounds per sq ft, abs (Psfa) (note units) ρ = the specific weight, lb/ft3 at T and p Table 15.14 Approximate k values for some common gases (68°F, 14.7psia). Gas
Chemical formula or symbol
Approximate molecular weight
k(Cp/Cv)
Acetylene (ethyne)
C2H2
26.0
1.30
Air
–
29.0
1.40
Ammonia
NH3
17.0
1.32
Argon
A
39.9
1.67
Butane
C4H10
58.1
1.11
Carbon dioxide
CO2
44.0
1.30
Carbon monoxide
CO
28.0
1.40
Chlorine
Cl2
70.9
1.33
Ethane
C2H6
30.0
1.22
Ethylene
C2H4
28.0
1.22
Helium
He
4.0
1.66
Hydrogen chloride
HCl
36.5
1.41
Hydrogen
H2
2.0
1.41
Methane
CH4
16.0
1.32
Methyl chloride
CH3Cl
50.5
1.20
Natural gas
–
19.5
1.27
Nitric oxide
NO
30.0
1.40
Nitrogen
N2
28.0
1.41
Nitrous oxide
N 2O
44.0
1.31
Oxygen
O2
32.0
1.40
Propane
C3H8
44.1
1.15
Propylene (propene)
C3H6
42.1
1.14
Sulfur dioxide
SO2
64.1
1.26
Fluid Flow 181 In SI units
v s = γ P′ /ρ
(15.185)
This sonic velocity occurs in a pipe system in a restricted area (for example, valve, orifice, venturi) or at the outlet end of the pipe (open-ended), as long as the upstream pressure is high enough. The physical properties in the above equations are at the point of maximum velocity. With a high velocity vapor flow, the possibility of attaining critical or sonic flow conditions in a process pipe should be investigated. These occur whenever the resulting pressure drop approaches the following values of ΔP as a percentage of the upstream pressure [6] where the properties are evaluated at the condition of sonic flow. This applies regardless of the downstream pressure for a fixed upstream pressure. This limitation must be evaluated separately from pressure drop relations, as it is not included as a built in limitation. Sonic velocity will be established at a restricted point in the pipe, or at the outlet, if the pressure drop is great enough to establish the required velocity. Once the sonic velocity has been reached, the flow rate in the system will not increase, as the velocity will remain at this value even though the fluid may be discharging into a vessel at a lower pressure than that existing at the point where sonic velocity is established. ΔP can be increased by continuing to lower the discharge pressure. But no additional flow rate will result. The usual pressure drop equations do not hold at the sonic velocity, as in an orifice. Conditions or systems exhausting to atmosphere (or vacuum) from medium to high pressures should be examined for critical flow, otherwise the calculated pressure drop may be in error. All flowing gases and vapors (compressible fluids) including steam (which is a vapor) are limited or approach a maximum in mass flow velocity or rate, i.e., lb/s or lb/h (kg/s or kg/h) through a pipe depending upon the specific upstream or starting pressure. This maximum rate of flow cannot be exceeded regardless of how much the downstream pressure is further reduced. The mean velocity of fluid flow in a pipe by continuity equation is [4]:
v=
0.0509 WV 0.0509 W or , ft /s d2 ρd 2
(15.186)
v=
354 W V 354 W , m/s or v = 2 d ρd 2
(15.187)
In SI units
where d = internal pipe diameter, in. (mm) W = rate of flow, lb/h (kg/h) V = specific volume of fluid, ft3/lb (m3/kg) ρ = fluid density, lb/ft3 (kg/m3) This maximum velocity of a compressible fluid in a pipe is limited by the velocity of propagation of a pressure wave that travels at the speed of sound in the fluid [4]. This speed of sound is specific for each individual gas or vapor and is a function of the ratio of specific heats of the fluid. The pressure reduces and the velocity increases as the fluid flows downstream through the pipe, with the maximum velocity occurring at the downstream end of the pipe. When or if the pressure drop is great enough, the discharge or exit or outlet velocity will reach the velocity of sound for that fluid. If the outlet or discharge pressure is lowered further, the pressure upstream at the origin will not detect it because the pressure wave can only travel at sonic velocity. Therefore, the change in pressure downstream will not be detected upstream. The excess pressure drop obtained by lowering the outlet pressure after the maximum discharge has been reached takes place beyond the end of the pipe [4]. This pressure is lost in shock waves and turbulence of the jetting fluid. See References [26], [28], and [29] for further expansion of shock waves and detonation waves through compressible fluids. In the case of a high pressure header, the flow may be sonic at the exit. Therefore, it is often necessary to check that the outlet pressure of each pipe segment is not critical. If Pc is less than terminal P2, the flow is subcritical. If however,
182 Petroleum Refining Design and Applications Handbook Volume 2 Pc is greater than P2, then the flow is critical. Although, it may be impractical to keep the flow in high pressure subheaders below sonic, Mak [30] suggests that the main flare header should not be sized for critical flow at the outlet of the flare stack. This would obviate the undesirable noise and vibration resulting from sonic flow. The equation for critical pressure can be expressed as:
RT Pc = 2 11400d k[k + 1] G
1/2
, psia
(15.188)
or
ZG T Pc = 2.45 × 10−3 2 d k M w
0.5
(15.189)
1544 R= , molar gas constant 29 Sp. Gr
(15.190)
where
Sp. Gr . =
Molecular weight of gas Molecular weight of air
Z = compressibility factor d = internal pipe diameter, in. Mw = fluid molecular weight. G = fluid flow rate, lb/h. T = fluid temperature, °R ρ = fluid density, lb/ft3
15.24.4 The Mach Number, Ma The Mach number, Ma is the velocity of the gas divided by the velocity of the sound in the gas and can be expressed as:
Ma = v/vs
(15.191)
The exit Mach number for compressible isothermal fluid has been shown to be Ma ≠ 1, but 1/ k , where k, is the ratio of the fluid specific heat capacities at constant pressure to that at constant volume. Table 15.14 shows the k values for some common gases. The following cases are such: 1. Ma 1/ k , the flow is supersonic
(15.191C)
Case 3 is produced under certain operating conditions in the throttling processes (e.g., a reduction in the flow cross-sectional area). Kirkpatrick [33] indicates that there is a maximum length for which continuous flow is applied for an isothermal condition, and this corresponds to Ma = 1/ k . The limitation for isothermal flow, however, is the heat transfer required to maintain a constant temperature. Therefore, when Ma 1/ k , heat must be rejected from the stream. Depending on the ratio of specific heats, either condition could occur with subsonic flow. Therefore, to maintain isothermal flow during heat transfer, high temperatures require high Mach numbers and low temperatures require low Mach numbers.
Flow Rate of Compressible Isothermal Flow The flow rate of compressible fluids and pressure drop of compressible isothermal flow are based on the following assumptions [31]: 1. 2. 3. 4. 5.
I sothermal compressible fluid. No mechanical work done on or by the system. The gas obeys the perfect gas laws. A constant friction factor along the pipe. Steady state flow or discharge unchanged with time.
The derivation is provided elsewhere [6, 30], and the maximum flow rate through the pipe is: 0.5
P12 − P22 A 2ρ1 g c G = P1 P1 K + 2 ln Total P2
(15.192)
where KTotal is the total resistance (loss) coefficient due to friction, fittings and valves:
K Total = f D
L + ∑ K f ( pipe fittings + valves) D
(15.193)
∑ K f ( pipe fittings + valves) is the sum of the pressure loss coefficient for all the fittings and valves in the line. Expressing the maximum fluid rate in pounds per hour, Eq. 15.192 becomes
P12 − P22 ρ1 G = 1335.6 d P1 P1 K + 2 ln Total P2 2
0.5 , lb/h
(15.194)
In SI units 0.5
2 2 ( p1′ ) − ( p′2 ) ρ1 −4 2 G = 2.484 × 10 d , kg /s p1′ p1′ K Total + 2 ln p′ 2
where d = pipe internal diameter, mm ρ1 = upstream gas density, kg/m3 p1′ = upstream gas pressure, bara p′2 = downstream gas pressure, bara
(15.195)
184 Petroleum Refining Design and Applications Handbook Volume 2
Pipeline Pressure Drop (ΔP)
P If ΔP due to velocity acceleration is relatively small compared with the frictional drop, then ln 1 may be neglected. P2 Therefore Eq. 15.194 becomes 0.5
ρ1 P12 − P22 G = 1335.6 d P K 1 Total
2
(15.196)
Putting C = 1335.6d2
(15.197)
G2 ρ1 P12 − P22 = C 2 K Total P1
(15.198)
That is,
P12 − P22 =
P1 G 2 K Total ρ1 C 2
(15.199) 0.5
2 P1G 2K Total P2 = P1 − 2 C ρ 1
(15.200)
Therefore, the pressure drop
ΔP ≅ P1 – P2
(15.201)
i.e. 0.5
P G 2K Total ∆P ≅ P1 − P12 − 1 ρ1C 2
(15.202)
In SI units 0.5
2 2 ( p1′ ) − ( p′2 ) ρ1 −4 2 G = 2.484 × 10 d p1′ p1′ K Total + 2 ln p′ 2
Putting C1 = 2.484 × 10−4 d2
(15.203)
(15.204)
G2 ρ1 P12 − P22 = C12 K Total P1
(15.205)
That is,
P12 − P22 =
P1 G 2 K Total ρ1 C12
(15.206) 0.5
P G 2K Total P2 = P12 − 1 ρ1C12
(15.207)
Fluid Flow 185 Therefore, the pressure drop
ΔP ≅ P1 – P2
(15.208)
That is 0.5
2 P1G 2K Total ∆P ≅ P1 − P1 − 2 ρ C 1 1
(15.209)
Janettt [32] shows how the back pressure in vent lines are calculated for compressible fluids. Table 15.15 shows the conditions at the pipe exit as a function of the Mach number. Ma1
Ma2
Ma = 1
d
L2
L1
The hypothetical Pipe length, L2 is such that its inlet Ma2 is the same as the exit from the actual pipe
Table 15.15 Conditions at the pipe exit as a function of the Mach number. Isothermal flow
Adiabatic flow
P = ρ × constant
1 G2 k P + = constant 2 ρ2 ( k − 1) ρ
Subsonic flow P2
P1 × Ma1/Ma2
T2
T1
ρ2
ρ1 × Ma1/Ma2
v2
v1 × Ma2/Ma1
2 + ( k − 1)Ma12 2 + ( k − 1)Ma 22
P1
Ma1 Ma 2
T1
2 + ( k − 1) Ma12 2 + ( k − 1) Ma 22
ρ1
Ma1 Ma 2
2 + ( k − 1) Ma 22 2 + ( k − 1) Ma12
v1 × ρ1/ρ2
Critical or sonic flow P2
P1 × Ma1 × k
T2
T1
ρ2
ρ1 × Ma1 × k
v2
v s/ k
P1 Ma1 T1
2 + ( k − 1) Ma12 ( k − 1)
ρ1 Ma1 vs
2 + ( k − 1)Ma12 k +1
( k + 1) 2 + ( k − 1) Ma12
186 Petroleum Refining Design and Applications Handbook Volume 2 Example 15.13 Case Study The vapor (C3, C4, and C5) from the debutanizer unit C1007 in Figure 15.20 is cooled via an air cooler E-1031 to the accumulator V1008. The overhead gas line 8”—P10170-3101C-P is 84.7 m long, and the debutanizer boil-up rate is 17 kg/s (1738 m3/h), which operates at 17.6 bara at the top. Calculate the pressure drop along the 8-in. pipe to the air cooler condenser E-1031. Other data obtained from the piping isometrics, piping data sheets and fluid characteristics are: Operating temperature = 86°C Fluid density = 35.2 kg/m3 Ratio of specific heats γ = (Cp/Cv) = 1.11 Kinematic viscosity = 0.2 cSt. = 0.2 × 10−6 m/s2 Fittings
Number
90° Ell (r/R = 1.5)
5
Ball valve
2
Tee (straight Thru)
3
Solution Dynamic viscosity = Density × Kinematic viscosity = 35.2 × 0.2 × 10−6 = 0.00704 × 10−3 kg/m.s = 0.00704 cP. The average molecular weight of the overhead vapor is: Composition
Molecular weight kg/kmol
Percentage in the vapor phase, %
Average molecular weight
C3H8
44
18.00
7.92
C4H10
58
80.00
46.4
C5H12
72
2.00
1.44
100.00
55.76
Total
Assume compressibility factor Z = 0.958 An 8 in. pipe size Sch. 40 (ID = 202.7 mm) Friction factor fT for 8 in. Schedule 40 CS material = 0.014. Assuming that the flow condition of the vapor through the 8-in. pipe is isothermal. Area of pipe
A = π(ID)2/4
= π(.2027)2/4
= 0.0323 m2 The velocity of gas in the line is:
G = ρ V A
Fluid Flow 187 where A = pipe area, m2 G = mass flow rate, kg/s v = fluid velocity, m/s ρ = fluid density, kg/m3 or
v = G/(ρA) 17 (35.2 × 0.0323) = 14.95 m/s =
Sonic velocity of vapor in the line is (Eq. 15.183):
v s = ZγRT m/s
J 1 8314 = (0.958)(1.11) (359.15) , kmol − K • kg • K = m/s 55.76 kmol = 238.63 m/s
The inlet Mach number Ma1 is (Eq. 15.191)
Ma1 =
v 14.95 = v s 238.63
= 0.0626
Therefore, the flow of gas through the pipe is subsonic, since the inlet Mach number is less than 1. Gas Reynolds number is:
Re = 354
W dµ
(354 )(17 )(3600) (202.7 )(0.00704 ) = 15,181975 (Fully turbulent ) =
(15.28)
= 1.52 × 107
Relative pipe roughness is:
ε 0.046 = = 0.000227 D 202.7 ε 1 5.02 = −4 log − log A 3.7 D Re fC
(15.35)
188 Petroleum Refining Design and Applications Handbook Volume 2 where
ε/D 6.7 A= + 3.7 Re
0.9
0.000227 6.7 A= + 15,181, 975 3.7
0.9
= 6.3657 × 10−5
0.000227 5.02 = −4 log − log(6.3657 × 10−5 ) 15,181, 975 3.7 fC
1
= 16.8884 fC = 0.003506
The Darcy friction factor fD = 4 fC
fD = 4 (0.003506)
= 0.01402 Using Darby’s 3-K method (Eq. 15.140) 0.3 25.4 K1 Kf = + Ki 1 + Kd Re Dn ,mm
Re = 15,181975 (turbulent) Pipe outside diameter of 8”, Dn = 203.2 mm Fittings
n
K1
nK1
Ki
nKi
Kd
Kf
90° Ell (r/D = 1.5)
5
800
4000
0.071
0.355
4.2
1.1543
Ball valve
2
300
600
0.017
0.034
4.0
0.1069
Tee (straight thru)
3
800
2400
0.14
0.42
4.0
1.3204
Total
2.5816
Total loss coefficient KTotal:
K Total = f D
L + ∑Kf : D
Pipe length, L = 84.7 m Diameter, d = 202.7 mm
K Total = 0.0140 × = 8.4316
84.7 + 2.5816 0.2027
Fluid Flow 189 Outlet pressure P2 is (Eq. 15.207): 0.5
2 P1G 2K Total P2 = P1 − ρ1C12
where
C1 = 2.484 × 10−4 d2
G2 17 2 = = 2.774 C12 (2.484 × 10−4 × 202.7 2 )2 P G 2K Total P2 = P12 − 1 ρ1C12
0.5
(17.6)(2.7774 )(8.4316) = 17.62 − 35.2
0.5
= 17.265 bara
Therefore, the pressure drop
ΔP = P1 – P2
= 0.34 bar The process is assumed to be isothermal, therefore outlet temperature, T2 is:
T2 = T1 , K T2 = 86 °C
= (273.15 + 86)K = 359.15 K
Density of the vapor at the exit is
105 P2 , kg /m3 ρ2 = (R /M w )T2 (105 )(17.26) ρ2 = (8314 /55.76)(359.15) = 32.23 kg /m3
Flow velocity at pipe exit is
v = G/(ρ A)
17 (32.23 × 0.0323) = 16.33 m / s =
190 Petroleum Refining Design and Applications Handbook Volume 2 The exit Mach number Ma 2 = 1/ k
Ma 2 = 1/ 1.11 = 0.949 Re-arranging Eq. 15.200 gives 0.5
2 P1G 2K Total P1 − − P2 = 0 2 ρ C 1 1
(15.210)
Eq. 15.210 involves a trial-and-error using a guess value for P2. This is substituted into Eq. 15.210 until the left side gives a value of zero. The Excel spreadsheet with a Goal Seek add in tool is the most convenient computational tool for Eq. 15.210. Eq. 15.210 is set to zero using a guess value of P2. The procedure involves setting the quadratic Eq. 15.210 to zero; with a guess value of the outlet pressure and the Goal Seek determines the outlet pressure after a set of iterations. The Excel spreadsheet (Example 15.13.xls) has been developed to determine the pressure drop of a compressible isothermal flow fluid using Eq. 15.210. Figure 15.24 shows the snap shots of the Excel spreadsheet calculations using the Goal Seek solver to determine the outlet pressure, P2. Example 15.14 Propane vapor at 90°F and an upstream pressure P1 = 20 psig flows at a rate of 24,000 lb/h in an 800 ft long, 6-in. Sch. 40 horizontal carbon steel pipe. Under these conditions, the viscosity (μ1 = 0.0094 cP) and the gas compressibility factor Z1 = 0.958. Calculate the total pressure drop under isothermal flow conditions. Check for critical flow. Solution Since the pipe is long, assume an isothermal condition for the compressible vapor flow. The density of propane at 90°F and pressure of 20 psig is
ρ=
Mw P 10.72 T
44 × 34.7 = = 0.2589 lb/ft 3 10.72 × 550 Velocity of the gas is: The 6 in. Sch. 40 pipe size (ID = 6.065 in.) Area of pipe
A = π(ID)2/4
= π(.5054)2/4
= 0.2006 ft.2 The velocity of gas in the line is:
where A = pipe area, ft2 G = mass flow rate, lb/s
G = ρ V A
Fluid Flow 191
Figure 15.24 The Excel spreadsheet snapshot of Example 15.13.
(Continued)
192 Petroleum Refining Design and Applications Handbook Volume 2
Figure 15.24 (Continued) The Excel spreadsheet snapshot of Example 15.13.
(Continued)
Fluid Flow 193
Figure 15.24 (Continued) The Excel spreadsheet snapshot of Example 15.13.
(Continued)
194 Petroleum Refining Design and Applications Handbook Volume 2
Figure 15.24 (Continued) The Excel spreadsheet snapshot of Example 15.13.
v = fluid velocity, ft./s ρ = fluid density, lb/ft.3 or
v = G/(ρ A) 24000 (0.2589 × 0.2006 × 3600) = 128.36 ft /s =
Sonic velocity is:
P v s = Z k g 144 , ft /s ρ
0.958 × 1.15 × 32 × 144 × 34.7 0.2589 = 824.87 ft /s =
(15.174)
Sonic flow would occur at the end of the pipe and not where the pressure is 20 psig. The inlet Mach number Ma1 is:
Ma1 =
v vs
128.36 = = 0.1556 824.87
(15.185)
Fluid Flow 195 Since the Mach number is less than 1, the flow of propane through the pipe is subsonic. Reynolds number Re:
Re =
Dvρ dvρ W = 123.9 = 6.31 µe dµ µ
24000 6.065 × 0.00094 = 2, 656, 329
Re = 6.31
= 2.66 × 106
Friction factor, f:
ε 5.02 log A = −4 log − 3.7D Re fC
1
(15.35)
where
ε/D 6.7 A= + 3.7 Re
and
0.9
ε = pipe roughness, ft. D = pipe internal diameter, ft.
ε 0.00015 = = 0.0002968 D 0.5054 0.0002968 6.7 A= + 2656329 3.7
0.9
= 8.9370 × 10−5
0.0002968 5.02 log(8.937 × 10−5 ) = −4 log − 2, 656, 329 3.7 fC
1
= 16.2247 fC = 0.00380
The Darcy friction factor fD = 4 fC
fD = 4(0.00380)
= 0.0152 Loss coefficient due to straight pipe:
L D 0.0152 × 800 × 12 = 6.065 = 24.052
K = fD
196 Petroleum Refining Design and Applications Handbook Volume 2 Outlet pressure, P2 is (Eq. 15.200) 0.5
P G 2K Total P2 = P12 − 1 ρ1C 2
Where in Eq. 15.197:
C = 1335.6 d 2 G2 240002 = C 2 (1335.6 × 6.065)2 = 0.2386
(34.7)(0.2386)(24.052) P2 = 34.7 2 − 0.2589
0.5
= 20.85 psia ∆P ≅ P1 − P2 = 34.7 − 20.85 = 13.85 psi
The critical pressure, Pc in Eq. 15.188 is:
G RT Pc = 11400d 2 k[k + 1] 24000 11400(6.065)2 = 4.752 psia
Pc =
1/ 2
, psia
34.085 × 500 1.15 × 2.15
The outlet temperature, T2 is the same as the inlet temperature, (i.e., isothermal condition)
T2 = T1, °R T2 = 550, °R Vapor density at the exit is:
ρ2 =
M w • P2 , lb/ft 3 10.72 T2
ρ2 =
(44 )(20.85) (10.72)(550)
= 0.1556 lb/ft 3
Flow velocity at pipe exit is:
v = G/(ρ A) 24000 (0.1556 × 0.2006 × 3600) = 213.59 ft /s. =
Fluid Flow 197 The exit Mach number Ma 2 = 1/ k
Ma 2 = 1/ 1.15 = 0.9325 The Excel program (Example 15.14.xlsx) provides calculations of Example 15.14.
15.24.5 Critical Pressure Ratio The maximum attainable mass flux for given supply conditions is when ψ is a maximum, and is represented by: 1/ 2
2 P 2 /γ P ( γ +1)/γ ψ = P − P γ − 1 1 1
(15.210)
Therefore, the pressure Pc causing the maximum flux can be found by differentiating ψ2 with respect to P and equating the result to zero, which gives
2 P 2 /γ γ + 1 P ( γ +1)/γ 1 c c − =0 P γ γ P Pc 1 1
(15.211)
The critical pressure ratio Pc/P1 is given by: γ
Pc 2 ( γ −1) = P1 γ + 1
(15.212)
For γ = 1.4, Pc/P1 = 0.528. For other values of γ, the value of the critical pressure ratio lies in the approximate range 0.5 to 0.6 where P1 = upstream pressure Pc = critical pressure γ = ratio of specific heat at constant pressure to specific heat at constant volume = (Cp/Cv). The maximum flux or maximum mass flow of the gas at sonic conditions is: 0.5
kgM 2 ( k +1))/k−1 W w = G max = Po A max ZRT o k +1
where W = mass flow rate, lb/s A = cross-sectional flow area, ft2 G = mass flux, lb/(s. ft2) Po = pressure at source condition, psia To = temperature at source condition, °R k = specific heat ratio constant
(15.213)
198 Petroleum Refining Design and Applications Handbook Volume 2 Recently, Kumar [34] develops a method using thermodynamic principles to determine the status of flow (i.e., whether choking flow exists or not), (ΔP/Po)cr and the flow rate. His method removes the use of plots as generated in Crane Manual [4], which have few limitations. For an adiabatic compressible fluid flow, he showed that ( γ +1) 2 2 2 2 + ( γ − 1)Ma12 {γ + 1}Ma1 + 1 2 ln +K =0 − 2 + ( γ 1 ) Ma Ma12 ( γ + 1) 2 + {γ − 1}Ma12 1
P and r = 2 = Po cr
(15.214)
1/ 2
0.5( γ + 1)Ma12
1 + 0.5( γ − 1)Ma 2 1
γ ( γ +1) 2( γ −1)
(15.215)
The critical expansion factor is:
Ycr =
K(1 + r ) 1 2 K + 2 ln r
(15.216)
and the mass flow rate at critical condition is:
where D L PA Po P1 P2 Ma1 Ma2 γ K Vo r Ycr W
W = 0.1265 D2 Ycr
Po − P2 KVo
(15.217)
= pipe internal diameter, mm = pipe length, m = ambient pressure, kPaa. = stagnation upstream pressure, kPaa = pressure at inlet tip of the pipe, kPaa = pressure at outlet tip of the pipe, kPaa = Mach number at inlet tip of the pipe = Mach number at outlet tip of the pipe = isentropic ratio of specific heat at constant pressure to specific heat at constant volume = loss coefficient = specific volume at upstream stagnation point, m3/kg = (P2/Po)cr = overall critical pressure ratio, dimensionless = critical expansion factor, dimensionless = mass flow rate, kg/h.
The simultaneous solution of Eqs. 15.214 and 15.215 eliminates Ma and yields a value for r, the critical pressure ratio. Table 15.16 shows a wide variation in the critical values with respect to γ (i.e., ratio of specific heats, Cp/Cv) and the loss coefficient K.
Fluid Flow 199 Table 15.16 Limiting critical values. γ = 1.2
γ = 1.3
γ = 1.4
γ = 1.5
γ = 1.6
K
(ΔP/Po)cr
Ycr
(ΔP/Po)cr
Ycr
(ΔP/Po)cr
Ycr
(ΔP/Po)cr
Ycr
(ΔP/Po)cr
Ycr
1
0.62
0.52
0.64
0.51
0.66
0.50
0.68
0.50
0.70
0.49
2
0.64
0.54
0.67
0.53
0.69
0.53
0.71
0.52
0.73
0.51
3
0.68
0.58
0.70
0.56
0.73
0.55
0.75
0.54
0.78
0.53
4
0.71
0.60
0.74
0.58
0.76
0.57
0.78
0.55
0.80
0.54
5
0.74
0.61
0.76
0.59
0.78
0.58
0.81
0.56
0.82
0.54
6
0.75
0.62
0.78
0.60
0.80
0.58
0.82
0.56
0.84
0.55
7
0.77
0.62
0.79
0.60
0.81
0.58
0.84
0.56
0.85
0.54
8
0.78
0.63
0.80
0.61
0.83
0.59
0.85
0.56
0.86
0.54
9
0.79
0.63
0.81
0.61
0.84
0.59
0.85
0.56
0.88
0.54
10
0.80
0.63
0.82
0.61
0.85
0.59
0.86
0.56
0.88
0.54
20
0.86
0.64
0.88
0.61
0.89
0.58
0.91
0.55
0.93
0.52
30
0.88
0.64
0.90
0.60
0.92
0.56
0.93
0.53
0.94
0.50
40
0.90
0.63
0.92
0.59
0.94
0.55
0.94
0.52
0.96
0.48
50
0.92
0.63
0.93
0.59
0.94
0.55
0.95
0.52
0.96
0.47
60
0.92
0.62
0.94
0.58
0.95
0.54
0.95
0.51
0.97
0.46
70
0.92
0.62
0.94
0.58
0.95
0.54
0.96
0.50
0.97
0.46
80
0.93
0.62
0.94
0.57
0.96
0.53
0.96
0.49
0.97
0.45
90
0.93
0.62
0.94
0.57
0.96
0.53
0.96
0.49
0.97
0.45
100
0.94
0.61
0.95
0.57
0.96
0.52
0.97
0.48
0.98
0.44
(Source: S. Kumar, Chem. Eng. Oct. 2002, p. 62).
Example 15.15 From the table listed below determine the status of flow (i.e., whether choking flow exists or not), (ΔP/Po)cr and the flow rate [34]. Data
Value
Upstream pressure, Po kPaa
6600
Downstream discharge pressure, PA kPaa
200
Upstream specific volume, Vo m3/kg
0.01724
Isentropic coefficient, γ
1.55
Internal pipe diameter, D mm
52.5
Length of pipe m
100
Number of elbows
4
Loss coefficient K
45
200 Petroleum Refining Design and Applications Handbook Volume 2 Solution For known isentropic coefficient γ and loss coefficient K, a guessed value of Ma1 is used in Eq. 15.214 until the left side of the equation approximates to a value of zero. Otherwise, a new guess value of Ma1 is used in Eq. 15.214. Once the required value is known, Eqs. 15.215, 15.216, and 15.217 are subsequently used to determine r, P2, ΔP, Ycr, and W, respectively. This procedure involves the use of the Excel spreadsheet with the Goal Seek or Solver add-in from the Tools menu and is given in Example 15.15.xlsx. Using the Excel spreadsheet Example 15.15.xlsx, the calculated overall critical pressure ratio r is:
r = 0.04804 The critical pressure P2 is
P2 = r × Po
= 0.04804 × 6600
= 317.06 kPa Test for choke flow Since P2 > PA, the pipe will choke The critical expansion factor Ycr from Eq. 15.216 is:
Ycr =
=
K(1 + r ) 2 K + 2 ln 1/r
(
)
45(1 + 0.04804 ) 2 ( 45 + 2[1/0.04804 ])
= 0.6795
The critical mass flow rate from Eq. 15.217 is:
W = 0.1265 D2 Ycr
Po − P2 KVo
= 0.1265(52.5)2 (0.6795) = 21320.96 kg /h
6600 − 317.06 (45 × 0.01724 )
15.24.6 Adiabatic Flow If there is no heat transfer or energy dissipated in the gas when traversing from state 1 to state 2, the process is adiabatic and reversible, i.e., isentropic. However, the actual flow conditions are somewhere between isothermal and
Fluid Flow 201 adiabatic, and as such, the flow behavior can be described by the isentropic equations, where the isentropic constant k replaced a polytropic constant γ (i.e., 1< γ < k). For isothermal condition, γ = 1, whereas truly isentropic flow corresponds to γ = k. The density and temperature as a function of pressure are:
P ρ = ρ1 P1
1k
P , T = T1 P1
( k −1) k
(15.218)
The mass flow rate, G by using Eq. 15.218 to eliminate ρ and T and solving for G gives: 1/ 2
k P ( k −1)/k 2 2 1 − k + 1 P1 G = P1 ρ1 P2 2 4 f L − ln D k P1
(15.219)
where f = Fanning friction factor. If the system contains fittings as well as straight pipe, the term 4 f L/D (= Kf, pipe) can be replaced by ΣK f , i.e., the sum of all loss coefficients in the system.
15.24.7 The Expansion Factor, Y The adiabatic flow Eq. 15.215 can be represented in a form:
2ρ P G = Y 1 ΣK f
12
2 (1 − P2 P1 ) = Y P1 ρ1 ΣK f 12
(15.220)
where ρ1 = P1Mw/RT1, ΔP = P1 – P2, and Y is the expansion factor. Note that Eq. 15.220 without the Y term is the Bernoulli equation for an incompressible fluid of density ρ1. Therefore, the expansion factor Y = Gadiabatic/Gincompressible, is the ratio of the adiabatic mass flux (Eq. 15.219) to the corresponding incompressible mass flux, and is a unique function of P2/P1, k, and Kf. Figure 15.25a shows values of Y for k = 1.3 and k = 1.4 as a function of ΔP/P1 and ∑K f (which is denoted by simply K on these plots). Figures 15.25b and 15.25c show the expansion factor Y for compressible flow through nozzles and orifices, and plots of the critical pressure ratio for compressible flow through nozzles and Venturi tubes, respectively. The conditions corresponding to the lower ends of the lines on the plots (i.e., the “nought”) represent the sonic (choked flow) state where P2 = P2* . These same conditions are shown in the tables accompanying the plots, thus allowing the relationships for choked flow to be determined more accurately than is possible from reading the plots. Note: It is not possible to extrapolate beyond the “nought” at the end of the lines in Figures 15.25a and 15.25b as this represents the choked flow state, in which P2 = P2* (inside the pipe), and is independent of the external exit pressure. Figures 15.25a and 15.25b provide an alternative method of solving compressible adiabatic flow problems for piping systems. However, this procedure requires some iterations because the value of Kf depends on the Reynolds number, which cannot be determined until G is obtained.
202 Petroleum Refining Design and Applications Handbook Volume 2 k = 1.3
(k = approximately 1.3 for CO2, SO2, H2O, H2S, NH3, N2O, CI2, CH4, C2H2, and C2H4)
1.0
Limiting Factors For Sonic Velocity
0.95
k = 1.3
0.90
K
∆P P’ 1
Y
1.2 1.5 2.0
.525 .550 .593
.612 .631 .635
3 4 6
.642 .678 .722
.658 .670 .685
8 10 15
.750 .773 .807
.698 .705 .718
20 40 100
.831 .877 .920
.718 .718 .718
0.85 0.80 Y 0.75
0 10 K= 40 K= 20 K = 15 K = 10 K = 8.0 K = 6.0 K=
0.70
K=
2.0 K = .5 1 K = 1.2 K=
0.60 0.55
0
0.1
0.2
0.3
0.4
0.5 ∆P
4.0 3.0
K=
0.65
0.6
0.7
0.8
0.9
1.0
P’ 1
k = 1.4 (k = approximately 1.4 for Air , H2, O2, N2, CO, NO, and HCI)
1.0
Limiting Factors For Sonic Velocity
0.95
k = 1.4
0.90
K
∆P P’ 1
Y
1.2 1.5 2.0
.552 .576 .612
.588 .606 .622
3 4 6
.662 .697 .737
.639 .649 .671
8 10 15
.762 .784 .818
.685 .695 .702
20 40 100
.839 .883 .926
.710 .710 .710
0.85 0.80 Y 0.75
0.65
0.2
0.3
0.4
0.5
1.5
0.1
2.0
0
K=
1.2
0.55
K=
K=
0.60
4.0 K = 3.0 K=
0 10 K= 0 4 K= 0 2 K = = 15 K 10 K = 8.0 K = 6.0 = K
0.70
0.6
0.7
0.8
0.9
1.0
∆P P’ 1
Figure 15.25a Net expansion factor, Y, for compressible flow through pipe to a larger flow area (reprinted/adapted with permission from “Flow of Fluids Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
Fluid Flow 203 Expansion Factor Y
1.00 0.95
Square edge orifice
d0 d1
0.90
0 to 0.2
0.4
0.85
0.6 0.7
0.80
Nozzle or venturi meter d0 = 0 to 0.2 0.75 d1
0.8
= 0.5 = 0.6
0.70
= 0.7 = 0.75
0.65
= 0.8 = 0.85
0.60 0.55 0.50 0.45 0.40 0.35 k = 1.45 k = 1.40 k = 1.35 k = 1.30 k = 1.25
0 0 0 0 0
.2 .2 .2 .2 .2
.4 .4 .4
.6 .6
1.0 1.0
.8 .8 .8
.6
1.0 1.0
.8
.6 .4 .4 .6 Pressure ratio, ∆ P / P’1
.8
1.0
Figure 15.25b Net expansion factor, Y, for compressible flow through pipe to a larger flow area (reprinted/adapted with permission from “Flow of Fluids Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
15.24.8 Misleading Rules of Thumb for Compressible Fluid Flow In general, compressible fluid flow calculations are much more complicated than incompressible fluid flow. Recently, Walters [35, 36] has shown that rules of thumb that are applied in the design calculations for compressible fluid flow can be grossly misleading and erroneous. These common rules are: 1. Th at adiabatic and isothermal flow bracket all flow rates. These conditions do not occur. 2. If the pipe pressure drop in a compressible flow system is less than 40% of the inlet pressure, then the incompressible flow calculation methods can be safely employed, with the average density along the pipe used in the equations. He further showed that this rule is invalid unless associated with a particular f L/D ratio. 3. Choked air flow at 50% pressure drop. An equation often used to determine the likelihood of sonic choking is:
p* 2 = po γ + 1
γ ( γ −1)
(15.221)
where p* is the critical static pressure at sonic velocity and po is the local stagnation pressure at the orifice/valve. Walters indicated that using Eq. 15.221 with γ = 1.4 results in 47% pressure drop to
204 Petroleum Refining Design and Applications Handbook Volume 2
.64
.62
β 0.85
rc = P’2/P’1
.60
0.80
.58
0.75
0.70
.56
0.65 0.60 .54
0.50 0.40 0.20 0
.52 1.25
1.3
1.35
1.4
1.45
k = Cρ/Cv
Figure 15.25c Critical pressure ratio, rc, for compressible flow through nozzles and venture tubes (reprinted/adapted with permission from “Flow of Fluids Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved, Note: P = psia, β = ratio of small –to – large diameter in orifice and nozzles and contraction or enlargements in pipes).
obtain choking. Furthermore, He stated that Eq. 15.221 cannot be used with the supply pressure if there is any significant pressure drop from the supply to the orifice/valve. For gases with different specific heat ratios, the pressure drop ratio will differ somewhat, in accordance with Eq. 15.221. In addition, Eq. 15.221 breaks down for pipe-system analysis when pipe friction becomes a factor. This is because the stagnation pressure in the equation is the pressure at the upstream side of the shock wave. However, if there is any pressure drop in the pipe from the supply pressure to the shock wave, then the supply pressure cannot be used in Eq. 15.221. Instead, the local stagnation pressure at the shock wave must be used, which is unknown unless the pressure drop is determined from alternative means. Therefore, Eq. 15.221 cannot be used to predict the supply and discharge pressures necessary for sonic choking unless the piping has negligible friction loss.
15.24.9 Other Simplified Compressible Flow Methods As shown earlier, most gases are not isothermal, and therefore it is impossible to know how much error is introduced by the assumption of constant temperature. Simplified equations typically do not address sonic-choking conditions, and cannot address the delivery temperature. These equations break down at high Mach numbers. The entire pipe is solved in one lumped calculation instead of coupling the governing equations in marching order. It is difficult to extend the equations to pipe networks. Walters developed compressible flow equations for single pipe [35]:
Fluid Flow 205 Adiabatic flow equation and integrated solution are:
∫
L
0
f dx = D
∫
Ma 2
Ma1
1 − Ma 2 dMa 2 − 1 γ Ma 2 γ Ma 4 1 + 2
(15.222)
γ −1 2 Ma 2 1+ 2 fL 1 1 1 γ +1 Ma1 2 ln = − + γ −1 D γ Ma12 Ma 22 2 γ Ma 22 2 1+ Ma1 2
(15.223)
Isothermal flow equation and integrated solution are:
∫
LT
0
f dx = D
∫
Ma 2
Ma1
(1 − γMa 2 ) dMa 2 4 γ Ma
Ma12 1− Ma 22 Ma 22 f LT − = ln Ma 2 D γMa12 1
(15.224)
(15.225)
Computer software has been developed that models pipe systems of compressible fluids and this can be obtained from the website: www.aft.com.
15.24.10 Friction Drop for Flow of Vapors, Gases and Steam See Figures 15.26a and 15.26b A. The Darcy Rational Relation for Compressible Flow for Isothermal Process [4]
∆Ρ 0.000336fW 2 V 0.000336 f W 2 = = 100 ft d5 d5 ρ
0.000001959f ( q ′h ) S g2 ∆Ρ or = 100 ft d5 ρ
(15.122)
2
(15.226)
In SI units
∆P/100 m =
62530 f W 2 V 62530 f W 2 = d5 d 5ρ 93650 f ( q ′h ) S g
(15.123)
2
or ∆P/100 m =
d 5ρ
(15.227)
The general procedures outlined previously for handling fluids involving the friction factor, f, and the Re chart are used with the above relations. This is applicable to compressible flow systems under the following conditions [4].
206 Petroleum Refining Design and Applications Handbook Volume 2
40 .04 .05
Index 2
30 20
− V−Specific Volume of Flowing Fluid, in Cubic Feet per Pound
.03
p −Weight Density, in Pounds per Cubic Foot
15 .1
.2 .3 .4 .5 .6 .7 .8 .9 1.0
1.5 2 3 4 5
3 2
1.5
.6 .7 .8 .9 1.0 1.5
10 9 8 7 6 5 4
1000 800 600 500 400 300
.5
2 3 4
(2) (3)
1.0 .9 .8 .7 .6 .5 .4
5 6 7 8 9 10 15 20
.3
30
.2
40
d
200
30
20
10 9 8 7 6 5 4 3 2
24 20 16 14 12 10 8 6 5 4 3½ 3 2½ 2 1½ 1¼
1 .9 .8 .7 .6 .5 .4 .3
1 ¾ ½ ¾ ¼ ½
.2
50
f .05
(1)
20
.04 .03 .02
100 80 60 50 40 30
.015
10 8 6 5 4 3 2
.01
W− Rate of Flow, in 1000 Pounds per Hour ΔP/100 feet = 0.000336 f W2/d5p
Index 1
f −Friction Factor
− V 50
Nominal Diameter, in Inches (Standard Pipe - Schedule 40)
P .02
W 1600
ΔP100 .4
ΔP100−Pressure Drop per 100 Feet, in Pounds per Square Inch d −Internal Diameter of Pipe, in Inches
∆ P/100 ft = 0.000 336 f W2/d5p
1 .8 .6 .5 .4 .3 .2 .1
Figure 15.26a Pressure drop in compressible flow lines (reprinted/adapted with permission from “Flow of Fluids, Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
where W = rate of flow, lb/h (kg/h) V = specific volume of fluid, cubic feet per pound (m3/kg) f = friction factor d = internal pipe diameter, in. (mm) ρ = fluid density, lb/ft3 (kg/m3) Sg = specific gravity of gas relative to air = the ratio of molecular weight of the gas to that of air. q ′h = rate of flow, cu ft/hr (m3/h) at standard conditions (14.7 psia and 60oF), SCFH (m3/s at metric standard conditions (MSC)—1.01325 bar absolute and 15°C) Babcock formula for steam flow at isothermal condition is
q ′h = 24 , 700[Yd 2 /Sg ](∆Ρ ρ1 /K )1/2 , CFH at 14.7 psia and 60°F
(15.228)
q ′h = 40, 700 Yd 2 (∆P)( P1′) /(KT1Sg )
(15.229)
or
1/ 2
Fluid Flow 207 Pressure Equivalent: 1bar = 10s Pa
W 103 800 600 500 400 300
= 100 kPa
3
.4
200
Index 2
1.5
.2
1.0
1.0
.3
1.5
.8 .7 .6 .5
600 500
.4
400
4 5
.2
6 7 8
.15
10
.1
15
.08 .07 .06 .05
20 30
.04 .03
40 50 60 70 80
.02
100
.01
.5 .6 .7 .8 .9 1.0 1.5 2
3 4 5 6 7 8 9 10
.015
300
d 24 20 16 12 10
200
8
150
6 5
100
4
80
3
60 50
2½ 2
40
1½ 1¼ 1
Interanl Diameter of Pipe, in millimeter
.3
Pressure Drop per 100 metres, in bar
.4 3
800
Specific Volume of Flowing Fluid, in cubic metres per kilogram
2 Density, in kilogram per cubic metre
.15
2
30 20 15 10 8 6 5
3/4 1/2 3/8 1/4 1/8
100 80 60 50 40 30 f
20 .05 10
.04 .03
.02
Friction Factor
.5 .6 .7 .8
.015 .01
8 6 5 4 3 2
1.0 .8 .6 .5 .4 .3
Rate of Flow, in thousands of kilograms per hour
Index 1
V
Normal Size of Schedule 40 Pipe, in inches
p .3
Δp100 .09 .1
.2 .1 .08 .06 .05 .04
Figure 15.26b Pressure drop in compressible flow lines (metric units) (reprinted/adapted with permission from “Flow of Fluids, Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
where d = internal pipe diameter, in. Y = net expansion factor for compressible flow through orifices, nozzles, or pipe (see Figures 15.25a–c) K = loss coefficient for all valves, fittings and pipe (resistance coefficient)
K = f L /D +
∑
Ki i = fittings+ valves
q ′h = flow rate, ft3/h at 14.7 psia and 60°F Sg = specific gravity of a gas relative to air = the ratio of the molecular weight of the gas to that of air P = pressure, lbf/in2 absolute ΔP = pressure drop, psi
208 Petroleum Refining Design and Applications Handbook Volume 2 W
C1
1500 2000 1500
ΔP100 − C2V
=
ΔP100ρ C2
C2 =
ΔP100 − C1V
C1 = discharge factor from chart at right. C2 = size factor, from table on next page. The limitations of the Darcy formula for compressible flow, as outlined on page 3-3 appy also to the simplified flow formula.
Example 1 Given: Stream at 345 psig and 500 F flows through 8-inch Schedule 40 pipe at a rate of 240, 000 pounds per hour Find: The pressure drop per 100 feet of pipe. Solution :
.1
10
6
ρ =
C1
5
4
V = 1.45
3 2.5
2
. . . . . . . . page 3-17 or A-16
ΔP100 = 57 × 0.146 × 1.45 = 12
Example 2
1.0
Given: Pressure drop is 5 psi with 100 psig air at 90 F flowing through 100 feet of 4-inch Schedule 40 pipe.
.9
Solution:
.5
.8
.6 .7 .8 .9 1.0
1.5
.005
2
.004
2.5
4
.002
5
.001 .0009 .0008 .0007 .0006
20
25 30
40
50
3
.0025
.0015
600
1000 900 800 700 600 500 400 300
.02
.006
700
15
.3 .4
.003 1.5
Find: The flow rate in standard cubic feet per minute.
.25
.025
.01 .009 .008 .007
800
.2
.04
3.5
C1 = 57 C1 = 0.146
.15
.05
.015
1000 900
.03
ΔP100ρ W - Rate of Flow, in Thousands of Pounds per Hour
C1 =
.1 .09 .08 .07 .06
7
C1C2
−
10
8
The simplified flow formula can then be written: ΔP100 = C1C2V =
C1
9
336 000 f d5
C2 =
C1
6 7 8 9 10
60 70
500
250 200
400
300 250
150
100 90 80 70 60 50
200
40 30
150
80
25 20 15
90 100
100
10
ΔP100 = 5.0 C2 = 5.17 ρ = 0.564
. . . . . . . . . . . page A-10
C1 = (5.0 × 0.564)÷ 5.17 = 0.545 W = 23000
For C2 values and an exmaple on “determining pipe size”, see the opposite page.
q'm = W ÷ (4.58 Sg) . . . . . . . . . . . page B-2 q'm = 23000 ÷ (4.58 × 1.0) = 5000 scfm
Figure 15.27a Simplified flow formula for compressible fluids (reprinted/adapted with permission from “Flow of Fluids, Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
Values of C1
C1 = W210–9
− V
336 000 f d5
W
Values of C1
− V = (W210–9)
0.00 336 f d5
ΔP100 = W2
W
W - Rate of Flow, in Thousands of Pounds per Hour
The Darcy formula can be written in following from:
Fluid Flow 209
9 8
62 530 × 103 f d5
7
The simplified flow formula can the bed written: Δp100 = C1 C2 V = C1 =
Δp100 C2 V
~
=
6
C1 C2 ρ Δp100ρ C2
C2 =
Δp100 C1 V
=
5
Δp100ρ C1
4
C2 = size factor from tables on pages 3-23 to 3-25
Given: Steam at 24 bar absolute and 250°C flows through an 8-inch Schedule 40 pipe at a rate pf 100 000 kilograms per hour. Find: The pressure drop per 100 metres of pipe. C2 V
= 100 = 0.257 . . . . . . . . . . . . . . . . . . facing page = 0.091 m3/kg . . . . . . .page 3-17 or A-15
Δp100 = 100 × 0.257 × 0.091 Δp100 = 2.34 bar Example 2 Given: Pressure drop is 1 bar with 7 bar guage air at 30°C flowing through 100 metres of 4 inch nominal size ISO steel pipe, 6.3 mm wall thickness. Find: The flow rate in cubic metres per minute at metric standard conditions (1.013 25 bar and 15°C). Solution:
Δp100
Rate of Flow, in thousands of kilograms per hour
Example 1
3 2.5
2.5
2
4
.2
5
.15
.04 .025
1.5
1.0 .9 .8 .7 .6
C2
= 9.42. . . . . . . . . . . . . . . . . . page 3-24
ρ
= 9.21. . . . . . . . . . . . . . . . . . page A-10
C2
= 1 ×09.21 9.42
W
= 9 900
q’m
= W ÷ (73.5 Sg) . . . . . . . . . . page B-2
q’m
= 9 900 ÷ (73.5 × 1) = 134.7 m3/min
= 0.978
.5
2000 25 30
40
20 25
50
30 60
50
70
60 70 80 90 100
80
.01 .009 .008 .007 .006
2500
20
.02
400
300 250
1500
1000 900 800 700 600 500 400 300
150
250 200 150
90 100
100
100
.005 .004
.0025 .002
.4
15
5000
3000
40
.003
=1
6 7 8 9 10
C1 6000
600
.025
.015
700
15
3
.25
.05
W 103 800
2
.4
.1 .09 .08 .07 .06
W 103 10
1.5
.5
3.5
The limitations of the Darcy formula for compressible flow, as outlined on page 3-3 apply also to the simplified flow formula.
C1
1
.3
C1 = discharge factor, from chart at right
Solution:
1 .9 .8 .7 .6
Values of C1
Let C1=
W2 and C = 2 108
~ V
C1
For C2 values see opposite page and pages 3-24, 3-25 For example on determining pipe size see opposite page.
.0015
.35
Figure 15.27b Simplified flow formula for compressible fluids (metric units) (reprinted/adapted with permission from “Flow of Fluids, Through Valves, Fittings and Pipe”, Technical Paper No. 410, 1999, Crane Co. All rights reserved).
Values of C1
The Darcy formula can be written in the folling form; 62 530 f W2 V W2 62 530 × 103 f Δp100 = = 8 5 10 d d5
C1
Rate of Flow, in thousands of kilograms per hour
W 103 10
210 Petroleum Refining Design and Applications Handbook Volume 2 T1 = inlet temperature, absolute (°R = °F + 460) ρ1 = upstream density of steam, lb/ft3 In SI units
q ′h = 1.0312[Yd 2 /S g ](∆p ρ1 /K )1/2
(15.230)
q ′h = 19.31 Yd 2 (∆p)( p1′ ) /(KT1S g )
(15.231)
or 1/ 2
where d = internal pipe diameter, mm Y = net expansion factor for compressible flow through orifices, nozzles, or pipe K = loss coefficient (resistance coefficient) p = pressure, bara q ′h = flow rate, m3/h at MSC (metric standard conditions—1.01325 bar at 15°C) Sg = specific gravity of a gas relative to air = the ratio of the molecular weight of the gas to that of air Δp = pressure drop, bar T1 = inlet temperature, absolute (K = °C + 273) ρ1 = upstream density of steam, kg/m3 B. Alternate Vapor/Gas Flow Methods Note that all specialized or alternate methods for solving are based on simplified assumptions or empirical procedures presented earlier. They are not presented as better approaches to solving the specific problem. Figures 15.27a and 15.27b (SI) are useful in solving the usual steam or any vapor flow problem for turbulent flow based on the modified Darcy friction factors. At low vapor velocities the results may be low; then use Figures 15.26a or 15.26b (SI). For steel pipe the limitations listed in (A) above apply. 1. D etermine C1 and C2 from Figures 15.27a or 15.27b (SI units) and Table 15.17 for the steam flow rate and assumed pipe size, respectively. Use Table 15.18 to select steam velocity for line size estimate. 2. Read the specific volume of steam at known temperature and pressure from steam tables. 3. Calculate pressure drop (Figures 15.27a or 15.27b) per 100 ft of pipe from
∆Ρ/100 feet = C1C 2 V 4. 5. 6. 7.
(15.232)
Determine the loss coefficient K of all fittings, valves, etc., and hence the equivalent length (K = f Leq/D). etermine expansion and contraction losses, fittings, and at vessel connections. D Determine pressure drops through orifices and control valves. Total system pressure drop
ΔPTOTAL = (L + Leq)(ΔP/100) + Item 5 + Item 6
(15.233)
8. I f pressure drop is too large (or greater than a percentage of the inlet system pressure), re-estimate line size and repeat calculations (see paragraph (A) above) and also examine pressure drop assumption for orifices and control valves.
Fluid Flow 211 C. Air For quick estimates for air line pressure drop and through an orifice, see Tables 15.19a and 15.19b [37]. D. Babcock Empirical Formula for Steam Comparison of results between the various empirical steam flow formulas suggests the Babcock equation is a good average for most design purposes at pressure 500 psia and below. For pipe lines smaller than 4 in., this relation may be 0–40% high [38].
p 1 − p2 = ∆Ρ = 0.000131(1 + 3.6 d )
w2 L ρd5
(15.234)
ΔP/100 feet = w2F/ρ
(15.235)
Figure 15.28 is a convenient chart for handling most in-plant steam line problems. For long transmission lines over 200 ft, the line should be calculated in sections is order to re-establish the steam specific density. Normally an estimated average ρ should be selected for each line increment to obtain good results. Table 15.20 is used to obtain the value for “F” in Eq. 15.235.
Table 15.17 Simplified flow formula for compressible fluid pressure drop, rate of flow, and pipe sizes (use with Figure 15.27a). Values of C2 Nominal pipe size Inches
Schedule number
1/8
3/4
3/8
1/2
Value of C2
Nominal pipe size Inches
Schedule number
Value of C2
Nominal Pipe Size Inches
Schedule Number
Value of C2
40 s
7,920,000.00
5
40 s
1.59
16
10
0.00463
80 x
26,200,000.00
80 x
2.04
20
0.00421
120
2.69
30 s
0.00504
40 s
1,590,000.00
160
3.59
40 x
0.00549
80 x
4,290,000.00
… xx
4.93
60
0.00612
40 s
319,000.00
40 s
0.610
80
0.00700
80 x
718,000.00
80 x
0.798
100
0.00804
120
1.015
120
0.00926
6
40 s
93,500.00
160
1.376
140
0.01099
80 x
186,100.00
… xx
1.861
160
0.01244
160
4,300,000.00
… xx
11,180,000.00
20
0.133
30
0.135
10
0.00247
40 s
0.146
20
0.00256
8
18
(Continued)
212 Petroleum Refining Design and Applications Handbook Volume 2 Table 15.17 Simplified flow formula for compressible fluid pressure drop, rate of flow, and pipe sizes (use with Figure 15.27a). (Continued) Values of C2 Nominal Pipe Size Inches
Schedule Number
Value of C2
3/4
40 s
1
Nominal Pipe Size Inches
Schedule Number
Value of C2
21,200.00
60
80 x
36,900.00
80 x
160
100,100.00
… xx
627,000.00
11/2
Value of C2
0.163
…s
0.00266
0.185
30
0.00276
…x
0.00287
40
0.00298
100
0.211
120
0.252
5,950.00
140
0.289
60
0.00335
80 x
9,640.00
… xx
0.317
80
0.00376
160
22,500.00
160
0.333
100
0.00435
…xx
114,100.00
120
0.00504
20
0.0397
140
0.00573
160
0.00669
10
0.00141
20 s
0.00150
40 s
1,408.00
30
0.0421
80 x
2,110.00
40 s
0.0447
160
3,490.00
60 x
0.0514
… xx
13,640.00
80
0.0569
20
100
0.0661
30 x
0.00161
40 s
627.00
120
0.0753
40
0.00169
80 x
904.00
140
0.0905
60
0.00191
160
1,656.00
160
0.1052
…xx
4,630.00
80
0.00217
12 2
Schedule Number
40 s
10 11/4
Nominal Pipe Size Inches
20
0.0157
100
0.00251
40s
169.00
30
0.0168
120
0.00287
80x
236.00
…s
0.0175
140
0.00335
160
488.00
40
0.0180
160
0.00385
… xx
899.00
…x
0.0195
60
0.0206 (Continued)
Fluid Flow 213 Table 15.17 Simplified flow formula for compressible fluid pressure drop, rate of flow, and pipe sizes (use with Figure 15.27a). (Continued) Values of C2 Nominal Pipe Size Inches
Schedule Number
Value of C2
21/2
40 s
66.70
80 x
91.80
80
160
146.30
… xx
380.00
3
31/2
4
Nominal Pipe Size Inches
Nominal Pipe Size Inches
Schedule Number
Value of C2
24
10
0.000534
0.0231
20 s
0.000565
100
0.0267
…x
0.000597
120
0.0310
30
0.000614
140
0.350
40
0.000651
160
0.0423
60
0.000741
10
0.00949
80
0.000835
Schedule Number
Value of C2
40 s
21.40
80 x
28.70
160
48.30
20
0.00996
100
0.000972
… xx
96.60
30 s
0.01046
120
0.001119
40
0.01099
140
0.001274
160
0.001478
14
40 s
10.00
…x
0.01155
80 x
37.70
60
0.01244
40 s
5.17
80
0.01416
80 x
6.75
100
0.01657
120
8.94
120
0.01898
160
11.80
140
0.0218
… xx
18.59
160
0.0252
Note: The letters s, x, and xx in the columns of Schedule Numbers indicate Standard, Extra Strong, and Double Extra Strong pipe respectively.
Table 15.18 Suggested steam pipe velocities in pipe connecting to steam turbines. Service-steam
Typical range, ft/s
Inlet to turbine
100–150
Exhaust, non-condensing
175–200
Exhaust, condensing
500–400
15.25 Darcy Rational Relation for Compressible Vapors and Gases 1. D etermine first estimate of line size by using suggested velocity from Table 15.13. 2. Calculate the Reynolds number, Re and determine the friction factor, f using Figure 15.5 or from Eqs. 15.33–15.36. 3. Determine the total straight pipe length, L
214 Petroleum Refining Design and Applications Handbook Volume 2 4. D etermine equivalent pipe length for fittings and valves, Leq. 5. Determine or assume losses through orifice plates, control valves, equipment, contraction and expansion and so on. 6. Calculate pressure drop, ΔP/100 ft (or use Figures 15.26a or 15.26b).
∆P 0.000336 fW 2 0.000000726 f TS g ( q ′h ) = = 100 ft ρd 5 P′ d 5
2
7. Total pressure drop, ΔP total
= (L + Leq)(ΔP/100) + Item 5
(15.236)
8. I f the total line or system pressure drop is excessive (or greater than a percentage of the inlet system pressure), examine the portion of pressure drop due to pipe friction and that due to other factors in Table 15.19a Flow of air through Sch. 40 pipe (used for estimating for detailed calculations, use friction factors f). For lenghts of pipe other than 100 feet, the pressure drop is proportional to the length. Thus for 50 feet of pipe, the pressure drop is approximately one-half the value given in the table ... for 300 feet, three times the given value, etc. The pressure drop is also inversely proportional to the absolute pressure and directly proportional to the absolute temperature. Therefore, to determine the pressure drop for inlet or average pressures other than 100 psi and at temperature other than 60 F, multiply the values given in the table by the ratio: 100 + 14.7 P + 14.7
460 + 1 520
where: “P” is the inlet or average gauge pressure in pounds per square inch, and, “t” is the temperature in degrees Fahrenheit under consideration. The cubic feet per minute of compressed air at any pressure is inversely proportional to the absolute pressure and directly proportional to the absolute temperature. To tdetermine the cubic feet per minute of compressed air at any temperature and pressure other than standard conditions, multiply the value of cubic feet per minute of free air by the ratio: 14.7 14.7 + P
460 + t 520
Free air q’m cubic ft per min at 60°F and 14.7 psia
Compressed air cubic ft Pressure drop of air in pounds per square inch per 100 feet of per min at 60°F and schedule 40 pipe for air at 100 lbs per square in. gauge pressure and 60°F temperature 100 psig 0.128 0.256 0.384 0.513 0.641
1/8ʺ 0.361 1.31 3.06 4.83 7.45
1/4ʺ 0.083 0.285 0.605 1.04 1.58
3/8ʺ 0.018 0.064 0.133 0.226 0.343
1/2ʺ
1 2 3 4 5
0.020 0.042 0.071 0.106
0.027
6 8 10 15 20
0.769 1.025 1.282 1.922 2.563
10.6 18.6 28.7 … …
2.23 3.89 5.96 13.0 22.8
0.408 0.848 1.26 2.73 4.76
0.148 0.255 0.356 0.834 1.43
0.037 0.062 0.094 0.201 0.345
0.019 0.029 0.062 0.102
0.026
25 30 35 40 45
3.204 3.845 4.486 5.126 5.767
… … … … …
35.6 … … … …
7.34 10.5 14.2 18.4 23.1
2.21 3.15 4.24 5.49 6.90
0.526 0.748 1.00 1.30 1.62
0.156 0.219 0.293 0.379 0.474
0.039 0.055 0.073 0.095 0.116
0.019 0.026 0.035 0.044 0.055
2”
50 60 70 80 90
6.408 7.690 8.971 10.25 11.53
0.019 0.023
28.5 40.7 … … …
8.49 12.2 16.5 21.4 27.0
1.99 2.85 3.83 4.96 6.25
0.578 0.819 1.10 1.43 1.80
0.149 0.200 0.270 0.350 0.437
0.067 0.094 0.126 0.162 0.203
0.019 0.027 0.036 0.046 0.058
100 125 150 175 200
12.82 16.02 19.22 22.43 25.63
0.029 0.044 0.062 0.083 0.107
0.021 0.028 0.036
33.2 … … … …
7.69 11.9 17.0 23.1 30.0
2.21 3.39 4.87 6.60 8.54
0.534 0.825 1.17 1.58 2.05
0.247 0.380 0.537 0.727 0.937
0.070 0.107 0.151 0.205 0.264
225 250 275 300 325
28.84 32.04 35.24 38.45 41.65
0.134 0.164 0.191 0.232 0.270
0.045 0.055 0.066 0.078 0.090
0.022 0.027 0.032 0.037 0.043
37.9 … … … …
10.8 13.3 16.0 19.0 22.3
2.59 3.18 3.83 4.56 5.32
1.19 1.45 1.75 2.07 2.42
0.331 0.404 0.484 0.573 0.673
350 375 400 425 450
44.87 48.06 51.26 54.47 57.67
0.313 0.356 0.402 0.452 0.507
0.104 0.119 0.134 0.151 0.168
0.050 0.057 0.064 0.072 0.081
0.030 0.034 0.038 0.042
… … … … …
25.8 29.6 33.6 37.9 …
6.17 7.05 8.02 9.01 10.2
2.80 3.20 3.64 4.09 4.59
0.776 0.887 1.00 1.13 1.26
475 500 550 600 650
60.88 64.08 70.49 76.90 83.30
0.562 0.623 0.749 0.887 1.04
0.187 0.206 0.248 0.293 0.342
0.089 0.099 0.118 0.139 0.163
0.047 0.052 0.062 0.073 0.086
… … … … …
11.3 12.5 15.1 18.0 21.1
5.09 5.61 6.79 8.04 9.43
1.40 1.55 1.87 2.21 2.60
3/4” 1”
2 1/2”
3”
3 1/2”
1 1/4”
1 1/2”
4”
5”
(Continued)
Fluid Flow 215 Table 15.19a Flow of air through Sch. 40 pipe (used for estimating for detailed calculations, use friction factors f). (Continued) Free air Compressed q’m cubic ft air cubic ft per min per min Pressure drop of air in pounds per square inch per 100 feet of at 60°F and at 60°F and schedule 40 pip for air at 100 lbs per square in. 14.7 psia 100 psig gauge pressure and 60°F temperature 700 750 800 850 900
Calculations for Pipe Other than Schedule 40 To determine the velocity of water, ot the pressure drop of water or air, through pipe other than Schedule 40, use the following formulas: υa = υ40 ∆Pa = ∆P40
d40 dα d40 dα
2
5
Subscript “α” refers to the Schedule of pipe through which velocity pr pressure drop is desired. Subscript “40” refers to the velocity or pressure drop through Schedule 40 pipe, as given in the tables on these facing pages.
89.71 96.12 102.5 108.9 115.3
1.19 1.36 1.55 1.74 1.95
0.395 0.451 0.513 0.576 0.642
0.188 0.214 0.244 0.274 0.305
0.099 0.113 0.127 0.144 0.160
0.032 0.036 0.041 0.046 0.051
24.3 27.9 31.8 35.9 40.2
10.9 12.6 14.2 16.0 18.0
3.00 3.44 3.90 4.40 4.91
6”
950 1 000 1 100 1 200 1 300
121.8 128.2 141.0 153.8 166.6
2.18 2.40 2.89 3.44 4.01
0.715 0.788 0.948 1.13 1.32
0.340 0.375 0.451 0.533 0.626
0.178 0.197 0.236 0.279 0.327
0.057 0.063 0.075 0.089 0.103
0.023 0.025 0.030 0.035 0.041
… … … … …
20.0 22.1 26.7 31.8 37.3
5.47 6.06 7.29 8.63 10.1
1 400 1 500 1 600 1 800 2 000
179.4 192.2 205.1 230.7 256.3
4.65 5.31 6.04 7.65 9.44
1.52 1.74 1.97 2.50 3.06
0.718 0.824 0.932 1.18 1.45
0.377 0.431 0.490 0.616 0.757
0.113 0.136 0.154 0.193 0.237
0.047 0.054 0.61 0.075 0.094
0.023
2 500 3 000 3 500 4 000 4 500
320.4 384.5 448.6 512.6 576.7
14.7 21.1 28.8 37.6 47.6
4.76 6.82 9.23 12.1 15.3
2.25 3.20 4.33 5.66 7.16
1.17 1.67 2.26 2.94 3.69
0.366 0.524 0.709 0.919 1.16
0.143 0.204 0.276 0.358 0.450
0.035 0.051 0.068 0.088 0.111
0.016 0.022 0.028 0.035
12”
5 000 6 000 7 000 8 000 9 000
640.8 769.0 897.1 1025 1153
… … … … …
18.8 27.1 36.9 … …
8.85 12.7 17.2 22.5 28.5
4.56 6.57 8.94 11.7 14.9
1.42 2.03 2.76 3.59 4.54
0.55 2 0.794 1.07 1.39 1.76
0.136 0.195 0.262 0.339 0.427
0.043 0.061 0.082 0.107 0.134
0.018 0.025 0.034 0.044 0.055
10 000 11 000 12 000 13 000 14 000
1282 1410 1538 1666 1794
… … … … …
… … … … …
35.2 … … … …
18.4 22.2 26.4 31.0 36.0
5.60 6.78 8.07 9.47 11.0
2.16 2.62 3.09 3.63 4.21
0.526 0.633 0.753 0.884 1.02
0.164 0.197 0.234 0.273 0.316
0.067 0.081 0.096 0.112 0.129
15 000 16 000 18 000 20 000 22 000
1922 2051 2307 2563 2820
… … … … …
… … … … …
… … … … …
… … … … …
12.6 14.3 18.2 22.4 27.1
4.84 5.50 6.96 8.60 10.4
1.17 1.33 1.68 2.01 2.50
0.364 0.411 0.520 0.642 0.771
0.148 0.167 0.213 0.260 0.314
24 000 26 000 28 000 30 000
3076 3332 3588 3845
… … … …
… … … …
… … … …
… … … …
32.3 37.9 … …
12.4 14.5 16.9 19.3
2.97 3.49 4.04 4.64
0.918 1.12 1.25 1.42
0.371 0.435 0.505 0.520
11.8 13.5 15.3 19.3 23.9
8”
10” 37.3
the system. If the line pressure drop is a small portion of the total, little will be gained by increasing the pipe size. Consider reducing losses through items in step 5 above. Recheck other pipe sizes as may be indicated.
15.26 Velocity of Compressible Fluids in Pipe See Figures 15.29a and 15.29b.
2.40 W V 3.06 WV 3.06W = = 2 a d2 dρ
(15.237)
16670 W V 21220 WV 21220 W = = a d2 d 2ρ
(15.238)
vm = In SI units
vm =
216 Petroleum Refining Design and Applications Handbook Volume 2 Table 15.19b Discharge of air through an orifice* in cubic feet of free air per minute at standard atmospheric pressure of 14.7 lb. per sq. in. absolute and 70°F. Diameter of orifice 1 " 64
1 " 32
1 " 16
1 " 8
1 " 4
3 " 8
1 " 2
5 " 8
3 " 4
7 " 8
1
16.2
28.7
45.0
64.7
88.1
115
10.1
22.8
40.5
63.3
91.2
124
162
3.10
12.4
27.8
49.5
77.5
111
152
198
0.892
3.56
14.3
32.1
57.0
89.2
128
175
228
0.248
0.993
3.97
15.9
35.7
63.5
99.3
143
195
254
0.068
0.272
1.09
4.34
17.4
39.1
69.5
109
156
213
278
7
0.073
0.293
1.17
4.68
18.7
42.2
75.0
117
168
230
300
9
0.083
0.331
1.32
5.30
21.2
47.7
84.7
132
191
260
339
12
0.095
0.379
1.52
6.07
24.3
54.6
97.0
152
218
297
388
15
0.105
0.420
1.68
6.72
26.9
60.5
108
168
242
329
430
20
0.123
0.491
1.96
7.86
31.4
70.7
126
196
283
385
503
25
0.140
0.562
2.25
8.98
35.9
80.9
144
225
323
440
575
30
0.158
0.633
2.53
10.1
40.5
91.1
162
253
368
496
648
35
0.176
0.703
2.81
11.3
45.0
101
180
281
405
551
720
40
0.194
0.774
3.10
12.4
49.6
112
198
310
446
607
793
45
0.211
0.845
3.38
13.5
54.1
122
216
338
487
662
865
50
0.229
0.916
3.66
14.7
58.6
132
235
366
528
718
938
60
0.264
1.06
4.23
16.9
67.6
152
271
423
609
828
1082
70
0.300
1.20
4.79
19.2
76.7
173
307
479
690
939
1227
80
0.335
1.34
5.36
21.4
85.7
193
343
536
771
1050
1371
90
0.370
1.48
5.92
23.7
94.8
213
379
592
853
1161
1516
100
0.406
1.62
6.49
26.0
104
234
415
649
934
1272
1661
110
0.441
1.76
7.05
28.2
113
254
452
705
1016
1383
1806
Gauge pressure before orifice in pounds per sq. in.
Discharge in cubic feet of free air per minute
1………….
0.028
0.112
0.450
1.80
7.18
2………….
0.040
0.158
0.633
2.53
3………….
0.048
0.194
0.775
4………….
0.056
0.223
5………….
0.062
6
(Continued)
Fluid Flow 217 Table 15.19b Discharge of air through an orifice* in cubic feet of free air per minute at standard atmospheric pressure of 14.7 lb. per sq. in. absolute and 70°F. (Continued) Diameter of orifice 1 " 64
1 " 32
1 " 16
1 " 8
1 " 4
3 " 8
1 " 2
5 " 8
3 " 4
7 " 8
1
274
488
762
1097
1494
1951
284
506
790
1138
1549
2023
Gauge pressure before orifice in pounds per sq. in.
Discharge in cubic feet of free air per minute
120
0.476
1.91
7.62
30.5
122
125
0.494
1.98
7.90
31.6
126
Table is based on 100% coefficient of flow. For well rounded entrance multiply values by 0.97. For sharp edged orifices a multiplier of 0.65 may be used for approximate results. Values for pressures from 1 to 15 lbs gauge calculated by standard adiabatic formula. Values for pressures above 15 lb. gauge calculated by approximate formula proposed by S.A. Moss. aCP1 where Wa = 0.5303 T1 W = discharge in lbs. per sed. a C = Coefficient of flow a = area of orifice in sq. in. P1 = Upstream total pressure in lbs per sq. in. absolute T1 = Upstream temperature in °F, abs. Values used in calculating above table were: C = 1.0, P1 = gauge pressure + 14.7 lbs/sq. in. Weights (W) were converted to volume using density factor of 0.07494 lbs/cu. ft. This is correct for dry air at 14.7 lbs per sq. in. absolute pressure and 70°F. Formula cannot be used where P1 is less than two times the barometric pressure. By permission ”Compressed Air Data”, F. W. O’Neil, Editor, Compressed Air Magazine, 5th Edition, New York, 1939 [49].
where vm = mean velocity in pipe, at conditions stated for V, ft/min (m/s) a = cross-sectional area of pipe, in2 (mm2) W = flow rate, lb/h (kg/h) V = fluid specific volume, ft3/lb (m3/kg) d = inside pipe diameter, in. (mm) ρ = fluid density, lb/ft3 (kg/m3) at T and P Note that determining the velocity at the inlet conditions to a pipe may create significant error when results are concerned with the outlet conditions, particularly if the pressure drop is high. Even the average of inlet and outlet conditions is not sufficiently accurate for some systems; therefore conditions influenced by pressure drop can produce more accurate results when calculations are prepared for successive sections of the pipe system (long or high pressure). Example 15.16: Steam Flow Using Babcock Formula Determine the pressure loss in 138 feet of 8-in. Sch, 40 steel pipe, flowing 86,000 pounds per hour of 150 psig steam (saturated). Use Figure 15.28, w = 86,000/60 = 1432 lb/min Reading from top at 150 psig, no superheat, down vertically to intersect the horizontal steam flow of 1432 lb/min, follow diagonal line to the horizontal pipe size of 8 in. and then vertically down to the pressure drop loss of 3.5 psi/100 ft. For 138 ft (no fittings or valves), total ΔP is 138 (3.5/100) = 4.82 psi. For comparison, solve by equation, using value of F = 587.1 × 10−9 from Table 15.20.
ΔP/100 ft = (1432)2(587.1 × 10−9)/0.364
218 Petroleum Refining Design and Applications Handbook Volume 2 AVERAGE PRESSURE – LB. PER SQ. IN. ABSOLUTE
1 0 100 200 300 400 500 600 700 100000
SUPERHEATDEGREES FAHRENHEIT
5 1. 2
3 4 5 6 8 10 15 20 30 40 50 60 80 0 10 0 15 0 20
0 30 0 400 500 60 0 80 00 10 00 15 00 20
400 500 FAHR. Q 600 E 700 RE - D 800 ERATU 900 EMP 1000 EAM T 1100 ST
300
200
60 50
600000 400000
40
20000 10000
30
6000 4000
60 40
3
20 10
2
6 4 2 1.0
1
0.6
COPYRIGHT-WALWORTH COMPANY -1937
.2 01
30 40
.5
20
0.8
4 5 6 8 10
.75
.6 .4
STEAM FLOW IN POUNDS PER MINUTE
4
.01
.5
1
100
5
2
.75
1.5 1.25
200
6
.3 .4 .5 .6 .8 1.0
1
2
8
.2
1.5
1.25
2.5
600 400
.03 .04 .05 .06 .08 0.1
2
5 4.5 4 3.5 3
1000
.02
2.5
12 11 10 9 8 7 6
2000
16 14 12 10 ACTUAL INSIDE DIAMETER OF PIPE - INCHES
12 11 10 9 8 7 6 5 4.5 4 3.5 3
NOMINAL SIZE – STANDARD WEIGHT PIPE
NOMINAL SIZE – EXTRA STRONG PIPE
20
PRESSURE LOSS IN LB. PER SQ. INCH PER 100 FEET w2L 3.6 ) 5 Based on Backcock Formula : – P = 0.00013 (1+ d pd
Figure 15.28 Steam flow chart (by permission from Walworth Co. Note: Used for estimating only (Ludwig [19])).
= 3.31 psi/100 ft
ΔPtotal = (3.31/100)(138) = 4.57 psi These values are within graphical accuracy. For the discharge of compressible fluids from the end of a short piping length into a larger cross-section, such as a larger pipe, vessel, or atmosphere, the flow is considered adiabatic. Corrections are applied to the Darcy equation
Fluid Flow 219 Table 15.20 Factor “F” for Babcock steam formula*. Nominal pipe size in.
*Standard weight pipe
#Extra strong pipe
½
955.1 × 10−3
2.051 × 10−3
¾
184.7 × 10−3
340.8 × 10−3
1
45.7 × 10−3
77.71 × 10−3
1¼
9.432 × 10−3
14.67 × 10−3
1½
3.914 × 10−3
5.865 × 10−3
2
951.9 × 10−6
1.365 × 10−3
2½
351.0 × 10−6
493.8 × 10−6
3
104.7 × 10−6
143.2 × 10−6
3½
46.94 × 10−6
62.95 × 10−6
4
23.46 × 10−6
31.01 × 10−6
5
6.854 × 10−6
8.866 × 10−6
6
2.544 × 10−6
3.354 × 10−6
8
587.1 × 10−9
748.2 × 10−9
10
176.3 × 10−9
225.3 × 10−9
12
70.32 × 10−9
90.52 × 10−9
14 O.D.
42.84 × 10−9
55.29 × 10−9
16 O.D.
21.39 × 10−9
27.28 × 10−9
18 O.D.
11.61 × 10−9
14.69 × 10−9
20 O.D.
6.621 × 10−9
8.469 × 10−9
24 O.D
2.561 × 10−9
3.278 × 10−9
*Factors are based upon I.D. listed as Schedule 40. #Factors are based upon I.D. listed as Schedule 80. By permission, The Walworth Co.
220 Petroleum Refining Design and Applications Handbook Volume 2 to compensate for fluid property changes due to the expansion of the fluid, and these are known as Y net expansion factors [4]. The corrected Darcy equation is: For valves, fittings, and pipe (vapors/gases): English Engineering units
w = 0.525 Yd i2 ∆Ρ ρ1 /K , lb/s
(15.239)
w = 1891Yd i2 ∆Ρ ρ1 /K , lb/h.
(15.240)
In SI units
600 500 400
.06 .08
8 .2
4
3
2–
1.5 1– .8 .6 .5 .4 .3 .2
.15
.10 200 300 400 500 600 700 800 900 1000 1100 1200
1.2
1¼
.3 .4 .5 .6 1.0
2 2.5 3 4 5 6 10
150
2
V 60 40 30 20 10 6 4 3 2 1 .6 .4 .2 .1
100 80 60 50 40 30 20 15 10 8 6 5 4 3
3
3
3.5
”
nal mi No e Size ” 2 Pip ” 2½
200 V – Velocity, in Thousands of Feet per Minute
6 5
1”
1.6 1.8 2
300
.10
.9 1.0
1.4
d - Internal Diameter of Pipe, in Inches
10
8
t - Temperature, in Degrees Fahrenheit
1000 800
.04 .05
W - Rate of Flow, in Thousands of Pounds per Hour
15
.03
ρ – Weight Density, in Pounds per Cubic Foot
Steam
1
20
–V – Specific Volume, in Cubic Feet per Pound
rated
Satu
10 uge 15 h ga - inc 20 e r a squ 25 -per 30 unds o p 40 50 60 70 80 100 120 14600 1 0 1800 2 250 300 350 400 500 600 7000 80 0 90000 1 0 120 0 140 0 16000 18 00 20 0 240 0 280 1.5 319
d
W 1500
Index
3.06 W 3.06 WV = d2 d2 ρ
5
(15.241)
V=
0
– V 30
ρ
∆Ρ , kg /s (KV1 )
–
w = 1.111 × 10−6 Yd i2
3” ” 3½
4”
4 5”
5
6”
6 7
8”
8 9 10
10”
15 20
12” 14” 15” 18” 20” 24”
2 1.5
30
0 160 140 120 10 80 60 40 20
1
25
Schedule Number
Figure 15.29a Velocity in compressible flow lines (reprinted/adapted with permission from “Flow of Fluids Through Valves, Fittings, and Pipe”, Technical Paper No. 410, 1999, Crane, Co. All rights reserved).
Fluid Flow 221
− V
1.5 1 .8
2
2.5
3 ge r gau a b 4 5 6 7
.6 .5 .4
eam d St rate Satu
8 9 10 12 14 16 18 20 25
.3 .2 .15
30
.10
35
40 45 50 60 0 7 80
.08 .06 .05 .04
100
120
.03
140 160 180 200
.02 .015 .01 .008
100
200
300
400
500
600
700
4 5 6 8 10
20 25 30 40 50 60
100 80 V 103 20 10 6 4 2 1 .6 .4 .2 .1 .06 .04 .02
80 100
60 50 40 30 20 15 10 8 6 5 4
40 45 50
80 90 100
3” 3½”
4” 5”
150 200 250
2
350 400
.6 .5
nal omi 2” Nipe Size P 2½”
70
3
1.0 .8
450 500
6” 8” 10” 12” 14” 16” 18” 20” 24”
600 700 800
20 40 60 80 100 120 140 160
200
60
300
1.5
1¼” 1½”
35
200
.006
Temperature °C
30
150 2 2.5 3
1”
25
Internal Diameter of Pipre, in millimetres
1.5
d 20
300
.8 1.0
Rate of Flow, in thousands of kilograms per hour
1.0
Velocity, in thousands of metres per minutes
0.5
.5 .6
2
Specific Volume, in cubic meters per kilogram
0
Index
ρ
Density, in kilograms per cubic metre
Specific Volume of Steam
W 103 700 600 500 400
Schedule Number
Figure 15.29b Velocity in compressible flow lines (metric units) (reprinted/adapted with permission from “Flow of Fluids Through Valves, Fittings, and Pipe”, Technical Paper No. 410, 1999, Crane, Co. All rights reserved).
w = 1.265 Yd i2
∆Ρ ρ1 , kg/h K
(15.242)
For nozzles and orifices (vapors/gases): English Engineering units
w = 0.525 Yd i2 C′
∆Ρ , lb/s. V1
(15.243)
In SI units
w = 1.111 × 10−6 Yd i2 C′ For valves, fittings, and pipe (liquids):
∆Ρ , kg /s (KV1 )
(15.244)
222 Petroleum Refining Design and Applications Handbook Volume 2 English Engineering units
w = 0.525 d12
∆Ρ(ρ1 ) , lb/s K
(15.245)
In SI units
w = 1.111 × 10−6 d i2
∆Ρ(ρ1 ) , kg /s K
(15.246)
For nozzles and orifices (liquids): English Engineering units
w = 0.525 d i2 C′ ∆P(ρ1 ), lb/s
(15.247)
w = 1.111 × 10−6 d i2 C′ ∆Ρ(ρ1 ), kg /s
(15.248)
In SI units
where di = pipe inside diameter, in. (mm) ρ1 = upstream fluid density, lb/ft3 (kg/m3) w = rate of flow, lb/s (kg/s) ΔP = pressure drop across the system, psi, (bar) (inlet-discharge) K = total resistance (loss) coefficient of pipe, valves, fittings, and entrance and exist losses in the line Y = net expansion factor for compressible flow through orifices, nozzles, and pipes [3] (see Figures 15.25a, b, or c) ΔP = pressure drop ratio in ΔP/P , used to determine Y from Figures 15.25a and 15.25b. The ΔP is the difference between the inlet pressure and the pressure in the area of larger cross-section. C = flow coefficient for orifices and nozzles (Figures 15.15 and 15.16) For example, for a line discharging a compressible fluid to atmosphere, the ΔP is the inlet gauge pressure or the difference between the absolute inlet pressure and atmospheric pressure absolute. When ΔP/P falls outside the limits of the K curves on the charts, sonic velocity occurs at the point of discharge or at some restriction within the pipe, and the limiting value for Y and ΔP must be determined from the tables on Figures 15.25a, 15.25b and used in the velocity equation, vs, above [4]. Figures 15.25a and 15.25b are based on the perfect gas laws and for sonic conditions at the outlet end of a pipe. For gases/vapors that deviate from these laws, such as steam, the same application will yield about 5% greater flow rate. For improved accuracy, use the charts in Figures 15.25a and 15.25b (SI) to determine the downstream pressure when sonic velocity occurs. Then use the fluid properties at this condition of pressure and temperature in:
v s = kgRT = kg144P′V , ft /s or v s = γRT = γP′V , m/s
(15.249) (15.250)
Fluid Flow 223 to determine the flow rate at this condition from
v = q/A = 183.3 q/d2 = 0.0509 W/(ρ)(d2), ft/s
(15.251)
In SI units
v = q /A = 1.273 × 106 q /d 2 = 21.22
Q W = 354 2 , m/s 2 ρd d
(15.252)
where d = internal diameter of pipe, in. (mm) A = cross-section of pipe, ft2 (m2) q = ft3/s (m3/s) at flowing conditions T = temperature, °R (K =273 + t) t = fluid temperature, °C k = γ = ratio of specific heats (Cp/Cv) P = pressure, psia (N/m2 abs) W = flow, lb/h (kg/h) v = velocity, mean or average, ft/s (m/s) These conditions are similar to flow through orifices, nozzles and venturi tubes. Flow through nozzles and venturi devices is limited by the critical pressure ratio, rc = downstream pressure/upstream pressure at sonic conditions (see Figure 15.25c). For nozzles and venturi meters, the flow is limited by critical pressure ratio and the minimum value of Y to be used. For flow of gases and vapors through nozzles and orifices:
q = YC′A
2g(144 )∆Ρ 3 , ft /s ρ
(15.253)
In SI units
or q = YC′ A
2 ∆Ρ 3 , m /s ρ
(15.254)
where β = ratio of orifice throat diameter to inlet diameter C = flow coefficient for nozzles and orifices (see Figures 15.15 and 15.16), when used as per ASME specification for differential pressure ρ = fluid density, lb/ft3, (kg/m3) A = cross-sectional flow area, ft2 (m2 ) Note: the use of C eliminates the calculation of velocity of approach. The flow coefficient C is C ′ = C d / 1 − β 4 , Cd = discharge coefficient for orifices and nozzles [4]. For compressible fluids flowing through nozzles and orifices use Figures 15.15 and 15.16, using hL or ΔP as differential static head or pressure differential across taps located one diameter upstream, and 0.5 diameters downstream
224 Petroleum Refining Design and Applications Handbook Volume 2 from the inlet face of orifice plate or nozzles, when values of C are taken from Figures 15.15 and 15.16 [4]. For any fluid: 1/ 2
2g(144 )∆Ρ 3 q = C′A , ft /s ρ
(15.255)
In SI units
q = C′ A
2 ∆Ρ 3 , m /s flow ρ
(15.256)
For estimating purposes for liquid flow with viscosity similar to water through orifices and nozzles, the following can be used [7]:
Q = 19.636 C′d12 h
1 d 1− o di
4
, gpm
(15.257)
, l/ min
(15.258)
In SI units
Q = 0.2087 C′d12 h
where
1 d 1− o di
4
do is greater than 0.3 di
Q = 19.636 C′d o2 h where
do is less than 0.3 di
(15.259) (15.260)
In SI units
Q = 0.2087 C′d o2 h , l/ min or [4], W = 157.6 d o2 C′ h Lρ2
= 1891d o2 C′ ∆Ρρ
(15.261) (15.262) (15.263)
In SI units
W = 0.01252 d o2 C h Lρ2 = 1.265 d o2C ∆p ρ
(15.264)
where Q = liquid flow, gpm (l/min) C = fl ow coefficient for orifices and nozzles = discharge coefficient corrected for velocity of approach = Cd / 1 − β4
Fluid Flow 225 do = diameter of orifice or nozzle opening, in. (mm) di = pipe inside diameter in which orifice or nozzle is installed, in. (mm) h = static pressure head existing at a point, in. (meters) of fluid. hL = loss of static pressure head due to fluid flow, m of fluid. C = flow coefficient (see Figure 15.30 for water and Figures 15.15 and 15.16 for vapors or liquids) q = ft3/s (m3/s) at flowing conditions rc = critical pressure ratio for compressible flow, = P2′ /P1′ ΔP = pressure drop, psi Δp = pressure drop, bar (hL and Δp measured across taps at 1 diameter and 0.5 diameter) W = flow rate, lb/h (kg/h) Flow of gases and vapors (compressible fluids) through nozzles and orifices [40]. (For flow field importance see [41]). From [4]: 1/ 2
2g(144 )∆Ρ q = YC′ A ρ
, ft 3/s
(15.265)
(at flowing conditions) In SI units
q = YC′A 2∆p/ρ, m3/s
(15.266)
Y = net expansion factor from Figures 15.25A or 15.25B ΔP = differential pressure (equal to inlet gauge pressure when discharging to atmosphere) ρ = weight density of fluid, lb/ft3 (kg/m3) at flowing conditions A = cross-section area of orifice or nozzle, ft2 (m2) C = flow coefficient from Figure 15.15 or 15.16
W = 1891 Yd o2C′
∆Ρ , lb/h. V1
(15.267)
or W = 1.265 d o2 C′ ∆p ρ1 , kg /h
(15.268)
where do = internal diameter of orifice, in. (mm) V1 = specific volume of fluid, ft3/lb (m3/kg) RE-ENRTANT TUBE
SHARPEDGED
LENGTH: 1/2 to 1 DIA
C= .52
C= .61C
SQUAREEDGED
RE-ENTRANT TUBE
STREAM CLEARS SIDES
LENGTH: 2-1/2 DIA.
C= .61
C= .73
SQUAREEDGED
WELL ROUNDED
TUBE FLOWER TUBE
C= .82
C= .98
Figure 15.30 Discharge coefficients for liquid flow (by permission, Cameron Hydraulic data, Ingersol – Rand Co., Washington, NJ, 1979).
226 Petroleum Refining Design and Applications Handbook Volume 2 ρ1 = density of fluid, lb/ft3. (kg/m3) Δp = pressure drop, pis (bar)
q ′ = 11.30 Yd o2 C′
∆Ρ P1′ 3 , ft /s T1S g
(15.269)
at 14.7 psia and 60°F where Sg = Sp Gr gas relative to air, = mol wt. gas/29 T1 = absolute temperature, °R = (460 + °F) P1′ = pressure, psia In SI units
q = 0.005363Yd o2 C′
p p1′ , m3/s T1S g
(15.270)
where T1 = 273.15 + t t = fluid temperature, °C p1′ = pressure, bara Δp = pressure drop, bar Sg = Sp Gr gas relative to air, = mol wt. gas/29 Y = net expansion factor compressibility flow through orifices, nozzles, or pipe. Example 15.17 What is the flow rate of natural gas through a ruptured exchanger tube assuming (1) a complete break near the tube sheet (see Figure 15.23), and (2) isothermal flow? Compare the flow rate to adiabatic condition. The following data are [39]: Pressure in exchanger tubes, P1, psig
1,110
Relief valve set pressure, P2, psig
400
Gas temperature, °F
100
Compressibility factor, Z
0.9
Molecular weight
18.7
Ratio of specific heats k = Cp/Cv
1.3
Gas viscosity, μ, cP
0.1
Exchanger tubes, ¾ in. Schedule 160, ID. in.
0.614
Tube length, ft.
20
Friction factor for complete turbulence, f
0.025
Fluid Flow 227 Solution Internal diameter of ¾ inch Schedule 160 = 0.614 in. = 0.051 ft.
Upstream density is ρ1 =
=
M w P1 10.72 ZT (18.7 )(1110 + 14.77 ) (10.72)(0.9)(100 + 460)
= 3.892 lb/ft 3
Total resistance (loss) coefficient KTotal is:
L K Total = f T + D
∑K + K f
entrance
+ K exit
20 + 0.5 + 1.0 = 0.025 0.0051
= 11.30
Pressure in the exchanger, P1 = 1110 + 14.7 = 1124.7 psia Relief valve set pressure, P2 = 400 + 14.7 = 414.7 psia Using Eq. 15.188, the gas flow rate through the ruptured tube is
P12 − P22 ρ1 G = 1335.6 d P1 P1 K + 2 ln Total P2 2
0.5 , lb/h
1124.7 2 − 414.7 2 3.892 G = 1335.6(0.614) 1124.7 11.3 + 2 ln (1124.7 / 414.7 ) = 8492.5 lb/h 2
Gas Reynolds number is:
Re = 6.31
G 8492.5 = 6.31 dµ (0.614 )(0.1)
= 8.7 × 105 (fully turbulence)
Gas velocity v is:
v = 0.0509
G 8492.5 = 0.0509 2 d ρ1 (0.614 )2 (3.892)
= 294.61 ft /s.
0.5
(15.153)
228 Petroleum Refining Design and Applications Handbook Volume 2 Sonic velocity vs is:
v s = 68.1 kP1 /ρ1
= 68.1 (1.3)(1124.7)/(3.892) = 1319.9 ft /s.
The Mach number at the inlet Ma1:
Ma1 =
v 294.61 = v s 1319.9
= 0.223
Since the Mach number is less than 1, the flow through the pipe is subsonic. At adiabatic condition, the net expansion factor Y is:
ΔP/P1 = (1124.7 – 414.7)/1124.7 = 0.63 The loss coefficient KTotal = 11.3 From Figure 15.25a, Y at ΔP/P = 0.63 and k = 1.3 by interpolation is Y = 0.77 Using Eq. 15.234, the gas flow rate is:
w s = 1891Yd i2 ∆Ρ ρ1 /K , lb/h. = (1891)(0.77 )(0.614 )2
(710)(3.892) (11.3)
= 8584.1 lb/h
The percentage deviation between the two conditions (isothermal and adiabatic) is 1.07%. As such, there is very little difference in flow rate between the two conditions. However, the adiabatic flow rate is always greater than the isothermal flow rate. Table 15.21 shows the computer result of Example 15.17 at isothermal condition using developed program prog33.for with data file DATA33.DAT. The Excel spreadsheet Example 15.17.xlsx gives calculations for isothermal and adiabatic conditions of the above example.
15.27 Procedure A. How to determine pipe size for given capacity and pressure drop. 1. A ssume a pipe diameter, and calculate velocity in ft/s using the given flow. 2. Calculate sonic velocity for fluid using Eqs. 15.174–15.179. 3. If sonic velocity of step 2 is greater than calculated velocity of step 1, calculate line pressure drop using usual flow equations. If these velocities are equal, then the pressure drop calculated will be the maximum for the line, using usual flow equations. If sonic velocity is less than the velocity of step 1, reassume line size and repeat calculations.
Fluid Flow 229 Table 15.21 Input data and computer results for maximum compressible fluid in a pipe line. DATA33.DAT 0.614 20.0 0.026 2.026 0.9 100.0 18.7 1124.7 414.7 1.3 0.1 Compressible fluid flow calculations in a pipe line Pipe internal diameter, in:
0.614
Straight length of pipe, ft:
20.000
Maximum fluid flow rate, lb/h:
8222.85
Fluid density, lb/ft^3:
3.893
Pipe friction factor:
0.0260
Fluid compressibility factor:
0.9000
Fluid temperature, °F:
100.000
Fluid molecular weight, Mw:
18.700
Ratio of specific heat capacities, Cp/Cv:
1.300
Fluid viscosity, cP:
0.1000
Resistance coeff. due to frictional loss:
10.163
Resistance coeff. due to fittings + valves:
2.026
Total resistance coefficient:
12.189
Inlet fluid pressure, psia:
1124.700
Outlet fluid pressure, psia:
414.700
Pressure drop, psi:
710.000
Fluid velocity, ft/s:
285.201
Fluid sonic velocity, ft/s:
1391.393
Mach number at inlet:
0.2050
Mach number at critical condition:
0.8771
Reynolds number:
845052.
Critical pressure, psia:
237.926
Fluid flow is:
SUBSONIC
230 Petroleum Refining Design and Applications Handbook Volume 2 B. How to determine flow rate (capacity) for a given line size and fixed pressure drop. This is also a trial and error solution following the pattern of (A), except capacities are assumed and the pressure drops are calculated to find a match for the given conditions of inlet pressure, calculating back from the outlet pressure. C. How to determine pressure at inlet of pipe system for fixed pipe size and flow rate. Determining the sonic flow rate involves knowledge of the local conditions at the exit. However, this is difficult to establish and highly complicated in practice as it requires extensive iterative computations. Example 15.18: Gas Flow Through Sharp-Edged Orifice A 1-in. Sch. 40 pipe is flowing methane at 40 psig and 50°F. The flange taps across the orifice (0.750 in. diameter) show a 3 psi pressure differential. Determine the flow rate through the orifice. Solution CH4; Sp Gr = Sg = 0.553 Gas Constant = R = 96.5 Ratio of specific heat = k = 1.26 Absolute system pressure = P = 40 + 14.7 = 54.7 psia ΔP/P1 = 3.0/54.7 = 0.0549 Pipe ID = 1.049 in. do/d1 = 0.750/1.049 = 0.7149 From Figure 15.25a, Y = 0.97; From Figure 15.16.
C′(assumed turbulent ) =
Cd
12
1 − ( d /d )4 o 1
where Cd = orifice discharge coefficient, uncorrected for velocity of approach.
C = 0.74 at est. Re ≥ 2000 Temperature = 460 + 50 = 510°F
Density = ρ =
144P 144(54.7 ) = RT (96.5)(510)
= 0.160 lb/ft 3
W = 1891 Yd C (∆Ρ ρ)
W = 1891(0.97)(0.750)2(0.74)[(3)(0.160)]1/2
W = 529 lb/h methane
2 o
1/ 2
Checking: Calculate Re to verify turbulence; if not in reasonable agreement, recalculate C and balance of solution, Viscosity of methane Re
= 0.0123 cP = 6.31 W/d μ = 6.31(529)/(0.750)(0.0123) = 36,1841
Fluid Flow 231 This is turbulent and satisfactory for the assumption. For helpful quick reference for discharge of air through an orifice, see Table 15.19b.
15.28 Friction Drop for Compressible Natural Gas in Long Pipe Lines Tests of the U.S. Department of the Interior, Bureau of Mines, reported in Monograph 6 Flow of Natural Gas Through High-Pressure Transmission Lines [42] indicate that the Weymouth formula gives good results on flow measurements on lines 6 in. in diameter and larger when operating under steady flow conditions of 30–600 psig. Long gas transmission lines of several miles length are not considered the same as process lines inside plant connecting process equipment where the lengths usually are measured in feet (meters) or hundreds of feet (meters). Some plants will transfer a manufactured gas, such as oxygen, carbon dioxide, or hydrogen, from one plant to an adjacent plant. Here the distance can be from 1 to 15 miles (kilometers). In such cases, the previously discussed flow relations for compressible gases can be applied in incremental segments, recalculating each segment, and then the results can be checked using one of the formulas that follow. However, there are many variables to evaluate and understand in the Weymouth, Panhandle, Panhandle-A and modifications as well as other flow relationships. Therefore, they will be presented for reference. However, the engineer should seek out the specialized flow discussions on this type of flow condition. The above mentioned equations are derived somewhat empirically for the flow of a natural gas containing some entrained liquid (perhaps 5–12%), and the results vary accordingly, even though they are not two-phase flow equations. Table 15.22 [26] tabulates the transmission factors of the various equations. Most of these are established as correction factors to the correlation of various test data. Dunning [43] recommends this formula (from [42]) for 4 to 24 in. (102–575 mm) diameter lines with specific gravity of gas near 0.60, and actual mean velocities from 15 to 30 fps (4.6–9.1 m/s) at temperature near 60°F (16°C). The Bureau of Mines report states that minor corrections for bends, tees, and even compressibility are unnecessary due to the greater uncertainties in actual line conditions. Their checks with the Weymouth relation omitted these corrections. The relation with pressure base of 14.4 psia is to be used with the Bureau of Mines multipliers [42]. Table 15.22 Dry-gas flow transmission factors. Title
Transmission Factor ( 1/f ) Ref.*
Weymouth
11.2D0.167
Blasius
3.56Re0.125
Panhandle A
6.87Re0.073
Modified Panhandle
16.5Re0.0196
Smooth pipe law (Nikuradse)
4 log Re f − 0.4
Rough pipe law (Nikuradse) Colebrook
(
)
4 log
( D) + 3.48 ( 2ε )
4 log
D ( D) + 3.48 − 4 log 1 + 9.35 ( 2ε ) 2 Re f
(Source: By permission, Hope, P.M. and Nelson, R.G., “Fluid Flow, Natural Gas,” McKetta, J.J. Ed., Encyclopedia of Chemical Processing and Design, vol. 22, M. Dekker,1985, p. 304 [15].) Note: D = inches *See listing of source references in Reference [15].
232 Petroleum Refining Design and Applications Handbook Volume 2 1/ 2
q h (at 14.4 psia and 60°F) = 36.926 d
2.667
P12 − P22 , scfh Lm
(15.265)
1/ 2
q ′h (at 14.4 psia and 60°F) = 28.0 d
2.667
P12 − P22 520 T Sg L m
, scfh
(15.266)
Weymouth’s formula [8] has friction established as a function of diameter and may be solved by using alignment charts. The Weymouth formula is also expressed (at standard condition) as: 1/ 2
q d = 433.9 E d
E d Ts Ps T1 qd P1′ P2′ Z L SCC
2.667
Ts P1′ 2 − P2′ 2 P S T L Z s g 1 m
(15.267)
= transmission factor, usually taken as: 1.10 × 11.2 d0.167 (omit for pipe size smaller than 24 in.) = Pipe, ID, in. = 520 °R = 14.7 psia = flowing temperature of gas, °R = flow rate, cu ft/day (at std conditions, SCC of 14.7 psia and 520°R) = inlet pressure, psia = outlet pressure, psia = compressibility factor = pipe length, miles = Standard condition (14.7 psia and 60°F)
or from [4]: 1/ 2
2 2 520 2.667 ( P1′) − ( P2′ ) q h = 28.0 d T S g L m
(15.268)
where d = pipe internal diameter, in. T = flowing temperature of gas, °R Sg = specific gravity of gas qh = gas flow rate ft3/h, at 60 °F and 14.4 psia P1′ = inlet pressure, psia P2′ = outlet pressure, psia Z = compressibility factor Lm = pipe length, miles In SI units 1/ 2
2 2 288 −8 2.667 ( P1′) − ( P2′ ) 3 q ′h = 2.61 × 10 d , m /h S g L m T
(15.269)
Fluid Flow 233 where d = internal pipe diameter, mm T = flowing temperature of gas, K = (273 + °C) qh = m3/h gas at metric std conditions (MSC) of Ps and Ts P1′ = inlet pressure, N/m2 abs P2′ = outlet pressure, N/m2 abs Z = compressibility factor Lm = pipe length, km MSC = Metric standard conditions (1.01325 bara and 15°C)
Example 15.19: Use of Base Correction Multipliers Tables 15.23–15.26 are set up with base reference conditions. In order to correct or change any base condition, the appropriate multiplier(s) must be used. A flow of 5.6 × 106 ft3/day has been calculated using Weymouth’s formula [8], with these conditions: measuring base of 60°F and 14.4 psia; flowing temperature of 60°F, and specific gravity of 0.60. Suppose for comparison purposes the base conditions must be changed to measuring base of 70°F and 14.7 psia; flowing temperature of 80°F, and specific gravity of 0.74. Multipliers from the tables are: Pressure base:
0.9796
Temperature base:
1.0192
Specific gravity base:
0.9005
Flowing temperature base:
0.9813
Table 15.23 Pressure-base multipliers for quantity. Multiplier =
14.4 New pressure base, lbs/sq . in. abs
New pressure base (lb/in.2 abs)
Multiplier
12.00
1.2000
13.00
1.1077
14.00
1.0286
14.40
1.0000
14.65
0.9829
14.7
0.9796
14.9
0.9664
15.4
0.9351
16.4
0.8780
(Source: By permission, Johnson, T. W. and Berwald, W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of Interior, Bureau of Mines, Washington, DC.)
234 Petroleum Refining Design and Applications Handbook Volume 2 Table 15.24 Temperature-base multipliers for quantity. Multiplier =
460 + new temperature base , °F 460 + 60
New pressure base, lbs/sq. in. abs
Multiplier
45
0.9712
50
0.9808
55
0.9904
60
1.0000
65
1.0096
70
1.0192
75
1.0288
80
1.0385
85
1.0481
90
1.0577
95
1.0673
100
1.0769
(Source: By permission, Johnson, T. W. and Berwald, W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of Interior, Bureau of Mines, Washington, DC.)
Table 15.25 Specific gravity multipliers for quantity. 0.600 Multiplier = actual specific gravity
1
2
Specific gravity
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.5
1.0954
1.0847
1.0742
1.0640
1.0541
1.0445
1.0351
1.0260
1.0171
1.0084
0.6
1.0000
0.9918
0.9837
0.9759
0.9682
0.9608
0.9535
0.9463
0.9393
0.9325
0.7
0.9258
0.9193
0.9129
0.9066
0.9005
0.8944
0.8885
0.8827
0.8771
0.8715
0.8
0.8660
0.8607
0.8554
0.8502
0.8452
0.8402
0.8353
0.8305
0.8257
0.8211
0.9
0.8165
0.8120
0.8076
0.8032
0.7989
0.7947
0.7906
0.7865
0.7825
0.7785
1.0
0.7746
0.7708
0.7670
0.7632
0.7596
0.7559
0.7524
0.7488
0.7454
0.7419
1.1
0.7385
0.7352
0.7319
0.7287
0.7255
0.7223
0.7192
0.7161
0.7131
0.7101
(Source: By permission, Johnson, T. W. and Berwald, W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of Interior, Bureau of Mines, Washington, DC.)
Fluid Flow 235 Table 15.26 Flowing-temperature multipliers for quantity. 460 + 60 Multiplier = 460 + actual flowing temperature Temp. °F
1
2
0
1
2
3
4
5
6
7
8
9
1.0632
1.0621
1.0609
1.0598
1.0586
1.0575
1.0564
1.0552
1.0541
1.0530
10
1.0518
1.0507
1.0496
1.0485
1.0474
1.0463
1.0452
1.0441
1.0430
1.0419
20
1.0408
1.0398
1.0387
1.0376
1.0365
1.0355
1.0344
1.0333
1.0323
1.0312
30
1.0302
1.0291
1.0281
1.0270
1.0260
1.0249
1.0239
1.0229
1.0219
1.0208
40
1.0198
1.0188
1.0178
1.0167
1.0157
1.0147
1.0137
1.0127
1.0117
1.0107
50
1.0098
1.0088
1.0078
1.0068
1.0058
1.0048
1.0039
1.0029
1.0019
1.0010
60
1.0000
0.9990
0.9981
0.9971
0.9962
0.9952
0.9943
0.9933
0.9924
0.9915
70
0.9905
0.9896
0.9887
0.9877
0.9868
0.9859
0.9850
0.9841
0.9831
0.9822
80
0.9813
0.9804
0.9795
0.9786
0.9777
0.9768
0.9759
0.9750
0.9741
0.9732
90
0.9723
0.9715
0.9706
0.9697
0.9688
0.9680
0.9671
0.9662
0.9653
0.9645
(Source: By permission, Johnson, T. W. and Berwald, W. B., Flow of Natural Gas Through High Pressure Transmission Lines, Monograph No. 6, U.S. Dept. of Interior, Bureau of Mines, Washington, DC.)
New base flow = (5,600,000) (0.9796) (1.0192) (0.9005) (0.9813) = 4,940,000 cu ft/day
15.29 Panhandle-A Gas Flow Formula The Panhandle equation assumes that the friction factor can be represented by a straight line of constant negative slope in the moderate Reynolds number region of the Moody diagram (Figure 15.5) [4]. The equation is considered to be slightly better than the ± 10 percent accuracy of the Weymouth formula and is given by
P2 − P22 q g = 0.028 E 0.9611 S g T L m Z
0.51
d 2.53
(15.270)
or
1.9607
qg P −P = 2.53 0.028 Ed
where d = pipe internal diameter, in. L = pipe length, miles P1 = upstream pressure, psia P2 = downstream pressure, psia Sg = gas specific gravity Z = gas compressibility factor
2 1
2 2
S0g.961TL m Z
(15.271)
236 Petroleum Refining Design and Applications Handbook Volume 2 qg = gas flow rate, MMscfd (at 14.7 psia and 60°F) T = gas flowing temperature, °R = 460oF + t E = efficiency factor for flow, use 1.00 for new pipe without bends, elbows, valves and change of pipe diameter or elevation 0.95 for very good operating conditions 0.92 for average operating conditions 0.85 for poor operating conditions In practice, the Panhandle equation is commonly used for longer pipe with a large pipe size (greater than 10 in.) where the Reynolds number is on the straight line portion of the Moody diagram (Figure 15.5). Neither the Weymouth nor the Panhandle represents a “conservative” assumption. If the Weymouth formula is assumed, and the flow is at moderate Reynolds number, the friction factor will be higher than the horizontal portion of the Moody curve, and the actual pressure drop will be higher than calculated. If the Panhandle formula is used and the flow is in a high Reynolds number, the friction factor will be higher than assumed and the actual pressure drop will be higher than calculated. For bends in pipe add to length [27]: Bend radius
Add*, as pipe diameter, de
1 Pipe dia.
17.5
1.5 Pipe dia.
10.4
2 Pipe dia.
9
3 Pipe dia.
8.2
*These must be converted to the unit of length used in the formula.
In SI units Panhandle formula for natural gas pipe lines 150 to 600 mm diameter and Re = (5 × 106) to (14 × 106) is:
( p′ )2 − ( p′ )2 2 −3 q ′h = 5.06 × 10 E 1 Lm
0.5394
d 2.6182 m3/h
(15.272)
or
( p1′ ) − ( p′2 ) 2
2
1.8539
103 q ′h = Lm 2.6182 E 5.06 d
(15.273)
where d = internal pipe diameter, mm L = length of pipe in km p1′ = inlet pressure, bara p′2 = outlet pressure, bara q ′h = rate of flow in m3/h at metric standard conditions (MSC) 1.103 bara and 15°C If a line is made up of several different sizes, these may be resolved to one, and then the equation solved once for this total equivalent length. If these are handled on a per size basis, and totaled on the basis of the longest length of one size of line, then the equivalent length, Le, for any size d referenced to a basic diameter, de is
Le = Lm (de/d)4.854
(15.274)
Fluid Flow 237 where Lm is the length of pipe size d to be used. Le is the equivalent length of pipe size d, length Lm after conversion to basis of reference diameter de. The calculations can be based on diameter de and a length of all the various Le values in the line plus the length of line of size de, giving a total equivalent length for the line system.
15.30 Modified Panhandle Flow Formula
T q DS = 737.2 E o Po
1.02
P12 (1 + 0.67 Z P1 ) − P22 (1 + 0.67 ZP2 ) T L m G0.961
0.51
d 2.53
(15.275)
where [26] Lm = length, mi d = inside diameter, in. T = flowing temperature, °R Z = gas deviation, compressibility factor To = base temperature, (520°R) G = gas specific gravity P = pressure, psia Po = base pressure, (14.73 psi, absolute) E = “Efficiency factor,” which is really an adjustment to fit the data qDS = flow rate, SCF/day
15.31 American Gas Association (AGA) Dry Gas Method See Uhl et al. [44] AGA, Dry Gas Manual. Some tests indicate that this method is one of the most reliable above a fixed Reynolds number.
15.32 Complex Pipe Systems Handling Natural (or Similar) Gas The method suggested in the Bureau of Mines Monograph No. 6 [42] has found wide usage, and is outlined here using the Weymouth Formula as a base. 1. Equivalent lengths of pipe for different diameters
L1 = L2(d1/d2)16/3
(15.276)
where L1 = the equivalent length of any pipe of length L2 and diameter, d2, in terms of diameter, d1.
d1 = d2(L1/L2)3/16
(15.277)
where d1 = the equivalent diameter of any pipe of a given diameter, d2 and length L2, in terms of any other length, L1. 2. Equivalent diameters of pipe for parallel lines
(
d o = d18 / 3 + d 82/ 3 ..... + d 8n/ 3
)
3/ 8
(15.278)
238 Petroleum Refining Design and Applications Handbook Volume 2 where do is the diameter of a single line with the same delivery capacity as that of the individual parallel lines of diameters d1, d2,…… and dn. Lines of same length. This value of do may be used directly in the Weymouth formula. Example 15.20: Series System Determine the equivalent length of a series of lines: 5 mi of 14-in. (13.25-in. ID) connected to 3 mi of 10 in. (10.136in. ID) connected to 12 mi of 8-in. (7.981-in. ID). Select 10-in. as the base reference size. The five-mile section of 14-in. pipe is equivalent to:
L1 = 5(10.136/13.25)5.33 = 1.199 mi of 10 in. The 12 mile section of 8 in. is equivalent to:
L1 = 12(10.136/7.981)5.33 = 42.9 mi of 10 in. Total equivalent length of line to use in calculations is:
1.199 + 3.0 + 42.9 = 47.099 mi of 10-in. (10.136-in. ID).
An alternative procedure is to calculate (1) the pressure drop series-wise one section of the line at a time, or (2) capacity for a fixed inlet pressure, series-wise. Example 15.21: Looped System Determine the equivalent length of 25 mi of 10-in. (10.136-in. ID) which has parallel loop of 6 mi of 8-in. (7.981-in. ID) pipe tied in near the mid section of the 10-in. line. Figure the looped section as parallel lines with 6 mi of 8-in. and 6 mi of 10-in. the equivalent diameter for one line with the same carrying capacity is:
do = [(7.981)8/3 + (10.136)8/3]3/8 = 11.9 in. This simplifies the system to one section 6 mi long of 11.9-in. ID (equivalent) pipe, plus one section of 25 minus 6, or 19 mi of 10-in. (10.136-in. ID) pipe. Now convert the 11.9-in. pipe to a length equivalent to the 10-in. diameter.
L1 = 6(10.136/11.9)5.33 = 2.53 mi Total length of 10-in. pipe to use in calculating capacity is 19+2.53 = 21.53 mi. By the principles outlined in the examples, gas pipe line systems may be analyzed, paralleled, cross-tied, and so on. Example 15.22: Parallel System: Fraction Paralleled Determine the portion of a 30 mi, 18-in. (17.124-in. ID) line which must be paralleled with 20-in. (19.00-in. ID) pipe to raise the total system capacity 1.5 times the existing rate, keeping the system inlet and outlet conditions the same.
x=
(q da /q db )2 − 1
1 2.667 1 + ( d b /d a )
{
}
1 2
(15.279)
Fluid Flow 239 For this example, qdb = 1.5 qda
x=
(1/1.5)2 − 1 1
− 1 2 {1 + (19.00/17.124)2.667 }
= 0.683
This means 68.3% of the 30 mi must be parallel with the new 19-in. ID pipe. Parallel System: New Capacity After Paralleling Solve this relation, rearranged conveniently to [42]
q db =
q da
1/ 2
1 − 1 + 1 2 x 2.667 1 + (d b /d a )
(15.280)
15.33 Two-Phase Liquid and Gas Flow in Process Piping An understanding of two-phase flow is necessary for sound piping design. This is because, almost refinery and chemical process plants encounter two-phase flow conditions. The concurrent flow of liquid and gas in pipelines has received considerable study [45–48]. However, pressure drop prediction is not extremely reliable except for several gas pipe line conditions. The general determinations of pressure drop for plant process lines can only be approximated. The latest two-phase flow research and design studies have broadened the interpretation of some of the earlier flow patterns and refined some design accuracy for selected situations. The method presented here serves as a fundamental reference source for further studies. It is suggested that the designer compares several design concept results and interprets which best encompasses the design problem under consideration. Some of the latest references are included in the Reference Section. However, no one reference has a solution to all two-phase flow problems. If two-phase flow situations are not recognized, pressure drop problems may develop which can prevent systems from operating. It requires very little percentage of vapor, generally above 7% to 8% (by volume), to establish volumes and flow velocities that must be solved by two-phase flow analysis. The discharge flow through a pressure relief valve or a process reactor is often an important example where two-phase flow exists, and must be recognized for its backpressure impact. Two-phase flow often presents design and operational problems not associated with liquid or gas flow. For example, several different flow patterns may exist along the pipeline. Frictional pressure losses are more difficult to estimate, and in the case of a cross-country pipeline, a terrain profile is necessary to predict pressure drops due to elevation changes. The downstream end of a pipeline often requires a separator to separate the liquid and gas phases, and a slug catcher may be required to remove liquid slugs. Static pressure losses in gas-liquid flow differ from those in single-phase flow because an interface can be either smooth or rough, depending on the flow pattern. Two-phase pressure losses may be up to a factor of 10 higher than those in single-phase flow. In the former, the two phases tend to separate and the liquid lags behind. Most published correlations for two-phase pressure drop are empirical and, therefore limited by the range of data for which they were derived [51, 78–80].
15.33.1 Flow Patterns In determining the (type of flow)phase distribution in a process pipeline, designers refer to a diagram similar to Figure 15.31a, which is known as the Baker map. Figure 15.32 shows the types of flow regimes that can exist in a
240 Petroleum Refining Design and Applications Handbook Volume 2 By 100,000 Dispersed Flow
Wave Flow
Bubble or Froth Flow
Annular Flow
10,000
Stratified Flow
Slug Flow
1,000
Plug Flow
100 0.1
0.2
0.4 0.6 0.8 1.0
2
4
6 8 10
2
4 6 8 100
2
4
6 8 1,000 2
4 6 8 10,000
R
Figure 15.31a Flow patterns for horizontal two-phase flow (based on data from 1, 2, and 4 in. pipe) (source: Baker, O., Oil & Gas Journal Nov. 10, p 156, 1958). By 100,000 DISPERSED C3
C2
WAVE
4”
BUBBLE OR
6”
ANNULAR
FROTH
10,000 C4 C1
SLUG C5
C6
STRATIFIED 1,000
PLUG 100
.1
1
10
100
1,000
10,000
Bx
Figure 15.31b Baker parameters for horizontal two-phase flow regimes with modified boundaries (based on data from 1, 2, and 4 in. pipe).
horizontal pipe, and Table 15.27 lists the characteristic linear velocities of the gas and liquid phases in each flow regime. Seven types of flow patterns are considered in evaluating two-phase flow, and only one type can exist in a pipeline at a time. But as conditions change (e.g., velocity, roughness, and elevation), the type of flow pattern may also change. The pressure drop can also vary significantly between the flow regimes. The seven types of flow regimes in order of increasing gas rate at a constant liquid flow rate given below:
Fluid Flow 241 SEGREGATED
Stratified
Wavy
Annular INTERMITTENT
Plug
Slug DISTRIBUTED
Bubble
Mist
Figure 15.32 Representatives forms of horizontal two-phase flow patterns, same as indicated in Figures 15.31a (Source: Hein, H., Oil & Gas J., Aug. 2, p. 132, 1982).
Table 15.27 Characteristics linear velocities of two-phase flow regimes. Regime
Liquid phase (ft/s)
Vapor phase (ft/s)
Bubble or froth
5–15
0.5–2
Plug
2
10
(15.317)
where Wm = Mass flow rate of liquid phase, lb/h ft2 (of total pipe cross section area). v = Flow velocity (mean) or superficial velocity in pipe lines at flowing conditions for entire pipe cross section, ft/s. or as an alternative:
Fe = 1.7156 v G−0.702
vG = gas velocity, ft/s
A. To determine most probable type of two-phase flow using Figure 15.31a. 1. Calculate Bx
(15.318)
Fluid Flow 251 2. C alculate By 3. Read intersection of ordinate and abscissa to identify probable type of flow. Since this is not an exact, clear-cut position, it is recommended that the adjacent flow types be recorded. B. Calculate the separate liquid and gas flow pressure drops. 1. F or general process application both PL and PG may be calculated by the general flow equation: PL or PG
=
3.36 f D L W 2 (10−6 ) d 5ρ
(15.319)
where fD is the Darcy friction factor obtained from (Reynolds) Moody-Friction Factor chart (Figure 15.5) for an assumed line size, d or from Chen’s explicit equation for friction factor (fD = 4 fC). 2. For gas transmission, in general form [33]
∆PG =
(q d 14.65 )LS g TZ f 20, 000 d 5 Pavg
(15.320)
where qd 14.65 is the thousands of standard cubic feet of gas per day, measured at 60°F and 14.65 psia, and Pavg is the average absolute pressure(psia) in the pipe system between inlet and outlet. This is an estimated value and may require correction and recalculation of the final pressure drop if it is very far off.
where d = internal pipe diameter, in. f = Friction factor, Moody. L = Pipe length, ft Sg = specific gravity of gas relative to air, ( = ratio of molecular weight gas/29) T = Absolute temperature, °R = 460 + °F Z = compressibility factor For oil flow in natural gas transmission lines [45]
f L Q 2b ρ 181, 916 d 5
(15.321)
X = (ΔPL/ΔPG)1/2
(15.322)
∆PL =
where Qb = Flow rate in bbl/day L = Pipe length, ft ρ = Fluid density, lb/ft3 3. Calculate
4. Calculate Φ for types of flow selected from Figure 15.31a and as summarized below [45].
252 Petroleum Refining Design and Applications Handbook Volume 2 Type flow
Equation for (FGTT)
Froth or bubble
Φ = 14.2X0.75 /(WL/A)0.1
Plug
Φ = 27.315X0.855 /(WL/A)0.17
Stratified
Φ = 15,400X /(WL/A)0.8
Slug
Φ = 1,190X0.185 /(WL/A)0.5
Annular*
Φ = (4.8 – 0.3125d)X0.343−0.021d
*set d = 10 for any pipe larger than 10-in.
X = [ΔPL/ΔPG]1/2 5. C alculate two-phase pressure drop, horizontal portions of lines. For all types of flow, except wave and fog or spray:
2 ∆PTP = ∆PG ΦGTT , psi /ft
(15.323)
f ( G′ ) ∆PTP = TP G , psi /ft 193.2 d ρG
(15.324)
For wave [54]. 2
where
W µ f TP = 0.0043 m L G µG
where fTP G′G G Wm μL μG
0.214
(15.325)
= Two-phase friction for wave flow = Mass rate, lb/s (ft2. cross section) = Mass flow rate of gas phase, lb/h ft2 = Mass flow rate of liquid phase, lb/h ft2 = Liquid viscosity, cP = Gas viscosity, cP
6. T otal two-phase pressure drop, including horizontal and vertical sections of line. Use calculated value times 1.1 to 2.0, depending upon critical nature of application.
PTPh = PTPL +
n h FeρL 144
(15.326)
where ρL is the density, lb/ft3, of the liquid flowing in the line, and Fe, elevation factor using gas velocity, vG. Use Figure 15.34 for v less than 10. Most gas transmission lines flow from 1 to 15 ft/s. For fog or spray type flow, Baker [45] suggests using Martinelli’s correlation and multiplying results by two [53]. (a) For gas pipe line flow, the values of ( GTT) may be converted to “efficiency E” values and used to calculate the flow for the horizontal portion using a fixed allowable pressure drop in the general flow
Fluid Flow 253 Liquid head factor, Fe 1.0 0.9 Natural Gas Condensate in 16N Pipeline Natural Gas, Oil and Water in 2N Oil Well Tubing Air and Water in lN Vertical Tubing Air and Lube Oil in 2N Inclined Tubing
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 Superficial Gas Velocity, ft/s
Figure 15.34 Estimating pressure drop in uphill sections of pipeline for two-phase flow (source: O. Flanigan, Oil & Gas J., Mar. 10, p 132, 1958).
equation [33]. The effect of the vertical component must be added to establish the total pressure drop for the pumping system.
(
)
0.5
38.7744 Ts P12 − P22 d 5 E 0.5 q d 14.65 = 1000 P L S TZ f g s m g
(15.327)
where 14.65 refers to reference pressure Ps. d = internal pipe diameter, in. E = Gas transmission “efficiency” factor, varies with line size and surface internal condition of pipe. fg = Moody or “regular” Fanning friction for gas flow Lm = Pipe length, mi P1 = initial pressure, psi P2 = final pressure, psi Ps = standard pressure for gas transmission, psia Sg = specific gravity of gas relative to air, ( = ratio of molecular weight gas/ 29) T = Absolute temperature, °R = 460 + °F Ts = Standard temperature for gas measurement, °R = 460 + °F Z = Gas compressibility factor or
(q d 14.65 )2 LS g TZ f g ∆PTP = 20, 000d 5Pavg E 2
(15.328)
where E = 1/ΦGTT (b) For the Panhandle equation, Baker [45] summarizes:
T q d 14.65 = 0.43587 s Ps
1.07881
P12 − P22 Z TL m
0.5394
d 2.618 S0.4606 (E ) g
.077 where E (Panhandle) = 0.9/Φ1GTT
(15.329)
(15.330)
254 Petroleum Refining Design and Applications Handbook Volume 2 Example 15.23: Two-Phase Flow A liquid–gas mixture is to flow in a line having 358 ft of level pipe and three vertical rises of 10 ft each plus one vertical rise of 50 ft. Evaluate the type of flow and expected pressure drop. Liquid
Gas
Flow W, lb/h
1000
3000
Density ρ, lb/ft3
63.0
0.077
Viscosity μ, cP
1.0
0.00127
Surface tension σ, dyn/cm
15
Pipe schedule is 40 stainless steel. Use maximum allowable gas velocity = 15,000 ft/min. Solution 1. Determine probable types of flow:
(
WL ρL ρG B x = 531 WG ρL2/3
)
1/3 µ L σ L 1000 [(63)(0.077 )]0.5 1 B x = 531 3000 (63)2/3 15
0.5
= 1.64
W 1 B y = 2.16 G A (ρL ρG )0.5
Try 3-in. Sch. 40 pipe, (I.D. =3.068-in.) Pipe cross sectional area, A = π D2/4 = π(0.2556)2/4 = 0.0513 ft2.
(WL/A) = 1000/0.0513 = 19,493 lbs/ft2h (WG/A) = 3000/0.0513 = 58,480 lbs/ft2h W 1 B y = 2.16 G A (ρL ρG )0.5 B y = 2.16(58, 480)
1 (63)(0.077 )
= 57 , 352
Reading Figure 15.31a, the flow pattern type is probably annular, but could be wave or dispersed, depending on many undefined and unknown conditions. 2. Liquid pressure drop
ΔPL = 3.36 fD L W2(10−6)/d5ρ
(15.319)
Fluid Flow 255 Determine Re for 3-in. pipe: From Figure 15.8; /d = 0.00058 for steel pipe The liquid velocity, vL is:
W 1000 vL = L = = 0.086 ft /ss ρL A (63)(3600)(0.0513)
e
= 1 cp/1488 = 0.000672 lbs/ft s
D = 3.068/12 = 0.2557 ft L
= 63.0 lb/ft3
Re = DvL L/
Re = 2061 (this is borderline, and in critical region)
e
= 0.2557 (0.086) (63.0)/0.000672
Relative pipe roughness is:
ε 0.00015 = = 0.000587 D 0.25567
Friction factor
ε 5.02 log A = −4 log − 3.7D Re fC
1
(15.35)
where
ε/D 6.7 A= + 3.7 Re
0.9
0.000587 6.7 A= + 2060 3.7
0.9
= 5.926 × 10−3
0.000587 5.02 = −4 log − log(5.926 × 10−3 ) 2060 3.7 fC
1
= 9.01154 fC = 0.01231
The Darcy friction factor fD = 4 fC
fD = 4 (0.01231)
= 0.0492 The liquid pressure drop ΔPL is
ΔPL = 3.36(10−6)(0.0492)(1000)2(1 ft)/(3.068)5(63)
= 0.96 (10 5) psi/ft
256 Petroleum Refining Design and Applications Handbook Volume 2 3. Gas pressure drop The gas velocity vG is:
W vG = G ρG A
vG = e
3000 = 211 ft/s 0.077(3600)(0.0513)
= 0.00127/1488 = 0.000000854 lbs/ft s
Re = D vG ρG/μe
= (0.2557)(211)(0.077)/(0.000000854)
= 4,900,000
Friction factor, f:
ε 5.02 log A s = −4 log − 3.7D Re fC
1
where
ε/D 6.7 A= + 3.7 Re
and
0.9
ε = pipe roughness, ft. D = pipe internal diameter, ft.
ε 0.00015 = = 0.000587 D 0.25567 ε 5.02 log A = −4 log − 3.7D Re fC
1
0.000587 6.7 A= + 4900000 3.7
0.9
= 1.6392 × 10−4
0.000587 5.02 = −4 log − log(1.6392 × 10−4 ) 4900000 3.7 fC
1
= 15.1563 fC = 0.004353
The Darcy friction factor fD = 4 fC
fD = 4(0.004353)
= 0.0174
Fluid Flow 257 The gas pressure drop ΔPG is:
ΔPG = 3.36(10−6)(0.0174)(3000)2(1 ft)/(3.068)5(0.077)
= 0.0251psi/ft 4. Lockhart–Martinelli two-phase flow modulus X:
X = (ΔPL/ΔPG)1/2 = (0.96 × 10−5/2.51 × 10−2)1/2
= 1.95 × 10−2 5. For annular flow:
ΦGTT = (4.8 – 0.3125d)X0.343−0.021d
= [4.8 – (0.3125)(3.068)](1.95 × 10−2)0.343−0.021(3.068)
= 1.28 6. Two-phase flow for horizontal flows:
2 ∆Ρ TP = ∆Ρ G ΦGTT = (0.0251)(1.28)2 = 0.0411 psi/ft
7.
Fe = 0.00967 (WL /A ) /vG0.7 0.5
= 0.00967(19, 494 )0.5 /(211)0.7 = 0.032
Vertical elevation pressure drop component:
= nh Fe ρL/144 = [(3)(10) + (1)(50)](0.032)(63)/144
= 1.12 psi total Total:
ΔPTPh = (0.0411)(358) + 1.12
= 15.8 psi, total for pipe line
Because these calculations are somewhat uncertain due to lack of exact correlations, it is best to calculate pressure drop for other flow patterns, and apply a generous safety factor to the results. Table 15.30 gives calculated results for other flow patterns in several different sizes of lines.
15.33.6 Pipe Sizing Rules All two-phase flow correlations shown have been developed for long, horizontal pipes where uniform flow types are more likely to develop. In general, the application of all two-phase flow correlations to process piping design is arbitrary. These correlations do not take into account the three dimensional reality of process piping. Process piping has varying alternating flow regions because of pipe configurations, elevation changes, offsets, branch connections,
258 Petroleum Refining Design and Applications Handbook Volume 2 manifolds, pipe components, reducers, and other restrictions. Large deviations can occur from friction loss predictions, compared to actual friction losses [56]. In using the correlations and graphs presented for two-phase flow, the following general pipe sizing rules are suggested: Dispersed Flow: Apply ΔPTP (two-phase) throughout three dimensional pipe (horizontal, up and downflow sections). Use dispersed flow correlation for pipe smaller than 2½ in. for all flow regions. Annular and Bubble Flow: Apply ΔPTP (two-phase) for 3 in. and larger pipe throughout the process line. Check the vertical upflow correlations and unit flow loss for long, vertical upflow runs. Use the upflow losses if they are greater than the annular or bubble flow unit losses. Stratified and Wave Flow: Use stratified and wave flow correlations only for long, horizontal runs. Use annular flow correlations for three dimensional, process pipe sizing where stratified and wave flow regions are determined. For stratified flow, use the Eq. 15.296 to determine the two-phase modulus and apply the Huntington correlation to determine the wave flow unit loss calculations. Plug Flow: The process piping designer rarely meets with plug flow conditions. (But slug flow is not uncommon.) Example 15.24 Figure 15.35 shows the piping configuration with a 4 in. control valve and a vessel. The available pressure difference ΔP is 10 psi, including that of the control valve. The two-phase flow data in the line after the control valve are given below [56]. Determine a reasonable pipe size downstream of the control valve. Liquid
Gas
Flow, W, lb/h
59,033
9336
Molecular weight, Mw
79.47
77.2
Density, lb/ft3
31.2
1.85
Viscosity, μ, cP
0.11
0.0105
Surface tension, σ, dyn/cm
5.07
(Source: R. Kern, Piping Design For Two-Phase Flow, Chem. Eng. June 23, 1975).
Table 15.30 Two-phase flow example. Horizontal flow pattern Pipe ID (in.)
Annular (psi/ft)
Stratified (psi/ft)
Wave (psi/ft)
Elevation factor, Fe
Gas velocity (ft/s)
3.068
0.0438
0.000367
0.131
0.032
210.9
4.026
0.0110
0.000243
0.0336
0.0465
122.5
6.065
0.00128
0.000131
0.00434
0.0826
53.9
7.981
0.00027
0.000087
0.00110
0.121
31.1
10.02
0.000062
0.000062
0.00035
0.166
19.7
Fluid Flow 259 Solution Table 15.31 shows a typical computer results of Example 15.24 using a pipe size of 3” and Table 15.32 shows the results of 3, 4, 6, and 8 in. Sch 40, respectively. The results of ΔPTotal exclude the vertical sections of the piping configuration. These results show that the overall pressure drop of the two-phase (ΔPTP) for both the 3 and 4 in. pipe sizes is greater than the available pressure drop of 10 psi (including that for the control valve: see Figure 15.35). The 6 and 8 in. pipe sizes show an overall pressure drop of the two-phase less than 10 psi, however the 8 in. pipe size indicates an undesirable slug flow pattern on the Baker map. The 6 in. pipe size gives a bubble flow on the Baker’s map, and thus this pipeline is the optimum size. Figure 15.35A and B provide alternate locations for the control valve for alternative A. Here, slug flow cannot develop. However, for alternate B, a shortened and self-draining pipe line improves the pipe configuration but at
C El . 14 8 ft
325 °F
165 psig
V-5603
C El . 14 8 ft 4-i n
175 psig
C
2 10 El . t f
de Gra
on ati
t
0f
10
10
ft
4-in Control valve
v
ele
C El . 14 8 ft
C El . ft
14
8
C El. 11 0 ft
C E l. 1 10 ft
Alternative A
Figure 15.35 Configurations of piping for sample problem of Example 15.24 [56].
Alternative B
260 Petroleum Refining Design and Applications Handbook Volume 2 Table 15.31 Computer results of two-phase flow pressure drop calculations of Example 15.24. Two-phase pressure loss calculation in a pipe line Pipe internal diameter, in.:
3.068
Equivalent length of pipe, ft:
152.307
Actual length of pipe, ft:
56.000
Total length of pipe, ft:
208.307
Liquid density, lb/ft^3:
31.200
Liquid viscosity, cP:
0.1100
Liquid surface tension, dyne/cm:
5.0700
Liquid flow rate, lb/h:
59033.000
Liquid Reynolds number:
1103764.
Liquid friction factor:
0.0177
Pressure drop of liquid, psi/100 ft:
2.6660
Gas density, lb/ft^3:
1.850
Gas viscosity, cP:
0.0100
Gas flow rate, lb/h:
9336.000
Gas Reynolds number:
1920149.
Gas friction factor:
0.0176
Pressure drop of gas, psi/100 ft.:
1.1137
Flow regime is:
Bubble
Lockhart–Martinelli two phase flow modulus:
1.5472
Velocity of fluid in pipe, ft./s:
37.521
Baker parameter in the liquid phase:
243.246
Baker parameter in the gas phase:
51702.714
Two-phase flow modulus:
23.8109
Pressure drop of two-phase mixture, psi/100 ft:
26.5171
Overall pressure drop of the two-phase mixture, psi:
55.2369
Index 13,888. is greater than 10,000 pipe erosion is possible
the expense of convenient access to the control valve. In both alternatives, there is considerable turbulence after the control valve, which helps to provide slug free liquid-phase carry over. The following are ways to adjust the pressure loss distribution in a pipe system [56]: 1. 2. 3. 4.
hange the pipe size. C Design a section of the pipe line with either an increase or a decrease in pipe diameter. Adjust the static head of elevated vessels. Change valve or orifice restrictions to consume more or less pressure drop (differentials).
Fluid Flow 261 Table 15.32 Computer results of two-phase pressure drop calculation of Example 15.24. Pipe internal diameter, in (Sch. 40).
3.068
4.026
6.065
7.981
Equivalent length of pipe, ft
76.445
98.822
145.729
188.042
Actual length of pipe, ft.
56.0
56.0
56.0
56.0
Total length of pipe, ft.
132.445
154.822
201.729
244.042
Liquid density, lb/ft3
31.2
31.2
31.2
31.2
Liquid viscosity, cP.
0.11
0.11
0.11
0.11
Surface tension, dyne/cm.
5.07
5.07
5.07
5.07
Liquid flow rate, lb/hr.
59033.
59033
59033
59033
Liquid Reynolds number
1103764
841120
58343
424301
Liquid friction factor.
0.0177
0.170
0.0162
0.0159
Pressure drop of liquid, psi/100 ft.
2.6685
0.6555
0.0805
0.0201
Gas flow rate, lb/hr.
9336
9336
9336
9336
Gas density, lb/ft3
1.85
1.85
1.85
1.85
Gas viscosity, cP.
0.01
0.01
0.01
0.01
Gas Reynolds number
1828713
1393565
925060
702981
Gas friction factor.
0.0176
0.0167
0.0157
0.0153
Pressure drop of liquid, psi/100 ft.
1.115
0.2723
0.033
0.0082
Flow regime:
Bubble
Bubble
Bubble
Slug
Lockhart-Martinelli two-phase modulus
1.547
1.5514
1.5617
1.5709
Velocity of fluid in pipe, ft/sec.
37.5212
21.7891
9.6012
5.5446
Baker parameter in the liquid phase.
243.246
243.246
243.246
243.246
Baker parameter in the gas phase.
51702.71
30024.55
13230.08
7640.28
Two-phase flow modulus.
23.806
26.652
31.711
17.400
Pressure drop of two-phase mixture, psi/100ft.
26.544
7.258
1.047
0.1418
Overall pressure drop of the two-phase, psi.
35.156
11.238
2.113
0.3461
Index for Pipe erosion.
13888
4684
909
303
15.33.7 A Solution for All Two-Phase Problems Dukler et al. [58] have pointed out that only three flow regimes are apparent in any piping configuration: segregated, intermittent, and distributed. Segregated flow occurs when the gas and liquid are continuous in the axial direction. Stratified flow is easily recognized as belonging to this category, as do the wavy and annular regimes (see Figure 15.32). Intermittent flow results when the phases form alternating pockets. Plug and slug flows therefore fall in this grouping. Flow is considered distributive when one fluid phase is continuous and flows to some degree in the directions which are both perpendicular and parallel to the pipe axis. The other phase may not necessarily be distributed uniformly over the same section of the pipe, but should be locally continuous. Mist flow and bubble flow are included in this type of regime.
262 Petroleum Refining Design and Applications Handbook Volume 2 These regimes, which completely characterize any flow type, simplify the analysis of a physical situation by resolving into three the numerous regimes described earlier. Erwin [59] expressed that in considering Baker’s froth zone flow regime, to have a froth or homogeneous flowing gas–liquid mixture, a high Reynolds number is required (Re ≥ 200,000). Every case of refinery, oil and gas, and chemical plant piping involves higher Reynolds number for economic pipe sizing; even pipelines are sized for higher Reynolds numbers. A pipe flowing at 3 ft/s would qualify for the minimum Reynolds number of 200,000. Dukler’s [60, 61] work resolved two-phase flow pipe sizing and configuration problems. The key to the success is maintaining Re ≥ 200,000, which is accomplished by making the pipe size small enough. Dukler’s work is summarized as follows: 1. F or pressure loss due to friction, first determine the homogeneous flow liquid ratio λ, volume of liquid per volume of mixed fluid flow.
λ=
where QLPL QGPL WL WG ρL ρG
Q LPL Q GPL + Q LPL
(15.331)
= volume of liquid flow WL/ρL, ft3/h = volume of gas flow, WG/ρG, ft3/h = liquid flow, lb/h = gas flow, lb/h = liquid density at flow pressure and temperature, lb/ft3 = gas density at flow pressure and temperature, lb/ft3
The calculated λ value is valid only over a range in which the pressure loss in the pipe does not exceed 15% of inlet value. For a large pressure loss, the pipe run is divided into several segments with each segment having different pressure inlet and different temperature due to the gas flashing cooling effect. 2. The ratio of two-phase friction factor to gas-phase friction factor in the pipeline is determined as:
S = 1.281 + 0.478(ln λ) + 0.444(ln λ)2 + 0.09399999(ln λ)3 + 0.008430001(ln λ)4
f TP lnλ = 1− fo S
(15.332) (15.333)
where fTP = two-phase flow friction factor in the pipe run fo = gas-phase friction factor in the pipe run ln = natural logarithm of base e, 2.7183 3. Th e Reynolds number is calculated. Dukler developed experimental data in calculating liquid holdup in two-phase flow systems. Re > 200,000 are free of liquid slugs and holdup. If Re is greater than 200,000 then the flow is in the froth regime, or it is homogeneous flow as a mixture. For homogeneous flow, the average density of the two-phase fluid mixture is:
ρm = ρLλ + ρG(1 − λ)
(15.334)
The average viscosity μm, lb/ft.s:
µm =
µL µ λ + G (1 − λ ) 1488 1488
(15.335)
Fluid Flow 263 where μL = liquid viscosity, cP μG = gas viscosity, cP Calculate the mixture flowing velocity vm, ft/s
vm =
Q GPL + Q LPL
(
3600 πD2 /4
)
(15.336)
where D = pipe internal diameter, ft. The two-phase Reynolds number is:
Re =
Dv m ρm µm
(15.337)
4. C alculate the two-phase flow friction factor fTP as follows. First define fTP from fo. In charting fTP/fo against λ, Dukler expressed
fo = 0.0014 +
f TP =
0.125 Re0.32
f TP fo fo
(15.338)
(15.339)
Re must be 200,000 or greater before these equations can be applied. Knowing Re and the ratio of fTP/fo, fTP is calculated from Eqs. 15.338 and 15.339. 5. The pipe friction pressure loss of straight pipe, ΔPf is:
∆Pf =
4 f TP L ρm v 2m , psi 144 g c D 2
(15.340)
where D = pipe diameter, ft. L = straight pipe length, ft. lb ft gc = conversion factor, = 32.174 m • 2 lbf s 6. C alculate the pressure drop due to elevation changes ΔPE, psi. First determine the superficial gas velocity vsg as:
v sg =
Q GPL
(
3600 πD2 /4
)
, ft /s.
(15.341)
vsg is the velocity of the gas alone in the full cross section area of the pipe, ft/s. A factor ϕ is related to the two-phase gas velocity vsg, and its value increases as the gas velocity decreases. A curve fit equation of ϕ vs. vsg is:
264 Petroleum Refining Design and Applications Handbook Volume 2
φ = 0.76844 − 0.085389 v sg + 0.0041264 v sg2
− 0.000087165 v 3sg + 0.00000066422 v 4sg
(15.342)
Eq. 15.342 determines the ϕ value, which is the correction to the static leg rise or fall of the gas phase. As the gas velocity approaches 0, ϕ approaches unity. Eq. 15.342 has a range limit that is: If vsg > 50, then ϕ = 0.04 If vsg < 0.5, then ϕ = 0.85
∆PE =
φρLH T 144
(15.343)
where HT = height of static leg, - for rise and + for fall, ft. Here the liquid density, ρL is used in Eq. 15.343, since ϕ corrects for the gas-phase static leg ΔP. 7. Calculate the pressure drop due to acceleration or pipe fittings and valves ΔPA, psi. The 90° standard elbow ΔPcell is calculated as follows:
∆PA =
∆Pcell =
2 ρGQ GPL ρ Q2 + L LPL 1− λ λ
PA
3.707 × 1010 + ( d/12 )4
(15.344)
(15.345)
where d = internal pipe diameter, in. If a 15% pressure loss in a pipe segment results, then a new pipe segment is required.
Tee angle
ΔPA = 3.0 ΔPcell
Tee straight ΔPA = 1.0 ΔPcell Check valve ΔPA = 2.5 ΔPcell For two-phase flow pipe entrance and exit,
∆PA = 4 f TP[6.469(ln d ) + 24]
ρm v 2m 144 g c 2
(15.346)
ρm v 2m 144 g c 2
(15.347)
Pipe sharp-edge exit:
∆PA = 4 f TP[14.403(ln d ) + 42]
Fluid Flow 265 8. The total two-phase pressure loss ΔPT is:
ΔPT = ΔPf + ΔPE + ΔPA
(15.348)
Note: All preceding seven steps are made on the assumption that Re is 200,000 or greater. Table 15.33 shows a glossary of two-phase flow. Example 15.25 Using the data in Example 15.24, determine the total two-phase pressure drop downstream of the control valve. Liquid
Gas
Flow, W, lb/h
59,033
9336
Molecular weight, Mw
79.47
77.2
Density, lb/ft3
31.2
1.85
Viscosity, μ, cP
0.11
0.0105
Surface tension, σ, dyn/cm
5.07
(Source: R. Kern, Piping Design For Two-Phase Flow, Chem. Eng. June 23, 1975).
Table 15.33 A glossary of two-phase flow. Critical Flow: When a point is reached in the system where the increase in specific volume for a small decrease in pressure is so great that the pressure and the enthalpy can no longer be simultaneously lowered across a cross section of pipe, it is called critical flow. It is analogous to sonic flow in a single phase flow. This does not imply, however, that the sonic velocity of a superficially flowing gas phase in a two-phase system is equal to the sonic or critical velocity of the two-phase system. Critical flow occurs in the so-called mist flow regime.
Plug Flow: This is a flow regime where most, but not all aggregates of the liquid phase occupy most of the cross section of the pipe for a given length of conduit. A similar length is occupied by all gas. The regimes alternate down the conduit.
Flow Regimes: A flowing two-phase fluid can exhibit several “patterns” of flow, such as the liquid occupying the bottom of the conduit with the gas phase flowing above, or a liquid phase with bubbles of gas distributed throughout. In essence, flow regimes are the physical geometry exhibited by the two-phase mixture in the conduit. They are influenced by pipe geometry as well as the physical properties of the fluid mixture and flow rate.
Slip: For the majority of the fluid’s history, one of the phases is flowing faster than the other. Thus, one phase seems to slip by the slower phase. Slip velocity is the difference in the phase velocities.
Flowing Volume Holdup: This term is given as the ratio of the superficial liquid velocity to the sum of the superficial gas and liquid velocities. The term arises in computing properties of homogeneous system and results naturally from the assumption of no slip flow.
Slug Flow: A flow regime characterized by each phase alternately occupying the entire cross section for a large length of the conduit: a “slug” of liquid or gas.
Homogeneous Flow: A mathematical model that considers a two-phase system as a single homogeneous fluid with properties representing the volumetric flow averages of the liquid and gas phases. Homogeneous flow does not exist in real physical situations.
Superficial Liquid Velocity: The velocity that the liquid phase would have in the pipe if there were no gas-phase flowing. Thus, it is the volumetric liquid flow rate divided by the cross sectional area of the pipe.
Mist Flow: At high gas flow velocities, the majority of the liquid becomes distributed as droplets in the gas phase. The liquid is said to be entrained and flow is described as mist flow.
Superficial Gas Velocity: Defined in a manner similar to superficial liquid velocity.
Source: (A. E. DeGance and R. W. Atherton, Chem. Eng. Mar. 23, pp. 135, 1970).
266 Petroleum Refining Design and Applications Handbook Volume 2 Data: Assume a 6 in. pipe size, ID = 6.065” Straight pipe length = 56 ft. Elevation (Static leg rise), HT = 0 ft. Fittings: 4 × 90° ells 1 pipe exit to vessel V-5603. Solution Two-phase flow after the control valve pressure drop. Step 1. Calculate the homogeneous flow liquid ratio λ
λ=
where QLPL QGPL WL WG ρL ρG
Q LPL Q GPL + Q LPL
(15.331)
= volume of liquid flow WL/ρL, ft3/h = volume of gas flow, WG/ρG, ft3/h = liquid flow, lb/h = gas flow, lb/h = liquid density at flow pressure and temperature, lb/ft3 = gas density at flow pressure and temperature, lb/ft3
59033 = 1892.1ft 3/h 31.2
Q LPL =
Q GPL =
λ=
9336 = 5046.5ft 3/h 1.85 1892.1 = 0.2727 (5046.5 + 1892.1)
2. Calculate fTP/fo The ratio of two-phase friction factor to gas-phase friction factor in the pipeline is determined as:
S = 1.281 + 0.478(ln 0.2727) + 0.444(ln 0.2727)2 + 0.09399999(ln 0.2727)3 + 0.008430001(ln 0.2727)4
S = 1.2274
f TP lnλ = 1− fo S
(15.332)
(15.333)
Fluid Flow 267 where fTP = two-phase flow friction factor in the pipe run fo = gas-phase friction factor in the pipe run ln = natural logarithm of base e, 2.7183
f TP ln(0.2727 ) = 1− = 2.0587 fo 1.2274
3. Calculate the Reynolds number Re of the mixture density ρm, lb/ft3:
ρm = ρLλ + ρG (1 − λ)
(15.334)
ρm = (31.2)(0.2727) + (1.85)(1 – 0.2727)
= 9.854 lb/ft3 The average viscosity μm, lb/ft.s:
µm =
µL µ λ + G (1 − λ ) 1488 1488
(15.335)
where μL = liquid viscosity, cP μG = gas viscosity, cP
0.11 0.0105 (0.2727 ) + (1 − 0.2727 ) 1488 1488 = 2.529 × 10−5 lb/ft•s
µm =
Calculate the mixture flowing velocity vm, ft/s
vm =
Q GPL + Q LPL
(
3600 πD2 /4
)
(15.336)
where D = pipe internal diameter, ft.
vm =
(5046.5 + 1892.1)
(
3600 π [ 0.5054 2 ] 4
)
= 9.607 ft /s
The two-phase Reynolds number Re is:
Re =
Dv m ρm µm
Re =
(0.5054 )(9.607)(9.854) (2.529 × 10−5 )
= 1.89 × 106 (Re > 200, 000)
(15.337)
268 Petroleum Refining Design and Applications Handbook Volume 2 4. Calculate the two-phase flow friction factor fTP as follows. First define fTP from fo. In charting fTP/fo against λ, Dukler expressed
0.125 Re0.32 0.125 fo = 0.0014 + = 0 . 00263 (1.89 × 106 )0.32 fo = 0.0014 +
f TP =
f TP fo fo
(15.338)
(15.339)
4 f TP L ρm v 2m ∆Pf = , psi 144 g c D 2
(15.340)
f TP = (2.0587 )(0.00263) = 0.0054 5. Calculate the friction pressure loss ΔPf of straight pipe:
where L = straight pipe length, ft. lb ft gc = 32.174 m • 2 lbf s
4(0.0054 ) 56 (9.854 )((9.607 2 ) 2 (144 )(32.174 ) 0.5054 = 0.235 psi
∆Pf =
6. Calculate the pressure drop due to elevation changes ΔPE, psi. First determine the superficial gas velocity vsg as:
v sg =
v sg =
Q GPL
(
3600 πD2 /4
(
)
5046.5 2
3600 π[0.5054] /4
)
(15.341)
= 6.99 ft /s φ = 0.76844 − 0.085389 v sg + 0.0041264 v sg2 − 0.000087165 v 3sg + 0.00000066422 v 4sg
φ = 0.76844 − 0.085389(6.99) + 0.0041264(6.99)2 − 0.000087165(6.99)3 + 0.00000066422(6.99)4 φ = 0.345
(15.342)
Fluid Flow 269
∆PE =
φρLH T 144
(15.343)
where HT = height of static leg, − for rise and + for fall, ft.
(0.345)(31.2)(0) psi 144 = 0 psi
∆PE =
7. Calculate the pressure drop due to acceleration or pipe fittings and valves ΔPA, psi. The 90o standard elbow ΔPell is calculated as follows: 2 ρGQ GPL ρ Q2 + L LPL 1− λ λ (1.85)(5046.5)2 (31.2)(1892.1)2 + ∆PA = (1 − 0.22727 ) 0.2727 = 474377476.2 psi
∆PA =
∆Pell =
∆PA
3.707 × 1010 + ( d/12 )4
(15.344)
(15.345)
where d = internal pipe diameter, in.
∆Pell =
(474377476.2)
3.707 × 1010 + ( 6.065/12 )4 = 0.0128 psi
4 × 90° standard elbow
ΔPA = 4 × 0.0128
= 0.051 psi. Pipe sharp – edge exit:
∆PA = 4 f TP[14.403(ln d ) + 42]
2 ρm v m 144 g c 2
∆PA = 4(0.0054 )[14.403(ln 6.065) + 42] = 0.144 psi
(9.854 ) 9.607 2 (144 )(32.174 ) 2
(15.347)
270 Petroleum Refining Design and Applications Handbook Volume 2 8. The total two-phase pressure loss ΔPT is:
ΔPT = ΔPf + ΔPE + ΔPA
(15.348)
ΔPT = 0.235 + 0 + (0.051 + 0.144) ΔPT = 0.43 psi The Excel spreadsheet Example 15.25.xlsx has been developed to determine the two-phase pressure drop of Example 15.25. Figure 15.36 shows snapshots of the calculations.
15.33.8 Gas–Liquid Two-Phase Vertical Down Flow Two-phase vertical downflow presents its own problems as often occurs in horizontal pipe line. In a vertical flow, large gas bubbles are formed in the liquid stream resulting in a flow regime known as slug flow. This flow regime (is associated with) can result in pipe vibration and pressure pulsation. With bubbles greater than 1 in. in diameter and the liquid viscosity less than 100cP, slug flow region can be represented by dimensionless numbers for liquid and gas phases respectively (Froude numbers, (NFr)L and (NFr)G). These are related by the ratio of inertial to gravitational force and are expressed as: 0.5
v L ρL (NFr )L = (gD)0.5 ρL − ρG
v G ρG (NFr )G = (gD)0.5 ρL − ρG
(15.349)
0.5
(15.350)
The velocities vG and vL are superficial velocities based on the total pipe cross-section. These Froude numbers exhibit several features in the range 0 < (NFr)L and (NFr)G < 2. Simpson [65] illustrates the values of (NFr)L and (NFr)G with water flowing at an increased rate from the top of an em pty vertical pipe. As the flow rate further increases to the value (NFr)L = 2, the pipe floods and the total cross-section is filled with water. If the pipe outlet is further submerged in water and the procedure is repeated, long bubbles will be trapped in the pipe below (NFr)L = 0.31. However, above (NFr)L = 0.31, the bubbles will be swept downward and out of the pipe. If large long bubbles are trapped in a pipe (d ≥ 1 inch) in vertically down flowing liquid having a viscosity less than 100cP and the Froude number for liquid phase, (NFr)L ≤ 0.3, the bubbles will rise. At higher Froude numbers, the bubbles will be swept downward and out of the pipe. A continuous supply of gas causes the Froude number in the range 0.31 ≤ (NFr)L < 1 to produce pressure pulsations and vibration. These anomalies are detrimental to the pipe and must be avoided. If the Froude number is greater than 1.0, the frictional force offsets the effect of gravity, and thus requires no pressure gradient in the vertical downflow liquid. This latter condition depends on the Reynolds number and pipe roughness. Figure 15.37 shows the flow patterns in a vertical liquid.gas flow and Figure 15.38 shows a correlation in a cocurrent vertical upflow of air-water mixture in terms of Froude numbers.
Fluid Flow 271
Figure 15.36 Excel spreadsheet calculation of Example 15.25.
(Continued)
272 Petroleum Refining Design and Applications Handbook Volume 2
Figure 15.36 (Continued) Excel spreadsheet calculation of Example 15.25.
(Continued)
Fluid Flow 273
Figure 15.36 (Continued) Excel spreadsheet calculation of Example 15.25.
Bubbly
Plug
Churn
Annular
Dispersed
Figure 15.37 Flow patterns in vertical liquid–gas flow (source: S. M. Walas, Chemical Process Equipment—Selection and Design, Butterworth Publishers, 1988).
The Equations The following equations will calculate Froude numbers for both the liquid and gas phases. A developed spreadsheet program will print out a message indicating if the vertical pipe is self-venting, if pulsating flow occurs, or if no pressure gradient is required.
d , ft 12 πD2 2 Area = , ft 4 D=
(15.351)
274 Petroleum Refining Design and Applications Handbook Volume 2 0.5
PG PL– PG
Slug flow eliminated with 24 in line 0.4
(NFr ) G =
VG gD
0.3 FROTH FLOW 0.2
0.1
Slug flow observed in 30 in line SLUG FLOW
0
0
0.1
0.2
0.3 (NFr ) L =
0.4 VL gD
0.5
0.6
0.7
PL PL– PG
Figure 15.38 Slug/forth transition in concurrent vertical upflow of air–water mixtures in terms of Froude numbers (source: L. L. Simpson, Chem. Eng., June 17, pp 192–214, 1968).
vL =
WL , ft /s (3600)(ρL )(Area )
(15.352)
vG =
WG , ft /s (3600)(ρG )(Area )
(15.353)
0.5
v L ρL FRNL = (gD)0.5 ρL − ρG
v G ρG FRNG = (gD)0.5 ρL − ρG
(15.354)
0.5
where Area d D FRNL, (NFr)L FRNG, (NFr)G g vL vG W L WG ρL ρG
= inside cross-sectional area of pipe, ft2. = inside diameter of pipe, in. = inside diameter of pipe, ft = Froude number of liquid phase, dimensionless = Froude number of gas phase, dimensionless = gravitational constant, 32.2ft/2 = liquid velocity, ft/s = gas velocity, ft/s = liquid flow rate, lb/h = gas flow rate, lb/h = liquid density, lb/ft3 = gas density, lb/ft3
(15.355)
Fluid Flow 275 The Algorithms If FRNL < 0.31, Vertical pipe is SELF VENTING ELSE 0.3 ≤ FRNL < 1.0, PULSE FLOW, and may result in pipe vibration. FRNL > 1.0, NO PRESSURE GRADIENT. Example 15.26 Calculate the Froude numbers and flow conditions for the 2, 4, and 6 in. (Schedule 40) vertical pipes having the following liquid and vapor flow rates and densities Liquid
Vapor
Mass flow rate, lb/h
6930
1444
Density, lb/ft3
61.8
0.135
Solution 2 in. pipe diameter (Schedule 40) I.D. = 2.067 in.
D = 2.067/12 = 0.17225 ft. π D2 π(0.17225)2 Area = = 4 4 2 = 0.0233 ft
(15.351)
Liquid velocity, vL is:
vL =
WL (ρL A)
6930 (61.8)(0.0233)(3600) = 1.34 ft /s =
(15.352)
Vapor velocity, vG is:
vG =
WG (ρG A)
1444 (0.135)(0.0233)(3600) = 127.52 ft /s =
(15.353)
276 Petroleum Refining Design and Applications Handbook Volume 2 Froude number for liquid phase is:
v L ρL FRNL = (gD)0.5 ρL − ρG
0.5
61.8 = (32.2)(0.17225) 61.8 − 0.135 (1.34)
0.5
(15.354)
(15.355)
= 0.57 The Froude number for vapor phase is:
v G ρG FRNG = (gD)0.5 ρL − ρG
0.5
0.135 = (32.2)(0.17225) 61.8 − 0.135 127.5
0.5
= 2.53 Since the Froude number for liquid phase is greater than 0.31 and less than 1.0, the 2 in. pipe can produce a pulse flow, which may result in pipe vibration. The Excel spreadsheet Example 15.26.xlsx calculates the Froude numbers for 4 and 6 in. pipe sizes. Table 15.34 shows typical computed results for 2, 4, and 6 in. pipe sizes.
Table 15.34 Gas–liquid two-phase downflow. Pipe internal diameter, in.
2.067
4.026
6.065
Liquid flow rate, lb/h
6930
6930
6930
Liquid density, lb/ft3
61.80
61.80
61.80
Gas flow rate, lb/h
1444
1444
1444
Gas density, lb/ft3
0.135
0.135
0.135
Pipe area, ft2
0.023
0.088
0.201
Liquid velocity, ft/s
1.337
0.352
0.155
Gas velocity, ft/s
127.504
33.609
14.810
Froude number for liquid phase
0.5682
0.1073
0.0385
Froude number for gas phase
2.5332
0.4784
0.1718
Flow is pulse and this may result in pipe vibration.
Line is self-venting. Therefore, no vibration problems would be expected.
Line is self-venting. Therefore no vibration problems would be expected.
Fluid Flow 277
15.33.9 Pressure Drop in Vacuum Systems Vacuum in process systems refers to an absolute pressure that is less than or below the local barometric pressure at the location. It is a measure of the degree of removal of atmospheric pressure to some level between atmospheric barometer and absolute vacuum (which cannot be attained in an absolute value in the real world), but is used for a reference of measurement. In most situations, a vacuum is created by pumping air out of the container (pipe, vessels) and thereby lowering the pressure. See Figure 15.3 to distinguish between vacuum gauge and vacuum absolute. This method [62] is for applications involving air or steam in cylindrical piping under conditions of (a) turbulent flow, (b) sub-atmosphere pressure, (c) pressure drop is limited to 10% of the final pressure (see comment to follow), and (d) the lower limit for application of the method is
W/d = 20
(15.356)
where W is the flow rate in lbs/h and d is the inside pipe diameter in inches. If the above ratio is less than 20, the flow is “streamlined” and the data do not apply. If the pressure drop is greater than 10% of the final pressure, the pipe length can be divided into sections and the calculations made for each section, maintaining the same criteria of (c) and (d) above. Method [62] The method solves the equation (see Figure 15.39)
∆Pvac =
(F1C D1C T1 ) + (F2C D 2C T 2 ) P1
(15.357)
where
Pvac = pressure drop, in. water/l00 ft of pipe P1 = initial pressure, inches mercury absolute F1 = base friction factor, Figure 15.39 F2 = base friction factor, Figure 15.39 CT1 = temperature correction factor, Figure 15.39 CT2 = temperature correction factor, Figure 15.39 CD1 = diameter correction factor, Figure 15.39 CD2 = diameter correction factor, Figure 15.39
Example 15.27: Line Sizing for Vacuum Conditions Determine the proper line size for a 350 equivalent feet vacuum jet suction line drawing air at 350°F, at a rate of 255 lbs/h with an initial pressure at the source of 0.6 in. Hg. Abs. Assume 10-in. pipe reading Figure 15.39. Note: watch scales carefully. Fl F2 CD1 CD2 CT1 CT2
= 0.0155 = 0.071 = 0.96 = 0.96 = 1.5 = 1.67
Pvac = [(0.0155) (0.96) (1.5) + (0.071) (0.96) (1.67)]/0.6
= (0.02232 + 0.1138)/0.6
= 0.2269 in. water/l00 ft.
(15.357)
278 Petroleum Refining Design and Applications Handbook Volume 2 Total line pressure drop:
0.2269 ∆Pvac = (350) = 0.794 in. water (for 350′ ): 100
= (0.794 /13.6) = 0.0584 in. Hg
Final calculated pressure = 0.6 + 0.0584 = 0.6584: in. Hg
10% of 0.658 = 0.0658 in. Hg STANDARDS OF THE HEAT EXCHANGE INSTITUTE, INC 1000 800 600
C r2
TEAM
FOR S
400 C r1
60
AIR
100 80
FOR
TEMPERATURE, °F
200
C r2 C r1
40
1.0
1.4
1.8
2.2 2.6 3.0 3.4 3.8 CrC1τ TEMPERATURE CORRECTION FACTORS
60
CD1
ACTUAL PIPE DIAMETER INCHES
40
4.2
4.6
5.0
CD2
20
10 8 6 4 F2
F1
2
10–5
10–4
10–3
10–2
10–1
Note: Friction Factors F1 and F2 are on rate flow, while Factors CD1 and CD2 are based on actual diameter.
Figure 15.39 Evaluation curves for friction losses of air steam flowing turbulently in commercial pipe at low pressure (source: Standards for Steam Jet Ejectors, 4th. Ed., Heat Exchange Institute, 1988). (Continued)
Fluid Flow 279 STEAM JET VACUUM SYSTEMS 105
RATE OF FLOW, POUNDS PER HOUR
104
w must be greater than 20 For turbulent flow — d F=
103
(F1 × CD1 × CT1) + (F2 × CD2 × CT2) P1
F = Pressure Drop, inches of water in 100 feet of pipe P = Initial Pressure, inches of mercury absolute
102
CD2
1
102
10
CD1
103
10 104
Figure 15.39 (Continued) Evaluation curves for friction losses of air steam flowing turbulently in commercial pipe at low pressure (source: Standards for Steam Jet Ejectors, 4th. Ed., Heat Exchange Institute, 1988).
Therefore the system is applicable to the basis of the method, since the calculated pressure drop is less than 10% of the final pressure, and w/d = 25.5, which >20.
15.33.10 Low Absolute Pressure Systems for Air For piping with air in streamline flow at absolute pressures in the range between 50 microns and one millimeter of mercury, the following is a recommended method. Calculation procedures in pressure regions below atmospheric are very limited and often not generally applicable to broad interpretations. For this method to be applicable, the pressure drop is limited to 10% of the final pressure. Method [62, 63] Refer to Figure 15.40 for low pressure friction factor and air viscosity of Figure 15.41 to correspond to Figure 15.40.
P1′− P2′ =
4 fLρv 2 , psi 2g c D(144 )
(15.127)
280 Petroleum Refining Design and Applications Handbook Volume 2
5.00 4.00 3.00 2.00 12 AND 18 INCH PIPE
FRICTION FACTORS f
1.00 0.80 0.70 0.60 0.50 0.40 0.30 0.20
6 AND 8 INCH PIPE
0.10 0.08 0.07 0.06 0.05 0.04 0.03 0.02
0.01 40 50 60
80
200
300
500
REYNOLDS NUMBER RD =
1000
2000
DvP µ
Figure 15.40 Friction factor for streamlined flow of air at absolute pressures from 50 µHg to 1 mm Hg (source: Standard for Steam Jet Ejectors, 3rd ed., Heat Exchange Institute, 1956 [62] and Standards for Steam Jet Vacuum Systems, 4th ed., 1988).
where P1′ = upstream static pressure, psia. P2′ = downstream static pressure, psia. f = friction factor, from Figure 15.5. L = length of pipe (total equivalent), ft, incl. valves and fittings ρ = average density, lbs/ft3 v = average velocity, ft/s lb ft gc = conversion factor = 32.174 m • 2 lb D = inside diameter of pipe, ft f s = abs. viscosity of air, lbs/ft-s.
Fluid Flow 281 24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
24
23
22
21
20
19
18 17
16
15
14
13
12
11
10
0
100
200
300
400 500 600 TEMPERATURE, °F
700
800
900
1000
25 25
ABSOLUTE VISCOSITY × 106 POUNDS PER FOOT·SECOND
ABSOLUTE VISCOSITY OF AIR
Figure 15.41 Absolute viscosity of air (source: Standard for Steam Jet Ejectors, 3rd ed., Heat Exchange Institute, 1956 [62] also, Standards for Steam Jet Vacuum Systems, 4th ed., 1988 [38]).
15.33.11 Vacuum for Other Gases and Vapors Ryans and Roper categorize [64] vacuum in process systems as: Category
Absolute vacuum (absolute pressure)
Rough vacuum
760 torr to 1 torr
Medium vacuum
1 to 10−3 torr
High vacuum
10−3 to 10−7 torr
Ultra high vacuum
10−7 torr and below
The majority of industrial chemical and petrochemical plants’ vacuum operations are in the range of 100 microns to 760 torr. This is practically speaking the rough vacuum range noted above. For reference: 1 torr = 1 mm mercury (mmHg) 1 in. mercury (in. Hg) = 25.4 torr 1 micron (µm Hg) = 0.0010 torr
282 Petroleum Refining Design and Applications Handbook Volume 2 In general, partially due to the size and cost of maintaining vacuum in a piping system, the lines are not long (certainly not transmissions lines), and there is a minimum of valves, fittings, and bends to keep the resistance to flow low. The procedure recommended by [64] is based on the conventional gas flow equations, with some slight modifications. The importance in final line size determination is to determine what is a reasonable pressure loss at the absolute pressure required and the corresponding pipe size to balance these. In some cases a trial/error approach is necessary. Method [64], by permission: 1. Convert mass flow rate to volumetric flow rate, qm.
qm = W (359/M) (760/Pt) (T/(32 + 460)) (1/60), ft3/min
(15.359)
where Pt = pressure, torr T = temperature, °R W = mass flow, lbs/h M = molecular weight 2. C alculate section by section from the process vessel to the vacuum pump (point of lowest absolute pressure). 3. Assume a velocity, v, ft/s consistent with Figure 15.42. Use Table 15.35 for short, direct connected connections to the vacuum pump. Base the final specifications for the line on pump specifications. Also the diameter of the line should match the inlet connection for the pump. General good practice indicates that velocities of 100 to 200 ft/s are used, with 300 to 400 ft/s being the upper limit for the rough vacuum classification.
Sonic velocity, vs = (kg [1544/M] T)1/2, ft/s. Use v from Figure 15.42, and qm from Eq. 15.359.
1000 8 6 4 3 Maximum
Velocity, ft/s
2
Design basis
100 8
Minimum
6 4 3 2 10
1
2
3
4 5 6 7 89
2 3 Pressure, torr
4 5 6 7 8 9 100
2
3 4 5 6 7 89 1000
Figure 15.42 Typical flow velocities for vacuum lines. Note: 1 torr = 1.33 mb = 133.3. Pa, 1 ft/s = 0.3048 m/s (source: Ryans, J. L., and D. L. Roper, Process Vacuum System Design and Operation, McGraw-Hill Book Co., Inc. 1986 [64]).
Fluid Flow 283 Table 15.35 Criteria for sizing connecting lines in vacuum service. Vacuum pump
Assumed flow velocity, ft/s
Steam jet: System pressure, torr 0.5 5
300
5 25
250
25 150
200
150 760
150
Liquid ring pump: Single-stage*
100
Two-stage
150
Rotary piston: Single-stage
50
Two-stage
25
Rotary vane:† Single-stage
200
Two-stage
400
Rotary blowers: Atmospheric discharge
50
Discharging to backing pump
100
*Assumes the pump features dual inlet connections and that an inlet manifold will be used. † Based on rough vacuum process pumps. Use 25 ft/s for high vacuum pumps. By permission, Ryans, J.L. and Roper, D.L., Process Vacuum Systems Design and Operation, McGraw-Hill Book Co. Inc., 1986 [18].
4. Determine pipe diameter, D, ft,
D = 0.146 q m /v
(15.360)
Round this to the nearest standard pipe size. Recalculate v based on actual internal diameter of the line. 5. Determine Reynolds Number, Re,
Re = ρDv/μe = density, lb/ft3 at flowing conditions D = pipe inside diameter, ft v = vapor velocity (actual),ft/s = viscosity of vapor, lb/ft-s e
(15.26)
284 Petroleum Refining Design and Applications Handbook Volume 2 6. Determine friction factor, f, from Moody Friction Factor Charts (Figure 15.5) or Chen’s explicit equation. or, calculate for turbulent flow using Blausius’ equation [64]:
f = 0.316/(Re)1/4, for Re < 2.0 × 105
(smooth pipe only)
7. T abulate the summation of equivalent lengths of straight pipe, valves, fittings, entrance/exit losses as presented in earlier sections of this chapter. 8. Calculate the pressure drop for the specific line section (or total line) from: 2 ∆Ρ T = 0.625 ρi f D Lq m /d 5 , torr
(15.361)
ΔPT = 4.31 ρi fD L v2/2gcd, torr
(15.362)
or
where
= density, lb/ft3 d = pipe inside diameter, in. qm = volumetric flow rate, cu ft/min fD = friction factor, (Moody Darcy) Figure 15.5 PT = pressure drop, torr
Calculate:
i
= PiM/555Ti, lb/ft3
(15.363)
Pi = pressure, torr M = average molecular weight of mixture flowing Ti = temperature, °R 9. I f the calculated pressure drop does not exceed the maximum given in Figure 15.43, use this calculated value to specify the line. If the P exceeds the limit of Figure 15.43, increase the pipe size and repeat the calculations until an acceptable balance is obtained. For initial estimates, the authors [64] recommend using 0.6 times the value obtained from Figure 15.43 for an acceptable pressure loss between vessel and the pump. The suction pressure required at the vacuum pump (in absolute pressure) is the actual process equipment operating pressure minus the pressure loss between the process equipment and the source of the vacuum. Note that absolute pressures must be used for these determinations and not gauge pressures. Also keep in mind that the absolute pressure at the vacuum pump must always be a lower absolute pressure than the absolute pressure at the process.
15.33.12 Pressure Drop for Flashing Liquids When a liquid is flowing near its saturation point (i.e., the equilibrium or boiling point) in a pipe line, decreased pressure will cause vaporization. The higher the pressure difference, the greater the vaporization resulting in flashing of the liquid. Steam condensate lines cause a two-phase flow condition, with hot condensate flowing to a lower pressure through short
Maximum acceptable pressure loss across component ΔP/P0
Fluid Flow 285 1.0 8 6 5 4
Primary condenser
3
Vent condenser Section 1 Section 2
0.1 8
Sections 3 and 4
6 5 4 3 2
0.01
1
2
3
4 5
7 8 10
2
3
4 5 6 7 8 100
2
3
4
6 7 8 1000
Pressure in vacuum vessel P0
Figure 15.43 Acceptable pressure losses between the vacuum vessel and the vacuum pump. Note: Reference sections on figure to system diagram to illustrate the sectional type hook-ups for connecting lines. Use 60% of the pressure loss read as acceptable loss for the system from process to vacuum pump, for initial estimate; P, pressure drop (torr) of line in question; Po, operating pressure of vacuum pressure equipment, absolute torr (source: Ryans, J. L., and D. L. Roper, Process Vacuum System Design and Operation, McGraw-Hill Book Co., Inc. 1986 [64]).
and long lines. For small lengths with low pressure drops, and the outlet end being a few pounds per square inch of the inlet, the flash will be assumed as a small percentage. Consequently, the line can be sized as an all liquid line. However, caution must be exercised as 5% flashing can develop an important impact on the pressure drop of the system [5]. Sizing of flashing steam condensate return lines requires techniques that calculate pressure drop of two-phase flow correlations. Many correlations have been presented in the literature [49, 50, 53, 66]. Most flow patterns for steam condensate headers fall within the annular or dispersed region on the Baker map. Sometimes, they can fall within the slug flow region, however the flashed steam in steam condensate lines is less than 30% by weight. Steam is the most common liquid that is flashed in process plants, but of course, it is not the only one as many processes utilize flash operations of pure compounds as well as mixtures. Although this presentation is limited to steam, the principles apply to other materials. Steam condensate systems often are used to generate lower pressure steam by flashing to a lower pressure. When this occurs, some steam is formed and some condensate remains, with the relative quantities depending upon the pressure conditions. Figure 15.44 is a typical situation. Percent incoming condensate flashed to steam:
% flash =
(h1 − h 2 )100 Lv
(15.364)
where h1 = enthalpy of liquid at higher pressure, Btu/lb h2 = enthalpy of liquid at lower or flash pressure, Btu/lb Lv = latent heat of evaporation of steam at flash pressure, Btu/lb Example 15.28: Calculation of Steam Condensate Flashing There are 79,500 lbs/h of 450 psig condensate flowing into a flash tank. The tank is to be held at 250 psig, generating steam at this pressure. Determine the quantity of steam produced.
286 Petroleum Refining Design and Applications Handbook Volume 2
Flash Pressure, Z psig (Lower than either X or Y)
Pressure: X, psig
Vapor
Pressure: Y, psig Condensate From Various Collection Headers
Flashing
Liquid
Liquid Level
Condensate Return to Collection Tank
Figure 15.44 Typical steam condensate flashing operation.
Enthalpy of liquid at 450 psig = 441.1 Btu/lb Enthalpy of liquid at 250 psig = 381.6 Btu/lb Latent heat of vaporization at 250 psig = 820.1 Btu/lb From Eq. 15.364
% flash into steam =
441.1 − 381.6 (100) = 7.25% 820.1
Steam formed = (0.0725) (79,500) = 5,764 lbs/h Condensate formed = 79,500 5,764 = 73,736 lbs/h
15.33.13 Sizing Condensate Return Lines Steam condensate lines usually present a two-phase flow condition, with hot condensate flowing to a lower pressure through short and long lines. As the flow progresses down the pipe, the pressure falls and flashing of condensate into steam takes place continuously. For small lengths with low pressure drops, and the outlet end being within a
Fluid Flow 287 few pounds per square inch of the inlet, the flash will be such a small percent that the line can often be sized as an all liquid line. However, caution must be exercised as even 5% flashing can develop an important impact on the pressure drop of the system. Calculation of condensate piping by two-phase flow techniques is recommended; however, the tedious work per line can often be reduced by using empirical methods and charts. Some of the best methods are proprietary and not available for publication; however, the Sarco method [67, 68 ] has been used and found to be acceptable, provided no line less than 1½ in. is used regardless of the chart reading. Under some circumstances, which are too random to properly describe, the Sarco method may give results too small by possibly a half pipe size. Therefore, latitude is recommended in selecting either the flow rates or the pipe size. Design Procedure Using Sarco Chart [67, 68] 1. E stablish upstream or steam pressure from which condensate is being produced and discharged into a return line through steam traps, or equivalent, psig. 2. Establish the steam condensate load or rate in lbs/h flow. 3. Establish the pressure of the condensate return line, psig. 4. The method is based on an allowable 5000 ft/min velocity in the return line (mixture). 5. Calculate load factor:
=
5, 000(100) 500, 000 = Condensate Rate, lbs / h C
(15.365)
6. E stablish condensate receiver (or flash tank) pressure, psig. 7. Referring to Figure 15.45, enter at steam pressure of (1) above, move horizontally to condensate receiver pressure of (6) above, and then up vertically to the “factor scale.” 8. Divide the load factor (step 1) by the value from the “factor scale” of (7) above, obtain ft/min/(100 lb/h load). 9. Enter chart on horizontal velocity line, go vertically up to the steam pressure of (1) above, and read pipe size to the next largest size if the value falls between two pipe sizes. 10. For pipe sizes larger than 3-in., follow the steps (1) thru (8) above. Then enter the vertical scale at the steam pressure of (1) above, and more to the 3-in. pipe size and down to the horizontal velocity scale. 11. Divide the result of step 8 above by the result of step (10). 12. Refer to the large pipe multipliers shown in the table on the chart, and select the pipe size whose factor is equal to or smaller than the result of step (11) above. This is the pipe size to use, provided a sufficient factor of safety has been incorporated in the data used for the selection of pipe size. 13. Calculation of “factor scale” for receiver pressures different than those shown on chart:
factor =
36.2(V )(h p − h r ) L v (h p − 180)
where 3 V = specific volume of steam at return line pressure, ft /lb hp = enthalpy of liquid at supply steam pressure, Btu/lb hr = enthalpy of liquid at return line pressure, Btu/lb Lv = latent heat of evaporation at return line pressure, Btu/lb Use the factor so calculated just as if read from the chart, i.e., in step (8) above.
(15.366)
288 Petroleum Refining Design and Applications Handbook Volume 2 VELOCITY AT PIPE EXIT WHEN DISCHANGING CONDENSATE AT SATURATION TEMPERATURES FROM VARIOUS PRESSURES TO ATMOSPHERE AT A RATE OF 100 POUND/HR. FOR LARGER PIPES MULTIPLY 3" PIPE VELOCITY BY FOLLOWING FACTORS: PIPE FACTOR 4" 0.58 5" 0.37 6" 0.25 8" 0.15 10" 0.095 12" 0.066 14" 0.054
WHEN DISCHARGING TO PRESSURE HIGHER THAN ATMOSPHERIC, MULTIPLY VELOCITY TO ATMOSPHERE BY FACTOR CORRESPONDING TO SUPPLY PRESSURE AND RECEIVER PRESSURE.
600
.1
FACTOR SCALE .3 .4 .5
.2
.6
.8
1.0
500
60 P SIG
400
SSU
RE
SIG
CIT
Y
. PRE 40
60
" 1/2
" 3/4
1·1
1"
/4"
/2" 1·1
2"
/2"
3"
5 PSIG
SIG 10 P
15 P
60
SIG
20 P
80
PIP
SIG
EV
ELO
REC
100
2·1
40 P
30 P
SUPPLY PRESSURE
200
SIG
50 P
SIG
300
40 30
20
10 10
20
30
80 100
200
400
600
1000
2000
4000
VELOCITY FT/MIN PER 100 POUNDS/H CONDENSATE
Figure 15.45 Sarco flashing steam condensate line sizing flow chart (source: Spirax Sarco. Inc., Allentown, PA [75]).
Example 15.29: Sizing Steam Condensate Return Line A 450 psig steam system discharges 9425 lb/h of condensate through traps into a return condensate line. The return header is to discharge into a flash tank held at 90 psig. The calculated total equivalent length of pipe, valves, and fittings is 600 ft. Using the Sarco chart, Figure 15.45, determine the recommended line size for the return line. 1. 2. 3. 4. 5.
pstream steam pressure = 450 psig U Condensate load = 9425 lbs/h Return line pressure = 90 psig Use the Sarco recommendation of 5000 ft/min Load factor
=
(5, 000)(100) = 53.05 9, 425
Fluid Flow 289 6. R eceiver pressure = 90 psig 7. Refer to Figure 15.45 and note that required receiver pressure is not shown, so calculate “factor scale” by previous formula:
Data: hp = 441 Btu/lb at 450 psig
hr = 302 Btu/lb at 90 psig Lr = 886 Btu/lb at 90 psig V = 4.232 ft3/lb at 90 psig
“factor scale” value =
36.2(4.232)(441 − 302) = 0.092 886(441 − 180)
(15.366)
53.05 = 576.1 0.092 9. R ead Chart: At 450 psig and 576.1, the line size shows just under 2-in. Recommend use 2-in. 8. Ft/min/l00#/h =
The procedure for using the convenient chart Figure 15.46 [76] is, for example: Step 1: Enter the figure at 600 psig below the insert near the right-hand side, and read down to the 200psig end pressure.
0
50 1.0 60 75 0.8 Step 7 100 0.6 150 200 0.4 End pressure 0.3 Step 6 psig
H4 0 H4
H4 10"
SC
ec
ft/s
ec
3.0
0
3"
SC H4 0 SC H4 0
0 ft/s
SC
20
ec
400 250 ft/se c ft 300 /sec ft/s ec
ec
0 2.0
SC
ec
Step 5
ec
ec
ec
ft/s
140 ft/sec f 120 t/sec ft/s ec
ft/s
ft/s
ft/s
160
ft/s
Step4
0 12 14" " SCH SC 10 H1 16" 0 18" SCH 1 20" SCH 0 SC 10 H1 0 24" SC H1 0
0.1 0.08 0.06 0.05
ft/s
ec
8"
0.2 1 0
30 15
200
6"
0.4
40
2"
1.0 0.3 0.6
60
f t/s
ec
4"
Pressure drop, psi/100 ft
1½
" SC
2.0
SC H4
1"
H8 0
SC
H8
0
H8
80
250
ft/s
0.2
100 200 400 600 Saturation pressure, psig 50 100 200 400 600 Step 1
Step 2
200 150 100 75 60 30 10 0 End pressure psig
ep St 3 1 1.5 2 100
3 4
Velocity-correction factor
100
0 H8
SC
SC
¼"
½"
10.0 8.0 6.0 4.0
6 8 1 1000
1.5 2.
3 4 6 8 1 1.5 2. Flowrate, lb/h 10,000
3 4
6 8 1
1.5 2
3 4
6 8 1
100,000
Figure 15.46 Flashing steam condensate line sizing chart (source: Ruskin, R. P., “Calculating Line Sizes For Flashing Steam Condensate”, Chem. Eng., Aug 18, p. 101, 1985).
290 Petroleum Refining Design and Applications Handbook Volume 2 Step 2: Proceed left horizontally across the chart to the intersection, with: Step 3: the 1000 lb/h flow rate projected diagonally up from the bottom scale. Step 4: Reading vertically up from this intersection, it can be seen that a 1-in. line will produce more than the allowed pressure drop, so a 1½-in. size is chosen. Step 5: Read left horizontally to a pressure drop of 0.28 psi/100 ft on the left-hand scale. Step 6: Note the velocity given by this line as 16.5 ft/s, then proceed to the insert on the right, and read upward from 600 psig to 200 psig to find the velocity correction factor as 0.41. Step 7: Multiply 0.41 by 16.5 to get a corrected velocity of 6.8 ft/s. The comparison between this method and that of Dukler [66] and others gives good agreement for reasonably good cross section of flow regimes. Because flashing steam-condensate lines represent two-phase flow, with the quantity of liquid phase depending on the system conditions, these can be designed following the previously described two-phase flow methods. Ruskin [69] assumes that a single homogeneous phase of fine liquid droplets is dispersed in the flashed vapor and the pressure drop is calculated using the Darcy equation:
0.00000336 f W 2 ∆P = , psi ρd5
(15.122)
Ruskin [69] developed a method for calculating pressure drop of flashing condensate. His method gave pressure drops comparable to those computed by the two-phase flow with good agreement with experimental data. The method employed here is based on a similar technique given by Ruskin. The pressure drop for flashing steam uses the average density of the resulting liquid-vapor mixture after flashing. In addition, the friction factor used is valid for complete turbulent flows in both commercial steel and wrought iron pipe. The pressure drop assumes that the vapor-liquid mixture throughout the condensate line is represented by mixture conditions near the end of the line. This assumption is valid since most condensate lines are sized for low pressure drop, with flashing occurring at the steam trap or valve close to the pipe entrance. If the condensate line is sized for a higher pressure drop, an iterative method must be used. For this case, the computations start at the end of the pipeline and proceed to the steam trap. The method employed determines the following: 1. Th e amount of condensate flashed for any given condensate header from 15 to 140 psia. Initial steam pressure may vary from 40 to 165 psia. 2. the return condensate header temperature. 3. the pressure drop (psi/100ft) of the steam condensate mixture in the return header. 4. the velocity of the steam condensate mixture and gives a warning message if the velocity is greater than 5000 ft/min, as this may present problems to the piping system. The Equations The following equations are used to determine the pressure drop for flashed condensate mixture [70].
WFRFL = B(ln Pc)2 − A
(15.367)
A = 0.00671(ln Ph)2.27
(15.368)
B = ex 10−4 + 0.0088
(15.369)
16.919 X = 6.122 − ln Ph
(15.370)
where
and
Fluid Flow 291
WG = (WFRFL) (W)
(15.371)
WL = W − WG
(15.372)
TFL = 115.68(Ph)0.226
(15.373)
ρG = 0.0029Ph0.938
(15.374)
ρL = 60.827 − 0.078Ph + 0.00048Ph2 − 0.0000013Ph3
(15.375)
ρM =
WG + WL WG WL ρ + ρ G L
(15-376)
For fully turbulent flow
f=
0.25 0.000486 − log d
2
(15.377)
where d = pipe diameter, in. Pressure drop
∆Ρ T =
v=
0.000336 f W 2 d 5 ρM
(15.378)
3.054 WG WL + d 2 ρG ρL
(15.379)
If v ≥ 5000 ft/min, print a warning message as condensate may cause deterioration of the process pipeline. where D = internal pipe diameter, in. F = friction factor, dimensionless = steam condensate pressure before flashing, psia Pc = flashed condensate header pressure, psia Ph V = velocity of flashed condensate mixture, ft/min W = total flow of mixture in condensate header, lb/h = flashed steam flow rate, lb/h WG = flashed condensate liquid flow rate, lb/h WL WFRFL = weight fraction of condensate flashed to vapor TFL = temperature of flashed condensate, °F = pressure drop of flashed condensate mixture, psi/100 ft ΔPT = flashed condensate liquid density, lb/ft3 ρL ρG = flashed steam density, lb/ft3 ρM = density of mixture (flashed condensate/steam), lb/ft3.
292 Petroleum Refining Design and Applications Handbook Volume 2 Example 15.30 Determine the pressure drop for the 4, 6, and 8 in. (Schedule 40) condensate headers under the following conditions Flow rate, lb/h
10,000
Steam condensate pressure, psia
114.7
Header pressure, psia
14.7
Solution For the 4 in. (Schedule 40) pipe size, I.D. = 4.026 in. The weight fraction of the condensate is:
16.919 X = 6.122 − ln Ph 16.919 = 6.122 − ln14.7
(15.370)
= −0.1726
A = 0.00671(ln Ph)2.27
(15.368)
= 0.0632
B = e(−0.1726)(10−4) + 0.0088
(15.369)
= 0.008884 The weight fraction of the condensate is:
WFRFL = B(ln Pc)2 − A
= 0.008884(ln 114.7)2.27 – 0.0632
= 0.1365
WG = (WFRFL) (W)
= 0.1365 × 10,000
= 1365 lb/h
WL = W − WG
= 10,000 – 1365
= 8635 lb/h.
(15.367)
(15.371)
(15.372)
Fluid Flow 293 The temperature of the flashed condensate is:
TFL = 115.68(Ph)0.226
(15.373)
= 212.4°F The flashed steam density, condensate and density of the mixture are:
ρG = 0.0029Ph0.938 = 0.0029(14.7 )0.938
= 0.0361 lb/ft
(15.374)
3
ρL = 60.827 − 0.078Ph + 0.00048Ph2 − 0.0000013Ph3
ρL = 60.8827 − 0.078(14.7 ) + 0.00048(14.7 )2 − 0.00000013(14.7)3 = 59.78 lb/ft
ρM =
ρM =
(15.375)
3
WG + WL WG WL ρ + ρ G L (1365 + 8635) 1365 8635 + 0.0361 59.78
(15.376)
= 0.2634 lb/ft 3 Assuming that the flow through the line is turbulent: For fully turbulent flow
f=
0.25 0.000486 − log d
2
(15.377)
where d = pipe diameter, in.
f=
0.25
0.000486 − log 4.026 = 0.01628
2
294 Petroleum Refining Design and Applications Handbook Volume 2 Pressure drop of the steam condensate mixture in the return header:
∆Ρ T =
0.000336 f W 2 0.000336(0.01628)(10, 000)2 = d 5 ρM (4.026)5 (0.2634 )
= 1.964 psi /100 ft.
Velocity of the flashed condensate mixture is:
v=
3.054 WG WL + d 2 ρG ρL
3.054 1365 8635 + (4.026)2 0.0361 59.78 = 7154 ft/ min. =
(15.379)
Since the velocity v ≥ 5000 ft/min, the condensate may cause deterioration of the 4 in. line. The Excel spreadsheet (Example15.30.xlsx) calculates the parameters for 6 and 8 in. Schedule 40 pipe sizes. Table 15.36 compares the results of 4, 6, and 8 in. pipe sizes. Table 15.37 shows the friction factor for pipes carrying water and constant accounting for surface roughness. Table 15.36 Computer results of line sizes for flashing steam condensate of Example 15.30. Pipe internal diameter, in.
4.026
6.065
7.981
Total flow of mixture in condensate header, lb/h
10000
10000
10000
Steam condensate pressure before flashing, psia
114.7
114.7
114.7
Flashed condensate header pressure, psia.
14.7
14.7
14.7
Weight fraction of condensate flashed to vapor
0.135
0.135
0.135
Flashed steam flow rate, lb/h
1346
1346
1346
Flashed condensate liquid flow rate, lb/h
8654
8654
8654
Temperature of flashed condensate, °F
212.36
212.36
212.36
Flashed steam density, lb/ft
0.036
0.036
0.036
Flashed condensate liquid density, lb/ft3
59.780
59.780
59.780
Density of mixture, lb/ft
0.267
0.267
0.267
Friction factor:
0.0163
0.0149
0.0141
Pressure drop of flashed condensate mixture, psi/100 ft
1.937
0.228
0.055
Velocity of flashed condensate mixture, ft/min.
7,055
3,109
1,795
Velocity is greater than 5000 ft/min. Deterioration of the pipe line is possible.
Velocity is less than 5000 ft/min. The Condensate header line will not deteriorate.
Velocity is less than 5000 ft/min. The Condensate header line will not deteriorate.
3
3
Fluid Flow 295 Table 15.37 Cameron hydraulic data.* Friction losses in pipes carrying water Among the many empirical formula e for friction losses that have been proposed that of Williams and Hazen has been most widely used. In a convenient form it reads: in which 100 1.85 q1.85 f = friction head in ft of liquid per f = .2083 C d4.8655 100 ft of pipe (if desired in lb per sq in. multiply f × .433 × sp gr) d = inside dia of pipe in inches q = flow in gal per min C = constant accounting for surface roughness This formula gives accurate values only when the kinematic viscosity of the liquid is about 1.1 centistokes or 31.5 SSU, which is the case with water at about 60F. But the viscosity of water varies with the temperature from 1.8 at 32F to .29 centistokes at 212F. The tables are therefore subject to this error which may increase the friction loss as much as 20% at 32F and decrease it as much as 20% at 212F. Note that the tables may be used for any liquid having a viscosity of the same order as indicated above. Values of C for various types of pipe are given below together with the corresponding multiplier which should apply to the tabulated values of the head loss, f, as given on pages 29 to 48.
Friction losses in pipe; C = 100 1/8 in Standard wt steel Flow US gal per min 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Extra strong steel
0.269” inside dia Velocity ft per s 0.565 1.13 1.69 2.26 2.83 3.39 3.95 4.52 5.08 5.65
Velocity head ft 0.00 0.02 0.04 0.08 0.12 0.18 0.24 0.32 0.40 0.50
0.215” inside dia Head loss ft per 100 ft 1.75 6.31 13.4 22.8 34.4 48.2 64.1 82.0 102 124
Commonly used value for design purposes 140 140 140 140 130 110 100 100 100 100 100 100 100 100 100 100 90 60 70 1.93
Velocity head ft 0.01 0.05 0.11 0.19 0.30 0.44 0.61 0.78 0.98 1.21
Head loss ft per 100 ft 5.21 18.8 39.8 67.7 102 147 191 244 303 369
1/4 Inch Standard wt steel
Values of C Range — high = best, Average smooth, value well laid for — good, low = clean, poor or new Type of Pipe corroded pipe Cement—Asbestos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160–140 150 Fibre. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . − 150 Bitumasc-enamel-lined iron or steel centrifugally applied. . 160–130 148 Cement-lined iron or steel centrifugally applied. . . . . . . . . . . − 150 Copper, brass, lead, n or glass pipe and tubing . . . . . . . .. . . 150–120 140 Wood-stave. . . . . . . . . . . . . . . . . . . . . .. . . . .. . . . . . .. . . . . . . . 145–110 120 Welded and seamless steel. . . . . . . . . . . . . . . . . . . . . . . . . . . . 150–80 140 Connuous-interior riveted steel (no projecng rivets or joints. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . − 139 Wrought-iron. . . . . . . . . . . . . . . . .. . . . . . . .. . . . . . . . . . . . . . . 150–80 130 Cast-iron. . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . 150–80 130 Tar-coated cast-iron. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145–80 130 Girth-riveted steel (projecng rivets in girth seams only). . . − 130 Concrete. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152–85 120 Full-riveted steel (projecng rivets in girth and horizontal seams). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. − 115 Vitrified. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . − 110 Spiral-riveted steel (flow with lap). . . . . . . . . . . . . . . . . . . . . . − 110 Spiral-riveted steel (flow against lap). . . . . . . . . . . . . . . . . . . . − 100 Corrugated steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . − 60 Value of C . . . . . . . . . . . . . . . . . . . . . 150 140 130 120 110 100 90 80 Mulplier to correct tables . . . . . . 0.47 0.54 0.62 0.71 0.84 1.0 1.22 1.50
Velocity ft per sec 0.884 1.77 2.65 3.54 4.42 5.32 6.29 7.08 7.96 8.84
60 2.57
Flow US gal per min 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.5
Extra strong steel
0.364” inside dia Velocity ft per s 1.23 1.85 2.47 3.08 3.71 4.33 4.94 5.55 6.17 7.71
Velocity head ft 0.02 0.05 0.09 0.15 0.21 0.29 0.38 0.48 0.59 0.92
0.302” inside dia Head loss ft per 100 ft 5.22 11.1 18.8 28.5 39.9 53.0 67.9 84.4 103 155
Velocity ft per sec 1.79 2.69 3.59 4.48 5.38 6.27 7.17 8.07 8.96 11.2
Velocity head ft 0.05 0.11 0.20 0.31 0.45 0.61 0.80 1.01 1.25 1.95
Head loss ft per 100 ft 13.0 27.4 46.7 70.6 98.9 132 168 209 254 385
3/8 Inch Standard wt steel
Extra strong steel
0.493” inside dia
0.423” inside dia
Flow US gal per min
Velocity Velocity ft per sec head ft
Head loss Velocity Velocity ft per 100 ft ft per sec head ft
Head loss ft per 100 ft
0.8 1.0 1.5 2.0 2.5 3.0 3.5 4.0 5.0 6.0
1.35 1.68 2.52 3.36 4.21 5.05 5.89 6.73 8.41 10.1
4.30 6.50 13.8 23.4 35.4 49.6 66.0 84.5 134 179
9.07 13.7 29.0 49.4 74.6 105 139 178 269 377
0.03 0.04 0.10 0.18 0.28 .40 .54 .70 1.10 1.58
1.83 2.28 3.43 4.57 5.71 6.85 8.00 9.14 11.4 13.7
0.05 0.08 0.18 0.32 0.51 0.73 0.99 1.30 2.0 2.9
*By permission G. V. Shaw and A. W. Loomis Cameron Hydraulic Data, 11th Edition, Ingresoll-Rand Co., 1942 [53].
Example 15.31 Pressure drop for Vapor System The calculations are presented in Figure 15.47. Line Size Specification Sheet. Figure 15.48 is convenient when using Dowtherm vapor.
15.34 UniSim Design PIPESYS Generally, pipelines transport fluids over a wide range of topography and diverse conditions. Normally, this is computed with a correctly sized pipeline that sufficiently accounts for pressure drop, heat losses and includes the properly specified and sized line facilities, such as compressor, pumps, heaters, or fittings. However, the complexity of pipeline
296 Petroleum Refining Design and Applications Handbook Volume 2 SHEET NO. __________
By
AKC
LINE SIZE SHEET
Job No.
Date
Charge No.
Line No.
Flow Sheet Drawing
LP – 61
No.
Line Description
Vent through Exchanger for Tower T – 3
Fluid in line
N2+ Hydrocarbon
Temperature
GPM (Calc.)
GPM (des.)
Pressure
CFM (Calc.)
2060
CFM (des.)
2270
lb/h.(Calc.)
10,841
lb/h. (des)
12,000
Recommended Velocity
ft/s
Straight pipe, fittings, valves No.
Unit Eq. Ft.
Total Eq. Ft.
St. Line
Tee –S0
5.3
psig
Sp. Gr.
0.975
Sp. Vol.
11.3
Viscosity
0.019
1
11
11
1
6
6
1
50
50
cP
in psi 0.0617
Orifice Motor Valve Miscellaneous Exchanger drop
1.50 Total
Total
cu.ft./lb
Pressure Drop
Pipe & Equivalent
5
Gate Valve.
°F
Item
expansions, contraction, etc Item
140
1.56
72 Estimated line size
10“ (existing)
Actual Velocity
4150 fps
Unit Loss per 100 ft.
0.0857 psi
Total head loss in feet of liquid
37
Cross-sect. area, 10” pipe = 0.547 sq. ft. Velocity = 2270 / 0.547 = 4150 ft/min.
Total pressure drop in psi
Selected pipe size
10”
Material & Weight
Calculations:
Re = 6.31 W/dµ = [ (6.31) (12,000) / (10.02) (0.019) ] = 3.98 × 105 f = 0.0158
∆P/100 feet
=
(0.000336) ( f ) (W)2 d5ρ
Schedule 40, Steel
ρ = 1/ V
=
(0.000336) (0.0158) (12,000)2 (10.02)5 (1/11.3)
= 0.0857 psi/100 equivalent feet of pipe (as pipe, fittings, valves, etc.) ∆P Total (friction) = (0.0857 / 100) (72) = 0.0617 psi
Checked by:
1.56
Date:
Figure 15.47 Line size sheet. Example of pressure drop for a vapor system, Example 15.31.
Fluid Flow 297 80 50 40 30
12"
10"
8"
6"
4"
3"
2½ "
1¼ "
1" SIZ E
5.0 4.0 3.0
PIP E
PRESSURE DROP — lb/in.2/100 ft
10 8.0
1½ "
20 15
2.0 1.5 1.0 0.6
TEMP.—°F 500 550 600 650 700 750
0.5 0.3 0.2
CORRECTION 8.95 5.77 3.72 2.40 1.55 1.00
All sizes are schedule 40 except 1" & 1¼" which are schedule 80 1
2
3
5
8 10
20
30
50 100 200 300 500 800 1000 2000 DOWNTHERM VAPOR FLOW — lb/h × 102
5000
10,000
Figure 15.48 Pressure drop, Dowtherm “A” ® vapor in steel pipe (by permission from Sruthers Wells Corp., Bull. D-45).
network calculations often proves the task to be arduous and difficulty. It is not uncommon that during the design phase, an over-sized pipe is chosen to compensate for inaccuracies in the pressure loss calculations. Additionally, with multi-phase flow, this can result to greater pressure and temperature losses, increased requirements for liquid handling and increased pipe corrosion. Accurate fluid modeling resolves these and other complications and results in a more economic pipeline system. Accomplishing this requires single and multi-phase flow technology that is capable of accurately and efficiently simulating the pipeline flow. PIPESYS is a powerful model pipeline hydraulics that uses the most reliable single and multi-phase flow technology available to simulate pipeline flow. It accesses UniSim Design features such as the component database and fluid properties; includes many inline equipment and facility options relevant to pipeline construction and testing. The extension models pipelines that stretch over varied elevations and environments. The software allows the user to [50]: • Rigorously model single phase and multi-phase flows. • Compute detailed pressure and temperature profiles for pipelines that traverse irregular terrain, both on shore and off. • Perform forward and reverse pressure calculations. • Model the effects of inline equipment such as compressors, pumps, heaters, coolers, regulators and fittings including valves and elbows. • Perform special analyses including: pigging slug prediction, erosion velocity prediction, and severe slugging checks. • Model single pipelines or networks of pipelines in isolation or as part of a UniSim Design process simulation. • Perform sensitivity calculations to determine the dependency of system behavior on any parameter. • Quickly and efficiently perform calculations with the internal calculation optimizer, which significantly increases calculation speed without loss of accuracy.
298 Petroleum Refining Design and Applications Handbook Volume 2 • Determine the possibility of increasing capacity in existing pipelines based on compositional effects, pipeline effects and environmental effects. In computing the PIPESYS extension, a wide variety of correlations and mechanistic models are employed such as horizontal, inclined and vertical flows, flow regimes, liquid holdup, and friction losses. There is also flexibility in performing calculations such as • Calculate the pressure profile using an arbitrarily defined temperature profile, or determine the pressure and temperature profiles simultaneously. • Given the conditions at one end of the pipe, perform pressure profile calculations either with or against the direction of flow to determine either upstream or downstream conditions. • Perform iterative calculations to determine the required upstream pressure and the downstream temperature for a specified downstream pressure and upstream temperature. • Calculate the flow rate corresponding to specified upstream and downstream conditions.
PIPESYS Features The PIPESYS extension is functionally equivalent to a UniSim Design Flowsheet operation and it is installed in a flowsheet and connected to material and energy streams. All PIPESYS extension properties are accessed and changed through a set of property views and amongst these and the starting point for the definition of a PIPESYS operation is the Main PIPESYS view, which is: Main PIPESYS View—Used to define the elevation profile, add pipeline units, specify material and energy streams, and choose calculation methods and check results. The PIPESYS extension includes these pipeline units, each of which is accessible through a property view: • Pipe—The basic pipeline component used to model a straight section of pipe and its physical characteristics • Compressor—Boosts the gas pressure in a pipeline • Pump—Boosts the liquid pressure in a pipeline • Heater—Adds heat to the flowing fluid(s) • Cooler—Removes heat form the flowing fluid(s) • Unit X—A “black box” component that allows you to impose arbitrary changes in pressure and temperature on the flowing fluid(s) • Regulator—Reduces the flowing pressure to an arbitrary value • Fittings—Used to account for the effect of fittings such as tees, valves, and elbows on the flowing system • Pigging Slug Size Check—An approximate procedure for estimating the size of pigging slugs • Severe Slugging Check—A tool for estimating whether or not severe slugging should be expected • Erosion Velocity Check—Checks fluid velocities to estimate whether or not erosion effects are likely to be significant. Case Study: Pressure Drop Through Pipeline Water at 25°C (density 1000 kg/m3) and 2.5 atm pressure is being transferred with a 0.45 kW pump that is 75% efficient at a rate of 8.025 m3/h. All the piping is 4 in. Sch 40 steel pipe and the last section is a 2 in. Sch. 40 steel pipe. Solution PIPESYS simulation software is used to determine the pressure drop and to calculate the pump duty at 75% adiabatic efficiency. Figure 15.49 shows a snap shot of the process flow diagram of PIPESYS (Pipeline-deltaP.usc) and Table 15.38 shows the results of the flow streams. Figures 15.50a–d show profiles of the pressure and temperature
Fluid Flow 299
Temperature Pressure Molar Flow Pipe-100
Temperature Pressure Molar Flow
S1 25.00 202.6 444.1
C kPa kgmole/h Pipe-101
Pump-100
PIPESYS-UniSim
S1
S3 25.02 356.3 443.3
PIPESYS-UniSim
S3
S2 E-100
C kPa kgmole/h
Q-100
Temperature Pressure Molar Flow
S4
Q-101 Temperature Pressure Molar Flow
Pump-100 Speed Energy Actual Vol. Flow Feed Pressure Product Pressure Product Temperature
S2 25.00 C 202.6 kPa 443.3 kgmole/h
rpm
S4 25.02 353.9 443.3
C kPa kgmole/h
1642 kJ/h 8.011 m3/h 202.6 kPa 356.3 kPa 25.02 C
Figure 15.49 Process flow diagram of the Case study using PIPESYS software (courtesy of Honeywell, UniSim ® Design Suite).
Table 15.38 Results of the simulation of the case study. Streams 1
2
3
4
Temperature, °C
25
25
25.02
25.02
Pressure, kPa
202.6
202.6
356.3
353.9
Molar flow, kg mole/h
444.1
443.3
443.3
443.3
Comp. mass frac (H2O)
1
1
1
1
Flowing Pressure : Pipe−100 202.7 202.6 202.6 Pressure (kPa)
202.6 202.6 202.6 202.6 202.6 202.6 202.6 0.000
1.000
2.000
3.000
4.000
5.000
Distance (m)
Figure 15.50a Flowing pressure profile of Pipe-100.
6.000
7.000
8.000
9.000
10.000
300 Petroleum Refining Design and Applications Handbook Volume 2 Flowing Temperatuare: Pipe-100 25.00
Fluid Temperature Ambient Temperature
Temperature (°C)
25.00 25.00 25.00 25.00 25.00 25.00 0.000
1.000 2.000
3.000
4.000
5.000
6.000
7.000
8.000 9.000 10.000
7.000
8.000 9.000 10.000
Distance (m)
Figure 15.50b Flowing temperature profile of Pipe-100. Flowing Pressure: Pipe-101 256.5
Pressure (kPa)
256.0 255.5 255.0 254.5 254.0 253.5 0.000
1.000 2.000
3.000
4.000
5.000
6.000
Distance (m)
Figure 15.50c Flowing pressure profile of Pipe-101.
for Pipe-100 and Pipe-101 respectively. Appendix E shows the flow diagram and results of the Pipeline-deltaP.usc simulation program of the case study.
15.35 Pipe Line Safety Process pipe lines are employed to transfer fluid types (liquids, gases, and solids) from one location to another and are considered as the safest, most economical, and environmental friendly mode of transportation of crude oil and gas and their products. This provides important link of petroleum supply chain management and cutting edge to petroleum/petrochemical industry. However, there have been significant losses due to failure of piping and piping leaks. Major hazards associated in pipeline safety are the results of the following: • • • •
Corrosion—internal/external. Human errors during pigging hot tapping, valve operation. System procedure failure—inspection, operation, start-up/shut down, material specification, and testing. External reason—accidental excavation, earthquake, flood, fire, lightening, rail/road accident.
Fluid Flow 301 Flowing Temperature; Pipe-101
25.02
Fluid Temperature Ambiemt Temperature
25.01
Temperature (°C)
25.01 25.01 25.01 25.01 25.00 25.00 25.00 0.000
1.000 2.000
3.000
4.000
5.000
6.000
7.000
8.000 9.000 10.000
Distance (m)
Figure 15.50d Flowing temperature profile of Pipe-101.
Major causes of pipeline external corrosion are poor/defective coatings, inadequate cathodic protection (CP), and coating defects combination with inadequate CP, interference due to external agencies, stress and bacterial corrosion. Major causes of internal corrosion is the corrosive nature of fluid being transported through pipeline, erosion–corrosion, localized chemical attack/bacterial corrosion. Nigam [71] provides the salient features in the transportation of petroleum: • No leaky tank or container that contains petroleum shall be tendered for transport. Barrels, drums, and other container filled with petroleum shall be loaded with their bung upwards. • No ship, vessel or vehicle shall carry petroleum (class A or B or C) in bulk, if it is carrying passengers or any other combustible cargo other than petroleum. • No person while engaged in loading/unloading or transporting petroleum shall smoke or carry matches, lighters, and any other appliances capable of producing ignition or explosion. • Petroleum shall not be loaded into or unloaded from any ship vessel or vehicle between the hours of sunset or sunrise unless adequate electric lighting is provided at the place of loading or unloading and adequate fire fighting facilities with trained personnel are kept in place for immediate action in the event of a fire.
15.36 Mitigating Pipeline Hazards The following are considered in mitigation hazards in pipeline: • • • • • • • •
Protection against external corrosion Protection against internal corrosion Protection against third party damages Protection of pipeline supports Leak detection system Pig based monitoring system Protection against overpressures Protection against detonation hazards.
15.37 Examples of Safety Design Concerns The following concerns are typically included in the design of piping systems and valves [75].
302 Petroleum Refining Design and Applications Handbook Volume 2
Piping Systems • Has all piping systems handling toxic or lethal materials been identified? (e.g., piping handling hydrogen cyanide, nitrogen, etc). • Does the piping need to be designed to contain a deflagration? A denotation? • Are special monitoring provisions provided for overflow lines which have a tendency to plug? (e.g., lines in caustic services) • Has the proper metallurgy been selected for the fluid transported? Has deleterious materials of construction been avoided? (e.g., has copper or brass been eliminated from ammonia service? Or has copper or iron been eliminated from benzyl chloride service? • Have high temperature shutdowns been provided for pumps which handle heat sensitive or reactive material? • Has the proper bolt design been provided for frangible flange systems to accurately control the break point? • Has a surge vessel been provided to contain thermal expansion of a hazardous liquid (like chlorine) instead of a pressure relief valve? • Has special insulation been used on Therminol or high temperature systems to prevent cracking of high molecular weight organics to a lower flash point material with subsequent auto-ignition? • If a bellows type expansion joint is used in flammable and/or pressure relief systems, has this type joint been correctly aligned during installation to maintain integrity? • If a hazardous condition exists when mechanical agitation is lost, has emergency gas agitation via a dip-pipe been provided? • Do dip pipes have weep holes to de-inventory the pipe during a plant shutdown? • Has a “deadman” start–stop station on a pump been provided to prevent overflow of flammable or very hazardous materials from the downstream vessel due to operator in attention? • Has a remote “stop” been provided on a pump which transports flammable material into an operating unit from the outside the battery limits? • Should uninsulated sections of pipe be added for planned heat loss? (e.g., the feed water regulator on a boiler). • Have the spring hanger settings for piping used in high temperature or high pressure service been documented during installation? • Has the proper gasket type and material been used in hazardous service? (e.g., lethal systems need spiral wound gaskets).
Valves • Have “air or open” control valves been selected for those remote valves which you want to activate closed during a fire event and has plastic air tubing been provided? • Are the valves which must be manually opened or closed during an emergency capable of remote operation? • Have the valves, nipples (open ended), etc., used in pressurized flammable, lethal gas or oxygen service been capped off? • Have the valves and piping, etc., in chloride or oxygen service been degreased before start up (and/or after repair)? • Have excess flow check valves been installed in pressured hazardous gas systems such as those involving ammonia, chlorine, hydrogen, etc.? • Has a hole been drilled in a butterfly valve to prevent overpressure due to thermal expansion? If this is not possible, has a pressure relief valve been provided? • Have “deadman” (spring or close) sampling valves been installed in high pressure, flammable, or lethal systems to prevent continued flow of material if the operator becomes incapacitated? • Has a manually activated water flush or quench system (if possible) been provided to stop an uncontrolled reaction or to provide internal fire fighting capability? • Have air-activated valves been locked out (defused) in the field while maintenance is in progress?
Fluid Flow 303 • Has a valve in a tank car and/or truck unloading line been provided which closes on disconnecting or which must be closed to disconnect? • Has a hazard analysis of the process been conducted to determine the fail safe position of control valves during a specific or total utility outage (electrical power, instrument air, etc.)? • Have special position indicators been provided for three way valves to clearly indicate which port is active?
Piping and Valves Used in ASME Section 1 Service • Have the piping systems been analyzed for stresses and movements due to thermal expansion? • Are the piping systems properly supported and guided? • Have the piping systems been provided with freezing protection, particularly cold water lines, instrument connections, lines in dead end service such as piping in standby pumps? • Have case iron valves and fittings been eliminated from piping which subjected to strain or shock service? • Have non rising stem valves been avoided where possible and has a visual indication of valve position been provided? • Have double block and bleed valves been provided on battery limit piping and/or emergency interconnections to ensure positive isolation and/or to prevent cross-contamination where this is undesirable? • Has a means of draining and trapping condensate from steam piping been provide?
15.38 Safety Incidents Related With Pipeworks and Materials of Construction Table illustrates various incidents involving pipeworks and materials of construction that occurred in refinery and chemical plants; investigations that determined the root causes of these incidents, findings and recommendations of mitigation/preventing similar occurrences in the future. Incidents
U.S. Chemical Safety and Hazard Investigation Board (CSB)
Tosco Refinery in Martinez, California February 23, 1999
Root Causes, Findings and Recommendations
This incident occurred on February 23, 1999 as workers attempted to remove and replace a leaky petroleum pipe, which was attached to an operating oil distillation/ fractionating tower. Over a 13 day period prior to the accident, workers had repeatedly tried to isolate and drain the pipe, but leaking and corroded shut-off valves hampered their efforts. At the same time of the incident, the pipe still contained a significant volume of pressurized naphtha, a highly flammable petroleum mixture similar to gasoline. While workers were in the process of replacing the pipe, the naphtha was released and burst into flame, and caused four fatalities. At the time of the fire, these workers were positioned on scaffolding up to an hundred feet off the ground and had limited means of escape.
Although the refinery procedures directed that the piping should be isolated and drained prior to attempting this kind of repair. This procedure was not followed, as opening a pipe containing naphtha in the presence of multiple ignition sources would result in a disaster. The entire process unit should have been shut down, which would have eliminated ignition source and allowed the naphtha to be fully and safely drained. The CSB concluded that management had a responsibility to ensure that work was halted, and should not have relied solely upon individual workers to stop an unsafe activity. Worker's error was not a root cause of this accident as a satisfactory management system is one that anticipates that humans will inevitably make mistakes, still ensures the safe conduct of work. An effective job planning, hazard review, and management oversight could have prevented the tragedy. The CSB recommended that the refinery implements a comprehensive system for safely managing hazardous maintenance work. Key provisions include a process for evaluating hazards before work is started and increased management oversight of ongoing work. The Board further recommended that the Corporation improves its safety auditing procedures and applies these to its seven refineries. (Continued)
304 Petroleum Refining Design and Applications Handbook Volume 2 Hydrogen Blast in 2009 Silver Eagle Refinery Accident in Woods Cross, Utah A massive explosion and fire at the Silver Eagle Refinery on November 4, 2009 damaged homes was caused by a rupture in a pipe that had become dangerously thin from corrosion was the finding by the CSB Board. The catastrophic rupture occurred in a 10-in. pipe at the bottom of a reactor in the distillated dewaxing unit. This led to a massive release of hydrogen, which caught fire immediately and exploded, sending a blast wave across the refinery into a subdivision. The blast wave damaged over 100 homes.
Root Causes, Findings and Recommendations Metallurgical failure and study analysis was carried out on the pipe segments recovered after the incident. The history of the pipe that ruptured was examined and the component that failed had no record of ever being inspected for corrosion as it thinned over the years. CSB inferred that metallurgical analysis details the same kind of sulfidation corrosion at the Silver Eagle Refinery as was found in the Bay Area Chevron refinery fire of 2012 and the Tesoro refinery explosion and fire that killed seven in Anacortes, Washington in 2010. The sulfur compounds in the process stream corroded a steel piping segment causing the pipe walls to become severely thin. The CSB noted that the examination of the ruptured pipe segment and adjacent piping clearly indicated wall thinning had occurred in the piping component. The elbow adjacent to the pipe segment that failed was noted to have an original thickness of 0.719 in. A 2007 thickness measurement of the elbow indicated a wall thickness of 0.483 in., indicating years of thinning had taken place. Furthermore, the adjacent straightrun segment that failed was found to have a wall thickness as low as 0.039 in. and there were no records of previous inspection.
Initial gas release on security video.
(Continued)
Fluid Flow 305 The CSB’s investigation noted records indicating other serious wide spread mechanical integrity deficiencies and gaps across the refinery at the time of the incident. The goal of mechanical integrity program is to ensure that process equipment is fabricated from the proper materials of construction and is properly installed, monitored and maintained to prevent failures and accidental releases.
Security video showing gas release rapidly expanding.
The hazardous nature of the materials in a refinery and the high temperatures and pressures that are frequently used should ensure a robust mechanical integrity program is essential to safe refinery operations. It is also a regulatory requirement for refineries and chemical plants under the OSHA Process Safety Management standard enacted in 1992.
Security video showing gas release rapidly expanding. R30101 discharge pipe bent around reactor support beam.
Ignition of released gas.
Rapture end of pipe still attached to R30101.
View from hydro pad towards adjacent residential community Ruptured end of pipe wrapped around support beam.
(Continued)
306 Petroleum Refining Design and Applications Handbook Volume 2
Pipe supports
Explosion damage to light structural elements, hydro pad.
Elevation view of surveyed pipe.
Plan view of surveyed pipe.
Images of the “SER 34” ruptured segment after sectioning and removal in March 2010.
Pipe support location. The fractured end SER 34, before and after sectioning in August 2010.
Approximate sectioning location
In-service location
SER 36 valve
Image of the downstream fractured section “SER 35” prior to sectioning for removal from the site in March 2010.
(Continued)
Fluid Flow 307 BP Texas City Refinery, TX Refinery Fire
Root Causes, Findings, and Recommendations
On July 28, 2005, the BP Texas City refinery experienced a major fire in the Residue Hydrotreater Unit (RHU) that caused a reported $30 million in property damage. One employee sustained a minor injury during the emergency unit shutdown and there were no fatalities.
The U.S. Chemical Safety and Hazard Investigation Board has now issued a safety Bulletin to focus attention on process equipment configuration control and positive material verification of critical alloy steel piping components. The CSB recommends that the refining, petrochemical and chemical industries review material verification programs to ensure that maintenance procedures include sufficient controls and positive material identification (PMI) testing to prevent improper material substitutions in hazardous process system.
Residual material from the crude oil processing unit is processed in the RHU to remove nitrogen, sulfur, and metals. Hydrogen is pressurized to about 3000 psi, and then preheated in the RHU heat exchangers (Figure 2) to about 600°F. The preheated hydrogen next passes through a furnace to increase the hydrogen temperature and then is injected into the reactor feedstock. Hydrogen combines with nitrogen compounds and sulfur within the feedstock in the presence of the catalyst inside the RHU reactors to hydrogen sulfide and ammonia. Light hydrocarbon, such as gasoline is then processed in downstream refinery units. On July 25, 2005, at about 6:00 pm, an RHU hydrogen gas heat exchanger process pipe ruptured. The venting hydrogen gas ignited and a huge fireball erupted in the unit. One employee sustained a minor injury while assisting with the RHU emergency shutdown. The RHU sustained major damage from the hydrogen-fed fire that burned for 2 h. There were no offsite impacts, but as a precaution Texas City ordered a shelter-in-place for nearby residents until the fire was contained. This incident occurred after a maintenance contractor accidentally switched a carbon steel elbow with an alloy steel elbow during a scheduled heat exchanger overhaul in February 2005. The alloy steel elbow was resistant to high temperature hydrogen attack (HTHA) but the carbon steel elbow was not. Metallurgical analyses of the failed elbow concluded that HTHA severely weakened the carbon steel elbow. BP personnel examined the extensively damaged unit and determined that an 8-in. diameter pipe elbow on an RHU heat exchanger hydrogen gas outlet pipe ruptured (Figure 3). The BP investigation team recovered the elbow segments that remained attached to the pipe and three pieces found in the debris (Figure 4).
Incidents involving HTHA are as far back as 1940s. Carbon steel in hydrogen service at temperatures above about 450°F and pressures above 100 psia is susceptible to HTHA. At these operating conditions, atomic and molecular hydrogen permeates the steel and reacts with dissolved carbons or carbides to form methane gas (CH4). The loss of carbon in the steel or “decarbonization,” significantly degrades the steel’s mechanical properties, including tensile strength and ductility. The CH4 gas creates high localized stresses, which combine with the normal piping system stresses to create voids and fissures in the steel, which ultimately causes the pipe to rupture (API, 2004). The American Petroleum Institute (API) Recommended Practice, 941, Steels in Hydrogen Service at Elevated Temperatures and Pressures in Petroleum Refineries and Petrochemical Plants, recommends operating limits for carbon steel and low alloy steel piping systems in hydrogen service. Experiments and operating plant data show that HTHA is typically avoided by using low alloy steels containing 1.25–3.0% chrome, as the chrome combines with carbon to form chromium carbide, which is resistant to reacting with hydrogen. Chemical analysis and microscopic examination determined that the elbow was made from carbon steel, and also that the segments were severely decarburized and had deep fissures on the inside surface (Figure 4). The decarburized steel and severe fissuring confirmed that HTHA caused the catastrophic elbow failure.
(Continued)
308 Petroleum Refining Design and Applications Handbook Volume 2
Figure 1. Carbon steel RHU heat exchanger outlet pipe (arrow) ruptured after operating only 3 months in high-temperature hydrogen service.
Figure 2. Dimentionally identical piping elbows on RHU heat exchangers A and B
A Elbow 3
A Elbow 2 Elbow 1
B
Detailed metallurgical examinations and micro-hardness testing quantified the extent of hydrogen damage to estimate the total time the elbow could have been in the hightemperature, high-pressure hydrogen service before it failed. The results compared to existing experimental data and empirical service life predictions, concluded that the elbow failed after being in service for fewer than 3000 h. Recommendations: To revise the maintenance quality control program to require positive material identification testing or another suitable material verification process for all critical service alloy steel piping components removed and reinstalled during maintenance, and inform work crews of special material and handling precautions. To develop/update the written piping component installation quality control procedure to require positive material identification testing or other suitable verification or tracking process for all alloy steel piping components removed during maintenance.
Upper left & top arrow-Allow steel elbows 2 and 3 Low left & top arrow-Carbon steel elbow I
Figure 3. Ruptured 8-inch pipe elbow on heat exchanger A outlet.
BP personal examined the extensively damaged unit and determined that an 8-inch diameter pipe elbow on an RHU heat exchanger hydrogen gas outlet pipe ruptured . . .
Figure 4. Ruptured 8-inch carbon steel pipe elbow pieces recovered after the fire
Figure 5. RHU hydrogen heat exchanger piping material requirements
The BP investigation team recovered the elbow segments that remained attached to the pipe and three pieces found in the debris . . . . 7/8"
Upper left -- Carbon steel elbow segments (view of inside surface) Above -Flange segments Lower left -- Close-up of fissure on middle elbow segement
High-temperature hydrogen to furnace
Preheat gas
Heat exchanger A Elbow 3 (failure location) Elbow 2
Bolted flange (typical)
1-¼ chrome alloy piping Elbow 1 Carbon steel Heat exchanger B
Preheat Gas to Separator Low temperature 3000 psig hydrogen feed
1-¼ chrome alloy pipe Carbon steel pipe
(Continued)
Fluid Flow 309 Valero’s McKee Refinery, Sunray Texas On February 16, 2007, a liquid propane release from cracked control station piping resulted in a massive fire in the propane deasphalting (PDA) unit injuring three employees and a contractor. The fire caused extensive equipment damaged and resulted in the evacuation and total shutdown of the refinery. The refinery remained shutdown for 2 months; the PDA unit was rebuilt and resumed operation nearly one year after the incident. Direct losses attributed to the fire were reported to exceed $50 million. The liquid propane under high pressure was released in the PDA, and the resulting vapor found an ignition source, and the subsequent fire injured workers, damaged unit piping and equipment and collapsed a major pipe rack. The fire grew rapidly and threatened surrounding units, including a Liquefied Petroleum Gas (LPG) storage area. Fire-fighting efforts were hampered by high and shifting winds and the rapid spread of the fire. A refinery-wide evacuation was ordered approximately 15 min after the fire ignited. The fire occurred in the refinery’s propane de-asphalting unit, which uses high-pressure propane as a solvent to separate gas oil from asphalt; gas oil is used as a feedstock in other gasoline-producing refinery processes. The propane leaked from an ice-damaged piping elbow that is believed to have been out of service since the early 1990s. PDA Unit
Root Causes, Findings, and Recommendations The CSB launched an investigation into the cause of the incident. The root cause was the water that leaked through a valve, froze, and cracked an out-of-service section of piping, causing a release of high-pressure liquid propane. The CSB report concluded the root causes of the incident were that the refinery did not have an effective program to identify and freeze-protect piping and equipment that was out of service or infrequently used; that the refinery did not apply the company’s policies on emergency isolation valves to control fires; and that current industry and company standards do not recommend sufficient fireproofing of structural steel against jet fires. CSB investigation reported that unknown to refinery personnel, a metal object had wedged under the gate of a manual valve above the piping elbow, allowing liquid to flow through the valve. Piping above the valve contained liquid propane at high pressure, and small amounts of water were entrained in the propane. The elbow was part of a dead-leg formed when the piping was taken out of service. This section of piping remained connected to the process but was not intended to have any flow of liquid through it. Dead-legs can pose special hazards in refineries that should be carefully managed. Hazards rising from the dead-leg when it was created in the 1990s were not identified and safeguards were not implemented such as removing the piping, isolating it from the process using metal plates known as blinds, or protecting it against freezing temperatures. Over time, water seeped past the leaking valve and built up inside the low point of the piping elbow. A period of cold weather in early February 2007 likely caused the water to freeze, expand and crack the piping. On February 16, the daytime temperature increased and the ice began to melt. At 2:09 p.m. high-pressure liquid propane flowed through the leaking valve and was released through the fracture elbow. Propane escaped from the pipe at an initial rate of 4500 lb/min, which quickly created a huge flammable vapor cloud that drifted toward a boiler house that acted as the ignition source.
Figure 1. PDA unit location in the McKee Refinery
(Continued)
310 Petroleum Refining Design and Applications Handbook Volume 2
High-Pressure C3 Accumulator
C3 = Propane
C3 to Extractor
Water drain
Pitch Feed No. 1 C3 Wash to Extractor Extractor
DAGO 1st Stage Flash Drum
Propane from Asphalt Flash Drums
Propane Makeup
(Not Used) Mix C3
Low-Pressure C3 Accumulater
DAGO Flash Drums
Failed elbow
To Asphalt Heater & Flash
Water drain DAGO to Processing
C3 to Extractor
Figure 2. No. 1 Extractor simplified process flow diagram
No. 1 Extractor
Figure 3. Approximately 90 seconds after ignition (from surveillance video)
Damaged Cooling Tower
Once the fire started, there was no way to shut off the supply of fuel, because the refinery had not implemented Valero procedures requiring the installation of remotely operable shutoff valves. These valves are critical in high-pressure service to prevent large inventories of flammable material inside process equipment from contributing to a fire. The CSB recommended that Valero and industry standards require fireproofing of structural steel supports to a maximum of 50 ft. from possible fuel sources. The collapse of a nonfireproofed pipe bridge 77 ft. away from the source of the jet fire indicated that industry practices require revision. The Board further called on the American Petroleum Institute (API) to develop a new recommended practice for freezeprotection of refinery equipment and to improve existing practices related to fireproofing, emergency isolation valves and water deluge systems. The CSB report called on Valero Energy Corporation to improve freeze protection, fireproofing, hazard analysis and emergency isolation procedures at its 16 refineries. Further, the CSB urged Valero to implement its strategic plan to eliminate the use of chlorine for water treatment in favor of inherently safer alternatives such as bleach. Also that staff work should upgrade hazard analysis procedures.
Heat Affected Butane Sphere
Chlorine Container Shed PDA Extractors
Collapsed Rack
Damaged Naphtha Column
Figure 4. Aerial photograph of damage from the PDA unit fire From Propane Pumps 10” Piping (Pressurized)
8" globe valve (closed)
10" inlet gate valve (closed, leaking) Cracke 10" diameter inlet elbow
To Extractor
10" gate valve (closed) 6" control valve
Figure 5. Propane mix control station schematic (not to scale)
(Continued)
Fluid Flow 311
Figure 6. Crack in the 10" diameter propane mix control station inlet elbow
Damaged Inlet to Extractor No. 1 Figure 8. Damaged 10" propane inlet on Extractor No. 1
Valve Body Foreign Material
Valve Gate
Collapsed Support
Edge of PDA Unit
Fireproofed Supports
Chlorine Shed
Figure 10. Extractor towers (upper right) and collapse pipe rack Figure 7. Downstream view of propane mix control station inlet block valve
Failed Pipe Rack Support (not fireproofed
- Remotely Operated Shutoff Valve (ROSOV) - Pump Column
Intact Pipe Rack Supports (fireproofed) Raw Materials Tanks
Reactor
Products Tanks
90 Feet Between Pipe Bridge Supports
Graphic based on FM Global Property Loss Prevention Data Sheet 7-14, 2004
Figure 12. Insurer-recommended locations for ROSOVs
Figure 9. Pipe bridge support fireproofing
(Continued)
312 Petroleum Refining Design and Applications Handbook Volume 2 Approx. location of Propanr Mix Valve & first release point No. 1 Extractor
Jet fire from failed flange
Non-affected rack supports
Deluge Valve Location
77 t fee
No. 2 Extractor
51
fee t
Heat Damaged Coating
Failed. nonfireproofed, pipe rack support
Figure 11. Distance between the E-W pipe rack supports the extractors
Extractors
Figure 14. Heat-damaged coating on sphere and location of sphere deluge valves
Figure 13. Ruptured one-ton chlorine container
ExxonMobil Baton Rouge Refinery, Louisiana, USA
Root Causes, Findings, and Recommendations
On November 22, 2016, two operators were preparing isobutane equipment for maintenance. As part of this preparation, the operators required to adjust valves to put a spare isobutane plug valve to the spare pump. The refinery used a type of valve known as a quarter-turn plug valve for many applications in the alkylation unit, including the inlet valves to these isobutane pumps.
The CSB learned that there were long – standing reliability issues with gearboxes used to operate plug valves in the refinery’s alkylation unit. Furthermore, when alkylation unit operators encountered a malfunctioning gearbox on a plug valve, it was an accepted practice for the operator to remove the gearbox to open or close the valve with a pipe wrench. Baton Rouge refinery management did not provide the workers performing this operations activity with a written procedure or training on safe gearbox removal from plug valves and its associated hazards.
During the removal of an inoperable gearbox, on a plug valve, the operator performing this activity removed critical bolts securing the pressure-retaining component of the valve known as the top-cap. When the operator then attempted to open the plug valve with a pipe wrench, the valve came apart and released isobutane into the unit, forming a flammable vapor cloud. The isobutane reached an ignition source within 30 s of the release causing a fire and severely burning four workers who were unable to exit the vapor cloud before it ignited.
There are a number of safety management deficiencies that led to the removal of the plug valve gearbox, and the inadvertently disassembly of its pressure-retaining top-cap resulted in an isobutane release and fire.
(Continued)
Fluid Flow 313 The deficiencies included:
Isobutane Vessel Spare Pump Suction Plug Valve
Normal Pump Suction Plug Valve
Spare Pump
Isobutane Recycle Pump
Figure 1. Simplified diagram of the alkylation unit equipment where the release occurred.
Gearbox
Handwheel
Valve Stem
Support Bracket
Equipment Design and Human Factor Analysis
Top-Cap
Figure 2. Illustration showing an alkylation unit plug valve with gearbox, valve stem, top-cap, support bracket, and handwheel similar to the one involved in the incident. Valve That Was Disassembled on Day of Incident
Method Gearbox Removed on Day of Incident
• Failure to identify and address the older model plug valve design and gearbox reliability issues. • Lack of a human factors evaluation to identify the older model plug valves’ design and reliability issues as well as the potential hazards associated with operating and maintaining these valves. • No written procedures detailing the steps needed to remove different models of gearboxes from plug valves to manually open or close the valve safely. • No training workers to safely remove the various plug valve gearbox models in the alkylation unit and the hazard associated with this type of work. • An organizational culture that accepted operators removing malfunctioning plug valve gearboxes despite the lack of detailed procedures and training for safe removal.
Designed Way to Remove Gearbox
Gearbox support brackets (Figure 2) on 15 plug valves located in the refinery’s alkylation unit were attached using the same four vertical bolts that secure the pressure retaining top-cap to the valve body. This gearbox (a 30-plus year old design), however could be removed by taking off the horizontal bolts without disturbing the critical top-cap bolts.
According to API Standard 599, Metal Plug Valves—Flanged, Threaded, and Welding Ends, Valves supplied with the capability of mounting actuators or gear operators shall be capable of doing so without removal of any [pressure-retaining parts (e.g., body bolts, bonnet bolts, flange bolts, packing gland bolts, packing retaining stem nut, etc.). The CSB provides the following key lessons for companies with chemical manufacturing facilities including refineries: Evaluate human factors associated with operational difficulties that exist in your machinery and other equipment, especially when the equipment is part of a process covered by the Process Safety Management (PSM) standard.
Figure 3. Gearbox removal. The left column illustrates how operators removed the plug valve gearbox on the day of the incident. Its shows operators removed the entire support bracket. By design, removing the gearbox did not require removing the four vertical bolts that secured the pressure-retaining top-cap. The right column illustrates how the plug valve gearbox should be disassembled. The gearbox and handwheel could be disconnected from the support bracket by removing the two horizontal bolts on the side of the support bracket.
Apply the hierarchy of controls to mitigate the identified hazards. In the case of Baton Rouge refinery, the personnel should have evaluated the fact that approximately three percent of the plug valves in the alkylation unit used a gearbox attachment design that could result in inadvertent disassembly of pressure-retaining components. Once identified the refinery should have applied the hierarchy of controls to establish a mitigating strategy for susceptible plug valves. (Continued)
314 Petroleum Refining Design and Applications Handbook Volume 2 Establish detailed and accurate procedures for workers performing potentially hazardous work, including job tasks such as removing an inoperable gearbox. In this case, establish procedures specific to removing malfunctioning gearboxes from plug valves.
Figure 4. Welding machine that was the likely ignition source of the isobutane vapor cloud.
Time = 2 seconds
Time = 27 seconds
Ignition
Isobutane Release
Time = 30 seconds
Time = 33 seconds
Time = 39 seconds
Time = 51 seconds
Figure 5. Sequence of events. Baton Rouge refinery security video clips showing the isobutane release and subsequent fire. This sequence does not show time zero − the first indication of the isobutane vapor cloud.
Provide training to ensure workers can perform all anticipated job tasks safely. This training should include a focus on processes and equipment to improve hazard awareness and help prevent chemical incidents. The CSB provides the following guidance to companies with chemical manufacturing facilities including refineries, which use manual gear-operated plug valves. Survey valves to identify gear-operated plug valves with gearbox support brackets attached to pressure-retaining components. Perform a hazard analysis and risk assessment on susceptible valves using the hierarchy of controls to establish a mitigating strategy. Consider upgrading to a newer plug valve design and attach the gearbox support bracket to the valve flanges (see Figure 9) or otherwise address the potential hazard. For susceptible valves identified for future upgrades, develop an interim hazard mitigation plan to control hazards and protect workers. Provide written procedures and training to ensure workers can safely remove the gearbox from the plug valve. Consider expanding this guidance beyond manual gear operators to evaluate other devices such as motors or actuators that may attach to pressure-retaining valve components.
Figure 6. Gearbox support bracket for the valve involved in the incident. The two horizontal bolts (yellow arrows) can be removed to take the gearbox off the support bracket and the valve without disturbing the top-cap bolts. By removing the gearbox in this manner, the pressure-retaining valve components are not disturbed.
Figure 7. The improved plug valve design includes an additional four gearbox support barcket holes separate from the pressure-retaining top-cap (located on the valve body flanges) as seen at the yellow arrows. The photo on the right shows a gearbox support bracket attached to the valve body flange for a valve that was not involved in the incident. The photo on the left shows a different plug valve in the alkylation unit with its gearbox removed.
(Continued)
Fluid Flow 315
Figure 8. Two of several designs of gearbox support brackets (yellow circles) used at the Baton Rouge refinery. These specific brackets attach to the valve body flange – they are not among the approximetly three percent of valves in the alkylation unit that attach the gearbox support bracket using the same four vertical bolts that secure the pressure-retaining top-cap.
Handwheel
Gearbox
Support Bracket Valve Stem
Support Bracket
Top-Cap
Figure 9. The post-1984 plug valve design, showing how one type of gearbox connects to all four dedicated attachment points on the valve flanges that are not pressure-retaining parts.
Newer Gearbox Design Not PressureRetaining Bolts
Gearbox Involved in Incident Pressure Retaining Bolts
Figure 10. Comparison of newer gearbox attachment designs, which are among the approximetly 97 percent of the gearboxes in the alkylation unit, and the gearbox involved in the incident. Importantly, the improved valve design connects the gearbox support bracket to dedicated connection points on the valve flange as shown in the left illustration.
VALVE COMPONENTS
Valve Bonnet (Continued) • Bonnet is connected to the body by a threaded, bolted, or welded joint • In all cases, the attached of the bonnet to the body is considered a pressure boundary • The weld joint or bolts that connect the bonnet to the body are pressure-retaining parts • Valve bonnets, although as necessity for most valves, represent a cause for concern. Bonnets can: – Complicate the manufacture of valves – Increase valve size/weight – Be a source for potential leakage
Figure 11. Slide from ExxonMobil Manual Valve Basics training course. The third bullet point in the slide, which is slide 24 of 153, indicates that the bolts connecting the bonnet to the body are pressureretaining parts.
(Continued)
316 Petroleum Refining Design and Applications Handbook Volume 2 Pump switch
Valve needed to be opened Gearbox Inoperable
Planned maintenance
Chronic reliability issues with plug valve gearboxes in alkylation unit Gearbox assembly removed to manually turn valve stem with a wrench
Past practice Described as a “learned habit”
Chronic gearbox reliability issues
Valve design Gearbox support bracket removed
Four bolts removed from pressure-containing top-cap
Support bracket could appear to be part of gearbox
Human factors No written procedures for gearbox removal. No training for gearbox removal
Plug valve disassembled Organizational culture Flammable Isobutane released
Hazard not identified
Work allowed without procedure or training Serious incidents at Torrance and Baton Rouge refineries Not evaluated in human factors analysis 30 year-old design
Gearbox attached to pressureretaining valve component
Isobutane release and fire critically injured four workers
Human factors
Valve Design
Only three percent of unit values with this gearbox attachment design Valves not upgraded Hierarchy of controls not applied
Isobutane used in alkylation
Process design
Post-turnaround maintenance Multiple ignition sources present
Oxygen present in atmosphere
Welding machine most likely source of ignition
Common practice at Baton Rouge refinery
Welding job permitted
Normal concentration
Figure 12. Simplified causal analysis of the November 22, 2016 incident at the Baton Rouge refinery.
Flixborough Chemical Plant Explosion, 1st June, 1974, U.K.
Health and Safety Executive (HSE) Root Causes, Findings, and Recommendations
On 1st June 1974, a vapor cloud explosion occurred in the reactor section of the caprolactam plant (a chemical used in the manufacture of Nylon 6) at the Flixborough Works (U.K.).
The cause of the disaster was a modification to a 28 in. pipe connection between two reactors. The modification involved the installation of a temporary 20 in. pipe with bellows at each end. The design of the pipe system was defective in that it did not take into account the bending moments on the pipe due to the pressure in it. The bellows were not installed in accordance with the manufacturer’s instructions. The pipework assembly was not adequately supported. The relevant British Standards, notably BS 3351 and 3974 were not followed.
The chemical works, owned by Nypro UK (a joint venture between Dutch State Mines (DSM) and the British National Coal Board (NCB)) had originally produced fertilizer from by-products of the coke ovens of a nearby steelworks. Since 1967, it had instead produced caprolactum, a chemical used in the manufacture of nylon 6. The caprolactam was produced from cyclohexanone. This was originally produced by hydrogenation of phenol, but in 1972 additional capacity was added, built to a DSM design in which hot liquid cyclohexane was partially oxidized by compressed air. The plant was intended to produce 70,000 tpa (tons per annum) of caprolactam but was reaching a rate of only 47,000 tpa in early 1974. Government controls on the price of caprolactam put further financial pressure on the plant. In the DSM process, cyclohexane was heated to about 155°C (311°F) before passing into a series of six reactors. The reactors were constructed from mild steel with a stainless steel lining; when operating they held in total about 145 tons of flammable liquid at a working pressure of 8.6 bar gauge (0.86 MPa gauge; 125 psig). In each of the reactors, compressed air was passed through the cyclohexane, causing a small percentage of the cyclohexane to oxidize and produce cyclohexanone, some cyclohexanol also being produced. Each reactor was slightly (approximately 14 in., 350 mm) lower than the previous one, so that the reaction mixture flowed from one to the next by gravity through nominal 28-in. bore (DN 700 mm) stub pipes with inset bellows. The inlet to each reactor was baffled so that liquid entered the reactors at a low level; the exiting liquid flowed
The disaster involved (and may well have been caused by) a hasty modification. There was no on-site senior manager with mechanical engineering expertise (virtually all the plant management had chemical engineering qualifications); mechanical engineering issues with the modification were overlooked by the managers who approved it, nor was the severity of the potential consequences of its failure appreciated. Failings in technical measures: • A plant modification occurred without a full assessment of the potential consequences. Only limited calculations were undertaken on the integrity of the bypass line. No calculations were undertaken for the dog-legged shaped line or for the bellows. No drawing of the proposed modification was produced. • Plant Modification /Change Procedures. Hazop. • Design Codes—Pipework. Use of flexible pipes. • No pressure testing was carried out on the installed pipework modification. • Maintenance Procedures. Recommissioning. • Those concerned with the design, construction and layout of the plant did not consider the potential for a major disaster happening instantaneously.
(Continued)
Fluid Flow 317 over a weir whose crest was somewhat higher than the top of the outlet pipe. The mixture exiting reactor 6 was processed to remove reaction products, and the unreacted cyclohexane (only about 6% was reacted in each pass) then returned to the start of the reactor loop. Although the operating pressure was maintained by an automatically controlled bleed valve once the plant had reached steady state, the valve could not be used during start-up, when there was no air feed, the plant being pressurized with nitrogen. During start-up the bleed valve was normally isolated and there was no route for excess pressure to escape; pressure was kept within acceptable limits (slightly wider that those achieved under automatic control) by operator intervention (manual operation of vent valves). A pressure-relief valve acting at 11 kg/cm2 (156 psi) gauge was also fitted. At about 16:53 on Saturday 1 June 1974, there was a massive release of hot cyclohexane in the area of the missing reactor 5, followed shortly by ignition of the resulting cloud of flammable vapor and a massive explosion in the plant. It virtually demolished the site. Since the accident took place at a weekend there were relatively few people on site: of those on-site at the time, 28 were killed and 36 injured. Fires continued on-site for more than 10 days. Off-site there were no fatalities, but 50 injuries were reported and about 2000 properties damaged. The occupants of the works laboratory had seen the release and evacuated the building before the release ignited; most survived. None of the 18 occupants of the plant control room survived, nor did any records of plant readings. The explosion appeared to have been in the general area of the reactors and after the accident only two possible sites for leaks before the explosion were identified: “the 20 in. bypass assembly with the bellows at both ends torn asunder was found jack-knifed on the plinth beneath” and there was a 50-in. long split in nearby 8-in. nominal bore stainless steel pipework. It was a failure of this plant that led to the disaster. A major leak of liquid from the reactor circuit caused the rapid formation of a large cloud of flammable hydrocarbon. When this met an ignition source (probably a furnace at a nearby hydrogen production plant) there was a massive fuel–air explosion.. The plant control room collapsed, killing all 18 occupants. Nine other site workers were killed, and a delivery driver died of a heart attack in his cab. Fires started on-site which were still burning 10 days later. Around 1000 buildings within a mile radius of the site (in Flixborough itself and in the neighboring villages) were damaged.
• Plant Layout: positioning of occupied buildings. • Control Room Design: structural design to withstand major hazards events. • The incident happened during start up when critical decisions were made under operational stress. In particular the shortage of nitrogen for inerting would tend to inhibit the venting of offgas as a method of pressure control/reduction The disaster was caused by “a well designed and constructed plant” undergoing a modification that destroyed its technical integrity. • Modifications should be designed, constructed, tested and maintained to the same standards as the original plant. When the bypass was installed, there was no works engineer in post and company senior personnel (all chemical engineers) were incapable of recognizing the existence of a simple engineering problem, let alone solving it. • When an important post is vacant special care should be taken when decisions have to be taken which would normally be taken by or on the advice of the holder of the vacant post. • All engineers should learn at least the elements of other branches of engineering than their own. HYDROGEN PLANT
LABORATORY
SECTION 25 A
MAIN OFFICE BLOCK
CONTROL ROOM
PLATE 1
Figure 1. Aerial photo of the plant.
Figure 2. Explosion and fire due to vapor cloud with an ignition source. (Continued)
318 Petroleum Refining Design and Applications Handbook Volume 2
Figure 3. Aerial photo of the plant site after explosion.
28” DIAMETER STUB PIPE
28” DIAMETER STUB PIPE
JACK KNIFE BY-PASS PIPE
PLATE
Figure 4. Damage to pipework, reactors, etc. Inside the plant, 28 people were killed and another 36 were injured. Injuries and damage were widespread outside the Works.
Trevor Kletz saw the plant as symptomatic of a general failure to consider safety early enough in process plant design, so that designs were inherently safe—instead processes and plant were selected on other grounds than safety systems bolted on to a design with avoidable hazards and unnecessarily high inventory. Health and Safety at Works Acts, 1974 (HASAWA) already required companies to have a safety policy, and a comprehensive plan to implement it. Advisory Committee on Major Hazards (ACMH) felt that for major hazard installations the plan should be formal and include: • the regulation by company procedures of safety matters (such as; identification of hazards, control of maintenance (through clearance certificates, permits to work etc.), control of modifications which might affect plant integrity, emergency operating procedures, access control). • clear safety roles (e.g., the design and development team, production management, safety officers). (Continued)
Fluid Flow 319 • training for safety, measures to foster awareness of safety, and feedback of information on safety matters. • Safety documents were needed both for design and operation. The management of major hazard installations must show that it possessed and used a selection of appropriate hazard recognition techniques, had a proper system for audit of critical safety features, and used independent assessment where appropriate.
15.39 Lessons Learned From Piping Designs There are case histories relating to system design that emphasize that accidents occur rapidly with inadequate time to manually return the system to control once the accident is in progress; the system designs required for preventing accidents or mitigating the consequences of accidents are precise that require only minor process changes and the time and effort required to develop a safe system design are justified. An engineer can be hired for a fraction of the cost most accidents. Howard [72] and Kletz [73] have emphasized the design features for safer plants and Crowl and Louvar [74] provide the following recommendations: • • • • • • • • • • • • • • • • • • • • • • •
Use the appropriate materials of construction, especially when using old systems for new applications. Do not install pipes underground. Be sure that the quality of construction (e.g., welds) meets the required specifications. Check all purchased instruments and equipment for integrity and functionality. Do not secure pipes too rigidly. Pipes must be free to expand so that they will not damage other parts of the system. Do not install liquid-filled flanges above electrical cables. A flange leak will douse the cables with liquid. Provide adequate supports for equipment and pipes. Do not allow spring supports to be completely compressed. Design doors and lids so that they cannot be opened under pressure. Add interlocks to decrease pressure before the doors can be opened. Also, add visible pressure gauges at the doors. Do not let pipes touch the ground. Remove all temporary supports after construction is completed. Remove all temporary startup or checkout branches, nipples, and plugs and replace them with properly designed welded plugs. Do not use screwed joints and fittings when handling hazardous chemicals. Be sure that all tracing is covered. Check to ensure that all equipment is assembled correctly. Do not install pipes in pits, trenches, or depressions where water can accumulate. Do not install relief tailpipes too close to the ground where ice blockage may make them inoperable. Be sure that all lines that can catch water can be appropriately drained. When welding reinforcement pads to pipes or vessels, ensure that trapped air can escape through a vent during heating. Do not install traps in lines where water can collect and develop a corrosion problem. Install bellows carefully and according to manufacturers’ specifications. Bellows should be used cautiously. If required, inspect frequently and replace when necessary before they fail. Make static and dynamic analyses of pipe systems to avoid excessive stresses and excessive vibrations. Design systems for easy operation and easy maintenance, e.g., install manual valves within easy reach of the operators, and design pipe networks for easy maintenance or with easy access to equipment requiring maintenance. Install bug screens on vent lines.
320 Petroleum Refining Design and Applications Handbook Volume 2 • Make structural analyses or relief system to avoid structural damage during emergency reliefs. • Safety technology must not work right the first time. Usually, there is no opportunity to adjust or improve its operation. • Critical safety instruments must have backups. • Provide hand-operated or automatic block valves, equivalent valves for emergency shutdowns. • Use electronic or mechanical level gauges, not glass sight glasses. • Add fail-safe block valves with a positive indication to the valve position (limit switches). Organizations should ensure that a good safety program is developed with personnel who can readily identify and eliminate safety problems. Better safety program can be developed by implementing management systems to prevent the existence of safety problems. The systems commonly employed in industry include safety reviews, operating procedures and maintenance procedures. The causes of major accidents can ultimately be attributed to a lack of management systems. Case histories have exemplified and recognized that the existence of procedures is inadequate. There must also be a system of checks in place to ensure that the procedures are used effectively and strictly followed. Process hazard analysis (PHA) and management systems are discussed later in Volume 4 of these series.
15.40 Design of Safer Piping Inherent safer piping design relies on abiding by several regulations and standard codes, such as the American Society of Mechanical Engineers (ASME) and American Petroleum Institute (API), British Standards (BS) and other international relevant standards and codes. A review of these codes and standards is presented as follows [81].
15.40.1 Best Practices for Process Piping ASME B31.3 Process piping provides best practices for compressor station air, hydraulic power and lube oil piping. This is applicable to piping that is typically found in petroleum refineries, chemical process industries, and related facilities. For these applications, the required thickness of straight steel piping is determined by [76]
tm = t + c
(15.380)
where tm = minimum required wall thickness, including mechanical, corrosion, and erosion allowances. t = nominal wall thickness based on the internal design pressure c = sum of the mechanical allowance (thread or groove depth) plus corrosion and erosion allowances.
t=
Pg D 2(S s Q f W + PgY )
(15.381)
where Pg = Internal design gage pressure Ss = Stress value for the material Qf = quality factor W = weld joint strength reduction factor Y = temperature coefficient and is calculated by:
t=
d + 2c D + d + 2c
(15.382)
Fluid Flow 321 where d = inside diameter of the pipe. D = nominal outside diameter of the pipe. Eqs. 15.381 and 15.382 apply when t < D/6, and is therefore checked after determining the thickness. For t ≥ D/6 or Ps/Ss Qf > 0.385, calculation of wall thickness for straight pipe requires special consideration of factors, namely: effects of fatigue, thermal stress, and theory of failure and so on. ASME B31.3 “Process Piping” and all gas compressing station, water and steam piping should be constructed in accordance with ASME B31.1 “Power Piping” [77]: • The discharge stacks, vents or outlet ports of all pressure relief devices must be located where gas can be discharged into the atmosphere without undue hazard. • Each pressure relief station, pressure-limiting station, or group of such stations installed to protect a piping system or pressure should have sufficient capacity and be set to operate to prevent the pressure from exceeding the level specified [78]. • Any safety device that consists of a series of regulators to control or limit the pressure in a piping system should be inspected to determine that the maximum allowable operating pressure of the system will not be exceeded should any one of the associated regulators malfunction or remain in the fully open position. • If the pressure < 10 psig, the steel service pipe should be designed for at least 100 psig pressure. • All pressure-relieving devices in compressor stations should be inspected and/or tested and all devices except rupture disks should be operated periodically to determine that they open at the correct set pressure. Any defective or inadequate equipment found must be promptly repaired or replaced. • All pressure-limiting stations, relief devices and other pressure regulating stations and equipment should be inspected and/or tested periodically. • Pressure relief or other suitable protective devices of sufficient capacity and sensitivity must be installed and maintained to ensure that the maximum allowable operating pressure of the station piping and equipment is not exceeded by more than 10%. • A pressure relief valve or pressure-limiting device, such as a pressure switch or unloading device, should be installed in the discharge line of each positive-displacement transmission compressor between the gas compressor and the first discharge block valve. • If the pressure relief valve is the primary overprotection device, the relieving capacity should be equal to or greater than the capacity of the compressor. • Vent lines provided to exhaust the gas from the pressure relief valves to atmosphere should be extended to a location where the gas may be discharged without due hazard. • Vent lines must be sufficient capacity so that they will not inhibit the performance of the relief valve. • At least once each calendar year, a review should be conducted to ensure that the combined capacity of the relief devices on a piping system or facility is adequate to limit the pressure at all times to the required values [78]. • An inspection and/or test of stop valves should be performed to determine that the valves will operate and are correctly positioned. ASME B31.3 provides a table for determining the minimum thickness of external threaded components. Chapter 22 in volume 4 reviews process safety and pressure relieving devices.
15.40.2 Designing Liquid Piping Designing safer hazardous liquid piping requires that the piping complies with 49 CFR part 195, “Transportation of Hazardous Liquids of Pipeline” [79]. If the piping has a maximum operating pressure (MOP) > 20% of the specified
322 Petroleum Refining Design and Applications Handbook Volume 2 maximum yield strength, or if the piping is carrying petroleum in a non rural area, it must comply with this standard. The internal design pressure for steel piping is calculated by [79].
P = 2S
t Fs E s D
(15.383)
where P = internal design pressure S = yield strength t = nominal wall thickness of the pipe D = nominal outside diameter of the pipe Fs = design factor for steel equal to 0.72 Es = seam joint factor. When designing for liquid piping, attention is given to these overlooked requirements [81]: • The minimum wall thickness of the pipe may not be 150
250+
15,000
300
3000
5000
Mixed flow (V)
100,000
75
Axial flow (V)
100,000
25
Centrifugal (V)
400±
5750
Single stage (V) Double suction Single stage (H) Multistage (H) Single and multistage (V)
*Not necessary at same point. (H), horizontal; (V), vertical.
Pumps 333 STATIONARY FACE ADAPTER
SEAT GASKET GLAND ROTATING FACE ROTATING FACE GASKET COIL SPRING
SHAFT SLEEVE DOUBLE MECHANICAL SEAL
IN
OUT
IMPELLER WASHER
CASING
STUFFING BOX COVER
IMPELLER
IMPELLER SCREW
Figure 16.1 Cross-sectional view of a vertical inline pump (by permission from Knoll, H. and S. Tinney, Hydroc. Proc., May 1971, p. 131 and Goulds Pump, Inc. Mechanical seal and seal venting details courtesy Borg-Warner).
The centrifugal pump (Table 16.2) develops its pressure by centrifugal force on the liquid passing through the pump and is generally applicable to high capacity, low to medium head installations. In order to satisfy pump discharge head (or pressure) requirements the unit may be a multistage (multiple impellers) instead of a single stage [1]. The conditions of pumping water vs. pumping hot light hydrocarbons require considerably different evaluation in pump design features for satisfactory operation, safety, and maintenance. The inline centrifugal process pump, Figure 16.1, is relatively new to general applications; however, it is finding many applications where space and installation costs are important. Each application must be carefully evaluated, as there are three basic types of pump construction to consider. Generally, for many applications the dimensions have been standardized through the American Voluntary Standard, American National Standard Institute (ANSI), or American Petroleum Institute (API)-610. The performance curves are typical of single stage centrifugal pumps. The turbine is a special type of centrifugal pump (Figure 16.2) and has limited special purpose applications. Generally, pumps can be used to move fluids that flow from regions of high pressure to regions of low pressure, by increasing the pressure of the fluid. A centrifugal pump increases the absolute pressure of a fluid by adding velocity 1 energy to the fluid mv 2 and then converting that to pressure or head energy (mgh) in the volute as illustrated 2 in Figure 16.3. The fluid is drawn into the impeller eye (Point 1) at a velocity v1, which is approximately equal to the volumetric flow rate divided by the cross-sectional area of the impeller eye. The rotation of the impeller increases the velocity and pressure of the fluid (Point 2). When the fluid reaches Point 3, it is slowed down by the increasing area of the volute, and the velocity is converted into the pressure head. Euler’s pump equation showing the relationship between the head to the change in velocity of the fluid through the impeller shroud is:
∆h = ηHY
∆(Uv u ) g
(16.1)
334 Petroleum Refining Design and Applications Handbook Volume 2
40 psi
10 psi
30 psi
Cross-section of Heads and Impellers Turbine – Pump Principle 20 psi
Figure 16.2 Turbine pump (courtesy of Roth Pump Co.).
Discharge
3
Impeller Eye 2
1
Suction
Volute Impeller
Figure 16.3 Centrifugal pump increases process head by adding energy to a fluid. Note: a rotating impeller imparts energy to the fluid moving through the pump.
where ηHY = hydraulic efficiency (excluding mechanical loss) U = centrifugal velocity component of the fluid (equal to the angular velocity (Ω) multiplied by impeller radius (r) vu = circumferential velocity component of the absolute velocity vector v g = acceleration due to gravity
Pumps 335 The pressure head (h) is related to the increase in height of a column of fluid that the pump would deliver if the velocity head were converted, without loss into the elevation head. The actual change in pressure resulting from this head can be determined by:
ΔP = hρg
(16.2)
where ΔP = change in pressure ρ = fluid density. Low density hydrocarbon fluids (e.g., ethylene (C2H4), propylene (C3H6) and diesel fuel) require more head to produce the same differential pressure as a higher-density fluid (e.g., water). Therefore, pumps for low-density fluids must incorporate more stages and/or larger impellers to achieve the same results as a pump producing the same differential pressure with a higher-density fluid. Table 16.3 and Figure 16.4 illustrate how fluid density affects the head required to produce the same change in pressure. The differential pressure required to produce flow is created by adding energy to the fluid through the spinning impeller. The amount of pressure increase is directly related to the density of the fluid (ρ) and the product of the impeller radius and the shaft rotation speed squared (rΩ)2. The Table 16.3 Effects of fluid density on head. Property
Ethylene (C2H4)
Propylene (C3H8)
Diesel fuel
Water
ρ (kg/m3)
440
614
850
1000
∆P (kPa)
2000
2000
2000
2000
h (m)
463
332
240
204
Note: h (m) =
10.2•∆P( bar ) Sp.Gr
800
700
Head, m
600
500
Ethylene
400
Propylene
300
Diesel fuel
200
100
0 0
500
1,000
1,500
2,000
2,500
Pressure, kPa
Figure 16.4 Head required to produce similar pressures is higher for lower density fluids.
3,000
336 Petroleum Refining Design and Applications Handbook Volume 2 radius of the impeller determines the pump’s size as well as the initial cost. At slow speeds, fewer problems and less net positive suction head (NPSH) often occur; however, a slow-speed pump requires a larger impeller and is more expensive than a higher-speed pump producing the same head.
16.2 Pump Design Standardization Certain pump designs have been standardized to aid manufacturer’s problems, and to allow the owners to take advantage of standardization of parts and dimensions, and consequently maintain a more useful inventory. The standards are sponsored through ANSI; however, many manufacturers also produce to the API standards and their own proprietary standards. These are special pumps that do not conform to all the standards, but are designed to accomplish specific pumping services. The primary pump types for the petroleum and chemical industries for horizontal and vertical inline applications have been standardized in ANSI B-123, ANSI Std # B73.1M for horizontal end suction centrifugal pumps, and ANSI B73.2M for vertical inline centrifugal pumps. The standards are in a continuous process of upgrading to suit requirements of industry and the manufacturers. The API-610 standard is primarily a heavy-duty application, such as is used for the refinery and chemical industry requirements. This is the only true world pump [2] standard, although the International Organization for Standardization (ISO) is studying such an improved design [3]. The standards are important because they allow the dimensional interchangeability of pumps and shaft packing of different manufacturers, but only as long as the manufacturers conform to the standard.
16.3 Basic Parts of a Centrifugal Pump Table 16.4 is a quick reference as to the function of the basic parts. Table 16.4 Basic parts of a centrifugal pump. Part
Purpose
Impeller
Imparts velocity to the liquid, resulting from centrifugal force as the impeller is rotated.
Casing
Gives direction to the flow from the impeller and converts this velocity energy into pressure energy which is usually measured in feet (meter) of head.
Shaft
Transmits power from the driver to the impeller.
Stuffing box
This is a means of throttling the leakage which would otherwise occur at the point of entry of the shaft into the casing; usually not a separate part, but rather made up of a group of small details, as “A” to “D”.
(A) Packing
This is the most common means of throttling the leakage between the inside and outside of the casing.
(B) Gland
Used to position and adjust the packing pressure.
(C) Seal gauge (also called water-seal or lantern ring)
Provides passage to distribute the sealing medium uniformly around the portion of the shaft that passes through the stuffing box. This is very essential when suction lift conditions prevail to seal against in-leakage of air.
(D) Mechanical seal
Provides a mechanical sealing arrangement that takes the place of the packing. Basically it has one surface rotating with the shaft and one stationary face. The minutely close clearance between these two faces prevents leakage of liquid out or air in.
Shaft sleeve
Protects the shaft where it passes through the staffing box. Usually used in pumps with packing but often eliminated if mechanical seals are employed. (Continued)
Pumps 337 Table 16.4 Basic parts of a centrifugal pump. (Continued) Part
Purpose
Wearing rings
Keeps internal recirculation down to a minimum. Having these rings as replaceable wearing surfaces permits renewal of clearances to keep pump efficiencies high. On small types only one ring is used in the casing and on larger sizes, composition rings are used in the casing and on the impeller.
Wearing plates
With open type impellers or end clearance wearing fits, these perform the same purpose as wearing rings do with radial clearances.
Bearings
Accurately locate shaft and carry radial and thrust loads.
Frame
Used to mount unit rigidly and support bearings. In most single suction pumps this is a separate piece. In many double suction pumps, the support is through feet cast as part of the casing. In some special suction pumps, the feet are also part of the casing and the bearing assembly is overhung. With close coupled single suction types, this support is provided by the motor or by special supporting adapters.
Coupling
Connects the pump to the driver.
Impellers The three common types of impellers that impart the main energy to the liquid for process applications are (see Figure 16.5) as follows: 1. F ully enclosed—used for high head, high pressure applications. 2. Semi-enclosed—used for general purpose applications, has open vane tips at entrance to break up suspended particles and prevent clogging. 3. Open—used for low heads, suspended solids applications, very small flows. Small radial vanes are usually provided on back shroud or plate of impeller to reduce the pressure on the stuffing box, and to prevent suspended solids from entering the backside and possibly causing clogging. The working or pumping vanes are backward in form relative to the impeller rotation. These impellers are available in nearly any material of construction as well as rubber, rubber-lined, glass-lined, and plastic. The lined impellers are of the open or semi-open type.
Casing The casing maybe constructed of a wide variety of metals, as well as may be lined to correspond to the material of the impeller. Operating pressures go to about 5000 psi (345 bar) for the forged or cast steel barrel-type designs. However, the usual process application is in the 75–1000 (5 - 69 bar) psi range, the latter being in light hydrocarbon and similar high vapor pressure systems. The removal of the casing parts is necessary for access to the impeller and often to the packing or seals. Some designs are conveniently arranged to allow dismantling the casing without removing the piping connections. There are proposed construction standards being considered which will allow easy maintenance of many of the types now being offered in a non-standard fashion.
Shaft Care should be given in selecting the shaft material. It must be resistant to the corrosive action of the process fluids, yet possess good strength characteristics for design. For some designs it is preferable to use a shaft sleeve of the proper corrosion-resistant material over the preferred structural shaft material. These sleeves may be metal, ceramic, rubber, and so on, as illustrated in Figure 16.6.
338 Petroleum Refining Design and Applications Handbook Volume 2
Enclosed single-suction impeller with sealing on suction and back sides (courtesy The Deming Co.).
Mixed flow semi-enclosed impeller (courtesy The Deming Co.).
Front
Enclosed double-suction impeller with sealing rings on both sides (courtesy The Deming Co.).
Semi-open or semi-enclosed impeller (courtesy Goulds Pumps Inc.).
Back
Figure 16.5 Impeller types. Open impeller for corrosive or abrasive slurries and solids (courtesy of Goulds Pumps, Inc.).
Bearings The bearings must be adequate to handle the shaft loads without excessive wear, provided lubrication is maintained. Usually this is not a point of real question provided the manufacturer has had experience in the type of loads imposed by the service conditions, and the responsibility for adequate design must be his/hers. In all cases, the bearings should be of the outboard type, that is, not in the process fluid, unless special conditions prevail to make this situation acceptable.
Packing and Seals on Rotating Shaft Conventional soft or metallic packing in a stuffing box (Figure 16.7) is satisfactory for many low pressures, noncorrosive fluid systems. Special packings such as Teflon, or mechanical seals are commonly used for corrosive fluids, since there can be leakage through the packing along the rotating shaft. However, for these conditions a mechanical seal is preferred. When the pressure becomes high (above about 50 psig (3.4 barg)) or the fluid is corrosive, additional means of sealing the shaft must be provided. Particular care must be taken in handling and using the mechanical seals, and these special instructions should be obtained from the seal manufacturer [4]. Generally, it is not wise to
Pumps 339 Turn Pump Over by Hand before Starting Motor to see that It Turns Freely. 3
44 74
1
No. 1. 3. 4. 7. 10. 12. 13. 44. 55. 56. 59. 60. 61. 62. 74. 79. 82. 88. 98. 106. 144. 164. 165. 167.
X&Y
165
Lubricate Stuffing Box By Circulating Water or Clear Solution thru Connections “X” & “Y”.
167
60
164
79
10
59 60
Tang of Key Must be Located Here
98
144
74 10
12 56
88
GAP
13 106
82 7
Impeller
3
Lining
62
55
Thrower Ring Must Not Cover Gap
61 4
Part Name Pump Casing Impeller Pump Frame Split Gland Shaft Lantern Ring Packing Nipple (2) Thrower Ring Rubber Ring Resilient Sleeve Shaft Sleeve Rubber Ring Retaining Ring Rubber Ring Capscrew (6) Gland Yoke Key Shaft Sleeve Ext’n. Tie Rod Key Washer (2) Retainer (2) Rubber Ring (2)
Figure 16.6 Stuffing box details lined pump with porcelain or Teflon® shaft sleeve (courtesy of Dorr-Oliver, Inc.).
Stuffing box packing Throat bushing
Lantern gland Longitudinal section with Lantern Gland
Stuffing box gland
Figure 16.7 Packed stuffing box (courtesy of Dean Brothers Pumps, Inc.).
have the mechanical seal installed at the pump factory, as the slightest amount of grit on the faces can cause permanent damage or destruction on only one or two revolutions at pump speed. The seals should be inspected and cleaned immediately prior to initial start-up. A mechanical seal system (Figures 16.8a [5] and b [6]) contains a rotating element attached to the rotating shaft by set screws (or a clamp) that turns against a stationary unit set in the gland housing. The necessary continuous contact between the seal faces (see Figure 16.8a) is maintained by hydraulic pressure in the pump from the fluid being pumped and by the mechanical loading with springs or bellows. To seal the mechanical seal elements to the rotating shaft to prevent leakage along the shaft, two basic types of seals are used: (a) pusher type using springs and seal “O” rings, wedge rings, and so on; and (b) non-pusher type using some form of bellows of elastomer or metal [7] (Table 16.5). The matching contact rubbing faces are made of dissimilar materials, precision finished to a mirror-like flat surface. There is little friction between these, and hence, they form a seal that is practically fluid tight. The rubbing materials may be some combination of low friction carbon, ceramics (aluminum oxide, silicon carbide), and/or tungsten.
340 Petroleum Refining Design and Applications Handbook Volume 2 Gland gasket
Flush connection
Driven pin
Gland ring
Stuffing-box housing
Pumped liquid
Shaft Primary seal elements
Mechanical-seal hardware
Secondary seals (a)
Figure 16.8a Basic components of all mechanical seals (by permission from Adams, W. H. Chem. Eng., Feb. 7, 1983, p. 48). C L
C L Stuffing Box
Stuffing Box
(1) Impeller
Impeller
(1)
(2)
(3) C L
(3)
(2)
Shaft
Internal Seal
External Seal (b)
Figure 16.8b The three sealing points in mechanical seals (by permission from Sniffen, T.G., Power and Fluids, Winter 1958, Worthington Corp.).
The choice of materials will depend on the service, as will the selection of the materials of construction for the other components, such as springs, “O” rings, other seal rings, and even the housing. The designer should consult the seal manufacturers for details of application not possible to include here. The “single” mechanical seal is made of a rotating element fixed to the shaft (or shaft sleeve), and a stationary element fixed to the pump casing [6]. The “double” seal is for severe sealing problems where out-leakage to the environment cannot be tolerated and must be controlled (Figures 16.9a and b). Depending upon the fluid’s characteristics, the vent between the double seals (Figures 16.9c and d) may be purged with process liquid, or a different liquid or oil, or it may be connected to a seal pot and vent collection to prevent leakage to the air/environment. There are techniques for testing for leakage of the inner seal by measuring the vent space pressure through the seal liquid surge port. This should be essentially atmospheric (depending on the vent system backpressure). This allows detection before the leakage breaks through the outer seal.
Pumps 341 Table 16.5 Requirements for mechanical seal installations. Feature
Description
Remarks
Cooling
Water jacketed stuffing box
Liquid must dead end in stuffing box
Cooling
Gland plate
Efficient to cool contact faces
Lubrication
Dead end
Good under vacuum, mild abrasives metal–metal, dry seals
Lubrication
Circulating
Good cooling of contact faces
Flushing
Inside seal
Good for volatile liquids, solutions tending to crystallize, steam
Flushing
Outside seal
Heating to prevent solidification
Quenching
Outside seals (only)
For oxidizing and corrosive liquids, seal liquid washes process fluid, for high temp.
Vent and drain
Inside seal
Safety feature, for venting to flare, draining
Flushing
Double
Requires circulation system
Flushing
Tandem
Requires circulation system
Two rotary
Double
For improved sealing
Four rotary
Double tandem
For improved sealing problem
Figure 16.10 illustrates a seal installed in a conventional stuffing box with cooling liquid flow path. Figures 16.8a, 16.8b, 16.8, to 16.12 identify the fundamentals of mechanical seals, even though there are many specific designs and details. These various designs are attempts to correct operational problems or seal weaknesses when used under various conditions in the wide variety of process fluids. The average unbalanced external seal is good for pressures of about 30 psig (2.1 barg), while the balanced design will handle 150 psig (10.3 barg). Special designs will handle much higher pressures. Actually the maximum operating process pressures are a function of the shaft speed and diameter for a given seal design fluid and fluid temperature. Figure 16.13 is an outside balanced seal designed for vacuum to 150 psig (10.3 barg) and −40°F to +400°F (−40 to 204°C) (see Table 16.5). The process fluid must be free of solids (as for practically all mechanical seals) and must not attack the material of the O-ring shaft packing. Many other designs are available, and the manufacturers should be consulted for advice on specific sealing problems.
16.4 Centrifugal Pump Selection The centrifugal pump is a versatile unit in the process plant, since its ease of control, non-pulsing flow; pressure- limiting operation fits many small and large flow systems [37].
Barrier Fluid In
Product Side
Atmosphere Side
Barrier Fluid In Product Side
Atmosphere Side A
(a)
B
A1
B2
(b)
Figure 16.9 (a) Double mechanical seal, two rotary elements against common stationary (by permission from Fischer, E. E., Chem. Process, Oct. 1983 [7]). (b) Tandem double seal (by permission from Fischer, E. E., Chem. Process, Oct. 1983 [24]).
342 Petroleum Refining Design and Applications Handbook Volume 2 PI
FLUSH
PI 3/8" 3/8" S.S. TUBING
PUMP END
PSV
DRIVE END
SHAFT
INNER SEAL
VIKING GEAR PUMP (1.5–2 GPM/ PROCESS PUMP)
OUTER SEAL
SEAL OIL TANK 25 GALLON (LOW MINERAL OR OTHER ACCEPTABLE LOW VISCOSITY OIL)
PUMP SEAL HOUSING
(c)
Figure 16.9 (c) Typical seal flush arrangement for double mechanical seals. PLANT VENT HEADER SYSTEM PI
½" SIGHT GLASS
½"
SEAL OIL SUPPLY TANK 2–3 GAL CAPACITY
FLUSH 3/8"
DISCHARGE PIPE FROM PUMP
½" ½" PUMPING RING IN SEAL
SHAFT
PUMP END
DRIVE END LOW PRESSURE SEAL (OUTSIDE)
INSIDE SEAL
PUMP SEAL HOUSING (d)
Figure 16.9 (d) Typical seal flush arrangement for tandem mechanical seals.
Pumps 343 SPRING HOLDER 11–13 CHROME SPRING–18-8 WORK HARDENED DRIVE COLLAR 11–13 CHROME
“U” CUP MOLDED HYCAR
ROTATING FACE MIN. CHROME 16% STATIONARY FACE-CARBON GLAND STUDS–18-8 AUXILIARY GLAND BRONZE ALL METAL SELF LOCKING NUTS
SHAFT SLEEVE 11–13 CHROME
SEAL COVER SAE 1020 CAD PLATED
DRIVE PINS–18-8
Figure 16.10 Typical single mechanical seal inside pump stuffing box (courtesy of Borg – Warner Co.). P = Pressure of Liquid in Box P' = Average Pressure Across Seal Faces
P P
P
B O
A CL
B
Shaft
Closing Force = P × Area “A” + Spring Loading Opening Force = P' × Area “B”
Figure 16.11 Area relationship for unbalanced seal construction (by permission from Sniffen, T. J., Power and Fluids, Winter 1958, Worthington Corp.). P = Pressure of Liquid in Box P' = Average Pressure Across Seal Faces
P P
P
B CL
B O
Shaft
Closing Force = P × (Area “A” – Area “C”) + Spring Loading Opening Force = P' × Area “B” “A”, “B” and “C” are Variable
Figure 16.12 Area relationship for balanced seal construction (by permission from Sniffen, T. J., Power and Fluids, Winter 1958, Worthington Corp.).
344 Petroleum Refining Design and Applications Handbook Volume 2
Figure 16.13 Single outside balanced seal (courtesy of Durametallic Corp.).
Generally, the centrifugal pump has these characteristics: 1. 2. 3. 4. 5. 6. 7.
ide capacity, pressure, and fluid characteristics range. w easily adapted to direct motor, V-belt or other drive. relatively small ground area requirements. relatively low cost. difficult to obtain very low flows at moderate to high pressures. develops turbulent conditions in fluids. turbine type: (a) offers very high heads at low flows, (b) self-priming, (c) limited to very clean, non-abrasive fluids with limited physical properties, (d) clearances can be problem on assembly and maintenance.
Single-Stage (Single Impeller) Pumps This type of pump (Figures 16.14 and 16.15) is the workhorse of the chemical and petrochemical industries. It also serves important functions in petroleum refining and almost every industry handling fluids and slurries. Although the performance characteristics may vary for specific applications, the general fundamental features are the same especially for manufacturers who standardize to some extent through the Hydraulic Institute [8] and ANSI. Figure 16.16 indicates the relative relationship for three of the centrifugal type pumps, with curves labeled “centrifugal” referring to the usual process (open or enclosed impeller) type unit. A similar set of curves is shown in Figure 16.17 for the turbine unit. Note that the flat head curve of the centrifugal unit has advantages for many process systems, giving fairly constant head over a wide range of flow. For some systems where changes in flow must be reflected by pressure changes, the turbine characteristic is preferred. The centrifugal impeller provides an ever rising horsepower requirement with increasing flow, while the horsepower of the turbine pump falls off with increasing
Pumps 345 5
3 77A 7
17
14
15
76
25
SECTION A–A 29
77B
3 5 7 9 10 10K 13 14 15 17 18 25 25A
56
77
55
10 13 10K 18
Impeller Casing Back Head Cradle Bearing Housing Foot Shaft Sleeve Shaft Sleeve Key Stuffing Box Gland Stuffing Box Gland Stud Stuffing Box Gland Stud Nut Seal Cage Splash Collar Shaft Bearing—Radial Shaft Bearing—Thrust
80
26 28 29 55 56 75 76 76A 77 77A 77B 80 105 105A
26
9
25A 76A 105
75
28
105A
Bearing Housing Bearing End Cover Pump Shaft Oil Disc. (Flinger) Casing Foot Retaining Ring Oil Seal—Front Oil Seal—Rear Gasket—Casing Gasket—Sleeve Gasket—Drain Plug Oil Vent Shaft Adjusting Sleeve Sleeve Lock Nut
Figure 16.14 General service centrifugal pump (courtesy of Dean Brothers Pumps, Inc.).
flow (and decreasing head); hence it is “overloading” at low flows and must be operated with ample horsepower for these conditions. Figure 16.18 shows typical pump performance curves created by the manufacturer based on actual tests. The diagram illustrates the relationship between volumetric flow rate and total dynamic head, efficiency, required net positive suction head (NPSHR) and required power (i.e., brake horsepower, BHP). The effects of impeller shape for the usual centrifugal process pump performance are given in Figure 16.19. The only part the process designer can play is in the selection of a manufacturer’s performance curve to fit the control requirements of the system. If the curve is too steep, select an impeller of necessary basic characteristics to move the curve in the proper direction, providing the manufacturer has an impeller pattern to fit that pump casing, and with the improved physical dimensions. This may require changing the make of pump to obtain the necessary range and characteristic. For conditions of (1) high suction side (or inlet) friction loss, from suction piping calculations or (2) low NPSHA (10 ft (3 m) or less), a large open eye on the impeller inlet is necessary to keep the inlet velocity low. Net Positive
346 Petroleum Refining Design and Applications Handbook Volume 2 Heavy Duty Volute (150,300 lb. Steel and 125 lb. Cast Iron)
Confined Type Gasket Quench Type Gland (Optional Construction) Constant Level Oiler
Wear Rings (Hardened Materials Optional) Seal Lantern
Oil Breather Thrust Bearing; Ball Type Shaft; Alloy Steel Slinger; Labyrinth Type
Enclosed Type Impeller; Hydraulically Balanced for Reduced Thrust Load
Vent and Drain (Optional)
Oil Reservoir Radial Bearing; Ball Type Water Jacket Cools Oil Reservoir Slinger; Labyrinth Type
Shaft Sleeve (Optional Construction) Back Plate; Water Cooled Type, Extra Deep Stuffing Box
Figure 16.15 Cut-a-way section of single-stage pump, Part 1 (above) enclosed type impeller, Part 2 (lower left) open type impeller (courtesy of Peerless Pump Div. FMC Corp.).
Suction Head is discussed in a later section. The manufacturer should be given the conditions in order to properly appraise this situation. In most instances the manufacturer has a series of impellers to use in one standard casing size. The impeller may be trimmed to proper diameter to meet head requirements and yet stay within the power range of a specified driver. It is not necessary to place a full size impeller in a casing unless the system requires this performance. It is good to know when larger impellers can be placed in the casing, and what their anticipated performance might be in order to adequately plan for future uses and changing loads on the pump. Although the previous discussion has pertained to single impellers, the principles are the same for the multi-stage units (impellers in series in the casing) and the casing with double inlets. The latter pump is used for the higher flows, usually above 500 gpm (114 m3/h), and this design serves to balance the inlet liquid load as it enters the impeller, or first stage (if more than one) from two sides instead of one as in the single impeller. The double suction pump has the liquid passages as a part of the casing, with still only one external suction piping connection.
Pumps 347
C
M
Head, ft of Liquid B.H.P. Efficiency Head (ft liquid)
Efficiency
A
A
A = Axial M = Mixed C = Centrifugal
M C
Rating Point Brake Horsepower A M C
Gallons per minute
Figure 16.16 Comparison of impeller types for centrifugal pump performance (adapted by permission from Pic-a-Pump. Allis – Chalmers Mfg. Co.).
The axial and mixed flow impellers are used primarily for very high capacities at relatively low heads as shown in Table 16.2. They are usually applied to services such as water distribution to a large system, waste water disposal, recirculating large process liquor flows, and so on. Many applications can be handled either by a horizontal or a vertical pump. In the range usually associated with process plants and the associated services, Tables 16.6 and 16.7 are helpful guides in making the selection [9].
Pumps in Series Sometimes it is advantageous or economical to use two or more pumps in series (one pump into and through the other) to reach the desired discharge pressure. In this situation the capacity is limited by the smaller capacity of any one of the pumps (if they are different) at its speed of operation. The total discharge pressure of the last pump is the sum of the individual discharge pressures of the individual pumps. For identical pumps, the capacity is that of one pump, and the discharge pressure of the last pump is the sum of the individual heads of each pump acting as a single unit. Thus, for two identical pumps the discharge head is twice that of the rated pressure of one pump at the designated flow rate (Figure 16.20). The pump casing of each stage (particularly the last) must be of sufficient pressure rating to withstand the developed pressure. Consider two centrifugal pumps in series as shown in Figure 16.20a. The total head for the pump combination ∆hT is the sum of the total heads for the two pumps, i.e.
348 Petroleum Refining Design and Applications Handbook Volume 2
1,000
Eff
y
Capacity-head at 30 ft NPSH
600 Head in Feet
Efficiency - %
30
nc
icie
800
40
20
Brake Horsepower
50
Capacity-head at 9 ft NPSH 400
Capacity-head at 6 ft NPSH
He
ad
-C ap
15.0
ac
200
10
ity
Horsep ower (B HP)
10.0 5.0
0
0 10
15
20
25
30
35
40
45
Capacity - GPM
Figure 16.17 Performance of turbine type centrifugal pump (courtesy of Roy E. Roth Co.). 40 20 0
380 360
80
340
70
320
Efficiency, %
60
Head, ft
90
NPSHR, ft
NPSHR
Efficiency
Rated Flow Head
BEP
300
50
280
40
260
30
240
Allowable Operating Region Preferred Operating Region
150 Power, BHP
100
20
Power
50
10
0
0 0
100
200
300
400
500
600 700 800 Flowrate, U.S gal/min
900
1000
Figure 16.18 Characteristics of a centrifugal pump are described by the pump performance curves [36].
1100
1200
1300
Pumps 349 Enclosed impeller characteristics Wide impeller
Head
Narrow impeller
Capacity, gpm
Enclosed or open impeller characteristics Less wrap of vanes
More vanes B
Head
A
C More wrap of vanes less vanes (some wrap as B above)
Capacity, gpm
Figure 16.19 Impeller performance guide. Wrap refers to curvature of vanes on impeller (adapted by permission from Pic-a-Pump. Allis Chalmers Mfg. Co.).
Table 16.6 Pump selection guide. Feature
Horizontal
Vertical
Space requirements
Less head room
Less floor area, more head room.
NPSH
Requires more*
Requires less
Priming
Require*
Usually not required
Flexibility (relative to future changes)
Less
More
Maintenance
More accessible
Major work project
Corrosion and abrasion
No great problem
Can be considerable problem
Cost
Less
More (requires more alloy to handle corrosive fluid)
*For some conditions.
350 Petroleum Refining Design and Applications Handbook Volume 2 Table 16.7 Type selection based on liquid handled. Liquid
Basic pump type
Type impellers
Water and other clear noncorrosive liquids at cold or moderate temperatures.
Single or double suction.
Closed except for very small capacities.
Water above 250°F
Single or double suction. This is usually boiler feed service at high pressures requiring multi-stage pumps.
Closed except for very capacities
Hydrocarbons, hot
Single suction, often of the special type called refinery pumps, designed particularly for high temperature service.
Closed with large inlets.
Corrosives: Mild acid or alkaline
Single or double suction
Strongly acid or alkaline
Single or double suction with single suction probably less expensive if available for the rating.
Hot corrosives
Single suction, with many refinery pump types also used here because of high temperatures and corresponding suction pressures.
Water with solids in suspension:
Coarse abrasives
Pulpy solids such as paper stock
Closed except for very small capacities or where liquid tends to form scale on surfaces of moving parts.
Single suction with end clearance wearing fits. If all particles pass through 1/8" mesh screen, rubber lined pumps are available which will give many times the life of metal pumps, providing no chemical action or excessive temperature will deteriorate the rubber. Special rubber compounds can be applied to improve resistance to certain chemicals.
Open, which allows better application of the rubber, except in larger sizes. Also made in closed type.
Single suction, Not available for full range of ratings, that is, small capacities not too easily obtained. Often have very large impellers operated at slow speeds for use when solids larger than 1" diameter are the standard diet. This would be of the type called dredge pumps handling sizeable rocks.
Closed
Single suction. Double suction only used on very slight solids concentrations and then with special end clearance wearing fits.
Closed. Open type used to be standard but change to end clearance wearing fits made closed impellers better suited.
ΔhT = Δh1 + Δh2
(16.3)
The volumetric flow rate or capacity for the pump combination QT is the same as the capacity for each pump, i.e.
QT = Q1 + Q2
(16.4)
The operating characteristics for two pumps in series are obtained as follows [38]: 1. D raw the ∆h against Q characteristic curves for each pump together with the system ∆hs against Qs curve on the same plot (Figure 16.20a).
Pumps 351 2. D raw a vertical constant capacity line which intersects the two pump curves at total heads ∆h1 and ∆h2 respectively and the system curve at total head ∆hs. 3. Add the values of ∆h1 and ∆h2 obtained in Step 2 to give
ΔhT = Δh1 + Δh2
(16.3)
4. C ompare ∆hT from Step 3 with ∆hs from Step 2. If they are not equal, then repeat Steps 2, 3, and 4 until ∆hT = ∆hs. This is the operating point of the two pumps in series. An alternative to the above trial and error procedure for two pumps in series is to calculate ∆hT from Eq. 16.3 for various values of the capacity from known values ∆h1 and ∆h2 at these capacities. The operating point for stable operation is at the intersection of the ∆hT against QT curve with the ∆h against Qs curve. The piping and valves may be arranged to enable two centrifugal pumps to be operated either in series or in parallel. For two identical pumps, series operation gives a total head of 2∆h at a capacity Q and parallel operations gives a capacity of 2Q at a total head ∆h. The efficiency of either the series or parallel combination is practically the same as for a single pump.
∆hT ∆h1
∆h2
QT QT
∆hT
∆hS Pump curve (1)
∆h
Pump curve (2)
∆h2 ∆h1 System curve
Q
Figure 16.20a Operating point for centrifugal pumps in series [38].
QT
352 Petroleum Refining Design and Applications Handbook Volume 2 300
Two Pu
mp s in
S er
Note: Systems illustrated Assume Duplicate Pumps.
ies
240 S
2
R H2
Head in Feet
180
150
Two Pumps in Pa
P–R
120
rallel
P–R Sing le P um p
S R H1
60
Q1
0
50
Q2
100
150
200
Capacity, gpm Pump in Series:
Pump in Parallel:
Q = Constant H (Total) = H1 + H2 + ’’’ 2O = S-R denotes Series-Rating Point, Total = Constant Q (Total) = Q1 + Q2 + ”’ (at H for each Single Pumps Curve) = P-R denotes Parallel-Rating Point 1O = Single pump rating
Figure 16.20b Operating curves of two duplicate centrifugal pumps in series and parallel.
Pumps in Parallel Pumps are operated in parallel to divide the load between two (or more) smaller pumps rather than a single large one, or to provide additional capacity in a system on short notice, or for many other related reasons. Figure 16.20 illustrates the operational curve of two identical pumps in parallel, each pump handling one half the capacity at the system head conditions. In the parallel arrangement of two or more pumps of the same or different characteristic curves, the capacities of each pump are added, at the head of the system, to obtain the delivery flow of the pump system. Each pump does not have to carry the same flow; but it will operate on its own characteristic curve, and must deliver the required head. At a common tie point on the discharge of all the pumps, the head will be the same for each pump, regardless of its flow.
Pumps 353 The characteristic curves of each pump must be continuously rising (right to left) as shown for the single pump of Figure 16.20, otherwise with drooping or looped curves they may be two flow conditions for anyone head and the pumps would “hunt” back and forth with no means to become stabilized. Figures 16.21a–d represent typical and actual performance curves showing discharge total head (head pressure at pump outlet connection for any fluid), required minimum water horsepower (for pumping water), and capacity or pumping volume of the pump (for any fluid) for several impeller diameters that would fit the same case (housing). Additionally, the important NPSHR characteristics of the pump’s design, impeller entrance opening and diameter, and the hydraulic operating efficiency of the pump at the fixed designated speed of the performance curves are shown on the chart. All of this performance is for one specific impeller diameter of the fixed rotating speed (rpm), and the fixed impeller design pattern proprietary to the manufacturer (number, shape and spacing of vanes, and wrap or curvature of vanes). Note that Figure 16.21c plots the NPSHR curve for this “family” of impellers (different diameters, but exact same design dimensions and features), while Figure 16.21a shows the NPSHR numbers printed at selected points on the curve. Figure 16.21c illustrates the change in performance for the exact same pump, same impellers, but for different rotating speeds of 1750 and 3550 rpm. (Note that the respective motor designated standard speeds are 1800 and 3600 rpm, but the pump manufacturer cannot count on these speeds under load in order to provide performance information the customer needs for the design of a system). Consider two centrifugal pumps in parallel as shown in Figure 16.21d. The total head for the pump combination ∆hT is the same as the total head for each pump, i.e.
ΔhT = Δh1 = Δh2
(16.5)
Horsepower for Liquid of Sp gr = 1.0 200
10 hp 6 1/2" Impeller Diameters
Total Head in Feet (for any Liquid)
160
6"
Performance
4 7/8"
60
5 hp Curves
67 70
3 hp
5 1/2"
120
Efficiency Values
7 1/2 hp
71
2 hp
4 1/2"
80
67 N.P.S .H. Re qu (Any Liquid ired, Fee t ) Suctio n Lift , Fee (Max t imu on W m Limit, B ater a t 70° ased F
40
60
12 17
21
22
16
0 Pump Speed: 3,500 rpm Pump Size: 2" × 2" Maximum Impeller Diameter: 6" Minimum Impeller Diameter: 4 1/2"
0
20
40
60
11
80
100
120
140
160
Capacity for any Liquid, gpm
Figure 16.21a Typical centrifugal pump curves (adapted by permission from Allis-Chalmers, Mfg. Co.).
180
200
220
Figure 16.21b Typical performance curves showing NPSHR in convenient form (by permission from Crane Co. Deming Pump Div.).
354 Petroleum Refining Design and Applications Handbook Volume 2
Pumps 355
Figure 16.21c Exact same pump casing and impellers at different shaft speeds (by permission from Goulds Pumps, Inc.).
356 Petroleum Refining Design and Applications Handbook Volume 2
∆hT Q1 QT
QT Q2
Pump curve (1) Pump curve (2) ∆hT
QS
∆hS
∆h
System curve
Q1
Q2
QT
Q
Figure 16.21d Operating point for centrifugal pumps in parallel [38].
The volumetric flow rate or capacity for the pump combination QT is the sum of the capacities for the two pumps, i.e.
QT = Q1 + Q2
(16.6)
The operating characteristics for two pumps in parallel are obtained as follows [38]: 1. D raw the ∆h against Q characteristic curves for each pump together with the system ∆hs against Qs curve on the same plot as shown in Figure 16.21d. 2. Draw a horizontal constant total head line which intersects the two pump curves at capacities Q1 and Q2 respectively, and the system curve at capacity Qs. 3. Add the values of Q1 and Q2 obtained in Step 2 to give
QT = Q1 + Q2
(16.6)
4. C ompare QT from Step 3 with Qs from Step 2. If they are not equal, repeat Steps 2, 3, and 4 until QT = Qs. This is the operating point of the two pumps in parallel. An alternative to this trial and error procedure for two pumps in parallel is to calculate QT from Eq. 16.6 for various values of the total head from known values of Q1 and Q2 at these total heads. The operating point for stable operation is at the intersection of the ∆hT against QT curve with the ∆hs against Qs curve. Figures 16.22 and 16.23 show photos of a motor driven centrifugal pump for the pre-flashed crude in the crude distillation unit; a main distillation tower and a mild vacuum column with associated centrifugal pumps, piping, and accumulators, respectively.
Pumps 357
Figure 16.21e Pump sizing calculation (SI units) for reflux centrifugal pump of Example 16.19.
358 Petroleum Refining Design and Applications Handbook Volume 2
Figure 16.22 Motor driven centrifugal pump for the pre-flashed crude in the crude distillation unit.
Figure 16.23 Main distillation tower and a mild vacuum column with associated centrifugal pumps, piping, and accumulators.
Pumps 359
16.5 Hydraulic Characteristics for Centrifugal Pumps Capacity: the rate of liquid or slurry flow through a pump. This is usually expressed as gallons per minute (gpm) or cubic meters per hour (m3/h) by pump manufacturers and design engineers in the chemical and refinery and petrochemical industries. A few convenient conversions are as follows: 1 Imperial gal/min = 1.2005 US gpm 1 barrel (42 gal)/day = 0.0292 US gpm = 0.04167 m3/h 1 m3/day 1 l/h = 1 × 10−3 m3/h 1 l/s = 3.6 m3/h For proper selection and corresponding operation, a pump capacity must be identified with the actual pumping temperature of the liquid in order to determine the proper power requirements as well as the effects of viscosity. Figures 16.18 and 16.21a illustrate typical manufacturers’ performance curves for centrifugal pumps as a function of capacity. Pumps are normally selected to operate in the region of high efficiency, and particular attention should be given to avoiding the extreme right side of the characteristic curve where capacity and head may change abruptly. Total Head: the pressure available at the discharge of a pump as a result of the change of mechanical input energy into kinetic and potential energy. This represents the total energy given to the liquid by the pump. Head, previously known as total dynamic head, is expressed as feet (meters) of fluid being pumped. The total head read on the pump curve is the difference between the discharge head (the sum of the gauge reading on the discharge connection on the pump outlet, for most pumps corrected to the pump centerline, plus the velocity head at the point where the gauge is attached) and the suction head (the sum of the suction gauge reading corrected to the pump centerline and the velocity head at the point of attachment of the suction gauge) [10]. Note that the suction gauge reading may be positive or negative, and if negative, the discharge head minus a minus suction (termed lift) creates an additive condition (See Section 16.6). This is shown on the curves of Figure 16.21a. This head produced is independent of the fluid being pumped and is, therefore, the same for any fluid through the pump at a given speed of rotation and capacity. Through conversion, head may be expressed in units other than feet of fluid by taking the specific gravity of the fluid into account.
(Head in feet), h = (psi) (2.31 ft/psi)/SpGr, for any fluid
(16.7)
(Head in meters), h = (bar) (10.2 m/bar)/SpGr, for any fluid
(16.8)
In Metric units
Note that pounds per square inch (psi) is the pressure on the system and is not expressed as absolute unless the system is under absolute pressure. Feet (meters) are expressed as head, not head absolute or gauge (see Example 16.1). Note the conversion of psi pressure to feet of head pressure.
or, (head in ft), h = (psi) (144/ )
(16.9)
(head in m), h = (bar) (10200/ )
(16.10)
In Metric units,
360 Petroleum Refining Design and Applications Handbook Volume 2 where
= fluid density, lb/ft3(kg/m3) = 2.31 ft of water at SpGr = 1.0 1 lb/in. = 2.31 ft of water/SpGr of liquid = ft liquid 1 lb/in.2 1 in. mercury = 1.134 ft of water = 1.134/SpGr liquid = foot liquid 1 bar = 10.2 m of water at SpGr = 1.0 2
For water, SpGr = 1.0 at 62°F (16.67°C), although for general use it can be considered 1.0 over a much wider range. For explanation of vacuum and atmospheric pressure see Chapter 15. The three main components illustrated in the example are (adapted [11]). 1. S tatic head 2. Pressure head 3. Friction in piping, entrance and exit head loss. The main influences on pump hydraulic efficiency are [43]: • • • •
Disc friction (secondary vortex between outer surface of impeller and casing). Surface roughness. Leakage (backflow from discharge through seating gaps) Mixing losses (flow direction changes).
These combine to reduce the hydraulic efficiency of the common centrifugal pump to typically 70–80% as the efficiencies result as heat. However, pump efficiencies exceeding 80% are possible by employing advanced manufacturing techniques, but these result in increased capital cost. Pumps produce pressure/head, and the consequence of head flow characteristic they produce higher pressures at low flow rates and to balance the hydraulics, the high discharge pressure is reduced across a control valve. This fitting wastes energy, and is better to employ a variable speed drive by matching speed to the required discharge pressure and thereby improving efficiency [43].
Example 16.1: Liquid Heads If a pump were required to deliver 50 psig (3.45 barg) to a system, for water, the feet (m) of head on the pump curve must read, 2.31 (50) = 115.5 ft In metric units 10.2 (3.45) = 35.19 m For a liquid of SpGr 1.3, the feet (m) of head on the pump curve must read, 115.5/1.3 = 88.8 ft of liquid. In metric units The head on the pump is 35.19/1.3 = 27.07 m of liquid For liquid of SpGr 0.86, the feet (m) of head on the pump curve must read, 115.5/0.86 = 134.2 ft of liquid. In Metric units The head on the pump is 35.19/0.86 = 40.92 m of liquid
Pumps 361 If a pump were initially selected to handle a liquid where SpGr = 1.3 at 88.8 ft (27.07 m), a substitution of light hydrocarbon where SpGr = 0.86 would mean that the head of liquid developed by the pump would still be 88.8 feet, but the pressure of this lighter liquid would only be 88.8/[(2.31)/(0.86)] or 33.06 psi. In Metric units, the pressure would be 27.07/[(10.2)/(0.86)] or 2.3 bar Note that for such a change in service, the impeller seal rings, packing (or mechanical seal) and pressure rating of casing must be evaluated to ensure proper operation with a very volatile fluid. For other examples, see Figures 16.24a and 24b. The total head developed by a pump is composed of the difference between the static pressure and velocity heads plus the friction entrance and exit head losses for the suction and discharge sides of the pump (Figures 16.25 and 16.26).
H = hd − hs
(16.11)
The sign of hs when a suction lift is concerned is negative, making H = hd − (−hs) = hd + hs. A pump is acted on by the total forces, one on the suction (inlet) side, and the other on the discharge side. By subtracting (algebraically) all the suction side forces from the discharge side forces, the result is the net force that the pump must work against. However, it is extremely important to recognize the algebraic sign of the suction side components, that is; if the level of liquid to be lifted into the pump is below the pump centerline, its algebraic sign is negative (−). Likewise, if there is a negative pressure or vacuum on the liquid below the pump centerline, then this works against the pump and it becomes a negative (−). (See discussion to follow.)
c) Butane sp gr = 0.6 b) Naphtha sp gr = 0.8
a) For water, sp gr = 1.0
166.7' d) Carbon tetrachloride sp gr = 1.50
125'
100' 66.6'
a) 43.3 psig
b) 43.3 psig
c) 43.3 psig
d) 43.3 psig
Pressure gauge attached at the bottom
Figure 16.24a Comparison of columns of various liquids to register 43.3 psig on pressure gauge at bottom of column.
362 Petroleum Refining Design and Applications Handbook Volume 2 c) Butane sp gr = 0.6 b) Naphtha sp gr = 0.8
a) For water, sp gr = 1.0
50.8 m d) Carbon tetrachloride sp gr = 1.50 38.1 m
30.48 m 20.3 m
a) 3.0 barg
b) 3.0 barg
d) 3.0 barg
c) 3.0 barg
Pressure gauge attached at the bottom
Figure 16.24b Comparison of columns of various liquids to register 3.0 barg on pressure gauge at bottom of column.
D'
Total static head
Liquid Exit loss
Atmospheric pressure
Suction head
Discharge head
Suction static head
Liquid
Discharge static head
D
Note: Suction: hS = S – hSL hSL = Pipe, fittings and other friction losses Discharge: hd = D + hdL
S
Entrance loss
Discharge piping
Note: Sw = Worst condition to empty this tank, ft
Sw Centerline pump
Suction piping Pump
Figure 16.25 Suction head system.
hdL = Pipe, fittings and other friction losses
Pumps 363 Exit loss D' Suction Discharge lift head Liquid
Total static head
Valve
Discharge static head
Exchanger
D
S
SL (Worst case) = SL + S1
Suction static lift
Centerline of pump Pump SL
S1
Entrance loss Liquid
Note: *Suction: h = –S – h S L SL hSL = Pipe, fitting, valves, exchanger, and other friction losses –hS = –SL + hSL **Discharge: h = D + h d dL
hdL = Pipe, fitting, valves, exchanger, and other friction losses *Suction: Worst case = S (substitute in L above) **Discharge: Worst case use = (D + D')
Figure 16.26 Suction lift system.
Static Head This is the overall height to which the liquid must be raised. For Figure 16.27a Discharge static head: H Suction static head: L (actually −L) Total system static head: H + L;
actually H − (−L)
(16.12)
H
L
Figure 16.27a Static head, overall = H + L (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
364 Petroleum Refining Design and Applications Handbook Volume 2
H
S
Figure 16.27b Static head, overall = H – S (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
For Figure 16.27b Discharge static head: H (from centerline of pump) Suction static head: S, (actually +S)
Total system static head: H − S; or H − (+S)
(16.13)
Pressure Head For Figure 16.27c Discharge pressure head = 100 psig Suction pressure head = 0 psig Total pressure head = 100 − (+0) = 100 psig = 100 2.31 ft/psi/SpGrH2O=1 * = 231 ft of water Note: The totals are differentials and neither gauge nor absolute values.
(
)
*Applies to water only. For the other fluids use appropriate specific gravity conversion. For Figure 16.27d Discharge pressure head = 100 psig Suction pressure head = +50 psig (=64.7 psia) Total pressure head = 100 (+50) = 50 psi not gauge or absolute = 50 2.31 ft/psi/SpGrH2O=1 = 115.5 ft of water
(
)
Atmosphere
100* GA.
The above examples purposely disregarded pressure head, friction, entrance, and exit losses.
Figure 16.27c Pressure head (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
Pumps 365 100* GA. ABS
100* GA.
Figure 16.27d Pressure head, positive suction (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
100* GA.
50
10 50* GA.
Figure 16.27e Pressure head with negative suction (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
Note that both the discharge and suction pressures must be on the same base/units. These illustrations are for static head only, while overall the pump has to work against the static and the pressure heads (discussed in Section 16.5). For Figure 16.27e Discharge pressure head = 100 psig = 100 2.31 ft/psi/SpGrH2O=1 = 231 ft water (system fluid) Discharge static head = 50 ft Total discharge head = 231 + 50 = 281 ft
(
)
(*Note that no flow friction losses or entrance/exit losses are included in this example) Suction pressure head = +50 psig = 50 2.31 ft/psi/SpGrH2O=1 = + 115.5 ft water (system fluid) Suction static head = −10 ft *Total suction head = +115.5 + (−10) = +105.5 ft *Total head on pump = 281 − 105.5 = 175.5 ft fluid
(
)
Friction Losses Due to Flow Friction, Entrance and Exit Heads, Valve Losses. These losses and calculation methods were presented in Chapter 15. Comments here will be limited. These losses are a function of the characteristics of the fluid flowing in the piping systems and the velocities of flow. Entrance and exit losses relate to the pipe and not the suction or discharge
366 Petroleum Refining Design and Applications Handbook Volume 2 connections at the pump. Usually they are very small, but cannot be ignored without checking. Velocity heads at the pump connections are considered internal losses. These are handled by the manufacturer’s design of the pump and are not considered with the external losses in establishing the pump heads.
Example 16.2: Illustrating Static, Pressure, and Friction Effects Refer to Figure 16.27f for basis of the example. 8" Check Valve 26* GA.
8" Gate Valve
60'
141'
10' of 10" pipe
of 8" Pipe
1500 GPM Capacity
10'
Figure 16.27f Pumping arrangement for Example 5-2 (adapted by permission, Centrifugal Pumps Fundamentals, Ingersoll-Rand Co., Washington, N.J. 07882).
To aid in speed of computation, the friction figures are taken from the Cameron Hydraulic Tables in Chapter 15 and water, which is suited to these tables, is used as an example fluid. Discharge head = 60 ft Discharge pressure head = 26 psig = 26(2.31 ft/psi/SpGr) = 60 ft gauge Discharge friction and exit head (at pipe/tank): 140 ft of 8-in. pipe: 6.32 ft/100(140) 3 8-in. 90° ell: (6.32/100) (3) (20.2) 1 8-in. gate valve 1 8-in. check valve *Exit loss: Assume 8-in. pipe
= 8.8 ft = 3.8 ft = 0.3 ft = 3.3 ft = velocity head = 1.4 ft
Subtotal, ft = 17.6 ft Total discharge head = 137.6 ft Suction static head (lift) = −10.0 ft Suction pressure head 0, psig (atm) = 0.0 Suction friction and entrance head: 10 ft of 10-in. pipe, (2.1 ft/100) (10) = 0.2 ft 1 10-in. suction 90° ell;(2.1/100) (25.3) = 0.5 ft *Entrance loss: 10-in. pipe assume = velocity head = 0.6 ft
Pumps 367 Subtotal = −1.3 ft Total suction head = 10 + (−1.3) = 11.3 ft Total pump head = 137.6 − (−11.3) = 148.9 ft *These are not velocity heads at pump connections, but are related to the piping connections. See earlier note in this regard.
16.6 Suction Head or Suction Lift, hs The total suction head, Figure 16.28, is the difference in elevation between the liquid on the pump suction side and the centerline of the pump (plus the velocity head). Note that the suction head is positive when above the pump centerline and that it decreased with an increase in friction losses through the suction piping system. Thus,
Total suction head (TSH) = static head −hSL
(16.14)
The total suction lift is defined as above except the level of the liquid is below the centerline of the pump or the head is below atmospheric pressure. Its sign is negative.
Total Suction Lift (TSL) = static lift plus friction head losses
In summary: 1. Th e pressure units (gauge or absolute) must be consistent for all components used in determining both suction side and discharge side conditions. Most designers use gauge as a reference, but this is not necessary.
P
SW
S
Pump
S P
Pump
hs = S – hSL + P (a)
hs = – S – hSL + P (b) Note: When P is expressed in absolute pressure units, hs will be in absolute units. If P is less than atmospheric pressure: P is (–) if expressed as a gauge reading and will be a negative feet of liquid. P is (+) if expressed in absolute units. The friction loss hSL includes any entrance or exit losses and other such fittings in the system.
Figure 16.28 Typical suction systems (adapted by permission, Carter, R. and Karassik, “R.P.-477.” Worthington Corp.).
368 Petroleum Refining Design and Applications Handbook Volume 2 2. S tatic head is positive pressure of fluid on pump suction above its centerline (S), (+). 3. Positive external pressure, P, on the surface of fluid on pump suction is used as a positive integer, expressed as feet (meters) of fluid, (+). 4. Partial vacuum, P, on the surface of liquid is a negative pressure. As a partial vacuum expressed as a gauge reading as feet (mm) of liquid below atmospheric, the pressure is negative and would be designated by a minus (−) sign. A partial vacuum, P, expressed as absolute vacuum or absolute pressure would be designated by a positive (+) sign. It is essential to be consistent for all pressure units. If absolute units are used, the total suction head would be in absolute units and the discharge head must be calculated in absolute units. 5. Suction lift is a negative suction head, S, used to designate a negative static condition on the suction of the pump (below atmospheric). The sign for suction head is positive (+), while its corresponding terminology of suction lift is negative (−), since the term “lift” denotes a negative condition. Note that the only difference in these terms is the difference in signs. This applies because the total head for a pump is total discharge head a(+), minus (−) the [suction head, a(+)], or [suction lift, a(−)]. For general service the average centrifugal pump should lift about 15 feet (5 m) of water on its suction side. However, since each process situation is different, it is not sufficient to assume that a particular pump will perform the needed suction lift. Actually, certain styles or models of a manufacturer’s pumps are often specially adapted to high lift conditions. On the other hand it is unnecessary to select a high lift pump when pressure head or flooded suction conditions prevail. Proper evaluation of suction lift conditions cannot be over emphasized. The theoretical maximum suction lift at sea level for water (14.7 psi) (2.31 ft/psi/SpGr) = 34 ft. However, due to flow resistance, this value is never attainable. For safety, 15 ft. (5 m) is considered the practical limit, although some pumps will lift somewhat higher columns of water. When sealing a vacuum condition above a pump, or the pump pumps from a vessel, a seal allowance to atmosphere is almost always taken as 34 ft. of water. High suction lift causes a reduction in pump capacity, noisy operation due to release of air and vapor bubbles, vibration and erosion, or pitting (cavitation) of the impeller and some parts of the casing. (The extent of the damage depends on the materials of construction.) (EL) Atmospheric Pressure
(EL)
Entrance Loss (EL) D
D
C Pump
D
C
C
Pump
Pump
hd = D + hdL + P
hd = D + hdL
hd = D + hdL
(a)
(b)
(c)
Note: For a system evaluation, icluding suction and discharge, the units of P must be the same either gage or absolute, expressed as feet of fluid. The friction losses from the pump to the vessel include any entrance or exit losses. Unless velocities are high, these losses are usually negligible.
Figure 16.29 Typical discharge systems.
Pumps 369
16.7 Discharge Head, hd The discharge head of a pump is the head measured at the discharge nozzle (gauge or absolute), and is composed of the same basic factors previously summarized: 1. S tatic head. 2. Friction losses through pipe, fittings, contractions, expansions, entrances, and exits. 3. Terminal system pressure. Some typical discharge systems are given in Figure 16.29. General practice is to express the terminal discharge pressure, P, at a vessel as in Figure 16.29 in terms of gauge pressure, and hence P = 0 for atmospheric discharge. If P is less than atmospheric or otherwise expressed in absolute units, then it must be added as equivalent feet (meters) of liquid to the value of hd ordinarily expressed as a gauge reading. Figures 16.25 and 16.26 illustrate the use of siphon action in pump systems. Theoretically, the head in the siphon should be recoverable, but actually it may not, at least not equivalent foot for foot. Usually not more than 20 ft. (60 m) of siphon action can be included [12] even though 34 ft. (10 m) are theoretical at sea level. The siphon length is D in the figures [13]. For some systems, the discharge head on the pump should be used as (D + D ), neglecting the siphon action. In any case, if air can be trapped in the loop, (and it usually can) it must be vented during start-up, otherwise the pump will be pumping against the head established using (D + D ). On start-up the flow can be gradually increased, making more head available from the pump to overcome the higher starting head of the system. This should not be overlooked nor underestimated in determining the specifications for the pump.
16.8 Velocity Head Velocity head is the kinetic energy of a liquid as a result of its motion at some velocity, v. It is the equivalent head in feet (meter) through which water would have to fall to acquire the same velocity.
hv= v2/2 g, ft.(m) of fluid
(16.15)
where hv = velocity head, ft (m) v = liquid velocity, ft/s (m/s) g = acceleration of gravity, 32.2 ft/s2 (9.81 m/s2) As a component of both suction and discharge heads, velocity head is determined at the pump suction or discharge flanges respectively, and added to the gauge reading. The actual pressure head at any point is the sum of the gauge reading plus the velocity head, the latter not being read on the gauge since it is a kinetic energy function as contrasted to the measured potential energy. The values are usually (but not always) negligible. Present practice is that these velocity head effects at the pump suction and discharge connections are to be included in the pump performance curve and pump design, and need not be actually added to the heads calculated external to the pump itself [11]. It is important to verify the effects of velocity head on the suction and discharge calculations for pump selection. In general, velocity head (kinetic energy) is smaller for high head pumps than for low head units. Sometimes the accuracy of all the other system calculations does not warrant concern, but for detailed or close calculations velocity head should be recognized. The actual suction or discharge head of a pump is the sum of the gauge reading from a pressure gauge at the suction or discharge and the velocity heads calculated at the respective points of gauge measurement. Regardless of their density, all liquid particles moving at the same velocity in a pipe have the same velocity head [14]. The velocity head may vary across a medium to large diameter pipe. However, the average velocity of flow (i.e., dividing the total flow as ft3/s (m3/s) by the cross-sectional area of the pipe) is usually accurate enough for most design purposes.
370 Petroleum Refining Design and Applications Handbook Volume 2 Using the example of Karassik and Carter [10], for a pump handling 1500 gpm, having a 6-in. discharge connection and 8-in. suction connection, the discharge velocity head is 4.5 ft and the suction is 1.4 ft, calculated as shown above. If the suction gauge showed 8.6 ft, the true head would be 8.6 + 1.4 = 10.0 ft. If the discharge head showed 105.5 ft head, the true total head would be 105.5 + 4.5 = 110.0 ft, less (8.6 + 1.4) or 100 ft. The net true total head would be 110 ft − 10ft = 100.0 ft. Looking only at the gauge readings, the difference would be 105.5 − 8.6 = 96.9 ft, giving an error of 3.1% of the total head. As an alternate example, if the discharge head were 45.5 ft, then the true total head = (45.5 + 4.5) − (8.6 + 1.4) = 40 ft, and the difference in gauge readings would be 45.5 − 8.6 = 36.9 ft, or an error of 7.8%. Most designers ignore the effects of velocity head, but the above brief examples emphasize that the effect varies depending on the situation and the degree of accuracy desired for the head determinations.
16.9 Friction The friction losses for fluid flow in pipe valves and fittings are determined as presented in Chapter 15. Entrance and exit losses must be considered in these determinations, but are not to be determined for the pump entrance or discharge connections into the casing.
16.10 Net Positive Suction Head (NPSH) and Pump Suction
Discharge
A pump is designed to handle liquid and not vapor. However, there are many instances where vapor easily gets into the pump if the design is not carefully done. As the fluid moves through the pump, pressure losses occur as shown in Figure 16.30, in the inlet passage (Point A to Point B), due to internal frictional (Point B to Point C), and at the blade and within the impeller (Point C to Point D). If the static pressure drops below the fluid’s vapor pressure, the fluid begins to boil, creating vapor bubbles and reducing the density of the fluid. When this occurs, the differential pressure created by the dynamic head of the
Impeller Eye
D
C
B
Suction
A
Volute Impeller
Figure 16.30 Static pressure losses occur as the fluid travels into the pump suction and moves in and out of the impeller [24].
Pumps 371 impeller decreases (Eq. 16.2). The lowest pressure occurs right at the impeller inlet where a sharp pressure dip occurs. The impeller rapidly builds up the pressure, which collapses vapor bubbles, causing cavitation and damage. Furthermore, if fluid enters a pump at its bubble point, it will start vaporizing inside the pump. The formation of these bubbles in the area of the impeller accounts for the noise associated with cavitation. The conversion of the pump’s suction pressure to velocity in the eye of the impeller is known as the required net positive suction head (NPSHR). This is the head of liquid that must exist at the edge of the inlet vanes of an impeller to allow liquid transport without causing undue vaporization. NPHSR is a function of impeller geometry and size, and is determined by factory testing. The NPSHR of an impeller can range from a few feet to a three-digit number. Figure 16.31 shows that as the flow-control valve on the discharge of the pump is opened, the velocity of liquid in the eye of the impeller rises. More of the pump’s suction pressure, or feet (meters) of head is converted to velocity or 1 kinetic energy mv 2 . This means that the NPSHR of a pump increases as the volumetric flow through the pump 2 increases [27].
The NPSHR of a pump is due primarily to the conversion of feet (meters) of head to velocity in the eye of the impeller. The NPSHA to a pump has the definition: Physical pressure pump at suction minus vapor pressure of liquid at pump suction. When the NPSHR of a pump equals the NPSHA to the pump, the pump will cavitate or slip. Figure 16.31 shows the liquid in the vessel is equilibrium with the vapor leaving the drum. This means that the liquid is at its bubble point pressure and the vapor is at its dew point temperature. The vapor pressure of the liquid is 10 psig (24.7 psia). The physical pressure at the suction of the pump is 15 psig (29.7 psia). Therefore, the physical pressure at the suction of the pump is 5 psia. Therefore, the NPSHA is 20 ft. (2.31 × 5/0.58). This matches the level of liquid in the drum above the suction line of the pump and equals the NPSHA. The NPSHR of the pump may be determined from Figure 16.32 (regardless of the specific gravity of the liquid being pumped). At 250 gpm, the NPSHR
Vapor and liquid P
Vapor 140°F
10 psig
Bubble-point liquid
P
20 ft
P 15 psig (s.g. = 0.58)
Figure 16.31 NPSHA equals 20 ft [27].
372 Petroleum Refining Design and Applications Handbook Volume 2
25 ft
20 ft
15 ft
Required net positive suction head
10 ft
5 ft
50
100
150
200
250
300
GPM
Figure 16.32 NPSHR increases with flow [27].
of 20 ft will equal the NPSHA of 20ft. Therefore, at a flow rate of 250 gpm, the pump will cavitate. This calculation neglects the frictional losses in the suction line, which should be subtracted from the NPSHA. At 300 gpm, if the flow control valve is opened, the flow will momentarily increase, but shortly afterwards; the flow will become erratically low as the pump begins to cavitate. This is because an additional 6 ft. of NPSH is required to increase the flow from 250 to 300 gpm (Figure 16.31). A way of achieving this extra suction pressure or NPSH is to raise the liquid level in the drum. For example, with SpGr = 0.58, with very 4 ft rise in the level in the drum, the suction pressure will increase by 1 psi (4 * 0.58/2.31) and the NPSHA will increase by 4ft. But the drum is almost full as shown in Figure 16.29. A way of remedying this is to cool the liquid by 5oF after it leaves the drum. The cooled liquid is not in equilibrium with the vapor in the drum. It has been subcooled by 5°F; this indicates that the bubble point liquid has been cooled without altering its composition. The vapor pressure of the liquid has been reduced as shown in Figure 16.33; and subcooling by 5°F reduces its vapor pressure by 2 psi. This is equivalent to an increase in the NPSH by 8 ft (with the same specific gravity of 0.58). However, as the objective is to increase the flow from 250 to 300 gpm, Figure 16.30 indicates that the NPSHR increases from 20 to 26 ft. But as the liquid is subcooled by 5°F, the NPSHA increases from 20 to 28 ft. This exceeds the NPSHR by 2 ft, and the flow can be increased without the risk of pump cavitation. Insufficient NPSH can also result in pump cavitation. This occurs when the vapor bubbles that have formed in low static pressure areas move along the impeller vanes into higher-pressure areas and rapidly collapse. The forces produced by these bubbles as, they implode to erode the impeller vane surfaces, causing progressive pitting damage as shown in Figure 16.34. Cavitation is associated with distinct crackling noise that resembles the sound of a fluid starting to boil. Note that at a 3% head loss, cavitation has already begun. This must be avoided by maintaining sufficient net positive suction head (NPSH) as specified by the manufacturer. API 610 defines Net Positive Suction Head required (NPSHR) as the amount of suction head needed to limit head loss at the first stage of the pump to 3% (using water as the test fluid). Although hydrocarbons generally require less NPSH than water, reduction factors for hydrocarbons are not allowed by API. Net positive suction head (in feet/meters of liquid absolute) above the vapor pressure of the liquid at the pumping temperature is the absolute pressure available at the pump suction flange, and is a very important consideration in selecting a pump which might handle liquids at or near their boiling points, or liquids at high vapor pressures.
Pumps 373
25
23
Vapor pressure, psia
Vap
21
e sur
res
p or
19
17
120°F
125°F
130°F
135°F
140°F
Temperature, °F
Figure 16.33 Subcooling increases NPSHA [27].
Le ad
ing
Ed
ge
of
Va n
e
Cavitation Damage
(a)
(b)
Figure 16.34 (a) Cavitation damage has occurred on an impeller and (b) erosion–corrosion of an impeller.
Do not confuse NPSH with suction head, as suction head refers to pressure above atmospheric [8]. If this consideration of NPSH is ignored the pump may well be inoperative in the system, or it may be on the border-line and become troublesome or cavitating. The significance of NPSH is to ensure sufficient head of liquid at the entrance of the pump impeller to overcome the internal flow losses of the pump. This allows the pump impeller to operate with a full “bite” of liquid essentially free of flashing bubbles of vapor due to boiling action of the fluid.
374 Petroleum Refining Design and Applications Handbook Volume 2 The pressure at any point in the suction line must never be reduced to the vapor pressure of the liquid (see Eqs. 16.14 and 16.15). Both the suction head and the vapor pressure must be expressed in ft (m) of the liquid, and as gauge pressure or absolute pressure. Centrifugal pumps cannot pump any quantity of vapor, except possibly some vapor entrained or absorbed in the liquid. The liquid or its gases must not vaporize in the eye/entrance of the impeller. (This is the lowest pressure location in the impeller.) For low available NPSH (less than 10 ft or 3 m) the pump suction connection and impeller eye may be considerably oversized when compared to a pump not required to handle fluid under these conditions. Poor suction condition due to inadequate NPSHA is one major contribution to cavitation in pump impellers, and this is a condition at which the pump cannot operate for very long without physical erosion damage to the impeller [14, 15]. Cavitation of a centrifugal pump, or any pump, develops when there is insufficient NPSH for the liquid to flow into the inlet of the pump, allowing flashing or bubble formation in the suction system and entrance to the pump. Each pump design or “family” of dimensional features related to the inlet and impeller eye area and entrance pattern requires a specific minimum value of NPSH to operate satisfactorily without flashing, cavitating, and loss of suction flow. Under cavitating conditions, a pump will perform below its head-performance curve at any particular flow rate. Although the pump may operate under cavitation conditions, it will often be noisy because of collapsing vapor bubbles and severe pitting, and erosion of the impeller often results. This damage can become so severe as to completely destroy the impeller and create excessive clearances in the casing (Figure 16.34). To avoid these problems, the following are a few situations to watch. 1. H ave NPSHA available at least 2 ft (0.6 m) of liquid greater than the pump manufacturer requires under the worst possible operating conditions (see Figures 16.22a–c) with pump curve values for NPSH expressed as feet (meter) of liquid handled. These are the pump’s minimum NPSHR. The pump’s piping and physical external system provide the NPSHA.
NPSHA must be greater than NPSHR
(16.16)
2. 3. 4. 5. 6.
I nternal clearance wear inside pump. Plugs in suction piping system (screens, nozzles, etc.). Entrained gas (non-condensable). Deviations or fluctuations in suction side pressures, temperatures (increase), low liquid level. Piping layout on suction, particularly tee-intersections, globe valves, baffles, long lines with numerous elbows. 7. Liquid vortexing in suction vessel, thus creating gas entrainment into suction piping. Figure 16.35 suggests a common method to eliminate suction vortexing. Since the forces involved are severe in vortexing, the vortex breaker must be of sturdy construction and firmly anchored to the vessel. 8. Nozzle size on liquid containing vessel may create severe problems if inadequate. Liquid suction velocities, in general, are held to 3 6.5 ft/s (0.9 2.0 m/s). Nozzle losses are important to recognize by identifying the exit design style (See chapter 15). Usually, as a guide, the suction line is at least one pipe size larger than the pump suction nozzle. The NPSHA available from or in the liquid system on the suction side of a pump is expressed (corrected to pump centerline) as:
(
)
NPSH A = p′a − p′vp ± S +
v s2 − h SL 2g
2.31 ft /psi v s2 NPSH A = (Pa − Pvp ) ± S + − h SL 2g SpGr
(16.17)
(16.18)
Pumps 375
Air/Vapor (non-condensed) entrained
Desired Liquid Level
Liquid Level Control Actual Liquid Flow Pattern
To Pump Suction
Note: (a) Dimension, “h”, min. of 5" to “h” = 1.25 x nozzle Dia.
(b) Dimension “L” approx. 3.5 to 5 times nozzle Dia.
Clearance, 2" min. to usual 4–6", except large nozzles require more clearance.
h L
(c) Bottom of vortex breaker may be attached to bottom or raised up 2" to 4". Vortex “cross” must be sturdy, welded of heavy plate (not light sheet metal). Vortex breaker must not restrict liquid flow into nozzle opening, but prevent swirling of liquid.
Figure 16.35 Liquid vortex in vessel and suggested design of vortex breaker.
With commonly used suction pipe diameters, the velocity head may be negligible, and the frictional head loss at the suction, hSL can be expressed in terms of velocity head loss by:
h SL = K
v2 2g
(16.19)
376 Petroleum Refining Design and Applications Handbook Volume 2 where
v=
Q
( πd 4 ) 2
(16.20)
and 2
K Q h SL = 2 2g πd 4
(16.21)
2
2.31 ft psi v s2 K Q NPSH A = (Pa − Pvp ) ±S+ − 2 2g 2g πd SpGr d 4
(16.22)
10.2 m bar v 2s NPSH A = (Pa − Pvp ) ± S + − h SL 2g SpGr
(16.23)
In Metric units
or
2
10.2 m bar v 2s K Q NPSH A = (Pa − Pvp ) ± S + − 2g 2g πd 2 4 SpGr
(16.24)
where p′a or Pa represents the absolute pressure in the vessel (or atmospheric) on the liquid surface on the suction side of the pump. p′vp or Pvp represents the absolute vapor pressure of the liquid at the pumping temperature. hSL is the suction line, valve, fitting, and other friction losses from the suction vessel to the pump suction flange. S may be (+) or (−) depending on whether static head or static lift is involved in the system. Figure 16.36 shows a typical relationship between the NPSHA in the system and the NPSHR by the pump as the volumetric flow rate of liquid or capacity Q is varied. The NPSHR by a centrifugal pump increases approximately with the square of the liquid throughput. Eqs. 16.22 and 16.24 show that the NPSHA in a system decreases as the liquid throughput increases because of the greater frictional head losses. A centrifugal pump will operate normally at a point on its total head against capacity characteristic curve until the NPSHA falls below the NPSHR curve. Beyond this point, the total head generated by a centrifugal pump falls drastically as illustrated in Figure 16.37 as the pump begins to operate in cavitation conditions. In centrifugal pump systems, a throttling valve is located on the discharge side of the pump. When this valve is throttled, the system ∆h against Q curve is altered to incorporate the increased frictional head loss. The effect of throttling is shown in Figure 16.38. Throttling can be used to decrease cavitation. A flow regulating valve or other constriction must not be placed on the suction side of the pump. This available value of NPSHA (of the system) must always be greater by a minimum of 2 ft (610 mm) and preferably three or more feet than the NPSHR stated by the pump manufacturer or shown on the pump curves in order to
Pumps 377
NPSH
Available NPSH in the system
NPSH required by the pump
Q
Figure 16.36 NPSHA and NPSHR vs. capacity in a pumping system [37].
∆h
Normal pump curve for adequate suction condition
Normal operating point System curve Pump curve for insufficient available NPSH
Q
Figure 16.37 Effect of insufficient NPSH on the performance of a centrifugal pump [37].
378 Petroleum Refining Design and Applications Handbook Volume 2
Pump curve
∆h
Operating point with throttling
System curve with throttling Normal operating point
Normal system curve
Q
Figure 16.38 Effect of throttling the discharge valve on the operating point of a centrifugal pump [37].
overcome the pump’s internal hydraulic loss and the point of lowest pressure in the eye of the impeller. The NPSHR by the pump is a function of the physical dimensions of casing, speed, specific speed, and type of impeller, and must be satisfied for proper pump performance. The pump manufacturers must always be given complete suction conditions if they are expected to recommend a pump to give long and trouble-free service. As the altitude of an installation increases above sea level, the barometric pressure, and hence p′a s or Pa decreases for any open vessel condition. This decreases the NPSHA. Figures 16.18 and 16.21 are present typical manufacturer’s performance curve. The values of NPSHR given are the minimum values required at the pump suction. As mentioned, good practice requires that the NPSHA available be at least 2 ft. (610 mm) of liquid above these values. It is important to recognize that the NPSHR and suction lift values are for handling water at about 70°F (21.1°C). To use with other liquids it is necessary to convert to the equivalent water suction lift at 70°F (21.1°C) and sea level. Total Suction Lift (as water at 70°F) = NPSHA (calculated for fluid system) −33 ft. The vapor pressure of water at 70°F is 0.36 psia.
16.11 General Suction System The suction system piping should be kept as simple as reasonably possible and adequately sized. Usually, the suction pipe should be larger than the pump suction nozzle. Furthermore, the suction system must maintain the pressure above the vapor pressure at all points. Usually, possible points of intermediate low pressure occur in the area of the vessel (drawoff) nozzle. Kern [26] provides some good rules of thumb on this. 1. Th e minimum liquid head above the drawoff nozzle must be greater than the nozzle exit resistance. Based on a safety factor of 4 and a velocity head “K” factor of 0.5:
Pumps 379
hL =
2u 2 2g
(16.24)
where hL = Liquid level above nozzle, ft. u = Nozzle velocity, ft/s. g = acceleration due to gravity, 32.2 ft/s2. 2. F or a saturated (bubble point) liquid, pipe vertically downward from the drawoff nozzle as close to the nozzle as possible. This gives maximum static head above any horizontal sections or piping networks ahead of the pump. A vortex breaker should be provided for the vessel drawoff nozzle (Figure 16.33). Some rules of thumb for the suction lines are: 1. K eep it short and simple. 2. Avoid loops or pockets that could collect vapor or dirt. 3. Use an eccentric reducer with the flat side up (to prevent trapping vapor) as the transition from the larger suction line to the pump suction nozzle. 4. Typical suction line pressure drops: Saturated liquids = 0.05 − 0.5 psi/100 ft Subcooled liquids = 0.5 − 1.0 psi/100 ft
Example 16.3: Suction Lift What is the Suction Lift value to be used with the pump curves of Figure 16.21a, if a gasoline system calculates an NPSH of 15 ft?
Solution Total Suction Lift (as water) = 15 − 33 = −18 ft. Therefore, a pump must be selected which has a lift of at least 18 ft. The pump of Figure 16.21a is satisfactory using an interpolated suction lift line between the dotted curves for 16 ft and 21 ft of water. The performance of the pump will be satisfactory in the region to the left of the new interpolated 18-ft line. Proper performance should not be expected near the line. If the previous system were at sea level, consider the same pump with the same system at an altitude of 6000 ft. Here the barometric pressure is 27.4 ft of water. This is 34 − 27.4 = 6.6 ft less than the sea level installation. The new NPSHA will be 15 ft − 6.6 ft = 8.4 ft. Referring to the pump curve of Figure 16.21a, it is apparent that this pump cannot do greater than 21 ft suction lift as water or 12 ft NPSHR of liquid (fluid). Total Suction Lift as water = 8.4 − 33 = −24.6 ft. The pump curves show that 21 ft suction lift of water is all the pump can do, hence the 24.6 ft is too great. A different pump must be used which can handle this high a suction lift. Such a pump may become expensive, and it may be preferable to use a positive displacement pump for this high lift. Normally lifts are not considered reasonable if over 20 ft.
Example 16.4: NPSHA in Open Vessel System at Sea Level Conditions: at sea level, atmospheric pressure, Pa = 14.7 psia (1.013 bara) (use Figure 16.25). Assume liquid is water at 85°F (29.4°C), vapor pressure, Pvp = 0.6 psia (0.04 bara). Assume tank liquid level is 10 ft above center line of pump, then S = + 10 ft (3.05 m). Assume that friction losses have been calculated to be 1.5 ft, hSL = 1.5 ft (0.46 m)
380 Petroleum Refining Design and Applications Handbook Volume 2 Using Eq. 16.18 gives: NPSHA = (14.7 − 0.6) (2.31/0.997) + 10 – 1.5 = 41.2 ft (good) In Metric units, by using Eq. 16.23 NPSHA = (1.013 − 0.04) (10.2/0.997) + 3.05 − 0.46 = 12.54 m (41.2 ft) Note: For the worst case, which is an empty tank, “S” becomes Sw on the diagram.
Example 16.5: NPSHA in Open Vessel Not at Sea Level Conditions: Vessel is at altitude 1500 ft (457.2 m), where atmospheric pressure is Pa = 13.92 psia (0.96 bara) (use Figure 16.26) Liquid: Water at 150°F (65.56°C), vapor pressure Pvp = 3.718 psia (0.256 bara) and specific gravity, SpGr = 0.982 Assume vessel liquid level is 12 ft (3.657 m) below centerline of pump, SL = −12(−3.657 m). Friction losses: Assume as calculated to be 1.1 ft (0.335 m) of liquid. Using Eq. 16.18 NPSHA = (13.92 − 3.718) (2.31/0.982) − 12 − 1.1 = 10.90 ft In Metric units, using Eq. 16.23 NPSHA = (0.96 − 0.256) (10.2/0.982) − 3.657 − 0.335 = 3.32 m (10.90 ft) The worst condition case should be calculated using S′L, since this represents the maximum lift.
Example 16.6: NPSHA in Vacuum System Conditions: Vessel is liquid collector at 28 in. Hg Vacuum (referred to a 30 in. barometer) (use Figure 16.28a). This is 30 − 28 = 2 in. Hg abs, or Pa = [(14.7/30)] (2) = 0.98 psia (0.067 bara). Liquid: Water at 101.2°F (38.4°C), vapor pressure = 0.98 psia (0.067 bara). Assume vessel liquid level is 5 ft (1.524 m) above centerline of pump, S = + 5ft, worst case, Sw = 2 ft (0.61 m) Friction losses: Assume to be 0.3 ft (0.091 m) of liquid Using Eq. 16.18 NPSHA = (0.98 − 0.98) (2.31/0.994) + 5 – 0.3 = 4.7 ft Worst case = 4.7 ft (not practical design) The pump selected for this application (water boiling at 0.98 psia) must have a NPSHR less than 4.7 ft preferably about 3–3.5 ft. This is a difficult condition. If possible the vessel should be elevated to make more head (S) available which will raise the NPSHA. In Metric units, using Eq. 16.23 NPSHA = (0.067 − 0.067) (10.2/0.994) − 1.524 − 0.091 =1.433 m (4.7 ft)
Pumps 381
Example 16.7: NPSHA in Pressure System Conditions: Vessel contains butane at 90°F (32.22°C) and 60 psia (4.138 bara) system pressure, Pa = 60 (4.138 bara) (use Figure 16.28b) Butane vapor pressure, Pvp at 90°F (32.22°C) = 44 psia (3.034 bara), SpGr = 0.58. Assume liquid level is 8 ft (2.438 m) below pump centerline, S = −8 ft (−2.438 m). Friction losses: Assume to be 12 ft (3.658 m) of liquid. Then from Eq. 16.18 NPSHA = (60.0 − 44) (2.31/0.58) − 8 − 12.0 = 43.72 ft In Metric units, from Eq. 16.23 NPSHA = (4.138 − 3.034) (10.2/0.58) − 2.438 − 3.658 = 13.32 m (43.7 ft) This presents no pumping problem.
Example 16.8: Closed System Steam Surface Condenser NPSH Requirements Refer to Figure 16.39 for this example This is a closed steam surface condenser system with condensate being pumped out to retreatment facilities. From the conditions noted on the diagram, Friction loss in suction line side = 2.92 ft Absolute pressure in condenser = pʹ = 1.5 in. Hg Abs = 1.5(1.137 ft/in. Hg) = 1.71 ft water
CONDENSER Abs = 1.50” Hg Vacuum = 28.42” Hg
Condensate 91.72°F
10'
Figure 16.39 Surface condenser condensate removal. Closed system steam surface condenser NPSH requirements (by permission, Cameron Hydraulic Data, 16th ed. Ingersoll-Rand Co., 1979, p. 1–12).
382 Petroleum Refining Design and Applications Handbook Volume 2 Water from steam tables at saturation = 1.5 in. Hg abs at 91.72°F Vapor pressure, p′vp, at 1.5 in. Hg Abs = 1.5(1.137) = 1.71 ft H2O NPSHA = (1.71 − 1.71) + 10 − 2.92 = + 7.08 ft The suction head or lift for the pump (separate calculation from NPSHA) is as follows: The 28.42 in. Hg vacuum (gauge) is equivalent to 1.5 in. Hg abs 28.42 in. vacuum (1.137 ft/in Hg) = 32.31 ft H2O Static submergence = 10.0 ft (see Figure 16.39) Friction/entrance losses = 2.92 ft Net static submergence = 7. 08 ft Equivalent suction lift = 25.23 ft (Note: 32.31 − 7.08) (= vacuum effect less net submergence) Note that the equivalent suction lift must be added to the total discharge head for the pump system to obtain the total system head. Keep in mind that the work the pump must accomplish is overcoming the suction losses (+ or −) plus the discharge losses, that is, + discharge loss (all) – (+ if head, or −if lift on suction losses all). Thus, the suction lift becomes a (−) (−) or a (+) to obtain the total system head. Keep in mind that a vacuum condition on the suction of a pump never helps the pump, but in effect is a condition that the pump must work to overcome.
Example 16.9: Process Vacuum System For this process example, again using water for convenience, a low pressure, low temperature water is emptied into a vented vessel, and then pumped to the process at a location at about 3000 ft. altitude where atmospheric pressure is approximately 13.2 psia, Water SpGr is at 200°F = 0.963 (refer to Figure 16.40 for this example) Determine the NPSHA for pump:
High altitude venting
P = 13.2 PSIA
Low pressure water
S = 10'
P = 11.5 PSIA 200°F Water
Friction/entrance type losses = 1', hSL
Figure 16.40 High altitude process vacuum system, NPSH requirements.
Pumps 383 Then From Eq. 16.18 NPSHA = (13.2 − 11.5) (2.31 ft/0.963) + 10 − 1.0 = 13.08 ft available For hydrocarbons and water significantly above room temperatures, the Hydraulic Institute [8] recommends the use of a correction deduction as given in Figure 16.41. This indicates that the NPSHR as given on the pump curves can be reduced for conditions within the range of the curve based on test data. 1000
500 400 300 10 9.5
E
N PA RO
200
P
8
150 100
4 3
VAPOR PRESSURE psia
50
2
40
1.5
30
NPSH REDUCTIONS-FEET
6 5
1.0
20
NE
TA
·BU
ISO
15
E
N TA BU
·1
10
1
0.5
RA
GE
I FR
R NT
TH YL AL CO H
OL
RE
5
ME
4
WA TE
R
3 2 1.5 1.0
0
50
100
150
200
250
300
400
TEMPERATURE °F
Figure 16.41 NPSH reductions for pumps handling hydrocarbon liquids and high temperature water (Note: Do not use for other fluids.) (by permission from Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pumps, Hydraulic Institute, 13th ed., 1975).
384 Petroleum Refining Design and Applications Handbook Volume 2 If the pump given in the curve of Figure 16.21a were being used to pump butane at 90°F and 0.58 gravity, the correction multiplier from the NPSH curve is about 0.99 by interpolation. This means that the values of Figure 16.21a should be multiplied by 0.99 to obtain the actual NPSH the pump would require when handling a hydrocarbon of these conditions. The correction does not apply to other fluids. If the system pressure were 46 psia, then NPSHA = (46 − 44) (2.31/0.58) − 8 − 12 = −12 ft, and this is an impossible and unacceptable condition. This means liquid will flash in the line and in the impeller, and cannot be pumped, and NPSH must always be positive in sign.
16.12 Reductions in NPSHR Limitations for use of the Hydraulic Institute NPSH reduction chart (Figure 16.41) are as follows [8]: 1. N PSH reductions should be limited to 50% of the NPSHR required by the pump for cold water, which is the fluid basis of the manufacturer’s NPSHR curves. 2. It is based on handling pure liquids, without entrained air or other non-condensable gases, which adversely affect the pump performance. 3. Absolute pressure at the pump inlet must not be low enough to release non-condensables of pure liquids. If such release can occur, then the NPSHR would need to be increased above that of the cold water requirements to avoid cavitation and poor pump performance. 4. For fluids, the worst actual pumping temperature should be used. 5. A factor of safety should be applied to ensure that NPSH does not become a problem. 6. The chart should not be extrapolated beyond NPSH reductions of 10 ft.
Example 16.10: Corrections to NPSHR for Hot Liquid Hydrocarbons and Water In Figure 16.41, use the dashed example lines at a temperature of 55°F for propane [8], and follow the vertical line to the propane vapor pressure dashed line, which reads 100 psia vapor pressure. Then follow the slant lines (parallel) to read the scale for NPSH reductions, that is, feet at 9.5 ft. Now the pump selected reads NPSHR on its pump performance curve of 12 ft for cold water service. Now, half of 12 ft = 6 ft But in Figure 16.41, the reduction = 9.5 ft Corrected value of NPSHR to use = 6 ft, since 9.5 ft is >half the cold water value.
Example 16.11: Alternate to Example 16.10 Assume that a boiler feed water is being pumped at 180°F. Read the chart in Figure 16.41 and the water vapor pressure curve, and follow over to read NPSH reduction = 0.45 ft. A pump selected for the service requires 6 ft cold water service NPSHR: Half of 6 = 3 ft Value from chart for 180°F = 0.45 ft reduction Then correct NPSHR to use = 6 ft − 0.45 ft = 5.55 ft required by the pump for this service
16.13 Charting NPSHR Values of Pumps For a given pump and speed, the NPSHR depends primarily on flow rate; thus NPSHR depends on the pump’s design characteristics such as the physical dimensions of the casing, pump speed and impeller types. The NPSHR of a pump
Pumps 385 at a given capacity-head rating increases with increasing pump speeds ( rpm is proportional to NPSH0.75). As a result, critical suction applications as those required to have a low NPSHA will have to employ pumps running at slow speeds. These include such applications as pumping of fluids in systems with restricted suction heads, relatively long suction pump lines or at the liquid boiling points. In critical suction installations, designers select pumps capable of running at 1750 rev/min or lower rather than units that operate at typically higher values, such as at 3500 rev/min. This requires that pumps have to be bigger for a given capacity requirement. Figure 16.42 shows a chart that provides a simple way of estimating the value of NPSHR of a pump and therefore helps to fix the conditions of the pipe system at pump suction so that the NPSHA can be sufficiently higher. Therefore, finding the NPSHR from the chart, designers can select the correct suction conditions (e.g., pipe diameter, length, static head, and so on), to ensure that the NPSHA will be higher than the NPSHR from the pump. Figure 16.42 requires using the following steps in determining NPSHR 1. D etermine the flow rate (gal/min) of the fluid to be pumped. Then by using the bottom half of Figure 16.40, a rough idea of the size of the pump that is required for the application is determined. For example, to pump 500 gal/min over 300 ft of head would require a pump with an 8-in. impeller running at about 3560 rev/min. 2. Looking at the upper section of the chart, the NPSHR is approximately 13 ft H2O. 3. Since the NPSHA of the system should be at least 3 ft higher than the NPSHR, the NPSHA should be a minimum of 16 ft. To prevent cavitation in a pumping system, NPSHA should be at least 3 ft above the NPSHR, read from the pump curve for the given total dynamic head (TDH) and pumping rate.
NPSHA ≥ NPSHR + 3ft
(16.25)
30
3,578
Required NPSH, ft
20
3,480 RPM
3,560
3,550
1,750
10 3,490 0 Pump sizes are in inches
1,300
3,578 3,560
173/16
Head, ft
1,000 13
500 3,480 RPM
3,490
14 15
3,560 7
6
100
14
16 3,550 8
19 18 1617 15 14
9¼
8
14 12
15¼
1,750
60 10
50
100 Flowrate, gal/min
Figure 16.42 Head, NPSHR vs. flow rate [25].
500
1,000
4,000
386 Petroleum Refining Design and Applications Handbook Volume 2 Based on Eq. 16.18, there are various ways to increase the NPSHA to make a pumping system feasible. These are [32]: 1. 2. 3. 4. 5.
aise the liquid level in the suction tank (increasing the S term). R Lower the pump location (increasing the S term). Reduce the frictional loss on the suction side (by reducing suction side velocity or pipe length). Pressurize the suction tank (increase Ps). Lower the vapor pressure by reducing pumping temperature (reduce pv).
16.14 Net Positive Suction Head (NPSH) NPSHR is normally specified by the pump supplier, while based on the installation of pump, NPSHA should be calculated and specified by the designer. Theoretically, NPSHA should be greater than zero to avoid cavitation. NPSHR depends on properties of liquid, the total liquid head; pump speed, capacity and impeller design. Practical curves of NPSHR vs. capacity and speed of the pump are supplied by the pump manufacturer. Figures 16.43 and 16.44 can be employed as a guideline to find the value of NPSHR for centrifugal pump handling water at temperatures below 100°C and above 100°C, respectively. When a pump installation is designed, the available net positive suction head, NPSHA can be calculated by the following equation.
NPSHA = hss – hfs – pv
(16.26)
30 25 20 15 e 70 abov rpm 0 5 5 3 bar to 70 en 35 e w t m be
(NPSH)R, m
10 7.5
3550 5
3550
4
r
rp
w 35
elo pm b
bar
bar
en 35 1750 rpm betwe
bar to 70
3
1750
2
min rpm/
w 35
belo
bar
Capacity, m3/h
Figure 16.43 Net positive suction head for high pressure centrifugal hot-water pumps (source: Hydraulic Institute (USA)).
400
350
300
250
200
175
150
125
100
90
80
70
60
50
45
40
35
30
25
1.5
Pumps 387 8
7
6
Additional (NPSH)R, m
5
4
3
2
1
0 100
150
200
Water Temperature, °C
Figure 16.44 Temperature correction chart for net positive suction head requirement for centrifugal hot water pumps (source: Hydraulic Institute).
where hss = Static suction head, m liquid column (LC) = p ± Z hfs = Friction loss in suction line, m of liquid column (LC) pv = Vapor pressure of liquid at suction temperature expressed in m of liquid column (LC) For existing installation, NPSHA can be determined by
NPSHA = atm pressure + hgs –pv + hvs
(16.27)
where hgs = suction gauge pressure, m of liquid column (LC) pv = vapor pressure of liquid at suction temperature expressed in m of liquid column (LC) As a general guide, NPSHA should preferably be above 3 m for pump capacities up to a flow rate of 100 m3/h and 6 m above this capacity. For a given system, if NPSHA is less than NPSHR, the following remedial measures are recommended as follows: 1. Change the location of the pump to improve NPSHA and increase the positive suction head.
388 Petroleum Refining Design and Applications Handbook Volume 2 2. P rovide jacketed cooling in the suction line to decrease the vapor pressure pv of the liquid. 3. Reduce the operating speed of the pump; thereby reducing the specific speed of the pump and subsequently its NPSHR is less.
16.15 NPSH Requirement for Liquids Saturation With Dissolved Gases There are instances where the liquid to be pumped is saturated with gases, which have definite solubility in the liquid. When a suction system for such a liquid is to be designed for a centrifugal pump, NPSHA calculations differ from Eq. 16.26 or 16.27. Pumping of cooling water (saturated with air), pumping of condensate from a knock-out drum of a compressor, pumping of solution from an absorber, and so on., are examples of situations where the liquids are saturated with gases. Dissolved gases start desorbing when the pump is started and suction is generated at the pump eyes. Generally, a pump can tolerate 2–3% flashed gases at the pump eye without encountering cavitations. If the design of the suction system is made to restrict about 2.5% flash, it is considered safe for the pump operation. In Eq. 16.26, suction source pressure is the system pressure minus the vapor pressure, pv for a normal liquid having practically no dissolved gases. For a liquid saturated with dissolved gases, pv is replaced by pva, which is called artificial liquid vapor pressure. For evaluation of pva, the following procedure is recommended as follows: 1. 2. 3. 4.
alculate molar mass of the gas mixture, dissolved in the liquid. C Calculate mass fraction (wo) of the dissolved gas mixture. Calculate pseudo-critical properties of the dissolved gas mixture, if system pressure is high. Calculate specific volume of the dissolved gas mixture (VGa) at the operating conditions. Steps 1–4 can be avoided if the solubility of the gas mixture in the liquid (such as air in water) is known. 5. Calculate volume fraction of the dissolved gas (GVP) in a hypo-theoretical gas – liquid mixture. Consider one unit mass of the liquid in which the gas mixture is dissolved. If GVP is less than or equal to 2.5%, Eq. 16.26 can be safely used to calculate NPSHA using vapor pressure, pv of the liquid at the operating temperature. If higher then calculate volume fraction a, of flashed gas mixture (as pressure is lowered) over the liquid, saturated with the dissolved gas mixture, using the following [39]:
a=
1 p p 2 p − v 1− v po po po + 1 V p p Ga 1 − VL po po
(16.28)
where p = liquid pressure at pump eye, kPa pv = vapor pressure of liquid at the operating temperature, kPa po = system pressure, kPa VGa = specific volume of the dissolved gas mixture, m3/kg VL = specific volume of the liquid at the operating conditions, m3/kg
Eq. 16.28 assumes that the dissolved gas mixture follows ideal gas law, Dalton’s law and Henry’s law.
6. C alculate a for different value of p. Draw a graph of a vs. p. Read p corresponding to a = 0.025 which is called pva. Alternatively, by trial and error, calculate pva, for a = 0.025. 7. Use Eq. 16.26 and insert pva in place of pv and calculate NPSHA.
Pumps 389
Example 16.12 Carbon dioxide gas (>99.5% pure) from ammonia plant at 1.01 atm a and 40°C is washed with cooling water in a packed column before compression in a plant [28]. Cooling water from bottom of the column is pumped back to cooling tower. Solubility of carbon dioxide in water is 0.0973 kg/100 kg water at 1 atm and 40°C. Find the volume fraction of carbon dioxide over cooling water. Also calculate the artificial liquid vapor pressure.
Solution Molar mass of carbon dioxide, Mw = 44 Vapor pressure of H2O at 40°C pv = 0.07375 bar = 0.0728 atm. Mass fraction of CO2 in the liquid
wo =
0.0973 × 1.01 (100 + 0.0973)
= 9.8177 × 10−4
9.8177 × 10−4 × 0.08206 × 313.15 44.0095 × 1.01 = 5.676 × 10−4 m3/kg H 2O
VGa =
From the Steam Tables,
VL = 1.0078 m3/kg H2O at 40°C 5.676 × 10−4 × 100 (1.0078 + 5.676 × 10−4 ) = 0.0563%
GVP =
Therefore, the volume per cent of carbon dioxide is only 0.0563% (4
Francis-Vane Area D2 D1
= 1.5 to 2
Mixed Flow Area D2 D1
< 1.5
Axial Flow Area D2 D1
Rotation
=1
Figure 16.45 Impeller designs and corresponding specific speed range (by permission from Standards of the Hydraulic Institute, 10th ed.) (also see [18]. Hydraulic Institute, 13th. Ed., 1975).
Pumps 391 600
400
300
200
150
100
80
60
50
40
30
20
30
20
4000
TO TAL SU TT CT OTA ION LS HE UC AD TIO 5F N TT H EA OTA D LS UC TIO NL IFT 15 FT TO TAL SU CT ION LIF T
3500
10
CT SU TO TAL
LIF ION CT SU TO TAL
FT
TO TAL
10
SU
FT
CT
T
ION
TO TAL
LIF
SU
T
CT
2000
25
ION
FT
HE
AD
ZER O
ION
LIF T
5F
2500
20
FT
1500
SPECIFIC SPEED, NS =
RPM
H%
GPM
FOR SINGLE SUCTION OVERHUNG IMPELLER PUMPS
16
FT
3000
1000
900
800
700
400
300
200
150
100
80
60
50
40
H = TOTAL HEAD IN FEET (FIRST STAGE)
Figure 16.46 Upper limits of specific speeds for single suction overhung impeller pumps handling clear water at 85°F at sea level (by permission from Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pumps, Hydraulic Institute, 13th ed. 1975).
392 Petroleum Refining Design and Applications Handbook Volume 2 20000
50
100
40
30
20
15
10
8
7
6
5
4
10
8
7
6
5
4
SUC 15
5 FT
FT T OTA L
TOT AL
SUC
TIO N
TIO N
LIFT
LIFT
EAD NH
SUC
TIO
SUC
5F
15
T TO TAL
FT T OTA L
H% RPM
10000
9000
ZER
O
8000
TIO
NH
EAD
7000
NL
IFT
SUC
TIO
TIO NL IFT
SUC
20
FT T OTA L
10
FT T OTA L
10 5000
SUC
6000
FT T OTA L
SPECIFIC SPEED, NS =
SINGLE SUCTION MIXED FLOW AND AXIAL FLOW PUMPS
GPM
TIO N
HEA
D
15000
4000
3500 100
50
40
30
20
15
H = TOTAL HEAD IN FEET (FIRST STAGE)
Figure 16.47 Upper limits of specific speeds for single suction, mixed, and axial flow pumps handling clear water at 85°F at sea level (by permission from Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pumps, Hydraulic Institute, 13th ed., 1975).
Pumps 393 A typical “operating specific speed” curve is shown in Figure 16.48 and represents a technique for plotting the specific speed on the operating performance curve. Figure 16.48 represents a 6-in. pump operating at 1760 rpm, with maximum efficiency at 1480 gpm and 132 ft head [10]. The operating specific speed is zero at no flow and increases to infinity at the maximum flow of 2270 gpm and zero head. Stable operations beyond about 1600 1700 gpm cannot be planned from such a curve with a sharp cutoff drop for head capacity. “Type specific speed” is defined as that operating specific speed that gives the maximum efficiency for a specific pump and is the number that identifies the pump type [10]. This index number is independent of the rotative speed at which the pump is operating, because any change in speed creates a change in capacity in direct proportion and a change in head that varies as the square of the speed [10]. The specific speed of the pump is reasonably close to the
HS FOR TYPING
SPECIFIC SPEED, NS
4000 3000 2000 D
IC SPEE
SPECIF
1000
180
90
HEAD-CA
PACITY
160
80
140
70 CY IEN
60
EF FIC
100
50
BHP 80
40
60
30
40
20
20
10
0
BHP, P
TOTAL HEAD, H, IN FEET
120
POINT OF MAXIMUM EFFICIENCY
EFFICIENCY, ?, PERCENT
0
0 0
2
4
6
8
10
12
14
14
18
20
22
CAPACITY, IN 100 GPM
Figure 16.48 Typical centrifugal pump characteristic curve with auxiliary specific speed curve. Double-suction, single stage, 6-in. pump, operating at 1760 rpm constant speed (by permission, Karassik, I. and R. Carter, Centrifugal Pumps, McGraw-Hill Book Co., Inc. 1960, p. 197).
394 Petroleum Refining Design and Applications Handbook Volume 2 conditions of maximum efficiency, and Figure 16.45 illustrates the range of typical specific speed index numbers for particular types of impellers.
Example 16.13: “Type Specific Speed” In Figure 16.48, where the pump operates at 1760 rpm (a standard motor speed under load) and has maximum efficiency at 1480 gpm and 132 ft head, the “type” specific speed using Eq. 16.29 is
1760 1480 = 1740 rpm Ns = 0.75 (132)
In metric units,
n Q Ns = 0.75 (gH)
(16.30)
where g = acceleration due to gravity, 9.81 m/s2 H = total head, m n = speed, rps Q = flow rate, m3/s Figure 16.45 indicates the general type of impeller installed. The specific speed of a given pump type must not exceed the specific speed values presented by the Hydraulic Institute [8]. This is based on a known or fixed condition of suction lift, and relates speed, head, and capacity. This index is a valuable guide in establishing the maximum suction lifts and minimum suction heads to avoid cavitation of the impeller with resultant unstable hydraulic performance and physical damage. For a given set of conditions on the suction and discharge of a pump, a slow rotative speed will operate safer at a higher suction lift than a pump of higher rotative speed.
16.17 Rotative Speed The rotative speed of a pump is dependent upon the impeller characteristics, fluid type, NPSHA and other factors for its final determination. The most direct method is by reference to manufacturer’s performance curves. When a seemingly reasonable selection has been made, the effect of this selected speed on the factors such as NPSHR, suction head or lift, fluid erosion and corrosion, and so on must be evaluated. For many systems these factors are of no concern or consequence. Normal electric motor speeds run from the standard induction speeds for direct connection of 3600, 1800, and 1200 rpm to the lower speed standards of the synchronous motors, and then to the somewhat arbitrary speeds established by V-belt or gear drives. For some cases, the pump speed is set by the type of drivers available, such as a gasoline engine. Electric motors in pump application never run at the “standard” rotative design speeds noted above, but rotate at about (with some deviation) 3450, 1750, and 1150 rpm, which are the speeds that most pump manufacturers use for their performance curves. If the higher numbers were used (motor designated or name plate) for pump performance rating, the pumps would not meet the expected performance, because the motors would not be actually rotating fast enough to provide the characteristic performance curves for the specific size of impeller.
Pumps 395
16.18 Pumping Systems and Performance It is important to recognize that a centrifugal pump will operate only along its performance curve [14, 16]. External conditions will adjust themselves, or must be adjusted in order to obtain stable operation. Each pump operates within a system, and the conditions can be anticipated if each component part is properly examined. The system consists of the friction losses of the suction and the discharge piping plus the total static head from suction to final discharge point. Figure 16.49 represents a typical system head curve superimposed on the characteristic curve for a 10 by 8-in. pump with a 12-in. diameter impeller. Depending upon the corrosive or scaling nature of the liquid in the pipe, it may be necessary to take this condition into account as indicated. Likewise, some pump impellers become worn with age due to the erosive action of the seemingly clean fluid and perform as though the impeller were slightly smaller in diameter. In erosive and other critical services this should be considered at the time of pump selection. Considering Figure 16.26 as one situation which might apply to the system curve of Figure 16.49 the total head of this system is:
H = D + hDL ( SL hSL)
(16.31)
The values of friction loss (including entrance, exit losses, pressure drop through heat exchangers, control valves, etc.) are hSL and hDL. The total static head is D − SL, or [(D + D′) − (−SL)] if siphon action is ignored, and (D + D′ ) − ( S′L ) s for worst case, good design practice. 45 10 Hp
15 Hp
40
Efficiency 70
35
75 80
12" Dia. Impeller, New Conditio n
Operating Points with New Impeller, H
Possible Operation of Worn Impeller
30
Total Head in Feet
Size Pump: 10" X 8" Speed: 860 rpm
p d”
25
Operating Points with Worn Impeller
e ip
e ge pip “A w” r e o f “N for rve ve Cu r u Suction plus Discharge Friction tem stem C Sys Sy
20
System Static Head 15
10
5
0
200
400
600
800
1000
1200
Capacity, gpm
Figure 16.49 System head curves for single pump installation.
1400
1600
1800
2000
2200
396 Petroleum Refining Design and Applications Handbook Volume 2 Procedure: 1. C alculate the friction losses hSL and hDL for three or more arbitrarily chosen flow rates, but rates which span the area of interest of the system. 2. Add [hSL + hDL+ (D ± S)] for each value of flow calculated. These are the points for the system head curve. 3. Plot the gpm values versus the points of step 2 above. 4. The intersection of the system curve with the pump impeller characteristic curve is the operating point corresponding to the total head, H. This point will change only if the external system changes. This may be accomplished by adding resistance by partially closing valves, adding control valves, or decreasing resistance by opening valves or making pipe larger, and so on. For the system of Figure 16.26, the total pumping head requirement is
H = (D + hDL) [ SL + ( hSL)] = (D + hDL) + (SL + hSL)
(16.32)
The total static head of the system is [D − ( S)] or (D + S) and the friction loss is still hDL + hSL, which includes the heat exchanger in the system. For a system made up of the suction side as shown in Figure 16.28a and the discharge as shown in Figure 16.29a, the total head is
50 Condition
45 Impeller Operating Head Curve
Friction + 22' Static Head
40 35
Friction + 15' Static Head Operating Points for System
Head in Feet
30 25 20 15 System Friction (See Figure 5.51)
10 5 0 0
10
20
30
40
50
60
70
80
90
System Flow, gpm
Figure 16.50 System head curves for variable static head.
100
110
120
130
140
150
160
Pumps 397
H = D + hDL + P1 [+ S − hSL + P2]
(16.33)
where P2 is used to designate a pressure different than Pl. The static head is [(D + P1) − (S + P2)], and the friction head is hDL + hSL. Figure 16.50 illustrates the importance of examining the system as it is intended to operate, noting that there is a wide variation in static head, and therefore there must be a variation in the friction of the system as the gpm delivered to the tank changes. It is poor and perhaps erroneous design to select a pump which will handle only the average conditions, for example, about 32 ft total head. Many pumps might operate at a higher 70-ft head when selected for a lower gpm value; however, the flow rate might be unacceptable to the process.
Example 16.14: System Head Using Two Different Pipe Sizes in Same Line The system of Figure 16.51 consists of the pump taking suction from an atmospheric tank and 15 ft of 6-in. pipe plus valves and fittings; on the discharge there is 20 ft of 4-in. pipe in series with 75 ft of 3-in. pipe plus a control valve, block valves, fittings, and so on. The pressure of the discharge vessel (bubble cap distillation tower) is 15 psig, with water as the liquid at 40°F in a 6-in. suction pipe (using Cameron Tables—see Chapter 15 Table 15.41). To simplify calculations for greater accuracy, use detailed procedure of Chapter 15. Pipe or fitting loss Loss, ft/100 ft For 15 ft
15.0 ft
Two 90° ells, eq.
22.8
Gate valve, open
3.2
Total
41.0 ft
200 gpm
300 gpm
0.584
1.24
0.24 ft
0.51 ft
The total suction head = hs = +7 − 0.24 = +6.76 at 200 gpm
hs = +7 − 0.51 = +6.49 at 300 gpm Discharge: 4 in. pipe
3 in. pipe
200 gpm
300 gpm
200 gpm
300 gpm
Loss, ft/100 ft
4.29
9.09
16.1
34.1
For 20 ft
20.0
20.0
For 75 ft
75.0
75.0
Two, 3”, 90° ells, eq.
8
8
1.7 ft
1.7 ft
One, 4”, 90° ells, eq.
4.6
4.6
One Gate valve open Total, equivalent ft
24.6
24.6
84.7
84.7
Friction loss, ft fluid
1.06
2.23
13.6
28.8
Control valve at 60% of total, ft
1.59
3.34
20.4
43.2
Total discharge friction loss, ft
2.65
5.57
34.0
72.0
398 Petroleum Refining Design and Applications Handbook Volume 2 15 psig
Bubble cap Distillation Tower
3" – 75 feet
Total Static Head
Atmospheric Vent D = 45' Max. 20' of 4"
Min.
Control Valve
S = 7' 6" 15"
Discrepancy due to Variation in Friction Loss Data
150
100
em yst
ad
He
(Av
lS ota
)
rve
Cu
Head in Feet
T
era
d ge
3" 50
0
0
sD plu
tio
Fric
e)
arg
sch
i n (D
Su
n(
tio
c Fri
on cti
e)
arg
h isc
ischarge)
Friction (D
100
200 Gallons per minute
Figure 16.51 System head using two different pipe sizes in same line.
300
4"
Pumps 399 Total static head = 45 − 7 + 15 (2.31 ft/psi/SpGr) = 72.65ft, SpGr = 1.0 Composite head curve at 200 gpm, head = 72.65 + 0.24 + 2.65 + 34.0 = 109.54 ft at 300 gpm, head = 72.65 + 0.51 + 5.57 + 72.0 = 150.73 ft Total head on pump at 300 gpm: H = 45 + 15(2.31) + 5.57 + 72.0
7 + 0.51 = 150.73 ft
The head at 200 gpm (or any other) is developed in the same manner.
Example 16.15: System Head for Branch Piping With Different Static Lifts The system of Figure 16.52 has branch piping discharging into tanks at different levels [17]. Following the diagram, the friction in the piping from point B to point C is represented by the line B–P–C. At point C, the flow will all go to tank E unless the friction in line C–E exceeds the static lift, b, required to send the first liquid into D. The friction for the flow in line C–E is shown on the friction curve, as is the corresponding friction for flow through C–D. When liquid flows through both C–E and C–D, the combined capacity is the sum of the values of the individual curves read at constant head values, and given on curve (C–E) + (C–D). Note that for correctness the extra static head, b, required to reach tank D is shown with the friction head curves to give the total head above the “reference base.” This base is an arbitrarily but conveniently selected point. The system curves are the summation of the appropriate friction curves plus the static head a required to reach the base point. Note that the suction side friction is represented as a part of B–P–C in this example. It could be handled separately, but must be added in for any total curves. The final total system curve is the friction of (B–P–C) + (C–E) + (C–D) plus the head a. Note that liquid will rise in pipe (C–D) only to the reference base point unless the available head is greater than that required to flow through (C–E), as shown by following curve (B–P–C) + (C–E) + a. At point Y, flow starts in both pipes, at a rate corresponding to the Y value in gpm. The amounts flowing in each pipe under any head conditions can be read from the individual system curves. The principles involved here are typical and may be applied to many other system types.
16.19 Power Requirements for Pumping Through Process Lines A fluid flows of its own accord as long as its energy per unit mass decreases in the direction of flow. Alternatively, it will flow in the opposite direction only if a pump is used to supply energy, and to increase the pressure at the upstream end of the system. From the energy balance equation in Chapter 15, (i.e., neglecting the internal energy and the heat input to the system).
∆P ∆v 2 +α + g∆z + Ws + e f = 0 ρ 2
(16.34)
The work done on unit mass of fluid (−Ws) is:
(− Ws ) =
∆P ∆v 2 +α + g∆z + e f 2 ρ
(16.35)
400 Petroleum Refining Design and Applications Handbook Volume 2 k-feet of L” pipe
D
b Ref. Base
E
a
g-feet of j”
m-feet of n” pipe C
A
B
e-feet of f” pipe
–P
–C
)+
(C
–E
Head Feet
)+
a
System Curves
(B
(B
[(C
–P
–
) –C
+
(C E) +
(C
) –D
+
)] +
–D
l Tota
a
P (B –
)+
–C
a Curves developed at points of equal head when combining individual parts of system
em
Syst
γ b
Friction Curves
a C
C–
–E
(C
D
+ – E)
(C –
D)
B–P–
b
C
Gallons per minute
Figure 16.52 System head for branch piping with different static lifts.
where ΔP = difference in the pressure between points 1 and 2. Δv = difference in the velocity between points and 2. Δz = difference in the distance/elevation between points 1 and 2. (−Ws) = work done on a unit mass of fluid ef = energy dissipated by friction in the fluid per unit mass (including all thermal energy effect losses due to heat transfer or internal generation) between points 1 and 2. α = 1 or 2 for turbulent or laminar flow, respectively.
Pumps 401 The total rate at which energy must be transferred to the fluid is G(−Ws), where G is the mass flow rate, and the power supplied P is
∆P ∆v 2 P = G(− Ws ) = G +α + g∆z + e f = Gh g 2 ρ
(16.36)
where G = mass flow rate, kg/s h = total head, m g = acceleration due to gravity, 9.81 m/s2 The overall power requirement taking into account the pump efficiency, e is
1 P = Gh g e
(16.37)
where e = pump efficiency, fraction.
Hydraulic Power Once the flow and corresponding system resistance have been established, the pumping hydraulic power can be calculated from the following:
hydraulic , hp =
(Q )(H)(SpGr ) 3960
(16.38)
(Q )(∆P) 1714
(16.39)
where Q = flow, gpm. H = total head, ft. SpGr = specific gravity of liquid. or
hydraulic , hp =
where ΔP = differential pressure, lbf/in.2 In metric units The pump hydraulic power output PQ is
where ρ = fluid density, kg/m3 Q = flow, m3/h H = total head, m
PQ =
(ρ)(Q )(H) , kW 367 × 103
(16.40)
402 Petroleum Refining Design and Applications Handbook Volume 2 or
(Q )(∆P) , kW 36
(16.41)
PQ (ρ)(Q )(H) = , kW e 367 e × 103
(16.42)
PQ =
where ΔP = differential pressure, bar. The pump power input
P=
where e = pump efficiency, fraction.
Relations Between Head, Horsepower, Capacity, Speed Brake Horsepower (BHP) Input at Pump
BHP = (Q )(H)
(SpGr ) (3960 e)
(16.43)
where H = total head, ft. Q = flow, gpm SpGr = specific gravity of liquid. e = hydraulic power, hp/Brake horse power. The efficiency, e (fraction) is the ratio of power out to power absorbed. Water or liquid horsepower [10]
WHP = (Q )(H)
(SpGr ) (3960)
(16.44)
The difference between the brake horsepower and the water or liquid horsepower is the pump efficiency. The requirement in either case is the horsepower input to the shaft of the pump. For that reason, the brake horsepower represents the power required by the pump, which must be transmitted from the driver through the drive shaft through any coupling, gearbox, and/or belt drive mechanism to ultimately reach the driven shaft of the pump. Therefore, the losses in transmission from the driver to the pump itself must be added to the input requirement of the driven pump and are not included in the pump’s BHP requirement.
Pump efficiency [8] =
Overall efficiency [8] =
LHP (energy delivered by pump to fluid ) BHP (energy to pump shaft)
(16.45)
WHP (energy delivered by pump to fluid ) EHP (energy supplied to input side of pump ' s driver )
(16.46)
Pumps 403 where EHP = electrical horsepower WHP = liquid horsepower For the rising type characteristic curve, the maximum BHP required to drive the pump over the entire pumping range is expressed as a function of the BHP at the point of maximum efficiency for any particular impeller diameter [18].
BHP (max.) = 1.18 (BHP at max. efficiency point)
(16.47)
Unless specifically identified otherwise, the BHP values read from a manufacturer’s performance curve represent the power only for handling a fluid of viscosity about the same as water and a specific gravity the same as water; i.e., SpGr = 1.0. To obtain actual horsepower for liquids of specific gravity other than 1.0, the curve values must be multiplied by the gravity referenced to water. Viscosity corrections are discussed in another section. Good design must allow for variations in these physical properties.
Example 16.16 3.5 m3/h water at 328 K is pumped through a 2-in. Sch. 40 (ID = 52.5 mm) stainless steel pipe, through a length 200 m in a horizontal direction and up through a vertical height of 20 m. In the pipe configuration, there is a control valve equivalent to 200 pipe diameters and other pipe fittings equivalent to 80 pipe diameters. Also in the line is a heat exchanger across which there is a loss in head of 2.5 m of water. What power must be required from the pump if it is 70% efficient?
Solution Viscosity of water at 328 K: μ = 0.511 × 10−3 Ns/m2 Flow rate = 3.5 m3/h = 9.72 × 10−4 m3/s
Area of flow =
Thus: velocity, v =
πD2 π(52.5 × 10−3 )2 = = 2.16 × 10−3 m 2 4 4 Q 9.72 × 10−4 = = 0.45 m s A 2.16 × 10−3
Reynolds number, Re:
Re =
ρvd µ
(1000)(0.45)(0.0525) (0.511 × 10−3 ) = 46233 (Turbulent) =
Friction factor: Using Chen’s explicit equation (15.35) to calculate the friction factor, fC: Pipe roughness of stainless steel, ε = 0.045 mm
404 Petroleum Refining Design and Applications Handbook Volume 2 Relative roughness is:
ε 0.045 = = 0.00857 D 52.5 ε D 6.7 A= + 3.7 Re
0.9
0.00087 6.7 = + 3.7 46233
0.9
= 5.85895 × 10−4
ε 5.02 = −4 log − log A 3.7D Re fC
1
0.00087 5.02 log10 (5.85895 × 10−4 ) − = −4 log10 46233 3.7 = 12.92815 fD = 4 fC
= 4 (5.9831 × 10−3)
= 0.02393
Equivalent length of pipe = 200 + (280 × 0.0525) = 214.7 m L v2 L v2 L v2 Head loss, h f = 4 fF : = 4fC = fD D 2g D 2g D 2g
214.7 0.452 h f = (0.02393) 0.0525 2 × 9.81
= 1.01 m
Total head to be developed = (1.01 + 10 + 2.5) = 13.51 m m3 kg Mass flow rate of water = 9.72 × 10−4 × 1000 . 3 s m = 0.972 kg/s From Eq. 16.36, the power supplied is: Power supplied = Ghg = (0.972)(13.5)(9.81) = 129 W 100 Power to be required = 129 × = 184 W 70
Pumps 405
Driver Horsepower The driver horsepower must be greater than the calculated (or value read from curves) input BHP to the shaft of the pump. The mechanical losses in the coupling, V-belt, gearbox, or other drive plus the losses in the driver must be accounted for in order that the driver-rated power output will be sufficient to handle the pump. Best practice suggests the application of a non-overloading driver to the pump. Thus a motor rated equal to or greater than the minimum required BHP of the pump, assuming no other power losses, would be non-overloading over the entire pumping range of the impeller. It is important to examine the pump characteristic curve and follow the changes in power requirements before selecting a driver. For example, referring to Figure 16.21a, if your pump were selected with a 6.in. diameter impeller for a rated normal pumping of 100 gpm, the pump would put out about 138 ft of head of any fluid (neglecting viscosity effects for the moment). The intersection of the 100-gpm vertical line with the 6-in. performance curve would indicate that 5.75 BHP would be required for water (between 5 hp and 7.5 hp). Therefore, to be non-overloading (that is, the motor driver will not overheat or lose power) at this condition would require a 7.5 hp motor (if no other losses occur between driver and pump), because there is no standard motor for direct connected service between the standard 5 and 7.5 hp. Now (1) if you know or project that you may need at some time to pump 160 gpm of any fluid with this pump at 160 ft head, then this pump could not be used because it will not physically take an impeller larger than 6.5-in. diameter. However, recognizing this, (2) if you change the external physical piping, valves, and so on, and reduce the head to fit the 6.5-in. impeller curve, at 160 gpm, you could handle 152 ft head (estimated from the curve for a 6.5-in. impeller). This condition would require a BHP from the pump curve between 7.5 and 10, that is, about 9.25 BHP for the pump’s input shaft (for water calculates at 9.03 BHP), estimating the spread between 7.5 and 10. Thus a 10 hp (next standard size motor) would be required, and this would satisfy the original condition and the second condition for water. It would still be satisfactory for any fluid with a specific gravity < 1.0, but if pumping a liquid of 1.28 SpGr (e.g., ethyl chloride), then (1), the original BHP would need to be 1.28 (5.75) = 7.36 BHP, and (2), the second condition would require 1.28(9.25) = 11.84 BHP (calculates 11.56). Whereas, a 10-hp motor would be non-overloading for the water pumping case, it would require a 15-hp (next standard above a 10 hp) motor direct drive to satisfy the ethyl chloride case under the 160 gpm condition. If you do not select a non-overloading motor, and variations in head and/or flow occur, the motor could overheat and stop operating. Study the pump-capacity curve shape to recognize the possible variations. Important note: Any specific pump impeller operating in a physical (mechanical) system will only perform along its operating characteristic curve. If there is a change in the system flow characteristics (rate or friction resistance or pressure head), the performance will be defined by the new conditions and the pump performance will “slide” along its fixed curve. Thus, the designer cannot arbitrarily pick a point and expect the pump to “jump” to that point. Refer to Figure 16.21a. Using a 6-in. impeller curve, for example, the designer cannot make this pump operate at a point of 100 gpm and 150 ft head. This would require about a 6¼-in. diameter impeller. The 6-in. curve will only put out 138 ft (approx.) at the intersection of 100 gpm and the 6-in. curve. A driver selected to just handle the power requirements of the design point (other than maximum) is usually a poor approach to economy. Of course, there are applications where the control system takes care of the possibilities of power overload.
16.20 Affinity Laws The affinity laws can be employed to estimate pump performance at off – design conditions. These relationships are based on the assumptions that pump efficiency is independent of speed n (which is mostly true for speeds greater than 50% of the maximum speed) and impeller diameter d (most true for diameters greater than 80% of the maximum diameter, dmax) and that the original pump design was close to the BEP. As the impeller diameter is reduced below 0.8dmax in the same casing, efficiency at the same speed falls off rapidly. Engineers can use the affinity laws with reasonable confidence to estimate the performance of a pump when the original impeller is trimmed < 15%, e.g., when a 12-in. impeller diameter is trimmed to 11 in. or 10.5 in based on the pump curve provided by the manufacturer at the time of purchase [35].
406 Petroleum Refining Design and Applications Handbook Volume 2 The affinity laws relate the performance of a known pump along its characteristic curve to a new performance curve when the speed is changed. This would represent the same “family” of pump curves. As an example, see Figures 16.21a–c. 1. F or change in speed with a geometrically similar family of fixed impeller design, diameter and efficiency, the following conditions and characteristics change simultaneously [10]:
n Q 2 = Q1 2 n1
n H 2 = H1 2 n1
H2 Q 2 = H1 Q1
Q H 2 = H1 2 Q1
n (BHP)2 = (BHP)1 2 n1
(16.52)
d Q 2 = Q1 2 d1
(16.53)
d H 2 = H1 2 d1
(16.48)
2
(16.49)
2
(16.50) 2
(16.51) 3
for a fixed speed [10]
2
(16.54) 3
d (BHP)2 = (BHP)1 2 d1
(16.55)
For geometrically similar impellers operating at the same specific speed, the affinity laws are [10, 14]: 3
Q2 n 2 d2 = Q1 n1 d1 2
(16.56)
2
H2 n 2 d 2 = H1 n1 d1
(16.57)
Pumps 407 3
5
BHP2 n 2 d 2 = BHP1 n1 d1
(16.58)
where subscript (1) represents the condition for which a set of conditions are known and subscript (2) represents the new non-cavitating or desired condition. There relations do not hold exactly if the ratio of speed change is greater than 1.5 to 2.0, nor do they hold if suction conditions become limiting, such as NPSH. Figure 16.53 illustrates the application of these performance laws to the 1750 rpm curves (capacity, BHP, and efficiency) of a particular pump to arrive at the 1450 rpm and 1150 rpm curves. Note that the key value is the constant efficiency of points (1) and (2). When the speed drops to 1450 rpm, capacity drops:
1450 Q 2 = 204 = 169 gpm 1750
Efficiency vs Capacity
50 40
Efficiency, %
1
2
14
175
0 rp m 50 rp 11 m 50 rp m
60
30 20 100
17
50
90
rp
m
80
14
50
60
m
50 115
40
2 0 rp m
1
30 1750
20
0
40
80
7.5
rpm
1450 rpm 1150 rpm
10 0
Head vs Capacity
1
rp
120
Brake Horsepower vs Capacity
2
160
Capacity, gpm
Figure 16.53 Relation of speed change to pump characteristics.
200
240
280
5 2.5 0
Brake Horsepower
Head in Feet
70
408 Petroleum Refining Design and Applications Handbook Volume 2 The head also drops: 2
1450 H 2 = 64 = 44 fts 1750
and 3
1450 (BHP)2 = 6.75 = 3.84 BHP 1750
d = diameter of impeller (ft, m) n = rotational speed (rpm, l/s) H = output head (ft, m) Q = volumetric discharge (gpm, m3/s) BHP = brake horse power (bhp, kW) 2. For changes (cut-down) in impeller diameter (not design) at fixed efficiency [14]
n d Q 2 = Q1 2 2 n1 d1
n d H 2 = H1 2 2 n1 d1
n d BHP2 = BHP1 2 2 n1 d1
2
(16.59)
2
3
(16.60)
3
(16.61)
where d1 is the original impeller diameter in inches, and subscript (1) represents the condition for which a set of conditions are known and subscript (2) designates the new or desired conditions corresponding to the new impeller diameter d2. All performance changes occur simultaneously when converting from condition (1) to condition (2), no single condition can be true unless related to its corresponding other conditions. An impeller can be cut from one size down to another on a lathe, and provided the change in diameter is not greater than 20%, the conditions of new operation can be described by the type of calculations above. A cut to reach 75–80% of the original diameter may adversely affect performance by greatly lowering the efficiency [12]. Most standard pump curves illustrate the effect of changing impeller diameters on characteristic performance (Figure 16.21a). Note: change as reflected in the different impeller diameters. However, the slight change in efficiency is not recorded over the allowable range of impeller change. Recognizing the flexibility of the affinity laws, it is better to select an original pump impeller diameter that is somewhat larger than required for the range of anticipated performance, and then cut this diameter down after in-service tests to a slightly smaller diameter. This new performance can be predicted in advance. Once the impeller diameter is too small, it cannot be enlarged. The only solution is to order the required large impeller from the manufacturer.
Pumps 409
Example 16.17: Pump Parameters For a centrifugal pump handling water, its manufacturer supplied the following data at 1150 rpm. Pump output = 60 m3/h Head developed = 50 m Power requirement = 12 kW Assuming that density of water is 1000 kg/m3, determine the pump efficiency head, output and power consumed at n = 1400 rpm taking into account that efficiency remains constant.
Solution Using Eq. (16.40), the pump hydraulic power output is
(ρ)(Q )(H) , kW 367 × 103 (1000)(60)(50) PQ = = 8.17 kW 367 × 103 PQ =
Pump efficiency e is
Power sup plied PQ (ρ)(Q )(H) 8.17 = = = Power required P 12 367 × 1003 = 0.68
e=
Head, output and power: Head, H2:
H2 n 2 = H1 n1
2
2
1400 n H2 = H1 2 = 50 1150 n1
2
= 74.1 m
Output, Q2
Q2 n2 = Q1 n1
1400 n Q 2 = Q1 2 = 60 1150 n1 = 73.04 m3/h
410 Petroleum Refining Design and Applications Handbook Volume 2 Power consumed, BHP2
BHP2 n 2 = BHP1 n1
3
3
1400 1400 BHP2 = BHP1 = 12 1150 1150
= 21.65 kW
3
Example 16.18: Specific Speed, Flow Rate, and Power Required by a Pump A centrifugal pump is required to deliver 1600 ton/h of water against a head of 30 m while operating at its maximum operating efficiency. Determine its specific speed if its driven at 20 Hz. If the same pump is operating at maximum efficiency under these conditions delivers at a head of 50 m, what should be its speed and its rate of discharge? Determine the power requirement of the pump if its overall efficiency is 75% while operating under the same operating conditions.
Solution Density of water = 1000 kg/m3 1 ton = 1000 kg. Specific speed by Eq. 16.30
n Q Ns = 0.75 (gH)
Mass rate, G = 1.6 × 106 kg/h Volume flow rate, Q =
Mass Density
Q=
1.6 × 106 = 0.44 m3 s 3 (10 × 3600)
20 0.44 = 0.187 Hz Ns = 0.75 (9.81 × 30)
Speed: 2
or
H2 n 2 = H1 n1
Pumps 411
n 2 H2 = n1 H1
0.5
H n 2 = n1 2 H1
0.5
50 = 20 30
0.5
= 25.82 Hz
Rate of discharge, Q: Also Q α n
Q2 n2 = Q1 n1 25.82 n Q 2 = Q1 2 = 0.44 20.0 n1 = 0.5568 m3/s (2044.8 m3/h )
Power required P is calculated using Eq. 16.40
(1000)(2044.8)(50.0) (0.75 × 367 × 103 ) = 371.4 kW
PQ =
Example 16.19: Ethylene Product Pump Consider an ethylene product pump under the following process conditions [24]: Rated flow, m3/h Rated suction pressure, kPag Discharge pressure, kPag Differential pressure, kPa Specific gravity Head, m (10.2 × ∆P, bar)/Sp.Gr NPSHA, m
= 400 = 1600 = 5700 = 4100 = 0.46 = 909 = 7.2
The specific speed, nq is:
nq =
nQ 0.5 ∆Η 0.75
(16.62)
412 Petroleum Refining Design and Applications Handbook Volume 2 where nq = specific speed Q = volumetric flow rate, m3/s ∆H = head, m n = pump speed, rpm. and in English units,
Ns =
nQ 0.5 ∆Η 0.75
(16.63)
nq =
Ns 51.64
(16.64)
where Q = flow rate, gal/min. ∆H = head, ft n = pump speed, rpm The relationship between nq and Ns is:
First determine the number of stages required assuming that each stage produces the same head and that a specific speed of Ns ≈ 1,100 (nq ≈ 21.3) is desired. Assume a shaft speed of 3,600 rpm, the head per stage is:
n Q H stg = nq
4 /3
0.5 400 3 3, 600 rpm m s 3, 600 = 1100 51.64
4 3
= 216 m stage
The number of stages is calculated by dividing the total head required, 909 m by 216 m/stage and round up to the next highest integer gives 5 stages. Each produces 181.8 m/stage (i.e., head/no. of stages). The specific speed is then:
Ns =
51.64 n Q H3 4
400 m3 51.64 × 3, 600 rpm × 3, 600 s = (181.8)0.75
0.5
(16.65)
= 1252
This is sufficiently close to the desired value of 1100. The size of the impeller can be approximated based on the impeller tip speed:
Pumps 413
H≈
(r Ω)2 2g
(16.66)
r≈
2gH Ω
(16.67)
d≈
2 2gH Ω
(16.68)
or
and
where d = diameter of the impeller, m r = radius of the impeller, m Ω = angular velocity, rad/s
(
2 181.8 m × 2 × 9.81 m s 2 d= 2 π rad 3, 600 × 60 s
)
0.5
= 317 mm
Since an NPSH margin of at least 1 m is desired, the assume NPSHR = 6.2 m The suction specific speed is an index describing the suction capabilities of a first stage impeller and can be calculated using Eq. 16.69. Use half of the flow for double suction impellers. The suction specific speed Nss is:
N ss =
N Q 51.64 n Q = 0.75 NPSHR NPSHR0.75
400 m 3 51.64 × 3, 600 rpm × 3, 600 s = (6.2 m)0.75
0.5
(16.69)
= 15, 772
This is too high; therefore the suction speed for a double-suction style pump is calculated based on one-half of the total flow:
200 m 3 51.64 × 3, 600 rpm × 3, 600 s (6.2 m)0.75
This is acceptable.
0.5
= 11,152
414 Petroleum Refining Design and Applications Handbook Volume 2 A pump supplier identifies a pump with the following specifications as being suitable for this application: number of stages =5 double suction? Yes rated impeller diameter = 318 mm = 5.2m NPSHR suction specific speed, Nss = 11,657 efficiency = 79.2% minimum flow rate = 193.7 m3/h shutoff head = 997.2 m Pumps with high suction speed tend to be susceptible to vibration (which may cause seal and bearing problems), when they are operated at other than design flow rates. As a result, some users restrict suction specific speed and a widely accepted maximum is 11,000. This exercise shows the importance of close coordination between the rotating equipment engineer and the process engineer throughout the design and specification phases of a project. Understanding the concept illustrated in this example and chapter enables the designer/process engineer to take on the task of adequately selecting a pump.
c
13.5 psi
Heater EI.,
48
ft
13
Suction drum
10
psi
ft
5f 15.
El., 1-in 4-in 10
4 ft
ft
2
30
dia 8-in
c
3 ft
Discharge vessel
r ate He
ft
.
1
EI.,
2f
Orifice
t
c
∆P
si .2 p
=5
El.,
4-i
nS
12
ch
1.5
10 EI.,
12
. 40
6- i n
5 ft
Sch 15
ft
ft
ft
4-in Control valve EI.,
t
. 40
ft
ft
ion uct t 5f , 1. n io vat Ele
S 4-in 3-in discharge
c
c Pump
Figure 16.54 Piping and equipment layout for the suction and discharge lines to the process pump (source: Kern, Robert, Chem. Eng., May 26, 1975).
Pumps 415
Example 16.20: Pump Sizing of Gas–Oil A centrifugal pump having a 4 in. nozzle and a 3 in. discharge nozzle will handle gas oil at a normal flow rate of 250 gpm through a piping and component system as described below (source: Kern, Robert, Chem. Eng., May 26, 1975). What is pump power requirement from Figure 16.54? Specific gravity and density are: = 1.18 Specific gravity, S60 = 73.6 lb/ft3. Density of gas oil, ρ60 Vapor pressure at operating temperature = 4.0 psia At flowing conditions, temperature T = 555°F Specific gravity SpGr = 1.04 Density of gas oil ρ = 64.87 lb/ft3 Viscosity of gas oil = 0.6cP The data for the suction and discharge sides are: Suction condition
Discharge condition
Nominal size, in.
6
Nominal size, in.
4
Inside diameter, in.
6.065
Inside diameter, in.
4.026
Pressure at equipment, psig
13.0
Pressure at equipment, psig
13.5
Static head, psi
5.403
Static head, psi
20.94
Pipe length, ft
39
Pipe length, ft
156
5 short radius elbows, ft
75
20 short radius elbows
210
1× Gate valve, ft
6.5
4× Gate valves
18
1 Reducer, ft
4.0
1 Reducer
3
1 Strainer, ft
30.0
1 Inlet
10
1 Inlet to pipe, ft
18.0
2 Exits
40
Total equivalent length, ft
172.5
Line loss ΔP, psi
0.26
437
Line loss ΔP, psi
4.89
1× Control valve, psi
10.71
Exchanger ΔP, psi
5.2
Orifice ΔP, psi
1.52
Solution The Excel spreadsheet (Example 16.20.xlsx) is developed to determine the estimated absolute power requirement for the sizing the gas oil centrifugal pump. Figure 16.55 shows the spreadsheet results of the pump sizing calculation for example 16.20, with an estimated pump efficiency of 70%. The power requirement for the design rate of 275 gpm is 8.64 hp.
416 Petroleum Refining Design and Applications Handbook Volume 2 A.K.C. TECHNOLOGY
PUMP CALCULATION SHEET Document No. Sheet of Item No. (s) No. Working Worki k ng
Job Item Name.
UNITS 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62
Liquid Pumped Corrosion/Erosion Due To Operating Temp. (T) Specific f Gravity at T Viscosity Vapor Pressure at T Normal mass Flowrate Normal Vol. Flowrate Min. Vol. Flowrate Design Vol. Flowrate SUCTION CONDITION Pressure at Equipment (+) Static Head (+/-) Total - Lines 14 + 15 P Suction Line (-) Filter/StrainerP (-)
Gas-oil
Gas-oil
555 1.04 0.6 4 130102.51 250
555 1.04 0.6 4 143111.42
F
m3/h
gpm
barg bar bar bar bar
psi g psi psi psi psi
13 6.3 19.3 0.26
13 6.3 19.3 0.26
bar g
psi g
19.04
19.04
bar g bar
psi g psi
13.5 20.94
13.5 20.94
P
bar
psi
5.2
5.2
P
bar bar bar
psi psi psi
1.52 10.71
1.52 10.71
bar
psi
4.89
4.89
bar g bar m
psi g psi fft.. ft
56.76 37.72 83.78
56.76 37.72 83.78
bar a psi a bar a psi a bar a psi a m fft.. ft m ft. f. ft m f. ft ft. kW k hp % % k hp kW Centrifu f gal Centrifugal
33.74 4 29.74 66.06
33.74 4 29.74 66.06
P
Total Discharge Press (+). Differe f ntial Pressure Differential Differe f ntial Head Differential NPSH Total Suction Pressure Vapor Pressure NPSH= Lines 37 - 38 = Safe f ty Margin Safety NPSH = Lines 40-41 Hydraulic Power Estimated Efficiency Effici f ency Estimated Abs. Power Type of Pump Drive
Total No. off f SKETCH OF PUMP HOOK-UP
o
cP psi a lb/h gpm
Furnace (+) Orifice Orific f e (+) P Control ValveP (+) Line loss (+)
C
CASE II
cP bar a kg/h 3 m /h
Total Suction Pressure (+) DISCHARGE CONDITION Pressure at Equipment (+) Static Head (+/-) Exchanger (+)
o
CASE I
Rev.
275
NOTES 1. Pum Pump ump shutshut-ff u ff head not to exceed……… 2. Relief valve on pump discharge to be set at 3. Pum Pump ump case design pressure pressure………… u ………… design temp m eratu ture…….. temperature……..
5.5 70 7.86
6.05 70 8.64
4. Sealing/flushing Sealing/fl f ushing fl ffluid uid available: 5. Cooling medium available: 6. Insulation required:
Material - Casing - Impeller -Shaft f
7. Start-up/commissioning fl ffluid uid SG.
Sour Service Yes/No m == 10.2 10.2 x bar /SpGr m == 10 10 x kg/cm2 / SpGr ft f = 2.31 x psi / SpGr HEAD m VOLUME m3/h x SpGr x 1000 = kg/h USgpm x SpGr x 600 = lb/h 3 3 2 k kW=m /h x ba kW=m k /h x kg/cm /36.71 POWER Hp = USgpm x psi/1714 3 Date 4 Date 1 Date 2 Date Description Made/Revised by Checked by Approved Process Approved
Figure 16.55 Pump sizing calculation of Example 16.20.
HP=Usgpm*head, ft f *SpGr/3960 5 Date
Pumps 417
Example 16.21: Debutanizer Unit of Example 15.2 What is the required power of pump P1017A for the controlled reflux rate of 1235.52 tons/day (106.56 m3/h) (see Figure 15.23)? Density of condensate, ρ = 488 kg/m3. Vapor pressure at operating temperature = 14.17 bara At flowing conditions, temperature T = 57°C Specific gravity, SpGr = 0.488 Viscosity of condensate = 0.112cP The data for the suction and discharge sides are: Suction condition
Discharge condition
Pressure at Equipment, barg
13.9
Pressure at equipment, barg
13.7
Static head, bar
0.335
Static head, bar
2.057
Line loss ΔP, bar
0.0128
Line loss ΔP, bar
0.447
Filter/Strainer ΔP bar
0
Exchanger ΔP, bar Furnace ΔP, bar Orifice ΔP, bar
Solution The Excel spreadsheet (Example 16.21.xlsx) is developed to determine the estimated absolute power requirement for sizing the reflux LPG centrifugal pump P1017A/B. Figure 16.56 shows the spreadsheet results of the pump sizing calculation for Example 16.21, with an estimated pump efficiency of 70%. The absolute pump power is 8.4 kW.
16.21 Centrifugal Pump Efficiency The design engineer must use the expected pump efficiency provided in the pump performance curve to evaluate the required BHP for a centrifugal pump. In the early stages of the design, it is customary to estimate a value for the efficiency. Final values depend on the specified pump, and at the operating conditions that are encountered. Branan [19] has developed an equation to calculate the centrifugal pump efficiency and pump horsepower. The equation was based on pump efficiency curves of the Natural gas transmitting pipeline safety act (NGPSA) Engineering Data Book. These efficiency curves give good results with the vendor data (the range of developed heads, 50–300 ft, and the flow rates, 100–1000 gpm). The equation for calculating the pump efficiency is expressed as follows:
EFF = (80 − 0.2855H + 3.78 × 10−4 HQ −2.38 × 10−7 HQ 2 + 5.39 × 10−4 H 2
1 −6.39 × 10−7 H 2 Q + 4 × 10−10 H 2 Q 2 ) 100
(16.70)
418 Petroleum Refining Design and Applications Handbook Volume 2 A.K.C. TECHNOLOGY
PUMP CALCULATION SHEET Document No. Sheet of Item No. (s) No. Working
Job Item Name.
UNITS
CASE I
1 2 Liquid Pumped 3 Corrosion/Erosion 4 Due To Operating Temp. (T) Specific Gravity at T Viscosity Vapor Pressure at T Normal mass Flowrate
56 57 58 59 60 61 62
C
F
cP bar a kg/h m3/h
cP psi a lb/h gpm
106.56
m3/h
gpm
barg bar bar bar bar
psi g psi psi psi psi
13.9 0.335 14.235 0.0128 0
13.9 0.335 14.235 0.0128 0
Total Suction Pressure (+) bar g DISCHARGE CONDITION Pressure at Equipment (+) bar g Static Head (+/-) bar
psi g
14.222
14.222
psi g psi
13.7 2.057
13.7 2.057
Design Vol. Flowrate SUCTION CONDITION Pressure at Equipment (+) Static Head (+/-) Total - Lines 14 + 15 P Suction Line (-) Filter/Strainer (-) P
Exchanger (+)
0.488 0.112 14.17 57201
117.216
P
bar
psi
0
0
P Furnace (+) P Orifice (+) Control Valve (+)
bar bar bar
psi psi psi
0 0 0
0 0 0
bar
psi
0.447
0.447
bar g bar m
psi g psi ft.
16.204 1.982 41.43
16.204 1.982 41.43
bar a psi a bar a psi a bar a psi a m ft. m ft. m ft. kW Hp % % kW Hp Centrifugal
15.235 14.17 1.065 22.26
Line loss (+)
SKETCH OF PUMP HOOK-UP
o
57 0.488 0.112 14.17 52000
10 Normal Vol. Flowrate 11 Min. Vol. Flowrate 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55
Total No. off
LPG
o
5 6 7 8 9
CASE II
Rev.
P
P
Total Discharge Press (+). Differential Pressure Differential Head NPSH Total Suction Pressure Vapor Pressure NPSH - Lines 37 - 38 = Safety Margin NPSH - Lines 40-41 Hydraulic Power Estimated Efficiency Estimated Abs. Power Type of Pump Drive
5.87 70 8.39
NOTES 15.235 1. Pump shut-ff head not to exceed……… 14.17 1.065 2. Relief valve on pump discharge to be set at …………… 22.26 3. Pump case design pressure………… de sign temperature…….. 6.45 70 4. Sealing/flushing fluid available: 9.21 5. Cooling medium available: 6. Insulation required:
Material - Casing - Impeller -Shaft
7. Start-up/commissioning fluid SG.
Sour Service Yes/No HEAD m = 10.2 x bar /SpGr m = 10 x kg/cm2 / SpGr ft = 2.31 x psi / SpGr VOLUME m3/h x SpGr x 1000 = kg/h USgpm x SpGr x 600 = lb/h kW=m3/h x bar/36.0kW=m3/h x kg/cm2 /36.71 POWER Hp = USgpm x psi/1714 1 Date 2 Date 3 Date 4 Date Description Made/Revised by Checked by Approved Process Approved
HP=USgpm*head,ft*SpGr/3960 5 Date
Figure 16.56 Pump sizing calculation (metric units) for reflux LPG centrifugal pump of Example 16.21.
Pumps 419 The actual Bhp required for pump operation is:
Bhp = Hp/EFF
(16.71)
where EFF = Pump efficiency (decimal fraction). H = Developed pump head, ft. Q = Flow rate through pump, gpm. Hp = Hydraulic brake horsepower, hp. Bhp = Actual brake horsepower required for pump operation, hp. A system resistance curve for the pump suction and discharge system configuration can be fixed by using the pump affinity laws in Eqs. 16.50 and 16.51, respectively. This procedure does not include any superimposed change in system resistance using methods such as manual valve pinching or control valve throttling. The point at which the system curve and the pump curve (pump head) is the operating point or duty point of the pump. To obtain the pump efficiency, power, and NPSH at the duty point, a vertical line is drawn at this point until it intersects these curves; horizontal lines are drawn to the vertical axis of these curves, which correspond to efficiency, power and NPSH. Using Eqs. 16.70, 16.50, and 16.51, Figure 16.57 shows typical characteristic curves for a 6-in. centrifugal pump operating at 1750 rpm. Eq. 16.70 gives results within 7% of the actual pump curves for the range of applicability (H = 50–300 ft, Q = 100–1000 gpm). For flows in the range of 25–99 gpm, an approximate efficiency can be obtained by solving Eq. 16.71 for 100 gpm and then subtracting 0.35% gpm times the difference between 100 gpm and the low flow rate in gpm. For low flow rates, near 25–30 gpm, this will give results within about 15% for the middle of the head range (pump ΔP) and 25% at the extremes. The horsepower at the 25–30 gpm level is normally below 10 [19]. The Excel spreadsheet program from the Scrivener-Wiley website (Pump-efficiency.xls) has been developed to determine the system curve, pump efficiency, hydraulic Bhp and actual Bhp for pumps at varying flow rate and developed head. Figures 16.57 and 16.58, respectively, show the plots for a 6 in. impeller at 1750 rpm and an 8 in. impeller at 1450 rpm. 180 160
Head, Eff, hp, Bhp, System curve
140 120 Pump Head, ft 100
Pump efficiency, Eff Hydraulic horsepower, hp
80
Brake horsepower, Bhp System curve
60 40 20 0 0
100
200
300
400
500
600
700
800
900
1000
Flow rate, Q (gpm)
Figure 16.57 Pump efficiency calculation at varying flow rate for a 6-in. impeller centrifugal pump at 1750 rpm.
420 Petroleum Refining Design and Applications Handbook Volume 2 200 180 160
Head ft, Eff, hp, Bhp, System curve
140 120
Pump head, H (ft) Pump efficiency, Eff
100
Hydraulic horsepower, hp Brake horsepower, Bhp
80
System curve
60 40 20 0 0
100
200
300
400
500
600
700
800
900
1000
Flow rate, Q (gpm)
Figure 16.58 Pump efficiency calculation at varying flow rate for an 8-in. impeller centrifugal pump at 1450 rpm.
Example 16.22: Reducing Impeller Diameter at Fixed RPM If you have a non-cavitating (sufficient NPSH) operating 9-in. impeller producing 125 gpm at 85 ft total head pumping kerosene of SpGr = 0.8 at 1750 rpm using 6.2 BHP (not motor nameplate), what diameter impeller should be used to make a permanent change to 85 gpm at 60 ft head, at the same speed? By using Eq. 16.53
d 85 = 125 2 9
d2 = 6.1 in. diameter (new) The expected head calculated using Eq. 16.54 is
6.1 H 2 = 85 9
2
= 39.0 ft (must check system new total head to determine if it will satisfy this condition.) The expected brake horsepower BHP calculated using Eq. 16.55 is
Pumps 421
6.1 BHP2 = 6.2 9
3
= 1.93 BHP (use a 2- or 3-hp motor)
16.22 Effects of Viscosity When viscous liquids are handled in centrifugal pumps, the brake horsepower is increased, the head is reduced, and the capacity is reduced as compared to the performance with water. The corrections may be negligible for viscosities in the same order of magnitude as water, but become significant above 10 cSt (10 cP for SpGr = 1.0) for heavy materials. While the calculation methods are acceptably good, for exact performance charts test must be run using the pump in the service. When the performance of a pump handling water is known, the following relations are used to determine the performance with viscous liquids [8]:
Qvis = CQ(Qw)
(16.72)
Hvis = CH(Hw)
(16.73)
Evis = CE(Ew)
(16.74)
BHPvis =
(Q vis )(H vis )(SpGr ) 3960(E vis )
(16.75)
Determine the correction factors from Figures 16.59a–c and Figure 16.58, which are based on water performance because this is the basis of most manufacturer’s performance curves (except, note that the “standard” manufacturer’s performance curves of head vs. gpm reflect the head of any fluid, water, or other non-viscous). Do not extrapolate these curves! Referring to Figure 16.59a–c [8]: 1. Th e values are averaged from tests of conventional single-stage pumps, 2–8 in., with capacity at best efficiency point (BEP) of less than 100 gpm on water performance. 2. Tests use petroleum oils. 3. The values are not exact for any specific pump. Referring to Figure 16.60 [8]: 1. T ests were on smaller pumps, 1-in. and below. 2. The values are not exact for any specific pump. The charts are to be used on Newtonian liquids, but not for gels, slurries, paperstock, or any other non-uniform liquids [8]. Figures 16.59a–c and 16.60 are used to correct the performance to a basis consistent with the conditions of the usual pump curves. In order to use the curves, the following conversions are handy:
422 Petroleum Refining Design and Applications Handbook Volume 2 Centistokes (cSt) = centipoises (cP)/SpGr SSU = Saybolt Seconds Universal = (cSt) (4.620) at 100°F = (cSt) (4.629) at 130°F = (cSt) (4.652) at 210°F
Example 16.23: Pump Performance Correction for Viscous Liquid When the required capacity and head are specified for a viscous liquid, the equivalent capacity when pumping water needs to be determined using Figure 16.59a or 16.60 in order to rate pump selection from manufacturer’s curves. Determine proper pump selection and specifications when pumping oil with SpGr of 0.9 and viscosity of 25 cP at the pumping temperature, if the pump must deliver 125 gpm at 86 ft total head (calculated using the viscous liquid). Viscosity conversion:
cSt = 25/0.9 = 27.8
Referring to Figure 16.59a: 1. E nter capacity at 125 gpm, follow vertically to 86 ft of head, then to right to viscosity of 27.8 cSt, and up to correction factors: Efficiency, CE = 0.80 Capacity, CQ = 0.99 = 0.96 (for 1.0 QN), Head, CH = head at best efficiency point QN Note: This represents a flow rate using water under maximum efficiency conditions [8]. 2. Calculate approximate water capacity:
Qw =
Q vis 125 = = 126.3 C Q 0.99
Hw =
H vis 86 = = 89.6 ft C H 0.96
(16.76)
3. Calculate approximate water head:
4. A pump may now be selected using water as the equivalent fluid with capacity of 126.3 gpm and head of 89.6 ft. The selection should be made at or very near to the point (or region) of peak performance as shown on the manufacturer’s curves. 5. The pump described by the curves of Figure 16.21 fits these requirements. The peak efficiency is 71% using water. 6. Calculate the viscous fluid pumping efficiency:
Evis = Ce(ew) = (0.8)(71) = 56.8% 7. Calculate Brake Horsepower for viscous liquid
(16.74)
Pumps 423
BHPvis = =
Q visH vis (SpGr ) 3960(E vis ) (125)(86)(0.9) = 4.3 BHP 39960(0.568)
(16.75)
Example 16.24: Corrected Performance Curves for Viscosity Effect When a pump performance is defined for water, the corrected performance for a viscous fluid can be developed using Figure 16.59a or 16.60. In order to develop the curves for viscosity conditions of 100 SSU or 1,000 SSU as shown in Figure 16.61, the following general procedure is used [8]. 1. S tarting with performance curve based on pumping water: a) Read the water capacity and head at peak efficiency. This capacity is the value of (1.0 QnW). b) Using this value of gpm, calculate 0.6, 0.8, and 1.2 times this value, giving 0.6, 0.8, and 1.2 QnW, respectively, and read the corresponding heads and water efficiencies. 2. Using Figure 16.59 or 16.60 enter gpm at value corresponding to peak efficiency, 1.0 QnW, and follow up to the corresponding head value, Hw, then move to the viscosity value of the liquid, and up to the correction factors CE, CQ, CH. 3. Repeat step 2 using gpm and head values of step (lb). 4. Correct head values:
Hvis = HwCH
(16.73)
5. Correct efficiency values:
Evis = ewCE
(16.74)
6. Correct capacity values:
Qvis = Qw CQ
(16.72)
7. C alculate the viscous BHP as indicated in the previous example. 8. Plot values as generally indicated on Figure 16.61 and obtain the performance curves corresponding to the viscous liquid conditions.
Example 16.25 Figure 16.62 shows the performance curve of a centrifugal pump operated on water [28]. If the same pump is used for handling oil with a specific gravity of 0.9 and a viscosity of 1000SSU (220cSt) at the pumping temperature, develop the performance curve of the same pump for oil.
Solution Using Figure 16.62 for water, locate the best efficiency point (BEP), it is 170 m3/h. For water with 170m3/h capacity, the pump efficiency is a maximum at 86%. Tabulate capacity, head, and efficiency for 0.6 × 170, 0.8 × 170, and 1.2 × 170 m3/h. From Figure 16.60, at 170 m3/h, the head developed by the pump is 31 m. Using Figure 16.59c, the viscosity correction factor at 170 m3/h capacity is obtained as follows: Draw a vertical line until it intercepts 31 m head. Then a horizontal line along the capacity axis until it intercepts 1000 SSU (220cSt) viscosity parallel lines.
424 Petroleum Refining Design and Applications Handbook Volume 2 PERFORMANCE CORRECTION CHART 100
C H
80 HEAD
0.6 × Qn 0.8 × Qn 1.0 × Qn 1.2 × Qn
60
CORRECTION FACTORS
CQ
80
60
CE
CAPACITY AND EFFICIENCY
100
40
20 00 33 00 22 0 6 17 20 13
8 88 0 66 0 44 0 33
0 22 6 17 2
13
43
88 65
32
20 15
10
CENTISTOKES
HEAD IN FEET
(FIRST STAGE)
600 400 300 200 150 100 80 60 40
10
15
20
0 00 15 0 00 10 0 0 80 00 60 00 40 00 30 00 20 00 15 00 10 0 80 0 60
8
0 40 0 30
6
0
4
0 20 0 15
2
10 80
1
60
VISCOSITY-SSU
40
600 400 300 200 150 100 80 60 40 30 20 15
40
60
80 100
CAPACITY IN 100 GPM
Figure 16.59a Viscosity performance correction chart for centrifugal pumps. Note: Do not extrapolate. For centrifugal pumps only, not for axial or mixed flow. NPSH must be adequate. For Newtonian fluids only. For multistage pumps, use head per stage (by permission from Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pumps, 13th ed., Hydraulic Institute, 1975).
Pumps 425 1.00
1.00 CH
Correction Factors
–0.90
–0.90
–0.80
–0.80
–0.70
–0.70
–0.60
–0.60 CO
–0.50
–0.50
CE
–0.40
–0.40
–0.30
–0.30
–0.20
–0.20
–0.10
–0.10
–0.0
–0.0
0 220 0 176
0 132
880
660
330
440
220
176
88
66
43
32
20
15
10
7.4
4.3
132
Viscosity - Centistokes
120 90
100 80 60 50 40
70 55 45 35
30
25 .5 20 17 5 1 12.5 9 7
10 8 6
5 5 4 3.5 3
2.2 2
00
10,0 8,00 0
120 90 70 55 45 35
0 0 4,00
25 20 7.5 1 15 12.5 10 9 8 1
2.5
2
0 2,00
5 4
0
6 4.5 3.5
3,00
Head, m
6,00
100 80 60 50 40 30
3
0
1,50
0
1,00
80 0
60 0
400
300
200
150
10 0
80
60
50
40
Viscosity - SSU
2.5
3
4
5
6
7
8
Capacity, m3/h
Figure 16.59b Viscosity correction chart (source: Hydraulic Institute, USA).
9
10 11
12 13 14 15
18
20
22
426 Petroleum Refining Design and Applications Handbook Volume 2 100
C H
80 Head
0.6 × QN 0.8 × QN 1.0 × QN 1.2 × QN
60
80
CO
Correction Factors Capacity and Efficiency
100
60 CE
40
20
00 33 00 22 0 6 17 0 2 13 0 88 0 36
0 44 0 33
0 22 6 17 2
88 65
43
32
20 15
10
13
Viscosity - Centistokes
Head, m (first stage)
180 150 125 100 80 60 50 40 28 24 20 16 12
Figure 16.59c Performance correction chart (source: The Hydraulic Institute, USA).
50
0 00 15 00 100 0 800 0 600
Capacity, qv × 10 m3/h
0 400 0 300
8 9 10 11 13 15 17 19 22 25 30 35 40
0 200 0 150 0
7
100 800 600
6
400 300
3 3.5 4 4.5 5
200 150
2.5
100 80
40
Viscosity - SSU
60
180 150 125 100 80 60 50 40 28 24 20 16 12 10 8 6 5
60 70 80 90 110 130 170190220
Pumps 427 1.00
CH
.90 .80
.60 CQ
.50
CE
CORRECTION FACTORS
.70
.40 Reproduced with permission of
.30
INGERSOLL-RAND COMPANY .20 .10 .0 0 220 0 176
0
880
660
440
330
220
132
176
88
66
43
22
15
20
10
7.4
4.3
132
CENTISTOKES
400 300 200 150 100 80 60 40 30 20 15 10 8 6 000
10, 8,0 00
400 300
6,0 00
200 150 4,0 3,0
40 30
00
HEAD IN FEET
00
100 80 60
20 2,0
15
00
10 8 6
80
90
00 1,5
70
00 1,0 800
600
60
400
50
300
200
40
150
30
80
25
100
20
60
15
50
10
40
VISCOSITY-SSU
100
CAPACITY, GALLONS PER MINUTE (at bep)
Figure 16.60 Viscosity performance correction chart for small centrifugal pumps with capacity at best efficiency point of less than 100 gpm (water performance). Note: Do not extrapolate. For small centrifugal pumps only, not for axial or mixed flow. NPSH must be adequate. For Newtonian fluids only. For multistage pumps, used head per stage (by permission from Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pumps, 13th ed., Hydraulic Institute, 1975).
428 Petroleum Refining Design and Applications Handbook Volume 2 200
15 10 5 0
90
180
80
160
Eff. Water
70
140
Eff. 100 S.S.U
60
120
50
100
40 30
Head in Feet
Brake Horsepower
20
Efficiency, %
100
Head, Water Head, 100 S.S.U Head, 1,000 S.S.U. Eff. 1,000 S.S.U
80
BHP 1,000 S.S.U. at 1.0 sp gr BHP 100 S.S.U. at 1.0 sp gr
60
20
40
10
20
0
BHP Water S.S.U. = Saybolt Sec. Univ.
0 0
100
200
300
400
500
Capacity, gpm
Figure 16.61 Typical curves showing the effect on a pump designed for water when pumping viscous fluids (by permission from Pic-aPump, 1959, Allis-Chalmers Mfg. Co.).
30 25 20
Brake Power Water
15 10 5
36
kW
0.90 SP 1000.SSU
0
1000.S
He
SU
30
ad
Wa ter
100
ter Wa
20
80 Efficiency
70 60
1000 SSU
50 40
10
30 20 10 0
0 0
50
100
150 Capacity, m3/h
Figure 16.62 Sample performance curve.
200
250
Efficiency
Head, m
90
Pumps 429 Then draw a vertical line to the correction factors scale on Figure 16.59c, and read the correction factors at CE, CQ and CH (0.6 × QN, 0.8 × QN, 1.0 × QN, and 1.2 × QN), respectively. CE
= 0.645
CQ
= 0.95
CH
= 0.96 (for 0.6 × QN)
CH
= 0.94 (for 0.8 × QN)
CH
= 0.92 (for 1 × QN)
CH
= 0.895 (for 1.2 × QN)
For 0.6 × QNW calculations QNW = 170 m3/h 0.6 × QNW = 0.6 × 170 = 102 m3/h Using 102 m3/h from Figure 16.62, The head Hw (from head vs. capacity curve for water) = 34.62 m The efficiency ηW (from efficiency vs. capacity curve for water) = 0.76 Viscosity of oil, µ = 220 cSt The corresponding capacity, head, and efficiency are:
Qoil = CQ × QW = 0.95 × 102 = 96.9 m3/h
Hoil = CH × HW = 0.96 × 34.62 = 33.235 m
ηoil = CE × ηW = 0.645 × 0.76 = 0.4902 The power required by the pump when it handles the oil is:
P=
Hoil × Q oil × ρoil 33.235 × 96.9 × 900 = = 16.11 kW 5 (3.67 × 10 × ηoil ) (3.67 × 105 × 0.4902)
Table 16.8 shows the results of the calculations for obtaining the performance curve for oil using Figure 16.59c. Microsoft Excel spreadsheet Example 16.25.xlsx gives the calculations of the revised performance of centrifugal pump for handling oil.
Example 16.26 A 20 in. (508 mm, 10-mm wall thickness) commercial steel pipe is used to transport LPG from a refinery to a storage tank 20 km away. Neglecting any difference in elevations, calculate the friction factor and pressure loss due to friction (kPa/km) at 1000 m3/h. Assume an internal pipe roughness of 0.05 mm. A delivery pressure of 2000 kPa must be maintained at the delivery point, and the storage tank is at an elevation of 250 m above that of the refinery. Calculate the pump pressure required at the refinery to transport the given volume of LPG to the storage tank location. Specific gravity of LPG = 0.5 and viscosity = 0.3 cSt.
430 Petroleum Refining Design and Applications Handbook Volume 2 Table 16.8 Revised performance calculations of centrifugal pump for handling oil (using Figure 16.57c). Capacity
0.6 × QNW
0.8 × QNW
1.0 × QNW
1.2 × QNW
Q, m3/h
102
136
170
204
H, m
34.62
33.21
31
27.3
W
0.76
0.84
0.86
0.85
CQ
0.95
0.95
0.95
0.95
CH
0.96
0.94
0.92
0.895
CE
0.645
0.645
0.645
0.645
Qoil = CQ × QW, m3/h
96.9
129.2
161.5
193.8
Hoil = CH × QW, m
33.325
31.217
28.52
24.434
0.4902
0.5418
0.5547
0.5483
16.11
18.26
20.363
21.18
oil
= CE ×
W
P, kW
Solution
ID = 508 − (2 × 10) = 488 mm
Reynolds number, Re = 1, 273, 000
ρq q = 1273 × 106 × dµ dν
where q = flow rate, m3/s ID = pipe internal diameter, mm ν = fluid viscosity, cSt. µ = fluid viscosity, cP ρ = fluid density, kg/m3 µ = cP = 10-3Pa.s The specific gravity, SpGr :
SpGr =
ρLPG ρH2O
The density of LPG, ρLPG = 0.5 × 1000 = 500 kg/m3 µ(cP) The viscosity, υ(cSt ) = ρLPG
µ = υρLPG = 0.3 × 500 = 150 cP
1000 × 500 3600 Re = 1, 273, 000 488 × 150 × 1000 6 = 2.42 × 10
Pumps 431 or
Re = 1273 × 106 ×
1000 1 1 × × 3600 488 0.3
= 2.415 × 106
Therefore, the flow is turbulent. Relative pipe roughness, ε/ID = 0.05/488 = 0.000102 Chen’s friction factor, f
ε 5.02 = −4 log − log A 3.7 D Re fC
1
where
A=
ε ID 6.7 + 3.7 Re
0.9
0.0001 6.7 A= + 2.42 × 106 3.7
0.9
= 3.764 × 10−5
ε 5.02 = −4 log − log A fC 3.7 ID Re
1
0.0001 5.02 − log(3.764 × 10−5 ) = −4 log 6 2.42 × 10 3.7 = 17.76 fC = 0.00316 The Darcy friction factor, fD = 4 × Chen friction factor
fD = 4 fC = 0.01267 The pressure drop per 100 m of pipe, ∆p100 is:
∆p100 = 81055 × 107
f D ρ q 2 bar , (ID)5 100m
1000 0.01267 × 500 3600 = 81055 × 107 (488)5 = 0.01436 bar /100 m
= 1.436 kPa /100 m
2
432 Petroleum Refining Design and Applications Handbook Volume 2 For a one km pipe, the pressure drop is 14.36 kPa. The pressure required at the pumping facility is calculated by adding the pressure drop due to the friction, the delivery pressure required and the static elevation head between the pumping facility and storage tank (all expressed in the same units). Pressure drop due to friction in the 20 km of pipe, Pf = 20 × 14.32 = 286.4 kPa The static head difference is 250 m, therefore the pressure due to elevation, Pelev is:
h(m) × Sp.Gr 10.2 250 × 0.5 = = 12.2549 bar 10.2 = 12225.5 kPa
P( bar ) =
The minimum pressure at the delivery point, Pdel = 2000 kPa Therefore the total pressure required at the refinery PTotal is:
PTotal = Pf + Pelev + Pdel where PTotal = total pressure require at refinery pump Pf = frictional pressure drop Pelev = pressure head due to elevation difference Pdel = delivery pressure at storage tank destination PTotal = 286.4 + 1225.5 + 2000 = 3512 kPa Therefore the pump pressure required at the refinery is 3512 kPa.
Example 16.27 Crude dichlorobenzene is being pumped from a storage tank to a distillation column. The tank is blanketed with nitrogen and the pressure above the liquid surface is held constant at 0.15 barg. The minimum depth of the liquid in the tank is 1.5 m. The distillation column operates at a pressure at 550 Torra. The feed point to that column is 15 m above the base of the tank. The tank and column are connected by a 2 in. (OD = 60.3 mm, 5 mm thickness) diameter commercial steel pipe, 250 m long. The pipe running from that tank to the column contains the following valves and fittings: 25 number of standard radius 90° elbows, two gate valves (fully open), an orifice plate and a flow control valve. If the maximum flow rate required is 25,000 kg/h, calculate the power required by the pump. Pump efficiency is 65%. Pressure drops across the control valve is 0.5 bar. The physical properties of dichlorobenzene are: Viscosity, mPas Density, kg/m3
1.4 1300
Flow meter/valve/elbows 25, Standard radius 90° elbows 2, Gate valves (Fully open) Orifice meter
Equivalent number of velocity heads, K 0.75 0.17 10
Pumps 433
Solution Power required by the pump is:
P=
HQ ρ , kW 3.67 × 105 × e ff
The volumetric flow rate in m3/h is:
Q=
25, 000 m3 = 19.23 1300 h
eff = 0.65 Pump differential head, H is:
H = hd – hs = (p + Zd + hfd) – (p + Zs – hfs) hd = discharge head hs = suction head p = operating pressure in distillation column Zd = static height of discharge hfd = total frictional loss in the discharge line p = pressure at the suction above the liquid surface Zs = static height at suction hfs = total frictional loss in the suction line
1 Torra = 1 mm Hg = 133.32 N/m2 (Pa) The operating pressure in the distillation column, p is:
p = 550 Torra = 73326.0 N/m2 (Pa) In liquid column h (m) is:
h=
10.2 P( bar ) 10.2 × 0.733 = Sp.Gr 1.3
= 5.75 m. LC
Zs = 1.5 m hfs = Assume negligible frictional loss in the suction line ≈ 0 m
h fd =
10.2 ∆PTotal disch arg e ( bar ) ,m Sp.Gr
434 Petroleum Refining Design and Applications Handbook Volume 2 where
∆p ∆PTotal = L + (∆p)cont . valve + (∆p)elbows+gate valve+orifice L frictional loss
The Reynolds number, Re
Re = 354
W 25, 000 = 354 dµ 50.3 × 1.4
= 1.25 × 105 (Turbulent flow)
Relative pipe roughness =
ε 0.045 = = 8.9 × 10−4 d 50.3
Chen friction factor, fC is:
ε 5.02 = −4 log − log A fC 3.7 ID Re
1 where
ε ID 6.7 A= + 3.7 Re
0.9
0.00089 6.7 A= + 1.26 × 105 3.7
0.9
= 3.825 × 10−4
ε 5.02 log A = −4 log − 3.7 D Re fC
1
0.00089 5.02 − log(3.825 × 10−4 ) = −4 log 5 1.26 × 10 3.7 = 13.6967 fC = 0.0053 The Darcy friction factor, fD = 4 × Chen friction factor
fD = 4 fC = 0.0213 The pressure drop per 100 m of pipe, ∆p100 is:
∆p100 = 62530
f D W 2 bar , (ID)5 ρ 100m
62530 × 0.0213 × 25, 0002 (50.3)5 × 1300 = 1.988 bar /100m =
Pumps 435 For 200 m length of pipe, ∆p = 2 × 1.988 = 3.976 bar = 397,600 N/m2 (Pa)
Pipe area, A =
=
π(ID)2 , m2 4
π(0.0503)2 = 0.001987 m 2 4
Fluid velocity, v is:
kg W 25, 000 1 1 h 1 1 = × × × × 2 × ρ A 3600 1300 0.001987 h 3600s kg m 3 m = 2.688 m/s
v=
2
ρv N , 2 m2 Using the 3-K method of Darby to calculate the resistance coefficient K of the fittings: The pressure drop of fittings, ∆p f (fittings) = K Total
0.3 25.4 Kl K f = + K i 1 + K d Re Dn , mm
Fitting types
Number
Kl
nKl
Ki
nKi
Kd
Kf
90° elbows
25
800
20,000
0.071
1.775
4.2
7.686
Gave valve
2
300
600
0.037
0.074
3.9
0.3015
Globe valve
1
1500
1500
1.70
1.70
3.6
6.432
Orifice
10.000
Total
24.4195
2 Velocity head = ρv , N2
2
m
1300 × 2.6882 N = = 4696.47 2 2 m ρv 2 N ∆p f (fittings) = K Total , 2 m2
= 24.419 × 4696.47 = 114685.449
N m2
436 Petroleum Refining Design and Applications Handbook Volume 2 The total pressure drop, ΔpTotal is:
ΔpTotal = (Δp)fric.loss + (Δp)fittings
= 397600 + 114685.449
= 512285.449 N/m2 (5.123 bar) Pressure drop in liquid height is:
h=
= Pump differential head, H is:
10.2 ∆pTotal ( bar ) ,m SpGr 10.2 × 5.123 = 40.195 m 1.3
H = h d − h s = ( p′ + Z d + h fd ) − ( p + Z s − h fs ) = (5.75 + 15 + 40.195)) − (9.127 + 1.5 + 0.) = 50.311 m (LC ) Power =
50.311 × 19.23 × 1300 = 5.27 kW 3.67 × 105 × 0.65
16.23 Temperature Rise and Minimum Flow When a pump operates near shut-off (low flow) capacity and head, or is handling a hot material at suction, it may become overheated and create serious suction as well as mechanical problems. To avoid overheating due to low flow, a minimum rate (gpm) should be recognized as necessary for proper heat dissipation. However, it is not necessarily impossible to operate at near shutoff conditions, provided (1) it does not operate long under these conditions, as temperature rises per minute vary from less than 1°F to 30–40°F, or (2) a by-pass is routed or recycled from the discharge through a cooling arrangement and back to suction to artificially keep a minimum safe flow through the pump while actually withdrawing a quantity below the minimum, yet keeping the flowing temperature down [20]. 1. Temperature rise in average pump during operation [21].
∆Tr =
42.4Pso , °F / min[25] W1c p
(16.77)
where [10] Tr = temperature rise, °F /min. Pso = brake horsepower at shutoff or no flow. W1 = weight of liquid in pump, lb. cp = specific heat of liquid in pump (Btu/lb °F) or, alternate procedure [21, 22]. For low capacity:
∆Tr =
H so (1 − e) 778(c p )(e)
(16.78)
Pumps 437 where Hso = total head of pump at no flow or shutoff or at any flow rate with corresponding efficiency from pump curve, ft. e = pump efficiency at the flow capacity involved (low flow), decimal. Another alternative procedure [16] is as follows:
∆Tr =
(gpm)(H so )(SpGr ) 3960
(16.79)
The following equation is used to estimate the temperature rise through the pump [42]:
∆t = CPshaft
(1 − e) cp q ρ
(16.80)
where ∆t = temperature rise through the pump, °F or °C C = conversion factor: C = 1 for SI units; C = 317 for US units. Pshaft = brake power to the pump, kw or hp e = pump efficiency, evaluated at the flow and head where the pump is operating cp = heat capacity of pumped fluid, Btu/lb °F, kJ/kg °C q = volumetric flow rate, gal/min, m3/s ρ = fluid density, lb/ft3, kg/m3 The absolute minimum flow through the pump is such that the fluid does not vaporize in the pump head. Use Eq. 16.80 to check that the boiling point (at suction pressure) is not reached. See Figure 16.63 and Figure 16.64 for a graphical solution to the equation above for temperature rise. Figure 16.63 illustrates the characteristics of a boiler feed water pump set to handle 500 gpm water at 220°F for a total of 2600 ft head. The temperature rise curve has been superimposed on the performance chart for the pump, and values of Tr are calculated for each flow–head relationship. Note how rapidly the temperature rises at the lower flows. This heating of the fluid at low flow or no flow (discharge valve shut, no liquid flowing through the pump) can be quite rapid and can cause major mechanical problems in the pump’s mechanical components. The maximum temperature rise recommended for any fluid is 15°F (can be a bit higher at times for the average process condition) except when handling cold fluids or using a special pump designed to handle hot fluid, such as a boiler feed water pump of several manufacturers.
∆Tr =
(BHP at shutoff )(42.4 ) , °F/ min ( weight of liquid in pump)(c p )
(16.81)
(BHP − WHP)(2545) , °F/ min ( pump capacity )
(16.82)
or
∆Tr =
where BHP = brake horsepower. WHP = Water or liquid horsepower.
438 Petroleum Refining Design and Applications Handbook Volume 2 36
100
34
45
90 B
40
32
35
30
HEAD CA PACITY
80
800
70
700
28
60
600
26
50
D
EFFICIENCY
15
24
R, sp
E POW ORSE AKE H
A BR
22
10
20
5
18
0
16
40
55
gr 0.9
30
P
ER H WAT
E
400 300
20
200
10
100
C
0 0
500 BRAKE HORSEPOWER
20
EFFICIENCY, PERCENT
WATER TEMPERATURE RISE, °F
25
TOTAL DYNAMIC HEAD, 100 FT
TEMPERATURE RISE 30
100
200
300
400
500
0
600
CAPACITY, GPM
Figure 16.63 Typical temperature rise for boiler feed water pump (by permission from Transamerica Delaval Engineering Handbook, 4th ed., H.J. Welch, ed. 1983. Transamerica Deleval, Inc., IMO Industries, Inc. Div.).
2. Minimum Flow (Estimate) [21] The validity of the method has not been completely established, although it has been used rather widely in setting approximate values for proper operation [16]. For multistage pumps use only the head per stage in temperature limit by this method. a) Determine NPSHA available at pump suction. b) Add the NPSH value to the vapor pressure of the liquid at suction conditions. This represents the vapor pressure corresponding to the temperature of the liquid at the flash point. Read temperature, t2, value from vapor pressure chart of liquid. c) Allowable temperature rise = t2 − (actual pumping temperature). Boiler feed water practice uses 15°F rise for average conditions [16]. d) Approximate minimum safe continuous flow efficiency:
eM =
H so , at shutoff from curve 778(∆Tr )c p + H so , at shutoff
where eM = minimum safe flowing efficiency, overall pump, fraction. Hso = head at no flow or shutoff, ft.
(16.83)
Pumps 439 50 40
PE
R
CE NT
30
CE NT
2
20
PE
R
MAXIMUM RECOMMENDED CE NT R
R
PE
R
PE
R NT
PE R
CE
25
20
PE
R
CE
CE
8
NT
NT
15
10
15
PE
6
R
CE
40
NT
PE
R
CE
NT
30
TEMPERATURE, IN °F
CE
CE
NT CE PE
R
10
NT
NT
PE 5
4
PE
R
CE NT
3
15
CE
NT
PU
M
P
EF
FI
CI
60
EN
PE
CY
3
R
CE
NT
50
PE
4
80
PE
R
CE
NT
70
PE
R
2
1 200
300
500
1,000
2,000
3,000
5,000
TOTAL HEAD, IN FEET
Figure 16.64 Temperature rise in centrifugal pumps in terms of total head and pump efficiency (by permission from Karrasik, I and R. Carter, Centrifugal Pumps, McGraw-Hill Book Co. Inc., 1960, p. 438).
cp = specific heat of liquid, Btu/lb °F. Tr = temperature rise in liquid, °F. e) Read minimum safe flow in gpm from pump performance curve at value of minimum efficiency calculated in (d).
440 Petroleum Refining Design and Applications Handbook Volume 2 The Hydraulic Institute [8] offers guidance for determining the minimum flow through a centrifugal pump: • Temperature rise of the liquid. This is usually established at 50°F (10°C) and results in a very low minimum flow limit. • Radial hydraulic thrust on impellers. This is most serious with single volute pumps, and even at a flow rate that is 50% of Best Efficiency Point (BEP) (see Figure 16.18), could be reduced bearing life, excessive shaft deflection seal failure, impeller rubbing, and shaft breakage. • Flow recirculation in the pump impeller. This can also occur below 50% of the BEP causing noise, vibration and cavitation, and mechanical damage. • Total head characteristic curve. Some curves droop toward shutoff. Operation in such a region should be avoided.
Example 16.28: Maximum Temperature Rise Using Boiler Feed Water Using the example of [23], assume a pump with characteristic curve and added temperature rise data as shown on Figure 16.63 is to handle boiler feed water at 220°F, with a system available NPSHA = 18.8 ft. The vapor pressure of water at 220°F is 17.19 psia from steam tables and the SpGr = 0.957. Correcting the 18.8 ft NPSHA: psia = 18.8 (1/[2.31/0.957]) = 7.79 psia at 220°F. The vapor pressure to which the water may rise before it flashes is 17.19 psia +7.79 psia = 24.98 psia. From steam tables (or fluid vapor pressure tables), read at 24.98 psia (for water of this example), temperature = 240°F. Therefore, allowable temperature rise of the water (this example) = 240° − 220°F = 20°F. A plotted curve as shown on Figure 16.63 [22] shows that at point A, a rise of 20°F on the temperature rise curve corresponds to a minimum of 47 gpm safe for the pump handling 220°F, with NPSHA of 18.8 ft. An alternate estimate for minimum flow [14]: Minimum flow (for water) through pump,
QM = 0.3 Pso, gpm
(16.84)
where Pso = shutoff horsepower For cold liquids, general service can often handle Tr of up to 100°F, a rule with approximately 20% factor of safety:
QM =
6Pso , gpm ∆Tr
(16.85)
Tr = permissible temperature rise, °F
The NPSHR at the higher temperature may become the controlling factor in order to avoid cavitation. The minimum flow simply means that this flow must circulate through the pump casing (not recirculation with no cooling) back to at least the initial temperature of the feed, if excessive temperatures are not to develop. The best practice is to request the manufacturer to state this value for the fluid handled and the calculated NPSHA condition. For NPSHR refer to corrections discussed earlier.
16.24 Centrifugal Pump Specifications Figures 16.65a and 16.65b present specifications and calculation for a centrifugal pump, respectively. Although the process engineer cannot or should not specify each item indicated, he/she must give the pertinent data to allow the pump manufacturer to select a pump and then identify its features. Pumps are selected for performance from
Pumps 441 the specific characteristic curves covering the casing size and impeller style and diameter. Often the process fluids are not well known to the pump manufacturer, therefore the materials of construction, or at least any limitations as to composition, must be specified by the engineer.
Example 16.29: Pump Specifications The pump specified identifies the design data, key portions of the construction materials and driver data as required information for the pump manufacturer (Figures 16.65a and b). If the pump is to be inquired to several manufacturers this is all that is necessary. The individual manufacturers will identify their particular pump selection, details of construction materials, and driver data. From this information a pump can be selected with performance, materials of construction, and driver requirements specified. In the example (Figure 16.65a) the manufacturer has been specified from available performance curves and the details of construction must be obtained. The pump is selected to operate at 22 gpm and 196–200 ft head of fluid, and must also perform at good efficiency at 18 gpm and a head which has not been calculated, but which will be close to 196–200 ft, say about 185 ft. Ordinarily, the pump is rated as shown on the specification sheet. This insures adequate capacity and head at conditions somewhat in excess of normal. In this case, the design gpm was determined by adding 10% to the capacity and allowing for operation at 90% of the rated efficiency. Often this latter condition is not considered, although factors of safety of 20% are not unusual. However, the efficiency must be noted and the increase in horsepower recognized as factors which are mounted onto normal operating conditions. Sometimes the speed of the pump is specified by the purchaser. However, this should not be done unless there is experience to indicate the value of this, such as packing life, corrosion/erosion at high speeds, and suspended particles; as the limitation on speed may prevent the manufacturer from selecting a smaller pump. In some cases it must be recognized that high heads cannot be reached at low speeds in single stage pumps. Table 16.9 presents suggestions for materials of construction for pump parts in the services indicated. The effect of impurities, temperature, analysis variations and many other properties make it important to obtain specific corrosion service data in the specific fluid being pumped. Sometimes this is not possible, and generalized corrosion tables and experience of other users must be relied on as the best information for the selection of materials.
Steps in Pump Sizing In order to size a pump, engineers need to estimate the temperature, density, viscosity, and vapor pressure of the fluid being pumped. Pump sizing can be accomplished as follows [32]: 1. F ind the total dynamic head, which is a function of the four key components of a pumping system, such as suction and discharge elevation; fluid velocity, friction loss and dynamic head; and tank pressure. 2. Correct for the viscosity of the fluid being pumped, since pump charts and data are given for water with a viscosity of 1cP. Viscosity of other process fluids can differ greatly. 3. Calculate the net positive suction head (NPSH) to select a pump that will not undergo cavitation. 4. Check the value of suction specific speed to see if a commercial pump is readily available. 5. Check the potential suitable pumps using a composite performance curve and an individual pump performance curve. 6. Compare the energy consumption and lifecycle cost of operating the selected pump.
16.25 Number of Pumping Units A single pump is the cheapest first-cost installation. However, if downtime has any value such as in lost production, in hazards created in the rest of the process, and so on, then a standby duplicate unit should be considered. A spare or standby can be installed adjacent to the operating unit, and switched into service on very short notice, provided it is properly maintained. Spare pumps which do not operate often should be placed in service on a regular schedule just to be certain they are in working order.
442 Petroleum Refining Design and Applications Handbook Volume 2
Figure 16.65a Centrifugal pump specification.
Pumps 443
Figure 16.65b Centrifugal pump calculation sheet.
Cast iron Misco C
Ni-resist*
Casing; C Steel-rings: C.I.
Cast iron
Misco C
Nickel
Cast iron
Brine (sodium chloride)
Butadiene
Carbon tetrachloride
Caustic, 50% (max. temp. 200oF)
Caustic, 50% (over 200oF)
Caustic, 10% (with some sodium chloride)
23% Cr 52% Ni Stainless steel
Nickel
Impeller: C.I.-rings C. steel
Ni-resist*
Cast iron
Cast iron
Benzene
Cast iron
Impeller and wearing rings
Cast iron
Casing and wearing rings
Ammonia, anhydrous and aqua
Liquid
Table 16.9 Pump materials of construction.
23% Cr 52% Ni stainless steel
Nickel or 18-8 stainless steel
18-8 stainless steel
Carbon steel
23% Cr 52% Ni Stainless steel
Nickel
Misco C
Carbon steel
13% Chrome steel
K Monel
K Monel
Carbon steel
Nickel moly. steel
Carbon steel
Shaft sleeves
Carbon steel
Carbon steel
Shaft
Ring packing
Ring packing
Ring packing
Cast iron
Nickel
Misco C
−
−
Mechanical
Mechanical
Ni-resist**
Cast iron
Ring packing Ring packing
−
Seal cage
Mechanical
Type of seal
−
Nickel
Carbon steel
Mall. iron
Carbon steel
Ni-resist**
Mall. iron
Mall. iron
Gland
(Continued)
Specifications for 50% Caustic (Max. Temp. 200oF) also used.
Misco C manufactured by Michigan Steel Casting Company. 29 Cr-9 Ni Stainless Steel or equal.
*Cast iron acceptable. **Mall. Iron acceptable.
NOTE: Materials of Construction shown will be revised for some jobs.
Remarks
444 Petroleum Refining Design and Applications Handbook Volume 2
Carbon Steel
Bronze
Impregnated carbon
Cast steel
Cast iron
Bronze
Impregnated carbon
Rubber lined C. iron Cast iron
Casing: C. Steel Rings: C.I.
Hard rubber Lined C.I.
Cast Si-iron
Cast iron
Cat iron
Casing, 1–2% Ni, Cr 3–0.5% cast iron rings: Ni-resist, 2B
Ethylene
Ethylene dichloride
Ethylene glycol
Hydrochloric acid, 32%
Hydrochloric acid, 32% (alternate) methyl chloride
Propylene
Sulfuric acid, Below 55%
Sulfuric acid, 55% to 95%
Sulfuric acid, Above 95%
Water, river
Water, sea
K Monel or Alloy 20 SS
K Monel (aged)
Ring packing
Mechanical
Bronze
18-8 Stainless steel
Ring packing
Ring packing
Mechanical
Ring packing
Mechanical
Ring packing
Si-iron
Hastelloy C
Carbon steel
Rubber or plastic
Impregnated carbon
Ring packing
Mechanical
Mechanical
Type of seal
13% Chrome steel bronze
Carbon steel
Type 316 Stn. Stl.
Carbon steel
Carbon steel
Carbon steel
18-8 Stainless steel
18-8 Stainless steel
K Monel
Steel 18-8 Stainless steel
Carbon Steel
Shaft sleeves
Carbon steel
Shaft
Note: Table materials are for general use, specific service experience is preferred when available.
Impeller: Monel Rings: S-Monel
Bronze
Cast iron
Si-iron
Special rubber
Imp.: C.I. rings C. Stl
Hard rubber
Cast iron
Impeller and wearing rings
Casing and wearing rings
Liquid
Table 16.9 Pump materials of construction. (Continued)
Monel or Alloy 20 SS
Cast Iron
Cast iron
Teflon
Special rubber
−
Rubber
−
−
−
Cast iron
Seal cage
Monel or Alloy 20 SS
Mall. iron
Mall. iron
Si-iron
Special rubber
Mall. iron
Rubber
Impregnated carbon
Bronze
K Monel
Mall. iron
Gland
Remarks
Pumps 445
446 Petroleum Refining Design and Applications Handbook Volume 2 If solids are carried in the fluid, this can present a difficult problem if they are not properly flushed from the pump on shutdown. Some spare or second pumps are selected for 100% spare; others are selected so that each of two pumps operate in parallel on 50% of the flow, with each being capable of handling 67–75% of total load if one pump should fall off the line. This then only reduces production by about 25% for a short period, and is acceptable in many situations. These pumps are usually somewhat smaller than the full size spares. When it is necessary to plan several pumps in parallel, the pump manufacturer must be advised, and care must be taken in arranging suction piping for the pumps, otherwise each may not carry its share of the flow. There are many flow conditions, and pumps should be selected to operate as efficiently as possible over the widest range of capacity. If the flow is expected to vary during the system operation, the high and low gpm (and corresponding heads) should be given to allow proper evaluation.
Fluid Conditions The manufacturer must be told the conditions of the liquid, percent suspended solids, physical properties, corrosive nature and maximum and minimum temperature ranges. For extremely hot liquids, special hot pumps must be used, and temperature effects taken into account.
System Conditions The manufacturer must know if the suction side of the pump is associated with vacuum equipment, or is to lift the liquid. This can make a difference as to the type of impeller suction opening he provides. If the system operates intermittently it should be noted. A piping diagram is often helpful in obtaining full benefit of the manufacturer’s special knowledge.
Type of Pump If there is a preference as to horizontal or vertical split casing, it should be stated. Also the suction and discharge connections should be stated as to top or end, or special, together with the preference as to flanged (rating) or screwed. Small pumps are commonly furnished with screwed connections unless otherwise specified.
Type of Driver Pumps are usually driven by electric motors, steam or gas turbine or gas (or gasoline) engines either direct or through V-belts or gears. The pump manufacturer should know the preferred type of drive. If the manufacturer is to furnish the driver, the data on the specification sheet under Driver [see Process Data Sheet (Appendix D)] should be completed as far as applicable. If a gas or gasoline engine is to be used, the type of fuel and its condition must be stated. Engine cooling water (if air not used) must be specified.
Sump Design for Vertical Lift The proper design of sumps for the use of vertical lift pumps or horizontal pumps taking suction from a sump is important to good suction conditions at the pump [23–25]. The arrangement and dimensions indicated in Figures 16.66 and 16.67 are satisfactory for single or multiple pump installations. (For more details, refer to [8]). A few key points in sump-pump relationships for good non-vortexing operation are as follows: 1. A void sudden changes in direction or elevation of flow closer than five bell diameters to pump. 2. Avoid sump openings or projections in water path close to pump. 3. Have water flowing parallel to sump walls as it enters pump. Water should enter pump suction with as low turbulence as possible. 4. Water velocity in sump must be low, 1.5 ft/s is good practice.
Pumps 447
Entering Water < 1 ft/s V=
S
S
Motor
Discharge Pump
Water Level
Submergence, Sb Bottom Level or 7° Max. Grade for 5Bd
Bd/2 Bd Bell Diameter
Water Depth in Pump, /
Bd/3 to Bd/2
Figure 16.66 Sump design. Note: S = (1.5 − 2)Bd.
5. I nlet channel width to each pump is considered optimum to 2 Bd to prevent secondary turbulence effects [24]. 6. Avoid placing several pumps in one open channel removing water in series fashion. If this must be done, velocity at each pump must be kept at same value as for a single pump. The channel width at each pump would be taken from [8]. A suction bell on the inlet of a vertical pump (or the inlet pipe of the suction side of a horizontal pump) is not necessary as far as pump or sump operation is concerned. If a bell is omitted, the entrance losses due to flow will be higher with only a straight pipe, and this must be considered in pump operation. An economic comparison will help decide the value of the bell. Strainers should not be placed on suction bells unless this is the only arrangement. Inlet water should be screened with trash racks, bars, and screens to keep the sump free of debris. Submergence of the inlet pipe column or bell inlet below the water level is necessary for good operation and to prevent vortexes and entrained air. The minimum submergence as recommended by the manufacturer must be maintained at all times. Generally, for 70°F (21.1°C) water, each 1000 ft (304.8 m) of elevation above sea level adds 14 in. (355.6 mm) to the required submergence. If the water is at 100°F (37.8°C) at sea level, approximately 17 in. (431.8 mm) must be added to the 70°F (21.1°C) submergence value [25].
448 Petroleum Refining Design and Applications Handbook Volume 2
Vertical Pumps Water Flow
d = Bd/3
Partition (see Ref. [17])
Figure 16.67 Acceptable sump arrangement for multiple pumps.
16.26 Rotary Pumps There are many different types of positive displacement rotary pumps [26] as illustrated in Figure 16.68 and Figures 16.69a–c. The majority of these types are capable of handling only a clean solution essentially free of solids. The designs using rubber or plastic parts for the pressure device can handle some suspended particles. In general, these pumps handle materials of a wide range of viscosity (up to 500,000 SSU), and can develop quite high pressures (over 1000 psi [69bar]). Additionally, the units can handle some vapor or dissolved gases mixed with the liquid being pumped. The capacity is generally low per unit, and at times, they are used for metering. For specific performance characteristics of any type consult the appropriate manufacturer. These pumps are low in cost, require small space, and are self priming. Some can be rotated in either direction, have close clearances, require overpressure relief protection on discharge due to positive displacement action; and have low volumetric efficiency [27].
Performance Characteristics of Rotary Pumps 1.
low proportional to speed and almost independent of pressure differential. F a) Internal slip reduces efficiency, and increases with pressure and decreasing viscosity. b) Entrained gases reduce liquid capacity and cause pulsations. c) Liquid displacement [21]
d′ =
d ′′(1 − E n ) , ft 3/ min (1 − E n ) + E n ( P P1 )
(16.86)
Pumps 449 Suction
Discharge
Discharge
Gear
Discharge Gear
Suction
Discharge
Suction
Three-Lobe Roter
Internal Gear
Gear
Crescent
External Gear Pump
Rotor Sliding Vanes
Suction
Four-Lobe Pump
Internal Gear Pump
Discharge
Suction
Discharge
Three-Lobe Pump
Driving Gear
Rotor
Sliding Vane Pump
Inlet Discharge Discharge
Discharge
Shaft
Seal Key
Suction Suction
Eccentric
Swinging Vanes
Single Screw Pump
Swinging Vane Pump
Roller
Power Rotor
Discharge Shuttle Block
Shaft
Rotor
Piston
Eccentric Squeeze Ring
Discharge
Three-Screw Pump
Cam-and-Piston Pump
Flexible Rubber Tube
Suction Idle Rotors
Eccentric
Cam or Roller Pump
Suction
Rotor Sleeve
Piston
Fluid Flow
Suction
Shuttle Block Pump
Squeegee Pump
Flexible Vane
Figure 16.68 Rotary pumps (by permission from Dolman, R. E., Chem. Eng., Mar. 1952, p. 159).
where P = the atmospheric pressure, and P1 is the inlet absolute pressure to the pump. dʺ = theoretical displacement, ft3/min dʹ = liquid displacement, ft3/min En = percent entrained gas by volume at atmospheric pressure 2. Volume displaced [8]
Q′ =
D′′ n − S′′ , gpm 231
(for no vapor or gas present)
where Q′ = capacity of rotary pump, fluid plus dissolved gases/entrained gases, at operating conditions, gpm. Dʺ = displacement (theoretical) volume displaced per revolution(s) of driving rotor, in.3/rev n = speed, revolutions per minute of rotor(s), rpm Sʺ = slip, quantity of fluid that leaks through internal clearances of pump per unit time, gpm
(16.87)
450 Petroleum Refining Design and Applications Handbook Volume 2 Disc Diaphragm
FLOW TO 1480 gph, PRESSURES TO 5000 psl
SUCTION
DISCHARGE
Figure 16.69a Diaphragm metering pump, “Pulsa” series. One of several styles/types (by permission from Pulsafeeder Inc.).
Positive-lock thrust control for precision rotor and shaft positioning to maintain original performance and minimize wear for extended pump life.
Choice of leak-resistant packing or standard mechanical seals (illustrated). Special mechanical seals available to fit equipment. Integral safety relief value works to protect system against excessive pressure.
Cushioned, positive-flow action of rotor and idler combination provides non-pulsating, low-shear transmission of liquid.
Figure 16.69b Typical rotary gear pump (by permission from Viking Pump, Inc., Unit of Idex Corp.).
Specially designed and machined revolvable casing provides eight port positions to suit application.
Pumps 451
FULL FLOW
PARTIAL FLOW
NEUTRAL OR NO FLOW
REVERSE FLOW
The liner (grey area) is in full-capacity position with the seal point at the top and the pumping chamber at the bottom. All liquid coming into the pump at the left is moving outt of it at the right.
Now the liner has been rotated counterclockwise, which opens the seal point at the top. This allows part of the liquid to be recirculated, reducing the net flow by a proportionate amount.
Rotating the liner still further, a point is reached where displacement volumes above and below the rotor are equalized. This causes as much liquid to be returned over the top as is brought forward across the bottom, resulting in zero flow.
When the liner is rotated past the “no flow” point, the volume above the rotor exceeds that below, and net flow reverses direction even though pump speed and rotation have not changed. (Limited to approx. 30% of full forward flow on all models.)
Figure 16.69c Sliding vane rotary pump (by permission from Blackmer Pump, Dover Resources Co.).
3. Pump power output (whp) [8]
whp1 =
(Q′Ptd ) 1714
where Ptd = differential pressure between absolute pressures at the outlet and inlet to pump, psi whp1 = power imparted by the pump to the fluid discharged (also liquid hp) Ev = volumetric efficiency, ratio of actual pump capacity to the volume displaced/unit time
(16.88)
452 Petroleum Refining Design and Applications Handbook Volume 2
Ev =
231Q′(100) (D′′n)
(16.89)
4. B hp varies directly with pressure and speed. 5. For speed and pressure constant, Bhp varies directly with viscosity.
Selection Suction and discharge heads are determined the same as for centrifugal pumps. Total head and capacity are used in selecting the proper rotary pump from a manufacturer’s data or curves. Since viscosity is quite important in the selection of these pumps, it is sometimes better to select a larger pump running at low speeds than a smaller pump at high speeds when dealing with viscous materials. As a general guide, speed is reduced 25–35% below rating for each tenfold increase in viscosity above 1000 SSU. Generally, the mechanical efficiency of the pump is decreased 10% for each tenfold increase in viscosity above 1000 SSU, and referenced to a maximum efficiency of 55 % at this point [28].
16.27 Reciprocating Pumps Reciprocating pumps are positive displacement piston units driven by a direct connected steam cylinder or by an external power source connected to the crankshaft of the pump piston. Being positive displacement, these pumps can develop very high pressures (10,000 psi [689.5 bar] and higher) for very low or high capacities (up to 1000 gpm [378.5 l/min]).
Significant Features in Reciprocating Pump Arrangements I. Liquid pump end A. Pump Pressure Component 1. Piston. 2. Plunger. B. Types 1. Simplex, one piston. 2. Duplex, two piston (Figure 16.70). 3. Triplex, three piston (not used as steam driven). C. Piston or plunger action 1. Single acting, one stroke per rpm. 2. Double acting, two strokes per rpm, cylinder fills and discharges each stroke (Figure 16.71). D. Packing for piston or plunger 1. Piston packed: packing mounted on piston and moves with piston; applied to comparatively low pressures. 2. Cylinder packed: packing stationary; plunger moves; applied to high pressures; more expensive than piston packed. II. Drive end: Steam A. Steam Cylinders 1. Simple: single cylinder per cylinder of liquid pump; uses more steam than compound. 2. Tandem Compound; high and low pressure cylinder on same centerline; usually requires 80 psi (5.5 bar) or greater steam to be economical. 3. Cross Compound: high and low pressure cylinder arranged side-by-side with cranks 90° apart. Need for crank and flywheel arrangement only; usually requires 80 psi or greater steam to be economical.
Pumps 453 Removable steam chest cover, permitting quick access to steam valves D type slide valves most simple and reliable
Rigid cast iron cradle of semicircular section, assuring strength and alignment
Stuffing boxes extra deep
Hammered iron piston rings self-adjusting assuring tightness and reducing friction
Disc type valve service
Drop forged steel valve-motion parts
Box type steam pistons
Piston rods divided at crossheads
Removable liners held in place by cylinder heads
Twin liquid cylinders machined in duplex boring mill assuring correct centers
Soft packing or hammered iron piston rings
Liquid pistons removable follower type
Figure 16.70 General service duplex steam-driven piston pump (courtesy of Worthington Corp.). Totally-Enclosed Dust-Proof Oil-Tight Power End Roller Main and Pinion-Shaft Bearings
Double Helical Gears Marine Type Connecting Rods Positive Flood Lubrication Provided by Oil Distributing Pump
Solid Forged Steel Cylinder–no Gaskets Under Discharge Pressure Wing Guided Valves
Figure 16.71 Duplex double-acting plunger pump, power driven (courtesy of Worthington Corp.).
Deep Stuffing Box Equipped for Packing Lubrication Flange and Screw Type Gland – Even Take-up on Packing
454 Petroleum Refining Design and Applications Handbook Volume 2 Percentage gain in compounding steam cylinders varies from 25–35% for non-condensing, and 25–40% for condensing [29]. B. Cylinder action 1. Direct: steam piston direct connected to liquid piston or plunger through piston rod. 2. Crank and Flywheel: flywheel mounted on crank shaft driven by steam cylinder. III. Drive end: Power General features same as steam, except drive always through crankshaft; speed gear increasers or reducers; V-belts, or direct coupling connection to drive shaft. IV. Designation Units are identified as steam cylinder diameter, inches; liquid cylinder diameter, inches; length of stroke, inches.
Application
Piston Type: used for low pressure light duty or intermittent service. Less expensive than the plunger design, but cannot handle gritty liquids.
Plunger Type: used for high pressure heavy duty or continuous service. Suitable for gritty and foreign material service, and more expensive than the piston design.
Performance The performance of reciprocating pumps provides for ease of operation and control. Depending upon the type of piston action, the fluid may be subjected to pulsations unless accumulator or surge drums are provided. The slip of a pump is fraction or percent loss of capacity relative to theoretical. Slip is (1 − evol), where evol is the volumetric efficiency. Volumetric efficiency is the actual liquid pumped (usually considered water) relative to that which should theoretically be pumped based on piston displacement. The NPSHR is approximately 3–5 psi of liquid above the vapor pressure of the liquid. The capacity of a pump is given in manufacturers’ tables as actual, after deducting for volume occupied by piston rod and slippage. Slip varies from 2–10% of displacement, with 3% being a fair average. Capacity: actual, for single acting pumps, single cylinder
Q=
(12a t )(e vol ) = 0.0204 d 2p te vol , gpm (231)(2)
(16.90)
For double acting pumps, single cylinder
Q = ( two times value for single acting ) − 0.0204 d r2 t , gpm
(16.91)
For multiple cylinders, multiply the capacities just obtained by the number of cylinders. If the piston rod does not replace pumping volume as in some arrangements, the last term of the double acting capacity equation is omitted.
Discharge Flow Patterns Figure 16.72 shows the discharge flow patterns for several reciprocating power pump actions which are essentially the same for steam pumps. The variations above and below theoretical mean discharge indicate the magnitude of the pulsations to be expected. Although not shown, the simplex double-acting discharge would follow the action of one piston on the duplex double acting curve from 0 to 360°. Its variation or pulsing is obvious by inspection, and
Pumps 455 QUINTUPLEX SINGLE-ACTING PUMP
Variation Above Mean, 1.8% Variation Below Mean, 5.2% Total Variation,
7.0% 8.5851 0°
72°
144°
216°
288°
360°
SEXTUPLEX SINGLE-ACTING PUMP
A 30572
Variation Above Mean, 4.82% Variation Below Mean, 9.22% Total Variation,
14.04%
0°
60°
120°
180°
300°
240°
360°
TRIPLEX SINGLE-ACTING PUMP Variation Above Mean, 6.1% Variation Below Mean, 16.9% Total Variation,
8.5852
23.0% 0°
120°
360°
240°
QUADRUPLEX SINGLE-ACTING PUMP
Variation Above Mean, 11.0% Variation Below Mean, 21.5% Total Variation,
32.5%
0°
90°
180°
270°
360°
DUPLEX DOUBLE-ACTING PUMP FORWARD STROKE
RETURN STROKE
Variation Above Mean, 24.1% Variation Below Mean, 21.5% Total Variation,
45.6%
8.8853
0°
90°
180°
Figure 16.72 Reciprocating pump discharge flow pattern (courtesy of the Aldrich Pump Co.).
270°
360°
456 Petroleum Refining Design and Applications Handbook Volume 2 accumulator bottles would be required to smooth the flow. The simplex single acting discharge would be one pumping stroke from 0 to 180°, then no pumping from 180° to 360°; and here again the pulse action is obvious.
Horsepower Hydraulic
HHP =
(Q actual )(H) 3960
(16.92)
HHP e
(16.93)
Brake
BHP =
where e represents the total overall efficiency, and is
e = em(evol),
and em is the mechanical efficiency and evol is the volumetric efficiency, fraction. Mechanical efficiencies of steam pumps vary with the types of pump, stroke and the pressure differential. Some representative values are 55–80% for piston pumps with strokes of 3 and 24 in., and pressure differential up to 300 psi. For the same strokes a plunger design varies from 50 to 78%, and at over 300 psi differential the efficiencies are 41.67% [30]. Steam required is approximately 120 lb/h per BHP.
16.28 Pump Selection Reciprocating pump selection follows the fundamentals of centrifugal pumps: 1. 2. 3. 4. 5. 6. 7.
valuate suction side head loss. E Evaluate discharge side head loss. Determine system static pressure. Determine total differential head across pump. Determine the NPSHA available on suction of pump. From manufacturer’s performance tables, select pump nearest to gpm and head requirements. Contact manufacturer for final recommendations, give complete system requirements, and physical properties of liquid. Figure 16.73 will serve this purpose.
16.29 Selection Rules-of-Thumb Every pump has a specific curve that relates flow, head, power, NPSHR, and efficiency for specific impeller diameters for that particular unit. This allows correct selection of the impeller diameter. Therefore, during specification, the objective is to select a pump with a rated or design point as close as possible to the best efficiency point (BEP), as determined by the pump manufacturer [31]. The following are general guidelines for proper selection [31]: 1. S elect the pump based on rated conditions. 2. The BEP should be between the rated point and the normal operating point.
Pumps 457
Figure 16.73 Horizontal direct-acting steam pump or power pump.
458 Petroleum Refining Design and Applications Handbook Volume 2 3. Th e head/capacity characteristic curve should continuously rise as flow is reduced to shutoff (or zero flow). 4. The pump should be capable of a head increase at rated conditions by installing a larger impeller. 5. The pump should not be operated below the manufacturer’s minimum continuous flow rate. The pump has a specific NPSHR, which varies, depending on the head and flow. Once the specific pump model and size have been determined from the basic process information, the materials of construction must be chosen. Selection depends on fluid properties (e.g., viscosity, corrosiveness, and erosiveness) and the presence of dissolved gases. In general, adequate knowledge of the chemical composition of the fluid helps to ensure proper material selection of the pump and its shaft seal. The following guide to pump types provides a better understanding and specifications for the selection of pumps [37]: • American Petroleum Institute (API) process pumps: Designed to meet the 610 Standard set by the API. • Boiler feed pumps: Built to control the amount of water that enters a boiler. They are centrifugal pumps and most are multistage. • Chemical pumps: Build to handle abrasive and corrosive industrial materials. They can be either centrifugal or positive displacement type. • Circulatory pumps: Used to circulate fluid through a closed or looped system. They are usually centrifugal pumps, but a few use positive displacement technologies. • Dewatering pumps: A dewatering process involves using a centrifugal pump (submersible or vertical turbine) to remove water from a construction site, pond, mine shaft or any other area. • Fire pumps: A type of centrifugal pump used for firefighting. They are generally horizontal split case, end suction or vertical turbine. • High-pressure pumps: Used in many applications including water blast, hydromining, and jet cutting. They can be a wide variety of pump types including positive displacement pumps, rotary pumps, and reciprocating pumps or centrifugal pumps. • Industrial pumps: Used in industrial applications such as slurry, wastewater, industrial chemicals, oil and gas. There are dozens of different industrial pumps both in positive displacement and centrifugal pump types. • Marine pumps: Built to pump seawater. They are often used in large saltwater tanks to continuously circulated water so it stays fresh. • Mixed flow pumps: Incorporate the features of both axial flow and radial flow pumps. Axial flow pumps operate on a vertical plane and radial flow pumps operate on a horizontal plane to the flow direction of water. • Mud pumps: Built to transfer heavy sludge or mud. They are sometimes used on oil rigs to pressure and circulate fluid. • Petrochemical pumps: Made to transfer petroleum products that are often very viscous and corrosive. They can be magnetic drive pumps, diaphragm pumps, piston pumps, and so on. • Pneumatic pumps: Use compressed air to pressure liquid through the piping system. • Pressure pumps: Used to create either high or low pressure. They can be metering pumps and sometimes booster pumps. • Process pumps: Are many times centrifugal pumps or positive displacement pumps used in process applications. The type of pump and construction details varies depending on the application in which these pumps are used. • Slurry pumps: A heavy-duty pump that is made to handle thick, abrasive slurries. They are made of durable materials and capable of handling abrasive fluids for long periods of time. • Solar pumps: Powered by the sun. They can be positive displacement or centrifugal pumps. • Water pumps: A type of equipment used to move water through a piping system. They rely upon principles of displacement, gravity, suction and vacuums to move water. They can be either positive displacement or centrifugal pumps. • Well pumps: Designed to draw water to the surface from an underground water source. Depending on the well depth and configuration, these pump types can be jet, centrifugal or submersible.
Pumps 459 In general, the final pump selection is influenced by several factors: • • • • •
Pump capacity (size) that is a function of the flow rate to be pumped. Fluid properties, both physical and chemical. Operating conditions. Type of power supply. Type of flow distribution.
16.30 Case Studies Case Study 1 Pump Simulation on a PFD A pump operation is used to increase the pressure of a fluid stream that is flowing from one process unit to another in a process flowsheet. Power (energy/time) in the form of electric energy drives a motor coupled to a steel drive shaft. The drive shaft connected to impellers imparts energy to the liquid in order to increase its pressure. The temperature of the liquid increases slightly, because of the effects of fluid friction. The conceptual model for the pump operation is given below for a steady state system. The system is a fluid mixture of chemical compounds (or components) passing into, through and from the pump. The mathematical model for the pump operation balances the material and energy flows of the system. This adiabatic unit operation occurs at steady state with no chemical reaction, and the kinetic and potential energy changes are negligible. The fluid is considered incompressible (i.e., at constant density); a good assumption for any fluid well removed from its critical point. The independent set of equations in the mathematical model contains the total and component material balances, the energy balance, the molar enthalpies of the two process streams, the adiabatic efficiency, the ideal work based on the mechanical-energy balance for a frictionless fluid, the pressure change, and the inlet mixture density and molecular weight. The adiabatic efficiency relates the ideal to the actual work and has a typical value of 75 % for most liquids. To solve this set of equations (nc + 5) variables must be specified, as indicated by the degrees-of-freedom analysis (DOF) in the model [33]. From this mathematical model, many mathematical algorithms can be derived for performing the process simulation calculations. These algorithms differ in their given variables and their solution procedures. To such algorithms are shown below for knowing the process state of the inlet stream and two additional variables. The unknown variables are calculated using the solution procedure define in the algorithm. The process state of material stream is its temperature, pressure, total flow rate and composition. Other possible simulation algorithms supported are summarized below.
Conceptual model
Model assumptions
WA Exit TI PI nI ZI
Inlet Pump
TE PE nE ZE
1. 2. 3. 4. 5. 6.
Continuous process Steady state No chemical reaction Negligible KE and PE changes Adiabatic Incompressible fluid
460 Petroleum Refining Design and Applications Handbook Volume 2
Mathematical Model 1
nI – nE = 0
2
nIZI,j – nEZE,j = 0
3
nI. HI – nE HE + WA= 0
4
Hl = hmix [TI, PI, ZI]
5
HE = hmix [TE, PE, ZE]
6
ε = 100. WI/WA
7
WI = ΔP.nI.MI/ρI
8
ΔP = PE − PI
9
ρI = liqden [TI, PI, ZI]
for j = 1, 2,…., nc
10 MI = mol wt [ ZI] vars = 2.nc + 14 eqns = nc + 9 DOF = 1.nc + 5
Variables Descriptions Ti Pi ni nc Zi zi,j Hi WA ε WI ΔP Mi ρi
Temperature of process stream, i, °C Pressure of process stream, i, kPa Bulk molar flow rate of process stream i, kg mol/h Number of chemical components or compounds in the mixture Bulk mole fractions of all nc-components or compounds in the mixture Bulk mole fractions of component j in process stream i; vector Zi means all elements zi,1, zi, 2,…….zi, nc Bulk molar enthalpy of process stream i, kJ/kg mol The actual work or power of the pump, kJ/h The adiabatic efficiency of the pump (0 to 100) percent The ideal work or power of the pump (ε = 100%), kJ/h The pressure drop between the exit and inlet streams, kPa Molecular weight of process stream i, kg/kg mol Liquid density of process stream i, kg/m3
Simulation Algorithm If the process state of the inlet stream is fully defined (i.e., TI, PI, nI, ZI are known) only two additional variables are required to calculate all unknowns as shown in the simulation algorithms below [33]:
Pumps 461 [TE, ∆P, nE, ZE, WA] [TE, PE, nE, ZE, WA] [TE, PE, nE, ZE, ∆P]
= = =
Pump A[TI PInI ZI, ε, PE] Pump B [TI, PI, nI, ZI, ε, PE] Pump C [TI, PI, nI, ZI, ε, WA]
[PE, nE, ZE, WA, ε] [∆P , nE, ZE, WA, ε]
= =
Pump D[TI, PI, nI, ZI, TE, ∆P] Pump E [TI, PI, nI, ZI, TE, PE]
[TE, PE, nE, ZE, ε] [TE, ∆P, nE, ZE, ε]
= =
Pump F [TI, PI, nI, ZI, WA, ∆P] Pump G [TI, PI, nI, ZI, WA, PE]
If the process state of the exit stream is fully defined (i.e., TE, PE, nE, ZE are known), only two additional variables are required to calculate all unknowns as shown in the UniSim simulation algorithms: [TI, PI, nI, ZI, WA] [TI, ∆P, nI, ZI, WA] [TI, PI, nI, ZI, ∆P]
= = =
Pump H[TE, PE, nE, ZE, ε, ∆P] Pump I [TE , PE, nE, ZE, ε, PI] Pump J [TE, PE , nE, ZE, ε, WA]
[PI, nI, ZI, WA, ε] [∆P , nI, ZI, WA, ε]
= =
Pump K [TE, PE, nE, ZE, TI, ∆P] Pump L [TE, PI, nE, ZE, TI, PI]
[TI, PI, nI, ZI, ε] [TI, ∆P, nI, ZI, ε]
= =
Pump M [TE, PE, nE, ZE, WA, ∆P] Pump N [TE, PE, nE, ZE, WA, PI]
Problem An equimolar mixture of n-hexane and n-octane at 25°C is pumped via P-100 from 1 atm to 4 atm. The liquid mixture is flowing at 100 lb-moles/h and its adiabatic efficiency is 70%. The conceptual diagram is shown in Figure 16.74. WA = ?
Exit
TE = ? PE = 4 atm nE = ? ZE, HX = ?
TI = 25°C PI = 1 atm nI = 100lb mol/h
ZE, OC = ? Hydroc. Feed
ZI, HX = 0.5
P- 100
ZI, OC = 0.5 ε = 70%
Figure 16.74 A conceptual diagram of P-100 with the variables.
462 Petroleum Refining Design and Applications Handbook Volume 2 The following procedures in constructing the simulation using UniSim software are as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
Start the UniSim software, load your preferences, and open your file Pump-Simulation-akc.usc. Enter the Basic Environment, complete the component list and note the fluid package. Return to the Simulation Environment, open the process stream and rename it HydrocFeed. Specify its process state; i.e., its temperature, pressure, flow rate, and composition. Add a pump to the process flow diagram (PFD) and name it P-100 Connect its Inlet stream, name its Exit material stream, and name its Wa energy stream. Specify its adiabatic efficiency to be 70% and its exit pressure to be 4 atm. Open the Workbook window using its icon in the UniSim button bar View the values in the Material Streams page of the Workbook window. Note that specified values appear in blue, while calculated values appear in black. View the Compositions, Component Flows, and Energy Streams pages in the Workbook.
After specifying the state of the inlet stream, the pump efficiency and the exit pressure, UniSim immediately calculates all of the other properties of the material streams (such as mass flow rate, volumetric flow rate, vapor fraction, heat flow, etc), using the Peng–Robinson (PR) equation of state. Also the pump power is calculated to be 804.4 W. A positive value for the pump power shows that energy must be added to the process stream to increase its pressure.
Discussion Figures 16.75 and 16.78 show process flow diagrams of the simulation exercise with an equimolar mixture of n-heptane and n-octane and water, respectively. Pumping of fluids increases their temperature and Figure 16.75 shows the pump was 70% efficient, and the outlet temperature is 25.15°C. Correspondingly, Figure 16.78 shows the pump was 75% efficient, and the outlet temperature is 25.02°C, which is negligible. Generally, the less efficient a pump, the greater the increase in the temperature of the fluid being pumped. This arises because in a low efficient pump, more energy is required to pump the fluid to obtain the same outlet pressure of a more efficient pump as the extra energy is transferred to the fluid. Figures 16.76 and 16.77 show plots of head vs. flow rate and efficiency vs. flow rate of the centrifugal pump respectively, and Figure 16.78 illustrates the process flow diagram of the pump for case study 1.
Exit Hydrc. Feed Temperature 25.00 C 101.3 kPa Pressure Molar Flow 45.36 kgmole/h
P-100 Exit
Hydrc. Feed
Temperature Pressure Molar Flow
25.15 C 405.3 kPa 45.36 kgmole/h
Wa P-100 Speed Energy Actual Vol. Flow Feed Pressure Product Pressure Product Temperature
3500 2896 6.668 101.3 405.3 25.15
rpm kJ/h m3/h kPa kPa C
Figure 16.75 PFD of a centrifugal pump (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
Pumps 463 Head Curves 3500 Pump Curve-1
3000
CummOperating Points Pump Curve-2 Pump Curve-3
2500
Head (m)
2000
1500
1000
500.0
0.0000 0.0000
2.000
4.000
6.000
8.000
10.00
12.00
14.00
16.00
18.00
20.00
Flow (m3/h)
Figure 16.76 Head vs. flow rate of a centrifugal pump (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
Efficiency Curves
70.00 Pump Curve-1 CummOperating Points
69.00
Pump Curve-2 Pump Curve-3
68.00
Efficiency
67.00
66.00
65.00
64.00
63.00 0.0000
2.000
4.000
6.000
8.000
10.00
12.00
14.00
16.00
18.00
20.00
Flow (m3/h)
Figure 16.77 Efficiency vs. flow rate of a centrifugal pump (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
464 Petroleum Refining Design and Applications Handbook Volume 2
S3 Temperature 25.02 C 356.3 kPa Pressure Molar Flow 443.3 kgmole/h Pipe-100 S1
PIPESYS-UniSim
S1 Temperature 25.00 C Pressure 202.6 kPa Molar Flow 444.1 kgmole/h
Pipe-101 PIPESYS-UniSim
Pump-100 S3
S2
Q-101
E-100 Q-100
S2 Temperature 25.00 C 202.6 kPa Pressure Molar Flow 443.3 kgmole/h
Pump-100 Speed
Energy 1642 Actual Vol. Flow 8.011 Feed Pressure 202.6 Product Pressure 356.3 Product Temperature 25.02
rpm kJ/h m3/h kPa kPa C
S4 S4 Temperature 25.02 C 353.9 kPa Pressure Molar Flow 443.3 kgmole/h
Figure 16.78 PFD of a centrifugal pump (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
The UniSim Design R460.2 simulation software (Pump Simulation-akc.usc) shows the results of case study 1. Figures 16.79 to 16.81 show the simulation results of PFD of Figure 16.75.
Case Study 2 Figure 16.82 shows a pumping network of 800 U.S gpm of water from a supply tank at atmospheric pressure (0 psig) having the suction size of 4 in. of carbon steel (CS) to the inlet of an N-S centrifugal pump. The discharge side of 3-in. from the pump is connected to a recycle line to the supply tank via pipe and fittings and a restriction orifice. The other discharge line consists of a filter, and a tee fitting in directing the flow to two tanks at the north and south of HGF-1 building. The total dynamic head is 245.2 ft. Figure 16.83 shows the piping isometric with results using Darcy–Weisbach method in calculating the pressure drop in the various sections of the piping segments, the filter, and the throttle valves. Figure 16.84 shows the pump characteristics with 8.125 in. impeller and at a Best Efficiency Point (BEP) of 70.6% corresponding to 710 US gpm. From Figure 16.85, the pump head is 261 ft, power is 66.5 hp, and NPSHR is 23.3 ft. Figure 16.85 shows the results from the calculations and gives the shutoff head of 295 ft. UniSim Design R471 (Pipeline-deltaP.usc) shows the simulation of Case Study 2.
16.31 Pump Cavitations We shall review some studies of pump cavitations in the process plants. Pumps cavitate for three reasons [27]. 1. L acking sufficient NPSHA to satisfy the conversion of pressure to velocity in the eye of the impeller (running NPSH). 2. Lacking sufficient NPSHA to satisfy the conversion of pressure to acceleration in the suction line as the pump is started (starting NPSH). 3. Lacking sufficient NPSHA to overcome the frictional losses in the suction piping and the drain or draw nozzle. 4. Partial plugging of drawoff nozzle. The case study is due to item 4 as illustrated from the experience of Lieberman. Figure 16.86 shows the side draw off from a fractionator. Slowly opening the pump’s discharge control valve increases flow up to a point. Beyond this
Pumps 465
Figure 16.79 Simulation results of centrifugal pump P-100 (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
466 Petroleum Refining Design and Applications Handbook Volume 2
Figure 16.80 Simulation results of centrifugal pump P-100 (Continued) (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
Pumps 467
Figure 16.81 Simulation results of centrifugal pump P-100 (Continued) (UniSim Design R443, Honeywell® and UniSim® are registered trademarks for Honeywell Inc. All rights reserved).
468 Petroleum Refining Design and Applications Handbook Volume 2 Building HGF-1
CS-05 Flow: 500 US gpm
Recirc 2
Recirc 1
CS-01 Flow: 800 US gpm
Supply Tank P Set: 0 psi g Level: 10 ft CS-04 Flow: 800 US gpm
Filter Flow: 800 US gpm dP: 10 psi
CS-02 CS-03 Flow: 800 US gpm N-01 Flow: 800 US gpm P: 116.6 psi g
N-02 P: 68.85 psi g CS-06 Flow: 300 US gpm
Recirc Orifice
CS-07 N-Throttle Valve Flow: 500 US gpm North Tank FCV @ 500 US gpm P Set: 20 psi g dP: 20.81 psi Level: 5 ft
N-S Pump TH: (245.2) ft
CS-08 Flow: 300 US gpm S-Throttle Valve South Tank FCV @ 300 US gpm P Set: 25 psi g dP: 14.33 psi Level: 3.002 ft
Lineup: System: FOF US tutorial Date: 28/03/18 9:42 pm Company: A.K.C TECHNOLOGY Project: XYZ PROJECT by: A.K. Coker
Darcy-Weisbach Flow of Fluids Premium 2009 Flow: US gpm Pressure: psi g Size: in Elevation: ft Velocity: ft/sec Length: ft Volume: gallons
Figure 16.82 Process flow diagram of the piping network from the supply tank via N-S centrifugal pump to the north and south tanks (source: Engineered Software, Inc.).
point, the pump’s discharge pressure and discharge flow become erratically low. This implies that the pump is cavitating. The fluid being pumped is hot water, and at the desired flow rate of 110 gpm, the manufacturer’s pump curve shows that the pump requires 14ft. NPSH. The elevation difference between the draw – off nozzle and the suction of the pump is 46 ft. By reducing the flow of water by 10% to 100 gpm, the cavitation stops. With a pressure gauge on the suction of the pump, and assuming that the suction line is full of 46 ft. H2O, the suction pressure is:
46 ft + 30 psig = 50 psig 2.31 ft psi
Assuming that SpGr of H2O = 1.0 However, the observed pressure is not 50 psig; it is only 47 psig indicating that 3 psig of 7 ft liquid is missing. i.e.,
(50 psig − 47 psig) × 2.31 ft/psig = 7 ft
The likely explanation for this head loss of 7 ft is frictional loss in the suction line. This reduces the NPSHA from 46 to 39 ft. But this is still a lot more NPSHA than the 14ft. of the NPSHR needed to pump 110 gpm. In opening the discharge flow-control valve sufficient to increase the flow from 100 to 110 gpm or by 10%, this would increase the frictional loss in the suction piping by 21%, or about 0.5 psi (∆P varies with Q2). However, this was not observed as the pressure in Figure 16.86 slips slowly down from 47 to 34 psig, at which point the pump begins to cavitate. How could a 10% increase in the flow rate through the pump cause a 400% increase in the pressure drop in the suction line? What has happened to the lost 13 psig (i.e., 47–36 psig) of suction pressure?
Pumps 469 ISOMETRIC EXAMPLE This example shows the completed US Tutorial example laid out on an isometric grid. To change your grid layout to isometric format, go to System / Settings and click on the Drawing Options tab. Under the Grid Options section, check the box next to Isometric, and click OK. You can change the grid layout at any time during the modeling of a PIPE-FLO system.
pm Sg -07 0 U CS : 50 w Flo lve Va gpm ttle US hro 500 psi T N V @ :81 FC : 20 dP
CS-01 Flow: 800 US gpm
Supply Tank P Set: 0 psi g Level: 10 ft
Fil CS Flo ter Flo -04 dP w: 8 w: : 10 0 0 80 0U psi US g Sg pm p
: 11
6.6
1 circ Re
psi
g
pm Sg 0U
-08 CS : 30 w Flo lve m Va S gp e l U tt hro 300 si S-T V @ :33 p FC : 14 P d
Re Flo circ 2 w: 0U Sg pm
NCS Flo -02 TH S Pu : (2 mp w: 800 U 45 Sg pm .2) Nft P 01
k Tan si g rth 0 p No et: 2 5 ft P S vel: Le
e ific Or circ i R e : 0 ps P d
CS Flo -03 w: 80 0
m NP: 6 02 8.8 5
k Tan si g uth 5 p ft So et: 2 3.002 S P vel: Le
pm Sg -05 0 U CS w: 50 Flo
psi
g
CS Flo -06 w: 30 0U Sg pm
US gp m Lineup: System: FOF US tutorial-isometric Date: 28/03/18 9:49 pm Company: A.K.C TECHNOLOGY Project: XYZ by: A.K. Coker
Darcy-Weisbach Flow of Fluids Premium 2009 Flow: US gpm Pressure: psi g Size: in Elevation: ft Velocity: ft/sec Length: ft Volume: gallons
Figure 16.83 Isometric diagram of the piping network from the supply tank via N-S centrifugal pump to the north and south tanks (source: Engineered Software, Inc.).
The boiling point pressure of the water is equal to 30 psig (the pressure in the tower shown in Figure 16.86); that can be assumed that the water drawoff is at its bubble point pressure. At 36 psig pump suction pressure, the NPSHA is:
(36 psig − 30 psig) × 2.31 = 14 ft
This matches the NPSHR at a flow of 110 gpm, such that the pump cavitates. Half of the 46 ft of liquid head to the pump is missing, and where could it be? Figure 16.87 illustrates the true situation. If 110 gpm is being pumped at the discharge, but only 109 gpm can be drain through the drawoff nozzle. If the water level in the suction line is slowly lowered, the water level would creep down as would the pump’s suction pressure. When the water level in the suction line dropped to 14 ft, the pump would cavitate or slip. The flow rate from the pump would drop, and the water level in the suction line to the pump would partially refill. The pump’s NPSHR would then be temporarily satisfied. Normal pump operation would be restored, but only temporarily. Although, it may seem that the draw-off nozzle is undersized as this may be determined by calculating the velocity, (v, ft/s) through the nozzle:
∆H = 0.34 v2
(16.94)
Head - ft
NPSHr - ft
0
100
0 200
25
50 50
100
150
200
250
7 in
300 8.125 in
350
400
10 in
100
100
100
200
200
200
300
300
300
500
500
500
600
600
600
65
60 Hz Centrifugal Demo Catalog Catalog: , Vers ESP - 3600
400
400
400
60
60
700
US gpm
700
700
70.6
70
Figure 16.84 Centrifugal pump characteristics (source: Pump-FLOTM, Engineered Software Inc.).
Company: A.K.C TECHNOLOGY Name: A.K. Coker 3/28/2018
Power - hp
450
800
800
800
73
65
1000
1000
1000
Size: 4×3-13 Speed: 3550 rpm Dia: 8.125 in Curve: ABC1062-1
900
900
900
74
1100
1100
1100
74
1200
1200
1200
73
1300
1300
1300
70
1400
1400
1400
470 Petroleum Refining Design and Applications Handbook Volume 2
Pumps 471 Pump Data Sheet - 60Hz Centrifugal Demo Catalog
Company: A.K.C TECHNOLOGY Name: A.K. Coker Date: 3/28/2018
Pump:
Operating Point:
Size: 4×3-13 Type: ESP Synch speed: 3600 rpm Curve: ABC1062-1
Flow: --Speed: 3550 rpm Dia: 8.125 in Impeller:
Specific Speeds:
Ns: 1190 Nss: 9700
Dimensions:
Suction: 4 in Discharge: 3 in
Pumps Limits: Temperature: 300 °F Pressure: 375 psi g Sphere size: 0.157 in
Fluid: Water Density: 62.37 lb/ft 3 Viscosity: 1.105 cP NPSHA: ---
Temperature: 60 °F Vapor pressure: 0.2563 psi a Atm pressure: 14.7 psi a
Motor: Power: --Eye area: ---
Flow:
710 US gpm
Head:
261 ft
Eff:
70.6%
Power:
66.5 hp
NPSHr:
23.3 ft
Shutoff dP:
128 psi
Min flow:
350 US gpm
65
70
73
74 74
400
73 350
Head - ft
295 ft
60
450
---- Design Curve ---Shutoff head:
Size: 100 hp Speed: 3600 Frame: 405TS
10 in
---- Data Point ----
BEP:
Head: ---
Preferred Operating Region: 70% – 120% BEP
70.6% @ 710 US gpm
300 8.125 in
70.6
250 200
NOL power: 81.7 hp @ 1048 US gpm
70
7 in
65 60
150
-- Max Curve -Max power: 165 hp @ 1336 US gpm
100
NPSHr - ft
50 50
500
750
1000
1250
250
500
750
1000
1250
250
500
750
1000
1250
25
0 200
Power - hp
250
100
0
US gpm
Figure 16.85 Pump data sheet of N-S centrifugal pump (source: Pump-FLOTM, Engineered Software Inc.).
472 Petroleum Refining Design and Applications Handbook Volume 2
30 psig 18 in
46 ft
Hot water
Figure 16.86 Partly plugged drawoff nozzle [27].
Partly plugged nozzle
18 in
32 ft
14 ft
Figure 16.87 Most common cause of cavitation [27].
Pumps 473 Where ∆H is the hydraulic head in inches of liquid required to push 110 gpm of liquid through the draw-off nozzle. Lieberman [27] found ∆H to be 9 in. of H2O; there is twice as much pressure loss through the nozzle than intended, indicating that the draw-off nozzle must be partly plugged. It has been shown that many draw nozzles and especially those in the bottom of vessels plug because of the presence of vortex breakers. Many designers routinely add complex vortex breakers to prevent cavitation in pumps. However, these are only required in nozzles operating with high velocities and low liquid levels. Corrosion products, debris, and products of chemical degradation can more easily foul and restrict nozzle equipped with vortex breakers. Furthermore, lack of NPSHA may be caused by high frictional loss in the suction piping. In this instance, a small reduction in flow will increase the pressure at the suction of the pump. A properly designed suction line to a centrifugal pump should have a frictional head loss of only a few feet of liquid. A large diameter suction line and a relatively small draw-off nozzle can result in excessive loss of NPSHA.
Case Study 3. Low NPSHA at Main Column Bottoms (MCB) An FCC main column bottoms (MCB) pump was specified with an NPSHA of 8 ft. Therefore a low NPSHR was selected and installed. When equipment reliability was poor, the contractor and operator blamed the original equipment manufacturer (OEM). However, the root cause was the design engineer’s ultra-conservative specification of the NPSHA. New pumps are frequently installed during revamps where they must fit into an existing process system and operate within constraints. The designer must ensure cost effective solutions working with the existing equipment, and when trying to install new pumps, plot space will often determine location rather than ideals such as minimum suction piping run. Figure 16.88 shows a typical FCC main column bottoms (MCB) system. Reactor effluent enters the column at temperature of 980–1015°F (527–546°C) where the MCB system must remove up to 35% of the heat so the reactor
Boiler feed water preheat
Reactor effluent
FRC R
690
Main column bottom pump
Figure 16.88 Main column bottom system [34].
Debutaniser reboiler
600 psig steam Temperature, °F
474 Petroleum Refining Design and Applications Handbook Volume 2 products can be fractionated. Fluid mixed with catalyst and coke fines is withdrawn from the bottom of the main column and pumped through heat exchangers, then back to the column as sub-cooled pump around return (PAR) and quenched. PAR liquid flows down the column through internals such as shed trays or grid where heat is transferred from reactor effluent to the PAR liquid. To prevent coke formation, must refiners maintain a constant temperature in the bottom of the main column by varying quench flow rate. The MCB circulation rate depends on the system design, and the operating philosophy can cause large flow variability from the start-of-run (SOR) to end-of-return (EOR). Before detail pump hydraulic calculations, the process engineer determines the total (PAR plus quench) heat removal requirements from the design basis heat and material balance around the column. Then, the MCB circulating rate is calculated based on the exchanger configuration and its tendency to foul. The pump design point matches the real operation, but the system design requires the flexibility to meet realistic process variability. Furthermore, operating philosophy influences SOR and EOR conditions, which determine maximum and minimum flow. If the main column bottoms temperature is held constant from SOR to EOR, then flow rate will be low at SOR when exchangers are clean and increase as the exchangers foul. The rate of exchanger fouling depends on velocity through the exchanger tubes and fluid temperature throughout the main column bottoms pool.
16.32 Pump Fundamentals As illustrated in the text, the selected pump requires a certain amount of NPSHR to operate properly and the NPSHA needs to be higher than NPSHR for stable and reliable pump performance through the run. NPSHA is the amount of head available at the pump suction above the fluid vapor pressure. Liquid level, suction piping pressure loss, and fluid vapor pressure determine the NPSHA. When the designer specifies the NPSHA, this value plays a critical role in the pump selection and it determines stable operating range for the MCB pump. Pump impeller eye design is characterized by:
Nss =
N Q NPSH0R.75
(16.69)
and NPSHR is equal to the inlet velocity head:
NPSHR =
v2 2g
(16.95)
At the pump inlet, the velocity is a function of the volumetric flow rate and inlet area:
v=
Q A
(16.96)
where A = inlet area of pump, ft2 Nss = suction specific speed (dimensionless) n = pump speed, rpm Q = flow at the best efficiency point (BEP), gpm A high value of Nss corresponds to a low NPSHR; it would be assumed that a higher suction specific speed is better. Eqs. 16.69 and 16.95 show that the easiest way to reduce the NPSHR for a particular flow is to increase the inlet
Pumps 475 area; however, this is not a good practice because it increases the minimum flow rate for the pump. The minimum continuous stable flow (MCSF) is the lowest flow rate at which the pump can operate without exceeding the vibration limits imposed by API 610. The BEP is the flow/head combination that corresponds to the highest efficiency. As the flow rate drops below that of the BEP, the NPSHR by the pump initially decreases until it reaches a minimum before beginning a steady increase (Figure 16.18). Pumps with high suction specific speed will have correspondingly high minimum flow requirements that will likely require a recirculation line to provide satisfactory operation. Common practice is to keep NSS between 7,000 and 12,000 depending on the fluid. When a pump cannot be found that meets the suction specific speed criteria for a particular project, a recirculation line can be added to meet the minimum flow requirement [24]. Another important parameter is the minimum continuous thermal flow (MCTF) and as defined by API 610: the lowest flow rate at which the pump can operate without being impaired by the temperature increase of the pumped liquid. MCSF refers to recirculation of the fluid that can result in cavitation and vibration; MCTF is concerned with temperature rise. From Figure 16.18, the efficiency drops corresponding to a decrease in flow below the best efficiency point (BEP). This decrease in efficiency is characterized by an increase in temperature. MCTF is the point at which this rate of temperature rise obstructs the operation of the pump. The minimum operating flow is the higher of MCTF and MCSF. Generally, MCSF occurs at a higher flow rate than MCTF and becomes the defining variable. For very-low flow pumps, MCTF may be prevalent.
NPSHA In Figure 16.88, the pump of the MCB was initially specified with an NPSHA of 8 ft. Therefore, a low NPSHR pump was selected and installed. However, when equipment reliability was poor, the contractor and operator blamed original equipment manufacturers (OEM). However, the root cause was the design engineer’s ultra-conservative specifications of the NPSHA [34]. NPSHA is based on the pump system configuration, the fluid flow rate and properties. Liquid level above pump centerline, fluid vapor pressure and system pressure drop all influence NPSHA. Once the plot space for the pump is selected and pipe routing set by pipe stress considerations, system ∆P cannot be materially lowered by the designer. First the designer requires defining the minimum acceptable level. It is the bottom of the head, vessel tangent line or low liquid level? Figure 16.89 illustrates how there can be a 6 ft (1.8 m) difference between the most conservative method, which uses the bottom of the head, and least that uses the low liquid level. This is the difference of between 8 ft and 14 ft (2.4 m and 4.3 m) NPSHA, and more importantly, allows the refiner to select a pump that turns down to 65–70% BEP flow rate rather than one that turns down to only 80–85% of BEP flow [34]. Furthermore, the fluid vapor pressure can be reduced by 5 ft (1.5 m) for every 10°F reduction in the main column bottoms pool liquid temperature. Reducing pool temperature from the 690°F to 680°F (366°C to 360°C) increases NPSHA from 14 ft to 19 ft (4.3–5.8m) (Figure 16.90).
16.33 Operating Philosophy Instead of maintaining constant MCB temperature, the exchanger velocity should be controlled by allowing the temperature in the bottom of the main column to vary from start-of-run (SOR) to end-of-run (EOR) by adjusting the quench flow rate. As the exchanger fouls, the PAR rate increases and the quench flow decreases so that exchanger velocity is maintained. Using this approach, the exchanger fouling is reduced, MCB pump erosion decreases, and the rate of coke formation in the bottom of the main column reduces. Furthermore, the MCB exchanger tube velocity should be maintained between 10 and 13 ft/s (3–3.9 m/s) to minimize fouling. When MCB system operates to maximize exchanger velocity at SOR, the NPSHA is very high because the main column temperature is low resulting in low fluid vapor pressure [34].
476 Petroleum Refining Design and Applications Handbook Volume 2
Reactor effluent
Normal liquid level
Quench
LC
Low liquid level Vessel tangent line
25'–0" 22'–0" 19'–0" Main column bottoms pumps
Figure 16.89 Main column bottom pump level [34].
Boiler feed water preheat
Reactor effluent
FRC R Quench distributor
680
Main column bottom pump
Figure 16.90 Main column bottom quench system [34].
Debutaniser reboiler
600 psig steam Temperature, °F
Pumps 477
Main Column Bottoms (MCB) Pump Specification MCB pump reliability is an important part of FCC unit profitability. Therefore, the process engineer should review opportunities to improve MCB system performance before selecting a MCB pump. Selecting a pump that operates as close as possible to the BEP flow minimizes pump erosion and maximizes operating flexibility. Furthermore, MCB pump specification should not use an ultra-conservative NPSHA. Since NPSHR value decreases in Eq. 16.69, suction specific speed increases, and consequently, pump stable flow range also decreases. When constant MCB temperature is the operating objective, MCB flow rate can be less than 60% of design flow rate at SOR when MCB exchangers are clean. A pump selected to operate with only 8 ft of NPSHA will not turn down to 60% of BEP. Process designs that permit higher quench flow rates at SOR minimize pump flow variation and maximize exchanger velocity. Operating philosophy should be changed from constant main column bottoms temperature to constant MCB recirculation rate by varying quench flow from SOR to EOR. Also, lowering bottoms temperature reduces the vapor pressure and subsequently increases NPSHA without significant changes in pump flow rate or the heat exchanger performance. Therefore, a more reliable lower Nss pump can be selected with better turndown [34].
Frequent Loss of Suction of Vacuum Residue Pumps at Vacuum Column Bottom A set of two pumps located in the vacuum column bottom in the fuel-type vacuum unit of a refinery were losing suction frequently [44]. They were vertical-split, multistage pumps with four stages in each. The suction line to each pump entered the pump vertically. A new set of pumps with a newer mechanical seal design was purchased to solve the problem and installed, but the loss of suction of the pump in the vacuum unit still continued. The supplier of the pump, along with the supplier of the mechanical seal and other pump experts inspected the unit but could not come to an agreement on what was causing the loss of suction, so the problem continued. To identify the root cause, a pressure survey was carried out with a compound gauge during the operation of the pump (see Figure 16.91). An equation can be written as P2 = P1 + h − DPf. The pressure was noted as P1 and P2 and the height h between points P1 and P2 was measured. The pressure drop due to flow (DPf value) was calculated using the above equation and found to be almost negligible, indicating that there was no choking in the suction line or the column bottom coke trap. The pressure at point 4 (P4 at the pump suction strainer downstream) was found to be 2 meters of liquid column (mLC), which is lower than the NPSHR of 3.5 mLC for the pump. Thus, it was concluded that lower NPSHA was causing the pump to lose suction. The lower NPSHA was attributed to higher pressure drop across the strainer. But no muck was found in the strainer to cause the pressure drop. The inlet and outlet pressures were measured again with a calibrated compound gauge. The openings of the strainer were also checked and found to be around five times the cross-sectional area of the suction pipe, which was acceptable. The reason for the higher pressure drop across the suction strainer continued
P1
Coke P4
P3
h
trap
P2
Suction strainer
Figure 16.91 Pressure survey of the pump suction line [44].
Suction valve
478 Petroleum Refining Design and Applications Handbook Volume 2
Coke PI
PI
Suction strainer
To pump
trap
Suction block valve
PI – Pressure indicator
Figure 16.92 Pressure survey of the pump suction line across strainer [44].
to be a mystery. The suction line oil-soaked insulation was removed by the maintenance group, possibly for a change of insulation, and the bare suction pipe section became visible. It was discovered that the strainer was too close to the reducer in the suction line. It was suspected that the convex part of the strainer may be leaving very low clearance for the flow due to its proximity to the reducer (Figure 16.92). The suction pool was taken out and a modification was carried out to move the reducer away from the suction strainer and toward the suction of the pump (a distance of more than five times the diameter of the pipe). The new spool was fabricated and fitted and the pump tested. The pressure drop disappeared across the strainer and the strainer downstream pressure was found almost equal to the strainer upstream pressure and higher than the NPSHR by the pump. The problem totally disappeared and performance of the pump became normal.
Failure of Pumps in a Hydrocracker In a newly commissioned hydrocracker, the bottom pumps of the stripper and the main fractionator as well as some other pumps were repeatedly losing suction, making the unit unstable [44]. The pumps had a suction line arrangement as illustrated in Figure 16.93. The strainers were installed from the side and the line to the pump suction was taken through a tee pipe below the strainer. It was decided to replace the pumps with new pumps with a lower suction NPSHR. A pressure survey was carried out in the suction line as shown by the points P1 and P2. The height between P1 and P2 is indicated by h. The calculation is: P2 = P1 + h x ρ − DPf. The pressure drop due to flow (DPf ) was calculated from the measured data and was found to be negligible, confirming no choking in the lines or column outlet nozzles. The strainer could not be inspected as the pump isolation could not be achieved because the suction valves were not holding properly. The drawing of the strainer was found, which showed that the strainer was similar to a straight pipe with perforations indicating that the design was incorrect. This is because the perforations on the straight portion of the strainer would provide less flow area through the perforations (lower than the cross-sectional area of the suction pipe). The strainer design was modified with a curved surface facing the pump suction to provide more holes in the curved surface and consequently greater flow area (Figure 16.94). The unit was replaced with the modified strainer arrangement and the problem was resolved. The performance of all the pumps became normal.
Oil Refinery Fire and Explosion at Ciniza Oil Refinery, New Mexico, USA A fire and explosion occurred at the Giant Industries’ Ciniza oil refinery in Jamestown, New Mexico, on April 8, 2004. The incident occurred in the refinery’s hydrofluoric acid (HF) alkylation unit. Alkylation is a standard oil refinery process that
Pumps 479 P1
h
P2
Figure 16.93 Column outlet piping scheme and pressure survey.
Original strainer
Fluid
Strainer
To Pump
Modified strainer
Fluid
Strainer
To Pump
Figure 16.94 Original and modified strainer.
combines olefins with isobutane using a catalyst HF to produce alkylate. Alkylate is a highly flammable gasoline blending component used to boost the octane of gasoline. It forms explosive vapor/air mixtures at above-ambient temperatures. The day before the incident, alkylation unit operators performed a regularly scheduled switch of the alkylate recirculation pumps in the iso-stripper section of the alkylation unit. The primary electric pump would be taken out of service and the spare steam-driven pump started up. The switch was scheduled because of recurring problems with the spare pump’s mechanical seal leaking. Mechanical seals are used to keep the contents of rotating equipment from escaping. This is done by sealing the shaft that protrudes from the casing. When operators attempted to put the spare pump in service, they discovered that it would not rotate. The next morning, an operator prepared a work permit that outlined the work to be done and the safeguards required for a safe repair*. The valve used to isolate the pump for maintenance was a ¼ turn plug valve. A plug valve is used primarily for on/off, and some throttling services. It controls the flow by a cylinder or tapered plug with a hole in the center that lines up with the flow path of the valve to permit flow. The valve is opened or closed with the use of a valve wrench as illustrated in Figure 16.95. During the preparation for maintenance, the operator relied on the valve wrench to determine that the suction valve was open. He moved the wrench to determine that the suction valve was open. He moved the wrench to what he believed was the closed position with the wrench perpendicular to the flow of product. The pump required to be
480 Petroleum Refining Design and Applications Handbook Volume 2
Valve Wrench
Valve Wrench Collar
Position Indicator
Figure 16.95 Suction valve and position indicator as found after the incident (source: www.csb.gov).
disassembled, and the rear pump housing assembly and impeller moved to the shop for repair. The mechanic noticed that the valve position indicator on the suction valve body showed that the valve was open (Figure 16.95). However, he did not relate this information to his co-workers. The plant operator placed tags on locks on the suction and discharge valves to prevent inadvertent opening and to indicate that the valves had been closed. The mechanic specialist then placed tags and locks on the suction and discharge valves. When the mechanic returned, the mechanic specialist told him that the valves had been closed, secured, tagged and locked per the facility’s lockout/tagout (LOTO) procedure, and that they could remove the pump. Neither mechanic observed the operator closing the valve as both mechanics believed the task had been completed because the wrench used to open and close the valve was positioned perpendicular to the flow, and the operator had affixed his tags (Figure 16.95). The operator then disconnected the pump’s vent hose to verify that no pressure was in the pump case. The low point drain plug was not used because it was not equipped with a valve to isolate it from the line used for depressuring the pump (Figure 16.96). After uncoupling the hose at the connection to the flare line, a stream of alkylate flowed from the pump housing through the hose and subsided after a few seconds. The operator and the maintenance mechanics believed the pump had been de-pressured and was ready for removal. However, the vent line was plugged, and the pump was not depressured. The pump shaft coupling and the flange connecting the pump to the pump case were unbolted (Figure 16.97). As the pump case flange was separated, alkylate was suddenly released at about 150 psig and 350°F. The release produced a loud roaring sound that could be heard throughout the refinery. The mechanic was blown over an adjacent pump and suffered broken ribs. Material was blown into the mechanic’s eyes and had to make his way to an eyewash station. Alkylate, which covered the plant operator’s clothing, quickly ignited seriously burning the operator in the ensuring fire. About 30–45 s after the initial release, the first of several explosions occurred. * The work permit is issued by the operator and contains information on hazards involved in the maintenance operation, the appropriate personal protective equipment to be worn, and lock-out–tag-out (LOTO) information. Lock out/tagout refers to a program to control hazardous energy during the servicing and maintenance of machinery and equipment. Lock out refers to the replacement of a locking mechanism on an energy – isolating device, such as a valve, so that the equipment cannot be operated until the mechanism is removed. Tagout refers to the secure placement of a tag on an energy-isolating device to indicate that the equipment cannot be operated until the tag is removed.
Pumps 481
Vent Hose
Low Point Bleeder
Figure 16.96 Depressurizing hose (source: www.csb.gov).
Rear Pump Assembly
Pump Case Flange
Figure 16.97 Damage to area of the pump (source: www.csb.gov).
The U.S. Chemical Safety and Hazard Investigation Board (CSB) findings related to this explosion are:
Mechanical Integrity Giant’s mechanical integrity program did not effectively prevent repeated failures of the pump seals. Problems were addressed when equipment broke down, not in a preventive manner. The design of the valve wrench made it easy to remove and reposition onto the valve stem in different directions, and this led to a potential hazard because operators sometimes determined whether the valve was open by its wrench position, rather than the valve position
482 Petroleum Refining Design and Applications Handbook Volume 2 indicator. In this incident, the valve wrench collar was installed in the wrong position. Operators depended on the wrench position and mistakenly determined the valve was closed. Plugging material was found in the pump discharge line, the depressuring line, pump housing, and the impeller (Figure 16.98). Giant’s approach to these frequent pump seal problems was an example of break down maintenance. Pump failures were addressed when the equipment finally broke down instead of identifying causes of breakdowns and preventing them before they occurred again. The Center for Chemical Process Safety (CCPS) recommends that maintenance program, troubleshoot and search for possible hidden or multiple reasons for frequent occurring problems [45]. An effective mechanical integrity program would have investigated and resolved the problems that were repeatedly causing the recirculation pump seals to fail.
Corrosion and Scale Formation At Giant, the isostripper pumps frequently had plugging problems. A number of factors contribute to the formation of corrosion, scale, and deposits, which in turn can lead to the fouling or scoring of pump seals. Although, the failure of the seals is not often caused by corrosion of the seal faces, but by the scoring/erosion (galling) of the seal faces from a solid fouling material. Carbon seal faces such as those used for the pump involved in this incident are prone to contaminant scoring. Some corrosion and scale products occur because of operating temperatures and pressures at which hydrofluoric acid (HF) alkylation unit is run. Many HF alkylation units operate at about 125 psig or lower. The isostripper column at the Giant refinery is normally operated at about 150 psig. Operation at higher pressures requires much higher temperatures, which can result in accelerated corrosion of equipment. Plugging and fouling material can also occur as soft iron fluoride scale develops and forms in the tower overhead and domes when preparing to shut down the unit for maintenance/cleanup activities. Some of this soft, water – laden scale comes off when the unit is running. The rest creates a site for extremely fast and extensive corrosion, resulting in a large amount of scale being added into the process. Scale many accumulate and settle at low points and orifices in equipment such as spare pumps. Management did not investigate why excessive iron fluoride generation in the process caused the mechanical seals on the alkylate recirculation pump to fail repeatedly.
Valve Design Examination of the 6 in., ¼ turn, plug valve after the incident determined that it was originally designed to be opened and closed by a gear-operated actuator. The gear-driver was removed and was replaced by a valve wrench. The wrench was a two-foot-long bar inserted into a collar. Because it had a square shape, the collar could be easily removed and repositioned on the valve stem in different directions (Figure 16.99). Giant did not consider the design or engineering safety implications of changing from a gear-operated valve actuator to using a wrench as a valve handle. Interviews with operators revealed that they would sometimes determine whether the valve was opened or closed by the valve wrench position. If the wrench was perpendicular to flow through the valve, it was considered closed. If the wrench was aligned parallel to the flow, the valve was thought to be open. There are instances where set screws would be loosened, and the wrench would be removed and placed on the pump based to provide better clearance for personnel walking nearby. When the valve was to be opened or closed, the wrench would be replaced on the valve stem. In the Ciniza oil refinery incident, the valve wrench collar was installed in the wrong position. Figure 16.100 illustrates a drawing of the plug valve in the open position with the wrench in the perpendicular or perceived “closed” position. Both the plant operator who attached the locks and the mechanics who removed the pump mistakenly believed the suction valve had been closed, in part because the valve wrench was perpendicular to the normal pipe flow. After he returned from obtaining materials needed for pump removal, the mechanic who earlier observed the position indicator in the open position began working on the opposite side of the pump, not realizing that the valve position indicator still indicated the valve was open. From his position, he observed the valve wrench in a perpendicular orientation and believed the valve was closed. The wrench was not intended to indicate valve position because the valve was equipped with the position indicator, which was located on the valve stem as shown in Figure 16.101.
Pumps 483
Figure 16.98 Plugging material found in discharge valve (source: www.csb.gov).
Figure 16.99 Spare pump valve wrench collar (source: www.csb.gov).
484 Petroleum Refining Design and Applications Handbook Volume 2
Figure 16.100 Plug valve in the open position (source: www.csb.gov).
Figure 16.101 Suction valve and position indicator (source: www.csb.gov).
Pumps 485
Management of Change The CSB urges management of change analyses for any valve modifications, effective “lock out tag out” programs to ensure equipment has been isolated, depressurized, and drained; and proper mechanical integrity programs to prevent breakdown maintenance. The study stated that Giant should have determined the cause of the frequent alkylate recirculation pump malfunctions and implemented a program to prevent them. A CSB member stated that proper mechanical integrity programs and effective management of change analyses are essential components of safe operations at any refinery. Occupational Safety and Health Administration (OSHA) Process Safety Management standard (1910.119) states that any change that may affect a process covered by that standard should trigger a management of change (MOC) analysis. The only exception to this is when the change is a “replacement in kind” [46].
16.34 Piping Pump requirements, nozzle size, type of fluid, temperature, pressure and economics determine materials and size of piping. Suction lines should be designed to keep friction losses to a minimum. This is accomplished by using an adequate line size, long radius elbows, full bore valves, and so on. Pockets where air or vapor can accumulate should be avoided. Suction lines should be sloped, where possible toward the pump when it is below the source and toward the source when it is below the pump. Vertical downward section pipes require special care to avoid pulsation and vibrations and could result in air or vapor entrainment. Adequate liquid height above the suction piping inlet, or a vortex breaker should be provided to avoid vortex formation which may result in vapors entering the pump. Furthermore, suction vessel tangential inlets and centrifugal pumps may induce a vortex in the vessel and pump suction line thereby opening a vapor core that feeds into the pump suction. All these can be eliminated with a straightening cross known as a vortex breaker, installed at the vessel outlet nozzle. For discharge piping, sizing is determined by the available head and economic considerations. Velocities range from 3–15 ft/s. A check valve should be installed between the discharge nozzle and the block valve to prevent backflow [30].
16.35 Troubleshooting Checklist for Centrifugal Pumps Centrifugal: Good practice: head-capacity curve should not be too flat if pump capacity is controlled by valve positioned. Select pump such that a larger diameter impeller could be installed later. An increase in flow rate causes an increase in NPSHR and a decrease in NPSHA [29]. Loss of prime is one of the common problems that will be encountered during daily trouble shooting. This may be caused by incomplete venting, a blocked suction line or entrained gases in the liquid. Entrainment can result from a number of situations including: 1. 2. 3. 4.
aving incoming lines in the suction vessel above the liquid’s surface. h vortexing in the suction vessel. air leaks through joints valve packing or pump packing. flashing of low boiling points.
The reason a centrifugal pump stops pumping liquid when the impeller becomes filled with air is that its output pressure in psi (bar) is drastically reduced. The impeller may be creating about the same head in feet of fluid but the specific gravity of the fluid has changed from that of liquid (1.0 for water at 20°C) to that of air (0.00123). Therefore, a water pump generating 100 ft of head or 43.3 psi (i.e., psi = (ft. of fluid) (Sp.Gr.)/2.31)) will only generate about 0.053 psi when it becomes “air bound”. As the liquid in the discharge line requires a given pressure to move it at a certain velocity, the sudden reduced pressure allows flow to stop.
486 Petroleum Refining Design and Applications Handbook Volume 2 The following problems associated with centrifugal pumps and corrective modes of action should be taken. Symptoms
Possible causes
No liquid delivery
Instrument error Not primed Cavitation Supply tank empty
Liquid flow rate low
Instrument error Cavitation Non-condensibles in liquid Inlet strained clogged
Intermittent operation
Cavitation Not primed Non-condensibles in liquid
Discharge pressure low
Instrument error Non-condensibles in liquid Speed too low Wrong direction of rotation (or impeller in backwards if double suction)
Power demand excessive
Speed too high High liquid density Required system head lower than expected High viscosity
Peripheral: No liquid delivery
Instrument error Pump suction problems Suction valve closed Impeller plugged
Liquid flow rate low
Instrument error Speed too low Incorrect impeller trim Loose impeller
Discharge pressure low
Instrument error Speed too low Incorrect impeller trim Loose impeller
Power demand excessive
Speed too high Improper impeller adjustment Impeller trim error
Cavitation
Liquid too hot Non-condensibles in liquid Air leakage into suction line Vortex entraining gas Decrease in density of the liquid Blockage or excessive ∆p on suction Suction velocity too high Increase in rpm
Pumps 487 Table 16.10 shows the various problems that often occur in the operation of centrifugal pumps and means of detecting and rectifying these occurrences [30].
Pump Installation Check List Production operators should be trained in pump fundamentals as this ensures good operation and low maintenance costs. Often some operators are accused after incidents of being careless with equipment when in truth; they have never had proper training. The specialist can be used to provide training sessions for plant operators. Furthermore, an individual or team in either maintenance or the engineering department should be assigned to perform trouble shooting as indicated in Table 16.10. A summary of the installation checkpoints is presented in Table 16.11.
Factors in Pump Selection The selection of a pump depends on several factors, which include the properties of the pumped fluids, the required capacity, and the desired location of the pump. Generally, high viscosity liquids are pumped with positive displacement pumps. Centrifugal pumps are not only very inefficient when pumping high viscosity liquids but their performance is very sensitive to changes in liquid viscosity. A high viscosity liquid also results in high frictional head losses and a reduced NPSHA. Since the latter must be greater than NPSHR by the pump, a low NPSHA imposes a severe limitation on the choice of a pump. Liquids with a high vapor pressure also reduce the NPSHA. If these liquids are pumped at a high temperature, this may cause the gears to seize in a close clearance gear pump. Correspondingly, if the liquid being pumped is shear thinning, its apparent viscosity will decrease with an increase in shear rate and thus the pumping rate. It is therefore an advantage to use high speed pumps to pump shear thinning liquids and in fact centrifugal pumps are frequently employed. In contrast, the apparent viscosity of a shear thickening liquid will increase with an increase in shear rate and thus the pumping rate. Therefore, it is advantageous to use large cavity positive displacement pumps with a low cycle speed to pump shear thickening liquids [37]. Some liquids can be permanently damaged by subjecting them to high shear in a high speed pump. For example, certain liquid detergents can be broken down into two phases if subject to too much shear; although these detergents may exhibit shear thinning characteristics, they should be pumped with relatively low speed pumps. Positive displacement pumps are more susceptible to wear than with centrifugal pumps. Liquids with poor lubricating qualities increase the wear on a pump. Furthermore, wear is a result of liquids containing suspended solids that are abrasive and by corrosion. In general, centrifugal pumps are less expensive, last longer and are more robust than positive displacement pumps. However, they are unsuitable for pumping high viscosity liquids and when changes in viscosity occur (see Table 16.5).
Pump Reliability Pumps are essential to the daily operation of refinery and industrial chemical processes, as they are the workhorses of the plant, and require constant maintenance to ensure the availability of the business. Pumps start to degrade the day they start up, and the challenge is to know when and how to intervene before they contribute to a critical plant shutdown or slowdown, impacting the company’s bottom line. A typical pump failure can result in a very high cost in thousands of dollars and a bigger incident could draw media attention. This can negatively impact public image as well as increase the possibility of fines and government involvement. Worse, serious problems may affect the safety of plant personnel and the environment. Few pumps that are critical to operation may have on-line condition based monitoring as these systems minimize unnecessary maintenance common with preventive (time-based) maintenance while avoiding catastrophic failures that result in expensive repairs, fire and plant downtime. In a typical refinery, these may account for approximately 10% of pumps. That leaves about 90% of pumps subject to manual rounds of a technician/operator to acquire data and assess pump condition, preventive maintenance tasks, or running-to-failure.
488 Petroleum Refining Design and Applications Handbook Volume 2 Table 16.10 Troubleshooting centrifugal pumps. Symptoms
Possible causes
Failure to deliver liquid
1. 2. 3. 4. 5. 6. 7. 8.
Pump does not deliver rated capacity
1. Wrong direction of rotation. 2. Suction line not filled with liquid. 3. Air or vapor pocket in suction line. 4. Air leaks in suction line or stuffing boxes. 5. Inlet to suction pipe not sufficiently submerged. 6. NPSHA not sufficient. 7. Pump not up to rated speed. 8. Total head greater than head for which pump designed. 9. Foot valve too small. 10. Foot valve clogged with trash. 11. Viscosity of liquid greater than that for which pump designed. 12. Mechanical problems. (a) Wearing ring worn. (b) Impeller damaged or worn out. (c) Internal leakage resulting from defective or damaged gaskets. 13. Discharge valve not fully opened.
Pump does not develop rated discharge pressure
1. Gas or vapor in liquid. 2. Pump not up to rated speed. 3. Viscosity of liquid greater than that for which pump designed. 4. Wrong direction of rotation. 5. Mechanical problems: (a) Wearing rings worn. (b) Impeller damaged or worn out. (c) Internal leakage resulting from defective or damaged gaskets.
Pump loses liquid after starting
1. 2. 3. 4. 5. 6. 7. 8.
Pump over loads driver
1. 2. 3. 4.
rong direction of rotation W Pump not primed Suction line not filled with liquid Air or vapor pocket in suction line. Inlet to suction pipe not sufficiently submerged. NPSHA not sufficient. Pump not up to rated speed. Total head required greater than head which pump is capable of delivering.
S uction line not filled with liquid. Air leaks in suction line or stuffing boxes. Gas or vapor in liquid. Air or vapor pocket in suction line. Inlet to suction pipe not sufficiently submerged. NPSHA not sufficient. Liquid seal piping to lantern ring plugged. Lantern ring not properly located in stuffing box.
S peed to high. Developed head greater than rated head. Excessive recirculation. Either or both the specific gravity and viscosity of liquid different from that for which pump is rated. 5. Mechanical problems: (a) Misalignment. (b) Shaft bent. (c) Rotating element dragging. (d) Packing too tight. (Continued)
Pumps 489 Table 16.10 Troubleshooting centrifugal pumps. (Continued) Symptoms
Possible causes
Excessive vibration
1. Starved suction. (a) Gas or vapor in liquid. (b) NPSHA not sufficient. (c) Inlet to suction pipe not sufficiently submerged. (d) Gas or vapor pockets in suction line. 2. Misalignment. 3. Worn or loose bearings. 4. Rotor out of balance. (a) Impeller plugged. (b) Impeller damaged. 5. Shaft bent. 6. Improper location of control valve in discharge line. 7. Foundation not rigid.
Stuffing boxes overheat
1. 2. 3. 4. 5. 6. 7.
Bearings overheat
1. Oil level too low. 2. Improper or poor grade of oil. 3. Dirt in bearings. 4. Dirt in oil. 5. Moisture contamination in oil. 6. Oil cooler clogged or scaled. 7. Failure of oiling system. 8. Insufficient cooling water circulation. 9. Insufficient cooling air. 10. Bearing too tight. 11. Bearing over-lubrication (when grease lubricants are used). 12. Misalignment.
Bearings wear rapidly
1. 2. 3. 4. 5. 6. 7. 8. 9.
acking too tight. P Packing not lubricated. Wrong grade of packing Insufficient cooling water to jackets. Packing provided improperly. Mechanical seal hydraulic balance improper. Uneven torquing of gland nuts.
isalignment. M Shaft bent. Vibration. Lack of lubrication. Bearings improperly installed. Dirt in bearings. Moisture contamination in oil. Excessive or insufficient cooling of bearings. Oil/lubricant viscosity not maintained as recommended.
It is estimated that pumps account for 7% of the total maintenance cost of a plant or refinery, and pump failures are responsible for 0.2% of lost production. These avoidable costs could be significantly reduced if the unmonitored pumps had online condition monitoring. This can be decided upon by reviewing those pumps that are at risk to cause process upsets and downtime, often taking hours or days to recover normal operations. Basic pump maintenance is normally performed either by manual rounds or online conditioned-based monitoring (CBM). The aim is to prevent failures that require expensive repairs and cause process slowdowns or shutdowns. However, the former is typically very expensive. Preventive maintenance and manual rounds are not always adequate
490 Petroleum Refining Design and Applications Handbook Volume 2 Table 16.11 Pump installation check list [31]. 1.
The foundation should be substantial to support the pump rigidly and absorb vibration.
2.
The base plate must be leveled on the foundation so the oil in the pump bearing housing will be at the correct level and the oil on the housing will feed properly.
3.
There are two types of coupling misalignment—angular and parallel. Wedges, feeler gauges, or dial indicators are generally used to check alignment.
4.
Factory alignment of the pump and driver must be checked after the complete unit has been leveled on the foundation, and again after the grout has set and the foundation bolts tightened.
5.
Suction piping is especially critical for proper pump operation and should not have pockets that may collect air or have sharp turns at the suction nozzle, which could cause an uneven flow to the impeller.
6.
Very little pipe loading should be imposed on the pump. Expansion joints and pipe supports should be used judiciously.
7.
The direction of rotation should be checked by jogging the motor before the coupling are joined.
8.
A final alignment should be made after the pump has run long enough for temperatures to stabilize. Some manufacturers’ manuals give cold alignment data that indicate the amount of misalignment the couplings should have when cold in order to be correctly aligned at the normal running temperature.
to identify degrading performance in time to act. Most pumps do not operate at optimal efficiency, as processes were not designed for the pump to run at its best efficiency pump (BEP). Some pumps may be over-sized to accommodate design contingencies and planned increase in throughput capacity or have been stretched beyond their design and capacity limits due to production demands. This introduces higher stress on pump system components, leading to higher maintenance requirements. Installing dedicated online monitoring systems as well as overfilled cable trays in highly congested areas has prevented online condition monitoring from being expanded beyond the most critical pumps. However, the relative ease of adding online pump condition monitoring with today’s technology, services now include [40]: 1. 2. 3. 4. 5.
umps without spares. P Pumps that can cause a fire or environmental incident. Pumps with repeat failures. Pumps that can lead to a significant process disturbance, process shutdown or slowdown. Any pumps that previously were not considered critical enough to have wired monitoring systems in place.
Plants may not have initially invested in online monitoring for pumps that have an installed spare. However, the reason for including an installed spare is that the pump will require maintenance before the next overall turnaround maintenance shutdown. If the operating pump runs-to-failure, it could make the condition worse by increasing maintenance time and cost to repair the damage. The spare pump may take time to start up, which can prevent an operator from making a transfer to a backup pump. This could create a process upset from which hours or days are required to return to normal operation, in addition to the costs of repairing the pump that was run-to-failure. Furthermore, it is more difficult to take a failed pump out of service and commission the spare pump if the failed pump is leaking process material or burning. Figure 16.102 shows typical root causes of pump failures and resulting impacts. Many pump failures can be predicted using modern monitoring techniques such as innovative reliability-centered practices for pump monitoring such as pump seal leaks, cavitation, and vibration related failures. These allow customers to minimize environmental and business impacts. Appendix D shows construction commissioning start-up checklists of rotary equipment such as pumps, compressors and others such as blower, fans and mixers.
Pumps 491 Process Change
Root Cause
Equipment Impact
Environmental Impact
Business Impact
Bad installation Shaft misalignment
Excessive vibration
Bearing wear
Premature bearing ware Reduced unit throughput
Pump failure
Restricted discharge flow
Restricted Suction flow/ Plugged suction strainer Undetected Conditions
Process upsets Excessive maintenance
High discharge pressure
Seal failure
Leaks
Cavitation
Impeller/shaft damage
Hazardous releases
Fires Evening news
Low suction pressure
Rebuild costs
Fires
Human impact Abnormal Situations
Avoidable Consequences
Figure 16.102 Typical root causes of pump failures and resulting impacts [40].
The following three innovative methods of increasing pump availability using predictive technologies and reliability-centered maintenance best practices are as follows [40]: Pump seal monitoring The latest edition of API Standard 682 now shows a preference for level and pressure transmitters instead of level and pressure switches in order to provide the signal to annunciate the level or pressure alarms. The use of transmitters provides an improved view of the pump seal flush reservoirs. A level signal also allows for monitoring the rate of change of a reservoir level for earlier indication of potential seal failure. Cavitation monitoring For high head multistage pumps that cannot tolerate cavitation even for a brief moment, users monitor discharge flow and pressure, the integrity of the level instrument on the suction vessel, and the differential pressure across the suction strainer to help prevent cavitation from occurring. Vibration monitoring Vibration monitoring allows for the detection of the presence of any one of many common causes of pump failure, such as worn bearings, worn shaft coupling, misalignment, impeller damage, cavitation, foundation, or frame faults.
Bloch [41] has provided reliability tips for centrifugal pumps and these are as follows: • Greater spacers lengths reduce angles of misalignment in case of the unavoidable differential temperature-related parallel offset between the centerlines of pump and driver. Greater-length spacers may be more expensive by reducing bearing loads, and vibration severity, they beneficially affect the likely cost of future maintenance.
492 Petroleum Refining Design and Applications Handbook Volume 2 • Pipe elbows located too close to the pump inlet nozzle may save money initially but they often create flow disturbances, which tend to reduce pump life. • The higher the peripheral impeller velocity, the greater the rate of erosion in solid-containing pumps. High-tip velocity pumps manage with fewer impellers than would be needed for equivalent-head lowtip velocity pumps, but lowest installed cost today usually equates with higher maintenance outlays later. • Operation at locations too far from BEP comes at a price. Inefficient operation increases power consumption or maintenance frequency or both. • Some non-hydrocarbon process fluids have properties that make it advisable to provide an NPSHA well in excess of the published NPSHR. Investigation of the applicable process experience is advised. • Operation at zero flow is not allowed and if over a minute’s duration, could cause temperature rise and internal recirculation effects that might destroy many pumps. • Occasional, high efficiencies are alluded to in the manufacturer’s literature when bearing, seal and coupling losses are not included in the vendor’s test reports. Ensure to install a large-enough motor. • Pump manufacturers often use modular casing construction. A given casing size may, however, accommodate several different impeller sizes or geometries. Once an existing plant has determined actual operating flows and heads, it may be cost-effective to purchase custom-built, optimized “upgrade” impellers from knowledgeable manufacturers. • Operation outside the design range will have some repercussions. There are no exceptions to this immutable rule. Pump internal recirculation can cause surging and cavitation, even when the NPSHA exceeds the manufacturer’s published NPSHR by considerable margins. Also, extensive damage to the pressure side of the impeller vanes has been observed in pumps operating at reduced flow rates.
New Technology With Screw Pumps Screw pumps classified as rotary pumps are positive displacement pumps and are increasingly employed to strip out even highly viscous media. When emptying an oil or fuel tank, quite often residual product of highly viscous media is retained. Tanks farms are storage tanks for storing media that are volatile, potentially hazardous or have to be protected from contamination. They are often used for oil products from diesel and petrol and other petrochemical products. However, all these materials could pose problems when being emptied such as when medium is pumped to the next step in the production chain. This may be due to comparatively high viscosity or a low vapor pressure. If a conventional centrifugal pump is used, the poor intake often retains the residual product that cannot be utilized resulting in a significant cost factor with diesel prices of $1,200 per m3. The wrong pump types can cause cavitation which could interrupt the flow resulting in pump failure on the long term. Screw pumps are therefore used as an alternative to centrifugal pumps because they have low NPSHR values, and therefore significantly better intake behavior, and are able to completely strip out a tank. The pumped fluid is moving axially without turbulence which eliminates foaming that would otherwise occur in viscous fluids. Furthermore, they are able to pump fluids of higher viscosity without losing the flow rate; changes in the pressure difference have little impact on the positive displacement pumps compared to centrifugal pumps. Screw pumps have self-priming property and therefore do not require any special arrangements for suction. However, because of this, they are suitable for pumping liquid as well as gases without any loss of suction. All types of rotary pumps continue to force fluid into the system regardless of the opposition to the transfer. Owing to this reason, additional protection to these pump types is required or else the pump will continue to build pressure, which could result in rupture or damage of the pump. This happens due to the building up of excessive pressure in the pump. This is avoided by fitting relief valves in the pumps. These relief valves are capable of bypassing entire throughput of the pump. The relief valves should operate only for short interval of time otherwise they would lead to increase in liquid and pump temperature. Screw pumps have several advantages as compared to centrifugal pumps: • They allow a wide range of flows and pressures. • They can also accommodate a wide range of liquid and viscosities.
Pumps 493 • • • •
They have high speed capability and this allows the freedom to driver selection. They have low internal velocities. They are self-priming which allows them to have a good suction characteristics. Due to the close arrangement of rotating parts, a high tolerance for entrained air and other gases is produced.
The disadvantages are: • Cost of manufacturing is high because of close tolerances and running clearances. • Any changes in the viscosity of the fluid results in high fluctuations in the performance. • A screw pump with high pressure capability will require high pumping elements which increase the overall size of the pump.
Nomenclature a B d BHP (BHP)vis CE CH CQ cp D D’ D” d dp dr d’ d” EHP E En Ev Ew Evis e ew evis evol eM g H H Hso Hvis Hw hd
= Area of piston or plunger, in.2 = Bell diameter of vertical sump pump, ft = Brake horsepower = Brake horsepower when handling viscous material = Viscosity correction for efficiency to convert to water performance = Viscosity correction for head, to convert to water performance = Viscosity correction for capacity, to convert to water performance = Specific heat of liquid, BTU/lb °F = Height of liquid (static) above (+) or below (−) the centerline of the pump on discharge side, ft = Incremental height of liquid (static) above normal D level, to establish “worst case” condition, ft. = Theoretical displacement volume displaced per revolution(s) of driving rotors, in3/rev = Impeller diameter, in. = Diameter of piston or plunger, in. = Diameter of piston rod, in. = Liquid displacement, ft3/min = Theoretical displacement, ft3/min = Electrical horsepower = Efficiency, % = Fraction entrained gas by volume at atmospheric pressure = Volumetric efficiency ratio of actual pump capacity to volume displaced per unit of time = Pump efficiency with water, % = Pump efficiency with viscous fluid, % = Pump efficiency, fraction = Pump efficiency with water, fraction = Pump efficiency with viscous fluid, fraction = Volumetric efficiency, fraction = Maximum safe flowing efficiency, overall pump fraction = Acceleration of gravity, 32 ft/s2 (9.81 m/s2) = Total head developed by a pump ft (m) of fluid; or total head/stage, ft, or, = Static head discharge ft (m) = Head at no flow, or shutoff, ft = Head of viscous fluid, ft = Water equivalent head, ft = Discharge head on a pump, ft of fluid
494 Petroleum Refining Design and Applications Handbook Volume 2 hs hSL, hDL hv L=S I Ns Np n P Pa Pso Ptd Pvp p’ p’a p′vp Q Q’ QM QN Qvis Qw S S S′L S” SpGr s t Tr t V v W W1 whp whp1
= Suction head (or suction lift) on a pump, ft of fluid = Friction losses in pipe and fittings: subscript SL for suction line; and DL for discharge line, ft of fluid = Velocity head, ft of fluid = Static head, suction side, ft = Water depth in sump, ft = Specific speed, dimensionless = Number of pumps = Rotative speed, revolutions per minute = RPM = rpm = Positive external pressure on surface of liquid (+) or partial vacuum on surface of liquid (−) = Atmospheric pressure or absolute pressure in vessel, psia = Brake horsepower at shutoff or no flow = Differential pressure between absolute pressures at outlet and inlet to pump, psi = Vapor pressure of liquid at pumping temperature, psia = Absolute pressure, in. Hg abs = Atmospheric pressure or absolute pressure in vessel expressed as ft of fluid = Vapor pressure of liquid at pumping temperature expressed as ft of fluid = Flow rate, gpm = Capacity of rotary pump, fluid plus dissolved gases/entrained gases at operating conditions, gpm = Minimum flow, gpm = Head at best efficiency point on pump curve, ft = Viscous liquid capacity, gpm = Water capacity, gpm = Suction static head, ft, or height of liquid (static) above (+) or below ( ) the center line of the pump on suction side, ft, or, = Suction lift, negative suction head, ft = Worst case suction side static lift, ft = Slip, quantity of fluid that leaks through internal clearances of rotary pump per unit time, gpm = Specific gravity of liquid at pumping temperature referred to water = 1.0 = Stroke, in. = Temperature rise, °F = Temperature rise, °F/min = Piston speed or travel, ft/min = Liquid velocity, ft/s = Average velocity, ft/s = Width of channel with series pump, ft = Weight of liquid in pump, lb = Water or liquid horsepower = Power imparted by pump to fluid discharged (also liquid horsepower)
Subscripts 1,2 A L d d1 s1 R s
= Refer to first and second condition, respectively = Available from pump system (NPSH) = Liquid = Discharge side of pump = Friction losses for pipe fittings and related items on discharge side of pump = Friction losses for pipe valves and other system losses, suction side of pump = Required by pump (NPSH) = Suction side of pump
Pumps 495
Greek Symbols
= Fluid mass gravity, lb/ft3
References
1. Yeddidah, A., Multistage Centrifugal Pump, Chem. Eng., Vol. 91, No. 24, p. 81, 1984. 2. Dufour, J. W., Revised API Pump Standard, Chem. Eng., Vol. 96, No. 7, p. 101, 1989. 3. Lahr, P. T., Better Standards: Better Pumps, Chem. Eng., Vol. 96, No. 7, p. 96, 1989. 4. Taylor, I., Cameron, B., and B. Wong., A Users Guide to Mechanical Seals, Chem. Eng., Vol. 95, No. 12, p. 81, 1988. 5. Adams, W. V., Troubleshooting Mechanical Seals, Chem. Eng., Vol. 90, No. 3, p. 48, 1983. 6. Sniffen, T. J., Mechanical Seals From A to Z, Power and Fluids, Winter, Worthington Corp., Harrison, NJ, 1958. 7. Fischer, E. E., Seals and Packings Selection Criteria., Chem. Proc., p. 60, Oct. 1983. 8. Hydraulic Institute Standards for Centrifugal, Rotary and Reciprocating Pump, 13th. Ed., 1975 and Engineering Data Book, 1st ed., 1979, Hydraulic Institute. Also see 14th. Ed., 1983, and 2nd. Ed. 1991 respectively. 9. Mann, M. S., Horizontal and Vertical Pumps, Pet. Ref., Vol. 32, No. 12, p. 111, 1953. 10. Karassik, I., and R. Carter, Centrifugal Pumps, McGraw-Hill Book Co, 1960. 11. Centrifugal Pump Fundamentals, Form 7287, Ingersoll-Rand Co., Cameron Pump Division, P. O. Box 636, Woodcliff Lake, NJ. 07657. 12. Carter, R. and I. J. Karassik, Basic Factors in Centrifugal Pump Application and Basic Factors in Preparing a Centrifugal Pump Inquiry, Reprint RP – 477, Worthington Corp., Harrison, NJ, reprinted from Water and Sewage Works magazine. 13. Kern, R., How Discharge Piping Affects Pump Performance, Hydroc. Proc., Vol. 51, No. 3, p. 89, 1972. 14. Karassik, I. J., Krutzsch, W. C., Fraser, W. H., and J. P. Messina, Pump Handbook, McGraw – Hill Book Co., NJ, 1976. 15. Karassik, I. J., Are You Short on NPSH?, Combustion, p. 37, July 1980. 16. Karassik, I. J, and T. W. Edwards, Centrifugal Pumps on Closed Discharge Industry and Power, No. 6, p. 54, 1955. 17. Mann, M., How to Use System-Head Curves, Chem. Eng., No. 2, p. 162, 1953. 18. Dean Brothers Pumps, Engineering Catalog Circular No. 190, Mar. Indianapolis, IN, 1958. 19. Branan, C. R., Pocket Guide to Chemical Engineering, Gulf Publishing Co., Houston, TX, 1999. 20. Taylor, L, Pump Bypass Now More Important, Chem. Eng., Vol. 94, No. 7, p. 53, 1987. 21. Church, A. H., and H. Gartmann, eds., DeLaval Handbook, 2nd ed., DeLaval Steam Turbine Co., Trenton, NJ, 1955. 22. Welch, H. J., ed., Transamerica DeLaval Engineering Handbook, 4th. Ed., McGraw-Hill Book, Co. Inc., 1970 and 1983 by Transamerica Delaval, Inc., IMO Industries Div. 23. Branan, Carl, Rules of Thumb for Chemical Engineers, 4th ed., Gulf Publishing Professional, 2005. 24. Kelly, J. Howard, Understand the Fundamentals of Centrifugal Pumps, CEP, pp. 22–28, Oct. 2010. 25. Durand, Alejandro Anaya, Charting NPSH Values of Pumps, Calculation & Shortcut Deskbook, Chemical Engineering, p. 100, Mc. Graw-Hill Co., 2018 26. Kern, R., How to Design Piping for Pump-Suction Conditions, Chem. Eng., April 28, 1975. 27. Lieberman, Norman, P. and E. T. Lieberman, A working Guide To Process Equipment, 3rd. ed., Mc Graw-Hill Co., 2008. 28. Thakore, Shchen B., and Bharat I. Bhatt, Introduction to Process Engineering and Design, 2nd. Ed., McGraw-Hill India, 2015. 29. Woods, Donald, R., Successful Trouble Shooting for Process Engineers—A Complete Course in Case Studies, Wiley-VCH, 2006. 30. Gas Processors Suppliers Association (GPSA) Engineering Databook, 12th Ed., 2004. 31. Reynolds, John, A., Pump Installation and Maintenance, Chem. Eng., Oct, 11, 1971. 32. Sarver, Joseph, Finkenauer, Blake and Y.A. Liu, Pump Sizing and Selection Made Easy, Chemical Engineering, www. CHEMENGONLINE.COM, p. 34, January 2018. 33. Hanyak, Jr. M. E., “Chemical Process Simulation and the Aspen Hysys Software, MEH, 1998 to 2012. 34. White, S., and S. Fulton, “Specifications—importance of getting them right”, www.digitalrefining.com/article/1000320, Petroleum Technology Quarterly (PTQ), pp 1–9, Q4, 2003. 35. Kumana, Jimmy D, and Maneul, R. Suarez, “Analyzing the Performance of Pump Networks: Part 2: Improving Pump Efficiency, CEP, pp 32–41, February, 2018. 36. American Petroleum Institute, “Centrifugal Pumps for Petroleum, Petrochemical and Natural Gas Industries”, 10th ed., API Standard 610 (ISO 13709), API, Washington, D.C., 2004.
496 Petroleum Refining Design and Applications Handbook Volume 2 37. Hydraulic Institute, http://www.pumps.org/, retrieved on May 25, 2011. 38. Holland, F. A., and R. Bragg, Fluid Flow for Chemical Engineers, 2nd ed., Edward Arnold, 1995. 39. Tsai, M. J., Chem. Eng., 75 (13), June 17, p. 190, 1968. 40. White paper – 3 Innovative Ways to Improve Pump Reliability, Emerson Process Management, www.emersonprocess. com/pumps. 41. Bloch, H. P., “Reliability tips for centrifugal process pumps, http://www.hydrocarbonprocessing.com/Article/3337288/, May 2014. 42. Hall, Stephen., “Rules of Thumb for Chemical Engineers, 5th ed., Butterworth-Heinemann, imprint of ButterworthHeinemann, imprint of Elsevier, 2012. 43. Tom Baxter, Energy Saviours—Ways for operations and design engineers to boost efficiencies, The Chemical Engineer, p 43, September, 2018. 44. Ashis Nag., Distillation & Hydrocarbon Processing Practices, PennWell Corporation, 2016. 45. Center for Chemical Process Safety, Guidelines for Safe Process Operations and Maintenance, 1995. 46. Occupational Safety and Health Administration (OSHA), Process Management Standard.
17 Compression Equipment
17.1 Introduction A compression system is a simple arrangement of equipment designed to produce clean, dry compressed gas or air for industrial applications. These equipment types are used to transfer or compress light hydrocarbon gases, hydrogen, nitrogen, carbon dioxide, chlorine, and a variety of specialty gases. Compressors are used at pipeline lift stations to add energy to compressed feed stocks. A compression system comprises: process piping, valves, a compressor, a receiver, heat exchangers, dryers, back pressure regulators, gages, and moisture removal equipment. In a refinery or petrochemical plant, compressors are employed to compress gases such as nitrogen, hydrogen, carbon dioxide, and chlorine. These gases are sent to headers from which they are distributed to a variety of applications. When compressors are used in a process system, various supporting equipment types are used, such as safety valves, motor, lubricating systems, control instruments, regulators, demisters, and pipe header. The primary purpose of a compressor is to compress gases to create energy to transfer the gas from one place to another. The compression of gases and vapors forms an important operation in refineries, chemical and petrochemical plants, as it is necessary to be able to specify the proper equipment type by its characteristic performance. The compression step is conveniently identified for the process design engineer by the principal operation of the equipment: 1. 2. 3. 4.
eciprocating R Centrifugal Rotary displacement Axial flow
Compression may be from below atmospheric pressure, as in a vacuum pump or above atmospheric as for the majority of process applications. The word done by Scheel [1–3] is useful. This chapter presents the process and mechanical engineer with the basic details of reciprocating, centrifugal, and other major types of process compressors. The chapter provides equations for determining the work done, power requirements for adiabatic and polytropic compression processes, specifies the required process performance and mechanical requirements, including the corrosive and hazardous nature and the moisture content of the fluids (gases/vapors) to be compressed. Finally, the chapter lists troubleshooting options for both reciprocating and centrifugal compressors, the comparison between these compression types, and their applications in the refinery and chemical process industries.
A. Kayode Coker Petroleum Refining Design and Applications Handbook Volume 2, (497–836) © 2021 Scrivener Publishing LLC
497
498 Petroleum Refining Design and Applications Handbook Volume 2
17.2 General Application Guide Figures 17.1A, 17.1B, 17.1C, and 17.1D present a general view of the usual ranges of capacity and speed operation for the types of compression equipment listed. For pressure conversions of Figure 17.1B, 1000 psi = 6.8947 mpa. Figure 17.1E presents a classification of various compressor types.
Compressor Speed, rpm
105
104
103
102 10
102
103 104 Compressor Inlet Capacity, cfm
105
106 3×106
Axial Flow
Centrifugal
Reciprocating
Blowers and Fans
Vane Rotary Compressor
Lobe and Screw Rotary Displacement
Figure 17.1A General areas of compressing equipment application (source: De Jardins, P. R. Chemical Engineering, V. 63, No. 6, © 1956. McGraw-Hill, Inc., New York, All rights reserved).
1,000
Discharge Pressure, MPA
100
Piston Compressors
10
Centrifugal Compressors Diaphragm Compressors
1
Axial Compressors
0.1 1
10
102
103 104 Inlet Flow Actual m3/hour
105
106
10
Figure 17.1B Approximate ranges of application for usual process reciprocating centrifugal, diaphragm, and axial flow compressors used in chemical/petrochemical processes. Note that ethylene gas reciprocating compressors in the low density, high pressure process can reach 50,000–65,000 psi; 1000 psi = 6.8947 mPa; for example, 500 mPa = 72,519 psi (used by permission: Livingston, E. H., Chemical Engineering Progress, V. 89, No.2, © 1993. American Institute of Chemical Engineers, Inc. All rights reserved).
Compression Equipment 499 1000 Centrifugal units with vertically split casing (barrel types)
500
100
50 Pressure, bar
High pressure integrally geared centrifugal compressors
10 Axial-centrifugal compressors
Centrifugal compressors with horizontally split casing
5
Axial compressors
1 0.1
0.5
0.5
1
0.5
5 10 Volume flow 5
1
10
5
50
50
10
100
50
500 m3/s
100
500
100
1000 × 1000 m3/h
500 × 1000 ft3/min
Figure 17.1C Typical application ranges for turbocompressor capabilities extend over wide ranges of volume flow and pressures. Note: Barx 14.5 = psi (used by permission: Nissler, K. H. Chemical Engineering, V. 98, No. 3, p. 104, © 1991, McGraw-Hill, Inc., All rights reserved).
Table 17.1 outlines the compression limits for this equipment type. The value of the chart and table is to aid in establishing the probable types of equipment suitable for an operation. However, as in many other process situations, equipment is designed to handle special cases that might not be indicated by the guide. Usually inlet suction flow rate of the fluid, temperature, and pressure, as well as the outlet conditions and nature of the fluid, are all involved in identifying the equipment type best suited for the application. See Monroe [4], Huff [5], and Patton [96] for comparison. Also see Leonard [6].
17.3 Specification Guides Compressor cylinders or other pressure-developing mechanisms are never designed by the process companies involved in their operation, except in rare instances in which special know-how is available or secret process
500 Petroleum Refining Design and Applications Handbook Volume 2 Compressors Dynamic
Ejector
Displacement
Radial
Axial
Rotary One rotor
Vane
Liquid ring
Reciprocating Two rotors
Trunk Screw
Screw
Crosshead
Roots
Labyrinth Diaphragm
Figure 17.1D Basic compressor types (used by permission: Coker, A. K. Hydrocarbon Processing, V. 73, No. 7, p 39, © 1994. Gulf Publishing Co, Houston, Texas. All rights reserved).
Compressors
Positive Displacement Type
Dynamic Type
Reciprocating
Rotary
Single-Stage Multi-Stage Integral Gas-Engine Driven Separable Balanced/Opposed
Straight Lobe Helical Lobe (Screw) Sliding Vane Liquid - Ring
Radial flow (Centrifugal)
Single-Stage Multi-Stage Horizontally Split Vertically Split (Barrel) Integral Gear
Diaphragm
Mixed Flow
Figure 17.1E Types of compressors.
Thermal Type
Axial Flow
Ejectors
Multi-Stage Fixed Stator Vanes Variable Stator Vanes
Single-Stage Multi-Stage
Compression Equipment 501 Table 17.1 General compression and vacuum limits. Compressor type
Approx. max. commercially used disch. press., psia
Approx. max. compression ratio per stage
Approx. max. compression ratio per case or machine
Reciprocating
35,000–50,000
10
As required
Centrifugal
3000–5,000
3–4.5
8–10
Rotary displacement
100–130
4
4
Axial flow
80–130
1.2–1.5
5–6.5
Vacuum pump type
Approx, suction pressure attainable, mm Hg abs
Centrifugal
6
Reciprocating
0.3
Steam jet ejector
0.05
Rotary displacement
10−5
Oil diffusion
10−7 (or 10−4 micron)
Mercury or oil diffusion plus rotary
Less than 10−7
Used by permission and compiled in part from: Dobrowolski, Z. Chemical Engineering, V. 63, p. 181, ©1956 and Des Jardins, P. R. Chemical Engineering, V. 63, p. 178, ©1956. McGraw-Hill, Inc. All rights reserved.
information is involved. In the latter case, the process company might have the compressor cylinders (or compression components) built in accordance with special plans, purchase standard frames or housings, and then assemble the driver, cylinder, and packing at the plant site. Usually the selection of the basic type of compression equipment for the operation can be determined prior to consulting the manufacturers. However, when in doubt or where multiple types may be considered, enquiries should be sent to all manufacturers offering the equipment. Preparation of complete and appropriate specifications is of paramount importance in obtaining the proper performance rating as well as price considerations. Preliminary design rating calculations are usually prepared as guides or checks. The final and firm performance information is obtained from the manufacturer of the specific equipment. No standards of design exist among manufacturers; therefore, the performance will vary according to the details of the specific equipment. All performance will be close to the requirement, but none may be exact. This is the point where knowledge of compressor types and details is important to the engineer. Bid evaluations must include detailed analysis of performance, power drivers, and materials of construction.
17.4 General Considerations for Any Type of Compressor Flow Conditions In establishing specifications, the first important items to identify from the plant process material balance are normal, maximum, and minimum intake or suction flow rates together with corresponding conditions of temperature and pressure. The required discharge pressure must be established. If it is necessary or important to be able to operate at reduced or over-normal flow rates, these should be identified for the manufacturer, together with the length of time of such expected condition; e.g., full time at one-half rate, 20 min out of every hour at 10% over normal, and so on. These operating requirements may separate the types of equipment. Because it is uneconomical to purchase horsepower that cannot be used by the fluid system, ask that the manufacturer state the maximum load and/or conditions that will fully load the available horsepower of the compressor-driver unit.
17.4.1 Fluid Properties Fluid properties are important in establishing the performance of compression equipment. Whenever possible, fluid analysis should be given, and where this is not available due to lack of complete information or secrecy, close
502 Petroleum Refining Design and Applications Handbook Volume 2 approximations are necessary. Under these last conditions, actual field performance may not agree with the design data due to the deviation in values of the ratio of specific heats and the average molecular weights. Identify, as to composition and quantity, any entrained liquids or solids in the gas stream. No manufacturer will design for entrained liquids or solids, although some machines will handle “dirty” gases. Solids are always removed ahead of any compression equipment, using suitable wet- or dry-scrubbing equipment, and liquid separators are recommended for any possibilities of liquid carry over.
17.4.2 Compressibility Gas compressibility has an important bearing on compressor capacity performance. Therefore, it is good practice to state compressibility values at several temperature and pressure points over the compression range under consideration. When possible, a compressibility curve or reference thereto is included in the inquiry. Where specific information is not available, but compressibility is anticipated as being a factor to consider, approximate values should be established and so presented for further study by the manufacturer. Compressibility is expressed as the multiplier for the perfect gas law to account for deviation from the ideal. At a given set of conditions of temperature and pressure:
where Z N R T P V
PV = ZNRT
(17.1)
= compressibility factor, usually less than 1.0 = number of lb-mol = gas constant, depends on units of pressure, volume, and temperature = 1545 lb/ft3/lbmole. °R or ft. lb/lbmole. °R, 10.73 psia. ft3/lbmole °R = 8.314 kJ/kmole.K, 8.314 kPa (abs) m3/kmole.K, 213.6 kg. m/kmole.K = absolute temperature, °R = °F + 460 (K = °C + 273.15) = pressure, absolute, psia (kPa, bara) = volume, ft3(m3) (see paragraph to follow)
Gas volumes are corrected at the intake conditions on the first and each succeeding stage of the compression step, and compressibility factors are calculated or evaluated at these individual intake conditions. Some manufacturers use the average value between intake and discharge conditions.
17.4.3 Corrosive Nature Corrosive fluids or contaminants must be identified to the manufacturer. The principal gas stream may or may not be corrosive under some set of circumstances, yet the contaminants might require considerable attention in cylinder design. For example, considerable difference exists between handling “bone-dry” pure chlorine gas and the same material with 5 ppm moisture. The corrosiveness of the gas must be considered when selecting fabrication materials for the compression parts as well as seals, lubricants, and so on.
17.4.4 Moisture Moisture in a gas stream might be water vapor from the air or a water scrubber unit, or it could be some other condensable vapor being carried in the gas stream. It is important in compressor volume calculations to know the moisture (or condensable vapor) condition of the gas.
17.4.5 Special Conditions Often the process may have conditions that control the flexibility of compression equipment selection. These might include limiting temperatures before polymer formation, chemical reaction, excess heat for lubrication materials,
Compression Equipment 503 explosive conditions greater than a certain temperature, and so on. Any limiting pressure drops between stages should be specified, in which the gas and vapors are discharged from one stage, pass through piping, cooling equipment and/or condensate knock-out equipment, and are then returned to the next higher stage of the compression process. Usually a reasonable figure of 3–5 psi (0.21–0.34 bar) can be tolerated as pressure drop between stages for most conditions. The larger this drop is, the more horsepower is required. Special situations might hold this figure to 0.5–1 psi (0.034–0.069 bar).
17.5 Reciprocating Compression Mechanical Considerations Fundamental understanding of the principles involved in reciprocating compression is important for proper application of compressors to plant problems. The reciprocating compressor is a positive displacement unit with the pressure on the fluid developed within a cylindrical chamber by the action of a moving piston. Figures 17.2A–U illustrate the assembly and arrangements of typical cylinders for various pressure ranges and types of services. Figures 17.3A and 17.3B show cross-sections of a cylinder and crankshaft arrangement for two different styles of compressors. Compressor types, components, and arrangements are designed as:
Cylinders 1. S ingle Acting: Compression of gas takes place only in one end of the cylinder. This is usually the head end (out-board end), but may be the “crank” end (inboard end or end of cylinder nearest crankshaft of driving mechanism) (see Figures 17.2A, 17.3A, and 17.4A). 2. Double Acting: Gas compression takes place in both ends of the cylinder, head end, and crank end (Figures 17.2I, 17.2J, and 17.4B). Large inspection plate
Heavy enclosed oil tight frame
Suction valve
Forged steel connecting rod
Oil wiper ring
Unloader cylinder
Suction
Breather
Discharge valve Oil gauge Grout
Discharge
Counter balanced crank shaft
Tapered roller main bearing
Box section crosshead
Frame oil head
Liberal cooling water jackets
Figure 17.2A Sectional assembly. Worthington single stage, belt-driven air compressor showing construction of air cylinder and running gear (used by permission: Bul. L-600-B9. Dresser-Rand Company).
504 Petroleum Refining Design and Applications Handbook Volume 2 Discharge Double Deck Valves
Piston Lubrication Inlet Suction Valves To Crank
Head End Clearance Pocket Crank End of Cylinder (Next to Driving Mechanism) Head End of Cylinder
Piston Rod Packing Vent or Purge Packing Piston with Rings Outlet Gas
Figure 17.2B Cutaway view of typical high pressure gas cylinder showing double-deck feather valves in place (used by permission: DresserRand Company).
Figure 17.2C Dry vacuum pump cylinder for very low absolute suction pressures. Valves in heads for low clearance and high volumetric efficiency (used by permission: Bul. L-679-BIA, © 1957. Dresser-Rand Company).
Compression Equipment 505
Figure 17.2D Standard air compressor cylinder for 125 psig discharge pressure. Suction valve unloaders for automatic capacity control (used by permission: Bul. L-679 – BIA, © 1957. Dresser-Rand Company).
Figure 17.2E A 250 psig working pressure cylinder used in refrigeration service. Auxiliary stuffing box for added sealing on shutdown. Manual fixed volume clearance pockets for capacity control (used by permission: Bul, L-679-BIA, © 1957, Dresser-Rand Company).
Note that the crank end (Figure 17.2F) always has the piston rod running through it, while the head end usually does not, but may, if a tail rod (Figure 17.2L) is used.
Frames The cylinders are arranged on the main frame of the compressor to provide balanced crankshaft power loading (when possible), access for maintenance, piping convenience, and floor space to suit plant layout. Common designations by position of the cylinder are as follows: 1. 2. 3. 4. 5.
orizontal. H Vertical. 90° angle, cylinders mounted both vertically and horizontally from same crankshaft, Figure 17.5A. V or Y angle, Figure 17.5A. Radial.
506 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.2F Typical linear-type cast iron cylinder. Cylinder barrel and heads are rugged castings made from an iron composition especially selected for the particular cylinder and service involved. Heads and barrel are thoroughly water-jacketed. Water piping is included from inlet valve to outlet sight-flow indicator. Dry-type liner is of cast-iron, shrunk into the barrel, and extending for the full length of the cylinder. It is positively locked by the heads against end movement and by a threaded dowel against rotation. Piston is cast-iron, hollow for light weight, and well ribbed for strength. The piston is locked on the rod between either a taper or a solid collar and a nut. Piston rings are normally cast-iron of the single piece snap-ring type, although other materials may be used when conditions require. Piston rod is carbon steel, flame-hardened over packing travel area. The rod is packed with full-floating metallic packing, force-feed lubricated, and vented when the gas composition requires. Vented packing is illustrated. Distance piece opening provides free access to packing. All head studs and nuts are external and accessible without removal of valves to reach internal bolting. There is no possibility of hidden internal leakage. Support is provided at outer end (see sketch) to carry weight of cylinder directly to foundation. The piping need not be designed to support the cylinder. Clearance pocket shown in the head is the manual fixed volume type. Recycle and other services frequently require absolute elimination of lubricating oil contamination of the gas. The cylinder must operate with no oil (used by permission: Ingersoll-Rand Company).
6. D uplex, cylinders mounted in parallel on two separate frames from common crankshaft, Figure 17.5B. 7. Balanced Opposed, cylinders mounted opposite (180°) and driven off same crankshaft, Figures 17.5C and 17.5E. Also see Figure 17.5F; the following results are used by permission (Bul. PROM 635/115/95II, Nuovo Pignone, S.P.A.): • Either zero or minimum unbalanced forces and moments on the foundation. • Minimum foundation size and expense. • Minimum drive-end peak torques, reducing drive train torsional stresses. • Reduce motor current pulsations and power costs. • Reduce harmonic torques on the foundation. With a bearing between each crank throw and an extra main bearing and an outboard bearing for extra support at the HHE’s drive end, the result is minimum crankshaft deflections, minimum crankshaft stress, minimum driveend bearing loads and maximum crankshaft and bearing life. For example, an HHE for a three cylinder application has three crank throws set at 120°. “Piston weights may be balanced or balance weights added to the active crossheads to obtain zero unbalanced primary forces. By comparison, a fixed angle crankshaft requires four crankthrows with either an additional compression cylinder or a balance weight dummy crosshead to obtain acceptable unbalanced forces.” The minimum number of cylinders
Compression Equipment 507
Figure 17.2G Typical non-lubricated recycle cylinder. Cylinder barrel and heads are rugged castings and may be either cast-iron or cast steel depending upon pressure and service requirements. In each case, water jackets are supplied, and piping is included from inlet valve to outlet sight-flow indicator. Dry-type Cast-iron liner is furnished in all nodular iron or cast steel cylinders for best wearing characteristics. Full cylinder-length liner is shrunk into the barrel and securely locked against any movement. Piston is an assembly of carbon rings held between steel end-plates. The piston is locked on the rod between a solid collar and a nut. When the bottom bearing surface of the carbon piston becomes worn, a new surface is easily made available by turning the assembly through an appropriate arc. This will be infrequent; the light solid piston design ensures maximum life. Piston rings are a special carbon material, of the segmental type, held to the cylinder wall by stainless steel expanders. Piston rod is carbon steel, flame-hardened over packing travel area unless otherwise specified. Two compartment distance piece. This illustration shows a typical design for maximum protection in two directions. First, to prevent crankcase oil particles from reaching the cylinder, and second, to prevent any contamination of the crankcase oil by gas constituents. The compartment nearest the cylinder is of sufficient length to prevent any part of the rod traveling from one stuffing box into another. Furthermore, a baffle collar stops crankcase oil from creeping along the rod. Normally, in this design, the distance piece toward the cylinder is enclosed with solid covers over the access openings and has provisions for venting or purging. The section on the crankcase end is open. Full-floating packing rings in the cylinder and middle partition are of a special carbon material. Only the cylinder packing is vented, unless a positive pressure is to be held in the cylinder end compartment, in which case the partition packing is also vented. They are vented in both cases in this design. Considerable heat is generated in the cylinder packing, particularly during break-in periods. It has been found desirable to remove this heat, which might otherwise result in an overheated and warped rod. Water-cooling is used, the water being circulated through a special packing cup. Support is provided at outer end (see sketch) to carry weight of cylinder directly to foundation. The piping need not be designed to support the cylinder. The channel valve, as designed especially for non-lubricated service (used by permission: Ingersoll-Rand Company).
Figure 17.2H Double-acting cast steel cylinder to 3500 psi pressure (used by permission: Cooper-Cameron Corporation. All rights reserved).
508 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.2I Double-acting cast Meehanite or ductile iron cylinder to 1250 psi pressure (used by permission: Cooper-Cameron Corporation. All rights reserved).
Figure 17.2J Double-acting Meehanite metal or ductile iron cylinder to 1000 psi pressure (used by permission: Cooper-Cameron Corporation. All rights reserved).
Figure 17.2K Forged steel single-acting for 6000 psi pressure (used by permission: Cooper-Cameron Corporation. All rights reserved).
Compression Equipment 509 LUBRICATOR CONNECTION
WATER OUTLET
WATER OUTLET LUBRICATOR CONNECTIONS
DRAIN
VENT
VENT WATER INLET
LUBRICATOR CONNECTION
WATER INLET
WATER INLET
VENT
Figure 17.2L Typical forged steel cylinder with tail-rod. Construction of forged steel, tail-rod cylinders (other than circulators) as shown here. Cylinderbarrel with integral packing boxes is a single steel forging or a material especially selected for the design, pressure, and service requirements of each case. There are no heads as such; the forged steel nose-pieces of the piston and tail-rod packing act as closures. Both cylinder barrel and stuffing boxes are water-jacketed. Water piping is included from inlet valve to outlet sight-flow indicator. Alignment of packing boxes and cylinder bore is ensured because all three are bored at a single machining setup. This is an important factor in packing life. Dry-type lineris a special cast-iron, full length of the cylinder, shrunk in place, and securely locked against any movement. Pistonis usually cast iron, locked on the rod between a solid collar and a nut. If a cylinder size is such that sufficient piston wall thickness is not available, the piston and rod are forged integrally as shown, and special inserted rider rings are included to improve wearing qualities. Pistonrings are of the single-piece, snap-ring type. Piston rod and tail-rodare one piece of carbon steel, flame-hardened over packing travel area. The rod is packed with full-floating metallic packing, force-feed lubricated, and vented when the gas composition requires. Vented packing is shown. Distance piece openingprovides free access to frame end packing. Packing casesare completely contained and supported in full depth boxes in cylinder. Unequal tightening of flange bolts cannot destroy the alignment. Supportis provided at outer end (see sketch) to carry weight of cylinder directly to foundation. Valvesare cushioned type valves. This design is used in 3000–15,000 psi ammonia, methanol, and hydrogenation plant services (used by permission: Ingersoll-Rand Company).
Figure 17.2M For low compression ratios, designed for 1000 psi discharge pressure equipped with hand-operated crank and heat-end fixed clearance pockets for capacity control. Air-cooled, cast semi-steel, double-acting (used by permission: Dresser-Rand Company).
510 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.2N Designed for working pressure up to 6500 psi. A similar design cylinder is used for pressures in excess of 6500 psi. Water cooler, forged-steel, double-acting (used by permission: Dresser-Rand Company).
Figure 17.2O Fourth and fifth-stage cylinder assembly of 3500 psi pressure hydrogen compressor. Opposed single-acting cylinders balance frame-bearing loading and minimize torque fluctuations (used by permission: Dresser-Rand Company).
Figure 17.2P Fifth and sixth-stage cylinder assembly of 15,000 psi gas compressor. “Bathtub” design of intermediate crosshead permits use of short opposed plungers ensuring operating alignment. A design for pressures as high as 35,000 psi (used by permission: Dresser-Rand Company).
Compression Equipment 511
Cast or nodular iron cylinders for pressures to 1,500 PSI.
Figure 17.2Q Cast or nodular iron cylinders for pressures to 1500 psi. Note double “distance pieces” (left vented or purged, to prevent oil and process gas from leaking past the shaft; and right-end fixed clearance pocket) (used by permission: Bul 85084, © 1992. Dresser-Rand Company).
(a) Dresser-Rand Standard FRAME END
VENT
CYLINDER END
WIPER RINGS
PACKING DRAIN
Long single-compartment distance piece (sufficient length for oil slinger travel). (b)
FRAME END
VENTS
CYLINDER END
OIL SLINGER
WIPER RINGS
DRAIN
PACKING DRAIN PARTITION PACKING
Two-compartment or double distance piece arrangement (in-board distance piece of sufficient length for oil slinger travel).
Figure 17.2Q (a) Long, single compartment distance piece (sufficient length for oil slinger travel). (b) Two-compartment or double distance piece arrangement (inboard distance piece of sufficient length for oil slinger travel) (used by permission: Bul. 85084 ©1992. Dresser-Rand Company).
512 Petroleum Refining Design and Applications Handbook Volume 2
Fabricated carbon or stainless steal cylinder for special applications.
Figure 17.2R Fabricated carbon or stainless-steel cylinders for special applications. Note double-distance piece, no clearance pocket (used by permission: Bul. 85084, © 1992. Dresser-Rand Company).
Forged steel cylinder with tail-rod design (right) for pressures to 7,500 PSI.
Figure 17.2S Forged steel cylinder with tail-rod design (right) for pressure to 7500 psi (used by permission: Bul. 85084, © 1992. DresserRand Company).
Figure 17.2T Medium or high pressure, double-acting cylinder with flanged liner. The liner is locked in place by a flange between head and cylinder barrel. A step on the liner O.D. permits easy insertion. The cylinder may be made of cast-iron, nodular iron, or cast steel, depending on operating pressure. Note: Optional two compartment distance piece (type D) designed to contain flammable, hazardous, or toxic gases is illustrated (used by permission: Bul. 33640, June 1985. © Dresser-Rand Company).
Compression Equipment 513
Figure 17.2U High pressure, circulator-type cylinder, double-acting. The steel cylinder and packing box are forged in one piece, and the one-piece piston and rod ensure positive alignment. The packing boxes are water cooled, and the packing is additionally cooled by internally circulated oil. Note tail-rod construction (used by permission: Bul. 3640, June 1985. © Dresser-Rand Company).
13
12
16
15
14
1
2
11
3
10
9
8
7
6
5
4
Figure 17.3A Typical cross-section of motor-driven, single-stage compressor. 1. Valves. 2. Piston sealed by two single-piece snap rings. Rod threaded and locked into piston. 3. Cylinder head. 4. Cylinder barrel, head, and air passages water-jacketed for cooling. 5. Air passages. 6. Distance piece allows access to packing and oil-wiper rings. 7. Crosshead guide. 8. Counterweights and permanently bolted in place. 9. Foundation. 10. Screened oil suction. 11. Crankpin and main bearings. 12. Frame. 13. Die-forged steel connecting rod has rifle-drilled oil passage. 14. Crosshead pressure-lubricated through drilled passages in crosshead body. Piston rod threaded and locked into crosshead. 15. Wiper rings keep crankcase oil out of cylinder. 16. Full-floating metallic packing is self-adjusting (used by permission: Ingersoll-Rand Company. All rights reserved).
514 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.3B Partial cross-section of balanced opposed compression cylinders. (A) Single-acting cylinder (B) Double-acting cylinder (used by permission: Bul. 85084, (c) 1992. Dresser-Rand Company. All rights reserved).
(a)
(b)
Stroke Suction Valve Cylinder Head End
End of Stroke Piston Rod
Drive
Crank End Discharge Valve Single Acting Cylinder
Stroke
Suction Valve
End of Stroke Piston Rod
Cylinder Head End
Drive
Crank End Discharge Valves Double Acting Cylinder
Figure 17.4 Cylinder action.
is not “locked” into even numbers to handle compression problems. The minimum number required is used; therefore, “balanced weight dummy crossheads are not necessary.”
8. F our-cornered, opposed, two cylinders mounted opposite (180°) at each end of crankshaft. 9. Tandem, two or more cylinders are on the same compressor rod, or one cylinder may be steam operated as drive cylinder with second as compressing cylinder. May also be duplex or multiple tandem (Figure 17.5D). Also see Figure 17.5E.
17.6 Suction and Discharge Valves Several of the types of valves used in compressor cylinders are shown in Figure 17.6. To function properly, a valve must seat uniformly and tightly, yet must not have “snap-action” on opening or closing. Until pressure builds up to the discharge point, the valve must remain closed, open at discharge pressure, and then reseat as the pressure in the cylinder drops below the discharge value. The same type of action is required for the suction valves. Valves must be made of fatigue-resistant carbon or alloy steel or 18-8 stainless steel, depending upon the service. The 18-8 stainless and 12-14 chrome steel are often used for corrosive and/or high temperature service. Any springs,
Compression Equipment 515 (A)
(B)
Rod
Vertical Cylinder
Crank Shaft
Horizontal Cylinder Crank Shaft
Rod
90° Angle Compressor
Driver may be Located either Position
Duplex Compressor Cylinders
For Y or V-type, Angle between Compressor Rods is still 90°, however Cylinders are Rotated to be about 45° from Vertical.
(C)
Crank Shaft
Crank Shaft
(D)
Rod Rod
Rod
Balanced Opposed Compressor Cylinders Additional Groups of Two Opposing Cylinders may be Fixed to the Same Crank Shaft
Tandem Compressor Cylinders Sometimes the Inboard Cylinder is Steam Operared as the Driver Cylinder
Figure 17.5A–D Cylinder arrangement.
as in the plate-type valves, are either carbon or nickel steel. Valve passages must be smooth, streamlined, and as large as possible. Cylinder efficiency depends to a certain extent upon the proper selection and sizing of the valves. Valves must be adequately cooled, so provision is usually made for water jackets immediately adjacent to the valves, particularly the discharge valves. Bauer [7] has studied losses in compressor cylinder performance associated with valve losses as they relate to overall efficiency. Bunn [8] examines poppet valves for retrofitting cylinders. Double-deck valves reduce valve velocities in large diameter cylinders. With these valves, high clearance volumes and clearance pockets can be added to give additional unloading of a cylinder as designed to maintain proper loading on the driver. For a typical d ouble-deck valve, the theoretical indicated horsepower loss may be 6% at a ratio of compression, Rc of 3.0 and 17% at an Rc of 1.5. The valve plates, discs, and springs are mounted in a valve cage, which is inserted in the cylinder. Valve breakage occurs due to fatigue of the metal or improper action. This requires replacement and an evaluation of the materials of construction as well as the basic type of valve. Each manufacturer presents his valve design to match his equipment. Only experience can determine which type of valve works best in a given application.
Piston Rods See Figure 17.7. Piston rods are usually forged and hardened steel or alloy.
516 Petroleum Refining Design and Applications Handbook Volume 2
HHE tree-throw crankshaft arrangement vs. Fixed-angle crankshaft design.
90°
120°
HHE
HHE three-throw crank angles are each 120.
Other
Fixed 90 angles require a dummy crosshead to balance the three throws.
Figure 17.5E Balanced arrangement for Dresser-Rand shaft system, 1–10 crank throws (used by permission: Bul. 85084, © 1992. DresserRand Company).
Figure 17.5F Lubricated and non-lubricated balanced opposed process reciprocating compressors, designed to API 618 code. Fixed and variable speed drives using gas or diesel engines, steam or gas turbines, or electric motor. Note power drive to connect to right side of crosshead box in center (used by permission: Bul. PROM 635/115/95-II. Nuovo Pignone S.P.A. Florence, Italy; New York; Los Angeles, and Houston, Texas. All rights reserved).
Compression Equipment 517
Figure 17.6A Double-deck feather valve (used by permission: Dresser-Rand Company. All rights reserved).
Figure 17.6B Double-deck valve with valve cap unloader (used by permission: Cooper-Cameron Corporation).
518 Petroleum Refining Design and Applications Handbook Volume 2
VALVE STRIPS CLOSED
VALVE STRIPS BEGIN TO OPEN
VALVE STRIP FULLY OPEN
Figure 17.6C Action of gas flow through strip-type feather valve (used by permission: Bul. S-550-B27. Dresser-Rand Company).
Ribbed Body Construction Hardened Valve Seats Fatigue Resistant Multiple Rings Valve Discs Positively Cushioned by Increasing Spring Tension as They Approach Stops Multiple Springs Assure Uniform Valve Action Valve Bodies of High Tensile Strength Semi-Steel
Figure 17.6D Plate-type valves (used by permission: Dresser-Rand Company).
Compression Equipment 519 Stop plate
Valve springs Valve channels
Valve guides
Seat plate Valve seat Valve closed
Valve open
Figure 17.6E Channel-type valves (used by permission: Ingersoll-Rand Company).
Figure 17.6F Ring channel valves (used by permission: Cooper-Cameron Corporation).
ANNUAL SAVINGS U.S. $(000)
520 Petroleum Refining Design and Applications Handbook Volume 2 kW Consumption
$.60/ kW
150
$.05/ kW $.04/ kW
100
$.03/ kW
200
50
0 50 100 150 200 250 300 350 400 450 500 VALVE LOSS HP SAVED
Fully machined 1141 steel, ductile iron or stainless steel seats and guards Poppets — available in PEEK® materials or fully machined nylon. Durable, self-seating, lightweight, low cost Springs — variable rate 17-7PH stainless steel. Custom-match rated. API-618 design. A broken stud can’t work its way into cylinder
Figure 17.6G AJAX® APV-100, high-efficiency compressor valve. Suction and discharge losses are 4–8% compared to conventional valves with losses of 6–20% as a percentage of the total indicated horsepower. A typical Worthington BDC plate valve in closed position. All passageways, lift clearances, springs, and plates have been dynamically designed and individually selected for maximum flow efficiencies. The individual spring plate valve offers both efficiency and reliability advantages over valves that have flexing strips or plates (used by permission: Bul. 2-241. Cooper-Cameron Corporation. Cooper Energy Services. AJAX Superior. All rights reserved).
A typical Worthington BDC plate valve in closed position. All passageways, lift clearances, springs and plates have been dynamically designed and individually selected for maximum flow efficiencies. The individual spring plate valve offers both efficiency and reliability advantages over valves that have flexing strips or plates.
Figure 17.6H Dresser-Rand specialized valve (used by permission: Bul. 3640. Dresser-Rand Company).
Piston See Figure 17.8. Pistons may be of aluminum, built-up carbon or graphite, cast iron, cast steel, fabricated and metalized steel, stainless steel, or forged carbon. The selection involves the corrosive nature of the gas plus the weight- balancing problem of the compressor manufacturer.
Piston Rings See Figures 17.8 and 17.8A. Piston rings are rings mounted on the piston that seal against the cylinder wall and allow the piston to develop required pressures. Many types and materials are available. There are usually at least two rings
Compression Equipment 521 Balanced spring system delivers equivalent motion to the valve plates.
Balanced flow paths through the valve provide minimum pressure drop.
Patented design prevents centerbolt from falling into the cylinder bore, meets API618.
The standard material for plates and buttons is “Hi-Temp”.
Figure 17.6I Dresser-Rand HPS proprietary valve design, using proprietary blend of “PEEK,” an advanced non-metallic valve plate material, allowing for temperature ranges greater than previously available non-metallic plate materials, for lubricated and non-lubricated applications for long life (used by permission: Form 85084, © 1992. Dresser-Rand Company).
(I)
(II)
Figure 17.6J Variety of standard and special valves designed and fabricated by the Hoerbiger Corporation. Many of these designs are used in compressor manufacturer’s cylinder designs, and some valves have been designed by the compressor manufacturer (used by permission: Bul. V-100a, © 1993, Hoerbiger Corporation of America).
522 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.7 Piston rods. Precision-manufactured rolled threads and induction hardening provide high fatigue strength and long life in heavy duty service. Standard rod material is AISI 4142 carbon steel; other materials are available as required. Tungsten carbide coatings are also available for maximum surface hardness. Pistons are locked securely onto the rods. For higher pressure, smaller bore cylinders, the piston may be integral with the rod (used by permission: Bul. 85084, © 1992. Dresser-Rand Company).
Figure 17.7A Stuffing box with rod packing direct and indirect cooling (used by permission: Bul. BCNA-3P100. Howden Process Compressors Incorporated).
Figure 17.8 Piston and rings. Lubricated and non-lubricated pistons with PTFE or other composite materials for the piston and rider rings. These designs prevent piston-to-bore contact and provide reliable service life, particularly during possible periods of lubrication interruption (used by permission: Bul. 85084, © 1992. Dresser-Rand Company).
Compression Equipment 523
Figure 17.8A Piston rings. The piston rod is manufactured from heat treated stainless steel and is coated with wear-resistant overlays, such as ceramic, chromium oxide, and tungsten carbide applied by plasma techniques. Piston-rod crosshead attachment has mechanical preloading system for the threads. Rider rings and seal rings are manufactured from PTFE filled resins; fillers are matched to the gas, piston speed, and liner specifications. Typical fillers are glass, carbon, coke, or ceramic (used by permission: Bul. BCNA- 3P100. Howden Process Compressors Incorporated. All rights reserved).
per cylinder for low pressure applications; six or more for high pressure services. Cast iron, bronze, micarta, aluminum, and carbon (graphite) are common ring materials.
Cylinders Cylinders are made of materials consistent with pressure range and gas service. Sometimes a liner is used to recognize and allow for wear (or corrosion) or possible future changes in capacity. Liners may be graphite, aluminum, cast iron, steel, tungsten carbide, or other suitable materials (see Figure 17.2F). Most cylinders have water jackets to remove some heat of compression and to maintain reasonable cylinder and/or liner temperatures. Any heat removed is reflected in a slight reduction in the compression horsepower. The cooler cylinder walls usually allow more efficient lubrication of the cylinder. When oil cannot be tolerated in the presence of the gas, nonlubricated cylinders are used with graphite (or carbon) liners or piston rings.
Piston Rod Packing See Figure 17.9. The pressure seal between the cylinder pressure and the crank case or atmosphere is maintained by a packing gland. The motion of the piston rod is reciprocating through this packing as contrasted to the rotating motion of the centrifugal compressor or pump shaft. In many applications, it is important to prevent any part of the shaft that has been inside the cylinder and exposed to the gas from being exposed to outside air. This is particularly important when handling such materials as hydrogen chloride, chlorine, hydrogen fluoride, and so on. This may be accomplished with a distance piece (Figure 17.2G). This packing may be arranged for vent or purge in a manner similar to that for reciprocating shaft glands (Figures 17.2A and 17.2Q (a, b)).
17.7 Specification Sheet Figure 17.10, Tables 17.11A, and 17.15A are convenient for summarizing the main specifications to the manufacturer. Any unusual conditions must be explained, and minimum and maximum ranges must be established.
524 Petroleum Refining Design and Applications Handbook Volume 2
Purge
Vent
Figure 17.9 Piston rod packing. To meet the latest environmental requirements for controlling packing gas leakage, special buffer gas sealing rings are installed in place of conventional ones. The packing design includes two seats of wedge rings with an inert buffer gas between them. The spring load on the rings forces them to slightly wedge in the cup. Two individual wedge rings seal the buffer gas in both directions. Packing cups with passages for lubrication, coolant, and venture are provided as required by the application and API 618. Packing can be supplied with special gasket designs if stringent emission requirements exist (used by permission: Bul. 85084, © DresserRand Company).
“Normal” conditions maybe the same as maximum, although this may not always be the case. Detailed driver specifications should be included unless the manufacturer is to make preliminary recommendations before final decisions are reached. Note that some of the data are to be supplied by the manufacturer, and the insistence on receipt of this information facilitates evaluation of competitive bids. When packings are purged with air or other gas, the manufacturer should specify the quantity passing into the cylinder and out to the air. Accessibility of packing, bearings, and valves should be identified on drawings for review at the time of bid evaluation.
17.8 Performance Considerations Cooling Water to Cylinder Jackets Most installations use water-cooled compressor cylinder jackets; however, some use air cooling (usually small horsepower units), and a few use no cooling. For water cooling of the cylinder.
Heat Rejected to Water Btu/bhp (h) Small Cyl 20 in. dia.
Temperature Difference tc − tw, °F
300
170
20
600
310
60
700
470
100
The usual temperature rise of the water is 10–15°F, and its inlet temperature to the cylinder is from 90–140°F, depending upon the manufacturer’s design and properties of the gas. The manufacturer can provide complete data on temperatures for the particular design together with the quantity of water circulated and the pressure drop through the jackets. This cooling water is usually arranged in a closed loop with the water being pumped through secondary coolers or over cooling towers and then returned to the jackets for reuse. Water quality must be good, with steam condensate being preferred, properly treated to prevent corrosion, and so on.
Compression Equipment 525
Figure 17.10 Reciprocating compressor specifications.
526 Petroleum Refining Design and Applications Handbook Volume 2
Pressure
(a) 3
P = C1
2 (Discharge Condition)
PVk = C 4
PVk = C P = C2
1 (Inlet Condition)
Volume Ideal Reciprocating Compression Diagram (b) Compressor Valves
Piston
Cylinder Piston and Valves at Start of Compression Stroke (c)
Piston and Valves at Start of Discharge (d)
Clearance Volume Piston and Valves at End of Discharge
Figure 17.11 Ideal pressure–volume cylinder action for single acting compressor cylinder with related piston positions. Note: DV, discharge valve; SV, suction valve; k or n may be exponent for gas (used and adapted by permission: Gas Engineers Handbook. 1st Ed. © 1977. Industrial Press. Inc. New York. All rights reserved).
Drivers Refer [107] to the chapter on drivers for mechanical equipment to obtain specification details. Reciprocating compressors are driven by the following: Electric motor
Directly connected, constant or variable speed, belt, gear, fluid coupling
Gas or diesel engine
Usually directly connected to crankshaft, jackshaft, belts, or gears
Steam turbine
Gear (not a usual application)
Compression Equipment 527
Ideal Pressure–Volume Relationship Although ideal conditions are not encountered in any compression operation, the actual condition is a series of particular deviations from this. Therefore, the theoretical ideal conditions can be practically considered as the building block of the operation (Figure 17.11A and Figure 17.12). The compression operation stepwise is as follows: Condition 1: Figure 17.12. Start of the compression stroke. The cylinder is full of gas at suction pressure and essentially suction temperature (neglecting valve loss). The piston moves during compression toward condition (2) with suction and discharge valves closed. Condition 2: Figure 17.12. Start of gas discharge from cylinder. Gas has slightly exceeded the system pressure, and the discharge valve opens releasing gas to the system. The piston begins to sweep the gas from the cylinder as it moves toward condition 3. Condition 3: Figure 17.12. Completion of gas discharge from cylinder. All the gas to be removed from the cylinder by the piston stroke has passed through the discharge valve. This also is the point where the return stroke of the piston starts, but not the beginning of the cylinder suction. As the piston starts its return stroke and the pressure in the cylinder is lowered slightly below the discharge pressure, the discharge valve closes. The volume of gas left in the cylinder between the end of the piston and the end of the cylinder (clearance volume) expands from condition (3) to condition (4) as the piston returns. Condition 4: Figure 17.12. Start of gas suction into cylinder. The cylinder pressure has dropped below the system suction pressure, and the suction valve opens to admit new gas into the cylinder as it returns to condition (1).
Actual Compressor Diagram The actual compression diagram naturally deviates from the ideal; the extent varying with the physical characteristics of the cylinder and the properties of the gas (Figures 17.12, 17.12A, and 17.12B).
Deviations From Ideal Gas Laws: Compressibility See Figures 17.13A–D. The thermodynamic processes that may occur during a compression operation are [9]: Adiabatic
– no heat added to or removed from the system
Isothermal
– constant temperature
Isometric
– constant volume
Isobaric
– constant pressure
Isentropic
– constant entropy
Isenthalpic
– constant enthalpy
For any gas compression,
PVn = constant = C
(17.2)
The “ideal” gas law or “perfect” gas law is combination of Boyle’s and Charles’ laws for any compressible fluid (gas/vapor).
528 Petroleum Refining Design and Applications Handbook Volume 2 Pressure and Friction Losses
3
Discharge
P 2 , T2 Discharge Pressure
2
Adiabatic Compression k = 1.4 (as for air)
ke ro St
ion
n sio es pr m Co
Clearance Volume
Isothermal Compression k = 1.0
s xpan R e -E
Pressure
Actual Compression
Suction Pressure 4
P1, T1
Intake Stroke Stroke or Displacement
Clearance
0
100
Volume, % Stroke
Figure 17.12 Reciprocating compressor compression diagrams. Actual losses and effect of k = cp/cv on performance.
T
2
P2
1
H2
P1
TH PA
RO
NT
ISE
C PI
H1
LOG PRESSURE
T
IDEAL GAS STATE H2–H1 –ΔH1
–ΔH2 H2°–H1°
Figure 17.12A Illustration of isentropic path on log pressure-enthalpy diagram, showing Mollier chart method of finding final temperature and calculation on H for reversible and adiabatic compression (used by permission: Edmister, W. C. Applied Hydrocarbon Thermodynamics, © 1961. Gulf Publishing Company, Houston, Texas. All rights reserved).
Compression Equipment 529 COMPRESSION PATHS FOR ETHANE 600
PATH
200
2.0
100
80
INITIAL CONDITION 500
1.0
ΔH
Q
∫VdP
B.T.U./LB. - MOLE
1.5 150
T = 420°
T = 180°F
PRESSURE, LB./SQ. IN ABS.
220 PATH A-ISENTHALPIC
400
V= 0.6 E/LB. CUF
200 PATH B-POL Y TRO PIC PAT HC -PO LY T 300 ROP IC PAT HD -IS EN PA TRO 340 TH PI C E-P OL YT RO PIC 380
500
300
FINAL CONDITIONS
0.4
A
O
– 2166
+ 2166
B
+ 908
– 1324
+ 2232
C
+ 1633
– 651
+ 2284
D
+ 2330
0
+ 2330
E
+ 2939
+ 553
+ 2386
.0
V=3
540
580 ENTHALPY, B.T.U./LB.
620
Figure 17.12B Section of ethane pressure-enthalpy diagram illustrating five compression paths (used by permission: Edmister, W. C., Applied Hydrocarbon Thermodynamics, © 1961. Gulf Publishing Company, Houston, Texas. All rights reserved).
Based on the prefect gas and the adiabatic equation:
P2/P1 = (V1/V2)k
(17.3)
T2/T1 = (V1/V2)k–1
(17.4)
P2/P1 = (T2/T1)k/(k-1) = r(k–1)/k
(17.5)
PV = WR T
(17.6)
PV = RoT
(17.7)
where Ro = 1545 ft-lbf/lbm-°R P = pressure, lbf/ft2 V = volume, ft3 T = temperature, abs, °R (°F + 460) R = universal gas constant, 1545/ (mol wt of gas) Boyle’s law: For constant temperature (isothermal, but not realized under actual conditions):
530 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.13A Enthalpy–entropy chart for natural gas, Sp.Gr. = 0.6 (used by permission: Engineering Data Book, 7th © 1957. Natural Gasoline Supply Men’s Association, Inc. All rights reserved).
Compression Equipment 531
Figure 17.13B Enthalpy–entropy chart for natural gas, Sp.Gr. = 0.7 (used by permission: Engineering Data Book, 7th © 1957. Natural Gasoline Supply Men’s Association, Inc. All rights reserved).
532 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.13C Enthalpy–entropy chart for natural gas, Sp.Gr. = 0.8 (used by permission: Engineering Data Book, 7th © 1957. Natural Gasoline Supply Men’s Association, Inc. All rights reserved).
Compression Equipment 533
Figure 17.13D Enthalpy–entropy chart for natural gas, Sp.Gr. = 0.8 (used by permission: Engineering Data Book, 7th © 1957. Natural Gasoline Supply Men’s Association, Inc. All rights reserved).
534 Petroleum Refining Design and Applications Handbook Volume 2
P1V1 = P2V2 = constant = C1
(17.8)
For the adiabatic condition of no heat lost or gained.
P1V1k = P2 V2k
k = n
(17.9)
A reversible adiabatic process is known as isentropic [10]. Thus, the two conditions are directly related. In actual practice, compressors generate friction heat, give off heat, have valve leakage and have piston ring leakage. These deviations from the true adiabatic condition result in the process known as “polytropic.” It is defined as an internally reversible change of state [10] where
P1V1n = P2 V2n = constant (different from preceding )
(17.10)
For a polytropic process the change of state does not take place at constant entropy, but for an adiabatic process, it does. Heat may be added to or rejected from a gas in a polytropic process. For a polytropic process, the correlating exponent for the P1V1n component is the exponent “n,” which becomes an important part of the compressor design. “n” values are determined from performance testing. An adiabatic process is never fully attained but can be closely approached for many processes and is the primary design basis for most positive displacement compression equipment. where k = ratio of specific heats, cp/cv n = polytropic coefficient P1 = inlet pressure, abs, lb(force)/ft2 abs P2 = discharge pressure, abs, lb(force)/ft2 abs R = gas constant, ft-lb(force)/(lbm-°R) = 1545/(mol wt) Rʹ = specific constant for the gas involved = 1545/gas mol wt Ro = universal gas constant = 1545 and is same for all gases = 10.729 when P is lbf/in.2 abs T1 = inlet gas temperature, °R = °F + 460 T2 = discharge gas temperature, °R = °F + 460 V = volume of gas, ft3/lb-mole W = weight of gas, lb 1 = inlet condition (intake) 2 = outlet condition (discharge) Adiabatic Calculations
k −1 P2 k k Adiabatic heat , Had = RT1 − 1 P1 ( k − 1)
where Had = adiabatic head, ft-lbf/lbm = ft
(17.11)
Compression Equipment 535 R T1 P1 P2 W
= gas constant, ft-lb(force)/lbm°R = inlet gas temperature, °R = °F + 460 = absolute inlet pressure, lb(f)/in.2 = absolute discharge pressure, lb(f)/in.2 = flow rate of gas, lb/s
1 hp = 550 ft-lbf/s ft − lbf lbm Work on the gas during compression = Had (W) • s lbm
HPad
WHad 550
k
(WRT1 ) [(P2 /P1 )( k (k 1) (550)
1)/ k
1]
(17.12)
HPad = horsepower, hp Q = volume flow of gas, ft3/min at inlet conditions
Charles’ Law at Constant Pressure [11] V2/V1 = T2/T1 or
(17.13)
V2/T2 = V1/T1
(17.14)
Amonton’s Law at Constant Volume [11] P2/P1 = T2/T1 or
(17.15)
P2/T2 = P1/T1
(17.16)
Combined Boyle’s and Charles’ Laws
P1V1 P2 V2 = T1 T2
(17.17)
Figures 17.12A and 17.12B illustrate various paths of the compression or expansion of gas that can take place, depending on the properties of the gas/vapor.
Mollier Chart Method After identifying the initial temperature (T) and pressure (P) values, the final temperature and both enthalpy values (H) can be read on the same entropy line of the appropriate gas Mollier chart [9]. For the adiabatic process, the work done on the gas is equal to H [9] (see Figures 17.13A–D). The following is reproduction by permission of Edminster, W.C., Applied Hydrocarbon Thermodynamics, Gulf Publishing Company [9].
536 Petroleum Refining Design and Applications Handbook Volume 2 “When a Mollier chart is available for the gas involved [12, 13] the first method, which is illustrated by Figure 17.12A is the most convenient. On the abscissa of Figure 17.12A, four enthalpy differences are illustrated. (H2 − H1) is the enthalpy difference for the isentropic path. (H2° − H1°) is the ideal gas state enthalpy difference for the terminal temperatures of the isentropic path. The other ΔH values are the isothermal pressures corrections to the enthalpy at the terminal temperatures. A generalized chart for evaluating these pressure corrections was presented previously. As Mollier charts are available for only a few pure components and practically no mixtures, this calculation method is very limited. For example, it cannot be used for most process calculations because these gases are usually mixtures. Some of the charts available for mixtures are the H-S charts presented by Brown [12] for natural gases of gravities from 0.6 to 1.01.”
Entropy Balance Method “Using generalized isothermal effects of pressure and the ideal gas state “S” and “H” values, the final temperature required to satisfy the condition S = 0 is found, and the value of H is determined for the path [9]. The second method can be applied to mixtures as well as pure components. In this method the procedure is to find the final temperature by trial, assuming a final temperature and checking by entropy balance (correct when Spath = 0). As reduced conditions are required for reading the tables or charts of generalized thermodynamic properties, the pseudo critical temperature and pressure are used for the mixture. Entropy is computed by the relation. See reference [12] for details.”
S = S° − RlnP + S
(17.18)
Isentropic Exponent Method [12] Using exponents defining temperature and volume changes with pressure for the gas, the final temperature and work are computed by simple equations.
Many gases deviate from the ideal state when pressures and/or temperatures are greater than 100–500 psia and 100°F. Some deviations yield a compressibility factor, Z, less than 1.0, and others give values greater than 1.0.
PV = ZNRT
(17.1)
PV = 10.73 ZNT
(17.19)
or
where P V T R Z N
= absolute pressure, psia = volume of gas, ft3/lb-mole = absolute temperature, °R (Rankine) = °F + 460 = universal gas constant = 10.729 psia ft3/lb mole .°R for units noted here = compressibility factor = number of lb-moles of gas
Values of gas constant, R, for the other units: R = 1545.3 when P = lbf/ft2, abs R = 0.7302 when P = atmosphere R = 10.729 when P = lbf/in.2, abs Generalized compressibility factors for gases are given in Figures 17.14A–E. These charts have been prepared to allow approximately the same accuracy in reading values over the entire range. Compressibility factors at low
Compression Equipment 537 pressure for several major hydrocarbons are presented by Pfennig and McKetta [14]. Compressibility charts for specific gases are given in Figures 17.14F–W. Figure 17.15 is a compressibility chart for natural gas based on pseudo-reduced pressure and temperature. The reduced pressure is the ratio of the absolute operating pressure to the critical pressure, Pr, and the reduced temperature is the ratio of the absolute operating temperature to the critical temperature, Tc, for a pure gas or vapor. The pseudo value is the reduced value for a mixture calculated as the sum of the mol percentages of the reduced values of the pure constituents.
pseudo Pr = y1Prl + y2Pr2 + y3Pr3, etc.
(17.20)
Similarly, the pseudo-reduced temperature can be determined.
pseudo Tr = y1Trl + y2Tr2 + y3Tr3, etc.
(17.20a)
Values of compressibility are available for many pure hydrocarbons and gases [15, 16, 28]. Figure 17.16A illustrates a compression path for deviation from the ideal that overestimates the actual power required (area of dotted portion is greater than solid line actual area). Actual volumetric efficiency and inlet volume is less than ideal due to the deviation on the re-expansion path [30]. Table 17.2 compares an example for propane; a compressor with 10% clearance, 1000 cfm piston displacement, compression from 100 psia and 80°F to 300 psia. For Figure 17.16B, as illustrated by a 24–76% (volume) mixture of nitrogen-hydrogen at around 5000 psia, the deviation is opposite to that of Figure 17.16A. The actual power requirements are greater than ideal; volumetric efficiency exceeds ideal gas laws. Figure 17.16C illustrates ethylene in the extreme high pressure range (30,000–40,000 psia) where the deviation is unpredictable without thermodynamic data. Figure 17.16D illustrates the type of reciprocating compressor performance problems that can develop from various mechanical details. To maintain peak efficiency in a compressor cylinder, a pressure–time indicator card of the cylinder during operation can be quite helpful in pointing to a problem and its nature [79]. These figures illustrate what takes place inside the cylinder during the compressor’s operation. When specifying performance, the actual capacity at suction and/or discharge conditions must be specified. Table 17.3 lists the variation of compressibility factor, Z, with pressure as read or computed from accepted charts. Compressibility must be taken into account along with the adiabatic coefficient, k (or, if known, the polytropic coefficient, n), and other losses, which will be presented in the following paragraphs. “k” value of Gas (Ratio of Specific Heats). The ratio cp/cv is known as the “k” value of a gas and is associated with adiabatic compression or expansion. The change in temperature during compression (for most average water cooled jackets) is related by
P1V1k = P2 V2k = P3V3k = constant
(17.21)
for the same weight of gas at three different states or conditions. Most compression and expansion curves are represented by the preceding relationship. The actual value of “n” for a polytropic compression is usually 1.0–1.5 and is a function of the gas properties, such as specific heats, degree of cooling during compression (external), and operating features of the cylinder [18]. Figure 17.12 shows the effect of change in “k” on the compression curve. A usual reciprocating compressor performance evaluation uses adiabatic cp/cv, and this is the representation here. With the k = 1.0, the compression is isothermal; with “k” = “n” greater than 1.0, the operation is actually polytropic. For air the adiabatic “k” = 1.4. In adiabatic compression or expansion, no release or gain of heat by the gas occurs, and no change occurs in entropy. This condition is also known as isentropic and is typical of most compression steps. Actual conditions often cause a realistic deviation, but usually these are not sufficiently great to give errors in the calculations. Table 17.4 gives representative average “k” values for a few common gases and vapors. The specific heat is the heat required to raise the temperature of a unit mass of material one degree. Specific heat varies with temperature, but essentially no variation occurs with pressure [19]. The ratio k, is important in most compression-related situations, i.e.,
538 Petroleum Refining Design and Applications Handbook Volume 2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
REDUCED PRESSURE, PR 0.8 0.9 1.0 1.1
1.2
CED T
RATU
R
1.7
1.8
1.1 0
8
1.0 6
4
COMPRESSIBILITY FACTOR, P V Po Vo T CONSTANT
1.6
1.0
1.02
0.7
1.0
0.6
1.5
2
1.00
0.5
1.8 1.00 0.98 0.96 0.94 0.92 0.90 0.88 0.86 0.84 0.82 0.80 0.78 0.76 0.74 0.72 0.70 0.68 0.66 0.64 0.62 0.60 0.58 0.56 0.54 0.52 0.50 0.48 0.46 0.44 0.42 0.40 0.38 0.36 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10
1.3 0 1.2 1.2 8 6 1.2 1.2 4 2 1.2 0 1.1 8 1.1 6 1.1 4
1.1
E
0.4
1.7
1.40
LIN
0.3
1.6
1.35
N TIO RA
0.2
1.5
1.50
RE, T
TU SA
0.1
1.4
1.60
EMPE
5
0
1.3
1.70 REDU
0.9
1.00 0.98 0.96 65 1.1 0.94 5 1.25 1.0 0.70 5 0.92 0.90 0.75 0.88 0.86 0.84 0.80 0.82 0.80 0.78 0.76 0.85 0.74 0.72 0.70 0.68 0.90 0.66 0.64 0.92 0.62 0.60 0.58 0.94 0.56 0.54 0.52 0.96 0.50 0.48 0.46 0.44 0.42 0.40 0.38 0.36 COMPRESSIBILITY FACTOR FOR GASES 0.34 0.32 p REDUCED PRESSURE PR = pc 0.30 T 0.28 REDUCED TEMPERATURE TR = Tc 0.26 p, pc, T AND Tc ARE IN ABSOLUTE UNITS 0.24 0.22 pV = 1 FOR IDEAL GASES po Vo T CONSTANT 0.20 0.18 0.16 0.14 COPYRIGHT 1949 0.12 WORTHINGTON PUMP AND MACHINERY CORPORATION 0.10
0.98
0.8 0.9 1.0 1.1 REDUCED PRESSURE, PR
1.2
1.3
1.4
Figure 17.14A Compressibility factor for gases, Part 1 of 5 (used by permission: Worthington research Bul. P – 7637 © 1949. Dresser-Rand Company. All rights reserved).
PV Po Vo T CONSTANT
COMPRESSIBILITY FACTOR,
1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00 0.98 0.96 0.94 0.92 0.90 0.88 0.86 0.84 0.82 0.80 0.78 0.76 0.74 0.72 0.70 0.68 0.66 0.64 0.62 0.60 0.58 0.56 0.54 0.52 0.50 0.48 0.46 0.44 0.42 0.40 0.38 0.36 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20
1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00
0
0
REDUCED
1
5
2
1.1
1.00
1
1.2 0
1.60
1.70
4 3 2.60 2.40 2.20 2.10 2.00 1.90 1.80
1.40
1.45
1.50
1.55
3
3
1.35
1.30 1.25
TURE, TR TEMPERA
2
4
4
40
7
8
REDUCED TE
9
REDUCED PRESSURE, PR
10
10
,T MPERATURE R
1.16 1.14 1.12 0 1.10 1.08 1.1 1.06 1.04 1.02 1.00 0 1.2 0.98 0.96 0.94 0.92 0.90 0.88 0.86 0.84 0.82 0.80 0.78 0.76 0.74 0.72 0.70 0.68 0.66 0.64 0.62 0.60 0.58 0.56 0.54 COMPRESSIBILITY FACTOR FOR GASES 0.52 p 0.50 REDUCED PRESSURE PR = pc 0.48 T 0.46 REDUCED TEMPERATURE TR = Tc 0.44 p, pc, T AND Tc ARE IN ABSOLUTE UNITS 0.42 0.40 pV = 1 FOR IDEAL GASES 0.38 po Vo T CONSTANT 0.36 0.34 0.32 0.30 COPYRIGHT 1949 0.28 0.26 WORTHINGTON PUMP AND MACHINERY CORPORATION 0.24 0.22 0.20 5 6 7 8 9
14 30
5 8
5 6 REDUCED PRESSURE, PR
4 6 10 12 20 40
4 the compressibility factor reaches a maximum, and then decreases with an increase in reduced temperature values, to avoid confusion in reading, the reduced temperature lines greater than 4 are offset on an identical scale.
1.0 5
NOTE: In this range, at reduced temperature approximately equal
8 10 12 14 20 30 40
4 5 6
11
11
12
1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00 12
Compression Equipment 539
1.10
1.05
Figure 17.14B Compressibility factor for gases, Part 2 of 5 (used by permission: Worthington research Bul. P – 7637 © 1949. Dresser-Rand Company. All rights reserved).
540 Petroleum Refining Design and Applications Handbook Volume 2 11
12
14
13
0 1.5 1.6 0 1 0 2 .80 2.2 .00 0
1.4 0
1.3
1.2
0
1.1
0
1.0
5
10
TR
0 2.4 60 2.
ED
TEM
PER
ATU
RE,
3
RED UC
NOTE: Lines are dotted to aid in reading.
4 3
0 2.6 0 2.4 0
2.2
0
2.0
COMPRESSIBILITY FACTOR FOR GASES 0
1.8
REDUCED PRESSURE PR =
p pc
0
T Tc
1.4
1.6
0
REDUCED TEMPERATURE TR =
1.5 0 1.1 0
p, pc, T AND Tc ARE IN ABSOLUTE UNITS pV
5
T CONSTANT
= 1 FOR IDEAL GASES
1.3 0
po Vo
1.0
1.20 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.10 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88
9
COPYRIGHT 1949
WORTHINGTON PUMP AND MACHINERY CORPORATION
0
1.40 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.30 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21
8
1.2
COMPRESSIBILITY FACTOR,
PV Po Vo
T CONSTANT
REDUCED PRESSURE, PR 7
7
8
9
10 11 REDUCED PRESSURE, PR
12
13
14
1.40 1.39 1.38 1.37 1.36 1.35 1.34 1.33 1.32 1.31 1.30 1.29 1.28 1.27 1.26 1.25 1.24 1.23 1.22 1.21 1.20 1.19 1.18 1.17 1.16 1.15 1.14 1.13 1.12 1.11 1.10 1.09 1.08 1.07 1.06 1.05 1.04 1.03 1.02 1.01 1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88
Figure 17.14C Compressibility factor for gases, Part 3 of 5 (used by permission: Worthington research Bul. P – 7637 © 1949. Dresser-Rand Company. All rights reserved).
2.50 2.48 2.46 2.44 2.42 2.40 2.38 2.36 2.34 2.32 2.30 2.28 2.26 2.24 2.22 2.20 2.18 2.16 2.14 2.12 2.10 2.08 2.06 2.04 2.02 2.00 1.98 1.96 1.94 1.92 1.90 1.88 1.86 1.84 1.82 1.80 1.78 1.76 1.74 1.72 1.70 1.68 1.66 1.64 1.62 1.60 1.58 1.56 1.54 1.52 1.50 1.48 1.46 1.44 1.42 1.40 1.38 1.36 1.34 1.32 1.30 1.28 1.26 1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00
12
12
PV Po Vo T CONSTANT
COMPRESSIBILITY FACTOR
14
15
16
17
T Tc
D
CE
TE
14
PE
M
,TR RE TU A R
15
16
17
COPYRIGHT 1949 WORTHINGTON PUMP AND MACHINERY CORPORATION
pV = 1 FOR IDEAL GASES po Vo T CONSTANT
p, pc, T AND Tc ARE IN ABSOLUTE UNITS
DU RE
13
p pc
REDUCED TEMPERATURE TR =
REDUCED PRESSURE PR =
COMPRESSIBILITY FACTOR FOR GASES
13
18
18
19
19
20
20
22
22
23
23
REDUCED PRESSURE, PR
21
21
REDUCED PRESSURE, PR
24
24
25
25
26
26
27
27
28
28
29
29
30
30
31
31
32
2.50 2.48 2.46 2.44 2.42 2.40 2.38 2.36 2.34 2.32 2.30 2.28 2.26 2.24 2.22 2.20 2.18 2.16 2.14 2.12 2.10 2.08 2.06 2.04 2.02 2.00 1.98 1.96 1.94 1.92 1.90 1.88 1.86 1.84 1.82 1.80 1.78 1.76 1.74 1.72 1.70 1.68 1.66 1.64 1.62 1.60 1.58 1.56 1.54 1.52 1.50 1.48 1.46 1.44 1.42 1.40 1.38 1.36 1.34 1.32 1.30 1.28 1.26 1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00
32
Compression Equipment 541
Figure 17.14D Compressibility factor for gases, Part 4 of 5 (used by permission: Worthington research Bul. P – 7637 © 1949. Dresser-Rand Company. All rights reserved).
542 Petroleum Refining Design and Applications Handbook Volume 2
30
REDUCED PRESSURE, PR
40
45
50
60
65
70
75
80
90
85
95
0
2.40
p pc
REDUCED TEMPERATURE TR =
2.6
2.00
2.20
REDUCED PRESSURE PR =
T Tc
p, pc, T AND Tc ARE IN ABSOLUTE UNITS
4.0
3.5
3.0
pV = 1 FOR IDEAL GASES po Vo T CONSTANT
5
4.5
COPYRIGHT 1949 WORTHINGTON PUMP AND MACHINERY CORPORATION
6
,TR RE
D CE
M
TE
7
TU RA
PE
DU
RE
9 10
12
14
20
25 30
40
35
40
45
50
55
60
65
70
REDUCED PRESSURE, PR
75
80
85
90
100 2.58 2.56 2.54 2.52 2.50 2.48 2.46 2.44 2.42 2.40 2.38 2.36 2.34 2.32 2.30 2.28 2.26 2.24 2.22 2.20 2.18 2.16 2.14 2.12 2.10 2.08 2.06 2.04 2.02 2.00 1.98 1.96 1.94 1.92 1.90 1.88 1.86 1.84 1.82 1.80 1.78 1.76 1.74 1.72 1.70 1.68 1.66 1.64 1.62 1.60 1.58 1.56 1.54 1.52 1.50 1.48 1.46 1.44 1.42 1.40 1.38 1.36 1.34 1.32 1.30 1.28 1.26 1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00
8
COMPRESSIBILITY FACTOR
55
COMPRESSIBILITY FACTOR FOR GASES
PV Po Vo T CONSTANT
2.58 2.56 2.54 2.52 2.50 2.48 2.46 2.44 2.42 2.40 2.38 2.36 2.34 2.32 2.30 2.28 2.26 2.24 2.22 2.20 2.18 2.16 2.14 2.12 2.10 2.08 2.06 2.04 2.02 2.00 1.98 1.96 1.94 1.92 1.90 1.88 1.86 1.84 1.82 1.80 1.78 1.76 1.74 1.72 1.70 1.68 1.66 1.64 1.62 1.60 1.58 1.56 1.54 1.52 1.50 1.48 1.46 1.44 1.42 1.40 1.38 1.36 1.34 1.32 1.30 1.28 1.26 1.24 1.22 1.20 1.18 1.16 1.14 1.12 1.10 1.08 1.06 1.04 1.02 1.00 30
35
95
100
Figure 17.14E Compressibility factor for gases, Part 5 of 5 (used by permission: Worthington research Bul. P – 7637 © 1949. Dresser-Rand Company. All rights reserved).
Compression Equipment 543
1.25
BROKEN LINES INDICATE EXTRAPOLATION
1.20
COMPRESSIBILITY FACTOR Z =
PV RT
1.15
1.10
°F 400 300° 200° 150°
1.05
100° 80° 60° 40° 20°
1.00
0° –20°
0.95
–40°
COMPRESSIBILITY CHART FOR AIR
0.90
BASED ON: DIN – THERMODYNAMIC PROPERTIES OF GASES BUTTERWORTHS SCIENTIFIC PUBLICATIONS TRANSACTIONS OF THE ASME – OCTOBER, 1954, HALL AND IBELE TABULATION OF IMPERFECT GAS PROPERTIES NATIONAL BUREU OF STANDARDS CIRCULAR 564, 1955 ISSUE CU. FT./POUND AT 14.696 PSIA AND 60°F = 13.106 Z AT 14.696 PSIA AND 60°F = 0.9985 © INGERSOLL – RAND COMPANY 1960
0.85
0.80
0
1000
2000
3000 PRESSURE – PSIA
4000
5000
6000
Figure 17.14F Compressibility factor for air (used by permission: From 3519 D (1981), © Ingersoll-Rand Company. All rights reserved).
544 Petroleum Refining Design and Applications Handbook Volume 2
1.00
0°
380° 20
0.95
340°
°
300°
COMPRESSIBILITY FACTOR Z =
PV RT
40
280° 260° 240°
° 60
°
SAT UR
0.90
220°
ATI ON L
80 °
200°
INE
180°
10
0°
160
°
12
0°
14
0°
0.85
COMPRESSIBILITY CHART FOR AMMONIA – (NH3) BASED ON BUREAU OF STANDARDS CIRCULAR NO. 142 – 1945 PRINTING CU FT./POUND AT 14.696 PSIA AND 60°F = 22.05 Z AT 14.696 PSIA AND 60°F = 0.989 © INGERSOLL – RAND COMPANY 1960
0.80
0
50
100
150 PRESSURE – PSIA
200
250
300
Figure 17.14G Compressibility chart for ammonia (used by permission: From 3519 D (1981), © Ingersoll-Rand Company. All rights reserved).
Compression Equipment 545
1.00
1.00 –50° –30° –10°
0° ° 40
10°
0.95
80 °
12
0°
16
0° 200°
240
°
280°
320°
360°
0.95
COMPRESSIBILITY FACTOR Z =
PV RT
30°
0.90
0.90
SA TU
RA TIO
NL
COMPRESSIBILITY CHART FOR CHLORINE – (Cl2)
0.85
INE
0.85
ADAPTED BY PERMISSION FROM R. M. KAPOOR AND J. J. MARTIN “THERMODYNAMIC PROPERTIES OF CHLORINE” ENGINEERING RESEARCH INSTITUTE PUBLICATIONS, UNIVERSITY OF MICHIGAN (1957) CU. FT/LB. AT 14.696 PSIA AND 60°F = 5.2834 Z AT 14.696 PSIA AND 60°F = 0.988 © INGERSOLL – RAND COMPANY 1967
0.80
0
50
100
150 PRESSURE – PSIA
200
250
300
0.80
Figure 17.14H Compressibility chart for chlorine (used by permission: From 3519 D (1981), © 1967. Ingersoll-Rand Company. All rights reserved).
546 Petroleum Refining Design and Applications Handbook Volume 2
1.30
COMPRESSIBILITY FACTOR Z =
PV RT
1.20
1.10 400°F 300°
200° 150°
100° 60° 40° 20°
1.00
0° –20°
COMPRESSIBILITY CHART FOR NITROGEN – (N2)
–40°
BASED ON SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” NATIONAL BUREAU OF STANDARDS CIRC, 564 — 1955 ISSUE TRANSACTIONS OF THE ASME, OCTOBER 1954, HALL AND IBELE, “TABULATION OF IMPERFECT GAS PROPERTIES CU. FT./POUND AT 14.696 PSIA AND 60°F = 13.55 Z AT 14.696 PSIA AND 60°F = 0.999 © INGERSOLL – RAND COMPANY 1960
1.90
0
1000
2000
3000 PRESSURE – PSIA
4000
5000
6000
Figure 17.14I Compressibility chart for nitrogen (used by permission: From 3519 D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
Compression Equipment 547
1.00 400° 350°
0.90
–4
0°
–2
0°
0°
20
°
40
°
60°
80°
COMPRESSIBILITY FACTOR Z =
PV RT
0.80
100
°
120
140
°
160
180°
200°
220°
240°
260°
300° 280°
°
°
0.70 SAT
0.60
UR
ATI O
NL
INE
0.50
0.40 COMPRESSIBILITY CHART FOR CARBON DIOXIDE (CO2) 0.30
BASED ON: DIN “THERMODYNAMIC FUNCTIONS OF GASES” SWEIGERT, WEBER AND ALLEN, “THERMODYNAMIC PROPERTIES OF GASES” − IND. B ENG. CHEM. FEB, 1946– NATIONAL BUREAU OF STANDARDS CIRC. 564-1955 ISSUE PERRY, “CHEMICAL ENG HANDBOOK.”
0.20
CRITICAL POINT
CU. FT./POUND AT 14.696 PSIA AND 60°F = 8.576 Z AT 14.696 PSIA AND 60°F = 0.994 © INGERSOLL RAND COMPANY 1960
0
100
200
300
400
500
600 700 PRESSURE – PSIA
800
900
1000
1100
1200
1300
Figure 17.14J Compressibility chart for low pressure carbon dioxide (used by permission: From 3519 D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
548 Petroleum Refining Design and Applications Handbook Volume 2
1.00
0.90
400° F 350°
0°
0.80
0.70
260°
SAT TIO URA
240°
IN NL
0.60
220°
E
COMPRESSIBILITY FACTOR Z =
PV RT
300° 280°
20°
40°
200° 60°
0.50
180°
80°
160° 140°
0.40 120°
0.30 CRITICAL POINT
COMPRESSIBILITY CHART FOR CARBON DIOXIDE (CO2)
100°
BASED ON “THERMODYNAMIC FUNCTIONS OF GASES” VOL. 2, DIN. “THERMODYNAMIC PROPERTIES OF GASES,” SWEIGERT, WEBER, AND ALLEN. INDUSTRIAL AND ENGINEERING CHEMISTRY–FEB. 1946 NATIONAL BUREAU OF STANDARDS-CIRCULAR 564-1955 ISSUE CHEMICAL ENGINEERING HANDBOOK–PERRY CU. FT./POUND AT 14.696 PSIA AND 60°F = 8.576 Z AT 14.696 PSIA AND 60°F = 0.994 © INGERSOLL RAND COMPANY – 1960
0.20
0
1000
2000
3000 PRESSURE – PSIA
4000
5000
6000
Figure 17.14K Compressibility chart for high pressure carbon dioxide (used by permission: From 3519 D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
Compression Equipment 549
1.15
1.10
DASH LINES ARE REFERENCE “A” SOLID LINES ARE REFERENCE “B”
1.05
400° 350° 300°
COMPRESSIBILITY FACTOR Z =
PV RT
1.00
250°
0.95
200° 180° 160°
0.90
140°
DOTTED LINES ARE EXTRAPOLATED
120° 100°
0.95
80°
COMPRESSIBILITY CHART FOR METHANE (CH4)
60°
0.80
REFERENCE “A” SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS.” MATTHEWS AND HURD, “THERMODYNAMIC PROPERTIES OF METHANE” – TRANSACTIONS, AMERICAN INSTITUTE OF CHEM. ENGINEERS, VOLUME 42–NO. 4. REFERENCE “B” AMERICAN GAS ASSOCIATION, “PAR RESEARCH PROJECT REPORT NX–19.” (COMPLETED DECEMBER 1962) CU. FT/LB. AT 14.696 PSIA & 60°F = 23.61 Z AT 14.696 PSIA & 60°F = 0.997 © INGERSOLL–RAND COMPANY 1966
40°
0.75
20°
0°
0.70
0.65
0
1000
2000
3000 PRESSURE – PSIA
4000
5000
6000
Figure 17.14L Compressibility chart for methane (used by permission: From 3519 D (1981), © 1966. Ingersoll-Rand Company. All rights reserved).
550 Petroleum Refining Design and Applications Handbook Volume 2
1.00
0.90
350 ° 300 ° 250 225 ° ° 200 ° 175 °
0° –2
–80°
15
–60°
0°
0.80
12
5°
°
20
0°
10
PV RT
–40°
°
80
0.70
°
TIO RA
60
TU SA N
°
40
E
LIN
COMPRESSIBILITY FACTOR Z =
0°
0.60
0.50 COMPRESSIBILITY CHART FOR ETHYLENE (C2H4) BASED ON: H. BENZLER AND A.V. KOCH TECHNISCHEN HOCHSCHULE KARLSRUHE 1954 CU. FT./LB AT 14.696 PSIA AND 60°F = 13.453 Z AT 14.696 AND 60°F = 0.9947 © INGERSOLL –RAND COMPANY 1967
0.40
0
100
200
300
400 500 PRESSURE – PSIA
600
700
800
900
1000
Figure 17.14M Compressibility chart for low pressure ethylene (used by permission: From 3519 D (1981), © 1967. Ingersoll-Rand Company. All rights reserved).
Compression Equipment 551
1.10
1.00
0.90
350° 300°
0.80
°
250 25°
0.70
2 °
200
40°
60°
0.50
0.40
E
0.30
CRITICAL POINT
1
5°
12 0°
10
80
0
°
COMPRESSILITY CHART FOR ETHYLENE (C2H4) BASED ON: H. BENZLER AND A.V. KOCH TECHNISCHEN HOCHSCHULE KARLSRUHE 1954 CU. FT/LB AT 14.696 AND 60°F = 13.453 Z AT 14.696 AND 60°F = 0.9947 © INGERSOLL − RAND COMPANY 1967
°
60 40
0.20
0.10
°
175 50°
80°
0.60
TION LIN SATURA
COMPRESSIBILITY FACTOR - Z =
RT
100°
PV
0°
°
1000
2000
3000
4000
5000
6000
PRESSURE − PSIA
Figure 17.14N Compressibility chart for high pressure ethylene. Note: special charts are available for pressures in the range 20,000–75,000 psi (used by permission from 3519 D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
552 Petroleum Refining Design and Applications Handbook Volume 2 1.00
400° 350°
0.90
16
14
RT
0°
0.70
TU SA N
TIO RA E
LIN
COMPRESSIBILITY FACTOR Z =
PV
0° 12 0° 10
80°
60 °
40°
20° 0°
0.80
0°
300° 280 260 ° 240 ° 22 ° 0° 20 0 ° 18 0°
0.60
0.50
COMPRESSIBILITY CHART FOR ETHANE (C2H6) BASED ON = SAGE AND LACEY, ” THERMODYNAMIC PROPERTIES OF HYDROCARBONS” AND “THERMODYNAMIC PROPERTIES OF ETHANE” BARKELEW, VALENTINE AND HURD, “TRANSACTIONS OF AICHE”− VOL. 43 NO. I, JANUARY 1947 CU. FT./LB. AT 14.696 PSIA AND 60°F = 12.52 Z AT 14.696 PSIA AND 60°F = 0.992 © INGERSOLL − RAND COMPANY 1960
0.40
0
200
400
600
800
1000
PRESSURE - PSIA
Figure 17.14O Compressibility chart for low pressure ethane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
Compression Equipment 553 1.10
1.00
0.90
0.80
0.70
350°
0°
300° 280°
20° 40° 60° 80°
260° 240° 220°
0.60
200°
N RATIO SATU
COMPRESSIBLITY FACTOR Z = PV RT
400°
0.50
180°
LINE
160°
0.40
COMPRESSIBILITY CHART FOR ETHANE (C2H6)
140°
BASED ON: SAGE AND LACEY, ” THERMODYNAMIC PROPERTIES OF HYDROCARBONS” AND, “THERMODYNAMIC PROPERTIES OF ETHANE” − BARKELEW, VALENTINE AND HURD, ”TRANSACTION OF AICHE,” VOL. 43 NO. I JANUARY 1947. CU. FT./LB. AT 14.696 PSIA AND 60°F = 12.52 Z AT 14.696 PSIA AND 60°F = 0.992 © INGERSOLL − RAND COMPANY 1960
120° 0.30
CRITICAL POINT
0
100°
1000
2000
3000 4000 PRESSURE − PSIA
5000
6000
Figure 17.14P Compressibility chart for high pressure ethane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
k = cp/cv
(17.22)
For monatomic gases, k is about 1.66; for diatomic gases, k is about 1.40; and for polyatomic gases, k is about 1.30. Details of values for specific heat of gases are available in many engineering tables. The ratio, k, may be calculated from the ideal gas equation:
k = c p /c v =
M cp M cp − 1.987
(17.23)
where Mcp = molal heat capacity at constant pressure, Btu/lb-mol (°F) M = molecular weight When values of Mcp are not available, they may be calculated:
Mcp = A + BT, Btu/mol/l °R
(17.24)
with T in °Rankine at compressor cylinder inlet. The constants A and B may be obtained from Table 17.5. Table 17.5A gives MCp values for gases at varying temperature in °R.
k = cp/cv = cp/(cp − 1.987)
(17.25)
554 Petroleum Refining Design and Applications Handbook Volume 2
1.00
0.90
0.80
60° 80°
COMPRESSIBLITY FACTOR Z =
PV RT
100° 120°
0.70
400°
140° 0.60
350°
160°
325° 180°
0.50
300° 280° 260°
0.40 240° 0.30
200°
COMPRESSIBILITY CHART FOR PROPYLENE (C3H6)
220°
BASED ON - CANJAR, GOLDMAN, AND MARCHMAN, “THERMODYNAMIC PROPERTIES OF PROPYLENE” INDUSTRIAL AND ENGINEERING CHEMISTRY VOL. 43, NO. 5, MAY 1951... CU. FT/LB. AT 14.696 PSIA AND 60°F = 9.021 Z AT 14.696 PSIA AND 60°F = 0.9836
NL
INE
CRITICAL POINT
SAT
URA TIO
0.20
0.10 0
500
© INGERSOLL − RAND COMPANY 1960
2000
1500
1000
2500
3000
PRESSURE − PSIA
Figure 17.14Q Compressibility chart for propylene (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
where cp and cv are specific heats at constant pressure and constant volume respectively, Btu/lb-mol-°F [20]. To obtain the average value of cp for a gas mixture, use the weighted mole fraction average, evaluating cp at the average of the suction and discharge temperatures of the compressor cylinder. Depending on the magnitude of the compression ratio, the cp at suction temperature can be used when the ratio is small. m = isentropic or adiabatic exponent = (k − 1)/k m = n = polytropic exponent = (k − 1)/kEp Ep = polytropic efficiency = m/m = [(k − 1)/k]/[(n − 1)/n]
m
1(k 1) E p (k )
m Ep
(n 1) (n)
Mcp = Mcv + 1.987 Btu/(lbmol.°F) Then; k =
Mc p c p Mc p = = Mc v c v Mc p −1.987
(17.26) (17.27) (17.28)
Compression Equipment 555
1.0 –50° 0.9
0° 400°F
COMPRESSIBLITY FACTOR Z =
PV RT
50°
350°
10
0.8
0°
12
300° 280° 260°
0° 14
0°
0.7
SAT
16 0° URA TIO NL
18
0°
INE
0.6
20
0°
24 0 22 230° ° 0 ° 21 0°
0.5
0.4
COMPRESSIBILITY CHART FOR PROPANE (C3H8)
0.3
CRITICAL POINT
BASED ON: SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” – STEARNS AND GEORGE, “THERMODYNAMIC PROPERTIES OF PROPANE” INDUSTRIAL AND ENGINEERING CHEMISTRY − VOL. 35, NO. 5 – MAY 1943... CU. FT/LB. AT 14.696 PSIA AND 60°F = 8.471 Z AT 14.696 PSIA AND 60°F = 0.9875 © INGERSOLL − RAND COMPANY 1962
0.2
0
100
200
300 PRESSURE − PSIA
400
500
600
Figure 17.14R Compressibility chart for low pressure propane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
where M = molecular weight of gas c = specific heat, Btu/lb-°F temperature rise Mcp = molar heat capacity, Btu/(lb mol-°F) ([60] see tables this reference), constant pressure; Mcv = molar heat capacity at constant volume, Btu/lb mol-°F 1.989 = constant for all hydrocarbon gases For mixtures of gases, calculate the average Mcp by multiplying the individual gas mol % of each component by its respective Mcp (see Reference [11] or other sources for tables) and sum to get the molar average, Mcp, for the mixture. For the ratio of specific heat, see Eq. 17.28. For a perfect gas:
cp – cv = R
(17.29)
For real gases, the relationship applies [19].
cp − cv = R/J
(17.30)
(cp/cv)ideal = cp/(cp − R)
(17.31)
556 Petroleum Refining Design and Applications Handbook Volume 2 1.2 1.1 1.0
100° .8 120° 140° 160° .7 180° 200°
400° 380° 360° 340° 320° 300°
ON ATI UR SAT
.6 .5
E LIN
COMPRESSIBILITY FACTOR Z = PV RT
.9
280°
.4
260°
COMPRESSIBILITY CHART FOR PROPANE (C3H8)
240°
.3
CRITICAL POINT
220°
BASED ON SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” STEARNS AND GEORGE, “THERMODYNAMIC PROPERTIES OF PROPANE” INDUSTRIAL AND ENGINEERING CHEMISTRY. VOL. 35, NO. 5, MAY 1943... CU. FT/LB. AT 60°F AND 14.696 PSIA = 8.471 Z AT 60°F AND 14.696 PSIA = 0.9875
.2 .1
© INGERSOLL − RAND CO. 1960
0
500
1000
1500
2000
2500
3000
3500
4000
PRESSURE − PSIA
Figure 17.14S Compressibility chart for high pressure propane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
where cp and cv are specific heats at constant pressure and constant volume respectively, Btu/lb-°R. R = gas constant, ft-lbf/lb-mol-°R J = Joules’ constant = 778 ft-lbf/Btu or, for a real gas [21].
cp cp = c v c p − (c p − c v )
(17.32)
From Edmister [16], Δcp = 1.44[(cp − cv°)/k2], where Δcp cp° R R'
= Btu/(lb-mol)(°R) = mol heat capacity at ideal gas state = universal gas constant = 1545 ft-lbf/lbm-°R. For dry air: R = 53.35 ft-lbf/lbm°R. = gas constant for a specific gas, 1545/(mol wt)
From combined Boyle’s and Charles’ Law Equation of State for Perfect Gas:
Pv = RT/Cp = RT
(17.33)
Compression Equipment 557
COMPRESSIBILITY CHART FOR N-BUTANE (C4H10)
1.00
.90
BASED ON: SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” CU FT./POUND AT 14.696 PSIA AND 60°F = 6.327 Z AT 14.696 PSIA AND 60°F = 0.975 © INGERSOLL − RAND COMPANY 1960
100°F
14
0° 18
0° 22 0° SAT 24 URA 0° TIO NL INE
RT
COMPRESSIBILITY FACTOR – Z =
PV
.80
.70
400°
380 ° 360 ° 34 0°
26
0° 28
.60
0°
32
30
0°
0°
.50
.40
.30 CRITICAL POINT .20
0
100
200
300
400
500
600
PRESSURE − PSIA
Figure 17.14T Compressibility chart for low pressure N-butane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
v = specific volume, ft3/lbm P = absolute pressure, lbf/ft2 abs R = gas constant, ft-lbf/lbm-°R T = absolute temperature, °R = (°F + 460) Cp = conversion factor = 1.0 For real gases: Pv = ZRT Z = compressibility factor
17.9 Compressor Performance Characteristics Piston Displacement Piston displacement is the actual volume of the cylinder displaced as the piston travels its stroke from the start of the compression (condition (1)) to the end of the stroke (condition (e)) of Figure 17.12 expressed as ft3 of volume displaced per minute. Displacement values for specific cylinder designs are available from the manufacturers (Table 17.6). Neerken [22] is a useful reference. Reciprocating compressors are usually rated in terms of piston displacement, which is the net volume in ft3 per minute displaced by the moving piston [23]. Note that the piston does not move through the clearance volume of Figure 17.12; therefore this volume is not displaced during the stroke.
558 Petroleum Refining Design and Applications Handbook Volume 2 1.20
°F 100 ° 5 1 00° 20 50° 2 ° 300
1.10
350°
.90
400°
.80 .70 .60 .50 .40 .30
0° 4075° 3 ° 35205° 3 0° NE 30 ° N LI 250 URATIO SAT
COMPRESSIBILITY FACTOR Z = PV RT
1.00
CRITICAL POINT
COMPRESSIBILITY CHART FOR N-BUTANE (C4H10)
.20
BASED ON: SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” CU FT/POUND AT 14.696 PSIA AND 60°F = 6.327 Z AT 14.696 PSIA AND 60°F = 0.975 © INGERSOLL − RAND COMPANY 1960
.10
0
500
1000
1500
2000 2500 PRESSURE − PSIA
3000
3500
4500
4000
Figure 17.14U Compressibility chart for high pressure N-butane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
For Single-acting Cylinder (Figure 17.4A)
PD = Aps(rpm)/1,728
(17.34)
where PD = piston displacement, cfm Ap = c ross-sectional net area of piston, in.2 If cylinder is head-end, Ap is total area of piston; if cylinder is crank-end, Ap is net area of piston area minus rod cross-section area. s = stroke length, in. rpm = revolutions per minute of crank shaft or number of compression strokes per minute For Double-acting Cylinder (Figure 17.4B): The displacement of the head end and crank end of the cylinder must be added for the total displacement. The displacement of the crank end is less than that of the head end by the volume equivalent to the piston rod displacement. For a multistage unit, the piston displacement is often only given for the first stage [24].
A p s(rpm) (A p − A r )s(rpm) + 1, 728 1, 728
PD =
PD = (Ap − Ar/2)2s(rpm)/1728
where Ar = cross-sectional area of piston rod, in.2
(17.35) (17.35a)
Compression Equipment 559
1.00 100° 0.90 120° 140°
18
0.80
0°
0.70
400°
20
0°
SAT
-400 ° 380° 360 °
22
0°
URA TIO
24
340
0°
N LI
NE
0.60
26 0°
COMPRESSIBILITY FACTOR Z =
PV RT
160°
°
30
28
°
320 0°
0°
0.50
0.40
0.30
COMPRESSIBILITY CHART FOR ISOBUTANE (C4H10)
CRITICAL POINT
BASED ON: SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” Z AT 14.696 PSIA AND 60°F = 0.992 CU. FT/LB. AT 14.696 PSIA AND 60°F = 6.339 © INGERSOLL – RAND COMPANY 1960
0.20
0
100
200
300
400
500
600
PRESSURE – PSIA
Figure 17.14V Compressibility chart for low pressure iso-butane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
Compression Ratio The compression ratio is the ratio, Rc, of the absolute discharge pressure to the absolute suction pressure of the cylinder:
Rc = P2/P1
(17.36)
where P1 = initial suction pressure, absolute units P2 = cylinder discharge pressure at cylinder flange, absolute units Compression ratios usually vary between 1.05 and 7 per stage; however, a ratio of 3.5–4.0 per stage is considered maximum for most process operations. Quite often temperature rise of the gas during the compression dictates a limit for the safe or reasonable pressure rise. The maximum temperature rise is governed either by the maximum operating temperature of the compressor cylinder or by the maximum temperature the gas can withstand before decomposition, polymerization, or even auto-ignition as for chlorine, acetylene, etc. Because the volumetric efficiency decreases with an increase in compression ratio, this also adds to the selection of a reasonable limiting
560 Petroleum Refining Design and Applications Handbook Volume 2
1.10
0.90 0.80 200° 0.70
220° 240° 260°
0.60
450°F
UR SAT
0.50
N ATIO
COMPRESSIBILITY FACTOR Z = PV RT
1.00
0.40
LINE
0.30
CRITICAL POINT 280°
400° 380° 360° 340° 320° 300° COMPRESSIBILITY CHART FOR ISOBUTANE ( C4H10)
0.20
BASED ON - SAGE AND LACEY, “THERMODYNAMIC PROPERTIES OF HYDROCARBONS” Z AT 14.696 PSIA AND 60°F = 0.992 CU. FT/LB AT 14.696 PSIA AND 60°F = 6.339 © INGERSOLL – RAND COMPANY 1960
0.10
0
500
1000
1500
2000
2500
3000
3500
4000
PRESSURE – PSIA
Figure 17.14W Compressibility chart for high pressure iso-butane (used by permission from 3519D (1981), © 1960. Ingersoll-Rand Company. All rights reserved).
discharge pressure. With a known maximum temperature, the maximum ratio of compression can be calculated from the adiabatic temperature rise relation. The optimum minimum horsepower occurs when the ratios of compression are equal in all cylinders for multistage units. With external cooling of the gas between stages, it is necessary to make reasonable allowances for pressure drops through the intercoolers and take this into account when setting the compression ratios: a. Ideal (no intercooling), for four stages (cylinders)
P2/P1 = P3/P2 = P4/P3
(17.37)
b. Actual (with intercooling)
Pi1 /P1 = Pi 2 /Pil′ = Pi 3 /Pi′2 = … Pfy /Piy
(17.38)
where 1,2,3,.…y = conditions of gas across a cylinder represented by (1) for first stage, (2) for second stage, etc. i = interstage discharge pressure condition, immediately at cylinder. Prime (') = interstage discharge condition, reduced by the pressure drop through the intercoolers, valves, piping, etc.; therefore, a prime represents actual pressure to suction of succeeding cylinder in multistage cylinder system. f = final or discharge pressure from multistage unit.
Compression Equipment 561 PSEUDO REDUCED PRESSURE 3
2
4
5
6
PSEUDO REDUCED TEMPERATURE 3.0 2.8 2.6 2.4 2.2 2.0 1.9 1.8
1.0
0.9
7
1.5
1.0 1.05 1.2 1.3 0.95 1.1
1.4
1.7
1.0
1.6 0.8
8 1.1
5
1.1
1.5 1.45
1.2
1.4 0.7
1.6
1. 3
1.35
COMPRESSIBILITY FACTOR Z
1.3 0.6
1.4 1.5 1.5 1.6
1.25
1.7
1.2 0.5 1.15
0.4
1.7
1.1
1.8 1.9 2.0 2.2
1.4
2.4 2.6 3.0
1.3
1.2
0.3 1.05
0.25
3.0 2.8
1.1
0.9
1.1
2.6 2.4 2.2 2.0 1.9 1.8 1.7 1.6
1.0
1.2 1.1
1.0
COMPRESSIBILITY OF NATURAL CASES JAN. 1, 1961
1.05
1.4 1.3 7
COMPRESSIBILITY FACTOR Z
1.1
1
0
0.9 8
9
10
11
12
13
14
15
PSEUDO REDUCED PRESSURE PR Pseudo-reduced temperature = Pseudo-reduced pressure =
absolute temperature molecular average critical temperature absolute pressure molecular average critical pressure
Figure 17.15 Compressibility factor for natural gas (used by permission: G.G. Oberfell, D. L. Katz, and R. C. Alden. Natural Gasoline Association of American, Inc. All rights reserved).
562 Petroleum Refining Design and Applications Handbook Volume 2 (A)
Note: Details of Losses not Shown, See Fig. –4 Discharge
Pressure
Ideal Gas Law Actual
Co m
pre
ssi
on
Intake Volume or Stroke Compressibility Factor Less than 1.0 (B)
Pressure
Ideal Gas Law Actual
Volume or Stroke Compressibility Factor Greater than 1.0 (C)
Pressure
Ideal Gas Law
Actual
Volume or Stroke Compressibility Factor Greater than 1.0: Extreme Deviation (Ethylene Discharging at 30,000 – 40,000 psig)
Figure 17.16A–C Deviations from the ideal gas law.
Compression Equipment 563
1
2
3
4
5
6
7
8
9
1–Suction Valve Chatters Probably due to weak valve springs, and may result in broken valve plate or a leaky valve. 2–Discharge Valve Chatters Shows weak springs in the discharge valves, and will result in a broken valve plate or a leaky valve, which in turn will result in cylinder heating and loss of horsepower. 3–Suction Passage Too Small In addition to too small a suction passage, too small a valve lift could also be indicated. 4–Discharge Passage Too Small In addition to too small a discharge passage, too small a valve lift could also be indicated. 5–Suction Valve Spring Too Sti Too stiff a suction valve means a loss of horsepower. Valve spring of proper tension should be installed. 6–Discharge Valve Spring Too Sti Too stiff a discharge valve spring likewise results in loss of horsepower. Valve spring of proper tension should be installed here, also. 7–Suction Valve Leaking Leak may be in either the valve or the valve gasket. 8–Discharge Valve Leaking Curve 1 indicates a badly leaking discharge valve; curve 2 a slightly leaking one. Leak may be in the valve or in the valve gasket. 9–Piston Ring Leaking Leaky piston rings may be due to worn rings, out of round compressor cylinders, or weak expander rings used with plastic-type piston rings.
Figure 17.16D Typical compressor aliments and how they look on P-T diagrams (used by permission: Palmer, E. Y. Petroleum Processing, p. 884, June 1954. © National Petroleum News, Adams Business Media).
Table 17.2 Comparison of performance for propane. Actual
Ideal
Volumetric efficiency
0.802
0.835
cfm at inlet conditions
0.802
835
Specific volume in inlet, ft3/lb
1.160
1.314
lb handled/min
691
635
Basic horsepower required
388
425
Horsepower/lb
0.561
0.670
Used by permission: Hartwick, W. Chemical Engineering, p. 204, Oct. 1956. ©McGraw-Hill, Inc. All rights reserved.
564 Petroleum Refining Design and Applications Handbook Volume 2 Table 17.3 Compressibility factors, Z. Propane
24% Nitrogen–76% Hydrogen
Pressure, Psia
Z
Psia
Z
Psia
Z
100
0.884
1,600
1.061
400
0.954
160
0.838
2,400
1.092
500
0.953
220
0.800
3,500
1.129
600
0.955
300
0.765
4,800
1.172
700
0.957
Used by permission: Hartwick, W. Chemical Engineering, p. 204, Oct.1956. ©McGraw-Hill, Inc. All rights reserved).
Table 17.4 Approximate ratio of specific heats (“k” values) for various gases. k @ 14.7 psia Gas
Symbol
Monatomic
He, Kr, Ne, Hg
1.67
Most diatomic
O2, N2, H2, etc
1.4
Acetylene
C2H2
Air
Mol wt
60°F
150°F
Density @ 14.7 psia & 60°F lb/ft3
26.03
1.3
1.22
0.0688
28.97
1.406
1.40
0.0765
17.03
1.317
1.29
0.0451
Ammonia
NH3
Argon
A
Benzene
C6H6
78.0
1.08
1.09
0.2064
Butane
C4H10
58.1
1.11
1.08
0.1535
Isobutane
iC4H10
58.1
1.11
1.08
0.1578
Butylene
C4H8
56.1
1.1
1.09
0.1483
Iso-butene
iC4H8
56.1
1.1
1.09
0.1483
Carbon dioxide
CO2
44.0
1.3
1.27
0.1164
Carbon monoxide
CO
28.0
1.4
1.4
0.0741
Carbon tetrachloride
C Cl4
153.8
1.18
0.406
Chlorine
Cl2
70.9
1.33
0.1875
Dichlorodifluoromethane
C Cl2F2
120.9
1.13
Dichloromethane
CH2Cl2
84.9
1.18
Ethane
C2H6
30.0
1.22
1.17
0.0794
Ethylene
C2H4
28.1
1.25
1.21
0.0741
Ethyl chloride
C2H5Cl
64.5
1.13
1.667
Flue gas Helium
0.1056
0.2245
0.1705
1.14 He
4.0
1.667
0.01058 (Continued)
Compression Equipment 565 Table 17.4 Approximate ratio of specific heats (“k” values) for various gases. (Continued) k @ 14.7 psia Gas
Symbol
Mol wt
60°F
150°F
Density @ 14.7 psia & 60°F lb/ft3
Hexane
C6H14
86.1
1.08
1.05
0.2276
Heptane
C7H16
100.2
1.04
0.264
Hydrogen
H2
2.01
1.41
1.40
0.0053
Hydrogen chloride
HCl
36.5
1.48
Hydrogen sulfide
H2S
34.1
1.30
1.31
0.0901
Methane
CH4
16.03
1.316
1.28
0.0423
Methyl chloride
CH3Cl
50.5
1.20
0.1336
19.5
1.27
0.0514 0.0793
Natural gas (approx.)
0.09650
Nitric oxide
NO
30.0
1.40
Nitrogen
N2
28.0
1.41
Nitrous oxide
N 2O
44.0
1.311
Oxygen
O2
32.0
1.4
1.39
0.0846
Pentane
C5H12
72.1
1.06
1.06
0.1905
Propane
C3H8
44.1
1.15
1.11
0.1164
Propylene
C3H6
42.0
1.16
0.1112
Sulfur dioxide
SO2
64.1
1.256
0.1694
Water vapor (steam)
H2O
18.0
1.33*
1.40
0.0743 0.1163
1.32
0.04761
*At 212°F. Used and compiled by permission: “Plain Talks on Air and Gas Compression,” Fourth of Series, Worthington. Dresser-Rand Corporation. Also compiled by permission from “Reciprocating Compressor Calculation Data,” ©1956. Dresser-Rand Corporation.
Compression ratios across stages:
R 1 = Pi1 /P1 R 2 = Pi 2 /Pi′1
R 3 = Pi 3 /Pi′2 R f = Pfy /Piy′
R 1 = R 2 = R 3 =…R f = y R t
(17.39)
where Rt= overall compression ratio of unit = Pi/P1. For two-stage, compression per stage is
R 1 = R 2 = Pf 2 / P1
(17.39a)
566 Petroleum Refining Design and Applications Handbook Volume 2 Table 17.5 Constants for molal heat capacity. Gas
Formula
Air
Molecular weight
Critical press, psia
Critical temp, °R
A
B
28.97
546.7
238.4
6.737
0.000397
Ammonia
NH3
17.03
1,638
730.1
6.219
0.004342
Carbon dioxide
CO2
44.01
1,073
547.7
6.075
0.005230
Carbon monoxide
CO
28.01
514.4
241.5
6.780
0.000327
Hydrogen
H2
2.016
305.7
72.47
6.662
0.000417
Hydrogen sulfide
H2S
34.07
1,306
672.4
7.197
0.001750
Oxygen
O2
32.00
730.4
277.9
6.459
0.001020
Sulfur dioxide
SO2
64.06
1,142
774.7
Water
H2O
18.02
3,200
1,165
7.521
0.000926
Methane
CH4
16.04
673.1
343.2
4.877
0.006773
Acetylene
C2H2
26.04
911.2
563.2
6.441
0.007583
Ethene
C2H4
28.05
748.0
509.5
3.175
0.013500
Ethane
C2H6
30.07
717.2
549.5
3.629
0.016767
Propene
C3H6
42.08
661.3
656.6
4.234
0.020600
Propane
C3H8
44.09
617.4
665.3
3.256
0.026733
1-Butane
C4H8
56.11
587.8
75.2
5.375
0.029833
Isobutene
iC4H8
56.11
580.5
736.7
6.066
0.028400
Butane
C4H10
58.12
530.7
765.3
6.188
0.032867
Isobutane
iC4H10
58.12
543.8
732.4
4.145
0.035500
Amylene
C5H10
70.13
593.7
853.9
7.980
0.036333
Isoamylene
C5H10
70.13
498.2
836.6
7.980
0.036333
Pentane
C5H12
72.15
485.0
846.7
7.739
0.040433
Isopentane
C5H12
72.15
483.5
829.7
5.344
0.043933
Neopentane
C5H12
72.15
485.0
822.9
4.827
0.045300
Benzene
C6H6
78.11
703.9
1,011
-0.756
0.038267
Hexane
C6H14
86.17
433.5
914.3
9.427
0.047967
Heptane
C7H16
100.2
405.6
976.8
11.276
0.055400
Used by permission: Hartwick, W. Chemical Engineering, p. 209, Oct. 1956. ©McGraw-Hill, Inc. All rights reserved.
26.038 28.054 30.070 42.081 44.097 56.108 56.108 56.108 58.124 58.124 72.151 72.151 78.114 86.178 100.205
C2H2 C2H4 C2H6 C3H6 C3H8 C4H8 C4H8 C4H8 C4H10 C4H10 C5H12 C5H12 C6H6 C6H14 C7H16
Methane
Ethyne (Acetylene)
Ethene (Ethylene)
Ethane
Propene (Propylene)
Propane
1-Butane (Butlyene)
cis-2-Butene
trans-2-Butene
iso-Butane
n-Butane
iso-Pentane
n-Pentane
Benzene
n-Hexane
n-Heptane
16.043
CH4
Gas
M
Chemical formula
142.943
123.401
66.435
105.133
101.897
85.277
83.476
77.329
67.598
73.359
64.176
55.878
47.131
38.254
39.888
34.301
−25
154.539
133.303
74.060
112.603
110.369
91.270
90.078
82.587
73.268
79.583
68.783
59.898
49.882
40.906
42.020
34.931
0
Temperature, °C
159.011
137.144
77.034
115.565
113.675
93.685
92.690
84.628
75.461
81.961
70.605
61.459
50.904
41.937
42.778
35.199
10
165.985
143.110
81.675
120.211
118.792
97.447
96.815
87.823
78.925
85.663
73.524
63.895
52.666
45.559
43.926
35.717
25
*Data source: selected values of properties of hydrocarbons, API Research Project 44
Table 17.5A Molal heat capacity MCp (ideal-gas state), kJ/mole °C.
177.141
152.709
89.224
130.686
127.335
105.326
103.624
92.979
84.508
91.509
78.561
67.832
55.723
46.115
45.650
36.744
50
188.293
162.308
96.761
136.160
135.581
110.334
110.408
98.174
90.154
97.310
83.585
71.789
58.819
48.695
47.235
37.870
75
199.400
171.884
104.324
144.452
144.029
117.024
117.340
103.387
95.851
103.111
88.820
75.762
62.114
51.283
48.720
39.201
100
210.046
181.080
111.321
152.182
152.011
123.326
123.932
108.434
101.323
108.493
93.820
79.584
65.294
53.753
49.981
40.529
125
(Continued)
220.585
190.194
118.202
161.448
159.999
130.400
130.521
113.464
106.800
113.860
98.838
83.395
68.556
56.214
51.168
41.986
150
Compression Equipment 567
29.131 29.079 28.290
31.999 28.013 2.016 34.076
O2 N2 H2 H2S CO CO2
Oxygen
Nitrogen
Hydrogen
Hydrogen sulfide
Carbon monoxide
Carbon dioxide
34.700
44.010
35.962
29.123
33.673
28.611
29.114
29.240
33.474
29.067
35.636
0
36.411
29.105
33.815
28.687
29.092
29.265
33.488
29.078
35.640
10
37.122
29.146
34.028
26.502
29.114
29.361
33.572
29.098
35.645
25
38.212
29.150
34.379
28.964
29.116
29.481
33.678
29.141
35.653
50
39.261
29.193
34.729
29.065
29.140
29.647
33.832
29.196
35.661
75
40.290
29.263
35.080
29.126
29.196
29.870
34.032
29.262
35.670
100
41.199
29.319
35.434
29.158
29.219
30.045
34.207
29.339
35.678
125
42.095
29.405
35.792
29.178
29.279
30.274
34.424
29.429
35.688
150
Source: Gas Processors Suppliers Association (GPSA) Engineering Data Book, SI Version, vol. 1, 12th ed., 2004.
*Exceptions: Air—Keenan and Keyes, Thermodynamic Properties of Air, Wiley, 3rd Printing 1947. Ammonia—Edw. R. Grabl, Thermodynamic Properties of Ammonia at High Temperatures and Pressures, Petr. Processing, April 1953, Hydrogen Sulfide—J. R. West. Chem. Eng. Progress, 44, 287, 1948.
29.087
28.010
33.313
33.388
18.015
H2O
29.048
35.626
17.031
Water
Ammonia
−25
M
28.964
NH3
Gas
Temperature, °C
Air
Chemical formula
*Data source: selected values of properties of hydrocarbons, API Research Project 44
Table 17.5A Molal heat capacity MCp (ideal-gas state), kJ/mole °C. (Continued)
568 Petroleum Refining Design and Applications Handbook Volume 2
Compression Equipment 569 Table 17.6 Typical reciprocating air compressor data. Single-stage horizontal
Size, in.
rpm
5×5
550
Two-stage, angle vertical
Piston Max. press., displacement, psi cfm
Size, in.
rpm
Two-Stage, Horizontal Duplex Piston displacement, cfm
Size, in.
rpm
Piston displacement, cfm
150
61
11¼/7 × 7
600
478
21/13 × 14
277
1546
6×5
100
88
13½/8½ × 7
600
690
23/14 × 14
277
1858
7×5
60
121
14½/9½ × 7
600
798
24/15 × 17
257
2275
8×5
40
157
16/10½ × 7
600
973
28/17 × 19
225
3031
10 × 5
20
248
18½/11½ × 8½
514
1351
30½/18 × 21
200
3704
20½/13 × 8½
514
1662
34½/21 × 25
180
4847
450
1975
225
6065
450
2412
30 1 2 18 1 2 / × 22 30 1 2 18 1 2
200
7396
34 1 2 21 / × 25 34 1 2 21
180
9673
36 1 2 22 / × 25 36 1 2 22
180
10,808
39 1 2 23 / × 25 39 1 2 23
164
12,189
6×7
150
100
7×7
100
138
8×7
60
180
10 × 7
35
283
12 × 7
20
410
8×9
450
1 17 3 4 / 16 × 9 2 17 3 4
28 17 / × 19 28 17
135
184
9×9
100
234
10 × 9
75
290
12 × 9
40
420
15 × 9
20
658
125
321
Designation numbers in table for multiple cylinders.
12 × 11
100
465
Bore of first stage/bore of second stage x stroke, all in inches.
14 × 11
60
635
15 × 11
50
730
17 × 11
30
940
19 × 11
20
1174
20 × 11
15
1300
125
502
14 × 13
100
686
17 × 13
55
1016
19 × 13
40
1270
10 × 11
12 × 13
360
1 16 /141 / 2 × 9 2 16
327
300
For example:
1 1 16 / 14 × 9 16 2 2
There are two first-stage cylinders, 16-in. dia. In parallel, one 14 ½ -in. second-stage cylinder and all on 9 ½ -in. stroke length.
(Continued)
570 Petroleum Refining Design and Applications Handbook Volume 2 Table 17.6 Typical reciprocating air compressor data. (Continued) Single-stage horizontal
Size, in.
rpm
20 × 13
23 × 13 26 × 13
277
Two-stage, angle vertical
Piston Max. press., displacement, psi cfm 35
1410
20
1717
12
2202
Size, in.
rpm
Two-Stage, Horizontal Duplex Piston displacement, cfm
Size, in.
rpm
Piston displacement, cfm
Used by permission: “Feather Valve Compressor Selection Handbook,” Worthington bul. L-600-B16. Dresser-Rand Company.
For five stages:
R 1 = R 2 = R 3 = R 4 = R 5 = 5 Pf 5 / P1
(17.39b)
It is common practice to use intercoolers on multistage machines. The function of the intercooler is to cool the gas to as near the original suction temperatures as practical with as little pressure drop as possible. With temperature sensitive material, this is essential. This cooling effects a savings in the required brake horsepower as it essentially is cooling at constant pressure and results in a smaller volume of gas to be handled by the next cylinder. To effect the greatest saving, the coldest cooling practically available should be used. In some cases, it is desirable to use a two-stage compression without intercooling. If the composition of the gas must remain constant throughout the compression and the temperature does not limit, intercoolers cannot be used if condensables are present. Sometimes two stages are used on low “k” or “n” value gases to improve the volumetric efficiency. When this is the case and high compression temperatures or economy of operation do not control, it may be advantageous to omit the intercooler. Note that when intercoolers are not used, the compressor jacket water should be 10–15°F greater than the interstage dew point. This will require warm jacket water through the preceding stage. The intercooler operation does not outwardly affect the theoretical optimum compression ratio per stage. However, it does affect the cumulative horsepower required to do the work of total compression, because all the pressure drop lost must be replaced as horsepower. There is also a gain in performance due to this intercooling as is shown in Figures 17.17A and 17.17B. The allowance for intercooler pressure drop is usually made by increasing the discharge pressure from the cylinder to include one-half of the intercooler pressure drop between stages, and the suction pressure on the following stage is reduced to the other one-half of the pressure drop, when compared to the theoretical pressures with no pressure drop allowance. Ratio of compression per stage may be calculated.
Pf = P1R − ( p1)R −1 − ( p2)R −2
– ( p3)Ry−3 − ( p4)Ry−4…..
(17.40)
Continue for number of terms on right side of equation equal to number of stages. This is usually best solved by trial and error and can be simplified if most of the P values are assumed equal. It assumes all the intercooler pressure drop is deducted from the suction pressure of the succeeding stage, i.e., first stage intercooler pressure drop is deducted from second stage suction pressure. Pf = final pressure of multistage set of cylinders = number of compression stages
Compression Equipment 571 Saving due to Jacketing of Cylinder Discharge Pressure Adiabatic Compression Average Compression Line had Compression Occurred in One Cylinder Saving Due to Intercooling
Pressure
H-p Card
If Perfect Intercooling between Stages, c c
L-p Card
Intercooler Pressure
b
Suction Temperature after Intercooling Saving due to Jacketing L-p Cylinder
Isothermal Compression Intake Pressure
Adiabatic Compression a
Volume
Clearance
Figure 17.17A Combined indicator cards from a two-stage compressor showing how cylinder water jackets and intercooler help bring compression line nearer to isothermal (used and adapted by permission: Miller, H. H. Power, © 1994. McGraw-Hill, Inc., New York, All rights reserved). C D
B
P2
10%
5%
10%
15%
Pressure
5%
15% P1
A
E F High Clearance Volume Moderate Clearance Volume Low Clearance Volume
Figure 17.17B Effects of clearance volume on performance efficiency of reciprocating compressor cylinder (valve design effect) (used by permission: Livingston, E. H. Chemical Engineering Progress, V. 89, No. 2 © 1993. American Institute of Chemical Engineers, Inc. All rights reserved).
p = pressure drop across interstage coolers, psi 1 = first stage 2 = second stage If one half p is added to discharge of one stage and one half deducted from suction of next stage:
Pf = P1R − (1/2 p1)R −1 − (1/2 p2)R −2
− (1/2 p3)R -3 − (1/2 p4)R -4 ….
(17.41)
572 Petroleum Refining Design and Applications Handbook Volume 2 In practice, the ratios for each stage may not work out to be exactly the same; however this does not keep the compressor from operating satisfactorily as long as all other factors are handled accordingly. Compressor Jacket Cooling. The compressor jacket cooling water does not have to be as warm as does the gas engine jacket water. Water 15–20°F warmer than the dew point of the gas being compressed will ensure against condensation. A maximum of 15–20°F rise in jacket water temperature is recommended. The flow of water to the jackets should never be throttled in order to maintain this temperature as the lowered velocity tends to facilitate fouling of the jackets. The amount of heat rejected by compressor jackets varies with the size and type of machine. This heat rejection is usually given as Btu/h/bhp. Heat rejection to the compressor cylinder will average about 500 Btu/h/bhp. Some are low as 130, and it is necessary to check with the manufacturer to obtain an accurate figure.
Example 17.1: Interstage Pressure and Ratios of Compression For two stages of compression, what should be the pressures across the cylinders if the intercooler and piping pressure drop is 3 psi?
Suction to first stage: P1 = 0 psig (14.7 psia)
Discharge from second stage: Pf2 = 150 psig (164.7 psia)
Per stage: Rc = 164.7 / 14.7 = 11.2 = 3.35 No intercooling:
R c = 3.35 Pi1 = 3.35(14.7 ) = 49.2 psia P1 = 14.7 psia
R c = 3.35 Pf 2 = 164.7 P2 = 49.2
With intercooling: First stage:
R c = 3.45 Pil = 49.2 + ( 1 2 )(3.0) = 50.7 psia P1 = 14.7 psia
Second stage:
Pi′1 = 49.2 − ( 1 2 )(3.0) = 47.7 psia R c = 3.445 Pf 2 = 164.7
The example shows that although the ratios per cylinder are balanced, they are each greater than the theoretical. This corresponds to actual operations.
Compression Equipment 573 It is important to note that quite often the actual compression ratios for the individual cylinders of a multistage machine will not be balanced exactly. This condition arises as a result of the limiting horsepower absorption for certain cylinder sizes and designs of the manufacturer. In the final selection, these will be adjusted to give compression ratios to use standard designs as much as possible.
Actual Capacity or Actual Delivery, Va This is the volume of gas measured at the intake to the first stage of a single or multistage compressor at stated intake temperature and pressure, ft3/min. Manufacturer performance guarantees usually state that this capacity is subject to 6% tolerance when intake pressure of first stage is 5 psig or lower, and may state a volume tolerance of about 3% for pressures above this 5 psig intake [25]. It is extremely important to state whether the capacity value has been corrected for compressibility. At low pressures, compressibility is usually not a factor, however, if conditions are such as to not require the use of compressibility, it is usually omitted and so stated. The actual required capacity may be calculated for process requirements, or if a known cylinder is being examined.
Va = PD(Ev); cfm cylinder will compress at suction pressure and temperature Ev = volumetric efficiency, is based on the characteristics of the cylinder.
(17.42)
Ev, or sometimes Ev, is the volumetric efficiency of a cylinder and is the ratio of the amount of gas that is actually compressed to the amount of gas that could be compressed if no clearance existed in the cylinder (see Figure 17.12). Ev can be obtained from Figures 17.18A–F.
(
)
%E v = 100 − R c − (Vpc ) R 1c/k − 1
(17.43)
Clearance Volume This is the total volume remaining in the cylinder at the end of the piston stroke. This consists of the volume between the end of the piston and the cylinder head, in the valve ports and the volume in the suction valve guards and the discharge valve seats [25] (see Figures 17.12, 17.17A, and 17.17B). The effect of clearance volume is shown in Figures 17.17A and 17.17B [98]. The illustrated volumes of 5%, 10%, and 15% usually satisfy a reasonable process compressor range. For example, in the 15% compression slope, ABC will reach pressure shown as P2 sooner than the slope of the 10% curve. Upon re-expansion at the end of the compression stroke DEF, the slope is steeper and allows the gas to enter the cylinder sooner during the suction or intake cycle [98]. The volumetric efficiency increases with a decrease in clearance volume and a decrease in compression ratio [98]. This is the most profound effect, although other design factors do influence the efficiency to a lesser extent. In attempting to balance volumetric and compression efficiency, Livingston [98] points out “for a high compression ratio (6 to 15) clearance volume is the key factor with valve design being secondary. For a compression ratio of 3 to 6, the clearance volume and valve design should be balanced. For low compression ratio, less than 3, the valve design is the primary factor.”
Percent Clearance Percent clearance is the volume % of clearance volume to total actual piston displacement [25].
Vpc = Calculate for each cylinder end.
Vc (100) PD′
(17.44)
574 Petroleum Refining Design and Applications Handbook Volume 2 where Vc = clearance volume, in.3 Vpc = percent clearance PD' = piston displacement, in.3 For double-acting cylinders, the clearance at the head end should be calculated separately from that of the crank end, because for small cylinders, the volume occupied by the piston rod is significant when cylinder unloading is considered. For double-acting cylinders, % clearance is based on total clearance volume for both the head end and crank end of the cylinder × 100 divided by the total net piston displacement. The head and crank end % clearance values will be different due to the presence of the piston rod in the crank end of the cylinder. The % clearance values are available from manufacturers for their cylinders. The values range from about 8% for large 36-in. cylinders to 40% for small 3- and 4-in. cylinders. Each cylinder style is different.
Cylinder Unloading and Clearance Pockets For a discussion of this important performance control topic, refer to “Cylinder Unloading,” later in this chapter and Figure 17.27. A reasonable average range is 7–22% with fixed valve pockets [98].
Volumetric Efficiency Volumetric efficiency is the efficiency of a cylinder performance based on operating experience and actual volume conditions.
(
)
%E v = 100 − R c − Vpc R 1c/k − 1
(17.45)
where Rc = ratio of compression across an individual cylinder. Volumetric efficiency may be expressed as the ratio of actual cylinder capacity expressed at actual inlet temperature and pressure conditions, divided by the piston displacement (see Figure 17.17B). Values of Ev may be read from Figures 17.18A–F for values of Rc and Vpc. For example, in a multistage compressor, such as a two-stage compressor, each cylinder does one half of the total work of compression. The low pressure or first stage of the two-stage unit controls the capacity of the overall compressor, i.e., the second stage can handle only the volume of gas passed to it from the discharge of the first-stage cylinder. That is, the gas that passes through the low or first-stage discharge valves must continue on through the compressor’s second (or final in this case) stage cylinder and be discharged at the specified pressure and calculated/actual temperature. Thus, in all compressors of two or more stages, the volumetric efficiency of the low-pressure cylinder determines the volumetric efficiency of the entire compressor (not recognizing packing leaks) [23].
Compression Efficiency (Adiabatic) The compression efficiency is the ratio of the work required to adiabatically compress a gas to the work actually done within the compressor cylinder as shown by indicator cards (Figures 17.12 and 17.16). The heat generated during compression adds to the work that must be done in the cylinder. Valves may vary from 50% to 95% efficient depending on cylinder design and the ratio of compression. Compression efficiency (or sometimes termed volumetric efficiency) is affected by several details of the systems. a. b. c. d. e.
rocess gas compression ratio across the cylinder. P Compressibility of the gas at inlet and discharge conditions, compression ratio. Compression valve action including friction and leakage. Nature of the gas, for example, its ratio of specific heat and compression exponent. Leakage across the piston rings during compression stroke.
Compression Equipment 575 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95
CO M
90
90
PRE S
SOR
CYL
IND
ER
85
CLE
ARA
85
NC
E5
%
80
80 10
70 15
60
40 30
25
40
30 35
30
40
20
60
1.5
2.0
70
80 90 100%
10
50
50
20
60
20
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N-1)
50
70
2.5 3.0 3.5 COMPRESSION RATIO
10
4.0
4.5
Figure 17.18A Compressor volumetric efficiency curve for gas with k or n of 1.15 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
f. T ype and loss through intake and discharge valves. g. Moisture or condensibles in the gas being compressed. h. Clearance volume of cylinder. The compressor manufacturer can control items a–c, e, f, and h; however, the control of clearance volume at high compression ratios for gases/vapors with low specific heat ratios is of great concern [98]. Compression efficiency is controlled by the clearance volume valves, and valve pocket design. A decrease in compression efficiency leads to increased power requirements [98].
Mechanical Efficiency Mechanical efficiency is the ratio of compressor cylinder indicated horsepower to the brake horsepower. Efficiency values range from 90%–93% for direct-driven cylinders to 87%–90% for steam engine units. The efficiency of the driver is not included.
576 Petroleum Refining Design and Applications Handbook Volume 2 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95
90
COM
PRE
90 SSO
RC
85
YLI
ND
ER C
LEA
RAN
CE
85 5%
80
80 10
70
70 15
60
40
40
35
30 20
60 70
1.5
2.0
80 90
100%
10
30
50
20
50
25
40
30
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N–1)
50
60 20
10
2.5 3.0 3.5 COMPRESSION RATIO
4.0
4.5
Figure 17.18B Compressor volumetric efficiency curve for gas with k or n of 1.20 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
Piston Speed Piston speed is a useful guide to set relative limits on compressor cylinder selection. It is difficult to establish acceptable and nonacceptable limits because this is best evaluated with operating experience and compressor manufacturer’s recommendations.
Piston speed =
(rpm)(s) , ft / min 6
(17.46)
This is of more significance in corrosive or polymer-forming services than in clean hydrocarbon or air applications. For example in hydrogen chloride and chlorine service using cylinders with either (a) cast iron liners or (b) carbon piston rings, a speed of around 600 ft per min is acceptable.
Compression Equipment 577 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95
COM PRE
90
90
SSO
RC
YLIN D
ER C
LEA RAN
85
CE 5
%
85
80
80 10
70
70 15
60
40
40
35
30
50
20
2.0
80
1.5
70
90 100%
10
30
60
20
50
25
40
30
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N–1)
50
60
20
2.5 3.0 3.5 COMPRESSION RATIO
10
4.0
4.5
Figure 17.18C Compressor volumetric efficiency curve for gas with k or n of 1.25 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
Horsepower Horsepower is the work done in a cylinder on the gas by the piston connected to the driver during the complete compression cycle. The theoretical horsepower is that required to isentropically (adiabatically) compress a gas through a specified pressure range. The indicated horsepower is the actual work of compression developed in the compressor cylinder(s) as determined from an indicator card [26]. The brake horsepower (bhp) is the actual horsepower input at the crankshaft of the compressor drive. It does not include the losses in the driver itself, but is rather the actual net horsepower that the driver must deliver to the compressor crankshaft. Single Stage Theoretical Hp
P ( k −1)/k 144 k 2 Hp = − 1 P1V1 33, 000 k − 1 P1
(17.47)
578 Petroleum Refining Design and Applications Handbook Volume 2 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95 COM
PRE S
90
SOR
90 CYL IND ER C LEA RAN
CE 5
85
%
85
80
80 10
70 60
40
50
30
40
35
30 50
20
60
20
25
40
30
60
20
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N–1)
50
70
15
70
1.5
2.0
80
100%
90
10
2.5 3.0 3.5 COMPRESSION RATIO
10
4.0
4.5
Figure 17.18D Compressor volumetric efficiency curve for gas with k or n of 1.30 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
Actual Brake Horsepower, Bhp
P ( k −1)/k 144 k 2 Bhp = − 1 (L o )(FL )Z1 P1V1 33, 000 k − 1 P 1
(17.48)
= suction pressure, psia = discharge pressure, psia = suction volumetric rate, ft3/min, at suction conditions = loss factor, comprised of losses due to pressure drop through friction of piston rings, rod packing, valves, and manifold (see Figure 17.19). FL = frame loss for motor-driven compressors only, values range 1.0–1.05 (note, this is not a driver efficiency factor) Z1 = compressibility factor, based on inlet conditions to cylinder (usually negligible, except at high pressures) (see Figures 17.14 and 17.15).
where P1 P2 V1 Lo
Figure 17.20 is convenient to use in solving the complete theoretical power expression, giving
Compression Equipment 579 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95
COM
90
PRE
90
SSO
R CY
LIND
ER C
LEA
85
80
20
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N–1)
60
25
50
30 35
40
40
30
50
60
20
85
70
45
30
%
80
15
60
40
CE 5
10
70
50
RAN
20
70
1.5
2.0
80
90 100%
10
2.5 3.0 3.5 COMPRESSION RATIO
10
4.0
4.5
Figure 17.18E Compressor volumetric efficiency curve for gas with k or n of 1.35 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
Btu lb mole Theoretical hp = Fw Z1T1N m / 2, 546 •o R • o h lbmole R
( k −1)/ k k P2 where Fw = R − 1 k − 1 P1
(17.49a)
(17.49b)
R = gas constant = 1.987 Btu/(lbmole.°R) Nm = lb-mol/h T1 = suction or inlet temperature, °R = (°F + 460) 1 hp (U.S.) = 2542.5 Btu/h To obtain actual brake horsepower, bhp, multiply the theoretical hp by Lo and FL. The actual brake horsepower, bhp is:
bhp = (Theoretical horsepower) (Lo) (FL)
(17.49c)
580 Petroleum Refining Design and Applications Handbook Volume 2 1.5
2.0
2.5
3.0
3.5
4.0
4.5
95
95
COM
90
90
PRE SSO
R CY LIND ER C
LEA R AN
CE 5
85
80
20
40
25
50
30 35
40
40
30
50
60
20
60
45
30
VOLUMETRIC EFFICIENCY PERCENT % Ev = 100–R-%CL.(R1/N–1)
50
70
15
60
85
80
10
70
%
20
70
1.5
2.0
80
90 100%
10
2.5 3.0 3.5 COMPRESSION RATIO
10
4.0
4.5
Figure 17.18F Compressor volumetric efficiency curve for gas with k or n of 1.40 (used by permission: Natural Gasoline Supply Men’s Association Data Book, © 1957. Origin Ingersoll-Rand Co. All rights reserved).
The values of n shown on the chart are Edmister’s isentropic exponents [27, 28]; however, the chart satisfactorily solves the preceding relation using “k” values. Loss Factor. The loss factor is a correction factor for standard horsepower curves for high suction pressures at low ratios of compression. The bhp (brake horsepower) is obtained from the curves (Figures 17.21A–C). These curves are a plot of the “n” or “k” value of the gas versus the required brake horsepower (required to compress 1 million ft3 of gas at 14.4 psia and suction temperatures) for various ratios of compression. The bhp curves give a value greater than the actual. At high ratios of compression, this deviation is not significant. However, when the compression ratio is less than 2.5 and the suction pressure is initially high, appreciable deviation may be encountered. From a percentage standpoint, this average is an appreciable part of the total bhp requirements, and a correction should be made by use of the “loss factor” multiplier (Figure 17.19). When the use of the “loss factor” is indicated and exact thermodynamic data on the gas handled is available, refer to a reliable compressor manufacturer for a more exact correction, which will indicate a lower bhp requirement. This case is justifiable but must be handled with caution and be accompanied by firm operating conditions. That is, there should be very little change of the suction and discharge temperatures, pressures, or capacity requirements. For approximate power requirements [29].
Compression Equipment 581 1.9 Actual Horsepower = (Adiabatic Hp) (Loss Factor, Lo) For Cp/Cv = 1.20 Adiabatic Hp/MMcfd = 262(R 0.2/1.2–1) c Loss Curve applies for all Cp/Cv
1.8
1.7
1.6
Loss Factor
1.5
1.4
1.3
Rc
Actual Bhp/MM See Fig.-21A,B,C
Adiabatic Hp/MM
1.1
10.0
4.19
2.39
1.2
15.4
8.07
1.91
1.4
24.4
15.1
1.616
1.7
34.9
24.2
1.44
2.0
43.0
32.1
1.34
2.5
54.5
43.2
1.26
3.0
65.1
52.6
1.24
3.5
74.1
60.8
1.22
4.0
82.5
68.1
1.21
5.0
98.5
80.7
1.205
6.0
113.0
91.1
1.20
Loss Factor
1.2
1.1
1.0
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
Pressure Ratio
Figure 17.19 Loss factor curve (used by permission: Cooper-Cameron Corporation).
ghp =
w 1H , gas horsepower (33, 000)( ηp )
(17.50)
Adjust for balanced piston leakage, 2% [29] Add losses for seal hp, (x) The shaft horsepower, shp is expressed by:
shp = [(ghp)(1.02)] + [losses, (x*)], shaft horsepower
* range (60–80) hp depends on number of stages and type of shaft seal
where ghp = gas horsepower shp = shaft horsepower H = head, (ft-lbf/lbm) =ft w1 = weight flow, lb/min ηp = polytropic efficiency, fraction, for selected units. See Table 17.9B (Centrifugal Compressors)
(17.51)
582 Petroleum Refining Design and Applications Handbook Volume 2 0.15
0.5
0.6
0.7
0.8
0.9 1.0
ISENTROPIC WORK FACTOR FOR COMPRESSION AND EXPANSION
–0.5
7.0
–1.0
6.5
–1.5
6.0 FOR EXPANDERS USE TOP AND LEFT SCALES
1.3
T AN ST
ES LIN
1
N CO F O
TA N
Tn
1.2
1.1
OF
1.0
1.0
4.0
ES
0.9
LIN
0.9
3.5
0.8
0.8
–4.0
4.5
CO NS
–3.5
5.0
1.
1.5 1.4 1.3
n
1.2
–2.5
–3.0
5.5
1 1.4 .5
–2.0
VALUES OF Fw FOR COMPRESSOR CALCULATIONS
RATIO: P2/P1 0.3 0.4
0.2
–4.5
3.0
–5.0
FOR COMPRESSORS USE BOTTOM AND RIGHT SCALES
–5.5
2.0
NOTES:
1. UNIT OF WORK: (Fw) (Zl) (Tl) = BTU/LB MOLE GAS 2. DIVIDE BTU/LB MOLE BY 2546 TO GET WORK IN HORSEPOWER.
–6.0
2.5
1.5
VALUES OF Fw FOR COMPRESSOR CALCULATIONS
0
0.1
∫ vdp = Zl Tl Fw n–1 P2 n n Fw = R –1 n–1 P1
1.0
WHEN: PVR = CONSTANT
1.0
1.5
2
3 4 RATIO: P2/P1
5
0.5
6
7
8
9
0 10
Figure 17.20 Chart for solving theoretical work of compression or expansion (used by permission: Edmister, W. C., Petroleum Refiner, V. 38, No. 5, © 1959. Gulf Publishing Co. All rights reserved).
Compression Equipment 583 60 30 0
75
25
80
0
PS
IG
&
AB
OV E
BHP/MMCFD CORRECTION FACTORS
200 150
90
100
50 95
4 1. W
1.5 2.0 COMPRESSION RATIO, R
2.5
0
100 1.0
PSIG & BELO
1 1. .2 1. 1 3
85
1.
BHP/MMCFD, BRAKE HORSEPOWER PER MILLION CUBIC FEET PER DAY AT 14.4 AND SUCTION TEMPERATURE
50
BHP/MMCFD CORRECTION FACTOR
70
40
K VALUES (RATIO OF SPECIFIC HEATS) Cp K= Cv 30
20
10 1.0
NOTE: FOR R = 1.0 TO 2.5 MULTIPLY BHP/MMCFD BY CORRESPONDING CORRECTION FACTOR OBTAINED FROM ABOVE CURVE. MECHANICAL EFFICIENCY OF COMPRESSOR CYLINDER = 95%
1.2
1.4
1.6
1.8
2.0
2.2
2.4 2.5
COMPRESSION RATIO, R
Figure 17.21A Brake horsepower required to deliver 1 million ft3 of gas per day, part 1 of 3 (used by permission: Cooper-Cameron Corporation. All rights reserved).
Actual Brake Horsepower, Bhp (Alternate Correction for Compressibility) The development of Hartwick [30] indicates an approach to correcting the ideal gas horsepower for the effects of compressibility. The results examined with high pressure (a maximum discharge pressure of 15,000 psia) systems give agreement within less than 6% of enthalpy methods. 1. Determine gas specific volume at inlet conditions to cylinder:
v = ZRT/(144P), ft3/lb
Obtain Z from compressibility charts (Figure 17.14) (or for specific gas or mixtures, if available).
(17.52)
584 Petroleum Refining Design and Applications Handbook Volume 2 90
BHP/MMCFD, BRAKE HORSEPOWER PER MILLION CUBIC FEET PER DAY AT 14.4 PSIA AND SUCTION TEMPERATURE
85
80
K
75
=
4 1.
1.3
1.2
70
1.1
65 1.0
60
NOTE: MECHANICAL EFFICIENCY OF COMPRESSOR CYLINDER = 95%
55
50 2.5
2.6
2.8
3.0
3.2 3.4 COMPRESSION RATIO, R
3.6
3.8
4.0
Figure 17.21B Brake horsepower required to deliver 1 million ft3 of gas per day, part 2 of 3 (used by permission: Cooper-Cameron Corporation. All rights reserved).
R = 1544/mol wt of gas
(17.53)
2. D etermine discharge temperature, T2, using adiabatic temperature rise Eq. 17.62. Use k values for gas or mixture or calculate them by Eq. 17.4. 3. Calculate the specific volume at discharge condition, v2, using Eq. 17.52. 4. Determine inlet volume, V1. a. Calculate volumetric efficiency from ideal equation:
E ′v = 1 − Vpc′ [(P2 /P1 )1/k − 1] = 1 − Vpc′ ( v 1 /v 2 − 1), fraction
(17.54)
Note that this requires or assumes that compressor cylinder clearance Vpc′ be established. For studies, it may be assumed and the effects calculated. Values range from 5% to 35% clearance on actual cylinders. Special designs are used for smaller or larger values.
Compression Equipment 585 130
BHP/MMCFD, BRAKE HORSEPOWER PER MILLION CUBIC FEET PER DAY AT 14.4 & SUCTION TEMPERATURE
120
110
K
=
1.4
1.3
100
1.2
1.1
90
1.0
80 NOTE: MECHANICAL EFFICIENCY OF COMPRESSOR CYLINDER = 95%
70 4.0
4.2
4.4
4.6
4.8 5.0 5.2 COMPRESSION RATIO, R
5.4
5.6
5.8
6.0
Figure 17.21C Brake horsepower required to deliver 1 million ft3 of gas per day, part 3 of 3 (used by permission: Cooper-Cameron Corporation. All rights reserved).
b. Calculate inlet volume.
V1 = (PD)( E ′v )
(17.55)
5. D etermine pseudo compression exponent, k , to reflect the actual shape of compression and re- expansion curves.
586 Petroleum Refining Design and Applications Handbook Volume 2
P1 v 1k ′ = P2 v 2k ′
(17.56)
k′ 144 bhp = P V [(P / P )( k ′−1)/k − 1](L o )(FL ) k ′ − 1 1 1 2 1 33, 000
(17.57)
6. Calculate horsepower required:
Another method to account for compressibility is given by Boteler [26]. c. Approximate Actual Brake Horsepower, Bhp, From Estimating Curves. Many process applications do not require the detailed evaluation of all factors affecting the actual horsepower requirements. A convenient and yet reasonably accurate calculation can be made using Figures 17.21A–C. Note that for low ratios of compression, a correction factor is to be multiplied by the curve reading as shown in Figure 17.21A. A mechanical cylinder efficiency of 95% is included. The curves use the ratio of heat capacities, k. The horsepower per stage may be calculated.
bhp = [bhp/(MMCFD)] (C/106)
(delivered to compressor crank shaft)
or, bhp = (PD) EvP1 (bhp/MMCFD)(10-4)
(17.58)
(17.59)
where Bhp/MMCFD = brake horsepower required to handle 1 × 106 ft3 of gas per day, measured at 14.4 psia and suction temperature to cylinder, from Figure 17.21, including correction. C = capacity of gas to be compressed, ft3/day, referenced to base conditions of 14.4 psia and suction temperature. Note that corrections for compressibility are not to be included unless the suction temperature at 14.4 psia requires such a correction, assuming most gases require no compressibility correction for pressures as low as 14.4 psia, the reference pressure. When used to approximate overall bhp for a multistage unit, the calculated results will be high due to the reduction in horsepower obtained from interstage cooling. For best results with these curves, evaluate total compression horsepower as the sum of the individual stages (see Figure 17.17A). The effect of compressibility has been omitted from these curves; however, if it is known to be appreciable, use the more exact calculation methods listed previously. An approximation can be made by multiplying the bhp from the curves by the Z factor at the actual stage inlet conditions. If Z is less than 1.0, it should be neglected for this approximation, and only values greater than 1.0 should be considered.
Multistage Multistage horsepower is the sum of the horsepower requirements of the individual cylinders on the compressor unit. Actual bhp:
= 0.0043664 FL k/(k − 1) {P1V1[(Pi1/P1)(k−1)/k − 1]Lo1
+ Pi1Vi1[(Pi2/Pi1)(k−1)/k − 1]Lo2 +
....PiiVii[(Pf/Pi1)(k−1)/ k – 1]Lof } where Pi
= interstage pressure, psia
(17.60)
Compression Equipment 587 Pf 1,2…i Lo1, Lo2…Lof V1
= final, or last-stage discharge pressure, psia = successive interstage designations = loss factors designated by cylinder stages = intake volume including effect of compressibility when applicable
Total bhp/stage and per multistage compressor assembly may be approximated using Figure 17.21. Corrections for compressibility (Z) may be incorporated as described for the single-stage cylinder, handling this on a per-cylinder basis.
Bhp Actually Consumed by Cylinders This horsepower is convenient to calculate when a known cylinder(s) exists on a compressor and when its performance is to be studied.
bhp = [(PD)(Ev)](P1)(bhp/MMCFD)(10-4)
(17.61)
Actual capacity handled = (PD) (Ev) (100), cfm, measured at suction conditions to the cylinder where (PD) (Ev) = ft3 per min (cfm) a cylinder must handle, measured at suction pressure and temperature to the cylinder. (PD) = compressor cylinder piston displacement in ft3/min (cfm). These values can be calculated from known cylinder data or obtained from the respective compressor manufacturer for the specific cylinder in question, operating at the designated rpm. Ev = Ev = volumetric efficiency from Figures 17.18A–F. Note that actual capacity at 14.4 psia and suction temperature = (PD) (Ev) (P1) (100). Therefore, the bhp value given previously is in the correct units for the curves of Figure 17.21.
Temperature Rise—Adiabatic The relation between the suction and discharge temperatures of a gas during any single compression step is
T2 = T1 (P2 /P1 )( k −1)/k = T1R (ck −1)/k
(17.62)
where T1 = initially suction temperature to cylinder, °Rankine = (460 + °F) T2 = discharge temperature from cylinder, °Rankine = (460 + °F) Rc = ratio of cylinder compression Figure 17.22 presents a convenient solution to this relation. The value of R (ck −1)/k read from the chart times the absolute suction temperature gives the discharge temperature T2. Thus,
T2 = T1 (value from graph, Figure 17.22) Temperature Rise—Polytropic Note that for reciprocating compressor work, values of “n” may be used as “k” up to 1.4. “n” represents the polytropic coefficient that is related to “k” by (n − 1)/n = (k − 1)/[(k) (Ep)], where (Ep) is the polytropic efficiency. Cylinder temperature rise is an important consideration, not only at the exit or discharge of the gases from the cylinder, but often for temperature-sensitive gas/mixtures. The temperature calculated or projected inside the cylinder
588 Petroleum Refining Design and Applications Handbook Volume 2
T2/T1 for Compression T2/T1 for Expansion
1.2 3.4 1.1 3.2
1.0 1.0
3.0
1.2
1.4
2.8
2.6 2.4
1.3 3.6
2.2
1.4 3.8
es
4.0
nV a lu
4.2
For Adiabatic Temperature Rise: T2/T1 = (P2/P1)(k-1)/k Read Chart as Values of “k” Parameter. For Polytropic Process: Exponent = n (n–1)/n = (k–1)/(kep) ep = Polytropic Efficiency 3.0 2.6 2.2 2.0 1.8 1.6 1.5 1.4 1.3 1.2 1.1
nV alu es
4.4
3.0
4.6
1.6 2.0
2.8 2.6
1.8
2.4
1.7
2.2
1.6
2.0
1.5 1.4
1.8
1.3
1.6
1.2
1.4
1.15 1.1
1.2
1.05 1.0
1
2
3 4 5 6 7 8 R = P2/P1 for Compression P1/P2 for Expansion
9
10
Figure 17.22 Compression temperature rise (used by permission: Rice, W. T., Chemical Engineering, April 1950, © McGraw-Hill, Inc., All rights reserved).
as compression proceeds must be calculated or measured to guard against too high a temperature developing during the compression process. Some gases such as chlorine, fluorine, bromine acetylene (and acetylene compounds), ethylene, and others must be carefully evaluated. If temperature rise is too high, conditions can lead to internal fires (actually consuming the metal of the compressor cylinder/liner) and explosions. Also for some gases/mixtures, polymerization can occur in the cylinder and, thereby, change the entire performance of the unit, as the cylinders and valves are not designed to handle polymers that are no longer gases.
Altitude Conversion Because all compressors do not operate at sea level pressure conditions, it is important to use the proper absolute pressure at the particular locality. Figure 17.23 is useful for converting altitude to pressure.
Compression Equipment 589
Example 17.2: Single-Stage Compression A compressor is to be installed at a location 2000 ft above sea level. It will handle a gas mixture with “k” = 1.25 at 5 psig suctions and discharge at 50 psig. Suction temperature is 90°F. The gas capacity is to be 5,250,000 SCFD measured at 14.7 psia and 60°F. Determine horsepower requirements and discharge temperature.
Solution 1. Calculate the altitude conversion (Figure 17.23)
Atmospheric pressure at 2000 ft = 13.68 psia 2. Ratio of compression,
Rc =
50 + 13.68 = 3.41 5 + 13.68
This is satisfactory for single-stage operation if temperature does not limit. 3. Discharge temperature (adiabatic rise)
T2 = T1R (ck −1)k = (90 + 460)(3.41)(1.25−1)/1.25
T2 = (550)(1.278) = 702.9°R T2 = 702.9 − 460 = 242.9°F
This temperature is safe.
4. H orsepower, bhp/MMCFD = 74.1 (at Rc = 3.41, k = 1.25, from Figure 17.21B). Note that these curves are for 14.4 psia and suction temperature; therefore, to this basis:
10,000
30
Barometer, Inches of Mercury 26 24 22
28
20
18
et
M er cu re ry , lb ./s q. in .
su
m os
ro m
0
At
2,000
ph er ic
er , In
4,000
Pr es
ch
es
of
6,000
Ba
Altitude Above Sea Level
8,000
15
14
13
12
Atmospheric Pressure, lb./sq.in.
Figure 17.23 Barometric and atmospheric pressure at altitudes.
11
10
590 Petroleum Refining Design and Applications Handbook Volume 2
14.7 460 + 90 Capacity = 5, 250, 000 14.4 460 + 60 = 5, 670, 000 CFD at 14.4 psia and 90°F
Capacity BHP required = ( bhp/MMCFD) (106 )
5, 670, 000 = (74.1) = 420 bhp 106
5. Compressor cylinders, Cylinder or cylinders volume must provide
(PD)(E v ) =
(17.63)
If the suction to a stage does not exceed 10 psig, use [31].
(PD)(E v ) =
( bhp) × 104 ( bhp MMCFD)(P1 )
( bhp) × 104 ( bhp MMCFD)(P1 − 0.5)
(17.64)
Note: Use the compression ratio calculated by
R c2 =
P2 (P1 − 0.5)
(17.65)
This equation is to be used only for this step in the evaluation of volumetric efficiency and should not be used for any other factor in the performance evaluation. Due to valve and other losses being a significant portion in low suction pressure (less than 10 psig) cylinder performance, this can result in reduced performance if not corrected as noted. Then, Rc2 = corrected compression ratio
50 psia + 13.68 psia (5 − 0.5) + 13.68 = 3.5 =
Reading chart Figure 17.18C, Ev = 84.5 at assumed 7% cylinder clearance. Required Piston displacement (PD)
=
bhp × 104 (0.845)( bhp/MMCFD)(18.68 − 0.5)
Compression Equipment 591
Required: (E v )(PD) =
(420)(104 ) = 3,118 cfm (or , cfm) (74.1)(18.18)
Because the manufacturer is the only firm qualified to design the actual cylinder, no further details will be presented here. Note that this is the horsepower available to the cylinders. It includes a 95% mechanical efficiency incorporated into the bhp/MMCFD curves. The rating of a gas engine must be such that its delivered horsepower is at least 420 hp, or if an electric motor drive is used, the mechanical losses of the intermediate frame (about 5%) must be added to arrive at the required motor shaft horsepower.
Example 17.3: Two-Stage Compression A natural gas compressor is required to handle 4 million SCFD (measured at 14.7 psia and 60°F) from a suction condition of 0 psig and 70°F to a discharge of 140 psig. The altitude at the location is 3000 ft. The cooling water for any interstage cooling is at 80°F for 95% of the peak summer months. Determine the horsepower requirements allowing 5 psi pressure drop through the intercooler. Solution 1. C ompression ratio At 3000 ft, atmospheric pressure = 13.14 psia (Figure 17.23)
P1 = 0 + 13.14 = 13.14 psia
Pf = 140 + 13.14 = 153.14 psia Rc =
153.14 = 11.65 13.14
This indicates a two-stage compression because Rc is greater than 5 or 6. Under some designs and for some capacities (not this) it can be satisfactorily handled in a single stage.
Approximate Rc per stage = 11.65 = 3.414 a. First stage: (allowing for one-half pressure drop handled by first stage)
P1 = 13.14 psia Pi1 = (3.41)(13.14) +
5 psia = 44.9 + 2.5 = 47.4 2
Rc = 3.61 b. Second stage:
Pi1 = 44.9 −
5 psia = 42.4 2
Pf2’ = 153.14
Rc = 3.61
592 Petroleum Refining Design and Applications Handbook Volume 2 2. Discharge temperature first stage
Ti1 = T1R (ck-1)/k
“k” for natural gas = 1.26
Ti1 = (70 + 460)(3.61)(1.26–1)/1.26 = (530)(1.305) Ti1 = 691°R Ti1 = 691° − 460° = 231°F
Showing the use of Figure 17.22,
Rc = 3.61, k = 1.26
Read T2/T1 = 1.30
Then, T2 = (1.30)(530) = 689°R
T2
= 689 − 460 = 229°F
This is usually as closed as needed. 3. Discharge temperature second stage Because the cooling water temperature is low enough to allow good cooling, cool the gas to 95°F. This will be the suction temperature to the second-stage cylinder.
Tf 2 = Ti′1R (ck −1)/k = (95 + 460)(3.61)(1.26−1)/1.26 Tf 2 = (555)(1.303) = 723°R
Tf 2 = 263°F
4. Horsepower First stage from Figure 17.21B. Bhp/MMCFD = 78.0 at Rc = 3.61 and k = 1.26 (Reference to 14.4 psia and suction temperature 70°F.) Suction volume @ 14.4 psia and 70°F,
14.7 460 + 70 = 4 , 000, 000 = 4 ,162, 000 CFD 14.4 460 + 60
Using Figure 17.21B,
suction volume capacity bhp = bhp/MMCFD 106
bhp = (78.0)(4,162,000/106) = 324.6 horsepower
Compression Equipment 593
Second stage,
bhp/MMCFD = 78.0 at Rc = 3.61 and k = 1.26 (Refer to 14.4 psia and 95°F.) Suction volume @ 14.4 psia and 95°F,
14.7 460 + 95 = 4 , 000, 000 = 4 , 358, 000 CFD 14.4 460 + 60
bhp = (78.0)(4,358,000/106) = 339.9 horsepower
Total bhp = 324.6 + 339.9 = 664.5 hp This is the horsepower consumed by the cylinders and does not contain any losses in transmitting the power from the driver to the point of use, such as belts or gears. It does contain 95% mechanical efficiency for the cylinder itself. 5. Cylinder Selection The general steps in cylinder selection will be outlined. However, actual selection can be accomplished only by referring to a specific manufacturer’s piston displacement and the volumetric efficiency is a function of the compression ratio and “k” value of gas (both independent of cylinder) and the % clearance, a function of cylinder design.
a.
[(Pd )(E v )] =
bhp(104 ) ( bhp/MMCFD)(P1 − 0.5)*
*Use 0.5 only when suction pressure less than 10 psig, and the Rc used for Ev selection must be corrected accordingly [25].
For a solution, use 325 bhp for first stage. However, it is quite likely that either a 660 bhp (overloaded) or a 750 hp driver may be available as “standard.” The available hp for the first stage is based on 750 hp.
325 First stage = (750) = 366 hp available 666
340 Second stage = (750) = 382 hp available 666 Total = 748 hp
Using these, the first-stage cylinder capacity is
(366)(104 ) Required [(PD)(E v )] = = 3, 600 cfm (78.0)(13.14 − 0.5) Usually this would be handled in two parallel cylinders.
Each cylinder, [(PD)(Ev)] =
3, 600 = 1, 800 cfm 2
594 Petroleum Refining Design and Applications Handbook Volume 2 Next select a type or class, diameter and PD of a cylinder that will meet the required volume and pressure conditions. This must be done with the manufacturers’ tables. No. Cyl
Diam.
Class of Type
% Clearance
PD
Ev
[(PD) (Ev)]
2
*
*
*
*
*
**
*Manufacturer table values **Calculated based on Cylinder
The calculated [(PD)(Ev)] should be equal to or greater than the required value of 1,800 cfm (in this example). Second stage:
(384 )(104 ) Required[(PD)(E v )] = = 1,170 cfm (78.0)(42.2)
For this stage also select a cylinder and check that its [(PD)(Ev)] is equal to or greater than the 1170 cfm. b. if the actual [(PD)(Ev)] is larger than the required, the actual horsepower loading with the cylinders selected must be calculated to be certain that the total cylinder load does not exceed the allowable horsepower operating rating of the driver.
17.10 Hydrogen Use in the Refinery Hydrogen has placed an essential role in converting poor-quality crude oil into modern-day products and to comply with strict environmental mandates. Although these heavy crudes are cheaper, refineries are faced with the additional expense of upgrading to sophisticated processes to refine them to the required standards and product slate that is meeting demand. The alternative is to pay a premium for the lighter crudes. This choice has impacted many refineries that were originally built to process light and sweet crudes and have had to shut down because they could not fund the technology upgrade required to process heavier crudes. The cost of hydrogen is part of the premium that the refiners must pay to process cheaper crudes economically. The challenge is made more complex by the fact that no two refineries are the same, and the naturally-occurring hydrocarbon distribution in crude does not always match customer demand. Various additional processing steps are required to re-adjust the molecules, reshape them and remove the contaminants to ensure the refinery products meet the requirements for the market as well as environmental performance. Hydrogen allows refineries to comply with the latest product specifications and environmental requirements for fuel production being mandated by market and governments and therefore, to reduce the carbon footprints of their facilities. While lighter sweet crudes require fewer processing, the heavier sour crudes contain higher levels of sulfur, other contaminants and fractions. Processing them usually starts with the same distillation process as for the sweet crudes to produce intermediate products; however, additional steps are required. This involves hydrotreating as one such process to remove sulfur, a downstream stream pollutant and other undesirable compounds, such as unsaturated hydrocarbons and nitrogen from the process stream as described in volume 1. Hydrogen is added to the hydrocarbon stream over a bed of catalyst that contains molybdenum with nickel or cobalt at intermediate temperature, pressure and other operating conditions. This process causes sulfur compounds to react with hydrogen to form hydrogen sulfide (H2S) while nitrogen compounds form ammonia (NH3). Aromatics (CnH2n−6, where n = 6, 7……) and olefins (CnH2n, where n = 2, 3, 4, …) are saturated by the hydrogen and lighter products are
Compression Equipment 595 created. The final product of the hydrotreating process is typically the original feedstock free of sulfur and other contaminants. The hydrocracking process is a much more severe operation to produce lighter molecules with higher value for diesel, aviation and petrol fuel. Heavy gas oils, heavy residues, or similar boiling-range heavy distillates react with hydrogen in the presence of a catalyst at high temperature and pressure. The heavy feedstocks are converted (cracked) into lighter distillates, e.g., naphtha, kerosene and diesel or base stock for lubricants. The hydrocracker unit is the top hydrogen consumer in the refinery. Hydrogen is the key source of the hydrocracking to reduce the product boiling range appreciably by converting the majority of the feed to lower boiling products. Hydrogen enables hydrotreating reactions in the hydrocracking process; the final fractionated products are free of sulfur and other contaminants. Other refinery processes such as isomerization, alkylation, and tail gas treatment also consume small amounts of hydrogen. A typical cost of a refinery expansion in the order of US$1 billion, with hydrogen supply representing about 10% of this investment. Therefore, the decision concerning the optimum way to utilize this hydrogen becomes a critical one. In many cases, refinery operators see the investment into hydrogen supply as a defensive outlay required to remain competitive in the market. Hydrogen is required in large volumes 10–200,000 Nm3/h on a refinery, but is also needed for various applications at varying scales of supply. An essential decision confronting operators is the supply method and there are three options for large-scale hydrogen supply [106]. Hydrogen can also be used on a smaller scale in the refinery that is less than 1000 Nm3/h. In these instances hydrogen can be purchased and stored on site in a liquefied form and further vaporized to be used as gaseous hydrogen. Other forms of supply involved a hydrogen tube trailer delivered by a third party supplier that it is hooked to the back of a truck or “bundle” delivery that requires a “bundle” of 12 or 15 cylinders being delivered to the refinery site. Typical bundle supply is used to support hydrogen needs during start-up and possibly an on-site pilot plant that requires research and development (R &D) activity, such as experiments into hydrogenation studies [106].
17.10.1 IsoTherming Technology for Kerosene, Vacuum Gas Oil, and Diesel Hydroprocessing Liquid stream hydroprocessing technology allowed refiners to attain fast-changing kerosene, vacuum gas oil and diesel sulfur specifications and process a wide variety of feedstocks. Hydroprocessing involves chemically treating a petroleum stream with hydrogen in the presence of a catalyst at elevated temperatures and pressures. The units allow refiners to improve product quality, comply with government regulations, and increase profitability by converting low valued streams into high margin and high-quality products. A conventional trickle bed technology as discussed in volume 1 depends on near perfect feedstock distribution throughout the catalyst bed to maximize reaction efficiency and to avoid overheating and coking. Additionally, the quench stages in a trickle bed reactor are used to manage temperature rise and require the injection of large volume of additional hydrogen using a compressor into the reactor (Figure 17.24). IsoTherming technology developed by DuPont as shown in Figure 17.25 has been applied in hydroprocessing of feedstocks that meets the increasing demand for cleaner-burning fuel while simultaneously maintain or increase profitability. Refiners need to increase their capacity and capabilities to produce more diesel and remove more sulfur while minimizing capital investment and operating costs. IsoTherming process provides the hydrogen necessary for hydroprocessing reactions using a liquid stream, rather than a recycle gas stream. Fresh reactor feed is saturated with hydrogen, and additional hydrogen can be added to the feed by a saturated product recycle stream using a recycle pump. This recycles a portion of the reactor product to the inlet to the reactor. This additional liquid volume is then saturated with hydrogen to ensure that sufficient levels of hydrogen are delivered to the reactor for required hydroprocessing chemical reactions. With an IsoTherming liquid full reactor, the catalyst is completely wetted. This draws the heat of reaction away from the catalyst surface and eliminates local hot spots that would otherwise promote coking and catalyst deactivation. All commercially operating IsoTherming gas oil hydrotreating units have experienced catalyst life in excess of four years, indicating that the technology can achieve lower catalyst deactivation rates than conventional trickle bed technology. Additionally, uniform liquid flow throughout the catalyst bed results in a uniform radial temperature profile and acts as a heat sink for exothermic chemical reactions. This results in a lower temperature rise across the
596 Petroleum Refining Design and Applications Handbook Volume 2 Make up hydrogen Make up compressor Feed
Recycle gas compressor
Amine scrubber
Reactor Condenser
Hot HP separator
Cold HP separator
Notes: · Two phase reactor. · Once through liquid. · Gas recycle. · Hydrogen diffuses into liquid as it is consumed in the catalyst bed. · Distribution critical.
To fractionation
Figure 17.24 Conventional Trickle Bed Process Flow Diagram. Make up compressor Make up hydrogen Feed
Off gas for recovery Reactor Condenser
Hot HP separator Notes: · Single phase reactor. · Once through hydrogen. · Liquid recycle. · All hydrogen in liquid phase within catalyst bed. · Distribution less critical.
Cold HP separator
Recycle pump
Figure 17.25 Iso Therming process flow diagram.
To fractionation
Compression Equipment 597 reactor and minimizes light ends generation. These different hydrogen addition options eliminate the need for a recycle gas compressor as shown in Figure 17.24. This process requires low head and high flow to pump high pressure and high-temperature liquid. Conventional pumps with mechanical seals do not offer the required safety and reliability; therefore, a canned motor pump is the only viable type for this process. Figure 17.26 shows the horizontal configuration of the recycle and the difference between the ordinary sealed and canned motor pump. Figure 17.27 illustrates the liquid flow path, and Figure 17.28 shows the top viewing of the recycle pump. This pump type is suitable because the high suction pressure and temperature make a conventional shaft sealing system unreliable, and the design guarantees there are no leaks in the seals. Figure 17.29 shows a photograph of recycle pump in IsoTherming hydroprocessing facility. IsoTherming technology employs a novel reactor system with a single liquid phase that uses hydrogen and catalyst more efficiently. Furthermore, all of the hydrogen required for the hydroprocessing chemical reactions is dissolved in a single liquid phase. This technology has successfully been used in kerosene, diesel hydrotreating, fluid catalytic cracking (FCC), pretreating, and mild hydrocracking. The technology has commercially processed a wide range of straight run and cracked feedstocks, including 100% light cycle oil (LCO), at capacities ranging from 2000 b/d to 78,500 b/d (13 m3/h to 520 m3/h) [169]. The major equipment differences between the two processes are [170]: IsoTherming
Conventional trickle bed
Hot high pressure separator
No
Yes
Cold high pressure separator
No
Yes
Recycle gas compressor
No
Yes
High pressure off gas cooler
No
Yes
Hot low pressure separator
Yes
No
High pressure amine contactor
No
Yes
Casing
Ordinary sealed Pump
Mechanical Seal Shaft
Eliminate Shaft Seal Casing Impeller
Leakage Coupling
Canned Motor Pump Casing Impeller
Frame
still two separate shafts
Stator Ass’y Bearing
Eliminate coupling DuPont Sustainable Solutions
Figure 17.26 Horizontal configuration of recycle pump [170].
Rotor with one piece shaft
598 Petroleum Refining Design and Applications Handbook Volume 2 High pressure amine pump
No
Yes
Stripper preheat exchangers
No
Yes
Reactor circulation pump
Yes
No
Feed/effluent heat exchange surface area
Low
High
Charge heater duty
Low
High
IsoTherming technology is inherently safer hydroprocessing since it eliminates the recycle gas compressor and associated treating equipment, and not only removes a large amount of high pressure equipment from the system, but also significantly reduces the hydrogen inventory in the process. Furthermore, the potential for runaway reaction
Process fluid out
CW out CW in
Process fluid in
Figure 17.27 Liquid flow path in IsoTherming ® Recycle pump [170].
Figure 17.28 Photograph: Top view of IsoTherming ® Recycle pump [170].
Positive Diesel flush
Compression Equipment 599
Figure 17.29 A photograph of IsoTherming recycle pump [170].
is significantly reduced and associated catalyst deactivation in the reactor. Figure 17.30 shows the operating data from a commercial unit that experienced a 4 power failure. A power failure with a conventional trickle bed unit would result in the process having to make up hydrogen and would trip the reactor charge pump and heater. To eliminate the potential for reactor temperature runaway and increased catalyst deactivation, operators would require depressurizing and draining the reactor in order to remove excess hydrogen and reactants. Once power is restored, the reactor bed unit would be refilled, reheated, and repressurized. This start-up process may take hours or days resulting in additional unit downtime. Figure 17.31 shows hydrogen consumption data during operation of the IsoTherming diesel hydrotreating unit, days 0–600, which varies from 60% to over 160% of the design hydrogen consumption. This flexibility enables the refiner to process a wide variety of feeds in order to maximize refinery profits [169]. Figure 17.32 shows a 3-D layout plot and Figure 17.33 shows a photography of IsoTherming hydroprocessing facility. The following advantages of IsoTherming are as follows [170]: Recycle of hydrogen accomplished by liquid recycle Elimination of recycle gas equipment in grassroots units Catalyst bed completely wetted
Reactor bed temperature, °C
1. A ll catalyst sites utilized • More effective use of catalyst.
Prior to power outage, WABT was ~347°C.
~4 h ~5 h
Power restored.
Power outage occurs.
Following power outage, WABT was equal or less for the same feed composition; desulphurization rates were the same as before the power outage.
~308°C Time, h
Figure 17.30 Feed sulfur content data during commercial operation of the IsoTherming Diesel Hydrotreating unit (source: [169]).
600 Petroleum Refining Design and Applications Handbook Volume 2 Initial operation (one reactor)
200
Full operation (two reactors)
Chemical hydrogen consumption, % of design
180 160 140 120 100
Design value
80 60 40 20 0
0
100
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Figure 17.31 Chemical hydrogen consumption data during commercial operation of the IsoTherming Diesel Hydrotreating unit [169].
2. M inimum hot spot formation • Decreased catalyst deactivation/coking/thermal cracking. • Decreased light end formation. • Increased liquid yield. 3. Heat of reaction absorbed by high liquid recycle rate • Adiabatic temperature rise less than conventional trickle bed technology. • Reactor temperature rise tends toward isothermal operation. • Lower heater firing rates in normal operation.
Figure 17.32 IsoTherming® 3-D layout plot [170].
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Figure 17.33 A photograph of IsoTherming® hydroprocessing unit [170].
Flexibility in reactor shape • L/D and bed depth not dependent on two phase flow regime. • Reactor fabrication • Often, reactor diameter and wall thickness can be minimized to allow use of cold-rolled plate instead of forged plate. • Reactor design can be optimized for local fabricator capabilities and transportation logistics. • Projects not dependent on just a few fabricators worldwide. • Shorter reactor delivery time possible. • Can be optimized based on fabricator’s capabilities. Plot plan will be smaller than conventional trickle bed reactor Heat integration may lead to smaller carbon/green house gas footprint
A Case Study Using UniSim Design R460.1 Software for a Two-Stage Compression Case Study 1. Consider the compression of a stream of 4000 lb/h hydrogen at 75°F from a feed pressure of 14.7 psia to a target pressure of 100 psia with an adiabatic efficiency of 72% for both compressors. Noting that the compression ratio is greater than the recommended value of 4. It is apparent that two-stage compression is desired with the pressure of the first stage set at 38.34 psia, thus setting a compression ratio of 2.6 for each stage. Coolers are installed to reduce the compression stage effluent temperature to 280°F. Determine the duties of the compressors and the coolers.
Solution The following steps are used to simulate the two-stage compression and are described as follows (Case Study-TwoStage-Compression-akc.usc):
602 Petroleum Refining Design and Applications Handbook Volume 2 1. Starting UniSim Design software 1. C lick on the UniSim Design R460.1 icon or from the Windows icon at the bottom on the left-hand side of the screen and then from the Honeywell UniSim suite design folder. Then click on the UniSim Design R460.1 icon and the unSim Desktop appears as shown in Figure 17.34 2. Creating a New Simulation To start a new case, select File/New/Case or press Crtl+N or click on the New Case. A simulation file is referred to as a “case”. This will open up the Simulation Basis Manager as shown in Figure 17.34B, which is where all of the components and their properties can be specified. Saving the Simulation Before proceeding any further, save the file in an appropriate location. Select File/Save As and select where to save the file. 3. Adding Components to the Simulation The first step in establishing the simulation basis is to set the chemical components that will be present in the simulation as follows: 1. T o add the components to the simulation, click on the Add button in the Simulation Basis Manager. 2. Clicking on Add will bring up the Component List View which is a list of all the components available in UniSim Design. 3. Select the desired components for the simulation. This can be done by searching through the list of components in one of three ways: a. Sim Name b. Full Name/Synonym c. Formula Select which match term you want of the three types by selecting the corresponding button above the list of components. Then type in the name of the component you are looking for. For example, typing Hydrogen for Sim
Figure 17.34A UniSim Design Desktop (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Compression Equipment 603
Figure 17.34B (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Name narrows the list down to a single component. If your search attempt does not yield the desired component, then either try another name or try searching under full name or formula (see Figures 17.34D and Figure 17.34E). 4. Once the desired component is located, either double click on the component or click Add Pure to add it to the list of components for the simulation. 5. At the bottom of the components page, you can give your component list a name. 6. Once this is complete, simply close the window by clicking the red X at the upper right hand corner of the component list view, which will return you the simulation basis manager.
Figure 17.34C (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
604 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.34D (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Figure 17.34E (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
4. Selecting a Fluid Package Once the components are specified, you can now set the fluid package for your simulation. The fluid package is used to calculate the fluid/thermodynamic properties of the components and mixtures in the simulation (such as enthalpy, entropy, density, vapor-liquid equilibrium, etc.). Therefore, it is important that you select the correct fluid package since this forms the basis for the results returned by your simulation. 1. F rom the simulation basis manager (Figure 17.34F), select the Fluid Pkgs tab. 2. Click the Add button to create a new fluid package as shown below.
Compression Equipment 605 3. F rom the list of fluid packages, select the desired thermodynamic package. The list of available p ackages can be narrowed by selecting a filter to the left of the list (such as EOSs, activity models, etc.). 4. Once the desired model has been located, select it by clicking on it once (no need to double click). For example, select Soave Redlich Kwong (SRK) property package for the simulation (Figure 17.34G). 5. You can give your fluid package a name at the bottom of the fluid package screen (e.g., the name in Figure 17.34G is Basis-1). 6. Once this is done, close the window by clicking on the red X on the upper right hand corner of the Fluid Packages window. Select Soave Redlich Kwong (SRK) from Property Package Selection and EOS button from Property Package Filter window. 5. Select the Units for the Simulation From the Simulation Basis Manager and in the Tools menu, select Preferences from the drop down menu as shown in Figure 17.34H. From the Session Preferences windows, Click on Variable tab and in the variables window, select the units required from Variable Unit Set window as shown in Figure 17.34I. 6. Enter Simulation Environment Click on the Enter Simulation Environment button to begin your simulation as shown in Figure 17.34J. Once you have specified the components and fluid package and entered the simulation environment, you will see the view as shown in Figure 17.34J. Here are few features of this simulation window. 1. U niSim solves the flowsheet after each addition/change to the flowsheet. This feature can be disabled by clicking the Solver Holding button (the red light button) located in the tool bar (Figure 17.34J). If this button is selected, then UniSim will not solve the simulation and it will not provide any results. In order to allow UniSim to return results, the Solver Active button (the green light button) must be selected.
Figure 17.34F (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
606 Petroleum Refining Design and Applications Handbook Volume 2
Figure 17.34G (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Figure 17.34H (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
2. U niSim simulation is capable of solving for information both downstream and upstream. Therefore, it is very important to pay close attention to your flowsheet specification to ensure that you are not providing UniSim with conflicting information. Otherwise, you will get an error and the simulation will not solve.
Accidentally Closing the PFD If the red X on the PFD is accidentally clicked, to revert back to the PFD, go to Tools PFDs, make sure Case is selected, then click View.
Compression Equipment 607
Figure 17.34I (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Figure 17.34J (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc., All rights reserved).
Object Palette On the right-hand side of Figure 17.34J, you will notice the vertical toolbar. This is known as the Object Palette. If for any reason this palette is not visible, go to the Flowsheet pulldown menu and select Palette or press F4 to display the palette. It is from this toolbar that you will add streams and unit operations to your simulation. 7. Adding Material Streams Material Streams are used to transport the material components from process units in the simulation. A material stream can be added to the flowsheet in one of three ways:
608 Petroleum Refining Design and Applications Handbook Volume 2 1. C lick on the blue arrow button on the Object Palette. 2. Select the “Flowsheet” menu and select “Add Stream” 3. Press F11. Using any of the above methods will create a new material stream (a blue arrow) on the flowsheet (Figure 17.34K). The UniSim default names the stream in increasing numerical order (i.e., the first stream created will be given the name “1”). This name can be modified at any time. 8. Specifying Material Streams To enter the information about the material stream, double click on the stream to show the window shown in Figure 17.34L. It is within this window that the user specifies the details regarding the material stream. For material stream that will be used as an input, we need to specify four variables. With UniSim environment, input material stream always has four degree of freedoms. Meaning we need to supply four information in order to fulfill the requirement for UniSim to start the calculations. Note: The four variables required for input stream are composition, flow rate, and two from temperature, pressure, or vapor/phase fraction. From Figure 17.34L, you will see the warning yellow message bar at the bottom of the window indicating what information is needed (unknown compositions). Just follow what the message wants. For example, the first thing that you need to supply is compositions. In order to specify the composition of the stream, select the “Composition” option from this list to display the window in Figure 17.34M. It is within this window that the user specifies the composition of the stream. Note that only the components that you specified in the simulation basis manager will appear in this list. You can specify the composition in many different ways by clicking on the “Basis..” button. The UniSim default is mole fractions, however the user can also specify mass fractions, liquid volume fractions, or flows of each component. If the user is specifying in fractions, all fractions must add up to 1. Enter mole fraction of 1 in the H2 section to indicate 1 mole fraction of hydrogen. Next, the warning yellow message bar indicates that you need to specify the input temperature for this stream. In order to specify the temperature of the stream, select the “Conditions” option from this list to display the window in Figure 17.34N. It is within this window that the user specifies the temperature of the stream. By clicking on this
Figure 17.34K (source: UniSim Design® R460.1. Honeywell® and UniSim® are registered trademarks of Honeywell Inc. All rights reserved).
Compression Equipment