Fluid and Thermal Sciences. A Practical Approach for Students and Professionals 9783030939397, 9783030939403

Table of contents :
Contents
Chapter 1: Fluid Properties and Units
1.1 Introduction
1.2 Systems of Units
1.3 Units of Force
1.4 Fluid Properties
1.4.1 Density
1.4.2 Specific Weight
1.4.3 Specific Gravity
1.4.4 Viscosity of Fluids
1.4.4.1 Dynamic Viscosity
1.4.4.2 Kinematic Viscosity
1.5 Further Reading
Practice Problems
Solutions to Practice Problems
References
Chapter 2: Fluid Statics
2.1 Pressure Due to a Fluid Column (Static Pressure)
2.2 Differential Manometers
2.3 Forces on Submerged Surfaces
Practice Problems
Solutions to Practice Problems
References
Chapter 3: Fluid Dynamics
3.1 Introduction
3.2 Conservation of Mass: Continuity Equation
3.3 Standard Pipe Sizes and Nomenclature
3.4 Laminar Flow and Turbulent Flow Through Pipes
3.4.1 Reynolds Number
3.4.2 Criteria for Laminar and Turbulent Flow in Pipes
3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes
3.5.1 Calculating the Friction Head Loss for Pipe Flow: Darcy Equation
3.5.2 Fanning Friction Factor
3.5.3 Hagen-Poiseuille Equation
3.5.4 Determining the Darcy Friction Factor: Moody Diagram
3.6 Flow Through Noncircular Cross Sections
3.7 Friction Head Loss Across Pipe Fittings and Valves
3.7.1 Velocity Head Method
3.7.2 Equivalent Length Method
3.7.3 Head Loss at Pipe Entrance and at Pipe Exit
3.7.4 Head Loss Due to Change in Pipe Cross Section
3.7.4.1 Sudden Expansion (Fig. 3.6)
3.7.4.2 Sudden Contraction
3.7.4.3 Gradual Expansion
3.7.4.4 Gradual Contraction
3.8 Flow Through Pipes in Series
3.9 Flow Through Pipes in Parallel
3.10 Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces
3.10.1 Boundary Layer Theory
3.10.2 Drag Force
3.10.2.1 Drag Coefficient for Different Objects
3.10.2.2 Terminal Velocity
3.10.3 Lift Force
3.11 Impulse-Momentum Principle
Practice Problems
Solutions to Practice Problems
References
Chapter 4: Energy Equation and Its Applications
4.1 Introduction
4.2 The Mechanical Energy Equation
4.3 Pump Power Equation
4.4 Bernoulli´s Equation
4.5 Pump Performance Parameters
4.5.1 System Curve and Operating Point
4.5.2 Pumps in Series
4.5.3 Pumps in Parallel
4.6 Affinity Laws (Also Known as ``Pump Laws,´´ ``Fan Laws´´)
4.7 Cavitation of Pumps
4.8 Net Positive Suction Head (NPSH)
Practice Problems
Solutions to Practice Problems
References
Chapter 5: Fluid Flow Measurements
5.1 Introduction
5.2 Pitot Tube
5.3 Orifice Meter
5.4 Venturi Meter
5.5 Orifice Meter and Venturi Meter Comparison
5.5.1 Calculation of Permanent Pressure Loss in an Orifice Meter
Practice Problems
Practice Problems Solutions
References
Chapter 6: Fundamentals of Compressible Flow
6.1 Introduction
6.2 Continuity Equation for Compressible Flow
6.2.1 Calculation of Density of Gases
6.3 Mach Number and Its Significance in Compressible Flow
6.4 Isentropic Gas Flow
6.4.1 Application of the Steady Flow Energy Equation for Isentropic Flows
6.4.2 Stagnation-Static Relationships
6.4.3 Isentropic Flow with Area Changes
6.5 Adiabatic Compressible Flow with Friction Loss
Practice Problems
Solutions to Practice Problems
References
Chapter 7: Dimensional Analysis and Similitude
7.1 Introduction
7.2 Dimensionless Parameters Used in Fluid Mechanics
7.2.1 Benefits of Using Dimensionless Parameters
7.2.2 Buckingham Pi Theorem
7.3 Similitude
7.3.1 Requirements for Successful Simulation
7.3.1.1 Geometric Similarity
7.3.1.2 Kinematic Similarity
7.3.1.3 Dynamic Similarity
Practice Problems
Solutions to Practice Problems
References
Chapter 8: Heat Transfer Principles
8.1 Introduction
8.2 General Equation for Heat Transfer Modeling
8.3 Heat Transfer Modes
8.3.1 Conduction Heat Transfer
8.3.2 Convection Heat Transfer
8.3.2.1 Free Convection
8.3.2.2 Forced Convection
8.3.3 Radiation Heat Transfer
8.4 Thermal Circuit Analogous to Electrical Circuit
8.5 Conduction-Convection Systems
References
Chapter 9: Conduction Heat Transfer
9.1 Introduction
9.2 Fourier´s Law of Heat Conduction
9.3 Conduction Through a Rectangular Slab
9.3.1 Multilayer Conduction
9.3.2 R-Values for Insulation and Building Materials
9.4 Conduction Through a Cylindrical Wall
9.5 Conduction Through a Spherical Wall
Practice Problems
Solutions to Practice Problems
References
Chapter 10: Convection Heat Transfer
10.1 Newton´s Law of Cooling
10.2 Convection Heat Transfer Resistance
10.3 Free and Forced Convection
10.4 Dimensionless Parameters Used in Heat Transfer
10.5 Correlations Used in Calculating Convection Heat Transfer Coefficients
10.6 Typical Range of Convection Heat Transfer Coefficients
10.7 Overall Heat Transfer Coefficients in Conduction-Convection Systems
10.7.1 Order of Magnitude Analysis to Determine the Value of Overall Heat Transfer Coefficients
10.8 The Relationship Between Fluid Flow and Heat Transfer
10.8.1 Colburn Analogy Between Fluid Friction Factor and Heat Transfer Coefficient
Practice Problems
Solutions to Practice Problems
References
Chapter 11: Radiation Heat Transfer
11.1 Introduction
11.2 Stefan-Boltzmann´s Law of Thermal Radiation
11.2.1 Nonideal Radiators: Gray Bodies
11.2.2 Absorptivity, Transmissivity, and Reflectivity
11.3 Radiation View Factor
11.3.1 View Factor Relationships
11.4 Calculation of Net Radiation Heat Transfer
11.4.1 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Black Bodies
11.4.2 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Gray Bodies
11.5 Heat Transfer Equilibrium in Radiation Systems
11.5.1 Correction for Thermocouple Readings
Practice Problems
Solutions to Practice Problems
References
Chapter 12: Heat Exchangers
12.1 Introduction
12.2 Heat Balance
12.3 Log Mean Temperature Difference (LMTD)
12.3.1 LMTD Correction Factors
12.4 Overall Heat Transfer Coefficient
12.5 Heat Exchanger Design Equation
12.6 Heat Exchanger Effectiveness
12.6.1 Effectiveness-NTU Method
Practice Problems
Solutions to Practice Problems
References
Chapter 13: Thermodynamics Fundamentals
13.1 Introduction
13.2 Thermodynamic Properties and Variables
13.2.1 Specific Properties
13.2.2 Intensive and Extensive Properties
13.3 Ideal Gas Law
13.3.1 Concept of Mole
13.3.2 Universal Gas Constant and Ideal Gas Equation in Molar Form
13.3.3 STP, NTP, SCF, ACF, and Molar Volume of Ideal Gas
13.3.3.1 Standard Temperature and Pressure (STP) and Molar Volumes
13.3.3.2 Normal Temperature Pressure (NTP)
13.3.3.3 Standard Cubic Feet (SCF) and Actual Cubic Feet (ACF)
13.4 Specific Heats of Gases
13.5 Nonideal Behavior of Gases
13.5.1 Generalized Compressibility Chart
13.6 Thermodynamic Processes Involving Ideal Gases
13.6.1 Isothermal Process
13.6.2 Isobaric Process
13.6.3 Isochoric Process
13.6.4 Isentropic Process
13.6.5 Constant Enthalpy/Throttling Process
13.7 Calculation of Work, Internal Energy Changes, Enthalpy Changes, and Entropy Changes for Processes Involving Ideal Gas
13.7.1 Work
13.7.1.1 Work for a Constant Pressure (Isobaric) Process
13.7.1.2 Work for a Constant Volume (Isochoric) Process
13.7.1.3 Work for a Constant Temperature (Isothermal) Process
13.7.1.4 Constant Entropy (Isentropic) Process
13.7.2 Internal Energy Change
13.7.3 Enthalpy Change
13.7.4 Entropy Change
13.8 Thermodynamic Phase Diagrams
13.8.1 Phase Diagram for Water
13.9 Properties of Steam
13.9.1 Saturated Steam Tables
13.9.1.1 Calculation of Properties of Liquid-Vapor Mixtures
13.9.2 Superheated Steam Tables
13.9.3 Properties of Compressed Liquid
13.10 Mollier Diagram
13.11 Pressure-Enthalpy (P - h) Phase Diagram
Practice Problems
Solutions to Practice Problems
References
Chapter 14: Conservation of Energy and First Law of Thermodynamics
14.1 First Law of Thermodynamics
14.1.1 First Law for a Closed System
14.1.2 I Law for Open Systems-Energy Balance
14.2 I Law Applied to Turbines and Compressors
14.2.1 Isentropic Efficiency of Turbines
14.2.2 Isentropic Efficiency of Compressors
14.3 I Law Applied to Heating and Cooling of Fluids
14.4 I Law Applied to Nozzles and Diffusers
14.5 Pumps
Practice Problems
Solutions to Practice Problems
References
Chapter 15: Ideal Gas Mixtures and Psychrometrics
15.1 Ideal Gas Mixtures
15.1.1 Key Definitions for Ideal Gas Mixtures
15.1.2 Laws Related to Ideal Gas Mixtures
15.1.2.1 Dalton´s Law
15.1.2.2 Amagat´s Law
15.2 Air-Water Vapor Mixture and Psychrometrics
15.2.1 Moist Air Properties and Definitions
15.2.2 Relationship Between Humidity Ratio and Relative Humidity
15.2.3 Use of Psychrometric Chart to Obtain Properties of Moist Air
15.3 Air-Conditioning Processes
15.3.1 Cooling and Dehumidification
15.3.2 Heating and Humidification
15.3.3 Sensible Heating
15.3.4 Other Air-Conditioning Processes
15.4 Cooling Towers
15.5 Mixing of Air Streams
15.6 Use of Psychrometric Formulas
Practice Problems
Solutions to Practice Problems
References
Chapter 16: Fuels and Combustion
16.1 Introduction
16.2 Fuels
16.2.1 Heating Value of Fuels
16.2.1.1 Higher Heating Value of Fuels
16.2.1.2 Lower Heating Value of Fuels
16.3 Combustion Fundamentals and Definitions
16.3.1 Stoichiometry of Combustion Reactions
16.3.2 Combustion in Air
16.3.3 Theoretical Air, Air-Fuel Ratio, and Excess Air
16.3.3.1 Theoretical Air
16.3.3.2 Air-Fuel Ratio
16.3.3.3 Combustion Using Excess Air
16.3.4 Analysis of Combustion Products-Flue Gas Analysis
16.3.4.1 Orsat Analysis of Flue Gases
16.4 Combustion of Coal: Use of Gravimetric Analysis of Coal
16.5 Dew Point of Combustion Products
Practice Problems
Solutions to Practice Problems
References
Chapter 17: Thermodynamic Cycles
17.1 Introduction
17.2 Carnot Cycle and Reversed Carnot Cycle
17.2.1 Carnot Cycle
17.2.1.1 Thermal Efficiency of Carnot Cycle
17.2.1.2 Second Law of Thermodynamics and Carnot Efficiency
17.2.2 Reversed Carnot Cycle
17.2.2.1 Performance Measure of Reversed Carnot Cycle: Coefficient of Performance (COP)
17.3 Rankine Cycle
17.3.1 Analysis of Rankine Cycle
17.3.2 Rankine Cycle with Regenerative Feedwater Heating
17.3.3 Rankine Cycle with Reheat
17.4 Brayton Cycle
17.4.1 Analysis of Brayton Cycle
17.4.2 Brayton Cycle with Regeneration
17.5 Combined Cycle
17.6 Cogeneration Power Plants: Combined Heat and Power
17.7 Otto Cycle
17.8 Vapor Compression Refrigeration Cycle
17.8.1 Analysis of Vapor Compression Cycle
17.8.2 Performance Measures for Vapor Compression Cycles
Practice Problems
Solutions to Practice Problems
References
Index

Citation preview

Nuggenhalli S. Nandagopal, PE

Fluid and Thermal Sciences

A Practical Approach for Students and Professionals

Fluid and Thermal Sciences

Nuggenhalli S. Nandagopal, PE

Fluid and Thermal Sciences A Practical Approach for Students and Professionals

Nuggenhalli S. Nandagopal, PE Department of Engineering Tech (retired) University of Houston - Downtown Bengaluru, India

ISBN 978-3-030-93939-7 ISBN 978-3-030-93940-3 https://doi.org/10.1007/978-3-030-93940-3

(eBook)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Contents

1

Fluid Properties and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Systems of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Units of Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Fluid Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Specific Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Specific Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.4 Viscosity of Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 3 3 3 4 6 10 10 10 13

2

Fluid Statics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Pressure Due to a Fluid Column (Static Pressure) . . . . . . . . . 2.2 Differential Manometers . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Forces on Submerged Surfaces . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

15 15 19 23 28 29 34

3

Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Conservation of Mass: Continuity Equation . . . . . . . . . . . . . 3.3 Standard Pipe Sizes and Nomenclature . . . . . . . . . . . . . . . . . 3.4 Laminar Flow and Turbulent Flow Through Pipes . . . . . . . . . 3.4.1 Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 Criteria for Laminar and Turbulent Flow in Pipes . . . 3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

35 35 35 37 40 41 42

.

42 v

vi

Contents

3.5.1

Calculating the Friction Head Loss for Pipe Flow: Darcy Equation . . . . . . . . . . . . . . . . 3.5.2 Fanning Friction Factor . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Hagen–Poiseuille Equation . . . . . . . . . . . . . . . . . . . 3.5.4 Determining the Darcy Friction Factor: Moody Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Flow Through Noncircular Cross Sections . . . . . . . . . . . . . . 3.7 Friction Head Loss Across Pipe Fittings and Valves . . . . . . . 3.7.1 Velocity Head Method . . . . . . . . . . . . . . . . . . . . . . 3.7.2 Equivalent Length Method . . . . . . . . . . . . . . . . . . . 3.7.3 Head Loss at Pipe Entrance and at Pipe Exit . . . . . . 3.7.4 Head Loss Due to Change in Pipe Cross Section . . . 3.8 Flow Through Pipes in Series . . . . . . . . . . . . . . . . . . . . . . . 3.9 Flow Through Pipes in Parallel . . . . . . . . . . . . . . . . . . . . . . 3.10 Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.1 Boundary Layer Theory . . . . . . . . . . . . . . . . . . . . . 3.10.2 Drag Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.3 Lift Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 Impulse-Momentum Principle . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

5

. . .

44 45 46

. . . . . . . . .

48 52 56 56 57 61 62 64 66

. . . . . . . .

69 69 71 77 78 83 86 99

. . . . . . . . .

101 101 102 106 112 113 114 117 119

. . . . . .

121 123 124 125 127 135

Fluid Flow Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Pitot Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Orifice Meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

137 137 137 139

Energy Equation and Its Applications . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Mechanical Energy Equation . . . . . . . . . . . . . . . . . . . . . 4.3 Pump Power Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Bernoulli’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Pump Performance Parameters . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 System Curve and Operating Point . . . . . . . . . . . . . 4.5.2 Pumps in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.3 Pumps in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Affinity Laws (Also Known as “Pump Laws,” “Fan Laws”) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Cavitation of Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Net Positive Suction Head (NPSH) . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Contents

vii

5.4 5.5

. 141 . 144

Venturi Meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Orifice Meter and Venturi Meter Comparison . . . . . . . . . . . . 5.5.1 Calculation of Permanent Pressure Loss in an Orifice Meter . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

144 145 145 146

Fundamentals of Compressible Flow . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Continuity Equation for Compressible Flow . . . . . . . . . . . . . . 6.2.1 Calculation of Density of Gases . . . . . . . . . . . . . . . . . 6.3 Mach Number and Its Significance in Compressible Flow . . . . 6.4 Isentropic Gas Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Application of the Steady Flow Energy Equation for Isentropic Flows . . . . . . . . . . . . . . . . . . 6.4.2 Stagnation-Static Relationships . . . . . . . . . . . . . . . . . 6.4.3 Isentropic Flow with Area Changes . . . . . . . . . . . . . . 6.5 Adiabatic Compressible Flow with Friction Loss . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 149 149 150 150 153

7

Dimensional Analysis and Similitude . . . . . . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Dimensionless Parameters Used in Fluid Mechanics . . . . . . . 7.2.1 Benefits of Using Dimensionless Parameters . . . . . . 7.2.2 Buckingham Pi Theorem . . . . . . . . . . . . . . . . . . . . 7.3 Similitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Requirements for Successful Simulation . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

171 171 172 173 173 175 175 177 178 181

8

Heat Transfer Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 General Equation for Heat Transfer Modeling . . . . . . . . . . . . 8.3 Heat Transfer Modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Conduction Heat Transfer . . . . . . . . . . . . . . . . . . . . 8.3.2 Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . 8.3.3 Radiation Heat Transfer . . . . . . . . . . . . . . . . . . . . . 8.4 Thermal Circuit Analogous to Electrical Circuit . . . . . . . . . . 8.5 Conduction-Convection Systems . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . .

183 183 183 184 184 184 185 185 186 187

6

154 155 157 161 163 164 169

viii

Contents

9

Conduction Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Fourier’s Law of Heat Conduction . . . . . . . . . . . . . . . . . . . . . 9.3 Conduction Through a Rectangular Slab . . . . . . . . . . . . . . . . . 9.3.1 Multilayer Conduction . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 R-Values for Insulation and Building Materials . . . . . 9.4 Conduction Through a Cylindrical Wall . . . . . . . . . . . . . . . . . 9.5 Conduction Through a Spherical Wall . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

189 189 189 190 192 194 194 197 199 200 204

10

Convection Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Newton’s Law of Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Convection Heat Transfer Resistance . . . . . . . . . . . . . . . . . . 10.3 Free and Forced Convection . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Dimensionless Parameters Used in Heat Transfer . . . . . . . . . 10.5 Correlations Used in Calculating Convection Heat Transfer Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Typical Range of Convection Heat Transfer Coefficients . . . . 10.7 Overall Heat Transfer Coefficients in Conduction-Convection Systems . . . . . . . . . . . . . . . . . . . . . 10.7.1 Order of Magnitude Analysis to Determine the Value of Overall Heat Transfer Coefficients . . . . 10.8 The Relationship Between Fluid Flow and Heat Transfer . . . . 10.8.1 Colburn Analogy Between Fluid Friction Factor and Heat Transfer Coefficient . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

205 205 205 207 207

11

. . . . .

. 207 . 212 . 212 . 215 . 217 . . . .

217 219 220 225

Radiation Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Stefan-Boltzmann’s Law of Thermal Radiation . . . . . . . . . . . . 11.2.1 Nonideal Radiators: Gray Bodies . . . . . . . . . . . . . . . . 11.2.2 Absorptivity, Transmissivity, and Reflectivity . . . . . . . 11.3 Radiation View Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3.1 View Factor Relationships . . . . . . . . . . . . . . . . . . . . 11.4 Calculation of Net Radiation Heat Transfer . . . . . . . . . . . . . . . 11.4.1 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Black Bodies . . . . . . . . . 11.4.2 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Gray Bodies . . . . . . . . . 11.5 Heat Transfer Equilibrium in Radiation Systems . . . . . . . . . . . 11.5.1 Correction for Thermocouple Readings . . . . . . . . . . .

227 227 228 229 230 231 231 231 232 233 235 237

Contents

ix

Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 12

Heat Exchangers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Heat Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Log Mean Temperature Difference (LMTD) . . . . . . . . . . . . . 12.3.1 LMTD Correction Factors . . . . . . . . . . . . . . . . . . . . 12.4 Overall Heat Transfer Coefficient . . . . . . . . . . . . . . . . . . . . . 12.5 Heat Exchanger Design Equation . . . . . . . . . . . . . . . . . . . . . 12.6 Heat Exchanger Effectiveness . . . . . . . . . . . . . . . . . . . . . . . 12.6.1 Effectiveness-NTU Method . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . .

247 247 247 251 252 254 258 262 262 271 274 291

13

Thermodynamics Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Thermodynamic Properties and Variables . . . . . . . . . . . . . . . . 13.2.1 Specific Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 Intensive and Extensive Properties . . . . . . . . . . . . . . . 13.3 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3.1 Concept of Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3.2 Universal Gas Constant and Ideal Gas Equation in Molar Form . . . . . . . . . . . . . . . . . . . . . . 13.3.3 STP, NTP, SCF, ACF, and Molar Volume of Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 Specific Heats of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Nonideal Behavior of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5.1 Generalized Compressibility Chart . . . . . . . . . . . . . . . 13.6 Thermodynamic Processes Involving Ideal Gases . . . . . . . . . . 13.6.1 Isothermal Process . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.2 Isobaric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.3 Isochoric Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.4 Isentropic Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.6.5 Constant Enthalpy/Throttling Process . . . . . . . . . . . . . 13.7 Calculation of Work, Internal Energy Changes, Enthalpy Changes, and Entropy Changes for Processes Involving Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7.2 Internal Energy Change . . . . . . . . . . . . . . . . . . . . . . 13.7.3 Enthalpy Change . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.7.4 Entropy Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.8 Thermodynamic Phase Diagrams . . . . . . . . . . . . . . . . . . . . . . 13.8.1 Phase Diagram for Water . . . . . . . . . . . . . . . . . . . . .

293 293 293 297 298 298 300 301 302 304 306 306 309 309 310 312 312 314

314 314 317 318 318 322 322

x

Contents

13.9

Properties of Steam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.9.1 Saturated Steam Tables . . . . . . . . . . . . . . . . . . . . . . 13.9.2 Superheated Steam Tables . . . . . . . . . . . . . . . . . . . . 13.9.3 Properties of Compressed Liquid . . . . . . . . . . . . . . . 13.10 Mollier Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.11 Pressure-Enthalpy (P – h) Phase Diagram . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . .

323 324 326 326 328 329 331 332 340

14

Conservation of Energy and First Law of Thermodynamics . . . . . 14.1 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 14.1.1 First Law for a Closed System . . . . . . . . . . . . . . . . . 14.1.2 I Law for Open Systems-Energy Balance . . . . . . . . . 14.2 I Law Applied to Turbines and Compressors . . . . . . . . . . . . . 14.2.1 Isentropic Efficiency of Turbines . . . . . . . . . . . . . . . 14.2.2 Isentropic Efficiency of Compressors . . . . . . . . . . . . 14.3 I Law Applied to Heating and Cooling of Fluids . . . . . . . . . . 14.4 I Law Applied to Nozzles and Diffusers . . . . . . . . . . . . . . . . 14.5 Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . .

341 341 341 345 346 348 351 353 354 357 359 360 369

15

Ideal Gas Mixtures and Psychrometrics . . . . . . . . . . . . . . . . . . . . . 15.1 Ideal Gas Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1.1 Key Definitions for Ideal Gas Mixtures . . . . . . . . . . . 15.1.2 Laws Related to Ideal Gas Mixtures . . . . . . . . . . . . . . 15.2 Air-Water Vapor Mixture and Psychrometrics . . . . . . . . . . . . . 15.2.1 Moist Air Properties and Definitions . . . . . . . . . . . . . 15.2.2 Relationship Between Humidity Ratio and Relative Humidity . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2.3 Use of Psychrometric Chart to Obtain Properties of Moist Air . . . . . . . . . . . . . . . . . . . . . . . 15.3 Air-Conditioning Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.1 Cooling and Dehumidification . . . . . . . . . . . . . . . . . . 15.3.2 Heating and Humidification . . . . . . . . . . . . . . . . . . . . 15.3.3 Sensible Heating . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3.4 Other Air-Conditioning Processes . . . . . . . . . . . . . . . 15.4 Cooling Towers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Mixing of Air Streams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.6 Use of Psychrometric Formulas . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

371 371 371 373 374 375 376 377 380 381 384 386 387 388 391 394 396 397 402

Contents

16

17

xi

Fuels and Combustion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Fuels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2.1 Heating Value of Fuels . . . . . . . . . . . . . . . . . . . . . . 16.3 Combustion Fundamentals and Definitions . . . . . . . . . . . . . . 16.3.1 Stoichiometry of Combustion Reactions . . . . . . . . . . 16.3.2 Combustion in Air . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.3 Theoretical Air, Air-Fuel Ratio, and Excess Air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3.4 Analysis of Combustion Products-Flue Gas Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Combustion of Coal: Use of Gravimetric Analysis of Coal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Dew Point of Combustion Products . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

403 403 403 403 404 405 406

. . . . .

413 416 418 419 422

Thermodynamic Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Carnot Cycle and Reversed Carnot Cycle . . . . . . . . . . . . . . . 17.2.1 Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2.2 Reversed Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . 17.3 Rankine Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3.1 Analysis of Rankine Cycle . . . . . . . . . . . . . . . . . . . 17.3.2 Rankine Cycle with Regenerative Feedwater Heating . . . . . . . . . . . . . . . . . . . . . . . . . 17.3.3 Rankine Cycle with Reheat . . . . . . . . . . . . . . . . . . . 17.4 Brayton Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4.1 Analysis of Brayton Cycle . . . . . . . . . . . . . . . . . . . 17.4.2 Brayton Cycle with Regeneration . . . . . . . . . . . . . . 17.5 Combined Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.6 Cogeneration Power Plants: Combined Heat and Power . . . . . 17.7 Otto Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.8 Vapor Compression Refrigeration Cycle . . . . . . . . . . . . . . . . 17.8.1 Analysis of Vapor Compression Cycle . . . . . . . . . . . 17.8.2 Performance Measures for Vapor Compression Cycles . . . . . . . . . . . . . . . . . . . . . . . . Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions to Practice Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

423 423 423 424 427 431 432

. . . . . . . . . .

437 442 445 446 450 452 457 461 465 466

. . . .

468 470 471 477

. 407 . 409

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479

Chapter 1

Fluid Properties and Units

1.1

Introduction

Units play a crucial role in engineering calculations. Every equation must satisfy dimensional homogeneity, that is, each term in the equation must have the same, consistent set of units. Keeping track of units during calculations will be useful in checking the correctness of the solution. Mistakes and inconsistencies in units, conversions, and calculations can be avoided by paying adequate attention to units.

1.2

Systems of Units

The two major systems of units used in engineering calculations are the United States Customary System (USCS) and Systeme Internationale (SI). It is important to be familiar with both systems of units and to have the ability to convert physical quantities from one system of units to another. There are two key aspects involved in understanding the units in different systems. The first is knowing the fundamental physical quantities and their units in both the systems. The second is understanding the definition and units of force. The four fundamental physical quantities and their corresponding units are shown here (Table 1.1). Conversion Factors 1 m ¼ 3.281 ft 1 kg ¼ 2.205 lbm   T F ¼ 1.8 T C + 32

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_1

1

2

1 Fluid Properties and Units

Table 1.1 Fundamental physical quantities and units

1.3

Physical quantity Length Mass Time Temperature

USCS units Feet (ft) Pound mass (lbm) Second (sec) Fahrenheit ( F)

SI units Meter (m) Kilogram (kg) Second (s) Celsius ( C)

Units of Force

The units of force are derived from Newton’s second law of motion. F ¼ ma

ð1:1Þ

In the US Customary System (USCS) of units, 1 pound-force is defined as the force required to accelerate 1 pound mass at 32.2 ft/sec2. Therefore,
 ft 1 lbf ¼ ð1 lbmÞ 32:2 ¼ 32:2 lbm‐ft= sec 2 sec 2 This leads to the conversion constant, gc, necessary for reconciling the units. gc ¼ 32:2

lbm  ft lbf  sec 2

ð1:2Þ

In the SI system, 1 Newton (N) is defined as the force required to accelerate 1 kilogram (kg) mass at 1 m/s2. Thus,   m 1 N ¼ ð1 kgÞ 1 2 ¼ 1 kg  m=s2 s Since 1 N ¼ 1 kg.m/s2, there is no need for a conversion constant to reconcile the units. It does not make any difference if we fail to multiply or divide by gc ¼ 1:0 kgm Ns2 to reconcile the units. However, it makes a huge difference if fail to multiply lbmft or divide by gc ¼ 32:2 lbf sec 2 to reconcile the units. Note: Sometimes, in the metric system, the unit of kilogram force (kgf) is used. Kilogram force is defined as the force required to accelerate 1 kg mass at 9.81 m/ s2. Therefore, 1 kgf ¼ 1 kg  9.81 m/s2 and the equivalencies are 1 kgf ¼ 9.81 N ¼ 2.205 lbf. Conversion Factor: 1 lbf ¼ 4.4482 N

1.4 Fluid Properties

1.4

3

Fluid Properties

The important properties that characterize a fluid are density (ρ), specific weight (γ), specific gravity (SG), dynamic viscosity (μ), and kinematic viscosity (ν). The definitions and units of fluid properties are presented in the following section. The standard values of certain fluid properties for water are also given in this section.

1.4.1

Density

The density (ρ) of a fluid is its mass per unit volume. The units for density are lbm/ft3 (USCS) and kg/m3 (SI). The standard density values of water are 62.4 lbm/ft3 (in the range 45–65  F) and 1000 kg/m3 (in the range 5–20  C). Standard Values of Density of Water ρw ¼ 62.4 lbm/ft3 (USCS) ρw ¼ 1000 kg/m3 (SI)

1.4.2

Specific Weight

The specific weight (γ) of a fluid is the weight per unit volume of the fluid. Weight, W, is the force experienced by a body due to the gravitational acceleration, g. On earth, g ¼ 32.2 ft/sec2 (USCS) and 9.81 m/s2 (SI). Units of specific weight are lbf/ft3 (USCS) and N/m3 (SI). Weight can be calculated from Newton’s second law by using “g” for acceleration. However, the formula for weight in USCS requires the use of the conversion constant, gc, to obtain proper units for force, which is lbf. The formulas for calculating weight are given here followed by a simple example illustrating the use of the formulas. mg gc

ðUSCSÞ

ð1:3aÞ

W ¼ mg

ðSIÞ

ð1:3bÞ

W¼

Example 1.1 A tank containing water has a mass of 10 lbm (4.54 kg). Calculate the weight of the tank in both USCS and SI units, and verify the answers by using the conversion factor for force.

4

1 Fluid Properties and Units

(Solution) USCS units: Using Eq. 1.3a   ft mg ð10 lbmÞ 32:2 sec 2 W¼ ¼ ¼ 10 lbf lbm‐ft gc 32:2 lbf‐ sec 2 SI units: Using Eq. 1.3b   m W ¼ mg ¼ ð4:54 kgÞ 9:81 2 ¼ 44:54 N s Convert pound-force to Newton using the conversion factor. W ¼ ð10 lbf Þ



 4:4482 N ¼ 44:482 N 1 lbf

The results are consistent within a small margin of difference due to roundoff. In USCS, the mass and weight have the same numerical value but different units. This is due to the use of the conversion constant gc. The formulas for specific weight of a fluid can be written using the definition.  

 
 m g ρg ðUSCSÞ ¼ ¼ V gc gc   W mg m γ¼ ¼ ¼ g ¼ ρg ðSIÞ V V V mg

gc W γ¼ ¼ V V

ð1:4aÞ ð1:4bÞ

The standard specific weight of water is 62.4 lbf/ft3 (in the range 45–65  F) and 9810 N/m3 (in the range 5–20  C). However, since the SI unit of pressure is usually kPa, it is preferable to use the value 9.81 kN/m3 as the standard specific weight of water. Standard Specific Weights of Water γ w ¼ 62.4 lbf/ft3 (USCS) and γ w ¼ 9.81 kN/m3 (SI)

1.4.3

Specific Gravity

Specific gravity (SG) is commonly used for liquids and is also known as relative density. The specific gravity (SG) of a liquid is the ratio of the density of the liquid to

1.4 Fluid Properties

5

Table 1.2 Specific gravity of common liquids at 20  C (68  F)

Liquid Pure water Sea water (10  C) Mercury Ethanol Ethylene glycol Glycerin Gasoline SAE 10 W-30 oil

Specific gravity 1.00 1.03 13.6 0.79 1.13 1.26 0.71 0.88

Source: Engineering Tool Box, (2003). Fluids—Specific Gravities. [online] Available at: https://www.engineeringtoolbox.com/ specific-gravity-liquid-fluids-d_294.html (Reproduced with permission)

the standard density of water. It is also the ratio of the specific weight of the liquid to the standard specific weight of water. SG ¼

ρliquid γ liquid ¼ ρwater,std γ water,std

ð1:5Þ

Most of the oils and hydrocarbon liquids are lighter than water and have specific gravity less than 1.0. Table 1.2 lists the specific gravities of common liquids. Example 1.2 A sample of oil has specific gravity of 0.85. Determine the density and specific weight of the oil in both USCS and SI units. (Solution) USCS units: From Eq. 1.5
   lbm ρoil ¼ ðSGoil Þ ρwater,std ¼ ð0:85Þ 62:4 3 ft 3 ¼ 53:04 lbm=ft
   lbf γ oil ¼ ðSGoil Þ γ water,std ¼ ð0:85Þ 62:4 3 ft ¼ 53:04 lbf=ft3 SI units: From Eq. 1.5,
   kg ρoil ¼ ðSGoil Þ ρwater,std ¼ ð0:85Þ 1000 3 m

γ oil

¼ 850 kg=m3
   kN ¼ ðSGoil Þ γ water,std ¼ ð0:85Þ 9:81 3 m ¼ 8:34 kN=m3

6

1 Fluid Properties and Units

1.4.4

Viscosity of Fluids

The viscosity of a fluid is a property that indicates the flow resistance of the fluid. If the flow of oil and water along an inclined surface is studied, we can observe that water flows with greater ease compared to oil. This is because the oil has higher viscosity and encounters greater resistance to flow. The oil particles tend to stick to the surface of the inclined plane, limiting the flow of oil.

1.4.4.1

Dynamic Viscosity

A clear understanding of viscosity can be obtained by considering the definition of the absolute or dynamic viscosity (μ) of a fluid, which is defined as the ratio of the applied shear stress to the rate of shear deformation (velocity gradient) of the fluid. τ μ¼  dv dy

ð1:6Þ

In Eq. 1.6, τ is the shear stress parallel to the direction of motion. The shear stress is the applied force per unit contact area of the fluid with the surface. The shear stress results in relative motion between different layers of the fluid. The layer adjacent to the surface has zero velocity to satisfy the “no-slip condition.” The velocity gradually increases in the y-direction perpendicular to the surface. This results in a fairly linear velocity profile from the surface to the outer edge of the boundary layer. The slope of the velocity profile is the velocity gradient, (dv/dy). For the same applied shear stress, the velocity gradient will be relatively small for high viscous fluids such as oils. In other words, the change in the fluid velocity will be small as we move away from the surface. Water has a relatively low viscosity because of the high velocity gradient in the direction perpendicular to the surface. The units of dynamic viscosity can be obtained by substituting the units for the physical quantities on the right side of Eq. 1.6.   lbf 32:2 lbm‐ft τ lbf  sec lbm sec 2 ð sec Þ ft2   μ    ft= sec  2 2 ft  sec dv ft ft ft dy

Conversion Factors 1 lbm/ft-sec ¼ 1.4881 kg/m.s ¼ 1.4881 N.s/m2 (Pa.s) 2 1 lbf-sec/ft ¼ 32.2 lbm/ft-sec ¼ 47.88 N.s/m2 (Pa.s)

1.4 Fluid Properties

7

τ

μ  dv dy

N m2 m=s m

Ns  2  m

kg‐m kg s 2 ðsÞ  ms m2

Since N/m2 is also known as pascal (Pa), the units of dynamic viscosity in the SI system can be represented as Pa.s. In industrial practice, the units of centipoise (cP) are commonly used for μ, the dynamic viscosity. Poise ¼ dyne.s/cm2 1 cP ¼ 102 Poise Conversion Factors 1 cP ¼ 0.001 N.s/m2, (Pa.s), (kg/m.s) 1 cP ¼ 0.000021 lbf-sec/ft2 ¼ 0.000672 lbm/ft-sec It is worth remembering that at room temperature, water has a dynamic viscosity of 1 cP. Table 1.3 lists the absolute (dynamic) viscosities of common fluids in different units.

1.4.4.2

Kinematic Viscosity

Kinematic viscosity (ν) is the ratio of absolute viscosity to the density of the fluid. ν¼

Table 1.3 Absolute viscosities of common fluids at 20  C (68  F)

μ ρ

Fluid Water Ethanol Ethylene glycol SAE 10 W 30 oil Gasoline Air Hydrogen Ethane

ð1:7Þ

Absolute or dynamic viscosity (μ) lbm/ft – sec cP N.s/m2 6.72  104 1.0 1.00  103 3 1.2 1.20  10 8.05  104 2 21.3 2.13  10 1.43  102 1 170 1.7  10 1.14  101 4 0.31 3.1  10 2.08  104 5 0.0182 1.82  10 1.22  105 0.0089 8.9  106 5.98  106 0.0091 9.1  106 6.11  106

Sources: https://en.wikipedia.org/wiki/List_of_viscosities, Crea tive Commons Attribution-ShareAlike License, Engineering ToolBox, (2011) Motor Oils – Dynamic Viscosity. [online] Available at: https:// www.engineeringtoolbox.com/dynamic-viscosity-motor-oils-d_1 759.html (Reproduced with Permission)

8

1 Fluid Properties and Units

The units of kinematic viscosity can be obtained by substituting the units for dynamic viscosity and density in Eq. 1.7. ν

μ ft‐lbm ft2 sec  lbm  ρ sec 3

ðUSCSÞ

ft

kg

ν

μ ms m2   ρ kg3 s

ðSIÞ

m

A unit of kinematic viscosity that is frequently used in the industry is centistokes (cSt), which is based on the unit Stoke, which is defined as cm2/s. Conversion Factors 1 m2/s ¼ 10.764 ft2/s 1 cSt ¼ 0.000001 m2/s ¼ 0.000011 ft2/s

Example 1.3 An oil has specific gravity (SG) of 0.90 and absolute viscosity of 15 cP. Calculate: A. B. C. D.

The density of the oil in lbm/ft3 and kg/m3 The specific weight of the oil in lbf/ft3 and N/m3 The dynamic viscosity in lbf-sec/ft2, lbm/ft-sec, N.s/m2, and kg/m.s The kinematic viscosity in m2/s, ft2/sec, and centistokes

(Solution) Use the definition of specific gravity to obtain the density and specific weight of the oil. A. From Eq. 1.5
 lbm ρoil ¼ ðSGoil Þðρwater Þ ¼ ð0:90Þ 62:4 3 ft 3 ¼ 56:16 lbm=ft
 kg ρoil ¼ ðSGÞðρwater Þ ¼ ð0:90Þ 1000 3 m ¼ 900 kg=m3

1.4 Fluid Properties

9

B. From Eq. 1.5
 lbf γ oil ¼ ðSGÞðγ water Þ ¼ ð0:90Þ 62:4 3 ft ¼ 56:16 lbf=ft3
 kN γ oil ¼ ðSGÞðγ water Þ ¼ ð0:90Þ 9:81 3 m ¼ 8:83 kN=m3 C. Given that absolute or dynamic viscosity, μ ¼ 15 cP. Use the conversion factors to obtain the dynamic viscosity in the desired units. 0:000021 lbf‐ft2sec μ ¼ ð15 cPÞ 1 cP

! ¼ 0:00032 lbf‐ sec =ft2


 0:000672 ft‐lbm sec μ ¼ ð15 cPÞ ¼ 0:0101 lbm=ft‐ sec 1 cP 0 1 Ns 0:001 2 B m C μ ¼ ð15 cPÞ@ A ¼ 0:015 N  s=m2 ðPa:sÞ 1 cP ¼ 0:015 kg=m  s D. Calculate the kinematic viscosity by using Eq. 1.7.

ν¼

μ 0:0101 ft‐lbm sec ¼ ¼ 1:798  104 ft2 = sec ρ 56:16 lbm3 ft

ν¼

μ ¼ ρ

kg 0:015 m:s 900 mkg3

¼ 1:667  105 m2 =s

Calculate the kinematic viscosity in centistokes using the relevant conversion factor.
ν¼

1:667  10

5

m2 s



! 1 cSt 2 1:0  106 ms

¼ 16:67 cSt

10

1 Fluid Properties and Units

1.5

Further Reading

For a good overview of basic concepts, fluid properties, and their units in the SI system, refer to the lecture notes in reference [2]. An introduction to fluid properties and their units can be found in any book on Fluid Mechanics [1, 3–5].

Practice Problems Practice Problem 1.1 The volume of water in Example 1.1 is 4.50 liters. Calculate the density and specific weight of water in both SI and USCS units. Practice Problem 1.2 500 cm3 of a liquid has a mass of 0.65 kg. Determine: A. B. C. D.

The density of the liquid in kg/m3 The specific gravity of the liquid The specific weight in kN/m3 The density and specific weight in USCS units

Practice Problem 1.3 The density of water at 50  C is 988 kg/m3, and the kinematic viscosity is 5.53  107 m2/s. Calculate: A. B. C. D. E.

The dynamic viscosity in Pa.s and lbm/ft-sec The specific gravity The specific weight in kN/m3 and lbf/ft3 The density in lbm/ft3 The kinematic viscosity in ft2/sec, and cSt

Solutions to Practice Problems Practice Problem 1.1 (Solution) From Example 1.1, the mass of the water is 10 lbm (4.54 kg). SI Units m ¼ 4.54 kg. The density is mass per unit volume.

Solutions to Practice Problems

11

ρ¼

m 4:54 kg  ¼ 1009 kg=m3 ¼
V 4:50 L 1000 L m3

Use Eq. 1.4b to determine the specific weight.  γ ¼ ρg ¼

  1009 mkg3 9:81 sm2






1N 1kg:m 2 s

1000 N 1 kN

¼ 9:898 kN=m3

USCS Units ρ¼

m 10 lbm  ¼ 62:96 lbm=ft3 ¼
V 4:50 L 28:33 L ft3

Use Eq. 1.4a to determine the specific weight.

γ¼

ρg ¼ gc

   ft 62:96 lbm 32:2 sec 3 2 ft lbm‐ft 32:2 lbf‐ sec 2

¼ 62:96 lbf=ft3

Practice Problem 1.2 (Solution) A. The definition of density is mass per unit volume. ρ¼

m 0:65 kg ¼ 1300 kg=m3 ¼ V ð500 cm3 Þ 1 m 3 100 cm

B. Use the definition of specific gravity (Eq. 1.5) to determine the specific gravity of the fluid.

SGliquid ¼

ρliquid 1300 mkg3 ¼ ¼ 1:3 ρwater,std 1000 kg3 m

C. Use the definition of specific weight for SI units (Eq. 1.4b) to determine the specific weight.

12

1 Fluid Properties and Units


γ liq: ¼ ρliq: g ¼

1300

kg m3

  
m 1 kN 9:81 2 1000 N s

¼ 12:75 kN=m3 Note: 1 N ¼ 1 kg.m/s2 D. Use the known value of SGliq. ¼ 1.3 and the equation for specific gravity (Eq. 1.5) to get the density and specific weight in USCS units.

ρliq: γ liq:


 lbm ¼ SGliq: ρwater,std ¼ ð1:3Þ 62:4 3 ¼ 81:12 lbm=ft3 ft
    lbf ¼ SGliq: γ water,std ¼ ð1:3Þ 62:4 3 ¼ 81:12 lbf=ft3 ft 





Practice Problem 1.3 (Solution) A. Use the definition of kinematic viscosity (Eq. 1.7) and conversion factors to obtain the dynamic viscosity in the desired units.

 2 kg 7 m μ ¼ νρ ¼ 5:53  10 988 3 s m ¼ 5:463  104 kg=m  s ðPa  sÞ
   1 lbm=ft  sec 4 μ ¼ 5:463  10 kg=m  s 1:4881 kg=m  s ¼ 3:671  104 lbm=ft  sec

B. Use the definition of specific gravity (Eq. 1.5) to obtain the specific gravity.

SGliquid ¼

ρliquid 988 mkg3 ¼ ¼ 0:988 ρwater,std 1000 kg3 m

C. Use the known value of specific gravity and the definition of specific gravity (Eq. 1.5) to calculate the specific weight in both the units.

References

13


    kg γ liq: ¼ SGliq: γ water,std ¼ ð0:988Þ 9:81 3 m

γ liq:

¼ 9:69 kN=m3
    lbf ¼ SGliq: γ water,std ¼ ð0:988Þ 62:4 3 ft 3 ¼ 61:6 lbf=ft

D. Use the known value of specific gravity and the definition of specific gravity (Eq. 1.5) to calculate the density in USCS units.
    lbm ρliq: ¼ SGliq: ρwater,std ¼ ð0:988Þ 62:4 3 ft 3 ¼ 61:6 lbm=ft E. Determine the kinematic viscosity in the desired units by using the relevant conversion factors.
  10:764 ft2 =s ν ¼ 5:53  10 m =s 1 m2 =s 

7

2

¼ 5:952  106 ft2 =s 

7

ν ¼ 5:53  10 m =s 2






1 cSt 0:000001 m2 =s



¼ 0:553 cSt

References 1. Cimbala, J.M., Cengel, Y.A.: Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Education (2017) 2. Kulkarni, V.: Basic Concepts and Properties of Fluids, Lecture 1 Notes (2013). Download from https://nptel.ac.in/content/storage2/courses/101103004/pdf/mod1.pdf 3. Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 10th edn. Wiley (2020) 4. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson (2015) 5. White, F.M.: Fluid Mechanics, 8th edn. McGraw Hill (2016)

Chapter 2

Fluid Statics

2.1

Pressure Due to a Fluid Column (Static Pressure)

Consider a fluid in a container (Fig. 2.1). The force that acts on the bottom surface of the container is the weight of the fluid. The weight of the fluid is the volume of the fluid multiplied by the specific weight of the fluid. The pressure or pressure intensity of the fluid on the surface is defined as the force per unit area [3]. The pressure intensity acts uniformly on the surface. P¼

F W γV γ ðAhÞ ¼ γh ¼ ¼ ¼ A A A A

Therefore, the static pressure due to a fluid column is the product of the specific weight of the fluid and the depth of the fluid from the free surface.

P ¼ γh

ð2:1Þ

Units of pressure: Since pressure is force per unit area, it has units of lbf/ft2 (USCS) and N/m2 (S I system). However, the more commonly used units in practice are lbf/in2 (psi) and kPa, where N/m2 is also known as pascal (Pa). Another commonly used unit for pressure is “bar” and 1 bar ¼ 105 N/m2 ¼ 100 kPa. Pressure increases with increase in depth and decreases with increase in elevation. The static pressure calculated by Eq. 2.1 is the pressure relative to atmospheric pressure, and it is also known as “Gage Pressure” since the pressure indicated by a pressure gage attached to the container at any depth is the static pressure due to the fluid column. In fluid mechanics, the term “pressure” always refers to gage pressure, unless specified otherwise. In thermodynamics, the pressure used is generally the © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_2

15

16

2 Fluid Statics

Fig. 2.1 Static pressure due to a fluid column

Atmospheric Pressure

Specific weight = γ h

F P

“absolute pressure,” which is the sum of the atmospheric pressure and the gage pressure. Thus, absolute pressure is the pressure of the fluid relative to zero pressure or perfect vacuum, and gage pressure is the pressure of the fluid relative to the pressure of the surroundings (usually the local atmospheric pressure). Note: In fluid mechanics, gage pressures are usually used, whereas in thermodynamics, absolute pressures are usually used. Conversion Factors: 1 psi ¼ 6.895 kPa ¼ 0.0689 bar ¼ 14.7 psi ¼ 101.3 kPa

Pabs ¼ Patm þ Pgage

ð2:2Þ

Atmospheric pressure (Patm) is the pressure exerted by local atmospheric air and is also known as barometric pressure, since it is measured by a mercury barometer. The value of this depends on the local elevation. The equivalent values of atmospheric pressure at sea level (zero elevation) are given here. Standard Atmospheric Pressure at Sea Level: 1 atm ¼ 760 mm Hg ¼ 29.92 in Hg = 14.7 psi ¼ 101.3 kPa ¼ 1.013 bar Using Eq. 2.1, pressure expressed in pressure units (psi, kPa) can be converted to an equivalent height of fluid column (ft., m). For example, standard atmospheric pressure can be converted to equivalent column of water by using Eq. 2.1 as shown here.
hwater ¼

P ¼ γ water




144 in2 ft2 62:4 lbf ft3

lbf 14:7 in 2

 ¼ 33:92 ft of water

2.1 Pressure Due to a Fluid Column (Static Pressure)

17

Pressure Equivalents: 1 atm ¼ 14.7 psi ¼ 101 kPa ¼ 33.92 ft of water The absolute pressure of a system in vacuum is lower than the atmospheric pressure. Vacuum gages measure vacuum relative to atmospheric pressure. Vacuum gauge readings must be subtracted from atmospheric pressure to obtain the absolute pressure. Example 2.1 A pressure gage attached to a tank indicates a reading 1.45 psi. The local atmospheric pressure is 14.45 psia. Calculate the absolute pressure in the tank in psia, kPa, bar, and inches Hg. (Solution) Use Eq. 2.2 to calculate the absolute pressure. Pabs ¼ Pgage þ Patm ¼ 1:45 psi þ 14:45 psi ¼ 15:90 psia Convert the absolute pressure to other units of pressure by using appropriate conversion factors.  Pabs ¼ ð15:90 psiÞ

6:895 kPa 1 psi

 ¼ 109:63 kPa

  0:0689 bar ¼ 1:0955 bar 1 psi   29:92 in Hg ¼ 32:36 in Hg ¼ ð15:90 psiÞ 14:7 psi

Pabs ¼ ð15:90 psiÞ Pabs

Example 2.2 The air pressure in the closed tank shown in the figure is 2 kPa (gage). The gage pressure at the bottom of the tank is 9 kPa. The specific gravity of the oil is 0.81. Determine the height of the oil column in meters.

18

2 Fluid Statics

(Solution) First, calculate the specific gravity of the oil using Eq. 1.5. 

γ oil ¼ ðSGoil Þ γ water,std



  kN ¼ ð0:81Þ 9:81 3 m ¼ 7:946 kN=m3

Take any point on the free surface of the oil as reference point 1 and the point vertically below at the bottom of the tank as reference point 2.The static pressure due to the oil column must be added to air pressure at the oil surface to obtain the pressure at the bottom of the tank. Therefore, P2 ¼ P1 þ γ oil hoil hoil ¼

P2  P1 γ oil

Substitute the known values into the preceding equation. hoil ¼

kN kN P2  P1 9 m2  2 m2 ¼ ¼ 0:881 m kN γ oil 7:946 m3

Example 2.3 An inclined tube open to the atmosphere is attached to a closed tank as shown in the figure. Determine the air pressure in the tank in psig.

Air

20 in.

Pair Liquid (SG = 1.12)

30°

(Solution) Label the reference points as shown in the figure.

2.2 Differential Manometers

19

1

20 in.

Air

h

Pair

2

Liquid (SG = 1.12)

30°

30°

Calculate the specific gravity of the liquid using Eq. 1.5.      lbf γ liq ¼ SGliq γ water,std ¼ ð1:12Þ 62:4 3 ft 3 ¼ 69:89 lbf=ft Calculate the vertical distance “h” shown in the figure. 

h ¼ ð20 inÞ sin 30 ¼ 10 in Since the pressures are equal at the same level in the same fluid [4], Pair ¼ P2 ¼ P1 þ γ liq hliq

0 1   lbf B10 inC 69:89 3 @ A 12 in ft ft ¼ 0ðgageÞ þ 12 in2 ¼ 0:40 psig

2.2

ft

Differential Manometers

Differential manometers are devices used in measuring the pressure difference between two points [2, 5]. The most common device is the “U-tube manometer,” shown in Fig. 2.2. The two legs of the manometer are connected to pressure taps upstream and downstream of the pipe in which a fluid is flowing. The manometer consists of a manometric fluid, which is usually heavier than the fluid flowing in the pipe. Mercury (Hg) is commonly used as a manometric fluid in water pipes. For air ducts, water is commonly used as a manometric fluid. Because of the pressure drop across the pipe, P1 > P2 and the equation for the pressure difference can be obtained by using the principles of static pressure head (Eq. 2.1). Determine the

20

2 Fluid Statics

Fig. 2.2 U-tube differential manometer

pressures at points 3 and 4 by using the principle that pressure increases as we move down a fluid column and pressure decreases as we move up a fluid column. P3 ¼ P1 þ γ f ðhf þ hm Þ P4 ¼ P2 þ γ f hf þ γ m hm Since the reference points 3 and 4 are at the same level in the same fluid, P3 ¼ P4 : Substitute for P3 and P4 from the preceding equations. P1 þ γ f ðhf þ hm Þ ¼ P2 þ γ f hf þ γ m hm Solve the preceding equation for the pressure difference and simplify. ΔP ¼ P1  P2 ¼ hm ðγ m  γ f Þ

ð2:3Þ

Multiply and divide the right-hand side of Eq. 2.3 by γ f.  ΔP ¼ P1  P2 ¼ γ f hm

γm 1 γf



If the flowing fluid is water, from the definition of specific gravity (Eq. 1.5),

2.2 Differential Manometers

21

SGm ¼

γm γf

Therefore, ΔP ¼ P1  P2 ¼ γ f hm ðSGm  1Þ

ð2:4Þ

Thus, the pressure difference is proportional to the manometer reading, hm, and it can be calculated by using either Eqs. 2.3 or 2.4. Example 2.4 A mercury manometer attached across a water pipe shows a reading of 5 in. The specific gravity of mercury is 13.6. Calculate the pressure drop across the pipe in psi. (Solution) Convert the manometer reading to feet. 5 in hm ¼ 12 in ¼ 0:4167 ft 1 ft

Use Eq. 2.4 to calculate the pressure drop across the pipe. ΔP ¼ γ f hm ðSGm  1Þ   lbf 62:4 3 ð0:4167 ftÞð13:6  1Þ ft ¼ 12 in2 ¼ 2:275 psi

ft

Example 2.5 An inclined manometer is used in measuring the pressure difference between the two water tanks, A and B as shown. The specific gravity of mercury is 13.6 Calculate the pressure difference.

22

2 Fluid Statics

(Solution) Designate the vertical distances as shown in the figure and convert them to meters.

hw1 ¼ hw2 ¼ hHg ¼

! 20 cm 100 cm 1m

! 25 cm 100 cm 1m

¼ 0:20 m ¼ 0:25 m

40 cm  sin 30 100 cm 1m



! ¼ 0:20 m

Use the principles of static pressure and Eq. 2.1 to obtain the pressures at points 1 and 3. P1 ¼ PA þ γ w hw1 P3 ¼ PB þ γ w hw2 þ γ Hg hHg Since pressures are equal at the same level in the same fluid, P1 ¼ P3 PA þ γ w hw1 ¼ PB þ γ w hw2 þ γ Hg hHg Calculate the required pressure difference by using the preceding equation.

2.3 Forces on Submerged Surfaces

23

PA  PB ¼ γ w ðhw2  hw1 Þ þ γ Hg hHg   kN ¼ 9:81 3 ð0:25 m  0:20 mÞ m   kN þ 13:6  9:81 3 ð0:20 mÞ m ¼ 27:17 kN=m2 ðkPaÞ

2.3

Forces on Submerged Surfaces

Any surface submerged in a fluid will experience a resultant force. The magnitude and direction of this resultant force depend on the orientation of the surface relative to the fluid. The force experienced by surfaces submerged in water is of practical interest, and this force is known as the “hydrostatic force.” Consider the tank with the liquid shown in Fig. 2.3. The vertical force on the bottom surface of the tank, FRV, is the weight of the liquid. Therefore, F RV ¼ W liq ¼ γV ¼ γhA

ð2:5Þ

In Eq. 2.5, γ is the specific weight of the liquid, and h is the total depth of the liquid column from the free surface. Now, consider the vertical surface of the tank. The pressure increases with the depth from the free surface according to the equation for static pressure, P ¼ γh. The force on a differential area, dA, of the vertical surface will be

Fig. 2.3 Hydrostatic static force on plane surfaces

24

2 Fluid Statics

dF ¼ PdA ¼ γhdA where h is the depth of the differential area from the free surface and keeps changing as we move deeper into the liquid. The total force on the vertical surface will be the sum of the forces on each differential area and is therefore the integral of the force on each differential area. Z F RH ¼

Z dF ¼

Zh γhdA ¼ γ

hdA ¼ γhc A

ð2:6Þ

0

Note that

Rh

hdA is the first moment of the area using the depth from the free

0

surface and is equal to the depth of the centroid of the area multiplied by the area of the surface [1]. Similarly, it can be shown that for an inclined surface submerged in water, the formula for the hydrostatic force is F RI ¼ γhc A

ð2:7Þ

The general formula for the magnitude of the hydrostatic force on any plane surface regardless of the orientation of the surface is:

F ¼ γhc A

ð2:8Þ

In Eq. 2.8, γ is the specific weight of the liquid, hc is the vertical depth of the centroid of the area from the free surface of the liquid, and A is the area of the surface. From the principles of static pressure, γhc ¼ Pc, the pressure at the centroid of the surface, it can be stated that the hydrostatic force on a plane surface is the product of the pressure at the centroid of the surface and the area of the surface. As shown in Fig. 2.3, the line of action of the resultant hydrostatic force on the vertical surface is through the “center of pressure (CP)” [5]. The location of the center of pressure can be determined by using Eq. 2.9. hp ¼ hc þ

Ic hc A

ð2:9Þ

In Eq. 2.9, Ic is the second moment of area of the surface about the centroidal axis of the surface. The formulas for the second moment of area of the surface about the centroidal axis for a rectangular surface and for a circular surface are given here.

2.3 Forces on Submerged Surfaces

25

Ic =

bh 3 12

Ic =

πD 4 64

Example 2.6 Calculate the tensile force, T, required to hold the gate in place. The width of the gate shown in the figure is 1.5 m.

(Solution) The hydrostatic force on the gate will turn the gate in the counterclockwise direction about the hinge. The moment due to the tensile force should counter the moment of the hydrostatic force. This moment can be obtained by determining the magnitude and location of the hydrostatic force by using the figure shown here.

26

2 Fluid Statics

C is the centroid of the gate, and P is the center of pressure through which the hydrostatic force, F, acts on the gate. The magnitude of the hydrostatic force is calculated using Eq. 2.8 F ¼ γhc A The standard specific weight of water is 9.81 kN/m3. From the triangle containing the distance hc, sin 60 ¼

hc hc ¼ 3:5 m þ 1 m 4:5 m

hc ¼ 4:5 m  sin 60 ¼ 3:90 m The area of the gate is A ¼ 1.5 m  2 m ¼ 3 m2. Substitute the known values into the equation for the hydrostatic force (Eq. 2.8) to obtain the magnitude of the hydrostatic force on the gate.

2.3 Forces on Submerged Surfaces

27

  kN F ¼ γhc A ¼ 9:81 3 ð3:90 mÞð3 m2 Þ m ¼ 114:78 kN Determine the location of the center of pressure, P, by using Eq. 2.9. hp ¼ hc þ

Ic hc A

For a rectangular surface, the second moment of area about the centroidal axis is 3

Ic ¼

bh3 ð1:5 mÞð2 mÞ ¼ ¼ 1 m4 12 12

Substitute the known values into Eq. 2.9 to obtain the location of P. hp ¼ hc þ

Ic 1 m4 ¼ 3:90 m þ hc A ð3:90 mÞð3 m2 Þ ¼ 3:985 m

Take the moments about the hinge to get the value of the tensile force, T. The variable L represents the moment arm from the hinge to the line of action of the force as shown in the figure. T  LT ¼ F  LF T¼

F  LF LT

Calculate the moment arm for the hydrostatic force, LF, as shown. If the total inclined distance from the water surface to the center of pressure is SP, then, SP ¼ LF þ 3:5 m ¼ LF ¼

hP sin 60

hp 3:985 m  3:5 m ¼ 1:10 m  3:5 m ¼ sin 60 sin 60

Substitute the known values into the equation for the tensile force to obtain the value of the tensile force. T¼

F  LF ð114:78 kNÞð1:10 mÞ ¼ 63:13 kN ¼ 2m LT

28

2 Fluid Statics

Practice Problems Practice Problem 2.1 The absolute pressure inside a pipeline is 150 kPa. The local barometric pressure is 755 mm Hg. Calculate the gage pressure in psig. Practice Problem 2.2 If the height of the oil column in Example 2.2 is 3 ft and the gage pressure at the bottom of the tank is 6 psi, calculate the air pressure in psig. Practice Problem 2.3 The pressure drop across an air duct is measured to be 3.5 in wg (water gage). The standard density of air is 0.075 lbm/ft3. Calculate the pressure drop in psi and kPa. Practice Problem 2.4 A U-tube manometer is used in measuring the pressure difference between the oil pipe and the water pipe as shown in the figure. The pressure in the oil pipe is 23 psig. The specific gravity of mercury is 13.6. Calculate the pressure in the water pipe.

Practice Problem 2.5 The density of air is 1.225 kg/m3, and the specific gravity of mercury is 13.6. Calculate the pressure difference (in kPa) between the water pipe and the pipe carrying glycol as shown in the figure.

Solutions to Practice Problems

29

Practice Problem 2.6 Determine the magnitude and location of the hydrostatic force on a 4 ft diameter circular gate located in the sidewall of a water tank. The top of the gate is 8 ft below the water surface, which is open to the atmosphere.

Solutions to Practice Problems Practice Problem 2.1 (Solution) Use Eq. 2.2 and conversion factors to calculate the gage pressure. 

Pgage ¼ Pabs  Patm

 14:7 psi ¼ ð150 kPaÞ 101:3  kPa  14:7 psi ð755 mm HgÞ 760 mm Hg ¼ 7:16 psig

Practice Problem 2.2 (Solution) Take any point on the free surface of the oil as reference point 1 and the point 3 ft vertically below the free surface at the bottom of the tank as reference point 2 as shown in the figure. The air pressure is P1.

30

2 Fluid Statics

The pressure at the bottom of the tank is the sum of the air pressure at the oil surface and the static pressure due to the oil column. P2 ¼ P1 þ γ oil hoil P1 ¼ P2  γ oil hoil Calculate the specific gravity of the oil using Eq. 1.5. γ oil

    lbf ¼ ðSGoil Þ γ water,std ¼ ð0:81Þ 62:4 3 ft ¼ 50:54 lbf=ft3

Substitute the known values into the equation for the air pressure P1.       lbf lbf 1 ft2 6 2  50:54 3 ð3 ftÞ in ft 144 in2 ¼ 4:95 psig

P1 ¼ P2  γ oil hoil ¼

Practice Problem 2.3 (Solution) The flowing fluid is air, and the manometric fluid is water. Calculate the specific weight of air by using Eq. 1.4a.
γ air

ρ g ¼ air ¼ gc

0:075 lbm ft3



ft 32:2 sec 2

lbm‐ft 32:2 lbf‐ sec 2

 ¼ 0:075 lbf=ft3

The standard specific weight of water is γ w ¼ 62.4 lbf/ft3. Calculate the pressure drop by substituting the preceding data into Eq. 2.3.

Solutions to Practice Problems

31

    3:5 lbf lbf 1 ft2 ft 62:4 3  0:075 3 ΔP ¼ hw ðγ w  γ air Þ ¼ 12 ft ft 144 in2 ¼ 0:126 psi Convert to kPa using the conversion factor.   6:895 kPa ΔP ¼ ð0:126 psiÞ ¼ 0:869 kPa 1 psi Practice Problem 2.4 (Solution)

Start with points A and B, and calculate the pressures at points 2 and 3 using the principle of static pressure due to a fluid column, P ¼ γh, γ is the specific weight of the fluid, and h is the depth of the fluid column. Subscript “w” represents water; subscript “m” represents mercury, and subscript “o” represents oil.  
  lbf 3 1 ft2 P2 ¼ PA þ γ w hw þ γ m hm ¼ PA þ 62:4 3 ft 12 ft 144 in2     lbf 6 1 ft2 ft þ 13:6  62:4 3 12 ft 144 in2 lbf ¼ PA þ 3:055 2 in

32

2 Fluid Statics

 
  lbf lbf 7 1 ft2 ft P3 ¼ PB þ γ o ho ¼ 23 2 þ 0:85  62:4 3 12 in ft 144 in2 lbf ¼ 23:21 2 in Equate the right-hand sides of the equations for P2 and P3, and solve for PA. PA þ 3:055

lbf lbf ¼ 23:21 2 in2 in PA ¼ 20:15 psi

Practice Problem 2.5 (Solution)

Start with points A and B, and determine the pressures at points 1 and 2 and at points 3 and 4 using the principle of static pressure due to a fluid column of depth h, P ¼ γh, γ is the specific weight of the fluid, and h is the depth of the fluid column. Pressure increases with increase in depth and decreases with increase in elevation. Subscript “w” represents water; subscript “m” represents mercury; subscript “a” represents air, and subscript “g” represents glycol.  
 kN 30 P1 ¼ PA þ γ w hw ¼ PA þ 9:81 3 m 100 m kN ¼ PA þ 2:943 2 m  
 kN 20 P4 ¼ P3 ¼ PB  γ g hg ¼ PB  1:13  9:81 3 m 100 m kN ¼ PB  2:217 2 m Calculate the specific weight of air in kN/m3 using Eq. 1.4b.

Solutions to Practice Problems

γ air

33

  
 kg m 1 kN ¼ ρair g ¼ 1:225 3 9:81 2 1000 N m s ¼ 0:012 kN=m3

Note: 1 kg.m /s2 ¼ 1 N    
 kN kN 40 m P2 ¼ P3 þ γa ha þ γm hm ¼ PB  2:217 2 þ 0:012 3 100 m m    kN 6 þ 13:6  9:81 3 m 100 m kN kN ¼ PB  2:217 2 þ 8:010 2 m m kN ¼ PB þ 5:793 2 m Equate the right-hand sides of the equations for P1 and P2 from the preceding results, and solve for the pressure difference. kN kN ¼ PB þ 5:793 2 m2 m kN kN PA  PB ¼ 5:793 2  2:943 2 ¼ 2:85 kPa m m PA þ 2:943

Practice Problem 2.6 (Solution) The situation described is shown in the figure. “C” is the centroid of the gate, and “CP” is the center of pressure.

34

2 Fluid Statics

Calculate the magnitude of the hydrostatic force by using Eq. 2.8.  F ¼ γhC A ¼


 lbf π 62:4 3 ð10 ftÞ ð4 ftÞ2 ¼ 7841 lbf 4 ft

The hydrostatic force acts at the center of pressure, which can be located by using Eq. 2.9.

hCP

π ð4 ftÞ4 I
64 ¼ hC þ C ¼ 10 ft þ π hC A ð10 ftÞ ð4 ftÞ2 4 ¼ 10:1 ft:

Thus, the hydrostatic force is located 0.1 ft below the center of the gate.

References 1. Ahmari, H., Shah, M.I.K.: Applied Fluid Mechanics Lab Manual. Mars Open Press (2019). Download from https://uta.pressbooks.pub/appliedfluidmechanics/chapter/experiment-1/#:~: text¼Hydrostatic%20forces%20are%20the%20resultant,fundamental%20subjects%20in%20 fluid%20mechanics 2. Benson, Mark: Manometer Types and Working Principle, (2019). Download from https:// engineeringclicks.com/manometer/ 3. Elgar, D.F., LeBret, B.A., Crowe, C.T., Roberson, J.A.: Engineering Fluid Mechanics, 12th edn. Wiley (2019) 4. Mortensen, T.: Manometer Explained. Download from https://realpars.com/manometer/# 5. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson (2015)

Chapter 3

Fluid Dynamics

3.1

Introduction

Fluid dynamics is the study of fluids in motion [4, 10, 11, 16]. Fluid dynamics includes flow through entities such as pipes, ducts, and open channels. Typically, the focus is on the pressure drop due to flows in different entities. The flow of fluids also generates forces on objects such as pipe bends. The energy generated by fluids due to flow over turbine blades is harnessed for power generation. Fluid dynamics also studies the flow of fluids around solid objects and the forces generated due to such flows [1, 3]. The mass flow rate of the fluid usually forms the basis of all calculations in fluid dynamics. Since mass of a species is always conserved, the mass flow rate of the fluid remains constant in the flow field for steady flow. The concept of conservation of mass in fluid flow is discussed in detail in Sect. 3.2.

3.2

Conservation of Mass: Continuity Equation

For steady flow of a fluid, the mass flow is constant. The law of conservation of mass states that the mass flow into a control volume is equal to the mass flow out of the control volume. The typical units for mass flow rates are pound mass per hour (lbm/hr) and kilogram per second (kg/s). Conversion Factor: 1 kg/s ¼ 7937 lbm/hr Consider the section of the pipe shown in Fig. 3.1. The following nomenclature is used: m_ ¼ mass flow rate of the fluid Q ¼ volume flow rate of the fluid © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_3

35

36

3 Fluid Dynamics

Fig. 3.1 Continuity equation

ρ ¼ density of the fluid A ¼ area of cross section of pipe v ¼ average velocity in the pipe The mass flow into the section is the same as the mass flow out of the section. m_ 1 ¼ m_ 2 ¼ m_

ð3:1Þ

The mass flow rate is the volume flow rate times the density of the fluid, and the volume flow rate is the cross-section area of the pipe times the average velocity of the fluid in that section [4, 7, 10, 11]. Therefore, m_ ¼ Q1 ρ1 ¼ Q2 ρ2

ð3:2Þ

m_ ¼ A1 v1 ρ1 ¼ A2 v2 ρ2

ð3:3Þ

Equation 3.3 is the most general form of the continuity equation. Due to pressure drop, the pressure at Section 2 is less than the pressure at Section 1. For gases, the density is inversely proportional to pressure, and the densities at Sections 1 and 2 are not equal because of the pressure drop. Since the density in the flow field is not constant, the flow of gases is described as “compressible flow.” Compressible flow is considered in detail in Chap. 6. The density of a liquid is constant and does not vary significantly with pressure. Liquids are essentially incompressible. Since ρ1 ¼ ρ2for liquids, the density terms cancel out in Eqs. 3.2 and 3.3. Hence, the continuity equation for liquids can be written for volume flow rate and for mass flow as: Q1 ¼ Q2 ¼ Q ¼ A1 v1 ¼ A2 v2 ¼ Av

ð3:4Þ

m_ ¼ ρ1 A1 v1 ¼ ρ2 A2 v2 ¼ ρAv ¼ ρQ

ð3:5Þ

In incompressible flow, the volumetric flow rate of a liquid remains constant in the flow field. The volumetric flow rate has units of cubic meters per second (m3/s) and cubic feet per second (ft3/sec). In practice, the units of gallons per minute (gpm) are widely used for the flow of liquids and the units of cubic feet per minute (cfm) for flow of gases.

3.3 Standard Pipe Sizes and Nomenclature

37

Conversion Factors: 1 m3/s ¼ 35.315 ft3/sec (cfs) 1 ft3/sec ¼ 449 gpm 1 m3/s ¼ 2119 cfm

3.3

Standard Pipe Sizes and Nomenclature

Pipes are manufactured in standard sizes and are identified by their “nominal diameter,” also designated as “DN” in the metric system. Accurate pipe data needs to be used while solving pipe flow problems in fluid mechanics. The nominal diameter/DN does not represent the actual diameter of the pipe. For greater accuracy, the internal diameter should be used while calculating velocities and flow rates. The pipe dimensions are available in standard tables in both systems of units [15]. Example 3.1 Fuel oil with specific gravity 0.92 flows through a 65 mm DN, Sch.40 pipe (62.7 mm ID) pipe at a mass flow rate of 25000 kg/hr. Calculate the average velocity of flow in m/s. (Solution) Calculate the density of fuel oil by using the definition of specific gravity (Eq. 1.5).


ρoil ¼ ðSGoil Þ ρwater,std



  kg ¼ ð0:92Þ 1000 3 m ¼ 920 kg=m3

Convert the pipe ID to m. D¼

62:7 mm 1000 mm m

¼ 0:0627 m

Calculate the average velocity by using Eq. 3.5. 

25000 kg=hr



3600 s=hr m_ v¼ ¼ 2:44 m=s ¼

 kg π ρA ð0:0627 mÞ2 920 3 m

4

38

3 Fluid Dynamics

Example 3.2 200 gpm of water at 70  F flows in a steel pipe with nominal diameter of 2 in (I D ¼ 2.07 in). It is proposed to increase the velocity of water by at least 20% by using a smaller pipe. Determine the nearest standard pipe size required to achieve the specified increase in velocity. (Solution) Subscript “1” refers to the original situation and subscript “2” refers to the new situation with 20% increase in velocity. Use the continuity equation (Eq. 3.4), with the volume flow remaining constant. Q ¼ v1 A1 ¼ v2 A2 Therefore,
π  2 D1 v2 A1 D 2 ¼ ¼
π4 2 ¼ 1 2 v1 A2 D2 4 D2 Solve the preceding equation for D2, and substitute the known values of the ratio v2/v1 and D1. D 2:07 in D2 ¼ q1ffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffi ¼ 1:89 in v2 1:20 v1

From the pipe dimension tables, the nearest standard size is 1.5 in nominal diameter with I D ¼ 1.61 in, as shown in excerpt from the pipe data table [15]. With this size, the velocity will increase by more than 20%, which is acceptable.

3.3 Standard Pipe Sizes and Nomenclature

39

Example 3.3 A branched water piping system with relevant data is shown in the figure. The nominal diameters may be used as approximations in calculating velocities. Determine the missing flow rates (gpm) and the missing velocities (ft/sec).

(Solution) Calculate the velocity in pipe A by using the continuity equation (Eq. 3.4)  3  1 ft = sec ð 600 gpm Þ 449 gpm Q ¼ 3:83 ft= sec vA ¼ A ¼  2 AA π 8 in 4

12 in=ft

Calculate the flow rate in pipe B using the continuity equation.    2 !  ft π 6 in 449 gpm Q B ¼ vB A B ¼ 4 sec 4 12 in=ft 1 ft3 = sec ¼ 353 gpm Apply the continuity equation at the junction of the pipes. Flow into the junction ¼ Flow out of the junction 600 gpm ¼ 353 gpm þ QC QC ¼ 600 gpm  353 gpm ¼ 247 gpm Calculate the velocity in pipe C by using the continuity equation.

40

3 Fluid Dynamics

 3  1 ft = sec ð 247 gpm Þ 449 gpm Q vC ¼ C ¼ ¼ 6:30 ft= sec  2 AC π 4 in 4

12 in=ft

Summary table of flow rates and velocities Pipe A B C

3.4

Flow rate, gpm 600 353 247

Velocity, ft/sec 3.83 4.00 6.30

Laminar Flow and Turbulent Flow Through Pipes

Laminar flow can be described as smooth and streamlined flow of fluid. In laminar flow, fluid particles remain in the same streamline and do not move to adjacent streamlines [4, 9, 10]. As a result, there is no exchange of momentum between adjacent fluid layers. The velocity distribution in a pipe is symmetrical about the centerline of the pipe. The velocity at the pipe wall is 0. The velocity gradually increases away from the wall and reaches a maximum value at the centerline of the pipe. The velocity distribution for laminar flow in a pipe is parabolic as shown in Fig. 3.2. In laminar flow, the average velocity, v, is one-half the maximum velocity, vmax. Laminar flow is sometimes encountered in oil pipelines because of the high viscosity values of the liquids. In turbulent flow, there is mixing of fluid particles from different streamlines. As a result, the faster-moving fluid particles situated near the center of the pipe will transfer some of their momentum to the slower moving fluid particles situated near the wall of the pipe. Hence the velocity profile for turbulent flow will be more uniform and flatter as shown in Fig. 3.3 [4, 9–11]. It can be observed from the velocity profile for turbulent flow that the average velocity is quite close to the maximum velocity. In some cases, the average velocity can be as high as 80% of the maximum velocity at the centerline of the pipe.

Fig. 3.2 Velocity profile for laminar flow in pipes

3.4 Laminar Flow and Turbulent Flow Through Pipes

41

Fig. 3.3 Velocity profile for turbulent flow in pipes

3.4.1

Reynolds Number

The Reynolds number is a dimensionless number used in fluid dynamics. It represents the ratio of inertial forces to viscous forces [4, 11]. The inertial force is responsible for the motion of the fluid due to the momentum it carries, and therefore it is proportional to the velocity and density of the fluid. The viscous force tends to resist the flow of the fluid, and it is proportional to the viscosity of the fluid. The Reynolds number is represented by Re and can be calculated using the following formula. Re ¼

Dvρ μ

ð3:6Þ

In Eq. 3.6, D is the internal diameter (ID) of the pipe, v is the average velocity of the fluid, ρ is the density of the fluid, and μ is the dynamic viscosity of the fluid. From Eq. 1.7, the ratio of dynamic viscosity to the density of the fluid is the kinematic viscosity of the fluid, ν. Therefore, the formula for the Reynolds number can also be written in terms of kinematic viscosity as shown in Eq. 3.7. Re ¼

Dvρ Dv Dv ¼ μ ¼ μ ν ρ

ð3:7Þ

Extreme care must be taken to ensure consistency of units while calculating the Reynolds number. All the units in the formula must cancel out leaving a dimensionless result. Example 3.4 illustrates the calculation of the Reynolds number in pipe flow. Example 3.4 100 gpm of water at 70  F flows in a steel pipe with nominal diameter of 2 in. (ID ¼ 2.07 in). At 70  F, the properties of water are ρ ¼ 62.3 lbm/ft3 and μ ¼ 2.05  105 lbf-sec/ft2. Calculate the Reynolds number. (Solution) Convert the pipe ID to feet.

42

3 Fluid Dynamics

D¼

2:07 in 12 in ft

¼ 0:1725 ft

Calculate the velocity of water by using the continuity equation (Eq. 3.4).

v¼

Q ¼ A

  ft3 1 sec 100 gpm  449 gpm π 4 ð0:1725

ftÞ2

¼ 9:53 ft= sec

From Eq. 3.6,    ft lbm ð0:1725 ftÞ 9:53 62:3 3 sec Dvρ ft   Re ¼ ¼ μ lbf‐ sec lbm‐ft 5 2:05  10 32:2 lbf‐ sec 2 ft2 ¼ 1:55  105

3.4.2

Criteria for Laminar and Turbulent Flow in Pipes

The Reynolds number is used as a criteria for determining if the flow in a pipe is laminar or turbulent [4, 9, 11]. Laminar flow occurs in a pipe if the Reynolds number, Re, is less than 2100. Generally, the flow is fully turbulent if Re > 10,000. The flow is in the “transient zone” if 2100 < Re < 10000 [6, 13].

3.5

Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

Fluid flow is always accompanied by a loss in pressure in the direction of flow. A fluid always flows from a point of higher pressure to a point of lower pressure. Any fluid flowing in a pipe has to overcome the resistance due to pipe friction [4, 10, 11, 16]. This results in a loss in energy of the fluid, which can be represented as a loss in static head corresponding to the pressure drop across the section of the pipe. Using Eq. 2.1, the friction head loss can be represented as hf ¼

ΔP γ

ð3:8Þ

3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

43

In Eq. 3.8, hf is the friction head loss (m or ft), ΔP is the pressure drop (kPa or lbf/ft2), and γ is the specific weight of the fluid (kN/m3 or lbf/ft3). In USCS, the unit commonly used for pressure drop is psi (lbf/in2), which needs to be converted to psf (lbf/ft2) by multiplying with 144 since 144 in2 ¼ 1 ft2. Hence Eq. 3.8 can be written as follows for USCS units. hf ¼

144  ΔPpsi ðUSCSÞ γ

ð3:9Þ

hf in ft, ΔP in psi, and γ in lbf/ft3. Further, if water is flowing in the pipe, γ water,std. ¼ 62.4 lbf/ft3. Substitute this into Eq. 3.9. 

2

hf ¼

lbf 144 inft2  ΔP in 2

62:4 lbf ft3

¼

 2:31 ft water  ΔPpsi ft of water 1 psi

ð3:10Þ

If a liquid with specific gravity, SG, is flowing in the pipe, then from the definition of specific gravity, γ liq ¼ SGliq  γ w,std ¼ SGliq  62:4 lbf=ft3 Substitute the preceding result into Eq. 3.9, and simplify as it was done in Eq. 3.10 to obtain an equation for head loss (in ft) for any liquid flowing in the pipe.  hf ¼

2:31 ft water 1 psi



 ΔPpsi

water SGliq in ftft liquid

in ft of liquid:

ð3:11Þ

Using Eq. 2.1, pressure expressed in pressure units (psi, kPa) can be converted to an equivalent height of fluid column (ft., m). For example, standard atmospheric pressure can be converted to equivalent column of water by using Eq. 2.1 as shown.

hwater ¼

P ¼ γ

   lbf 144 in2 14:7 in 2 2 ft 62:4 lbf ft3

¼ 33:92 ft of water

Conversion Factors 1 atm ¼ 14.7 psi ¼ 101.3 kPa ¼ 33.92 ft of water ¼ 10.33 m of water

44

3 Fluid Dynamics

Example 3.5 The pressure drop for oil (SG ¼ 0.82) across a section of a pipe is reported as 2.25 psi. Calculate the friction head loss in feet of oil. (Solution) Use Eq. 3.11 for any fluid flowing in a pipe.  hf ¼

3.5.1

   2:31 ft water 2:31 ft water ΔPpsi ð2:25 psiÞ psi psi ¼ ft water SGliq 0:82 ft oil ¼ 6:34 ft of oil

Calculating the Friction Head Loss for Pipe Flow: Darcy Equation

Intuitively, it can be hypothesized that the head loss due to fluid flow in a pipe would be directly proportional to the roughness of the pipe surface, the diameter of the pipe, the length of the pipe section, and the velocity of flow. The head loss due to pipe friction can be calculated using the Darcy equation [4, 6, 10, 11, 16]. hf ¼ f

  2  L v D 2g

ð3:12Þ

In the Darcy equation, the following nomenclature is used. hf ¼ friction head loss (ft, m) f ¼ Darcy friction factor (dimensionless) L ¼ length of the pipe (ft, m) D ¼ internal diameter of the pipe (ft, m) v ¼ velocity of flow (ft/sec, m/s) g ¼ acceleration due to gravity (32.2 ft/sec2, 9.81 m/s2) Note that the term (L/D) is dimensionless, and the term (v2/2g) is known as the “velocity head,” and it has the units ft or m. Example 3.6 illustrates the use the Darcy equation for calculating the friction head loss in a pipe. Example 3.6 200 gpm of water at 70  F flows through a 6 in nominal steel pipe (ID ¼ 6.06 in). The Darcy friction factor is 0.02. At 70  F, the properties of water are ρ ¼ 62.4 lbm/ft3 and μ ¼ 2.05  105 lbf-sec/ft2. Calculate the pressure drop (psi) across 100 ft of this pipe.

3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

45

(Solution) Convert the pipe ID to feet. D¼

6:06 in 12 in ft

¼ 0:505 ft

Calculate the velocity of water using the continuity equation (Eq. 3.4).  v¼

Q ¼ A



200 gpm 449 gpm=ðft3 = sec Þ π 4 ð0:505

ftÞ2

¼ 2:224 ft= sec

Calculate the head loss across 100 ft of the pipe by using the Darcy equation (Eq. 3.12). 0 1   
  2  ft 2 L v 100 ft B 2:224 sec C hf ¼ f ¼ ð0:02Þ @ A ft D 2g 0:505 ft 2  32:2 2 sec ¼ 0:304 ft of water Use Eq. 3.10 to calculate the pressure drop in psi. ΔPpsi ¼

3.5.2

hf 0:304 ft water ¼ 2:31 ft water 2:31 ft water psi psi ¼ 0:132 psi

Fanning Friction Factor

In chemical engineering practice and literature, the Fanning friction factor [16] is more commonly used rather than the Darcy friction factor. The Darcy friction factor is four times the Fanning friction factor. f Darcy ¼ 4  f Fanning or

fD ¼ 4  fF

ð3:13Þ

46

3 Fluid Dynamics

The Darcy equation for calculating the friction head loss (Eq. 3.12) is based on the Darcy friction factor, and it can be modified for the use of the Fanning friction factor as shown here. hf ¼ f D

3.5.3

  2    2  L v L v ¼ ð4 f F Þ D 2g D 2g

ð3:14Þ

Hagen–Poiseuille Equation

For laminar flow, it can be shown that the Darcy friction factor is a linear function of the Reynolds number as per Eq. 3.15. f ¼

64 Re

ð3:15Þ

Substitute the preceding relationship into Eq. 3.12.   2     2  L v 64 L v hf ¼ f ¼ ¼ D 2g Re D 2g hf ¼

! 64 Dvρ μ

  2  L v ) D 2g

32μLv ρgD2

ð3:16aÞ

The friction head loss for laminar flow can be calculated using Eq. 3.16a. From Eq. 3.8, the pressure drop across the pipe is the friction head loss multiplied by the specific weight of the fluid. Therefore, the pressure drop for laminar flow will be     32μLv 32μLv 32μLv ΔP ¼ γhf ¼ γ ¼ ðρgÞ ¼ 2 2 ρgD ρgD D2 ΔP ¼

32μLv D2

ð3:16bÞ

Equation 3.16b is known as the Hagen–Poiseuille equation [14], and it can be used to calculate the pressure drop for laminar flow in pipes. Example 3.7 40000 kg/hr of oil flows through a DN 150 mm pipe (154 mm ID). The properties of the oil at the flow temperature are SG ¼ 0.85 and dynamic viscosity μ ¼ 53 cP. Determine the pressure drop across 1 kilometer of the oil pipe and the corresponding head loss and friction factor.

3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

47

(Solution) Determine the density and the specific weight of oil by using the definition of specific gravity (Eq. 1.5)   kg ρoil ¼ ðSGÞ ρwater,std ¼ ð0:85Þ 1000 3 ¼ 850 kg=m3 m  
 kN γ oil ¼ ðSGÞ γ water,std ¼ ð0:85Þ 9:81 3 ¼ 8:34 kN=m3 m




Convert the pipe ID to m. D¼

154 mm 1000 mm m

¼ 0:154 m

Determine the velocity of oil in the pipe by using the continuity equation (Eq. 3.5). m_ v¼ ¼ ρA





 1 hr 40000 kg hr  3600 s 
 154 mm 2 ¼ 0:702 m=s 850 mkg3 π4 1000 mm=m

Calculate the Reynolds number by using Eq. 3.6. Re ¼



 kg m Dvρ ð0:154 mÞ 0:702 s 850 m3   ¼ ¼ 1734 2 μ ð53 cPÞ 0:001 Ns=m 1 cP

Since Re < 2100, the flow is laminar, and the Hagen–Poiseuille equation (Eqs. 3.16a and 3.16b) can be used for calculating the pressure drop across the pipe. ΔPLaminar ¼

¼

32μLv 2 D

0:001 N  s=m2 32 53 cP  1 cP



  m ð1000 mÞ 0:702 s

ð0:154 mÞ2

¼ 50202 N=m2 ðPaÞ  50:2 kPa Calculate the corresponding friction head loss by using Eq. 3.8.

48

3 Fluid Dynamics

hf ¼

kN ΔP 50:2 m2 ¼ ¼ 6:02 m of oil γ oil 8:34 kN m3

Calculate the friction factor by using Eq. 3.15 for laminar flow. f ¼

3.5.4

64 64 ¼ ¼ 0:037 Re 1734

Determining the Darcy Friction Factor: Moody Diagram

From the Darcy equation (Eq. 3.12), it is clear that the friction head loss is directly proportional to the Darcy friction factor. It can be hypothesized that the Darcy friction factor, f, will be a function of the following five variables: the roughness of the pipe surface (ε), the inside diameter of the pipe (D), the velocity of the fluid (v), the density of the fluid (ρ), and the dynamic viscosity of the fluid (μ). The three fundamental variables are mass, length, and time. As per the Buckingham Pi Theorem (discussed in detail under Dimensional Analysis, Ch.7), the number of dimensionless parameters needed to study the functional dependence of the friction factor will be 5 – 3 ¼ 2. The two dimensionless variables are the Reynolds number (Re) and the relative roughness (ε/D). Relative roughness, r ¼

f D

ð3:17Þ

Most of the piping in the industry is commercial steel and pipe roughness, ε, for commercial steel is 0.00015 ft (in S I Units, ε ¼ 0.05 mm). The Darcy friction factor can be determined from the “Moody diagram” shown in Fig. 3.4. The Moody Diagram represents the Reynolds number on the x-axis on a log scale and the Darcy friction factor on the y-axis, also on log scale. The relative roughness, r ¼ ε/D, is represented as parametric curves. To read the Moody diagram, locate the Reynolds number on the x-axis and move along the corresponding ordinate to the appropriate relative roughness curve and locate the intersection of the Reynolds number ordinate and the relative roughness curve. From this intersection point, move horizontally to the left to the y-axis to obtain the Moody friction factor. The procedure to obtain the Darcy friction factor from the Moody diagram is illustrated in Example 3.8.

3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

49

Fig. 3.4 Moody diagram

The following observations can be made in the Moody diagram (Fig. 3.4): The flow regimes are: Laminar (Re < 2100) Critical (2100 < Re < 4000) Transient (4000 < Re < 40000) Fully Turbulent (Re > 40000) Example 3.8 The Reynolds number for water flowing in a 6 in nominal commercial steel pipe (ID ¼ 6.06 in) is 1.63  105. Determine the Darcy friction factor for the flow. (Solution) Convert the pipe ID to feet. D¼

6:06 in 12 in ft

¼ 0:505 ft

Calculate the relative roughness of the pipe surface. The roughness of a commercial steel pipe is ε ¼ 0.00015 ft. Using Eq. 3.17, the relative roughness is r ¼ Dε ¼ 0:00015 ft 0:505 ft ¼ 0:0003 Locate the intersection of the ordinate at Re ¼ 1.63  105 and the curve for r ¼ 0.0003, as shown in the accompanying figure. At the point of intersection, f ¼ 1.8  102 ¼ 0.018.

50

3 Fluid Dynamics

The equations [3, 4, 6, 10, 16] for calculating the Darcy friction factor (with constraints shown in parenthesis) are: 64 Laminar flow : f ¼ ð Re < 2100Þ Re   1 r 2:51 pffiffiffi ð Re > 2300Þ þ Colebrook equation : pffiffiffi ¼ 2 log 3:7 Re f f

ð3:18Þ ð3:19aÞ

Since the Colebrook equation is implicit in f, it is modified to enable explicit calculation of the friction factor, f Modified Colebrook equation:   1:11  1 6:9 r pffiffiffi ¼ 1:8 log þ ð Re > 2300Þ Re 3:7 f

ð3:19bÞ

Fully turbulent flow (rough pipes): 1 pffiffiffi ¼ 1:14  2 log r Re > ð3500=r Þ f

ð3:20Þ

Turbulent flow, smooth pipes (r ¼ 0):   1 6:9 pffiffiffi ¼ 1:8 log ð Re > 2300Þ Re f

ð3:21Þ

Example 3.9 150 kg/s of water flows through a 250 mm DN commercial steel pipe (I D ¼ 254 mm, ε ¼ 0.05 mm). The properties of water at the flow temperature are ρ ¼ 1000 kg/m3 and μ ¼ 1 cP. Determine the Reynolds number and Darcy friction factor for this flow. Compare the answer obtained for f by Moody diagram with the answer obtained by using the Colebrook equation and the modified Colebrook equation.

3.5 Friction Head Loss and Pressure Drop for Fluid Flow in Pipes

51

(Solution) Convert the pipe ID to m. D¼

254 mm 1000 mm m

¼ 0:254 m

Calculate the velocity of water by using the continuity equation (Eq. 3.5). kg

v¼

150 m_ ¼ 2:96 m=s ¼

 s kg π ρA 1000 3 ð0:254 mÞ2 m

4

Calculate the Reynolds number (Eq. 3.6) and the relative roughness (Eq. 3.17).

 kg m Dvρ ð0:254 mÞ 2:96 s 1000 m3 ¼ Re ¼ ¼ 7:52  105 0:001Ns μ m2 ð1 cPÞ 1 cP r¼

ε 0:05 mm ¼ ¼ 0:0002 D 254 mm

Using the Reynolds number and relative roughness as parameters, determine the Darcy friction factor from the Moody diagram as shown in the accompanying figure.

Example 3.9 Solution

From the Moody diagram, f ¼ 1.5  102 ¼ 0.015, as shown. Calculate the Darcy friction factor using the Colebrook equation (Eq. 3.19a). Try the value f ¼ 0.015   1 r 2:51 pffiffiffi ¼ 2 log pffiffiffi þ 3:7 Re f f Colebrook Equation

52

3 Fluid Dynamics

1 1 pffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ 8:165 f 0:015 Value of Left Hand Side Calculate the value of the right-hand side of the Colebrook equation and compare with the value obtained for the left-hand side. !  r 2:51 0:0002 2:51 pffiffiffi ¼ 2 log þ þ

pffiffiffiffiffiffiffiffiffiffiffi 2 log 3:7 Re f 3:7 0:015 7:51  105 ¼ 8:18 

Since the calculated values on both sides of the Colebrook equation are equal, we can conclude that the friction obtained from the Colebrook equation would be f ¼ 0.015. Calculate the Darcy friction factor using the modified Colebrook equation (Eq. 3.19b), which can be explicitly solved for the Darcy friction factor.   1:11  1 6:9 r pffiffiffi ¼ 1:8 log þ Re 3:7 f     6:9 0:0002 1:11 ¼ 1:8 log þ 3:7 7:51  105 ¼ 8:223 Solve the preceding equation for f, f ¼ 0.0148 Thus, the results from the Moody diagram and from the Colebrook equations are in excellent agreement.

3.6

Flow Through Noncircular Cross Sections

Fluid flow often occurs in noncircular sections such as rectangular ducts used extensively in air conditioning systems. Other examples of flow through noncircular sections are open channels and flow through annular sections between concentric pipes. Flow through noncircular sections is handled by using an equivalent diameter known as the “hydraulic diameter” [4, 11, 12], represented by the symbol DH. The diameter, D, in circular pipe formulas is replaced by DH for flow through noncircular sections. For example, the equivalent formula for Reynolds number for flow through a noncircular section is:

3.6 Flow Through Noncircular Cross Sections

Re ¼

DH vρ DH v ¼ μ ν

53

ð3:22Þ

The hydraulic diameter is based on the concept of equivalent hydraulic radius (RH) being equal to the cross-section area (A) divided by the wetted perimeter (P). The wetted perimeter is the total length of the duct in contact with the fluid. The following result is obtained by applying this concept to a circular pipe.
π  2 D A D RH ¼ ¼ 4 ¼ P 4 πD The hydraulic diameter can be calculated by using Eq. 3.23. DH ¼ 4RH ¼ 4

    Cross SectionArea A ¼4 Wetted Perimeter P

ð3:23Þ

Example 3.10 Determine the hydraulic diameter of the shaded section between the rectangle and the circle as shown in the figure.

(Solution) The cross-section area of flow is the area between the rectangle and the circle. Therefore, 2   π 9 in A ¼ ð3 ft  4 ftÞ  ¼ 11:56 ft2 4 12 in=ft The fluid flowing in the shaded area will touch all the four sides of the rectangle and the outside perimeter of the pipe. The wetted perimeter will be the sum of the perimeter of the rectangle and the circumference of the pipe.

54

3 Fluid Dynamics



9 in P ¼ 2ð3 ft þ 4 ftÞ þ π 12 in=ft

 ¼ 16:36 ft

Calculate the hydraulic diameter of the shaded area using Eq. 3.23 and the preceding results. DH ¼ 4

    A 11:56 ft2 ¼4 ¼ 2:826 ft P 16:36 ft

Example 3.11 Air flows through an 800 mm  500 mm rectangular duct at a flow rate of 1.45 m3/s. The properties of air at the flow conditions are density, ρ ¼ 1.21 kg/m3 and kinematic viscosity, ν ¼ 1.25  105 m2/s. The equivalent length of the duct is 200 m, and the surface roughness of the duct is 0.00015 m. Calculate the following: A. B. C. D. E. F. G.

Mass flow rate of the air in kg/s Velocity of air in m/s Hydraulic diameter of the duct in m Reynolds number for airflow The Darcy friction factor The friction head loss in millimeter water gage (mm w g) The pressure drop across the duct in kPa

(Solution) A. Calculate the mass flow rate using the continuity equation (Eq. 3.5)  m_ ¼ ρQ ¼

kg 1:21 3 m



m3 1:45 s

 ¼ 1:754 kg=s

B. Calculate the cross-section area of duct.  A¼

800 mm 1000 mm=m



500 mm 1000 mm=m

 ¼ 0:4 m2

Calculate the air velocity using the continuity equation (Eq. 3.4). 3

v¼

Q 1:45 ms ¼ ¼ 3:625 m=s A 0:4 m2

C. When flowing through the duct, the air touches all the four sides of the duct. Therefore, the wetted perimeter is P ¼ 2(0.8 m + 0.5 m) ¼ 2.6 m

3.6 Flow Through Noncircular Cross Sections

55

Calculate the hydraulic diameter by using Eq. 3.23. DH ¼ 4

    A 0:4 m2 ¼4 ¼ 0:615 m P 2:6 m

D. Calculate the Reynolds number using Eq. 3.7. Use the hydraulic diameter for D.
 DH v ð0:615 mÞ 3:625 ms Re ¼ ¼ 1:78  105 ¼ 2 ν 1:25  105 m s

E. Calculate the relative roughness using Eq. 3.17. r¼

ε 0:00015 m ¼ 0:00024 ¼ DH 0:615 m

Using r and Re as parameters, determine the Darcy friction factor using the Moody diagram (Fig. 3.4) as shown in the figure. f ¼ 0.0175.

F. Calculate the head loss by using the Darcy equation (Eq. 3.12). 

L hf ¼ f DH



v2 2g

0 2 1 
200 m @ 3:625 ms A ¼ ð0:0175Þ 0:615 m 2  9:81 m s2 ¼ 3:811 m of air  3811 mm of air column





Convert the head loss into equivalent height of water column. ΔP ¼ γ a ha ¼ γ w hw

56

3 Fluid Dynamics



 kg m 1:21 3  9:81 2 ð3811 mmÞ m s ðρ gÞh γ h hw ¼ a a ¼ a a ¼ N γw γw 9810 3 m ¼ 4:611 mm of water column Note: kg.m/s2  N G. Calculate the pressure drop using Eq. 3.8.

ΔP ¼ γ w hw ¼

3.7

   kN 4:611 m ¼ 0:0452 kPa 9:81 3 1000 m

Friction Head Loss Across Pipe Fittings and Valves

Piping systems typically consist of fittings (elbows, tees, reducers, flanges) and valves in addition to straight pipe. The friction losses across fittings and valves must be considered since they can contribute significantly to the overall friction head loss for piping systems with tight layouts in confined spaces. The friction losses across fittings and valves are sometimes known as Minor Losses. However, the term “minor losses” needs to be used with caution because the losses across fittings and valves can be “minor” only if the piping system has significant lengths of straight pipe and relatively few fittings and valves. There are two methods to calculate the friction head loss across fittings and valves [2, 4–6, 13]: 1. Velocity head method 2. Equivalent length method

3.7.1

Velocity Head Method

The velocity head for fluid flow is calculated by using the equation: h¼

v2 2g

ð3:24Þ

As per the Darcy equation (Eq. 3.12), the friction head loss in a segment of a straight pipe is directly proportional to the velocity head, v2/2g. Similarly, the friction head loss across a fitting or valve will also be proportional to the velocity head. The velocity head is multiplied by an appropriate Loss Coefficient, K, to get the

3.7 Friction Head Loss Across Pipe Fittings and Valves Table 3.1 Typical values of loss coefficient, K, for flanged valves and fittings

Valve/fitting Gate valve

57 Status/type Fully open Half open Fully open Half open Fully open Half open 5 closed 40 closed 5 closed 40 closed Open Open Standard (R ¼ D) Long radius (R ¼ 1.5D) Standard (R ¼ D) Long radius (R ¼ 1.5D) Flanged Line flow Branch flow

Globe valve Diaphragm valve Ball valve Butterfly valve Angle valve Swing check valve 90 Elbow 45 Elbow Return bend Tee

K 0.17 4.5 10 36 2.5 4.5 0.05 17.5 0.25 11 2 2 0.3 0.2 0.4 0.2 0.2 0.2 1.0

friction head loss across the fitting/valve. The friction head loss across fittings/valves can be calculated by the following equation:  hfitting ¼ K fitting

v2 2g

 ð3:25Þ

The value of the loss coefficient, K, depends on the nature of the fitting, and it can be obtained from any standard reference in fluid mechanics [4, 5, 11]. Typical values of K for fittings and valves are shown in Table 3.1. Combining Eq. 3.25 with the Darcy equation (Eq. 3.12) for friction loss for a straight pipe, an equation for calculating the total friction head loss for the piping system can be obtained.  htotal ¼

3.7.2

v2 2g

h   X i L þ K f D

ð3:26Þ

Equivalent Length Method

In the equivalent length method of determining the minor losses, each fitting is represented by an equivalent length of straight pipe represented by Le [6, 10, 13]. In

58

3 Fluid Dynamics

terms of the equivalent length, the head loss for fittings can be calculated by substituting Le for the length in the Darcy equation (Eq. 3.12).  hf ¼ f

v2 2g

  Le D

ð3:27Þ

The relationship between the loss coefficient and the equivalent length for fittings can be obtained by combining Eqs. 3.25 and 3.27. 

v2 h¼K 2g





v2 ¼f 2g

  Le D

Therefore, Le ¼

KD f

ð3:28Þ

Equation 3.28 provides a method for calculating equivalent length for fittings and valves. Example 3.12 illustrates a sample calculation for equivalent length. Example 3.12 Using the results from the solution to Example 3.9 (pipe ID ¼ 254.5 mm), calculate the equivalent length in mm for a long radius 90 elbow (K ¼ 0.2). (Solution) From the solution to Example 3.9, the friction factor is f ¼ 0.015. Substitute the known values into Eq. 3.28 to calculate the equivalent length of the long radius elbow. Le ¼

KD ð0:2Þð254:5 mmÞ ¼ 3393 mm ð3:39 mÞ ¼ 0:015 f

Thus, the equivalent length of a fitting depends on the diameter of the pipe. Table 3.2 shows typical values of Le/D, (dimensionless) for various valves and fittings, from which the equivalent lengths can be easily calculated. For example, the (Le/D) value for a 90 long radius bend is 13. For a pipe with ID ¼ 254.5 mm, the equivalent length will be Le ¼ 13  254.5 mm ¼ 3309 mm. The total length of the piping system including the equivalent lengths of fittings will be: Ltotal ¼ Lpipe þ

X

Le,fittings

ð3:29Þ

The total head loss for a piping system including fittings can be calculated by using Ltotal in the Darcy equation (Eq. 3.12) for friction head loss:

3.7 Friction Head Loss Across Pipe Fittings and Valves Table 3.2 Equivalent lengths of valves and fittings

Valve/fitting Gate valve Globe valve Butterfly valve Angle valve Swing check valve 90 Elbow 45 Elbow Return bend Tee

htotal

59 Status/type Fully open Half open Fully open Open Open Open Standard (R ¼ D) Long radius (R ¼ 1.5D) Standard (R ¼ D) Flanged Line flow Branch flow

   Ltotal v2 ¼f D 2g

(Le/D) 13 260 340 40 150 135 30 20 16 50 20 60

ð3:30Þ

Example 3.13 illustrates the calculation of the total head loss for a piping system. Example 3.13 A piping system consists of 250 ft of 10 in nominal (I D ¼ 10.02 in, ε ¼ 0.00015 ft) steel pipe. In addition, the piping system consists of the following fittings (all flanged): Fittings 90 LR elbows Fully open gate valves Fully open globe valve Fully open check valve

Qty 6 2 1 1

Le/D 15 10 340 135

2500 gpm water at 70  F (ρ ¼ 62.3 lbm/ft3, ν ¼ 1.059  105 ft2/sec) is flowing through the piping system. Calculate the total pressure drop across the piping system in psi. (Solution) Determine the friction factor by calculating the Reynolds number. Convert the inside diameter to feet. D¼

10:02 in 12 in ft

¼ 0:835 ft

Calculate the velocity of water using the continuity equation (Eq. 3.4).

60

3 Fluid Dynamics

 v¼

Q ¼ A



2500 gpm 449 gpm=ðft3 = sec Þ π 4 ð0:835

ftÞ2

¼ 10:17 ft= sec

Calculate the Reynolds number using Eq. 3.7.
 ft Dv ð0:835 ftÞ 10:17 sec Re ¼ ¼ ¼ 8:02  105 2 ν 1:059  105 ft sec

Calculate the relative roughness using Eq. 3.17. r¼

ε 0:00015 ft ¼ ¼ 0:00018 ’ 0:0002 D 0:835 ft

Using Reynolds number and relative roughness as parameters, determine the friction factor by using the Moody diagram (Fig. 3.4) as shown in the figure. From the Moody diagram, f ¼ 0.015.

Calculate the sum of the equivalent lengths of the fittings by using the given data. X

Le ¼ D

XL

e ¼ ð0:835 ftÞð6  10 þ 2  15 þ 1  340 þ 1  135Þ D ¼ 472 ft

Calculate the total equivalent length of the piping system using Eq. 3.29. Ltotal ¼ Lpipe þ

X

Le,fittings ¼ 250 ft þ 472 ft ¼ 722 ft

Calculate the head loss across the piping system using the Darcy equation (Eq. 3.30).

3.7 Friction Head Loss Across Pipe Fittings and Valves

htotal

61

0 1 2  
  2  ft L v 722 ft B 10:17 sec C ¼ f total ¼ ð0:015Þ @ A ft D 2g 0:835 ft 2  32:2 sec 2 ¼ 20:83 ft of water

Convert the head loss in feet of water to psi using Eq. 3.10. ΔPpsi ¼

3.7.3

hf ðft of waterÞ 2:31 ft of water 1 psi

¼

20:83 ft of water 2:31 ft of water 1 psi

¼ 9:02 psi

Head Loss at Pipe Entrance and at Pipe Exit

The fluid will experience a loss in energy as it enters into a pipe from the source entity and also when it exits from a pipe into the destination entity as shown in Fig. 3.5. The entrance and exit losses are calculated by using the velocity head method (Eq. 3.25). The loss coefficient used for pipe entrance is typically 0.5 based on a sharp-edged entrance [2, 11], although it could be lower in case smooth or rounded entrances are used. Equation 3.31 is used for calculating entrance loss into a pipe.  hent ¼ K ent

v2 2g



 ¼ ð0:5Þ

v2 2g

 ð3:31Þ

The velocity head represents the kinetic energy per unit weight of the fluid. When the fluid leaves a pipe and enters a destination tank, its velocity falls to zero within the tank. This is due to the static nature of the fluid in a tank. Thus, the kinetic energy of the fluid is completely dissipated as it exits the pipe into the tank. Thus, the loss coefficient for pipe exit into a tank is 1.0 [2, 11], which will result in the velocity

Fig. 3.5 Entrance and exit losses for pipes

62

3 Fluid Dynamics

head being completely dissipated. Equation 3.32 is used for calculating exit loss from a pipe into a tank.  hexit ¼ K exit

3.7.4

v2 2g



 ¼ ð1:0Þ

v2 2g

 ð3:32Þ

Head Loss Due to Change in Pipe Cross Section

In piping systems, there will be a need to change the pipe size for various reasons. For example, in a typical piping layout for pumps, reducers are used to connect larger-sized suction piping with smaller-sized pump nozzles. Sudden contraction and sudden expansion will also result in head losses. The head loss due to change in flow cross section is calculated using the velocity head method [10, 11, 16] (Eq. 3.25).  hm ¼ K

v2 2g



The value of the loss coefficient, K, depends on the configuration. The different configurations along with the equations and values of the associated loss coefficients are shown here.

3.7.4.1

Sudden Expansion (Fig. 3.6)

"

K SE

 2  2 #2 D1 v ¼ 1 and hSE ¼ K SE 1 D2 2g

Fig. 3.6 Sudden expansion in pipes

ð3:33Þ

3.7 Friction Head Loss Across Pipe Fittings and Valves

3.7.4.2

63

Sudden Contraction

The value of K depends on the ratio (D2/D1), and it varies from 0 to 0.50 and can be taken as 0.5 for purposes of conservative estimation (Fig. 3.7). hSC ¼ K SC

3.7.4.3

 2  2 v2 v ¼ 0:5 2 2g 2g

ð3:34Þ

Gradual Expansion

The value of K depends on the ratio (D2/D1) and the cone angle, θ. The most commonly used cone angle is 20 , and the maximum value of K for this angle is 0.3 (Fig. 3.8).  hGE ¼ K GE

3.7.4.4

v1 2 2g



 ¼ 0:3

v1 2 2g

 ð3:35Þ

Gradual Contraction

The value of K depends on the ratio (D1/D2) and the cone angle, θ. The most commonly used cone angle is 20 , and the maximum value of K for this angle is 0.05 (Fig. 3.9).

Fig. 3.7 Sudden contraction in pipes

Fig. 3.8 Gradual expansion in pipes

64

3 Fluid Dynamics

Fig. 3.9 Gradual Contraction in pipes

Fig. 3.10 Pipes in series

hGC ¼ K GC

3.8

 2  2 v2 v ¼ 0:05 2 2g 2g

ð3:36Þ

Flow Through Pipes in Series

For pipes in series, the flow rate is constant in each section due to the conservation of mass (continuity equation). However, the head loss is additive as shown in Fig. 3.10. Example 3.14

For the piping system shown in the figure, the flow rate of water is 3000 gpm, and the inside diameters of the pipes are DA ¼ 11.94 in and DB ¼ 7.98 in. The Darcy

3.8 Flow Through Pipes in Series

65

friction factor for both the pipes can be assumed to be 0.015. The length of section A is 100 ft and length of section B is 50 ft. Calculate: A. The pressure drop in psi due to sudden contraction B. The total pressure drop in the system including the pressure drop due to sudden contraction C. The percentage of the pressure drop due to sudden contraction (Solution) A. Convert the pipe diameters to feet. DA ¼

11:94 in 12 in ft

¼ 0:995 ft and DB ¼

7:98 in 12 in ft

¼ 0:665 ft

Calculate the velocities in pipes A and B using the continuity equation (Eq. 3.4). The volume flow rate is the same in both the pipes.  vA ¼

vB ¼

Q ¼ AA  Q ¼ AB

 3000 gpm 449 gpm=ðft3 = sec Þ π 4 ð0:995

ftÞ2

3000 gpm 449 gpm=ðft3 = sec Þ π 4 ð0:665

ftÞ2

¼ 8:59 ft= sec

 ¼ 19:24 ft= sec

Calculate the head loss due to sudden contraction by using Eq. 3.34.

hSC


 !  2 ft 2 19:24 sec vB ¼ 0:5 ¼ K SC ¼ 2:87 ft water 2g 2  32:2 secft 2

Convert the head loss to pressure drop using Eq. 3.10. ! ΔPSC ¼

hSC 2:31 ft water 1 psi

! ¼

2:87 ft water 2:31 ft water 1 psi

¼ 1:24 psi

B. Calculate the head loss across the pipes A and B using the Darcy equation (Eq. 3.12).

66

3 Fluid Dynamics



 2    2 

LA vA LB vB þ hf,pipes ¼ f DA 2g DB 2g    

2 2 f LA vA LB vB ¼ þ 2g DA DB 3 2
 ! ft 2 ð100 ftÞ 8:59 sec 0 1 7 6 0:995 ft 7 6 0:015 7 B C6 ¼@ A6 !
2 7 ft 7 6 ft ð50 ftÞ 19:24 sec 2  32:2 5 4 sec 2 þ 0:665 ft ¼ 8:21 ft water Convert the head loss in feet of water to psi using Eq. 3.10. ΔPpipes ¼

hf ft of water 2:31 ft of water 1 psi

¼

8:21 ft of water 2:31 ft of water 1 psi

¼ 3:55 psi

Calculate the total pressure drop in the system. ΔPtotal ¼ ΔPpipes þ ΔPSC ¼ 3:55 psi þ 1:24 psi ¼ 4:79 psi C. Calculate the percentage of pressure drop due to sudden contraction. %pressure drop due to sudden contraction ¼

ΔPSC 1:24 psi  100  100 ¼ 4:79 psi ΔPtotal

¼ 25:89%

3.9

Flow Through Pipes in Parallel

Figure 3.11 shows two parallel pipes in a simple pipe network. Apply the law of conservation of mass (continuity equation) at junctions and B. Fig. 3.11 Pipes in Parallel

3.9 Flow Through Pipes in Parallel

67

Flow into the junction ¼ Flow out of the junction Q ¼ Q1 þ Q2

ð3:37Þ

However, the pressures and the heads at points A and B remain constant regardless of path taken from point A to reach point B. Therefore, the pressure drop and head loss across pipe 1 will be the same as the pressure drop and head loss across pipe 2. The total flow distributes between pipes 1 and 2 in such a way that the pressure drops, and head losses are equal [4, 10, 11]. Use Darcy equation (Eq. 3.12) for calculating head loss. hL1 ¼ hL2 f 1 L1 v1 2 f Lv 2 ¼ 2 2 2 2gD1 2gD2

ð3:38Þ

Hence, in parallel flow networks, both Eqs. 3.37 and 3.38 must be satisfied. Example 3.15 Use the data shown in the figure for the parallel piping system, and calculate the flow rates (L/s) in each branch and the pressure drop, ΔPA–B (kPa).

(Solution) Convert the total volume flow rate, Q, to m3/s and the pipe diameters to m (1 m3 ¼ 1000 L and 1 m ¼ 1000 mm). Q¼ D1 ¼

425 Ls ¼ 0:425 m3 =s 1000 mL3

303 mm 154 mm ¼ 0:303 m, and D2 ¼ ¼ 0:154 m 1000 mm 1000 mm m m

Calculate the cross-section areas of the pipes. π π A1 ¼ D1 2 ¼ ð0:303 mÞ2 ¼ 0:0721 m2 4 4 π 2 π A2 ¼ D2 ¼ ð0:154 mÞ2 ¼ 0:0186 m2 4 4

68

3 Fluid Dynamics

From conservation of mass, Flow into junction A ¼ Flow out of junction A Q ¼ Q1 þ Q2 , therefore, Q2 ¼ Q  Q1 In the preceding equation, substitute for the volume flow rates, Q1 and Q2, using the continuity equation (Eq. 3.4). A2 v2 ¼ 0:425

m3  A 1 v1 s

Solve the preceding equation for v2 in terms of v1.

v2 ¼

m3   m3   0:425 A 0:0721 m2 1 s  s  v ¼ v A2 1 0:0186 m2 A2 0:0186 m2 1 m ¼ 22:85  3:876v1 s

0:425

Use the criteria of equal head loss in each branch (Eq. 3.38), and substitute the known values.   

  

f 1 L1 v1 2 f L v2 ¼ 2 2 2 2 gD1 2 gD2

Substitute the known values into the preceding equation including the expression for v2.
2 ð100 mÞv1 2 ð170 mÞ 22:85 ms  3:876v1 ¼ 0:303 m 0:154 m Simplify and solve the resulting quadratic equation for the velocity, v1.  2 m 0:299v1 2 ¼ 22:85  3:876v1 s 0:299v1 2 ¼ 22:852 þ ð3:876v1 Þ2  2ð22:85Þð3:876v1 Þ ¼ 522 þ 15:02v1 2  177v1 14:721v1 2  177v1 þ 522 ¼ 0 v1 2  12:02v1 þ 35:46 ¼ 0 Solve the quadratic for v1.

3.10

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

69

v1 ¼ 5:21 m=s Calculate v2 using the expression for v2 in terms of v1. v2 ¼ 22:85

m m m  3:876v1 ¼ 22:85  3:876  5:21 ¼ 2:66 m=s s s s

Calculate Q1 and Q2 using the continuity equation (Eq. 3.4). 
 m ¼ 0:376 m3 =s  376 L=s Q1 ¼ A1 v1 ¼ 0:0721 m2 5:21 s 
 m Q2 ¼ A2 v2 ¼ 0:0186 m2 2:66 ¼ 0:049 m3 =s  49 L=s s As a check, Q ¼ 425 L/s ¼ Q1 + Q2 ¼ 376 L/s + 49 L/s and the law of conservation of mass is satisfied. The head loss/pressure drop across pipe 1 is the same as the head loss/pressure drop between points A and B. Calculate the head loss across pipe 1 using the Darcy equation (Eq. 3.12).  h

f1

¼ f1

L1 D1



0 2 1 
100 m @ 5:21 ms A ¼ 0:017 0:303 m 2  9:81 m s2 ¼ 7:762 m water

v1 2 2g





Calculate the pressure drop using Eq. 3.8.  ΔP ¼ γh ¼

3.10

9:81

 kN ð7:762 mÞ ¼ 76:14 kN=m2 ðkPaÞ m3

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

3.10.1 Boundary Layer Theory The boundary layer is a thin layer of fluid adjacent to a solid boundary over which the fluid is flowing. Consider the flow of a fluid over a flat plate as shown in Fig. 3.12. The free stream velocity of the fluid as it approaches the flat plate is v1. The fluid particles will have zero velocity at the solid boundary because of stagnation and the “no slip condition,” which states that the fluid particles at the surface will remain

70

3 Fluid Dynamics

Fig. 3.12 Boundary layer: flow over a flat plate

stationary at the surface due to the viscous forces. The velocity of the fluid increases gradually in the y-direction, which is perpendicular to the flow direction. The velocity reaches the free stream value at the outer edge of the boundary layer. Thus, the hydrodynamic or fluid boundary layer can be defined as a thin layer of fluid adjacent to the solid boundary within which the fluid velocity keeps increasing before attaining the free stream value [10, 16]. The y-distance from the plate to the edge of the boundary layer is the thickness of the boundary layer represented by the symbol δ. The local Reynolds number at any point along the length of the plate is defined in Eq. 3.39. Re x ¼

xv1 ρ xv1 ¼ μ ν

ð3:39Þ

In Eq. 3.39, x is the distance from the leading edge, and ρ, μ, and ν are fluid properties as defined by the symbols in Chap. 1. As shown in Fig. 3.13, the boundary layer is laminar with layered, streamlined flow near the leading edge of the plate. The flow is laminar at lower Reynolds numbers (Rex < 5  105). At higher Reynolds numbers, the boundary layer (Rex > 106) is turbulent and is characterized by unsteady swirling flows inside the boundary layer. There is a transition region between the laminar and turbulent region.

Fig. 3.13 Laminar and turbulent flow over a flat plate

3.10

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

71

Example 3.16 Air flows over a flat plate at a free stream velocity of 10 m/s. The kinematic viscosity of air at the flow temperature is ν ¼ 1.3  105 m2/s. Determine if the flow is laminar at a distance of 10 cm from the leading edge. (Solution) Calculate the Reynolds number at 10 cm from the leading edge using Eq. 3.39.  Re x ¼

xv1 ¼ ν

10 cm 100 cm=m



1:3  105

10

m s



m2 s

¼ 7:69  104 Since Rex < 5  105, the flow is laminar at 10 cm from leading edge.

3.10.2 Drag Force The drag force is a resistance force that acts parallel and opposite to the relative velocity of motion of a fluid past a solid object. Consider the flow of air past a flat plate shown in Fig. 3.13. As the air moves across the plate, it experiences a resistance force acting in a direction opposite to the flow direction. The magnitude of the drag force depends on the shape and orientation of the object, the relative velocity of the fluid, the frontal area of the solid object that is resisting the flow, and the fluid properties such as density and dynamic viscosity [1, 4, 11]. Consider two contrasting flows as shown in Fig. 3.14. Intuitively, it can be observed that the resistance to flow and, therefore, the drag force will be much higher when the plate is oriented vertically than when it is oriented horizontally with respect to the flow direction of the fluid. There is easy separation of streamlines when the plate is oriented horizontally and the fluid encounters resistance to flow only at the plate surfaces. In sharp contrast, the fluid particles have to traverse long distances to get over the vertically oriented plate as shown by the streamlines. There are some

Fig. 3.14 Flow over horizontal and vertical plates—drag force

72

3 Fluid Dynamics

areas in the vicinity of the vertical plate surfaces at the center where there are no streamlines. Because of the high drag force, it takes much more force and momentum from the fluid to flow past the plate. It is clear that the shape and orientation of the object have a significant influence on the drag force. The following conclusions can be drawn about the magnitude of the drag force: • It is proportional to the frontal area of the object relative to the flow direction. The frontal area is the projected area of the object surface oriented perpendicular to the flow direction. • It is proportional to the momentum of the fluid, which is in turn proportional to the density and velocity of the fluid. • It depends on the shape of the object and the viscosity of the fluid. The general equation used for calculating the drag force (Eqs. 3.40a and 3.40b) supports the preceding statements concerning the magnitude of the drag force. FD ¼

CD ρAp v2 ðUSCSÞ 2gc

ð3:40aÞ

CD ρAp v2 ðS IÞ 2

ð3:40bÞ

FD ¼

The nomenclature in Eqs. 3.40a and 3.40b is as follows: FD: Drag force CD: Drag coefficient (dimensionless), which is a function of the Reynolds number and the shape of the solid object ρ: Density of the fluid at the flow temperature Ap: The projected frontal area of the object surface oriented perpendicular to the flow direction v: Relative velocity of the fluid with respect to the object velocity

3.10.2.1

Drag Coefficient for Different Objects

The drag coefficient, CD, is a function of the Reynolds number and the geometry of the object. The Reynolds number is calculated based on the characteristic dimension of the object. The characteristic dimension of an object depends on the geometry of the object and its orientation with respect to the flow direction of the fluid. For example, the characteristic dimension of a sphere is the diameter of the sphere, and the projected area is the area of the circle with the diameter being equal to the sphere diameter. Consider two possible orientations for the flow of a fluid past a cylinder as shown in Figs. 3.15 and 3.16. If the direction of fluid flow is perpendicular to the axis of the cylinder as shown in Fig. 3.15, then the characteristic dimension is the diameter of the cylinder, and the

3.10

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

73

Fig. 3.15 Flow past cylinder I

Fig. 3.16 Flow past cylinder II

projected area is the area of a rectangle with length equal to the cylinder length and height equal to the diameter of the cylinder. On other hand, if the direction of flow is parallel to the axis of the cylinder as shown in Fig. 3.16, then the characteristic dimension will be the length of the cylinder over which the flow occurs, and the projected area will be the area of a circle with the diameter being the same as the cylinder diameter. In terms of the characteristic dimension Dc, the formula for the Reynolds number is Re ¼

Dc vρ Dc v ¼ μ ν

ð3:41Þ

The drag coefficient can be obtained using formulas and graphs. For parallel flow over a flat plate (as shown in Figs. 3.12 and 3.13), the formulas for drag coefficient are  1:33
Re < 5  105 , Laminar Flow 0:5 Re  0:031
6 CD ¼ 10 < Re < 109 , Turbulent Flow 1 Re 7 CD ¼

ð3:42aÞ ð3:42bÞ

Example 3.17 Calculate the drag force on a flat plate with dimensions of 0.75 in thick  1.5 feet width  3.0 ft length. Air flows parallel to the plate, on both sides, at a free stream velocity of 4 ft/sec. The kinematic viscosity of air at the flow temperature is ν ¼ 1.6  104 ft2/s.

74

3 Fluid Dynamics

(Solution) Calculate the Reynolds number for air using Eq. 3.41. The length of the plate is the characteristic dimension.
ft  ð3 ftÞ 4 sec Dc v Lv Re ¼ ¼ ¼ 2 ¼ 75, 000 ν ν 1:6  104 ft sec

Since Re < 5  105, the flow is laminar. Calculate the drag coefficient by using Eq. 3.42a. CD ¼

1:33 1:33 ¼ ¼ 0:0049 0:5 Re ð75000Þ0:5

Calculate the drag force using Eq. 3.40a. Multiply by 2 to include both the top and bottom surfaces. The projected area perpendicular to the flow direction is the width of the plate multiplied by the thickness of the plate.



   lbm 0:75 in
ft 2 4 sec ð0:0049Þ 0:075 3 1:5 ft  12 in=ft 2C D ρAp v2 ft FD ¼ ¼ lbm‐ft 2gc 32:2 lbf‐ sec 2 ¼ 1:7  105 lbf 

Figure 3.17 is a graph of the drag coefficient vs Reynolds number for spheres, for cylinders with flow perpendicular to the cylinder axis, and for disks with the flow parallel to the disk surface. The straight line represents the region of “Stokes drag” for a small sphere falling in a viscous fluid. Example 3.18 Air flows past an 18 in diameter sphere at a velocity of 25 ft/sec. The properties of air at the flow temperature are density  0.075 lbm/ft3 and kinematic viscosity  1.61  104 ft2/sec. Calculate the drag force experienced by the sphere. (Solution) Calculate the Reynolds number for the sphere using Eq. 3.41. The characteristic dimension for a sphere is its diameter.    18 ft ft 25 12 sec D v Dv Re ¼ c ¼ ¼ ν ν 1:61  104 ¼ 2:33  105

3.10

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

75

Fig. 3.17 Drag Coefficients as a function of Reynolds number for various objects. (Source: FE Reference Handbook (v 9.5), NCEES, USA, (2013). Reprinted with permission from NCEES)

Determine the drag coefficient for the sphere from Fig. 3.17 as shown in the figure (excerpt of Fig. 3.17).

Keeping in mind that this graph is on a log scale, the drag coefficient from graph is CD ¼ 0.38. The projected area of a sphere is a circle. Calculate the drag force on the sphere by using Eq. 3.40a.     lbm π
18 2
ft 2 ð 0:38 Þ 0:075 ft 25 sec 2 12 3 4 CD ρAp v  ft  FD ¼ ¼ 2gc lbm‐ft ð2Þ 32:2 lbf‐ sec 2 ¼ 0:489 lbf

76

3 Fluid Dynamics

3.10.2.2

Terminal Velocity

As an object falls through a fluid, including atmospheric air, the force of gravity initially causes it to accelerate. However, as the velocity increases, so does the drag resistance force. The drag force slows down the object since it acts in the upward direction. Eventually, the drag force exactly counterbalances the gravity force causing the fall. At this point, there is no net force, and therefore the acceleration on the object is zero. Consequently, the object will continue falling with some constant velocity known as the terminal velocity [4, 10, 16]. The equation for the terminal velocity can be obtained by equating the gravity force to the drag force. CD ρAp v2 mg ¼ )v¼ 2

rffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2mg : C D ρAp

ð3:43Þ

Example 3.19 A sphere of diameter 2 in is dropped in oil (SG ¼ 0.87 and kinematic viscosity 4.84  104 ft2/sec). The specific gravity of the sphere material is 2.8, and the drag coefficient is 0.55. Calculate the terminal velocity of the sphere. (Solution) Calculate the density of the oil and the density of the sphere material using the definition of specific gravity (Eq. 1.5).   lbm ρoil ¼ SGoil  ρwater,std ¼ ð0:87Þ 62:4 3 ft 3 ¼ 52:29 lbm=ft   lbm ρsphere ¼ SGsphere  ρwater,std ¼ ð2:8Þ 62:4 3 ft 3 ¼ 175 lbm=ft Convert the diameter of the sphere to feet, and calculate the volume of the sphere. 3      2 in 4 4 0:1667 ft 3 D¼ ¼ 0:1667 ft ) V ¼ πR ¼ ðπ Þ 3 3 2 12 in=ft ¼ 0:0024 ft3 Calculate the mass of the sphere. 

  lbm
m ¼ ρsphere  V ¼ 175 3 0:0024 ft3 ft ¼ 0:42 lbm

3.10

Flow Past Solid Objects: Boundary Layer Theory, Drag and Lift Forces

77

Calculate the terminal velocity of the sphere using Eq. 3.43. The projected area of the sphere is the area of a circle with the diameter of the sphere. vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ffi u ft u rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 2ð0:42 lbmÞ 32:2 sec 2 u 2mg   v¼ ¼u
 C D ρoil Ap t lbm ft 2 ð0:55Þ 52:29 3 ðπ Þ 0:1667 2 ft ¼ 6:56 ft= sec

3.10.3 Lift Force When there is relative motion between a fluid and a solid object, the object experiences a drag force parallel to the direction of motion as explained in the previous section. In addition, there is a lift force acting on the body perpendicular to the direction of motion as shown in Fig. 3.18. The lift force opposes the weight of the object [1, 8, 16]. For example, the lift force holds the airplane in the air. Most of the lift force is generated by the wings of an airplane. The lift acts through the center of pressure of the object and is directed perpendicular to the flow direction. The magnitude of the lift force can be calculated by using Eqs. 3.44a and 3.44b). FL ¼

CL ρAv2 ðUSCSÞ 2gc

ð3:44aÞ

C L ρAv2 ðS IÞ 2gc

ð3:44bÞ

FL ¼

The nomenclatures in the preceding equations are as follows: FL: Lift force CL: Coefficient of lift, a function of Reynolds number, the shape of the air foil, and the angle of attack (the angle between the direction of the velocity vector of the relative wind and a reference line on the airfoil, usually the chord of the airfoil) ρ: Density of the fluid Fig. 3.18 Lift and drag forces

78

3 Fluid Dynamics

A: Surface area of the airfoil or wing, usually equal to the chord length multiplied by the width of the wing v: Velocity of the aircraft or airfoil Example 3.20 The typical weight of a high-capacity commercial airliner is 6.2  105 lbf. The total surface area available is 5500 ft2. The speed of the aircraft is 600 mph, and at the flight altitude, the density of air is 0.0190 lbm/ft3. Determine if a lift coefficient of 0.52 is suitable to generate a sufficient minimum lift force. (Solution) Convert the speed of the aircraft to ft/sec.     mi 5280 ft 1 hr v ¼ 600 hr mi 3600 sec ¼ 880 ft= sec Calculate the lift force by using Eq. 3.44a.   
 lbm
ft 2 ð0:52Þ 0:0190 3 5500 ft2 880 sec 2 C ρAv ft FL ¼ L ¼ lbm‐ft 2gc 2  32:2 lbf‐ sec 2
¼ 653, 430 lbf 6:53  105 lbf Since FL (6.53  105 lbf) > W (6.2  105 lbf), the lift coefficient has a suitable value.

3.11

Impulse-Momentum Principle

The impulse-momentum principle is based on Newton’s second law which states that the sum of unbalanced external forces in any particular direction, on a control volume, is equal to the rate of change of momentum in the same direction [10, 11, 16] (Fig. 3.19). Consider the 90 elbow shown in Fig. 3.19. The fluid within the elbow is the control volume (CV) since this fluid flowing through the elbow undergoes a change in the direction of flow. The rate of change of momentum in any direction is the mass flow rate times the velocity change in the same direction. The mass flow rate is a scalar quantity that remains constant. However, the velocity is a vector quantity that can change due to a change in the direction of flow. As shown in Fig. 3.19, FR is the resultant force acting on the CV (the body of the fluid within the elbow). FRx and FRy are the x and y components, respectively, of the force acting on the CV. The force

3.11

Impulse-Momentum Principle

79

Fig. 3.19 Impulsemomentum principle, flow through an elbow

acting on the elbow will be equal to FR in magnitude but will act in the opposite direction according to Newton’s third law: for every action force, there is an equal and opposite reaction force. Other forces acting on the CV are the pressure forces P1A1 and P2A2. The impulse-momentum equation in the x-direction can be stated as the net force acting on the CV in the x-direction will be equal to the rate of change of momentum in the x-direction, which is the mass flow rate of the fluid times the change in velocity in the x-direction. It can be written as follows: _ x ¼ m_ ðv2x  v1x Þ P1 A1  F Rx ¼ mΔv

ð3:45Þ

The subscript “1” represents the initial (or inlet) condition, and the subscript ‘2’ represents the final (or exit) condition. Solve Eq. 3.45 for FRx. F Rx ¼ P1 A1  m_ ðv2x  v1x Þ The impulse-momentum equation in the US Customary System must include the conversion constant, gc to keep the units consistent in the equation. The impulsemomentum equations in the x-direction for both the systems of units are given here. F Rx ¼ P1 A1 

m_ ðv2x  v1x Þ ðUSCSÞ gc

F Rx ¼ P1 A1  m_ ðv2x  v1x Þ ðS IÞ

ð3:46aÞ ð3:46bÞ

Similarly, the net force acting on the CV in the y-direction will be equal to the rate of change of momentum in the y-direction, which is the mass flow rate of the fluid times the change in velocity in the y-direction. Assuming that the weight of the elbow and the weight of the fluid in the CV is relatively small, it can be written as follows:

80

3 Fluid Dynamics


 _ y ¼ m_ v2y  v1y F Ry  P2 A2 ¼ mΔv Solve the preceding equation for FRy.
 F Ry ¼ m_ v2y  v1y þ P2 A2 The impulse-momentum equations in the y-direction for both the systems of units are given here.
 m_ v2y  v1y F Ry ¼ þ P2 A2 ðUSCSÞ gc
 F Ry ¼ m_ v2y  v1y þ P2 A2 ðSIÞ

ð3:47aÞ ð3:47bÞ

The resultant reaction force, FR, on the body of the fluid can be calculated by using Eq. 3.48. FR ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F Rx 2 þ F Ry 2

ð3:48Þ

The force exerted by the fluid on the elbow will be equal to FR in magnitude but will be opposite to FR in direction. Example 3.21 6000 gpm water flows through the 45 elbow connecting pipe sections 1-1 and 2-2 as shown in the figure. The weight of the elbow is 25 lbf, and the volume of water within the elbow is 8 ft3. The pressures at the elbow inlet and at the elbow outlet are 17 psi and 15 psi, respectively. Determine the horizontal and vertical forces required to anchor the elbow.

Example 3.21

3.11

Impulse-Momentum Principle

81

(Solution) The elbow between sections 1-1 and 2-2 is the control volume (CV). The forces acting on the CV are shown on the free body diagram (FBD) in the figure.

Example 3.21 Solution

FH and FV represent the horizontal and vertical forces required to anchor the elbow. Calculate the cross-section areas A1 and A2.  2 π π 16 in A1 ¼ D1 2 ¼ ¼ 1:3963 ft2 4 4 12 in=ft  2 π 2 π 8 in A2 ¼ D2 ¼ ¼ 0:3491 ft2 4 4 12 in=ft Calculate the velocities,v1 and v2 using the continuity equation (Eq. 3.4).  v1 ¼

Q ¼ A1

 6000 gpm 449 gpm=ðft3 = sec Þ

1:3963 ft2

¼ 9:57 ft= sec

82

3 Fluid Dynamics

 v2 ¼

Q ¼ A2

 6000 gpm 449 gpm=ðft3 = sec Þ

0:3941 ft2

¼ 33:91 ft= sec

Calculate the mass flow rate of water using Eq. 3.5.  m_ ¼ ρQ ¼

lbm 62:4 3 ft



! 6000 gpm
 449 gpm= ft3 = sec

¼ 834 lbm= sec The impulse-momentum principle states that the sum of the unbalanced external forces in any particular direction on a control volume is equal to the rate of change of momentum in the same direction. Apply the impulse-momentum principle in the x-direction using the free body diagram of the elbow as the reference. 

m_ ðv2x  v1x Þ g
c   m_ v2 cos 45  ðv1 Þ ¼ gc

F H  P1 A1  P2 A2 sin 45 ¼

Solve the preceding equation for FH, and substitute the known values.
  m_ v2 cos 45  ðv1 Þ  þ P1 A1 þ P2 A2 sin 45 FH ¼ gc     lbm ft ft 33:91  cos 45 þ 9:57 834 sec sec sec ¼ lbm‐ft 32:2 lbf‐ sec 2   lbf 144 in2
þ 17 2  1:3963 ft2 2 in ft     lbf 144 in2
þ 15 2  0:3941 ft2 sin 45 2 in ft ¼ 5138 lbf Calculate the weight of water in the CV.  W w ¼ γw V w ¼

 lbf
3  62:4 3 8 ft ¼ 499 lbf ft

Similar to the x-direction, apply the impulse-momentum principle in the y-direction using the free body diagram of the elbow as a reference.

Practice Problems

83


 m_ v2y  v1y F V  W b  W w  P2 A2 cos 45 ¼ gc
  m_ v2 sin 45  0 ¼ gc 

Solve the preceding equation for FV, and substitute the known values.
 m_ v2 sin 45  þ W b þ W w þ P2 A2 cos 45 FV ¼ gc     lbm ft 834 33:91  sin 45 sec sec ¼ þ 25 lbf þ 499 lbf lbm‐ft 32:2 2  lbf‐ sec    lbf 144 in2
þ 15 2  0:3941 ft2 cos 45 2 in ft ¼ 1747 lbf

Practice Problems Practice Problem 3.1 The mass flow rate of the fuel oil in Example 3.1 decreases to 15000 kg/s. Determine the nearest standard pipe size required to obtain a minimum average velocity of 2.50 m/s. Practice Problem 3.2 In Example 3.3, the total flow in pipe A is increased to 800 gpm, and it is to be distributed in the ratio 3:2 between pipes B and C. Determine the required pipe sizes of branches B and C if the same velocity is to be maintained in all the pipes. Practice Problem 3.3 Ethylene glycol solution (SG ¼ 1.10) flows in a 100 mm nominal, schedule 40 pipe (ID ¼ 102 mm) at a velocity of 1.25 m/s. Determine the mass flow rate of the glycol solution in kg/min. Practice Problem 3.4 Oil (SG ¼ 0.80) flows in a 600 mm DN pipeline (ID ¼ 575 mm) and is reported to have a Reynolds number of 8000. At the flow temperature, the dynamic viscosity of the oil is 35 cP. Determine the flow rate of oil in L/s. Practice Problem 3.5 The friction head loss for ethylene glycol (SG ¼ 1.10) flowing in a pipe is 22 ft of glycol. Calculate the pressure drop in psi.

84

3 Fluid Dynamics

Practice Problem 3.6 The pressure drop across 30 m of a 225 mm (ID ¼ 227 mm) steel pipe is measured to be 10 kPa. The pipe carries ethylene glycol solution at a temperature of 20  C (SG ¼ 1.07, dynamic viscosity ¼ 7.74 cP). The Darcy friction factor for the flow is 0.016. Calculate the volume flow rate and the Reynolds number for this flow. Practice Problem 3.7 The pressure drop across 50 ft of an oil (SG ¼ 0.81) pipeline is measured to be 0.05 psi. The dynamic viscosity of the oil at the flow temperature is 383 cP. The velocity of the oil is measured to be 4 ft/sec. Determine the diameter of the pipe assuming laminar flow. Verify the validity of the assumption by calculating the Reynolds number. Practice Problem 3.8 The pressure drop across 8 m of a 500 mm DN (ID ¼ 478 mm) oil (SG ¼ 0.85) pipeline is measured to be 250 Pa. The dynamic viscosity of the oil at the flow temperature is 237 cP. Assuming laminar flow, determine the mass flow of the oil in kg/s. Verify the validity of the assumption by calculating the Reynolds number. Practice Problem 3.9 30,000 gpm of water flows in a 3 ft I D rough concrete pipe (ε ¼ 0.005 ft). The properties of water at the flow temperature are density ¼ 62.4 lbm/ft3 and dynamic viscosity ¼ 1 cP. Determine the Reynolds number and Darcy friction factor for this flow. Compare the answer obtained for f by Moody diagram with the answer obtained by using the equation for completely turbulent flow, rough pipes (Eq. 3.20). Practice Problem 3.10 10 L/s of water flows in a 50 mm I D smooth plastic pipe. The properties of water at the flow temperature are ρ ¼ 1000 kg/m3 and μ ¼ 1 cP. Determine the Reynolds number and Darcy friction factor for this flow. Compare the answer obtained for f by the Moody diagram with the answer obtained by using the equation for smooth pipes (Eq. 3.21). Practice Problem 3.11 A double pipe heat exchanger is constructed by using a 12 in nominal (I D ¼ 11.94 in) outer steel pipe and a 6 in nominal inner steel pipe (O D ¼ 6.63 in and I D ¼ 6.06 in). Hot oil (SG ¼ 0.77, ν ¼ 14.4 cSt) flows in the annular region at a mass flow rate of 250 lbm/sec. The length of the heat exchanger is 15 ft. Calculate the pressure drop of the oil in psi. Practice Problem 3.12 A water gage attached to a 3 ft  2 ft duct shows a reading of 3 in over a length of 500 ft. Air flowing through the duct has density of 0.075 lbm/ft3 and kinematic viscosity of 1.61  104 ft2/sec. The Darcy friction factor is 0.018. Determine the airflow rate in cfm.

Practice Problems

85

Practice Problem 3.13 Using the results from the solution to Example 3.8 (pipe ID ¼ 6.06 in), calculate the equivalent length in feet for a swing check valve (K ¼ 2). Practice Problem 3.14 Using the same data as in Example 3.13 for flow, pipe, and fittings, recalculate the total pressure drop across the piping system by using the velocity head method. Use the values for loss coefficients provided in Table 3.1. Practice Problem 3.15 Water is flowing in the piping system shown in the figure. The pressure drop across section A is 3.5 kPa.

Calculate: A. The volume flow rate of water in L/s B. The pressure drop across the expander C. The total pressure drop and the percent pressure drop due to expander Practice Problem 3.16 Use the data associated with parallel piping network shown in the figure, with water flowing, and calculate the flow rate in pipe 2 (gpm).

Practice Problem 3.17 A vertical cylindrical pole with a diameter of 25 cm and height of 3 m is subjected to a wind speed of 70 km/hr. The properties of air at the flow temperature are density, ρ ¼ 1.205 kg/m3 and kinematic viscosity, ν ¼ 1.3  105 m2/s. Calculate the drag force experienced by the pole.

86

3 Fluid Dynamics

Solutions to Practice Problems Practice Problem 3.1 (Solution) Calculate the density of fuel oil by using the definition of specific gravity (Eq. 1.5).  
 kg ρoil ¼ ðSGoil Þ ρwater,std ¼ ð0:92Þ 1000 3 m ¼ 920 kg=m3 Calculate the required pipe diameter by using the equation for mass flow rate (Eq. 3.5).

1 hr  15000 kg m_ hr 3600 s A¼ ¼

 ¼ 0:0018 m2 ρv 920 mkg3 2:50 ms   π ðDÞ2 ¼ 0:0018 m2 A¼ 4 Solve the preceding equation for the diameter. D ¼ 0.048 m ¼ 48 mm From the pipe data table, the closest standard pipe size available is 40 mm DN, schedule 40 pipe with I D of 40.9 mm. This will result in a velocity greater than minimum velocity of 2.50 m/s, which is acceptable. Practice Problem 3.2 (Solution) Apply the continuity equation at the junction of the pipes. Flow into the junction ¼ Flow out of the junction 800 gpm ¼ QB þ QC QB 3 ¼ ¼ 1:5 QC 2 800 gpm ¼ 1:5QC þ QC ¼ 2:5QC QC ¼ 320 gpm QB ¼ 1:5  320 gpm ¼ 480 gpm Use the same velocities as in Example 3.3 for pipes B and C, and calculate the pipe sizes using the continuity equation (Eq. 3.4) π Q AB ¼ DB 2 ¼ B 4 vB

Solutions to Practice Problems

87

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi u u 480 gpm rffiffiffiffiffiffiffiffiffi u u4  449 gpm=
ft3 = sec  4QB u DB ¼ ¼t ft πvB π4 sec ¼ 0:5833 ft ð7:0 inÞ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi u u 320 gpm rffiffiffiffiffiffiffiffiffi u u4  449 gpm=
ft3 = sec  4QC u DC ¼ ¼t ft πvC π  6:3 sec ¼ 0:3795 ft ð4:6 inÞ Practice Problem 3.3 (Solution) Determine the density of ethylene glycol using the definition of specific gravity (Eq. 1.5).  
 kg ρEG ¼ ðSGEG Þ ρwater,std ¼ ð1:10Þ 1000 3 m ¼ 1100 kg=m3 Calculate the mass flow rate using Eq. 3.5. 2     kg π 102 mm m 1:25 s m3 4 1000 mm=m ¼ 11:24 kg=s ð674:4 kg= min Þ 

m_ ¼ ρAv ¼

1100

Practice Problem 3.4 (Solution) Calculate the density of oil using the definition of specific gravity (Eq. 1.5).  
 kg ρoil ¼ ðSGoil Þ ρwater,std ¼ ð0:80Þ 1000 3 m ¼ 800 kg=m3 Calculate the velocity of the oil using the equation for Reynolds number (Eq. 3.6). 

N  s=m2 ð8000Þð35 cPÞ 0:001 1 cP ð Re ÞðμÞ   ¼  v¼ Dρ 575 mm kg 800 3 1000 mm=m m ¼ 0:6087 m=s



88

3 Fluid Dynamics

Calculate the volume flow rate of the oil using the continuity equation (Eq. 3.4).  2 πD Q ¼ Av ¼ v 4      π m 1000 L ð0:575 mÞ2 0:6087 ¼ 3 4 s m ¼ 158 L=s Practice Problem 3.5 (Solution) Calculate the pressure drop in psi using Eq. 3.11. ΔP ¼

hf  SGliq 2:31 ft psi

¼

water 22 ft glycol  1:1 ftft glycol 2:31 ft water psi

¼ 10:48 psi

Practice Problem 3.6 (Solution) Determine the specific weight of ethylene glycol by using the definition of specific gravity.  
 kN γ EG ¼ ðSGEG Þ γ water,std ¼ ð1:07Þ 9:81 3 m ¼ 10:50 kN=m3 Convert the pressure drop into equivalent head loss in meters of ethylene glycol using Eq. 3.8. hf ¼

ΔP 10 kPa ¼ ¼ 0:9524 m of EG γEG 10:50 kN m3

Calculate the velocity of ethylene glycol using the Darcy equation (Eq. 3.12). vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi u   u m 227 mm rffiffiffiffiffiffiffiffiffiffiffiffiffi u2 9:81 ð0:9524 mÞ 1000 mm=m s2 2ghf D t v¼ ¼ fL ð0:016Þð30 mÞ ¼ 2:973 m Calculate the volume flow rate using the continuity equation (Eq. 3.4)     π m Q ¼ Av ¼ ð0:227 mÞ2 2:973 4 s 3 ¼ 0:1203 m =s Calculate the Reynolds number by using Eq. 3.6.

Solutions to Practice Problems

89

   m kg 1:07  1000 3 ð0:227 mÞ 2:973 s m Dvρ 0 1 Re ¼ ¼ μ kg B0:001 m  sC ð7:74 cPÞ@ A 1 cP ¼ 93296 Practice Problem 3.7 (Solution) Solve the Hagen – Poiseuille equation (Eq. 3.16b) for the diameter and calculate the diameter of the pipe. vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u 0 1 lbf‐ sec u   0:000021 u ft u B ft2 C uð32Þ@383 cP  Að50 ftÞ 4 sec 1 cP rffiffiffiffiffiffiffiffiffiffiffiffiffi u 32μLv u u     D¼ ¼ ΔPLam u lbf in2 t 0:05 2 144 2 in ft ¼ 2:67 ft Calculate the density of oil by using the definition of specific gravity (Eq. 1.5)  
 lbm ρoil ¼ ðSGÞ ρwater,std ¼ ð0:81Þ 62:4 3 ¼ 50:54 lbm=ft3 ft Calculate the Reynolds number using Eq. 3.6. 
ft  lbm 50:54 ð 2:67 ft Þ 4 3 sec ft Dvρ   ¼ 2097 Re ¼ ¼ 0:000672 ft‐lbm μ sec ð383 cPÞ 1 cP Since Re < 2100, the flow is laminar and the use of Hagen–Poiseuille is justified. Practice Problem 3.8 (Solution) Rearrange the Hagen–Poiseuille equation (Eq. 3.16b) to calculate the velocity of the oil. ΔPLaminar D2 v¼ ¼ 32μL

  2 N
1m 250 2 478 mm  1000 mm 0 m 1 Ns 0:001 2 B m C 32@237 cP  Að 8 m Þ cp

¼ 0:941 m=s

90

3 Fluid Dynamics

Verify if the flow is laminar by calculating the Reynolds number using Eq. 3.6.    m kg ð0:487 mÞ 0:941 0:85  1000 3 s m Dvρ ¼ Re ¼ Ns μ 0:237 2 m ¼ 1644 Note: Substitute kg.m/s2 for Newton (N) to get the units to cancel out. Since Re < 2100, the flow is laminar. Calculate the mass flow rate of oil by using the continuity equation (Eq. 3.5).     kg π m 850 3 ð0:487 mÞ2 0:941 4 s m ¼ 149 kg=s

m_ ¼ ρAv ¼

Practice Problem 3.9 (Solution) Calculate the velocity of water using the continuity equation (Eq. 3.4).  v¼

Q ¼ A



30000 gpm 449 gpm=ðft3 = sec Þ π 4 ð3

ftÞ2

¼ 9:45 ft= sec

Calculate the Reynolds number (Eqs. 3.6) and the relative roughness (Eq. 3.17). 
 ft lbm 62:4 ð 3 ft Þ 9:45 3 s ft Dvρ   ¼ 2:63  106 ¼ Re ¼ 6:72104 ft‐lbm μ sec ð1 cPÞ 1 cP and r¼

ε 0:005 ft ¼ ¼ 0:0017 D 3 ft

Using the Reynolds number and relative roughness as parameters, determine the Darcy friction factor from the Moody diagram as shown in the accompanying figure. The result is f ¼ 0.023.

Solutions to Practice Problems

91

First, check if the criteria for using Eq. 3.20 is satisfied. The criteria for using Eq. 3.20 is Re > 3500/r. 3500 3500 ¼ ¼ 2:06  106 r 0:0017 Since Re (2.63  106) is greater than 3500/r, the criteria is satisfied. Calculate the friction factor by using Eq. 3.20. 1 pffiffiffi ¼ 1:14  2 log r ¼ 1:14  2 log ð0:0017Þ ¼ 6:6791 f  2 1 ¼ 0:0224 f ¼ 6:6971 The results from the Moody diagram ( f ¼ 0.023) and from Eq. 3.20 ( f ¼ 0.0224) are in excellent agreement. Practice Problem 3.10 (Solution) Calculate the velocity of water by using the continuity equation (Eq. 3.4).   10 L=s 1000 L=s

Q 1 m3 =s v ¼ ¼
 2 ¼ 5:09 m=s A π 50 mm 4

1000 mm m

Calculate the Reynolds number (Eq. 3.6).  

 50 mm 5:09 ms 1000 mkg3 1000 mm Dvρ m Re ¼ ¼ ¼ 2:54  105 0:001Ns μ m2 ð1 cPÞ 1 cP Using the Reynolds number and the curve for smooth pipes (r ¼ 0), determine the Darcy friction factor from the Moody diagram as shown in the accompanying figure. f ¼ 1.4  102 ¼ 0.014

92

3 Fluid Dynamics

Calculate the friction factor using the equation for smooth pipe (Eq. 3.21).     1 6:9 6:9 pffiffiffi ¼ 1:8 log ¼ 1:8 log Re f 2:54  105 ¼ 8:219   1 2 f ¼ ¼ 0:0148 8:219 Thus, the results from the Moody diagram and from the equation for smooth pipes are in excellent agreement. Practice Problem 3.11 (Solution) Determine the hydraulic diameter of the annular region using Eq. 3.23. Dio represents the inner diameter of the outer pipe, and Doi represents the outer diameter of the inner pipe. The area of the annular region is the area between the inner surface of the outer pipe and the outer surface of the inner pipe. The wetted perimeter of the annular region is the sum of the circumferences of the outer surface of the inner pipe and the inner surface of the outer pipe. Convert the given diameters to feet. Doi ¼

6:63 in 12 in 1 ft

¼ 0:5525 ft

Dio ¼

11:94 in 12 in 1 ft

¼ 0:9950 ft

0 π 
1   Dio 2  Doi 2 A A DH ¼ 4 ¼ 4@ 4 P π ðDio þ Doi Þ ¼ Dio  Doi ¼ 0:9950 ft  0:5525 ft ¼ 0:4425 ft Calculate the density of oil by using the definition of specific gravity (Eq. 1.5)  
 lbm ρoil ¼ ðSGÞ ρwater,std ¼ ð0:77Þ 62:4 3 ¼ 48:05 lbm=ft3 ft

Solutions to Practice Problems

93

Calculate the velocity of oil by using the continuity equation (Eq. 3.5). The crosssection area of flow is the area of the annual region. v¼

m_ m_ ¼   ρA ρ π
D 2  D 2  io oi 4 250

¼

lbm ft3 ¼ 9:67 ft=s 48:05

lbm sec

   π ð0:9950 ftÞ2  ð0:5525 ftÞ2 4

Calculate the Reynolds number using Eq. 3.7.   ft ð0:4425 ftÞ 9:67 sec D v  Re ¼ H ¼  ν 0:000011 ft2 = sec 14:4 cSt  1 cSt
 4 ¼ 27, 014 2:70  10 Calculate the relative roughness using Eq. 3.17. r¼

ε 0:00015 ft ¼ 0:00034 ¼ DH 0:4425 ft

Using r and Re as parameters, determine the Darcy friction factor using the Moody diagram (Fig. 3.4) as shown and f ¼ 0.024.

Calculate the head loss by using the Darcy equation (Eq. 3.12).

94

3 Fluid Dynamics

 hf ¼ f

L DH



0 1
  ft 2 v 15 ft B 9:67 sec C ¼ ð0:024Þ @ A ft 2g 0:4425 ft 2  32:2 sec 2 ¼ 1:18 ft of water 2





Convert the friction head loss to pressure drop using Eq. 3.10. ΔPpsi ¼

hf 1:18 ft water ¼ 2:31 ft water 2:31 ft water 1 psi 1 psi ¼ 0:511 psi

Practice Problem 3.12 (Solution) Convert the water gage reading to feet of air. ρw ghw ρa gha ¼ gc gc 0 1 lbm    62:4 3  ρw 3 in B C ft ha ¼ h ¼@ A lbm ρa w 12 in=ft 0:075 3 ft ¼ 208 ft of air ΔPduct ¼

Calculate the hydraulic diameter of the duct. When flowing through the duct, the air touches all the four sides of the duct. Therefore, the wetted perimeter is P ¼ 2ð3 ft þ 2 ftÞ ¼ 10 ft Calculate the hydraulic diameter by using Eq. 3.23.     A 3 ft  2 ft DH ¼ 4 ¼4 ¼ 2:4 ft P 10 ft Calculate the velocity of air using the Darcy equation (Eq. 3.12). vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u0 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ft    u   2  208 ft  32:2 u 2 C 2:4 ft 2ha g DH B sec u v¼ ¼ t@ A f 500 ft L 0:018 ¼ 59:77 ft= sec Calculate the volume flow rate of air using the continuity equation (Eq. 3.4).

Solutions to Practice Problems

95

 Q ¼ vA ¼

 ft 60 sec  ð3 ft  2 ftÞ 59:77 sec min

¼ 21, 517 ft3 = min ðcfmÞ Practice Problem 3.13 (Solution) From the solution to Example 3.8, the friction factor is f ¼ 0.018. Substitute the known values into Eq. 3.28 to calculate the equivalent length.
 KD ð2Þ 6:06 12 ft ¼ 56:11 ft Le ¼ ¼ 0:018 f Practice Problem 3.14 (Solution) Calculate the total head loss using Eq. 3.26. Use all the data available from Example 3.13, and use the loss coefficients provided in Table 3.1.  2 h   X i v L htotal ¼ þ K f D 2g 2 3  2 L v 4 f D þ 6ðK EL Þ þ 2ðK GTV Þ 5 ¼ 2g þK GLV þ K CHKV 3 2 0 1 0 1 7 6 B 250 ft C
 7 6 B C ft 2 B 10:17 sec C6 0:015@10:02 inA þ 6ð0:2Þ þ 2ð0:17Þ 7 ¼@ 7 A6 ft 7 6 12 in=ft 2  32:2 5 4 sec 2 þ10 þ 2 ¼ 28:91 ft of water Convert the head loss in feet of water to psi using Eq. 3.10. ΔPpsi ¼

hf ft of water 2:31 ft of water 1 psi

¼

28:91 ft of water 2:31 ft of water 1 psi

¼ 12:52 psi

Practice Problem 3.15 (Solution) A. Convert the pressure drop across pipe A to equivalent head loss (m of water) using Eq. 3.8. hfA ¼

ΔPA 3:5 kPa ¼ 0:3568 m ¼ γ 9:81 kN m3

Note: Substitute N/m2 for Pa to reconcile units.

96

3 Fluid Dynamics

Calculate the velocity in pipe A using the Darcy equation (Eq. 3.12). vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   u  154 mm m rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi u u2ð0:3568 mÞ 9:81 2 1000 mm=1 m s 2hfA DA g t vA ¼ ¼ f A LA ð0:02Þð50 mÞ ¼ 1:038 m=s Calculate the volume flow rate in pipe A using the continuity equation (Eq. 3.4).      π m 1000 L ð0:154 mÞ2 1:038 3 4 s m
 ¼ 19:33 L=s 0:01933 m3 =s

QA ¼ AA vA ¼

B. Calculate the head loss across the expander using Eq. 3.35 for head loss due to gradual expansion.

hGE


2 !  2 1:038 ms vA ¼ 0:3 ¼ 0:3 ¼ 0:0165 m of water 2g 2  9:81 sm2

Convert the head loss (m of water) to equivalent pressure drop across the expander using Eq. 3.8.  ΔPGE ¼ γhGE ¼

9:81

 kN ð0:0165 mÞ ¼ 0:1619 kPa m3

C. Calculate the velocity in pipe B using the continuity equation (Eq. 3.4). QB ¼ QA ¼ AB vB 0:01933 m3 =s Q Q vB ¼ B ¼ π  B ¼   2 AB π 255 mm ðDB Þ2 4 4 1000 mm=m ¼ 0:3785 m=s Calculate the head loss in pipe B using the Darcy equation (Eq. 3.12)  hfB ¼ f B

LB DB



0

1

0
 1 m 2 B 100 m C 0:3785 sec C@ A ¼ ð0:02ÞB m @ 255 mm A 2  9:81 sec 2 1000 mm=m ¼ 0:0573 m of water

 2

vB 2g

Solutions to Practice Problems

97

Convert the head loss (m of water) to equivalent pressure drop across pipe B using Eq. 3.8.  ΔPB ¼ γhfB ¼

9:81

 kN ð0:0573 mÞ ¼ 0:5621 kPa m3

Calculate the total pressure drop across the system. ΔPtotal ¼ ΔPA þ ΔPGE þ ΔPB ¼ 3:5 kPa þ 0:1619 kPa þ 0:5621 kPa ¼ 4:2240 kPa Calculate the percent pressure drop due to the expander.    ΔPGE 0:1619 kPa 100 ¼ 3:8% 100 ¼ 4:2240 kPa ΔPtotal

 %Pressure drop ¼ Practice Problem 3.16 (Solution)

Convert the pipe diameters to feet. D1 ¼

6:06 in ¼ 0:505 ft 12 in=ft

D2 ¼

10:02 in ¼ 0:835 ft 12 in=ft

Calculate the velocity of water in pipe 1 using the continuity equation (Eq. 3.4).  v1 ¼

Q1 ¼ A1

 900 gpm 449 gpm=ðft3 = sec Þ π 4 ð0:505

¼ 10:01 ft= sec

ftÞ2

Calculate the velocity of water in pipe 2 by equating the head losses in pipes 1 and 2 (Eq. 3.38) and using the relationship L2 ¼ 1.5L1. f 1 L1 v1 2 f Lv 2 ¼ 2 2 2 2gD1 2gD2 



f ð1:5 L1 Þv2 2 f 1 L1 v1 2 ¼ 2 2gD1 2gD2 



Solve the preceding equation for v2, and calculate the velocity in pipe 2.

98

3 Fluid Dynamics

s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi  ffi    f1 D2 ft 0:01 0:835 ¼ 10:01 v 2 ¼ ð v1 Þ sec 1:5  0:015 0:505 1:5 f 2 D1 ¼ 8:58 ft= sec Calculate the volume flow rate in pipe 2 using the continuity equation (Eq. 3.4).      ft π 449 gpm 8:58 ð0:835 ftÞ2 sec 4 ft3 = sec ¼ 2110 gpm

Q2 ¼ v2 A2 ¼

Practice Problem 3.17 (Solution) Convert the wind speed to velocity in m/s.     km 1000 m 1 hr v ¼ 70 ¼ 19:44 m=s hr km 3600 s Calculate the Reynolds number using Eq. 3.7  Dv ¼ Re ¼ ν

25 cm 100 cm=m




19:44 ms

1:3  105 ms

2

 ¼ 3:74  105

Determine the drag coefficient, CD, for a cylinder at this Reynolds number from Fig. 3.17 as shown in the figure.

References

99

From the graph, CD ¼ 0.40. The projected area of the cylindrical pole with axis perpendicular to the direction of flow will be area of the rectangle 3 m height and 25 cm width. Therefore, the projected area is AP ¼ 3 m  0.25 m ¼ 0.75 m2 . Calculate the drag force using Eq. 3.40b. 


2 kg ð0:40Þ 1:205 3 ð0:75 m2 Þ 19:44 ms m C D ρAp v2 FD ¼ ¼ 2 2 ¼ 68:31 N Note: N  kg.m/s2

References 1. Arizona State University: Fluid Mechanics Forces. Download from https://www.asu.edu/ courses/kin335/documents/Fluid%20mechanics.pdf 2. Balsiger, A., Bastos, L., Behm, J.: Minor Losses in Pipes, Colorado State University. Download from http://www.tfd.chalmers.se/~lada/MoF/assignment_2-starccm/Minor-Losses-in-PipesBalsiger-Bastos-Behm.pdf (2014) 3. Blevins, R.D.: Applied Fluid Dynamics Handbook. Krieger Publishing Company, Malabar (2003) 4. Cimbala, J.M., Cengel, Y.A.: Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Education, London (2017) 5. Connor, N.: What Is Resistance Coefficient Method, Thermal Engineering. Download from https://www.thermal-engineering.org/what-is-resistance-coefficient-method-k-method-excesshead-definition/ (2019) 6. Crane Technical Paper 410: Flow of Fluids through Valves, Fittings, and Pipe. Crane Co., Stamford (1988) 7. DOE Fundamentals Handbook: Continuity Equation, Thermodynamics, Heat Transfer, and Fluid Flow. U.S. Department of Energy. Download from https://engineeringlibrary.org/ reference/continuity-equation-fluid-flow-doe-handbook (1992) 8. Mpower UK: Aerodynamic Lift and Drag and the Theory of Flight. Download from https:// www.mpoweruk.com/flight_theory.htm 9. Mishra, P.: Difference Between Laminar and Turbulent Flow, Mechanical Booster. Download from https://www.mechanicalbooster.com/2016/08/difference-between-laminar-and-turbulentflow.html (2016) 10. Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 10th edn. Wiley, Hoboken (2020) 11. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson, New York (2015) 12. Neutrium: Hydraulic Diameter. Download from https://neutrium.net/fluid-flow/hydraulicdiameter/ (2020)

100

3 Fluid Dynamics

13. Nuclear Power.Net: Equivalent Pipe Length Method. Download from https://www.nuclearpower.net/nuclear-engineering/fluid-dynamics/minor-head-loss-local-losses/equivalent-pipelength-method/ 14. Tec-Science.com: Hagen Poiseuille Equation for Pipe Flows with Friction. Download from https://www.tec-science.com/mechanics/gases-and-liquids/hagen-poiseuille-equation-for-pipeflows-with-friction/ (2020) 15. Tioga Pipe: Pipe Data Chart. Download from https://www.tiogapipe.com/assets/files/tiogapipe-chart.pdf (2020) 16. White, F.M.: Fluid Mechanics, 8th edn. McGraw Hill, New York (2016)

Chapter 4

Energy Equation and Its Applications

4.1

Introduction

The energy equation is based on the law of conservation of energy, which is the same as the I law of thermodynamics for open systems [2]. The control volume referenced in the application of the I law can be any entity (pipe, open channel, pump, hydraulic turbine) through which the fluid is flowing. The different forms of energy associated with a flowing fluid are pressure energy, potential energy, kinetic energy, and internal energy. The internal energy of a substance changes only if there is a change in the temperature of the substance. Since the temperature of a flowing fluid typically remains constant, the internal energy of the fluid remains constant. Equations for each form of energy and the corresponding equations for energy per unit weight of the fluid are given here. Note that the weight of the fluid is the specific weight times the volume of the fluid, that is, W ¼ γ V. The weight of the fluid is also the mass of the fluid times the acceleration due to gravity, that is, W ¼ m g (Table 4.1).

Table 4.1 Energy forms and equations Energy form Pressure energy

Equation for total energy PV

Potential energy

mgz

Kinetic energy

mv2/2

Equation for energy per unit weight Pr:Energy PV PV P Unit wt: ¼ W ¼ γV ¼ γ Potential Energy mgz ¼ mgz W ¼ mg ¼ z Unit wt: 2 Kinetic Energy mv2 v2 ¼ mv 2W ¼ 2mg ¼ 2g Unit Wt:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_4

101

102

4.2

4 Energy Equation and Its Applications

The Mechanical Energy Equation

Fig. 4.1 Control volume for energy equation

Consider the control volume of the fluid between points 1 and 2 in the pipe shown in the figure. Application of the I law of thermodynamics for the control volume (CV) results in the following equations. Total energy of the fluid at point 1 þ Energy added to the control volume  Energy removed from the control volume ¼ Total energy of the fluid at point 2 ð4:1Þ Energy can be added to the fluid in the form of pressure head added by a pump between points 1 and 2. The pressure head added by a pump is typically represented by hPump with units of feet (ft) or meter (m). As the fluid moves from point 1 to point 2, it has to overcome the friction at the pipe surface, which results in a loss of energy of the fluid [5, 13]. This loss in energy is termed as the friction head loss, represented as hf (in feet or meter). The friction head loss for flow through a pipe can be calculated using the Darcy equation (Eq. 3.12). The energy terms in Eq. 4.1 can be written in terms of energy per unit weight (as shown earlier) resulting in the following equation. v2 P1 v21 P þ þ z1 þ hPump  hf ¼ 2 þ 2 þ z2 γ 2g γ 2g

ð4:2Þ

Equation 4.2 is the mechanical energy equation [2, 6, 7, 13], also known as the “extended Bernoulli’s equation” [8] for fluid flow between points 1 and 2 in a flow field. The kinetic energy per unit weight, that is v2/2g, is also known as “velocity head.” Each term in the mechanical energy equation has the units of head of the liquid (in feet or meter). The energy lines can be plotted with respect to an arbitrary datum line. The sum of pressure head and elevation is known as “hydraulic grade line (HGL),” and when the velocity head is added to the hydraulic grade line, the “energy grade line (EGL) is obtained [3]. The mechanical energy equation can also be written in terms of energy per unit mass with each term in the equation having the units of ft – lbf/lbm (USCS) or N.m/ kg (SI). The energy terms per unit mass in USCS and SI units are:

4.2 The Mechanical Energy Equation

103 lbf

Pressure Energy PV P P ft  lbf ft2 ¼ ¼ m ¼  lbm ðUSCSÞ  Unit mass m ρ lbm 3 V ft

N

Pressure Energy PV P P m2 N:m J ¼ ¼ m ¼  kg   ðS IÞ Unit mass m ρ kg kg V m3


 Potential Energy mgz ft ft lbm ¼ ¼ gz  ð ft Þ  ð ft Þ  Unit mass m lbm sec 2 sec 2 


lbm  ft ft ft ft  lbf   ðlbf Þ  ðUSCSÞ lbm lbm lbm sec 2   Potential Energy mgz m ¼ ¼ gz  2 ðmÞ Unit mass
 m
s 
 ðS IÞ   m kg kg  m m Nm J     2 ðm Þ  kg kg kg kg s s2

Therefore,

ft2 ft‐lbf m2 N  m J and 2    2 lbm kg kg sec s

Kinetic Energy mv2 v2 ft2 ft  lbf ¼ ðUSCSÞ  ¼  Unit mass lbm 2m 2 sec 2 Kinetic Energy mv2 v2 m2 N  m J ¼  ðS IÞ ¼  2  Unit mass kg kg 2m 2 s The units for the energy added by the pump per unit mass of the fluid are determined as shown here. mghPump  ghPump  m mghPump  ghPump m




 ft ft2 ft  lbf ðUSCSÞ  ðftÞ  2 lbm sec sec 2   m m2 N  m J  ðS IÞ  2 ðm Þ  2  kg kg s s

The energy terms in Eq. 4.1 can be written in terms of energy per unit mass using the preceding results. v2 P1 v21 P þ þ z1 g þ hPump g  hf g ¼ 2 þ 2 þ z2 g ρ 2 ρ 2 Example 4.1 illustrates the application of the mechanical energy equation.

ð4:3Þ

104

4 Energy Equation and Its Applications

Example 4.1 Fuel oil at 60  F is pumped to an open tank by using a 12 in., Sch. 40 steel pipe [ID ¼ 11.94 in.]. The equivalent length of piping system is 4000 ft. The flow rate of the oil is 3000 gpm. The oil surface in the tank is 300 ft above the pump suction line. The specific gravity of the oil is 0.86, and its kinematic viscosity is 5.547  105 ft2/ s. The suction pressure of the pump inlet is 2 psig. Determine the head added by the pump in feet and the discharge pressure of the pump. (Solution) Determine the friction head loss in the pipe by using the following steps. in Convert the pipe diameter to feet. D ¼ 11:94 ¼ 0:995 ft 12 in ft

Calculate the velocity in the pipe by using the continuity equation (Eq. 3.4).

v¼

Q Q ¼ ¼ A πD2 4

ð3000 gpmÞ



π 4 ð0:995

1 ft3 = sec 449 gpm 2

ftÞ

 ¼ 8:593 ft= sec

Calculate the Reynolds number for the flow by using Eq. 3.7.   Dv ð0:995 ftÞ 8:593 fts   Re ¼ ¼ ¼ 1:54  105 2 ν 5:547  105 ft s

Calculate the relative roughness of the pipe surface. From the Moody diagram (Fig. 3.4), the roughness of a commercial steel pipe is ε ¼ 0.00015 ft. Using Eq. 3.17, the relative roughness is r¼

ε 0:00015 ft ¼ ¼ 0:00015 D 0:995 ft

Locate the intersection of the ordinate at Re ¼ 1.54  105 and the curve for r ¼ 0.00015, as shown in the accompanying figure. At the point of intersection, f ¼ 1.75  102 ¼ 0.0175.

4.2 The Mechanical Energy Equation

105

Calculate the friction head loss in the piping system using the Darcy equation (Eq. 3.12).  
2  L v hf ¼ f D 2g 2 3 2
  ft 7 8:593 s 4000 ft 6 6
7 ¼ ð0:0175Þ 4 0:995 ft ft 5 ð2Þ 32:2 2 s ¼ 81 ft of oil Solve the mechanical energy equation (Eq. 4.2) for the head added by the pump, hPump. hPump ¼

P2  P1 v22  v21 þ þ ð z 2  z 1 Þ þ hf γ 2g

ð4:4Þ

Take reference point 1 as the suction of the pump and reference point 2 as the oil surface in the open tank. Calculate the specific weight of the oil by using the definition of specific gravity, (Eq. 1.5).
   lbf γ oil ¼ ðSGoil Þ γ water,std ¼ ð0:86Þ 62:4 3 ft 3 ¼ 53:66 lbf=ft Since the oil surface is open to the atmosphere, P2 (gage) ¼ 0 psig. The pressure at the pump suction is P1 ¼ 2 psig. Since the cross section of the tank is large and since the level of oil in the tank is constant, the velocity of the oil in the tank is zero, that is, v2 ¼ 0. The velocity of oil on the suction side of the pump is v1 ¼ 8.593 ft/sec. The oil surface is 300 ft above the suction pipe, therefore, z2  z1 ¼ 300 ft The friction head loss in the pipe is hf ¼ 81 ft Substitute all the known values into Eq. 4.4. P2  P1 v22  v21 þ þ ðz2  z1 Þ þ h f γ

2g  lbf 144 in2   0 2 2 ft 2 0  8:593 sec in ft2 ¼ þ lbf ft 53:66 3 2  32:2 2 sec ft þ300 ft þ 81 ft ¼ 375 ft

hPump ¼

106

4 Energy Equation and Its Applications

Convert the head added by the pump to equivalent pressure difference across the pump by using Eq. 3.11.
  ft water hPump SGliq in ft liquid
 ¼ 2:31 ft water 1 psi 

ΔPPump,psi


 ft water ð375 ft oilÞ 0:86 ft oil
 ¼ 2:31 ft water 1 psi ¼ 139:6 psi

Calculate the discharge pressure of the pump. Pdisch: ¼ Psuction þ ΔPpump ¼ 2 psi þ 139:6 psi ¼ 141:6 psi

4.3

Pump Power Equation

Pumps add energy to the flowing fluid. The rate of energy transferred to the fluid by the pump, expressed in horsepower (hp), is the “hydraulic horsepower (HHP)” of the pump. Hydraulic horsepower is also referred to as “water horsepower” [6, 7, 12, 13]. As noted earlier, the head added by the pump represents the energy added per unit weight of the fluid. Since power is the rate of consumption of energy, the hydraulic power consumed by the fluid is the weight flow rate of the fluid times the head supplied by the pump. The weight flow rate of the fluid is the volume flow rate multiplied by the specific weight of the fluid. Recall that the specific weight of a fluid is the weight per unit volume. Also, horsepower is the consumption of energy at the rate of 550 ftlbf/sec. If the volume flow rate, Q, is in ft3/sec and the specific weight, γ, is in lbf/ft3, and the head added by the pump, hPump, is in ft, then the hydraulic horsepower can be calculated by Eq. 4.5. Q γhPump HHP ¼ 550cfsftlbf= sec

ð4:5Þ

hp

Typically, the volume flow rate is in gallons per minute (gpm) and 1 ft3/ sec ¼ 449 gpm. The specific weight of the fluid is the specific gravity (SG) of the fluid multiplied by standard specific weight of water, 62.4 lbf/ft3. Making these substitutions into Eq. 4.5 results in the following equation for calculating the hydraulic horsepower for any fluid, with, the volume flow rate, Q, in gpm [3]. HHP ¼

Qgpm ðSGÞhPump 3960 ftgpm hp

ð4:6Þ

4.3 Pump Power Equation

107

Since pumps are not 100% efficient, the power supplied to the pump should be more than the required hydraulic horsepower. The actual mechanical horsepower supplied to the pump is called the “brake horsepower (BHP).” Therefore, BHP ¼

Qgpm ðSGÞhPump HHP   ¼ ftgpm ηPump η 3960 Pump hp

ð4:7Þ

Pumps are driven by electric motors, which operate at less than 100% efficiency. Therefore, the electrical power to be supplied to the pump must factor in the efficiency of the motor. The electrical power is typically calculated in kilowatts (kW), and therefore the energy consumption will be in kilowatt hours (kWH). Since 1 hp ¼ 0.746 kW, the electrical power consumption can be calculated by using Eq. 4.8.

 kW HHP PðkWÞ ¼ 0:746 hp ηPump  ηMotor

ð4:8Þ

The pump power equations in SI units [12] are given here. Hydraulic Power ðkWÞ ¼ QγhPump

ð4:9Þ

In Eq. 4.9, Q is in m3/s, γ is in kN/m3, and hPump is in m. Explanation of units in Eq. 4.9:
3 
 m kN kN  m kJ   kW Hydraulic Power ¼ QγhPump  ðmÞ  s s s m3 Brake Power ðkWÞ ¼ Electrical Power ðkWÞ ¼

QγhPump ηPump

ð4:10Þ

QγhPump ηPump  ηMotor

ð4:11Þ

Example 4.2 Determine the following for the pump and the fluid given in Example 4.1: A. The hydraulic horsepower. B. The power consumption (kW) if the efficiency of the pump is 70% and efficiency of the motor is 90%. C. The annual cost of power at the rate of $0.09 per kWH. The average usage of the pumping system is predicted to be 20 hours/day, 340 days/year. (Solution) A. Using the data in Example 4.1, calculate the hydraulic horsepower by using Eq. 4.6.

108

4 Energy Equation and Its Applications

Qgpm ðSGÞhPump ð3000 gpmÞð0:86Þð375 ftÞ ¼ ft  gpm ft  gpm 3960 3960 hp hp ¼ 244:32 hp

HHP ¼

B. Calculate the power consumption using Eq. 4.8.


 
kW HHP kW 244:32 hp ¼ 0:746 PðkWÞ ¼ 0:746 hp ηPump  ηMotor hp ð0:70Þð0:90Þ ¼ 289:31 kW C. Calculate the annual cost of power by using the given cost of power.


 Hrs days $0:09 Annual CostPower ¼ ð289:31 kWÞ 20 340 day yr kWH ¼ $177, 058=yr Example 4.3

The piping system shown in the figure handles a liquid with SG ¼ 0.85 and dynamic viscosity, μ ¼ 0.3 cP. The mass flow rate of the liquid is 45 lbm/sec. The pressure in the suction line is maintained at 0.50 psig. The pressure in the overhead storage tank is 10 psig. Assume the Darcy friction factor to be 0.015. Calculate: A. The head added by the pump (ft) B. The pump discharge pressure (psig) C. The power supplied to the motor (kW) if the efficiency of the pump is 75% and the efficiency of the motor is 90%

4.3 Pump Power Equation

109

(Solution) A. Determine the friction head loss in the pipe by using the following steps. Convert the pipe diameter to feet. D¼

6:065 in 12 in ft

¼ 0:5054 ft

Calculate the velocity in the pipe by using the continuity equation (Eq. 3.5). lbm 45 m_ m_ sec  v¼ ¼
2 ¼
ρA πD lbm π 0:85  62:4 3 ð0:5054 ftÞ2 ρ 4 4 ft ¼ 4:229 ft= sec Calculate the velocity head for the flow.   ft 2 4:229 sec v2 ¼ 0:2777 ft velocity head ¼ ¼ ft 2g 2  32:2 sec 2 Calculate the head loss in the piping system using the Darcy equation (Eq. 3.12).  
2  L v hf ¼ f D
2g  1600 ft ¼ ð0:015Þ ð0:2777 ftÞ 0:5054 ft ¼ 13:19 ft of liquid From Eq. 4.4, the equation for head added by the pump is hPump ¼

P2  P1 v22  v21 þ þ ð z 2  z 1 Þ þ hf γ 2g

Apply the energy equation between the suction of the pump (reference point 1) and the discharge of the pipe into the tank (reference point 2) as shown in the figure.

110

4 Energy Equation and Its Applications

Typically, the velocity heads are exceedingly small in magnitude compared to the other terms in the energy equation. In this case, the velocity head is just 0.2777 ft, and it can be neglected. Calculate the specific weight of the liquid using Eq. 1.5.
   lbf γ liquid ¼ ðSGÞ γ water,std ¼ ð0:85Þ 62:4 3 ft ¼ 53:04 lbf=ft3 Other known values are P1 ¼ 0.50 psig, P2 ¼ 10 psig, z2 ¼ 65 ft, and z1 ¼ 3 ft. Substitute all the known values into the equation for the head added by the pump. P2  P1 v22  v21 þ þ ð z  z 1 Þ þ hf 2g 
2
γ  lbf lbf 144 in2 10 2  0:50 2 in in ft2 ¼ þ 0 þ ð65 ft  3 ftÞ þ 13:19 ft lbf 53:04 3 ft ¼ 101 ft:

hPump ¼

B. Calculate the differential pressure across the pump using Eq. 2.1.

ΔPpump ¼ γ liquid hpump



 lbf 1 ft2 ¼ 53:04 3 ð101 ftÞ ft 144 in2 ¼ 37:20 lbf=in2 ðpsigÞ

Calculate the discharge pressure of the pump. Pdisch: ¼ Psuction þ ΔPpump ¼ 0:5 psig þ 37:2 psig ¼ 37:7 psig C. Calculate the volume rate of flow by using Eq. 3.5.

Qcfs ¼

45 lbm m_ sec ¼ 0:8484 ft3 = sec ¼ ρ 0:85  62:4 lbm3 ft

Calculate the hydraulic horsepower by using Eq. 4.5.

4.3 Pump Power Equation

111





 ft3 lbf 0:8484 53:04 3 ð101 ftÞ Qcfs γ liq hPump sec ft HHP ¼ ¼ 550 ft  lbf= sec 550 ft  lbf= sec hp hp ¼ 8:263 hp Calculate the power supplied to the motor using Eq. 4.8.


 
kW HHP kW 8:263 hp ¼ 0:746 PðkWÞ ¼ 0:746 hp ηPump  ηMotor hp 0:75  0:90 ¼ 9:13 kW Example 4.4 60 gpm of an organic liquid (SG ¼ 0.80, dynamic viscosity ¼ 0.25 cP) is pumped using a centrifugal pump. The pump is driven by a motor rated at 1.5 kW, operating at 90% efficiency. The efficiency of the pump is 80%. What will be the reading (psi) of a differential pressure gage connected across the pump? (Solution) Rearrange Eq. 4.8 to calculate the hydraulic horsepower (HHP) delivered by the pump.   ðPÞ ηPump  ηMotor ð1:5 kWÞð0:80  0:90Þ HHP ¼ ¼ kW kW 0:746 0:746 hp hp ¼ 1:448 hp Calculate the head delivered by the pump by using Eq. 4.6.

hPump



 ft  gpm ft  gpm ðHHPÞ 3960 ð1:448 hpÞ 3960 hp hp ¼ ¼ Qgpm ðSGÞ ð60 gpmÞð0:80Þ ¼ 119:46 ft liquid

Convert the differential head to differential pressure (psi) by using Eq. 3.11.

ΔPpsi



   ft water ft water hpump SGliq in ð119:46 ft liquidÞ 0:80 ft liquid ft liquid

 ¼ ¼ 2:31 ft water 2:31 ft water 1 psi 1 psi ¼ 41:37 psi

112

4.4

4 Energy Equation and Its Applications

Bernoulli’s Equation

For frictionless flow between points 1 and 2, and in the absence of turbo-machinery such as pumps, hf and hPump can be set to zero in the mechanical energy equation (Eq. 4.2). This simplification results in the Bernoulli’s equation (Eq. 4.12): v2 P1 v21 P þ þ z1 ¼ 2 þ 2 þ z2 γ 2g γ 2g

ð4:12Þ

The Bernoulli’s equation [6, 7, 13] states that the sum of the pressure head, the velocity head, and the potential head (elevation) is constant in a frictionless flow field without any energy-producing or energy-consuming devices. Bernoulli’s equation has important applications in deriving equations for flow meters such as orifice meters and venturi meters discussed in Chap. 5. Example 4.5

An orifice opening in the side of a tank is shown in the figure. Derive an equation for the velocity of the jet through the orifice in terms of the height, h, of the free surface above the centerline of the orifice. (Solution) The potential energy of the liquid due to the elevation, h, of the free surface is converted to kinetic energy of the jet flowing through the orifice. Hence, the velocity of the jet will be a function of “h.” Apply the Bernoulli’s equation (Eq. 4.12) between points 1 and 2 shown in the figure with the following simplifications: gage pressures P1 and P2 are zero since points 1 and 2 are open to the atmosphere, considering the large cross section of the tank and the constant level of the free surface at steady state, the velocity, v1 ¼ 0 and z1 – z2 ¼ h.

4.5 Pump Performance Parameters

113

v22 ¼ z1  z2 2g Therefore, v2 ¼

4.5

pffiffiffiffiffiffiffiffi 2gh

Pump Performance Parameters

Several parameters are used in evaluating and representing the performance of pumps. The parameters are obtained by testing the pump, and they are plotted to generate the “pump performance curves” [2, 6, 7, 13]. While plotting the performance curves of a centrifugal pump, the volume flow rate, Q (gpm or m3/s), is used as the independent variable. The typical parameters plotted to obtain the centrifugal pump performance curves are: 1. Head (H ) vs. capacity (Q) for different impeller diameters. Note: The head supplied by a pump is also known as “total dynamic head (TDH)” [4, 14]. 2. Power required to drive the pump (BHP) vs. capacity (Q). 3. Pump efficiency (η) vs. capacity (Q). 4. Net positive suction head required (NPSHR) vs. capacity (Q). Figure 4.2 illustrates typical performance curves for a centrifugal pump.

Fig. 4.2 Pump performance curves

114

4.5.1

4 Energy Equation and Its Applications

System Curve and Operating Point

The system curve, hPump vs. Q, is generated by calculating and plotting the head to be supplied by the pump at different volumetric flow rates. Consider the pumping of a liquid from the lower reservoir to the upper reservoir as shown in the figure. Both the reservoirs are open to the atmosphere, and therefore the gage pressures P1 and P2 are zero. At steady-state operating conditions, the liquid levels in the reservoirs are constant. Hence, the velocities v1 and v2 are also zero. Make these substitutions in Eq. 4.4 for calculating the head to be supplied by the pump.

Use the Darcy equation (Eq. 3.12) and the continuity equation (Eq. 3.4) to calculate the friction head loss, hf, in the piping system. !  
2    Q2 L v L A hf ¼ f ¼f D 2g D 2g Simplifying, hf ¼ f

 
2  L Q D 2gA2

Substitute for hf into the equation for the head supplied by the pump, hPump. hPump

 
2  L Q ¼ ðz2  z1 Þ þ f D 2gA2

ð4:13Þ

4.5 Pump Performance Parameters

115

Equation 4.13 can be used to generate the system curve, hPump vs. Q, by varying the volume flow rate. The term (z1 – z2) represents the static head to be overcome by the pump. The second term in Eqn. 4.13 represents the friction head to be overcome by the pump. The intersection of the pump performance curve (H vs. Q) and the system curve (hPump vs. Q) represents the operating point, O, of the pump as shown in Fig. 4.3. Fig. 4.3 Operating point for a pump

At the operating point, the pump can supply a head 230 ft at a capacity of 730 gpm. Example 4.6 Generate the system curve for the piping system shown in the figure. Determine the operating point for a pump with the performance curve given here.

(Solution) Use Eq. 4.13 to generate the system curve by varying the volume flow rate, Q. z2  z1 ¼ 250 ft  50 ft ¼ 200 ft

116

4 Energy Equation and Its Applications

  πD2 π 6:04 in A¼ ¼ 12 in 4 4 ft

!2 ¼ 0:1989 ft2

Substitute the known values into Eq. 4.13 to obtain an expression for hPump in terms of Q.  
2  L Q hPump ¼ ðz2  z1 Þ þ f D 2gA2 0 1
 C 7000 ft B Q2 B
C  ¼ 200 ft þ ð0:018Þ A   0:5033 ft @ ft 2 2 2  32:2 0:1989 ft 2 sec Simplify the preceding equation to obtain the following equation. hPump ¼ 200 ft þ 98:26Q2 Note that the volume flow rate, Q, in the preceding equation should have the units, cubic feet per second (cfs). Since 1 cfs ¼ 449 gpm, divide the flow rate in gpm by 449 to obtain the flow rate in cfs. The following table can be generated by using the equation for the head supplied by the pump. Q (gpm) 0 200 400 600 800 1000

Q (ft3/sec) 0 0.4454 0.8909 1.336 1.782 2.227

hPump (ft) 200 220 278 376 512 688

Plot the system curve using the data from the table as shown in the graph.

4.5 Pump Performance Parameters

117

The intersection of the given pump performance curve and the system curve, at 450 gpm capacity and 300 ft head, represents the operating point of the pump.

4.5.2

Pumps in Series

Pumps are arranged in series, as shown in Fig. 4.4(a), if a single pump is inadequate to provide the required head. The head provided by each pump is added together to obtain the total head required. However, the same flow rate is handled by each pump. Equations 4.14 and 4.15 are applicable for two pumps in series. Q ¼ Q1 ¼ Q2

ð4:14Þ

hP,total ¼ hP1 þ hP2

ð4:15Þ

Fig. 4.4a Pumps in series

Fig. 4.4b Pump curves and operating point for pumps in series

Figure 4.4(b) illustrates the pump curve and the operating point for two pumps operating in series [1, 7, 11].

118

4 Energy Equation and Its Applications

Example 4.7

Given the pump curve and the system curve shown in the figure, determine the number of pumps in series needed to deliver 700 gpm at system conditions. Sketch the pump curve for the combined pumps in series. (Solution) From the pump curve for a single pump, the head delivered at a capacity of 700 gpm is 60 ft as shown in the figure. Locate the operating point “O” on the system curve by drawing a vertical line at 700 gpm capacity. At the operating point, the corresponding head added is 180 ft. Therefore, to satisfy the system requirements of 700 gpm and 180 ft head, three identical pumps in series will be required, with each pump delivering a head of 60 ft. To obtain the pump curve for the combination of three pumps in series, calculate the head delivered by three pumps in series by using the data from the original pump curve for a single pump. For example, at a capacity of 200 gpm, the head delivered by a single pump is 190 ft. At the same capacity of 200 gpm, the head delivered by three identical pumps in series will be 3  190 ft ¼ 570 ft. Similarly, at zero capacity, the head delivered by three pumps in series is 3  200 ft ¼ 600 ft. Therefore, two points on the pump curve for three pumps in series will be (0 gpm, 600 ft) and (200 gpm, 570 ft). More points can be obtained in a similar manner to generate the pump curve for three identical pumps in series as shown in the figure.

4.5 Pump Performance Parameters

4.5.3

119

Pumps in Parallel

When multiple pumps are operating in parallel, the capacities of the pumps are additive since the total flow is distributed between the multiple pumps. However, the head provided by the system of multiple pumps is the same as the head provided by a single pump. The schematic for pumps in parallel is shown in Fig. 4.5(a) Fig. 4.5(a) Pumps in parallel

The continuity equation can be applied at the junctions. Flow into the junction ¼ Flow out of the junction Q ¼ Q1 þ Q2

ð4:16Þ

The two pumps in parallel provide identical heads, which are also the head provided by the system (assuming small friction losses in the piping). hsystem ¼ hP1 ¼ hP2

ð4:17Þ

The pump curves and the system curve for pumps operating in parallel are shown in Fig. 4.5(b) [1, 13].

120

4 Energy Equation and Its Applications

Fig. 4.5b Pump curves and operating point for pumps in parallel

Example 4.8

Given the pump curve and system curve shown in the figure, how many pumps in parallel are required to operate at a head of 300 ft at system conditions. (Solution) Using the given pump curve, determine the capacity that can be handled by a single pump when the head added is 300 ft. At point “A” shown in the figure, the capacity of the pump while providing a head of 300 ft is 170 gpm.

4.6 Affinity Laws (Also Known as “Pump Laws,” “Fan Laws”)

121

The operating point at system conditions will be the intersection of the system curve with 300 ft head. At the operating point B, the capacity is 900 gpm. To deliver a capacity of 900 gpm, the number of pumps needed in parallel are: N¼

4.6

900 gpm ¼ 5:3 ð6 pumpsÞ 170 gpm

Affinity Laws (Also Known as “Pump Laws,” “Fan Laws”)

The affinity laws mathematically relate the size of impeller/the speed of the impeller to pump parameters such as capacity, head, and power. The affinity laws can be stated as follows [7, 14]: 1. Pump capacity is proportional to the impeller diameter/speed. 2. Pump head is proportional to the square of the impeller diameter/square of the speed. 3. Pump power is proportional to the cube of the impeller diameter/cube of the speed. Equations 4.18, 4.19, and 4.20 are mathematical expressions for the pump affinity laws. Q2 D2 N 2 ¼ ¼ Q1 D1 N 1
2
2 h2 D2 N2 ¼ ¼ h1 D1 N1
3
3 P2 D2 N2 ¼ ¼ P1 D1 N1

ð4:18Þ ð4:19Þ ð4:20Þ

122

4 Energy Equation and Its Applications

Example 4.9 The requirement of a liquid supplied to a process unit has increased by 33%. Since the increase in impeller diameter is limited to 15%, it has been proposed to use a combination of increase in the impeller diameter and increase in speed in order to meet the new demand. The original speed of the pump is 1500 rpm, and the original power is 75 hp. Calculate A. The new speed of the pump B. The new shaft power of the pump (Solution) A. Calculate the new speed of the pump required to meet the new capacity by modifying Eq. 4.18 as shown. Q2 D2 N 2 ¼ ¼ Q1 D1 N 1 Therefore,
2

 Q2 D2 N2 ¼ Q1 D1 N1 Substitute the following known relationships into the preceding equation. Q2 ¼ 1:33 Q1

D2 ¼ 1:15 D1
 N 2 1:33 ¼ 1:15 2 N1

Calculate the new speed by solving for N2.
N2 ¼

 1:332 N 1 ¼ 1:538  1500 rpm ¼ 2307 rpm 1:15

B. Calculate the pump power required to meet the new capacity by modifying Eq. 4.20 as shown. P2 ¼ P1 Therefore,




D2 D1

3 ¼


3 N2 N1

4.7 Cavitation of Pumps

123




P2 P1

2 ¼


3
3 D2 N2 D1 N1

Solve the preceding equation for P2, and substitute the relationships, and

N2 N1

D2 D1

¼ 1:15

¼ 1:538 to calculate the new power required. P2 ¼

! rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi     D2 D1

3

N2 N1

3

P1
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ð1:15Þ3 ð1:538Þ3 ð75 hpÞ

¼ 176:42 hp

4.7

Cavitation of Pumps

The pressure profile of the fluid across a pump is quite complex. The fluid being pumped enters through the suction pipe into the eye of the impeller, which has a much smaller flow area compared to the other parts of the pump. This results in a high velocity at the impeller eye. Bernoulli’s equation states that the sum of pressure head, velocity head, and potential head (elevation) should be constant in a frictionless flow field. The higher velocity head at the impeller eye results in a smaller pressure head. Thus, the pressure at the eye of the impeller has the lowest value when compared with the pressure in different parts of the pump. If the pressure of the fluid drops below the vapor pressure of the fluid at the pumping temperature, the liquid being pumped vaporizes, causing the formation of vapor bubbles. This is most likely to happen at the impeller eye since the pressure at this location is at its lowest value within the pump. The vapor bubbles are swept across the pump by the flowing fluid into the discharge section of the pump where the pressure is much higher than the saturation or vapor pressure of the liquid at the pumping temperature. This causes the vapor bubbles to condense and collapse on the impeller surface releasing high amounts of energy and possibly eroding the impeller surface. The generation and subsequent collapse of the vapor bubbles is called cavitation [6, 7, 9]. Besides damage to the impeller, cavitation also results in noise, vibration, fluctuation in throughput, discharge pressure, and increased power consumption. To avoid cavitation, the net value of the pressure at the impeller eye must be greater than the vapor pressure of the fluid at the pumping temperature [9]. The quantity “net positive suction head (NPSH)” is used in determining if conditions within the pump can cause cavitation.

124

4.8

4 Energy Equation and Its Applications

Net Positive Suction Head (NPSH)

NPSH is the net head/net pressure existing at the impeller eye. To avoid cavitation, the net positive suction head available (NPSHA) must be greater than the minimum net suction head required (NPSHR). The criteria to avoid cavitation is mathematically stated in the inequality 4.21. NPSHA > NPSHR

ð4:21Þ

The net positive suction head available is a function of pump layout and the vapor pressure of the liquid being pumped [10, 15]. The items that contribute to the suction head are the head due to the absolute pressure of the liquid being pumped at the source (habs) and the static head due to the elevation of the source tank (helev). The items that cause reduction in the suction head are the friction head losses in the suction pipe (hf) and the head equivalent to the vapor pressure (or saturation pressure, hsat) of the liquid being pumped taken at the pumping temperature. Equation 4.22 can be used to calculate the net positive suction head available. NPSHA ¼ habs þ helev  hf  hsat

ð4:22Þ

To ensure the availability of sufficient net positive suction head, any of the following strategies or a combination of them can be used. • Increase the absolute pressure at the liquid surface in the source tank. • Increase the elevation of the liquid in the source tank. • Decrease the friction losses in the suction piping by minimizing the length of the suction line and/or by increasing the diameter of the suction piping. The net positive suction head required (NPSHR) is a parameter specified by pump vendors, and it is usually presented as part of the pump characteristic curves. Example 4.10 Hexene (SG ¼ 0.68) is being pumped from a closed tank maintained at a gage pressure 0.5 psig. The level of hexene in the tank varies from 10 to 8 ft above grade, and the eye of the pump impeller is 2 ft above grade. Friction and other losses in the suction line amount to 1.3 psi. At operating conditions, the vapor pressure of hexene is 6.1 psi, and the atmospheric pressure at the location is 14.7 psi. Can a pump with NPSHR of 7.5 psia be used in this application? (Solution) In this example, the data is given in terms of pressure rather than in terms of head of liquid. Therefore, it is more convenient to work in terms of units of pressure. Calculate the absolute pressure of hexene in the storage tank.

Practice Problems

125

Pabs ¼ Patm þ Pgage ¼ 14:7 psi þ 0:5 psi ¼ 15:2 psia Using a conservative approach, the minimum elevation of the liquid level above the eye of the impeller is helev ¼ 8 ft  2 ft ¼ 6 ft Convert the elevation into equivalent static pressure by using Eq. 2.1.

 lbf 1 ft2 Pelev ¼ γhelev ¼ 0:68  62:4 3 ð6 ftÞ ft 144 in2 ¼ 1:77 psia C. Write Eq. 4.22 in terms of pressure, and substitute the known values. NPSHA ¼ Pabs þ Pelev  Pf  Psat ¼ 15:2 psia þ 1:77 psia  1:3 psia  6:1 psia ¼ 9:57 psia The net positive suction head required is given to be 7.5 psia. NPSHR ¼ 7:5 psia Since NPSHA > NPSHR, the pump is suitable for this application.

Practice Problems Practice Problem 4.1 Fuel oil with specific gravity 0.92 and kinematic viscosity 1.6  105 m2/s flows through a 65 mm DN and Sch.40 horizontal steel pipe (62.7 mm ID, roughness, ε ¼ 0.05 mm) at a mass flow rate of 25000 kg/hr. Calculate the head (m) to be added by the pump to handle this flow across 100 m of the pipe. Practice Problem 4.2 The pump used in Practice Problem 4.1 consumes 1.75 kW of power. The motor driving the pump has an efficiency of 80%. Calculate the efficiency at which the pump is operating.

126

4 Energy Equation and Its Applications

Practice Problem 4.3

The total friction loss in the piping system shown in the figure is equivalent to 1.50 psi. The liquid being pumped has a specific gravity of 1.27. Both the reservoirs are open to the atmosphere. The efficiency of the pump is 73%, and the brake horsepower used is measured to be 60 hp. Determine the maximum flow rate (gpm) of the liquid that can be handled by the pumping system. Practice Problem 4.4

Items A and B shown in the figure are pressure gages attached to the suction and discharge of the pump. The readings in the gages are PA ¼ 0.1 bar(g) and PB ¼ 2.7 bar(g). The pump is sourcing water from an open reservoir. The water flow rate is 25 L/s, and it discharges into an open tank, where the water surface is 15 m above the water level in the source tank. The Darcy friction factor for the flow is 0.022. Determine the equivalent length (m) of 100 mm DN (ID ¼ 102.3 mm) piping that can be handled by this system. Practice Problem 4.5 30,000 bpd (1 barrel ¼ 0.1590 m3) of crude oil (SG ¼ 0.85) flows through DN 500 mm (ID ¼ 478 mm), Sch.40 pipeline. The pumping stations along this line are

Solutions to Practice Problems

127

located 50 km apart. The friction factor for the pipeline is 0.02, and the efficiency for the pump is 75%. Calculate the pump power required in kW. Practice Problem 4.6

Given the pump and system curves shown in the figure, determine the number of pumps and the configuration (series, parallel, etc.) required to operate at system conditions represented by point A. Practice Problem 4.7 300 gpm of water is pumped using a centrifugal pump supplying a head of 200 ft. The horsepower consumed by the pump is 20 hp. The impeller speed is 1200 rpm. Determine the efficiency, capacity, head supplied, and power consumption of the pump when the speed is increased to 1500 rpm, assuming that the efficiency of the pump remains constant. Practice Problem 4.8 A centrifugal pump is used in lifting water from a sump where the lowest possible water level is 12 ft below the centerline of the pump. The water in the sump is at a temperature of 80  F and at atmospheric pressure. The friction losses in the suction line are equivalent to 5 ft of water. Calculate the net positive suction head available (NPSHA) for the pump.

Solutions to Practice Problems Practice Problem 4.1 (Solution) Calculate the density of fuel oil by using the definition of specific gravity (Eq. 1.5).
   kg ρoil ¼ ðSGoil Þ ρwater,std ¼ ð0:92Þ 1000 3 m ¼ 920 kg=m3

128

4 Energy Equation and Its Applications

Convert the pipe ID to m. D¼

62:7 mm 1000 mm m

¼ 0:0627 m

Calculate the average velocity by using Eq. 3.5.  v¼

25000 kg=hr 3600 s=hr



m_ ¼ 2:44 m=s ¼   ρA 920 kg3 π ð0:0627 mÞ2 m

4

Calculate the Reynolds number for the flow by using Eq. 3.7.   Dv ð0:0627 mÞ 2:44 ms  ¼ 9:56  103 Re ¼ ¼  2 ν 1:6  105 m s

Calculate the relative roughness of the pipe surface. Using Eq. 3.17, the relative roughness is r¼

ε 0:05 mm ¼ ¼ 0:0008 D 62:7 mm

Locate the intersection of the ordinate at Re ¼ 9.56  103 and the curve for r ¼ 0.0008, as shown in the accompanying figure. At the point of intersection, f ¼ 0.034

The head to be added by the pump is equivalent to the head loss in the pipe. Calculate the head loss in the pipe using the Darcy equation (Eq. 3.12).

Solutions to Practice Problems

hPump

129

 
2  L v ¼ hf ¼ f D 2g 2 3   2:44 m2 100 m 4 s  5 ¼ ð0:034Þ 0:0627 m ð2Þ 9:81 m s2 ¼ 16:45 m of oil

Practice Problem 4.2 (Solution) Calculate the volume flow rate of oil by using the data in Practice Problem 4.1. and Eq. 3.5.  Q¼

m_ ¼ ρoil

25000 kg=hr 3600 s=hr

920 mkg3

 ¼ 0:0075 m3 =s

Calculate the specific weight of oil by using Eq. 1.5.
   kN γ oil ¼ ðSGoil Þ γ water,std ¼ ð0:92Þ 9:81 3 ¼ 9:025 kN=m3 m From Practice Problem 4.1, hPump ¼ 16.45 m of oil. Solve Eq. 4.11 for the pump efficiency, and substitute the known values

QγhPump PkW ηMotor ¼ 0:7953 ð79%Þ ηPump ¼



 m3 kN 9:025 3 ð16:45 mÞ 0:0075 s m ¼ ð1:75 kWÞð0:80Þ

Practice Problem 4.3 (Solution) Calculate the specific weight of the liquid using Eq. 1.5.
   lbf γ liquid ¼ ðSGÞ γ water,std ¼ ð1:27Þ 62:4 3 ft ¼ 79:25 lbf=ft3 From Eq.4.4, the head added by the pump is hPump ¼

P2  P1 v22  v21 þ þ ð z 2  z 1 Þ þ hf γ 2g

130

4 Energy Equation and Its Applications

Select the reference points 1 and 2 as shown in the figure.

Since the liquid surfaces are open to the atmosphere, P1 ðgageÞ ¼ 0 psig and P2 ðgageÞ ¼ 0 psig: Since the cross sections of the tanks are large and since the level of the liquid in each tank is constant, the velocity at the reference points 1 and 2 are zero, that is, v1 ¼ 0, and v2 ¼ 0. The difference in elevation between points 1 and 2 is z2 – z1 ¼ 190 ft. Convert the friction loss from pressure units to equivalent head of the liquid by using Eq. 2.1. 2

lbf 144 in ΔPf 1:50 in2  ft2 hf ¼ ¼ ¼ 2:726 ft liquid γ 79:25 lbf3 ft

Substitute the known values into the equation for the head added by the pump. P2  P1 v22  v21 þ þ ðz2  z1 Þ þ hf γ 2g ¼ 0 þ 0 þ 190 ft þ 2:726 ft ¼ 192:73 ft

hPump ¼

Solve Eq. 4.7 for the volume flow rate, Qgpm, and substitute the known values.

Solutions to Practice Problems

131

 ft‐gpm ðBHPÞ ηPump 3960 hp   ¼ ðSGÞ
hPump  ft‐gpm ð60 hpÞð0:73Þ 3960 hp ¼ ð1:27Þð192:73 ftÞ ¼ 708:62 gpm 

Qgpm




Practice Problem 4.4 (Solution) Convert the differential pressure across the pump into equivalent head added by the pump (in meters) by using Eq. 3.8.

hpump ¼

ΔPpump γ


 100 kPa ð2:7 bar  0:1 barÞ 1 bar ¼ kN 9:81 3 m ¼ 26:50 m of water

Note: kPa  kN/m2 Apply the mechanical energy equation (Eq. 4.4) between the water surfaces in the source tank (reference point 1) and in the discharge tank (reference point 2). Since the water surfaces are open to the atmosphere, P1 (gage) ¼ 0 psig and P2 (gage) ¼ 0 psig. Since the cross sections of the tanks are large and since the level of the liquid in each tank is constant, the velocity at the reference points 1 and 2 are zero, that is, v1 ¼ 0, and v2 ¼ 0. The difference in elevation between points 1 and 2 is z2 – z1 ¼ 15 m. Simplify Eq. 4.4 using the preceding substitutions. P2  P1 v22  v21 þ þ ðz2  z1 Þ þ hf γ 2g ¼ 0 þ 0 þ 15 m þ hf

hPump ¼

As calculated earlier, the head added by the pump is equivalent to 26.50 m of water. Therefore, the friction head loss is hf ¼ hpump  15 m ¼ 26:50 m  15 m ¼ 11:50 m Calculate the velocity in the pipe by using the continuity equation (Eq. 3.4).

132

4 Energy Equation and Its Applications

 v¼

Q Q ¼ ¼ A πD2 4

25

L s





1 m3 1000 L 2

π 4 ð0:1023

mÞ

¼ 3:04 m= sec

Calculate the equivalent length of the piping system that can be handled by the pump by using the Darcy equation (Eq. 3.12). hf ¼ f 2gDhf Leq ¼ f v2



2  Leq v D 2g

m  0:1023 m  11:50 m 2 s ¼  2 ð0:022Þ 3:04 ms ¼ 113:53 m 2  9:81

Practice Problem 4.5 (Solution) Convert the volume flow rate to m3/s.


barrels day ¼ 0:0552 m3 =s

Q ¼

30000




 0:1590 m3 1 day 1 hr 1 barrel 24 hrs 3600 s

Calculate the velocity in the pipeline by using the continuity equation (Eq. 3.4). 3

v¼

0:0552 ms Q Q ¼ πD2 ¼ π ¼ 0:3076 m=s 2 A 4 ð0:478 mÞ 4

Calculate the friction head loss in the pipeline using the Darcy equation (Eq. 3.12). 0 2 1
   
2  L v 50000 m @ 0:3076 ms A hf ¼ f ¼ ð0:02Þ m D 2g 0:478 m 2  9:81 2 s ¼ 10:09 m The head to be supplied by the pump is equivalent to the friction head loss in the pipeline. hPump ¼ hf ¼ 10:09 m of oil Calculate the pump power required using Eq. 4.10.

Solutions to Practice Problems

133

Qγ oil hPump η
Pump 3 
 m kN 0:0552 0:85  9:81 3 ð10:09 mÞ s m ¼ 0:75 ¼ 6:19 kW

Power ðkWÞ ¼

Note: kN.m/s  kJ/s  kW since N.m  J, and J/s  W Practice Problem 4.6 (Solution)

Determine the capacity at operating point A as shown in the figure. Therefore, at operating point A, the required flow rate is 750 gpm, and the head added is 200 ft. From the pump curve, at a capacity of 400 gpm, the head added is 100 ft. To meet the flow requirement of 750 gpm, two pumps in parallel will be required. To meet the head requirement of 200 ft, two additional pumps in series will be required in conjunction with the first set of parallelly operating pumps as shown in the figure.

Each pump will handle 400 gpm, providing a head of 100 ft. The combined flow from the second set of pumps can be throttled to 750 gpm as shown in the figure.

134

4 Energy Equation and Its Applications

Practice Problem 4.7 (Solution) Calculate the hydraulic horsepower required by using Eq. 4.6. HHP ¼

Qgpm ðSGÞhPump ð300 gpmÞð1:0Þð200 ftÞ ¼ ft  gpm ft  gpm 3960 3960 hp hp ¼ 15:15 hp

Calculate the efficiency of the pump by using Eq. 4.7. ηPump ¼

HHP 15:15 hp ¼ ¼ 0:76 ð76%Þ BHP 20 hp

The efficiency of the pump remains at 76% when the speed is increased to 1500 rpm. Determine the capacity, head supplied, and power consumption at 1500 rpm by using the affinity laws. The capacity at 1500 rpm can be determined from Eq. 4.18.

 N2 1500 rpm Q2 ¼ Q1 ¼ ð300 gpmÞ 1200 rpm N1 ¼ 375 gpm The head supplied at 1500 rpm can be determined from Eq. 4.19. h2 ¼ h1

 2 N2 N1

 2 rpm ¼ ð200 ftÞ 1500 1200 rpm ¼ 312:5 ft

The power consumed at 1500 rpm can be determined from Eq. 4.20. P2 ¼ P1

 3 N2 N1

 3 rpm ¼ ð20 hpÞ 1500 1200 rpm ¼ 39:1 hp

Practice Problem 4.8 (Solution) Convert the absolute pressure of water, which is atmospheric pressure, to equivalent column of water by using Eq. 2.1 or alternatively use the known pressure equivalents.

References

135

 habs ¼

P ¼ γ



144 in2 ft2 lbf 62:4 ft3

lbf 14:7 in 2

 ¼ 33:92 ft of water

Pressure Equivalents: 1 atm ¼ 14.7 psi ¼ 101 kPa ¼ 33.92 ft of water Find the saturation pressure (vapor pressure) of water at 80  F from steam tables, and convert it into equivalent height of water column.

hsat ¼



Psat at 80 F ¼ γ

   lbf 144 in2 0:51 in 2 2 ft 62:4 lbf ft3

¼ 1:18 ft of water

Substitute the known values into Eq. 4.22. Since the water level in the sump is below the centerline of the pump, the elevation will be negative. NPSHA ¼ habs þ helev  hf  hsat ¼ 33:92 ft  12 ft  5 ft  1:18 ft ¼ 15:74 ft

References 1. CheCalc: Pumps Operating in Series and Parallel, Chemical Engineering Calculations. Download from https://checalc.com/solved/pumpOp.html (2020) 2. Cimbala, J.M., Cengel, Y.A.: Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Education, London (2017) 3. Crane Technical Paper 410: Flow of Fluids through Valves, Fittings, and Pipe. Crane Co., Stamford (1988) 4. Engineering 360, https://www.globalspec.com/pfdetail/pumps/flow 5. Mechanical Energy Balances, https://www.engr.colostate.edu/CBE101/topics/mechanical_ energy_balances.html 6. Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 10th edn. Wiley, Milton (2020) 7. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson, Hoboken (2015) 8. Nuclear Power.Net: Conservation of Energy in Fluid Mechanics – Bernoulli’s Principle. Download from, https://www.nuclear-power.net/laws-of-conservation/law-of-conservation-ofenergy/conservation-of-energy-in-fluid-mechanics-bernoullis-principle/ 9. Pump Cavitation and How to Avoid It, White Paper by Xylem Applied Water Systems. Download from, https://www.xylem.com/siteassets/support/case-studies/case-studies-pdf/cavi tation-white-paper_final-2.pdf (2015) 10. Rankin, D.R.: What is NPSH?, Petroleum Refiner. Download from, https://www.gsengr.com/ uploaded_files/NPSH%20Calculations.pdf

136

4 Energy Equation and Its Applications

11. Rodriguez, M.Elena.: Theory Bites – System Curves for Pumps in Series, Engineering Toolbox. Download from, https://empoweringpumps.com/theory-bites-system-curves-pumps-in-series/ (2020) 12. Sugar Process Technologies: Pump Power Calculations. Download from, https://www. sugarprocesstech.com/pump-power-calculation/ 13. White, F.M.: Fluid Mechanics, 8th edn. McGraw Hill, USA (2016) 14. Whitesides, R.W.: Basic Pump Parameters and the Affinity Laws, PDH Online. Download from, https://www.pdhonline.com/courses/m125/m125content.pdf (2012) 15. Whitesides, R.W.: Understanding NPSH, PDH Online. Download from, https://pdhonline.com/ courses/m124/npsh.pdf (2012)

Chapter 5

Fluid Flow Measurements

5.1

Introduction

In fluid flow systems, it is necessary to measure and monitor fluid flow parameters such as velocity of flow, volume flow rate, and pressure drop. These parameters are measured by using suitable devices based on the applications of Bernoulli’s equation (Eq. 4.12). Pitot tubes are used in measuring point velocities in a flow field. Orifice meters and venturi meters are used in measuring volume flow rates. Pressure drops in piping systems are measured using differential manometers (discussed in detail in Chap. 2) and differential pressure gages.

5.2

Pitot Tube

A pitot tube is used in measuring fluid velocities in ducts and pipes [6, 7, 9]. As shown in Fig. 5.1, it consists of a bent stagnation tube that is inserted into the flow system. The velocity at the stagnation point is zero. This results in a higher pressure, P0, at the stagnation point. The pressure downstream of the stagnation point is the static pressure, Ps. The velocity of the fluid is proportional to the pressure difference between the stagnation pressure and the static pressure. ΔP ¼ P0  Ps

ð5:1Þ

The pressure difference can be measured by a differential manometer or by a differential pressure gage. In this case, a differential manometer shows a reading of hm. The equation for the velocity of the fluid can be obtained by applying Bernoulli’s equation (Eq. 4.12) between the stagnation point and the downstream static point, assuming negligible friction loss between the two points. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_5

137

138

5 Fluid Flow Measurements

Fig. 5.1 Pitot tube with differential manometer

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ΔP v ¼ 2g γ

ð5:2Þ

In Eq. 5.2, γ ¼ specific weight of the flowing fluid ¼ γ f. Example 5.1 The velocity of air in a duct is 80 ft/sec. The density of air is 0.075 lbm/ft3. What will be the reading of a differential water manometer connected across the pitot tube in inches of water. (Solution) The flowing fluid is air. Calculate the specific weight of air by using Eq. 1.4a.  ρg γf ¼ f ¼ gc

0:075 lbm ft3



ft 32:2 sec 2

sec 2 32:2 lbmft= lbf

 ¼ 0:075 lbf=ft3

Calculate the difference between the stagnation and static pressure by using Eq. 5.2.    ft 2 lbf 80 0:075 3 sec v γf ft   ¼ ΔP ¼ ¼ 7:453 lbf=ft2 ft 2g ð2Þ 32:2 sec 2 

2

Using the standard specific weight of water (62.4 lbf/ft3) and using Eq. 2.3, calculate the reading of the differential water manometer.

5.3 Orifice Meter

139 lbf

hm ¼

7:453 ft2 ΔP ¼ ¼ 0:1196 ft lbf γ m  γ f 62:4 3  0:075 lbf3 ft ft

0.1196 ft water  1.43 in. water.

5.3

Orifice Meter

An orifice meter [1, 2, 4, 6, 7] is used in measuring the flow rate of a fluid through a pipe. It consists of an orifice plate with a circular opening that is concentric with the pipe. The diameter of the orifice, Do, is usually half the diameter of the pipe. The streamlines shown in Fig. 5.2 indicate that the smallest flow area is downstream of the orifice plate. The section with the smallest flow area (section 2 in Fig. 5.2) is known as “vena contracta.” Pressure taps are provided far upstream (section 1) of the orifice plate and downstream at the vena contracta as shown in Fig. 5.2. Due to the reduction in area, the velocity increases at the vena contracta resulting in a decrease of pressure at the vena contracta. The velocity and hence the flow rate through the pipe are proportional to the pressure difference between sections 1 and 2 shown in Fig. 5.2. The pressure difference can be measured by a differential manometer or by a differential pressure gage [10]. To obtain the equation to calculate the volume flow rate, apply Bernoulli’s equation (Eq. 4.12) between the upstream point (1) and the vena contracta (2).





v2 P1 v21 P þ þ z1 ¼ 2 þ 2 þ z2 γ 2g γ 2g Apply the continuity equation (Eq. 3.4) to Sections “1,” “2,” and “o.”

Fig. 5.2 Orifice meter with differential manometer

ð5:3Þ

140

5 Fluid Flow Measurements

Q ¼ A1 v1 ¼ A2 v2 ¼ Ao vo

ð5:4Þ

The coefficient of contraction, Cc, is used in expressing the velocity at the vena contracta in terms of the velocity at the orifice. Using Eq. 5.4,  v2 ¼

   Ao 1 v ¼ v A2 o Cc o

ð5:5Þ

The ratio of the area at vena contracta to the area of the orifice is known as the coefficient of contraction, Cc. The equation for the volume flow rate of the fluid can be obtained by combining Eqs. 5.3, 5.4, and 5.5. Cc Ao Q ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi A2 1 A1

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi  ffi P1  P2 C c Ao 2gΔP ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2g  2ffi γ γ A2 1 A1

ð5:6Þ

The orifice coefficient, Co, also known the discharge coefficient is defined as Cc C o ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi A2 1 A1

ð5:7Þ

Combining Eqs. 5.6 and 5.7 results in Eq. 5.8, which is used in calculating volume flow rate. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffi  ffi P1  P2 2gΔP Q ¼ Co Ao 2g ¼ C o Ao γ γ

ð5:8Þ

For a sharp-edged orifice, the orifice coefficient (discharge coefficient) is typically taken as 0.61. Example 5.2 A differential mercury manometer connected across an orifice meter shows a reading of 6 cm of mercury (SGm ¼ 13.6). The pipe diameter is 254 mm, and the orifice diameter is 127 mm. Calculate the flow rate of water through the pipe in liters per second. (Solution) Convert the manometer reading into an equivalent differential pressure by using Eq. 2.4.

5.4 Venturi Meter

141

ΔP ¼ P1  P2 ¼ γ f hm ðSGm  1Þ 0 1   kN B 6 cm C ð13:6  1Þ ¼ 9:81 3 @ 100 cmA m 1m ¼ 7:42 kN=m2 ðkPaÞ Calculate the cross-section area of the orifice.

Ao ¼

πDo ¼ 4 2

ðπ Þ



127 mm

2

1000 mm 1 m

4

¼ 0:01267 m2

Calculate the volume flow rate by using Eq. 5.8. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ΔP Q ¼ Co Ao 2g γw

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1ffi u kN u    u m B7:42 m2 C ¼ ð0:61Þ 0:01267 m2 u A tð2Þ 9:81 s2 @ kN 9:81 3 m ¼ 0:02977 m3 =s

Convert the volume flow rate to liters per second.  Q¼

0:02977

m3 s



 1000 L 3 m

¼ 29:77 L=s

5.4

Venturi Meter

A venturi meter [5–8] consists of a narrow section (called the “throat”) that is concentric with the pipe. Pressure taps are provided upstream of the venture meter and at the throat as shown in Fig. 5.3. The velocity of the fluid is higher at the throat due to the constriction. As per Bernoulli’s equation, higher velocity results in lower pressure at the throat. The velocity and hence the flow rate through the pipe are proportional to the pressure difference between the pipe and the throat. Applying Bernoulli’s equation (Eq. 4.12)

142

5 Fluid Flow Measurements

Fig. 5.3 Venturi meter with a differential manometer

between upstream section (1) and the throat (2) and simplification results in the following equations.





v2 P1 v21 P þ þ z1 ¼ 2 þ 2 þ z2 γ 2g γ 2g

ð5:9Þ

Apply the continuity equation (Eq. 3.4) at Sections “1” and “2.” Q ¼ A1 v1 ¼ A2 v2   A2 Therefore, v1 ¼ v A1 2

ð5:10Þ ð5:11Þ

Combining Eqs. 5.9, 5.10, and 5.11 and applying the velocity correction factor (coefficient of velocity, Cv) result in the following equation for the volume flow rate of the fluid. C v A2 Q ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2ffi 1  AA21

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    ffi P1 P2 Cv A2 ΔP ffi 2g 2g  ¼ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  2 γ γ γ 1  AA21

ð5:12Þ

The ratio of the throat diameter to the pipe diameter is represented by β. Hence, D β¼ 2 D1

and

 2  2  4 ðπ=4ÞD2 2 A2 D2 ¼ ¼ ¼ β4 A1 D1 ðπ=4ÞD1 2

ð5:13Þ

5.4 Venturi Meter

143

Combining Eqs. 5.12 and 5.13 results in Eq. 5.14, which is commonly used for calculating the volume flow rate in a venturi meter. Cv A2 Q ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  β4

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ΔP 2g γ

ð5:14Þ

In Eq. 5.14, the velocity coefficient, Cv, is usually taken as 0.98. Example 5.3 A differential pressure gage connected across a venturi meter shows a reading of 0.90 psi. The pipe diameter is 12 in, and the throat diameter is 6 in. Determine the flow rate of water through the pipe in gpm. (Solution) Calculate the ratio of the throat diameter to the pipe diameter. β¼

D2 6 in ¼ 0:50 ¼ D1 12 in

Calculate the cross-section area of the throat. 2

A2 ¼

πD2 2 π ð0:50 ftÞ ¼ ¼ 0:1963 ft2 4 4

Calculate the volume flow rate using Eq. 5.14. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ΔP 2g γ vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  1 0 u u lbf 144 in2    u  0:90 2 B C ð0:98Þ 0:1963 ft2 u in ft2 C u2 32:2 ft B pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 @ A t 4 lbf sec 1  0:5 62:4 3 ft

Cv A2 Q ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi 1  β4

¼ 2:298 ft3 = sec ðcfsÞ Convert the volume flow rate to gallons per minute.    ft3 449 gpm Q ¼ 2:298 ¼ 1032 gpm sec ft3 = sec

144

5.5

5 Fluid Flow Measurements

Orifice Meter and Venturi Meter Comparison

The advantage of orifice meter over a venturi meter is its simple construction and ease of dismantling for servicing and maintenance purposes [3, 5]. However, due to the sudden constriction at the orifice plate, the pressure recovery is extremely poor in an orifice meter resulting in a higher permanent pressure loss. In contrast, a venturi meter has gradually convergent and divergent sections into and out of the throat area resulting in a lower permanent pressure loss. The permanent pressure loss in an orifice meter depends on the ratio of the orifice diameter to the pipe diameter, o represented by β. β ¼ D D1 . Depending on the value of β, the permanent pressure loss in an orifice meter can range from 50% to 90% of the total pressure loss across the meter. In a venturi meter, the permanent pressure loss is usually about 10% of the total pressure loss.

5.5.1

Calculation of Permanent Pressure Loss in an Orifice Meter

The permanent pressure loss in an orifice meter can be calculated by using the following formula:   ΔPPerm ¼ 1  β2 ΔPTotal

ð5:15Þ

Example 5.4 Calculate the permanent pressure loss in the orifice meter used in Example 5.2. (Solution) From Example 5.2, the total pressure drop across the orifice meter is 7.42 kN/m2. Calculate the ratio of the orifice diameter to the pipe diameter. β¼

Do 127 mm ¼ 0:5 ¼ 254 mm D1

Calculate the permanent pressure drop by using Eq. 5.15. 





ΔPPerm ¼ 1  β ΔPTotal ¼ 1  0:5 2

2





 kN 7:42 2 m

¼ 5:565 kPa Thus, the permanent pressure drop is 75% of the total pressure drop.

Practice Problems Solutions

145

Practice Problems Practice Problem 5.1 A differential manometer using mercury (SG ¼ 13.6) as the manometric fluid is connected to the pressure taps of a pitot tube. The manometer shows a reading of 20 cm. Calculate the velocity of water flowing in the pipe in m/s. Practice Problem 5.2 A differential pressure gage is connected between the upstream point and the vena contracta of an orifice meter attached to a 12 in. pipe. The diameter of the sharpedged orifice is 6 in. The flow rate of water in the pipe is measured to be 200 gpm. Determine the reading of the differential pressure gage in psi. Practice Problem 5.3 In Example 5.3, what will be the reading of a differential mercury (SG ¼ 13.6) manometer in cm Hg, if it is used in place of a differential pressure gage?

Practice Problems Solutions Practice Problem 5.1 (Solution) Convert the manometer reading to the difference between the stagnation pressure and the static pressure by using Eq. 2.4. 0 1   kN B 20 cm C ΔP ¼ P0  Ps ¼ γ f hm ðSGm  1Þ ¼ 9:81 3 @ ð13:6  1Þ 100 cmA m 1m ¼ 24:72 kN=m2 ðkPaÞ Calculate the velocity of water by using Eq. 5.1. vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi  ffi u  24:72 kN u  ΔP m 2 m ¼ 7:03 m=s ¼ tð2Þ 9:81 2 v ¼ 2g γf s 9:81 kN m3 Practice Problem 5.2 (Solution) Calculate the cross-section area of the orifice.

146

5 Fluid Flow Measurements



πDo 2 ðπÞ 121 ftin Ao ¼ ¼ 4 4

6 in

2 ¼ 0:1963 ft2

Calculate the pressure difference by using Eq. 5.8. 2   γw Q C o Ao 2g 1 0 12 0 lbf 200 gpm 62:4 3 3 449 gpm=ðft = sec Þ C ft A B ¼@ @ A ft 0:61  0:1963 ft2 2  32:2 2 sec ¼ 13:41 lbf=ft2 

ΔP ¼

Convert the pressure difference to psi.    lbf 1 ft2 ΔP ¼ 13:41 2 ft 144 in2 2 ¼ 0:0931 lbf=in ðpsiÞ Practice Problem 5.3 (Solution) From Example 5.3, the reading of the differential pressure gage is 0.9 lbf/in2. Using the standard specific weight of water (62.4 lbf/ft3) and using Eq. 2.3, calculate the reading of the differential mercury manometer.

hm ¼

   lbf 144 in2 0:9 in 2 2 ft

ΔP ¼ 0:1648 ft ¼ γ f ðSGm  1Þ 62:4 lbf3 ð13:6  1Þ ft

Convert the manometer reading to cm Hg, hm ¼ 5.02 cm Hg. Note: A handy reference for conversion factors is https://www.unitconverters.net/ length-converter.html

References 1. Ashlin: What Is Orifice Meter and Uses of Orifice Meter?, Automation Forum.co, (2020). https://automationforum.co/what-is-orifice-meter-and-what-is-the-use-of-orifice-meter/ 2. Instrumentation Tools: How an Orifice Meter Measures Flow, instrumentationtools.com, (2016). Download from https://instrumentationtools.com/how-a-orifice-measures-flow/ 3. Instrumentation Tools: Comparison of Venturi and Orifice Flow Meters, instrumentationtools. com, (2016). Download from https://instrumentationtools.com/comparison-venturi-orificeflow-meter/

References

147

4. Kumar, Amrit: Orifice Meter—Definition, Working, Formula, Advantages, and Disadvantages. The Mechanical Engineering.com. Download from https://themechanicalengineering.com/ orifice-meter/ 5. Kumar, A.: Venturi Meter—Definition, Parts, Working Principles, Formula, Advantages, and Disadvantages, Learn Mechanical, (2020). Download from https://learnmechanical.com/ venturi-meter/ 6. Mitchell, J.W.: Fox and McDonald’s Introduction to Fluid Mechanics, 10th edn. Wiley (2020) 7. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson (2015) 8. Reader-Harris, M.J.: Venturi Meters, Thermopedia, (2011). Download from https:// thermopedia.com/content/1241/ 9. Sivaranjith: Pitot Tube—Working, Advantages, Disadvantages, Automation Forum, (2018). Download from https://automationforum.co/pitot-tube-working-advantages-disadvantages/ 10. Visaya Solutions: How to Calculate Flow Rate for an Orifice Plate With a Differential Pressure Transmitter, Automation 24, (2019). Download from https://visaya.solutions/en/qa/flow-rangeorifice-plate

Chapter 6

Fundamentals of Compressible Flow

6.1

Introduction

Flow of gases through pipes and ducts [1, 2, 4] is also known as compressible flow since the density of the fluid is not constant during the flow. This is because the density of a gas is sensitive to pressure (unlike liquids) and hence keeps changing as pressure drop occurs during the flow. In contrast to incompressible flow, density and temperature are additional variables in compressible flow. Compressible flow is also known as variable density flow, and the subject matter is sometimes referred to as gas dynamics [1, 3]. At low velocities, gas flow can be considered incompressible. The criteria for gas flow to be incompressible is lower ranges of Mach number (defined as the ratio of the velocity of the fluid to the speed of sound). The density variations in gas flow are usually small (less than 5%), when the Mach number is less than 0.30 [7]. Compressible flow is commonly encountered in high-speed aircraft, in gas pipelines, and in flow of gases through nozzles and orifices, where the constricted flow area can result in high fluid velocities [1–3]. It also becomes relevant in the study of the impact of high-speed winds in tall structures. The study of compressible fluid flow requires the knowledge and application of thermodynamic principles in addition to the knowledge base of fluid dynamics.

6.2

Continuity Equation for Compressible Flow

The law of conservation of mass must always be satisfied regardless of the type of flow or flow conditions. The continuity equation, applied to fluid flow, is based on the principle of conservation of mass, that is, the mass flow rate of the fluid remains constant throughout the flow field. In the case of liquid flow, it is reasonable to assume constant density, and hence the volume flow rate is also conserved. However, in gas flow since the density can vary in the flow field, the volume flow also © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_6

149

150

6 Fundamentals of Compressible Flow

varies (density is mass per unit volume and hence volume flow rate is mass flow rate divided by the density). Because of the variation of density with flow, the continuity equation for gas flow [4] must account for variation in density as shown in Eq. 6.1 (also the same as Eq. 3.3). m_ ¼ ρ1 A1 v1 ¼ ρ2 A2 v2

6.2.1

ð6:1Þ

Calculation of Density of Gases

The density of a gas can be calculated using the ideal gas law (same as Eq. 13.12 and reproduced here for reference) at moderate pressures and temperatures. ρ¼

m P ¼ V RT

ð6:2Þ

However, the ideal gas law does not give accurate density values at very high pressures and also at very low temperatures. In such situations, it is necessary to use the compressibility factor as a correction factor for the application of the ideal gas law. The modified ideal gas equation includes the compressibility factor, Z, as shown in Eq. 6.3. The compressibility factor can be obtained from the generalized compressibility chart given in the thermodynamics section and reproduced here for reference (Fig. 6.1). ρ¼

6.3

m P ¼ V ZRT

ð6:3Þ

Mach Number and Its Significance in Compressible Flow

Consider the flow of gas from a source such as a compressor into a pipe. At steady state, the gas is already moving in the pipe at a constant velocity. As more gas gets into the pipe, it pushes the gas existing in the pipe and the pressure pulse results in a compression wave. The infinitesimally small pressure wave (disturbance) propagates through a fluid at the speed of sound [1, 7, 8]. If the fluid is incompressible, the pressure pulse will not have any effect on the fluid density since the density remains constant. In other words, the fluid adjacent to the pressure pulse will not get compressed, and the compression wave is felt instantaneously throughout the pipe. If the fluid is compressible, then the pressure wave affects the fluid density by compressing (or piling up) the fluid layers adjacent to the wave. Thus, the speed

6.3 Mach Number and Its Significance in Compressible Flow

151

Fig. 6.1 Generalized compressibility chart

of sound in any medium depends on the compressibility of the medium [1, 3]. The lower the compressibility of the medium, the higher the speed of sound in the medium. In an ideal incompressible medium, the speed of sound approaches infinity. Solids have very small compressibility, whereas gases have relatively high compressibility. As a result, the speed of sound in solids will be typically 10–15 times higher than in a gas at ambient temperature. The local speed of sound in a gas depends only on the absolute temperature of the gas, and it is calculated by using Eqs. 6.4a and 6.4b. S I Units USCS Units

pffiffiffiffiffiffiffiffiffi kRT pffiffiffiffiffiffiffiffiffiffiffiffi c ¼ kRTgc

c¼

ð6:4aÞ ð6:4bÞ

In Eqs. 6.4a and 6.4b, k is the ratio of specific heats of the gas (k ¼ cp/cv), R is the individual gas constant, and T is the absolute temperature of the gas. Example 6.1 Determine the speed of sound in hydrogen at 30  C. (Solution) Calculate the absolute temperature of hydrogen using Eq. 13.3.

152

6 Fundamentals of Compressible Flow

T ¼ 273 þ 30 C ¼ 303 K Calculate the individual gas constant of hydrogen using Eq. 3.16. R¼

kJ R 8:314 kmolK ¼ ¼ 4:157 kJ=kg  K M 2 kg kmol

From Table 13.2, the ratio of specific heats for hydrogen is k ¼ 1.4. Substitute the known values into Eq. 6.4a to obtain the speed of sound in hydrogen at 30  C. pffiffiffiffiffiffiffiffiffi c ¼ kRT ¼ qffiffiffiffi Note:

J kg

¼

qffiffiffiffiffiffi N:m kg

ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   J ð1:4Þ 4:157  103 ð303 KÞ ¼ 1328 m=s kg  K rffiffiffiffiffiffiffiffiffiffiffi

¼

kg:m m s2

kg

¼

qffiffiffiffiffi m2 s2

¼ m=s.

The Mach number, Ma, is the ratio of the average fluid velocity to the local speed of sound in the gas. Ma ¼

v c

ð6:5Þ

Since the pressure disturbance in a gas propagates at the speed of sound, it follows that the Mach number has a significant impact on the nature of flow. The physical significance of the Mach number can be summarized as follows [1, 2, 8]: 1. If Ma < 0.3, then the flow can be considered incompressible. 2. The Mach number is proportional to the ratio of kinetic energy of the fluid to the internal energy of the fluid as shown here. Ma 

v v v2 v2 v2 kinetic energy  α  pffiffiffiffiffiffiffiffiffi α α  α c kRT kðk  1Þu k cp  cv T k ðk  1Þcv T kRT

In the preceding equation, u is the specific internal energy of the fluid. The Mach number is the single most important parameter in understanding and analyzing compressible flows [5]. If the velocity of the fluid is greater than the velocity of sound (Ma > 1.0), the flow is characterized as supersonic flow. If the velocity of the fluid is less than the velocity of sound (Ma < 1.0), the flow is characterized as subsonic flow. Flows with Ma ¼ 1.0 are characterized as sonic flows. Example 6.2 An aircraft is flying at an altitude of 10,700 m where the temperature is 220 K. The Mach number of the aircraft is 0.93. Calculate:

6.4 Isentropic Gas Flow

153

A. The speed of the aircraft. B. The Mach number at sea level where the temperature is 20  C. (Solution) A. Calculate the individual gas constant of air using Eq. 3.16.

R¼

kJ R 8:314 kmolK ¼ ¼ 0:287 kJ=kg  K M 29 kg kmol

From Table 13.2, the ratio of specific heats for air is k ¼ 1.4. Calculate the speed of sound in air at 220 K using Eq. 6.4a. pffiffiffiffiffiffiffiffiffi c ¼ kRT ¼

ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   J 3 ð1:4Þ 0:287  10 ð220 KÞ ¼ 297 m=s kg  K

Calculate the speed of the aircraft using Eq. 6.5. v ¼ Ma  c ¼ 0:93  297

m ¼ 276 m=s s

B. The speed of the aircraft remains the same. However, the speed of sound at sea level will be different due to the change in temperature. At sea level, the absolute temperature is TSL ¼ 273  + 20  C ¼ 293 K. Calculate the speed of sound in air at 293 K using Eq. 6.4a. pffiffiffiffiffiffiffiffiffiffiffiffiffi c ¼ kRT SL ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi   J ð293 KÞ ¼ 343 m=s ð1:4Þ 0:287  103 kg  K

Calculate the Mach number of the aircraft at sea level using Eq. 6.5. Ma ¼

6.4

v 276 ms ¼ ¼ 0:80 c 343 ms

Isentropic Gas Flow

During isentropic flow of an ideal gas in a duct, pipe, or nozzle, the entropy remains constant [1, 7]. Since an isentropic process is a reversible adiabatic process, there is no heat transfer between the gas and the surroundings. In Chap. 13, the following P-V-T relationships were presented for isentropic processes.

154

6 Fundamentals of Compressible Flow

P2 ¼ P1

ρ2 ¼ ρ1

 k  k ρ2 v1 1 ¼ since ρ ¼ v v2 ρ1 k  k1 P2 T2 ¼ P1 T1

 1k P2 ¼ P1

!1k   1 k  k1 T2 T 2 k1 ¼ T1 T1

ð6:6Þ ð6:7aÞ

ð6:7bÞ

Eq. 6.6 is particularly useful for compressible fluid flow where the density varies in the flow field.

6.4.1

Application of the Steady Flow Energy Equation for Isentropic Flows

The energy relationship between two points in the isentropic flow field (no heat transfer, no work, no elevation change, only enthalpy and kinetic energy terms need to be considered in the steady flow energy equation) can be obtained by energy balance between the two points. Energy in at reference point 1 ¼ Energy out at reference point 2 h1 þ

v1 2 v2 ¼ h2 þ 2 2 2

ð6:8Þ

Example 6.3 Air flows in an adiabatic divergent nozzle at a velocity of 560 ft/sec. The temperature of air at this location is 80  F. At the divergent section, the air speed reduces to 210 ft/s. Calculate the temperature of air at the divergent section. (Solution) The kinetic energy reduces when the air speed reduces. However, since the total energy is conserved, the enthalpy and hence the temperature must increase to compensate for the loss in kinetic energy. Let reference point 1 represent the normal section of the nozzle and reference point 2 represent the divergent section of the nozzle. From Table 13.2, the specific heat of air at constant pressure is cp ¼ 0.24 Btu/ lbm ‐  R. The absolute temperature at reference point 1 is T1 ¼ 460  + 80  F ¼ 540  R. Apply Eq. 6.8 between reference points 1 and 2, simplify, and substitute the known values to obtain the temperature of air at the divergent section.

6.4 Isentropic Gas Flow

h1 þ

155

v21 v2 v2  v22 v2  v22 ¼ h2 þ 2 ) h2  h1 ¼ 1 ) cp ðT 2  T 1 Þ ¼ 1 ) 2 2 2 2

T2 ¼ T1 þ

v21  v22 2cp



   ft 2 ft 2 560 sec  210 sec     ¼ 540 R þ   Btu 778 ft‐lbf lbm‐ft 32:2 ð2Þ 0:24 lbm‐ R Btu lbf‐ sec 2   ¼ 562:4 R ð102:4 FÞ 

6.4.2

Stagnation-Static Relationships

Consider a compressed gas in a reservoir. Since the velocity of the gas is zero in the reservoir, the conditions in the reservoir are referred to as stagnation (zero velocity) conditions. Stagnation conditions (P0,T0,ρ0, s0,h0) are identified with subscript “0,” and they can occur at any point in the flow field. If the reservoir is connected to a pipeline, the gas starts flowing in the pipeline and the conditions within the pipeline flow field are static conditions (P, T, ρ, s, h), and they are represented as variables without subscripts. Apply Eq. 6.8 between the stagnation point (v0 ¼ 0) and another point in the isentropic flow field and use the relationship h ¼ cpT for an ideal gas to obtain the relationship in Eq. 6.9 [1, 3, 7].

T0 v2 ¼1þ T 2cp T

ð6:9Þ

The preceding equation can be written in terms of Mach number and simplified using the following relationships to obtain Eq. 6.10. pffiffiffiffiffiffiffiffiffi v2 kR ¼ Ma2 , c ¼ kRT , cp ¼ 2 k1 c   T0 k1 Ma2 ¼1þ 2 T

ð6:10Þ

156

6 Fundamentals of Compressible Flow

Equation 6.10 can be combined with the P-V-T relationships for isentropic processes (Eqs. 6.6, 6.7a, and 6.7b) to obtain additional stagnation-static relationships for isentropic flow [7, 8]. k  

k1 P0 k1 Ma2 ¼ 1þ 2 P 1  

k1 ρ0 k1 2 Ma ¼ 1þ 2 ρ

ð6:11Þ ð6:12Þ

Example 6.4 Air flows from a reservoir maintained at 75  C into a pipe. Just before the pipe exit, the Mach number is determined to be 1.7. At this location, determine: A. The temperature of air B. The velocity of air (Solution) The conditions within the reservoir are the stagnation (zero velocity) conditions. Calculate the absolute temperature of the reservoir using Eq. 13.3. T 0 ¼ 273 þ 75 C ¼ 348 K Calculate the stagnation temperature to static temperature ratio using Eq. 6.10, and then calculate the static temperature at the location as shown.     T0 k1 1:4  1  2  Ma2 ¼ 1 þ 1:7 ¼ 1:578 ) ¼1þ 2 2 T T¼

T0 348 K ¼ 220:5 K ð52:5 CÞ ¼ 1:578 1:578

Alternative Solution Using Isentropic Compressible Flow Functions for Air Determine the ratio T/T0 from the table as shown here.

6.4 Isentropic Gas Flow

157

T ¼ 0:6337  T 0 ¼ ð0:634Þð348 KÞ ¼ 220:6 K ð52:4∘ CÞ B. Calculate the velocity of air using Eq. 6.9. (Note: From Table 13.2, for air, cp ¼ 1 kJ/kg.K ¼ 1000 J/kg.K) rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   T 0 v¼ 2cp T 1 T ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s  J ¼ 2  1000  220 K ð1:578  1Þ kg  K ¼ 504:3 m=s

6.4.3

Isentropic Flow with Area Changes

Any type of fluid flow must satisfy both the continuity and momentum equations. In addition, compressible flows must satisfy the energy equation and the relevant P-V-T relationships. Applying the preceding principles to isentropic flows with area changes, it can be shown that for subsonic flows (Ma < 1.0), decrease in flow area in the flow direction results in increase in velocity and for supersonic flows (Ma > 1.0), increase in flow area in the flow direction is required for the velocity to increase [1, 7, 8]. Compressible flows are often accelerated or decelerated through a nozzle or diffuser. The flow is sonic when Mach number is one and the point at which sonic flow occurs known as the critical point or throat [7, 8]. The throat area is represented by the symbol A*. The ratio between the critical temperature and the stagnation temperature can be obtained by substituting Ma ¼ 1 in Eq. 6.10 as shown here.   T0 k1 Ma2 ¼1þ 2 T

Ma¼1, T¼T

!

  T 2 ¼ kþ1 T0

ð6:13Þ

Substituting k ¼ 1.4 for air into the preceding equation, the critical temperature ratio for air is   T 2 ¼ 0:8333 ¼ 1:4 þ 1 T0

ð6:14Þ

Equation 6.14 implies that for air, sonic flow occurs when the temperature in the duct reaches 83% of the stagnation temperature. Similar critical ratios can be obtained by using isentropic P-V-T relationships as shown.

158

6 Fundamentals of Compressible Flow k   k1 1:4  k  1:41 T 2 k1 2 ¼ ¼ ¼ 0:5283 T0 kþ1 1:4 þ 1 1   k1 1  1  1:41 ρ T 2 k1 2 ¼ ¼ ¼ ¼ 0:6339 kþ1 1:4 þ 1 ρ0 T0

P ¼ P0

ð6:15Þ ð6:16Þ

Once sonic velocity is achieved in a duct or a nozzle throat, the mass flow rate remains constant, and it is at its maximum value, and the flow condition is referred to as choked flow [6]. The mass flow rate for compressible flow can be written in terms of Mach number as shown here. m_ ¼ ρAv ¼



   pffiffiffiffiffiffiffiffiffi P P AðMaÞðcÞ ¼ AðMaÞ kRT RT RT rffiffiffiffiffiffi! k ¼ PAðMaÞ RT

ð6:17Þ

In Eq. 6.17, substitute for T from Eq. 6.10 and for P from Eq. 6.11 to obtain the following equation for compressible flow mass flow rate [6, 8].   rffiffiffiffiffiffiffiffi 2ðkþ1 1kÞ k k1 2 m_ ¼ ðP0 AÞðMaÞ Ma 1þ RT 0 2

ð6:18Þ

For sonic flow, Ma ¼ 1.0 and the area at the sonic point is A*. Substitute Ma ¼ 1.0 into the preceding equation.   kþ1 rffiffiffiffiffiffiffiffi k k  1 2ð1kÞ m_ ¼ ðP0 A Þ 1þ RT 0 2 

ð6:19Þ

Equate the right-hand sides of Eqs. 6.18 and 6.19 and simplify to obtain the following equation for the area ratio between the throat area and the area at any location in the flow field [1, 5, 7].

kþ1 A 1 2 þ ðk  1ÞMa2 2ðk1Þ ¼ kþ1 A Ma

ð6:20Þ

Example 6.5 Air flows through a convergent-divergent nozzle. When the Mach number is 0.88, the nozzle area is 225 in2. The initial pressure and temperature of air are 75 psia and 250  F, respectively. Determine: A. The throat area required to achieve sonic velocity B. The maximum mass flow rate of air when the flow is choked

6.4 Isentropic Gas Flow

159

(Solution) A. Solve Eq. 6.20 for the throat area, A*, and substitute the known values to obtain the throat area.

kþ1 A 1 2 þ ðk  1ÞMa2 2ðk1Þ ¼ ) kþ1 A Ma

kþ1 2 þ ðk  1ÞMa2 2ð1kÞ A ¼ ðA  MaÞ kþ1

1:4þ1   2 þ ð1:4  1Þ0:882 2ð11:4Þ 2 ¼ 225 in  0:88 1:4 þ 1 ¼ 222 in2 Alternative Solution Using Isentropic Compressible Flow Functions for Air Determine the ratio A/A* from the table as shown here.

A A 225 in2  ¼ ¼ 222 in2  ¼ 1:0129 ) A ¼ 1:0129 1:0129 A B. Calculate the absolute value of the initial temperature using Eq. 13.2. T 0 ¼ 460 þ 250 F ¼ 710 R Calculate the individual gas constant of air using Eq. 13.16. R¼

ft‐lbf R 1545 lbmol R ¼ ¼ 53:28 ft‐lbf=lbm  R lbm M 29 lbmol

Calculate the maximum mass flow rate using Eq. 6.19 for choked flow. Note that for reconciling the units in USCS, the conversion constant gc needs to be used in the formula as shown.

160

6 Fundamentals of Compressible Flow

rffiffiffiffiffiffiffiffi!   kþ1 kgc k  1 2ð1kÞ  m_ ¼ ðP0 A Þ 1þ 2 RT 0 0vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 u lbm‐ft u   B  u 1:4  32:2 lbf‐ sec 2 C lbf  2 Bu C ¼ 75 2 222 in Bu C @t A ft‐lbf in  53:28 R  710  lbm‐ R    1:4þ1 1:4  1 2ð11:4Þ 1þ 2 ¼ 332:62 lbm= sec Alternate Solution Calculate the sonic temperature using Eq. 6.14. T  ¼ 0:8333T 0 ¼ 0:8333  710 R ¼ 591:6 R Calculate the sonic pressure using Eq. 6.15. P ¼ 0:5283P0 ¼ 0:5283  75 psia ¼ 39:62 psia Since the flow is choked, Ma ¼ 1.0. Calculate the sonic velocity using Eq. 6.4b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ c ¼ kRT  gc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    ffi ft‐lbf lbm‐ft ð591:6 RÞ 32:2 ¼ ð1:4Þ 53:28 lbm‐ R lbf‐ sec 2 ¼ 1192 ft= sec Determine the individual gas constant for air using Eq. 13.16. psia‐ft3

R¼

R 10:73 lbmol‐ R ¼ ¼ 0:3700 psia‐ft3 =lbm  R lb M 29 lbmol

   ft 1 ft2 2 222 in ð 39:62 psia Þ 1192  sec P  v A  144 in2   m ¼ ρ v A ¼  ¼ 3 RT psia‐ft ð591:6 RÞ 0:37 lbm‐ R ¼ 332:62 lbm= sec

6.5 Adiabatic Compressible Flow with Friction Loss

6.5

161

Adiabatic Compressible Flow with Friction Loss

The Darcy friction factor can also be used for compressible flow. Let L* be the length of the duct required for the flow to develop from given Mach number to sonic conditions (Ma ¼ 1), and if f is the average friction factor between L ¼ 0 to L ¼ L*, then the following equation can be obtained by applying the principles of conservation of mass, momentum, and energy [1, 7, 8].  

ðk þ 1ÞMa2 f L 1  Ma2 kþ1 ln ¼ þ 2k D kMa2 2 þ ðk  1ÞMa2

ð6:21Þ

If ΔL is the length between the locations of Ma1 and Ma2, then the following relationship can be written. f

      ΔL fL fL  ¼ D D 1 D 2

ð6:22Þ

The following equations can be written for the ratios between flow variables along the duct [7].

12 P 1 kþ1 ¼ P Ma 2 þ ðk  1ÞMa2  2 T c kþ1 ¼  ¼ T c 2 þ ðk  1ÞMa2

ð6:23Þ ð6:24Þ

Example 6.6 illustrates the use of the preceding equations in adiabatic compressible flow with friction loss. Example 6.6 Air flows through a circular duct (ID ¼ 102 mm). The pressure, temperature, and Mach number at Section 1 of the duct are: P1 ¼ 800 kPa, T 1 ¼ 147 C, Ma1 ¼ 0:17: The average friction factor for flow in this duct can be taken as 0.023. Determine the Mach number, pressure, and temperature 35 m downstream from Section 1. (Solution) Substitute the known values into Eq. 6.21, and calculate the parameter f L =D at Section 1 as shown.

162

6 Fundamentals of Compressible Flow



f L D



 

ðk þ 1ÞMa1 2 1  Ma1 2 kþ1 ln ¼ þ 2k kMa1 2 2 þ ðk  1ÞMa1 2 1

  ð1:4 þ 1Þ0:172 1  0:172 1:4 þ 1 ln ¼ þ 2  1:4 1:4  0:172 2 þ ð1:4  1Þ0:172 ¼ 21:11

Substitute the known values into Eq. 6.22, and obtain the value for the parameter f L =D at Section 2 as shown. !       fL fL ΔL 35 m ¼ 21:11  ð0:023Þ 102 ¼ 13:22 ¼ f D D 2 D 1 1000 m Substitute Ma2 for the Mach number into Eq. 6.21, and solve for Ma2 using the   preceding value obtained for f L =D 2. Note: For subsonic flow, as is the case here, Mach number increases downstream of the initial section.    

ðk þ 1ÞMa2 2 k¼1:4 fL 1  Ma2 2 kþ1 ln ¼ þ ¼ 13:22 ! 2 2 2k D 2 kMa2 2 þ ðk  1ÞMa2  

1 2:4Ma2 2 0:7143  1 þ ð0:8571Þ ln ¼ 13:22 2 þ 0:4Ma2 2 Ma2 2 After solving the preceding equation for Ma2, Ma2 ¼ 0.2082. Derive an expression for the pressure at Section 2 by using pressure ratios involving sonic pressures at Sections 1 and 2, which are equal. P2 ¼ P1

          P1 P2 P2 P1 P2 ¼ P ð1Þ 1 P1 P1  P2  P1 P2 

Use Eq. 6.23 to obtain the values for the ratios between sonic pressure and static pressure at Sections 1 and 2 as shown here.

12

12 P1 1 kþ1 1 1:4 þ 1 ¼ ¼ ¼ 6:425 0:17 2 þ ð1:4  1Þ0:172 P1  Ma1 2 þ ðk  1ÞMa1 2 P  1 ) 1 ¼ ¼ 0:1556 6:425 P1

12

12 P2 1 kþ1 1 1:4 þ 1 ¼ ¼ ¼ 5:239 0:2082 2 þ ð1:4  1Þ0:20822 P2  Ma2 2 þ ðk  1ÞMa2 2 Substitute the preceding results into the equation derived for P2.

Practice Problems

163

    P P2 P2 ¼ P1 1 ð1Þ ¼ ð800 kPaÞð0:1556Þð1Þð5:239Þ ¼ 652 kPa P1 P2  In a similar manner, obtain the temperature at Section 2 by using temperature ratios. The absolute temperature at Section 1 is T1 ¼ 273  + 147  C ¼ 420 K T2 ¼ T1

          T1 T2 T2 T1 T2 ¼ ð 420 K Þ ð 1Þ T1 T 1 T 2 T1 T 2

Use Eq. 6.24 to obtain the values for the ratios between sonic temperature and static temperature at Sections 1 and 2 as shown here. T1 kþ1 1:4 þ 1 T  ¼ ¼ 1:193 ) 1 ¼ 0:8382  ¼ 2 2 T1 T1 2 þ ðk  1ÞMa1 2 þ ð1:4  1Þ0:17 T2 kþ1 1:4 þ 1 ¼ ¼ ¼ 1:190 T 2  2 þ ðk  1ÞMa2 2 2 þ ð1:4  1Þ0:20822 Substitute the preceding results into the equation for T2. T 2 ¼ ð420 KÞð0:8382Þð1Þð1:190Þ ¼ 419 K

Practice Problems Practice Problem 6.1 Air flows in a duct at a velocity of 500 ft/s. The temperature of air is 80  F. Determine: A. The speed of sound in air B. The Mach number and the type of flow Practice Problem 6.2 Air flows into an adiabatic, frictionless duct from a large supply tank where the pressure is 70 psia and the temperature is 255  F. At a certain location downstream, the pressure reduces to 50 psia. At this location, determine: A. The Mach number of air B. The temperature, density, and velocity of air Practice Problem 6.3 A pitot tube is inserted into a duct where air is flowing. A differential manometer connected to the pressure taps of the pitot tube shows a reading of 10 cm of water.

164

6 Fundamentals of Compressible Flow

The static pressure at the location of the pitot tube is 4.5 kPa. At this location, determine: A. The Mach number B. The state of flow (subsonic or supersonic) Practice Problem 6.4 A propulsion system has a combustion chamber that produces a gas with the properties, k ¼ 1.3 and M ¼ 22 lbm/lbmol. The pressure and temperature in the combustion chamber are 800 psia and 4600  F. The Mach number at the exit of the nozzle of the propulsion system is 9. Determine: A. The velocity, density, pressure, and temperature of the gas at the nozzle throat B. The velocity, density, pressure, and temperature of the gas at the nozzle exit

Solutions to Practice Problems Practice Problem 6.1 (Solution) Calculate the absolute temperature of air using Eq. 13.2. T ¼ 460 þ 80 F ¼ 540 R Calculate the individual gas constant of air using Eq. 13.16. R¼

ft‐lbf R 1545 lbmol R ¼ ¼ 53:28 ft‐lbf=lbm  R lbm M 29 lbmol

From Table 13.2, the ratio of specific heats for air is k ¼ 1.4. Substitute the known values into Eq. 6.4b to obtain the speed of sound in air at 80  F. pffiffiffiffiffiffiffiffiffiffiffiffi kRTgc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi    ffi ft‐lbf lbm‐ft ð540 RÞ 32:2 ¼ ð1:40Þ 53:28 lbm  R lbf‐ sec 2

c¼

¼ 1139 ft=s Calculate the Mach number using Eq. 6.5.

Solutions to Practice Problems

165

Ma ¼

ft 500 sec v ¼ 0:4390 ¼ ft c 1139 sec

Since Ma < 1, the flow is subsonic. Practice Problem 6.2 (Solution) Since the velocity in the supply tank is practically zero, the pressure and temperature in the supply tank are the stagnation pressure (P0 ¼ 70 psia) and the stagnation temperature (T0 ¼ 255  F  460 + 255  F ¼ 715  R). A. Determine the Mach number where the static pressure is P ¼ 50 psia by solving Eq. 6.11 for Ma, and then substitute the known quantities. k    

k1  k1 P0 k1 k1 P k Ma2 Ma2 ¼ 0 ¼ 1þ )1þ ) 2 2 P P vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! ffi u   k1 u 2  70 psia1:41 1:4 k 2 P0 t 1 ¼ 1 Ma ¼ k1 1:4  1 50 psia P

¼ 0:71 B. Determine the static temperature using Eq. 6.10 (or the table for isentropic compressible flow functions).      T0 k1 1:4  1  Ma2 ¼ 1 þ 0:712 ¼ 1:1008 ) ¼1þ 2 2 T T¼

T0 715 R ¼ 649:5 R ð189:5 FÞ ¼ 1:1008 1:1008

Determine the individual gas constant for air using Eq. 13.16. psia‐ft3

R 10:73 lbmol‐ R R¼ ¼ ¼ 0:3700 psia‐ft3 =lbm  R lb M 29 lbmol Determine the static density using the ideal gas law (Eq. 13.12). ρ¼

P 50 psia  ¼ 0:2081 lbm=ft3 ¼ psia‐ft3 RT  0:37 lbm‐ R ð649:5 RÞ

Calculate the individual gas constant of air using Eq. 13.16.

166

6 Fundamentals of Compressible Flow

R¼

ft‐lbf R 1545 lbmol R ¼ 53:28 ft‐lbf=lbm  R ¼ lbm M 29 lbmol

From Table 13.2, the ratio of specific heats for air is k ¼ 1.4. Substitute the known values into Eq. 6.4b to obtain the speed of sound in air at the  static location where T ¼ 649.5 R. pffiffiffiffiffiffiffiffiffiffiffiffi kRTgc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi     ft‐lbf lbm‐ft  ð649:5 RÞ 32:2 ¼ ð1:40Þ 53:28 lbm  R lbf‐ sec 2

c¼

¼ 1249 ft=s Determine the velocity by using the Mach number and Eq. 6.5. v ¼ Ma  c ¼ 0:71  1249

m ¼ 887 m=s s

Alternative Solution Using Isentropic Compressible Flow Functions for Air psia Calculate the ratio PP0 ¼ 50 70 psia ¼ 0:7143. Using this pressure ratio, determine the Mach number, the temperature ratio, and the density ratio from the table as shown, and then calculate the static temperature and density.

P 50 psia T ρ ¼ ¼ 0:9084, ¼ 0:7865 ¼ 0:7143 ) Ma ¼ 0:71, P0 70 psia T0 ρ0 T ¼ 0:9084T 0 ¼ 0:9084  715 R ¼ 649:5 R 0 1   B C P0 70 psia C  ρ ¼ 0:7865ρ0 ¼ ð0:7865Þ ¼ ð0:7865ÞB 3 @ A RT 0 psia‐ft  0:37 ð715 RÞ  lbm‐ R ¼ 0:2081 lbm=ft3 The velocity can be calculated as before.

Solutions to Practice Problems

167

Practice Problem 6.3 (Solution) A. The static pressure is P ¼ 4.5 kPa. The pitot tube reading indicates the difference between the stagnation pressure and the static pressure. Therefore, P0  P ¼ ΔP ¼ γ w hw ¼ 9:81

kN 10 cm  ¼ 0:981 kPa m3 100mcm

P0 ¼ P þ ΔP ¼ 4:5 kPa þ 0:981 kPa ¼ 5:481 kPa Calculate the Mach number by solving Eq. 6.11 for Ma, and then substitute the known quantities. k    

k1  k1 P0 k1 k1 P k 2 Ma Ma2 ¼ 0 )1þ ) ¼ 1þ 2 2 P P v ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !ffi ffi u  1:41   k1   u 1:4 2 P0 k 2 5:481 kPa 1 ¼t 1 Ma ¼ k1 1:4  1 4:5 kPa P

¼ 0:54 B. Since Ma < 1.0, the flow is subsonic. Practice Problem 6.4 (Solution) Within the combustion chamber, the velocity is zero, and the variables represent stagnation conditions. Therefore, P0 ¼ 800 psia, T 0 ¼ 460 þ 4600 F ¼ 5060 F A. Calculate the individual gas constant for the gas using Eq. 13.16. psia‐ft3

R 10:73 lbmol‐ R R¼ ¼ ¼ 0:4877 psia‐ft3 =lbm  R lbm M 22 lbmol R¼

R 1545 ¼ M 22

ft‐lbf lbmol‐ R lbm lbmol

¼ 70:23 ft‐lbf=lbm  R

Calculate the temperature at the throat, where Ma ¼ 1.0, using Eq. 6.13.

168

6 Fundamentals of Compressible Flow

T ¼ T0



   2 2 ¼ ð5060 RÞ ¼ ð5060 RÞð0:8696Þ ¼ 4, 400 R kþ1 1:3 þ 1 

From the preceding equation, TT 0 ¼ 0:8696. Calculate the pressure and density at the throat using isentropic P-V-T relationships. k   k1 1:4 P T ¼ ¼ ð0:8696Þ1:41 ¼ 0:6132 ) P0 T0 P ¼ 0:6132P0 ¼ 0:6132  800 psia ¼ 491 psia 1   k1 1 T ¼ ð0:8696Þ1:41 ¼ 0:7052 ) T0 0 1   B C P0 800 psia C   ρ ¼ 0:7052ρ0 ¼ 0:7052 ¼ 0:7052B 3 @ A RT 0 psia‐ft  0:4877 ð R Þ 5060 lbm‐ R

ρ ¼ ρ0

¼ 0:2286 lbm=ft3 Since Ma ¼ 1.0 at the throat, calculate the velocity at the throat using Eq. 6.4b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ c ¼ kRT  gc sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi     ft‐lbf lbm‐ft  ð R Þ 32:2 ¼ ð1:3Þ 70:23 4400 lbm‐ R lbf‐ sec 2 ¼ 3597 ft= sec B. Calculate the pressure at the nozzle exit using Eq. 6.11. k  

k1   i 1:4 h P0 k1 1:3  1 2 1:41 2 Ma 9 ¼ 1þ ¼ 8246 ) ¼ 1þ 2 2 P P 800 psia ¼ 0:097 psia P¼ 0 ¼ 8246 8246

Calculate the temperature at the nozzle exit using Eq. 6.10.  

h   i T0 k1 1:3  1 2 2 Ma ¼ 1 þ 9 ¼ 13:15 ) ¼ 1þ 2 2 T T0 5060 R ¼ 385 R ¼ T¼ 13:15 13:15 Calculate the density at the nozzle exit using the ideal gas equation (Eq. 13.12).

References

169

ρ¼

P 0:097 psia  ¼ ¼ 0:0005 lbm=ft3 psia‐ft3 RT  0:4877 lbm‐ R ð385 RÞ

Calculate the velocity of sound at the nozzle exit temperature using Eq. 6.4b. pffiffiffiffiffiffiffiffiffiffiffiffi c ¼ kRTgc ¼

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi     ft‐lbf lbm‐ft  ð385 RÞ 32:2 ð1:3Þ 70:23 lbm‐ R lbf‐ sec 2

¼ 1064 ft= sec Calculate the velocity at the nozzle exit by using the given Mach number and the velocity of sound just calculated. v ¼ Ma  c ¼ 9  1064

ft ¼ 9576 ft= sec sec

References 1. Bar-Meir, G.: Fundamentals of Compressible Flow Mechanics. GNU Free Documentation License (2013). Download from https://mountainscholar.org/bitstream/handle/20.500.11785/2 61/OTL_BookId 86_gasDynamics.pdf?sequence¼1 2. Blevins, R.D.: Applied Fluid Dynamics Handbook. Krieger Publishing Company (2003) 3. Cimbala, J.M., Cengel, Y.A.: Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Education (2017) 4. Crane Technical Paper 410: Flow of Fluids Through Valves, Fittings, and Pipe. Crane Co (1988) 5. Fandom Wiki Aero Engineering Notes: Compressible Flow Example Problems, USA. Download from https://aeroengineeringnotes.fandom.com/wiki/Compressible_Flow_Example_Problems 6. Hall, N. (ed.): Mass Flow Choking. NASA Glenn Research Center (2021). Download from, https://www.grc.nasa.gov/www/k-12/airplane/mflchk.html 7. Pardyjak, E.: Introduction to Compressible Flow. University of Utah, Mechanical Engineering Department (2018). Download from, https://my.mech.utah.edu/~pardyjak/me3700/ IntroCompFlow.pdf 8. White, F.M.: Fluid Mechanics, 8th edn. McGraw Hill (2016)

Chapter 7

Dimensional Analysis and Similitude

7.1

Introduction

The three fundamental physical quantities are mass (M ), length (L), and time (T ). All other physical quantities/variables can be expressed in terms of the three fundamental quantities, M, L, and T. The dimensions of derived physical quantities can be easily obtained by examining the typical units of a given physical quantity provided in Chap. 1 [1, 3, 4]. A few examples are shown here. The square parenthesis is read as “dimensions of.” Velocity: ½ v 

m L   LT 1 s T

Force: ½F   N 

kg:m ML  2  MLT 2 s2 T

Pressure:

½P 

½F  MLT 2  ML1 T 2  ½A  L2

Example 7.1 Express the dynamic viscosity of a fluid in terms of the three fundamental dimensions.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_7

171

172

7

Dimensional Analysis and Similitude

(Solution) From Chap. 1, the units of dynamic viscosity are kg/m.s (lbm/ft-s). Therefore, the dimensions of dynamic viscosity are ½μ 

kg M   ML1 T 1 m  s LT

Example 7.2 Express power in terms of the fundamental dimensions M, L, and T. (Solution) From Chap. 3, the units of power are N.m/s (J/s or W), (ft–lbf/s). Therefore, the dimensions of power are ½Power ¼

7.2

½F   L MLT 2  L ¼ ¼ ML2 T 3 T T

Dimensionless Parameters Used in Fluid Mechanics

Several dimensionless parameters are used in fluid mechanics [1, 4, 5]. The dimensionless parameters are expressed as force ratios. The different types of forces experienced by a fluid in a flow field are: 1. Inertial force, FI, which is the force driving the motion of the fluid particles. 2. Viscous force, FV, which is the force that provides viscous resistance to fluid flow. 3. Gravity force, FG, which is present when the fluid flows under the influence of gravity. The different force ratios and the corresponding dimensionless numbers are shown here. Reynolds Number, Re ¼

Inertial Force F Lvρ ¼ I ¼ Viscous Force F V μ

ð7:1Þ

In Eq. 7.1, L is the characteristic dimension involved in the flow situation. For example, for boundary layer flow of a fluid past a flat plate, L is the length of the plate in the flow direction from the leading edge. For the flow of a fluid through a circular pipe, L is the internal diameter of the pipe. Froude Number, Fr ¼

Inertial Force F v2 ¼ I ¼ Gravity Force F G Lg

ð7:2Þ

7.2 Dimensionless Parameters Used in Fluid Mechanics

173

In Eq. 7.2, the characteristic length relevant to the flow situation is the depth of the channel. Flow in a channel occurs due to the gravity force. The other dimensionless parameters used in fluid mechanics are Cauchy number (Ca) and Weber number (We). However, these dimensionless parameters are not important in fluid flow. The Reynolds number is widely used in full flow through pipes and ducts, while the Froude number is primarily used in open channel flow.

7.2.1

Benefits of Using Dimensionless Parameters

Experiments are widely used in obtaining graphs that describe the parametric dependence of fluid flow variables. For example, if we examine the functional dependence of friction factor in fluid flow through pipes, we can state that the value of friction factor depends on the velocity of the fluid, the density of the fluid, the dynamic viscosity of the fluid, the diameter of pipe and the surface roughness of the pipe material (five variables). Graphical representation of the functional dependence of the friction factor will result in five different graphs for each of the five variables with limited ranges for the independent variables. This makes the experimentation and presentation of results a cumbersome process. It is quite possible that some of the values of the independent variables may be outside the range of the graphs. To overcome these problems, the five independent variables are grouped into dimensionless parameters [2, 5] as per the Buckingham Pi theorem.

7.2.2

Buckingham Pi Theorem

The Buckingham Pi theorem is used in deriving the number of dimensionless parameters that can be used in describing a physical phenomenon involving “n” variables [1, 4–6]. The theorem states that (n – r) dimensionless groups can be employed in parametric studies of the physical phenomenon. r represents the number of fundamental dimensions, that is, mass, length, and time. Therefore, r ¼ 3. Applying the Buckingham Pi theorem to the functional dependence of friction factor for fluid flow through pipes, number of dimensionless groups ¼ n – r ¼ 5–3 ¼ 2. From Chap. 3, we know that the dimensionless groups are Reynolds number (Re) and relative roughness, r, which are used in the parametric dependence of the friction factor as shown in the Moody diagram (Fig. 3.4). Example 7.3 Use the Buckingham Pi theorem to establish the parametric dependence of the friction factor given that the friction factor for flow through a pipe depends on the internal diameter of the pipe, the velocity of the fluid, the density of the fluid, the dynamic viscosity of the fluid, and the roughness of the pipe surface.

174

7

Dimensional Analysis and Similitude

(Solution) Express the functional dependence of the friction factor in terms of the variables specified. f ¼ fnðD, v, ρ, μ, εÞ There are five variables (n ¼ 5) and three fundamental dimensions (r ¼ 3). According to the Buckingham Pi theorem, the number of dimensionless groups required are Y

¼nr ¼53¼2

Write down the fundamental dimensions of each variable specified. ½D ¼ L, ½v ¼ LT 1 , ½ρ ¼ ML3 ½μ ¼ ML1 T 1 , ½ε ¼ L By inspection, the ratio of pipe roughness to the internal diameter is dimensionless, and this will be the first dimensionless parameter. Y 1

¼

ε ¼r D

Create the second dimensionless parameter by combining the remaining variables, with the diameter, velocity, and density as the scalable variables. Y Y

¼ μDa vb ρc

2

 b  c ¼ μDa vb ρc ¼ ML1 T 1  La  LT 1  ML3

2

¼ M 1þc L1þaþb3c T 1b Π2 is a dimensionless parameter. Y

¼ M 0 L0 T 0

2

Combine the preceding equations, and equate the exponents to solve for a, b, and c.

7.3 Similitude

175

M 0 L0 T 0 ¼ M 1þc L1þaþb3c T 1b M : 1 þ c ¼ 0 ) c ¼ 1 T : 1  b ¼ 0 ) b ¼ 1 L : 1 þ a þ b  3c ¼ 0 ) 1 þ a  1  3ð1Þ ¼ 0 ) a ¼ 1 Y μ 1 ¼ ¼ μDa vb ρc ¼ μD1 v1 ρ1 ¼ Dvρ Re 2 f ¼ fnðr, Re Þ Therefore, the two dimensionless parameters are the relative roughness and the Reynolds number.

7.3

Similitude

Similitude involves simulation of an actual flow situation, called the prototype ( p), by using a small-scale model (m) [5]. The ratio of the size of the model to the size of the prototype is called the scale ratio of the model. For example, if the model is 1=10th 1 the actual size, then the scale ratio can be written as: S ¼ LLmp ¼ 10 .

7.3.1

Requirements for Successful Simulation

For the simulation to work successfully, the model and prototype must satisfy three important criteria [4–6]: • Geometric similarity • Kinematic similarity • Dynamic similarity

7.3.1.1

Geometric Similarity

Geometric similarity requires that the model and prototype must have similar shapes and that the proportions of the key dimensions must be the same both in the model and in the prototype. For example, if the length of the prototype is twice its width, the same proportion between lengths should be maintained in the model as well.

176

7.3.1.2

7

Dimensional Analysis and Similitude

Kinematic Similarity

Kinematic similarity requires that the model testing must be done under similar fluid conditions encountered by the prototype. For example, if the normal temperature of air encountered by an automobile is 15  C, then the lab studies should also use air at 15  C. Also, the velocities at corresponding points on the model and prototype have a constant scale factor. This ensures that both the model and prototype encounter similar flow conditions.

7.3.1.3

Dynamic Similarity

Dynamic similarity requires that the relevant dimensionless parameters should have the same value for both the model and the prototype. For example, in the simulation of flow through a pipeline, the Reynolds number should be the same for both the model and the prototype. Mathematically, for flow through pipes, the dynamic similarity criteria is: Re m ¼ Re P

ð7:3Þ

Dm vm Dp vp ¼ νm νp

ð7:4Þ

For the simulation of flow through an open channel, the Froude numbers of the model and prototype should be equal. For dynamic similarity of flow through open channels, the mathematical equations are: Frm ¼ FrP v2p

v2m ¼ Lm g Lp g

ð7:5Þ ð7:6Þ

Example 7.4 The flow of oil in a pipeline with an internal diameter of 900 mm is to be simulated in a 1:10 scale model pipe. If the velocity of oil in the pipeline is 5 m/s, what should be the velocity in the model to achieve dynamic similarity? (Solution) Use Eq. 7.4 for dynamic similarity in flow through pipes.





Dm vm Dp vp ¼ νm νp

Practice Problems

177

 vm ¼

 Dp   10 m 5 ¼ 50 m=s vp ¼ 1 s Dm

Example 7.5 Dimensional analysis of the pressure drop across a valve results in the following dimensionless functional relationship:   ΔP ho dvρ ¼ fn , d μ ρv2 In the preceding equation, d is the diameter of the valve and ho is the height of the opening, v is the fluid velocity, and ρ, μ are fluid properties. A one-tenth scale model of an oil pipeline is used in the lab with water as the test fluid. The specific gravity of oil is 0.85, and its dynamic viscosity is 21 cP and the standard room temperature properties of water are density ¼ 1000 kg/m3 and dynamic viscosity ¼ 1 cP. If the velocity in the oil pipeline is 1.15 m/s, what should be the velocity of water in the test bench. (Solution) For dynamic similarity, the Reynolds number of the prototype and model should be equal. d p vp ρp d m vm ρm ¼ μp μm        ρp dp μm 1 cP m vm ¼ 1:15 vp ¼ ð10Þð0:85Þ 21 cP s dm ρm μp ¼ 0:4655 m=s

Practice Problems Practice Problem 7.1 Express kinematic viscosity and density of a fluid in terms of the three fundamental dimensions. Practice Problem 7.2 Surface tension, σ, is a property of a fluid, and it is defined as the upward force per unit length of the surface with which the fluid is in contact. Determine the dimensions of surface tension in terms of M, L, and T.

178

7

Dimensional Analysis and Similitude

Practice Problem 7.3 The drag force experienced by an object due to fluid flow around it is a function of the velocity of the fluid, v; the density of the fluid, ρ; the dynamic viscosity of the fluid, μ; the characteristic dimension of the object, L; and the projected area of the object perpendicular to the flow direction, Ap. Determine the dimensionless parameters required to characterize the drag force due to fluid flow past solid objects. Practice Problem 7.4 The flow of water in an open channel 20 ft deep is simulated using a lab model. What should be the width of lab channel in order if the velocity in the model is 20 ft/sec and the velocity in prototype is 10 ft/sec.? Practice Problem 7.5 It is desired to predict the drag force experienced by a slim chemical processing tower with dimensions of 40 ft height and diameter 2 ft, using a 1:20 scale model. The average wind velocity is 25 mi/hr. Determine: A. The velocity required for model testing B. The scale of the magnitude of the drag force experienced by the tower in terms of the drag force experienced by the model

Solutions to Practice Problems Practice Problem 7.1 (Solution) From Chap. 1, the units of kinematic viscosity are m2/s, (ft2/s). Therefore, the dimensions of kinematic viscosity are ½ν 

m2 L2   L2 T 1 s T

From Chap. 1, the units of density are kg/m3, (lbm/ft3). Therefore, the dimensions of density are ½ρ 

kg M   ML3 m3 L3

Practice Problem 7.2 (Solution) Use the definition of surface tension to obtain its dimensions. σ¼

½F  MLT 2 ¼ ¼ MT 2 L L

Solutions to Practice Problems

179

Practice Problem 7.3 (Solution) The variables involved are v, ρ, μ, L, and Ap. Therefore, n ¼ 5. As per the Buckingham Pi theorem, the number of Π dimensionless parameters involved in this situation is 5–3 ¼ 2. Represent the functional dependence of the drag force on the variables specified.   F D ¼ fn L, v, ρ, μ, Ap Write down the fundamental dimensions of the variables. ½L ¼ L, ½v ¼ LT 1 , ½ρ ¼ ML3  ½μ ¼ ML1 T 1 , Ap ¼ L2 and ½F D  ¼ MLT 2 Use L, v, and ρ as the scalable variables, and create a Π dimensionless parameter. F D ¼ μAp La vb ρc Q 1

¼

FD MLT 2 a b c ¼ b c μAp L v ρ ðML1 T 1 ÞðL2 ÞðLa ÞðLT 1 Þ ðML3 Þ ¼ M þ3c Labþ3c T 1þb

Π1 is a dimensionless parameter. Y

¼ M 0 L0 T 0

1

Combine the preceding equations, and equate the exponents to solve for a, b, and c. M 0 L0 T 0 ¼ M þ3c Labþ3c T 1þb M : þ3c ¼ 0 ) c ¼ 0 T : 1 þ b ¼ 0 ) b ¼ 1 L : a  b þ 3c ¼ 0 ) a  1 þ 3ð0Þ ¼ 0 ) a ¼ 1 Substitute the preceding results into the equation for Π1. Y 1

¼

FD FD ρ1 v2  1 2 a b c ¼ 1 1 0 ρ v μAp L v ρ μAp L v ρ    FD Lvρ ¼ μ Ap ρv2

180

The ratio

7 2F D Ap ρv2

Dimensional Analysis and Similitude

is known as the “drag coefficient,” CD, and

Lvρ μ

is the Reynolds

number, Re. The dimensionless relationship can be written as: CD ¼ fnð Re Þ Hence, the two dimensionless parameters required are the drag coefficient and the Reynolds number. Practice Problem 7.4 (Solution) Use Eq. 7.6 for dynamic similarity in flow through open channels.





v2p v2m ¼ Lm g Lp g !2  2 ft 10 sec vp Lp ¼ ð20 ftÞ ¼ 5 ft Lm ¼ ft vm 20 sec Practice Problem 7.5 (Solution) Convert the wind speed to ft/sec.     mi 5280 ft 1 hr ¼ 36:67 ft= sec vp ¼ 25 hr mi 3600 sec A. For dynamic similarity, the Reynolds number of the prototype and model should be equal. d p vp ρp d m vm ρm ¼ μp μm     ρp dp μm vm ¼ v dm ρm μp p The properties of air in the model and in the prototype can be assumed to be same by using testing air under the same conditions of temperature as the wind. ρp μ ¼ 1:0 and m ¼ 1:0 ρm μp

References

181

    dp 20 ft 36:67 ¼ 733:4 ft= sec v ¼ 1 sec dm p

 vm ¼

B. The dimensionless parameter associated with the drag force is the drag coefficient. CD ¼

2F D ρv2 A

For dynamic similarity, the drag coefficients of the model and the prototype should be equal.

F Dp F Dm

2F Dp 2F Dm ¼ ρm v2m Am ρp v2p Ap   2     2  2 ρp ρp vp Ap vp Lp ¼ ¼ ρm vm Am ρm vm Lm  2  2 1 20 ¼ ð1:0Þ 20 1 F Dp ¼ F Dm

Thus, the model and prototype will experience the same drag force.

References 1. Aspley, D.: Dimensional Analysis Notes. Manchester (2013). Download from https:// personalpages.manchester.ac.uk/staff/david.d.apsley/lectures/hydraulics2/t3.pdf 2. Cimbala, J.M., Cengel, Y.A.: Fluid Mechanics: Fundamentals and Applications. McGraw-Hill Education (2017) 3. Elger, D.F., LeBret, B.A., Crowe, C.T., Roberson, J.A.: Engineering Fluid Mechanics, 12th edn. Wiley (2019) 4. Mott, R.L., Untener, J.A.: Applied Fluid Mechanics, 7th edn. Pearson (2015) 5. Munson, B.R., Young, D.F., Okiishi, T.H., Huebsch, W.W.: Fundamentals of Fluid Mechanics, 6th edn. Wiley (2009) 6. White, F.M.: Fluid Mechanics, 8th edn. McGraw Hill (2016)

Chapter 8

Heat Transfer Principles

8.1

Introduction

Heat transfer is flow of heat from a high-temperature zone to a low-temperature zone. Heat is a form of energy, and therefore heat transfer rates are represented by units of energy flow per unit time. The symbol used for heat transfer rate is q, and the units for heat transfer rate are Btu/hr (USCS) and kW (S I). Note that watt (W) is the rate of energy transfer and W
Joules
Js. s Conversion Factor: 1 kW ¼ 3412 Btu/hr

8.2

General Equation for Heat Transfer Modeling

The equation for current flow is: Current Flow ¼

Potential Difference Resistance

I¼

V R

Similarly, the equation for heat transfer can be written in terms of driving force and resistance to heat flow [1–3]. The driving force for heat transfer is the temperature difference between the hot and cold regions. The resistance to heat flow is known as thermal resistance, and it is represented by Rth.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_8

183

184

8 Heat Transfer Principles

Heat Flow ¼

Temperature Difference Resistance to heat flow ΔT q¼P Rth

ð8:1Þ

The denominator in Eq. 8.1 indicates summation of all the thermal resistances since there can be multiple thermal resistances in the same system (explained in Sect. 8.4). The equation for thermal resistance depends on the particular mode of heat transfer. The different modes of heat transfer are discussed in the next section. The units for thermal resistance can be obtained by rearranging Eq. 8.1. Rth ¼

  ΔT F C

q Btu=hr kW

Sometimes, it is convenient to write Eq. 8.1 in terms heat flow per unit crosssection area perpendicular to the heat flow direction. Correspondingly, the thermal resistance becomes thermal resistance per unit cross-section area, and it can be written as Rth,ua, where “ua” represents unit cross-section area. q PΔT ¼ A Rth,ua

8.3

ð8:2Þ

Heat Transfer Modes

The principal modes of heat transfer are conduction, convection, and radiation.

8.3.1

Conduction Heat Transfer

The transfer of heat through solid media such as a metal pipe wall, building walls, and insulation is called conduction heat transfer. Consider a hot fluid flowing in an insulated pipe. The heat will flow through the pipe wall and the covering insulation by means of conduction mechanism.

8.3.2

Convection Heat Transfer

Convection heat transfer is heat transfer between a solid surface and an adjoining fluid in contact with the solid surface. Consider a flat hot plate exposed to ambient

8.4 Thermal Circuit Analogous to Electrical Circuit

185

air. Heat is transferred from the hot plate surface to the cooler air by means of convection mechanism. There are two types of convection heat transfer mechanisms, free or natural convection and forced convection.

8.3.2.1

Free Convection

In free convection, the ambient fluid media remains stationary relative to the solid surface. When a hot plate is exposed to still air, the air particles adjacent to the hot surface get heated up and rise above the surface due to lower density. The rising air particles displace the cooler, heavier air particles farther away from the surface, and this cycle continues, setting up “convection current” pathways until all the air particles are sufficiently heated up. Thus, the primary mechanism of heat flow in free convection is the density difference between the hot and cold fluid particles.

8.3.2.2

Forced Convection

In forced convection, the fluid adjacent to the solid surface keeps moving resulting in a faster rate of heat transfer. Consider air being blown across the hot surface of the plate. The heated air particles continue to rise above the surface. However, they are replenished by the advancing colder air particles approaching the plate resulting in faster cooling of the plate surface.

8.3.3

Radiation Heat Transfer

Unlike conduction and convection, radiation heat transfer does not require any media (solid or fluid) for heat transfer to occur. The radiation heat from the sun reaches earth after traveling through vast, empty space (vacuum). Thus, the only requirement for radiation heat transfer is a temperature difference between the hot and cold entities.

8.4

Thermal Circuit Analogous to Electrical Circuit

Heat flow can be represented by a thermal circuit similar to an electrical circuit for current flow as shown in Fig. 8.1.

186

8 Heat Transfer Principles

Fig. 8.1 Similarity between electrical and thermal circuits

Fig. 8.2 Combination of conduction and convection

8.5

Conduction-Convection Systems

In many practical applications, conduction and convection occur together to accomplish the heat transfer [1, 3]. Consider heat flow through a building window in a heated building during the winter season as shown in Fig. 8.2, along with the equivalent thermal circuit. At steady state, the heat flow is constant. Heat flow takes place from room air to the inside surface of the window by convection. The same heat flow occurs in the window glass due to conduction, and finally the same heat is transferred from the outside surface of the window to outside air by convection. The room air temperature drops from a bulk value of Troom to T1 at the inside surface of the window. This temperature drop is due to convection resistance of room (or inside) air. The temperature in the window glass slab drops from T1 to T2 due to the conduction resistance of the window glass. The thickness of the window glass is ΔX. The temperature drops from T2 at the outside surface of the glass to Tair due to convection resistance of outside air. The overall temperature drop is ΔToverall ¼ Troom – Tair. The equivalent thermal circuit with the three thermal resistances is shown in Fig. 8.2. The heat transfer equation for this system can be written by using Eq. 8.1 and the equivalent thermal circuit as shown here. ΔT T room  T air q ¼ Poverall ¼ Rconv,i þ Rcond þ Rconv,o Rth

ð8:3Þ

References

187

The details of calculating thermal resistances for conduction and convection will be presented in Chaps. 9 and 10. Example 8.1 Consider an insulated furnace wall, which typically consists of an insulating brick wall (refractory) in contact with the hot gases. The insulating brick wall is followed by the steel shell of the furnace, and the steel shell is covered by an external insulation exposed to the ambient air. The heat from the hot gases in the furnace is transferred to the insulating brick wall by convection, and the heat flows through the insulating brick, the steel shell, and the outer insulation by conduction. Finally, the heat is transferred from the outer insulation to ambient air by convection. A. Draw the equivalent thermal circuit for heat flow through the furnace. B. Write down the heat transfer equation for the furnace in terms of the thermal resistances. (Solution) A. This situation consists of two convection resistances and three conduction resistances as described in the problem statement. The equivalent thermal circuit will consist of five resistances as shown in the figure. Subscript “g” represents the hot gases in the furnace, and subscript “a” represents ambient air. Subscript “b” represents the insulating brick, subscript “s” is for the steel shell, and subscript “i” is for the external insulation.

T T

g a Poverall ¼ B. q ¼ ΔT Rconv,g þRcond,b þRcond,s þRcond,i þRconv,a R th

References 1. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P.: Fundamentals of Heat and Mass Transfer, 8th edn. Wiley, Hoboken (2019) 2. Flynn, A.M., Akashige, T., Theodor, L.: Kern’s Process Heat Transfer, 2nd edn. Wiley, Hoboken (2019) 3. Holman, J.P.: Heat Transfer, 10th edn. McGraw Hill, New York (2009)

Chapter 9

Conduction Heat Transfer

9.1

Introduction

Conduction is heat flow through a solid medium due to the temperature difference across the solid medium [5]. Flow of heat through an insulated building wall is a typical example of conduction heat transfer.

9.2

Fourier’s Law of Heat Conduction

Conduction heat transfer is governed by Fourier’s law of heat conduction [3, 5, 7]. According to Fourier’s law for one-dimensional heat flow, the rate of heat transfer per unit cross-section area is proportional to the temperature gradient in the direction of heat flow. The heat transfer per unit area (q/A) is also known as heat flux. Fourier’s law can be mathematically written as q dT ¼ k A dx

ð9:1Þ

Since the temperature gradient in the direction of heat flow is always a negative quantity, a negative sign is necessary to keep heat flow as a positive quantity. The variable “k” in Fourier’s law is called thermal conductivity. Thermal conductivity is a physical property of materials, and it has the following units: k  Btu=hr  ft  F

ðUSCSÞ

k  W=m:K

ðS IÞ

Conversion Factor: 1 Btu/hr – ft –  F ¼ 1.731 W/m.K © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_9

189

190

9 Conduction Heat Transfer

Table 9.1 Thermal conductivities of materials Material Carbon steel Stainless steel Copper Aluminum Glass PVC Fiber glass insulation Water (20  C) Air

Thermal conductivity (Btu / hr – ft –  F) 31 9.2 231 144 0.64 0.11 0.023 0.35 0.014

Thermal conductivity (W / m.K) 54 16 400 250 1.1 0.19 0.04 0.6 0.024

Fig. 9.1 Fourier’s law of heat conduction

Typically, metals have relatively high thermal conductivities while insulating materials have low thermal conductivities. Approximate values of thermal conductivities of a few common materials are presented in Table 9.1.

9.3

Conduction Through a Rectangular Slab

Consider heat flow across a solid slab of thickness ΔX as shown in Fig. 9.1. The heat flow is normal to the cross-section area of the slab. Fourier’s law (Eq. 9.1) can be integrated to obtain the following result:

9.3 Conduction Through a Rectangular Slab

q A

191

ZX 2

ZT 2 dX ¼ k

X1



dT T1

q ΔT ¼k A ΔX



ð9:2Þ

Where ΔT ¼ T1 – T2 and ΔX ¼ X2 – X1. Eq. 9.2 is the integrated form of Fourier’s law for one-dimensional heat flow through a slab [2, 4, 6]. Equation 9.2 can be written in the form of thermal resistance per unit area as shown here.   q ΔT ΔT ¼k ¼ ΔX  A ΔX k

ð9:3Þ

Compare Eq. 9.3 with the basic equation (Eq. 8.2, reproduced here for reference) for heat flow per unit area from Chap. 8. q PΔT ¼ A Rth,ua From the comparison of the two equations, the conduction thermal resistance per unit area is: Rcond,ua ¼

ΔX k

ð9:4Þ

Equation 9.3 can be written in terms of thermal resistance per unit area. q ΔT ¼ A Rcond,ua

ð9:5Þ

Example 9.1 A steel furnace is lined on the inside with a 6 in. thick insulating brick. The thermal conductivity of the insulating brick is 0.069 Btu / hr – ft –  F. The temperature on the hot side of the brick is 1500  F. Determine: A. The thermal resistance per unit area of the insulating brick B. The rate of heat transfer required per unit area of the furnace wall if the temperature of the cooler side of the insulating brick is to be limited to 700  F

192

9 Conduction Heat Transfer

(Solution) A. Calculate the thermal resistance per unit area by using Eq. 9.4. 

Rcond,UA ¼

ΔX ¼ k



6 in 12 in=1 ft Btu 0:069 hrft F

¼ 7:25 hr  ft2  F=Btu

B. Calculate the required heat transfer per unit area by using Eq. 9.5. q ΔT 1500  F  700  F ¼ ¼ ¼ 110:34 Btu=hr  ft2 2  A Rcond,UA 7:25 hrft  F Btu

9.3.1

Multilayer Conduction

Conduction through multiple layers can be modeled by adding up the thermal resistances of each layer [3, 5, 7]. Consider heat flow through two solid layers, layer 1 and layer 2 as shown in Fig. 9.2. At steady state, the overall heat flow across the multiple layers as well as the heat flow across each layer will be constant and identical in value. The heat transfer rate can be calculated by dividing the overall temperature difference by the sum of the resistances of all layers (Eq. 9.6). The heat transfer rate can also be calculated by dividing the temperature difference of a particular layer by the thermal resistance of the same layer [2, 6]. Once again, it is useful to write the equations in terms of heat transfer per unit cross-section area normal to the heat flow direction. Fig. 9.2 Conduction through multiple layers

9.3 Conduction Through a Rectangular Slab

193

ðT  T Þ q P ΔT overall ¼ ΔX11 ΔX32 ¼ A Rcond,ua þ k1

ð9:6Þ

k2

q ðΔT Þ1 T 1  T 2 ðΔT Þ2 T 2  T 3 ¼ ¼ ΔX 1 ¼ ¼ ΔX 2 A R1,ua R2,ua k1

ð9:7Þ

k2

Example 9.2 The wall of a home consists of a 30 mm thick common brick (k ¼ 0.6929 W/m.K) on the outside followed by 4 mm in layer of plaster (k ¼ 0.4750 W / m.K). During the winter months, the inside temperature is maintained at 21  C, and the outside temperature averages around 7  C. Calculate: A. The heat loss per unit area of the wall B. The temperature at the interface of the brick and the plaster (Solution) A. Calculate the heat loss per unit area of the wall by using Eq. 9.6. Layer 1 is the plaster adjacent to the inside air, and layer 2 is the brick exposed to the outside air. ðT 1  T 3 Þ q 21  C  7  C ¼0 ¼ Δx1 Δx2 4 mm 1 0 30 mm 1 A þ k1 k2 B 1000 mm C B 1000 mm C C B C B m m CþB C B C B C B W W A @0:6929 A @0:4750 m:K m:K ¼ 271 W=m2 B. Calculate the temperature at the interface of the brick and the plaster by using Eq. 9.7. 0

4 mm 1     B 1000 mm C C q ΔX 1 W B  m C T2 ¼ T1  ¼ 21 C  271 2 B B W C A k1 m @ A 0:4750 m:K ¼ 18:7  C

194

9 Conduction Heat Transfer

Table 9.2 Typical R-values of materials Material Fiber glass (Batt) Rock wool (Batt) Cellulose (Blown) Polyurethane (Foam) Brick, concrete Glass Plywood Air

R-value per inch thickness hr – ft2 –  F / Btu 3.67 3.53 3.63 6.15 0.14 0.15 1.25 6.04

R-value per cm thickness m2 . K / W 0.2543 0.2446 0.2516 0.4262 0.0097 0.0104 0.0866 0.4186

Note: To obtain the R-value per centimeter thickness in m2 .K/W multiply the USCS values in the table by 0.0693

9.3.2

R-Values for Insulation and Building Materials

The building industry as well as the heating, ventilation, and air conditioning (HVAC) industry commonly use the R-value for insulating and building materials [2, 3]. The R-value is the same as the conduction resistance per unit cross-section area as specified in Eq. 9.4. Typical R-values for some insulating materials are presented in Table 9.2. R  value ¼ Rcond,ua ¼

9.4

ΔX k

ð9:8Þ

Conduction Through a Cylindrical Wall

The heat transfer through a cylindrical wall occurs in the radial direction [1, 3, 5, 8] as shown in Fig. 9.3. L is the length of the cylinder and T1 > T2 as per the heat flow direction shown in Fig. 9.3. Consider a small element of the cylindrical wall at a distance r from the center and with thickness dr. Fourier’s law of heat conduction can be applied in the radial direction and integrated as shown here. qr ¼ kAr

  dT dr

ð9:9aÞ

In Eq. 9.9, Ar is the surface area of the cylindrical surface perpendicular to the radial heat flow direction, and k is the thermal conductivity of the cylindrical wall material. Ar ¼ 2πrL

ð9:9bÞ

9.4 Conduction Through a Cylindrical Wall

195

Fig. 9.3 Conduction through a cylindrical layer

Combine Eqs. 9.9(a) and 9.9(b), and integrate the resulting equation after separating the variables T and r.   dT qr ¼ kð2πrLÞ dr ZT 2 Zr2   dr qr ¼ ð2πkLÞ dT r r1

T1

qr ¼ qcyl ¼

2πkLðT 1  T 2 Þ   ln rr21

ð9:10Þ

Equation 9.10 is the general equation for steady-state heat conduction through a cylindrical layer. The most common application of heat conduction through a cylindrical wall is pipes covered with insulation. In a process plant, there can be several pipes with similar heat flow conditions but of varying lengths. Hence, it is of practical use to write Eq. 9.10 in terms of heat flow per unit length of the cylinder. qcyl 2πk ðT 1  T 2 Þ   ¼ L ln r2

ð9:11Þ

r1

Equation 9.11 can be written in terms of temperature difference and thermal resistance per unit length (ul) of a cylindrical wall. qcyl ΔT ΔT ¼  ¼ R L r2 cyl,ul ln r1

2πk

ð9:12Þ

196

9 Conduction Heat Transfer

From Eq. 9.12, the thermal resistance per unit length of a cylindrical wall is:   Rcyl,ul ¼

ln

r2 r1

2πk

ð9:13Þ

Example 9.3 A steel pipe (ID ¼ 5.76 in, OD ¼ 6.63 in, k ¼ 11 Btu / hr – ft –  F) carries steam at 400  F, which is approximately the temperature of the inside pipe wall. The pipe is covered with a 2-in layer thick fiberglass insulation (k ¼ 0.023 Btu / hr – ft –  F). A. Calculate the thermal resistances per unit length of the pipe wall and the insulation layer, and determine the percentage contribution of each to the total thermal resistance per unit length. B. If the temperature of the insulation surface is at 150  F, calculate the heat loss per foot length of the pipe. (Solution) A. Calculate the thermal resistances per unit length using Eq. 9.13. Pipe wall: OD 6:63 in ¼ ¼ 3:315 in 2 2 ID 5:76 in r1 ¼ ¼ ¼ 2:88 in 2 2     r2 3:315 in ln ln 2:88 in r1 ¼ ¼   Btu 2πk pipe ð2π Þ 0:11  hr  ft  F 2  ¼ 0:2035 hr  ft  F=Btu  ft length r2 ¼

RPipe,ul

Insulation: r2 ¼ 3.315 in, r3 ¼ r2 + Δrins ¼ 3.315 in + 2 in ¼ 5.315 in

Rins,ul

  5:315 in ln 3:315 in   Btu ð2π Þ 0:023  hr  ft  F ¼ 3:267 hr  ft2  F=Btu  ft length

  r3 r2 ¼ ¼ 2πkins ln

Calculate the total thermal resistance per unit length.

9.5 Conduction Through a Spherical Wall

197

Rtotal,UL ¼ Rpipe,UL þ Rins,UL hr  ft2  F hr  ft2  F þ 3:267 Btu  ft length Btu  ft length 2  ¼ 3:4705 hr  ft  F=Btu  ft length   0:2035 100 % contribution from pipe wall ¼ 3:4705 ¼ 5:86%   3:267 % contribution from insulation ¼ 100 3:4705 ¼ 94:14% ¼ 0:2035

Comment: The contribution of metal pipe wall to the overall thermal resistance is extremely small and can be ignored. This is because metal pipe walls have relatively much higher thermal conductivity compared to insulation materials and since thermal resistance is inversely proportional to thermal conductivity, insulating layers have much higher thermal resistances compared to metal layers. B. Calculate the heat loss per foot length of the pipe by using Eq. 9.12. qcyl ΔT overall ¼ L Rtotal,UL 400 F  150 F ¼ hr  ft2  F 3:4075 Btu  ft length ¼ 73:37 Btu=hr  ft

9.5

Conduction Through a Spherical Wall

Heat conduction through a spherical shell also occurs in the radial direction as shown in Fig. 9.4 [3, 6, 7, 9]. Consider a cross section of the spherical wall with a small element of thickness dr at a distance r from the center of the sphere. Apply Fourier’s law in the radial direction to the spherical section (Eq. 9.9(a)). For the spherical element, the surface area perpendicular to the heat flow direction is: Ar ¼ 4πr 2 From Eqs. 9.9(a) and 9.14,

ð9:14Þ

198

9 Conduction Heat Transfer

Fig. 9.4 Conduction through a spherical wall

    dT qr ¼ k 4πr 2 dr

ð9:15Þ

Separate the variables in Eq. 9.15 and integrate with appropriate limits. ZT 2  q  Zr2 dr r ¼ k dT 4πk r2 r1

T1

  q  1 1 r  ¼ ðT 1  T 2 Þ 4πk r 1 r 2 qr ¼ qsphere ¼

ð4πkÞðT 1  T 2 Þ 1 1 r1  r2

ð9:16Þ

Equation 9.16 can be written in terms of temperature difference and thermal resistance spherical wall. qsphere ¼

T1  T2 1 1 r 1 r 2

¼

ΔT Rsphere

ð9:17Þ

4πk

From Eq. 9.17, the thermal resistance due to a spherical wall is: 1 1 Rsphere ¼ r1 r2 4πk

ð9:18Þ

Example 9.4 A spherical tank with ID 178 cm is used in storing liquid ammonia at a temperature of 60  C, which can be considered as the inside surface temperature of the polyurethane insulation (k ¼ 0.028 W/m. K). The average outside surface temperature of the insulation is 15  C. Determine the thickness of insulation required to limit the heat gain of the sphere to 100 W.

Practice Problems

199

(Solution) Calculate the inside radius of the sphere in meters.

r1 ¼

ð178 cmÞ



D1 ¼ 2 2 ¼ 0:89 m

 1m 100 cm

Since the heat transfer is from outside the sphere to the inside, T2 > T1. Accordingly, the equation applicable in this situation is: qsphere ¼

ΔT 1 1 r 1 r 2

¼

T2  T1

4πk

1 1 r 1 r 2

4πk

Substitute the known values into the preceding equation. 100 W ¼

15  C  ð60  CÞ 1 1 0:89 mr2

W ð4π Þð0:028mK Þ

Solve the preceding equation for r2. r 2 ¼ 1:16 m Calculate the thickness of the insulation required. Δr ¼ r 2  r 1 ¼ 1:16 m  0:89 m ¼ 0:27 m ð27 cmÞ

Practice Problems Practice Problem 9.1 Use the data in Example 9.1, and assume constant, steady-state heat flow through the system. The temperature of the inside surface of the insulation is 680  F. Determine in inches the thickness of fiberglass insulation (k ¼ 0.023 Btu / hr – ft – F) required to keep the temperature of the outside surface of the insulation below 70  F. Practice Problem 9.2 A window in a home consists of a double glass pane each with a thickness of ¼ in. The air gap between the window panes is (1/16) in. The thermal conductivity of glass is 0.61 Btu / hr – ft –  F and that of air is 0.015 Btu / hr – ft –  F. Compare the heat loss through the double-pane glass window with the heat loss from a single

200

9 Conduction Heat Transfer

glass pane window of ½ in thickness, and calculate the percentage reduction in heat loss due to the use of the double-pane window. Practice Problem 9.3 The insulation surface in Example 9.3 can be a source of heat hazard at 150  F. Calculate the thickness of an additional layer of calcium silicate insulation (k ¼ 0.035 Btu / hr – ft –  F) required to limit the surface temperature to 80  F. Practice Problem 9.4 The heat transfer to liquid ammonia in Example 9.4 will cause the heating of liquid ammonia to its boiling point 33  C and subsequently vaporization of liquid ammonia. The relevant physical properties of liquid ammonia are density ¼ 686 kg/m3; specific heat ¼ 4.744 kJ/kg.K; and enthalpy of vaporization ¼ 1369 kJ/kg. Determine the time required in hours for liquid ammonia to reach its boiling point.

Solutions to Practice Problems Practice Problem 9.1 (Solution) Under steady-state conditions, the heat flow through the insulation layer will be the same as the heat flow through the insulating brick. From Example 9.1, the heat flow per unit area through the insulating brick is 110.34 Btu / hr – ft2 –  F. Calculate the required thermal resistance per unit area of the fiberglass insulation by using Eq. 9.5. Rcond,ua ¼

ΔT 680  F  70  F ¼ ¼ 5:53 hr  ft2  F=Btu Btu q=A 110:34 hrft 2

Calculate the required thickness of fiberglass insulation by using Eq. 9.4.    hr  ft2  F Btu ΔX ¼ ðRcond,ua Þðk Þ ¼ 5:53 0:023  Btu hr  ft  F ¼ 0:1272 ft

Convert the insulation thickness to inches. ΔX ¼ ð0:1272 ftÞ

  12 in ¼ 1:53 in ft

Solutions to Practice Problems

201

Practice Problem 9.2 (Solution) The thermal circuits for the double-pane and single-pane windows are shown in the figure. The overall temperature will be the same in both cases, and any convenient value, say, 20  F, can be taken for calculation purposes. The air is treated as a conductive layer.

Convert all thicknesses from inch to feet. 

 1 ft ¼ 0:0208 ft 12 in   1 ft ΔX Air ¼ ð0:0625 inÞ ¼ 0:0052 ft 12 in   1 ft ΔX SP ¼ ð0:50 inÞ ¼ 0:0417 ft 12 in

ΔX P1 ¼ ΔX P2 ¼ ð0:25 inÞ

Calculate the heat loss per unit area for the double-pane window by using Eq. 9.6. Include conduction resistance of air. q P ΔT overall ΔT overall ¼ ¼ ΔX ΔX Air ΔX P2 A Rcond,ua P1 þ þ kG k Air kG 20  F 1 0 ¼ 0 B 2@

1

0:0208 ft 0:0052 ft C B C Aþ@ A Btu Btu 0:61 0:015   hr  ft  F hr  ft  F ¼ 48:21 Btu=hr  ft2

202

9 Conduction Heat Transfer

Calculate the heat loss per unit area for the single-pane window by using Eq. 9.6. q ΔT overall ΔT overall ¼ ¼ ΔX SP A RSP,ua kG 20  F ¼0

1

B @

0:0417 ft C A Btu 0:61  hr  ft  F ¼ 292:57 Btu=hr  ft2 Calculate the percentage reduction in heat loss as shown.   ðq=AÞSP  ðq=AÞDP % reduction in heat loss ¼  100 ðq=AÞSP 0 1 Btu Btu 292:57  48:21 B hr  ft2 hr  ft2 C ¼@ A  100 Btu 292:57 hr  ft2 ¼ 83:52% Conclusion: 83% reduction in heat loss is achieved by using a double-pane window, and the air gap makes a significant contribution to the thermal resistance of the system. Practice Problem 9.3 (Solution) At steady state, the heat flow through the different layers will be constant at 73.37 Btu / hr – ft (calculated in Example 9.3). Let Rcs,ul be the thermal resistance of the calcium silicate insulation per unit length of the pipe. From Eq. 9.12 and using the results from Example 9.3: qcyl Btu ΔT overall ΔT overall ¼ ¼ ¼ 73:37 hr  ft L Rtotal,ul Rpipe,ul þ Rins1,ul þ Rcs,ul 400 ∘ F  80 ∘ F ¼ hr  ft2  ∘ F 3:4075 þ Rcs,ul Btu  ft length Solve the preceding equation for the thermal resistance of calcium silicate insulation.

Solutions to Practice Problems

203

   Btu hr  ft2  F 3:4075 73:37 hr  ft Btu  ft length   Btu þRcs,ul 73:37 ¼ 320  F hr  ft Rcs,ul ¼ 0:9540 hr  ft2  F=Btu  ft length Let r4 be the outer radius with the calcium silicate insulation in place. Calculate r4 by using Eq. 9.13.   

ln

r4

r3 hr  ft  F ¼ 2πkcs Btu  ft length       r4 hr  ft2  F Btu ð2π Þ 0:035 ln ¼ 0:9540 Btu  ft length hr  ft  F 5:315 in

Rcs,ul ¼ 0:9540

2

Solve the preceding equation for r4, r4 ¼ 6.556 in Therefore, the thickness of calcium silicate insulation required is: Δr cs ¼ r 4  r 3 ¼ 6:556 in  5:315 in ¼ 1:241 in Practice Problem 9.4 (Solution) Calculate the volume of the spherical tank in Example 9.4. V¼

    4 4 πr 1 3 ¼ ðπ Þð0:89 mÞ3 3 3 ¼ 2:953 m3

Calculate the mass of liquid ammonia in the tank using the definition of density (mass per unit volume).    kg  m ¼ ρ  V ¼ 686 3 2:593 m3 m ¼ 1779 kg Calculate the heat required to raise the temperature of liquid ammonia to its boiling point.

204

9 Conduction Heat Transfer

Q ¼ mcpl ðT b  T Þ   kJ ð33  C  ð60  CÞÞ ¼ ð1779 kgÞ 4:744 kg  K   ¼ 227869 kJ 2:279  108 J The time required to reach the boiling point is the heat required divided by the rate of heat transfer. Q 2:279  108 J ¼ q 100 W ¼ 2:279  106 s ð633 hrsÞ

t bp ¼

References 1. Aldossary, M.N., Baumann, R.L.: Heat Transfer Through a Cylinder, Wolram Demonstrations Project, USA (2017). Download from https://demonstrations.wolfram.com/ HeatTransferThroughACylinder/ 2. Arch Toolbox: R – Values of Insulation and Other Building Materials, USA (2021). Down load from https://www.archtoolbox.com/materials-systems/thermal-moisture-protection/rvalues.html 3. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P.: Fundamentals of Heat and Mass Transfer, 8th edn. Wiley, Hoboken (2019) 4. Connor, N.: What is Fourier’s Law of Thermal Conduction – Definition, Thermal Engineering, UK (2019).. Download from https://www.thermal-engineering.org/what-is-fouriers-law-ofthermal-conduction-definition/ 5. Engineering Library: DOE Fundamentals – Thermodynamics, Heat Transfer, and Fluid Flow, US Department of Energy (1992). Download from https://engineeringlibrary.org/reference/ conduction-heat-transfer-doe-handbook 6. Flynn, A.M., Akashige, T., Theodor, L.: Kern’s Process Heat Transfer, 2nd edn. Wiley, Hoboken (2019) 7. Holman, J.P.: Heat Transfer, 10th edn. McGraw Hill, New York (2009) 8. Jayanta, S.: Conduction of Heat through Cylindrical Wall, Engineering Notes. Download from https://www.engineeringenotes.com/thermal-engineering/heat-conduction/conduction-of-heatthrough-cylindrical-wall-thermal-engineering/30168 9. Jayanta, S.: Conduction of Heat through a Sphere, Engineering Notes. Download from https:// www.engineeringenotes.com/thermal-engineering/heat-conduction/conduction-of-heat-througha-sphere-thermal-engineering/30154

Chapter 10

Convection Heat Transfer

10.1

Newton’s Law of Cooling

As was mentioned in Chap. 8, convection heat transfer is heat transfer between a solid surface and a fluid adjacent to the solid surface. The heat transfer mechanism in convection heat transfer involves the motion and circulation of fluid particles. The equation governing convection heat transfer is called Newton’s law of cooling, which states that the heat transfer per unit area of the surface (also known as heat flux) is proportional to the difference between the surface temperature Ts and the temperature of the fluid far away from the surface, T1, also known as the bulk temperature of the fluid [1, 6]. The proportionality constant in Newton’s Law of Cooling is known as the heat transfer coefficient, h, also known as film coefficient or film conductance. The heat transfer coefficient h has the units Btu/hr-ft2- F (USCS) and W/m2. C (SI). Equation 10.1 is the mathematical representation of Newton’s law of cooling. q ¼ hAs ðT s  T 1 Þ

ð10:1Þ

In Eq. 10.1, As is the area of the surface in contact with the fluid. Conversion factor: 1 W/m2. C ¼ 0.17611 Btu/hr-ft2 –  F.

10.2

Convection Heat Transfer Resistance

Equation 10.1 can be rearranged to obtain the following equation.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_10

205

206

10

q¼

Ts  T1 1 hAs



ΔT Rconv

Convection Heat Transfer

ð10:2Þ

Comparing the terms in Eq. 10.2, the resistance to convection heat transfer [1, 4, 5] can be written as shown in Eq. 10.3. Rconv ¼

1 hAs

ð10:3Þ

Equation 10.1 can be written in terms of heat transfer rate per unit surface area. q T T ΔT conv ¼ s 1 1 As R conv,ua h

ð10:4Þ

Comparing the terms in Eq. 10.4, the resistance per unit surface area to convection heat transfer can be written as shown in Eq. 10.5. Rconv,ua ¼

1 h

ð10:5Þ

Thus, the resistance to convection heat transfer per unit surface area is the reciprocal of the heat transfer coefficient. Resistance is always inversely proportional to the transfer coefficient.

Example 10.1 Air at a bulk temperature of 70  F flows over a rectangular flat plate maintained at a temperature of 120  F. The heat transfer coefficient for air is 11.8 Btu/hr – ft2 –  F. The length of the plate is 3.7 ft, and the width is 1.4 ft. Calculate the heat transfer from the plate. (Solution) Calculate the surface area of the plate. As ¼ L  W ¼ 3:7 ft  1:4 ft ¼ 5:18 ft2 Calculate the heat transfer from the plate using Eq. 10.1.

10.5

Correlations Used in Calculating Convection Heat Transfer Coefficients

207

q ¼ hAs ðT s  T 1 Þ
   Btu ¼ 11:8 5:18 ft2 ð120 F  70 FÞ 2  hr  ft  F ¼ 3056:2 Btu=hr

10.3

Free and Forced Convection

As described in Sect. 8.3.2, there are two types of convection heat transfer mechanisms—free convection and forced convection [1, 3, 4]. In free convection, heat transfer takes place due to the density differences of the fluid particles, and there is no external force applied on the fluid. In forced convection, heat transfer takes place due to external forces applied on the fluid which causes the fluid to flow past the solid boundary. Forced convection heat transfer rates are proportional to the fluid velocity and involve dimensionless flow parameters like the Reynolds number.

10.4

Dimensionless Parameters Used in Heat Transfer

Calculation of convection heat transfer involves extensive use of heat transfer correlations. The heat transfer correlations typically use several dimensionless parameters [1, 3, 7]. Table 10.1 lists dimensionless parameters used in heat transfer.

10.5

Correlations Used in Calculating Convection Heat Transfer Coefficients

Convection heat transfer coefficients are usually calculated using heat transfer correlations [1, 3, 7], which are equations typically involving dimensionless parameters. Most of the heat transfer correlations are developed based on results from experimental studies. Hence, they are empirical and situation specific (laminar flow through pipes, turbulent flow through pipes, liquid metal heat transfer, etc.). While using heat transfer correlations, it is important to satisfy the constraints associated with the correlations. Correlations for convection heat transfer involve dimensionless parameters outlined in Table 10.1. Dimensionless parameters are calculated using fluid properties. Typically, the fluid properties are evaluated at the average bulk or mixing cup temperature of the fluid. A few important heat transfer correlations for convection heat transfer are presented in Table 10.2.

208

10

Convection Heat Transfer

Table 10.1 Dimensionless parameters used in heat transfer Dimensionless Parameter Reynolds number Grashof number Nusselt number Prandtl number Stanton number Peclet number Rayleigh number

Symbol Re

Formula Lvρ μ

Physical significance Ratio of inertial force to viscous force

¼ Lv ν

Gr

gL3 βðT s T 1 Þ ν2

Nu

hL k

Pr

cp μ k

St

Nu Re Pr

Pe

Re  Pr ¼ Lv α

Ra

Ra ¼ gL

Ratio of buoyancy force due to density difference to the resisting viscous force in free convection Ratio of heat transferred by convection to heat transferred by conduction Ratio of momentum diffusivity to thermal diffusivity Ratio of heat transfer from the fluid to the heat capacity of the fluid Ratio of heat transfer due to fluid motion to heat transfer due conduction Relative importance of buoyancy forces compared to viscous forces and thermal conduction

¼ αν ¼ ρchp v

3

βðT s T 1 Þ αν

Note: ν ¼ kinematic viscosity ¼ μρ; α ¼ thermal diffusivity ¼ ρck p; cp¼ specific heat (heat capacity) of the fluid, L ¼ characteristic dimension (diameter for a pipe, length for a flat plate), β ¼ coefficient of thermal expansion Table 10.2 Heat transfer correlations for convection heat transfer Situation Laminar flow in pipes Turbulent flow In pipes

Turbulent flow in pipes

Correlation with name Nu ¼ 3:66 þ

0:065 Re PrDL

ð10:6Þ

2

1þ0:04ð Re PrDLÞ3

Dittus-Boelter equation Nu ¼ 0.023Re0.8Prn (10.7) n ¼ 0.4 if fluid is being heated n ¼ 0.3 if fluid is being cooled Seider-Tate equation  0:14 Nu ¼ 0:027 Re 0:8 Pr1=3 μμ ð10:8Þ

Constraints Re < 2100 Re > 10,000

Re > 10,000

w

Laminar flow over flat plates Turbulent flow over flat plates Free convection: Immersed vertical plates

NuL ¼ 0.664ReL0.5Pr1/3

(10.9)

ReL < 5  105

NuL ¼ 0.037ReL0.8Pr1/3

(10.10)

5  105  ReL  107 0.6  Pr  60 All flow regimes, laminar (RaL < 109) turbulent (RaL > 1010)

Churchill 2 and Chu correlation 3 6 NuL ¼ 40:825 þ 

1þð

Free convection: Horizontal plates facing up and heating up the fluid

7 278 5

1

0:387RaL 6 9 0:492 16 Pr

Þ

ð10:12Þ NuL ¼ 0:54Ra0:25 L NuL ¼ 0:15Ra0:3333 ð10:13Þ L

ð10:11Þ

104  RaL  107RaL  107

10.5

Correlations Used in Calculating Convection Heat Transfer Coefficients

209

Example 10.2 Air at a bulk temperature of 70  F flows over a rectangular flat plate maintained at a temperature of 130  F. The velocity of air is 15 ft/sec. The length of the plate is 3.7 ft, and the width is 1.4 ft. Calculate the heat transfer coefficient for the air. The properties of air at the average bulk temperature of air (100  F) are density ¼ 0.0709 lbm/ft3; kinematic viscosity ¼ 1.8  104 ft2/sec; specific heat ¼ 0.2401 Btu/lbm –  F; and thermal conductivity ¼ 0.0156 Btu/hr – ft –  F. (Solution) Calculate the Reynolds number for the flow of air based on the length of the plate.   ft Lv ð3:7 ftÞ 15 sec 5 Re L ¼ ¼ 2 ¼ 3:083  10 ν 1:8  104 ft sec

Calculate the dynamic viscosity of air using Eq. 1.7. 

2 lbm 4 ft μ ¼ ρν ¼ 0:0709 3 ¼ 1:276  105 lbm=ft  sec 1:8  10 sec ft Calculate the Prandtl number for air.  Pr ¼

cp μ ¼ k

Btu lbm  F




lbm 1:276  105 ft  sec   
Btu 1 hr 0:0156 hr  ft  F 3600 sec

0:2401



¼ 0:707 Since ReL < 5  105, the flow is laminar. Calculate the Nusselt number using Eq. 10.9. NuL ¼ 0:664 Re L 0:5 Pr1=3  0:5 ¼ ð0:664Þ 3:083  105 ð0:707Þ1=3 ¼ 328:44 Calculate the heat transfer coefficient for air by using the definition of Nusselt number (Table 10.1). NuL ¼ 328:44 ¼

hair L k

210

10

hair

Convection Heat Transfer

 ð328:44Þ 0:0156

 Btu 328:44  k hr  ft  F ¼ ¼ L 3:7 ft ¼ 1:385 Btu=hr  ft2  F

Example 10.3 1200 cfm air enters a 2 ft W  1 ft. H rectangular duct at a temperature of 50  F, and it is to be heated to a temperature of 70  F. For this purpose, the walls of the duct are maintained at 90  F. The properties of air relevant to this situation are density ¼ 0.0744 lbm/ft3; kinematic viscosity ¼ 1.66  104 ft2/sec; specific heat ¼ 0.2399 Btu/lbm –  F; and thermal conductivity ¼ 0.0148 Btu/hr – ft –  F, Pr ¼ 0.713. Calculate the length of the duct required to accomplish this heating. (Solution) Energy (heat) balance across a differential length, dL, results in the following equation: heat lost by the wall ¼ heat gained by air ðhÞðdAÞðT w  T air Þ ¼ m_ air cp,air dT air The differential surface area corresponding to a differential length, dL, of the duct is the perimeter of the duct multiplied by the differential length. dA ¼ P  dL ¼ 2ð2 ft þ 1 ftÞdL ¼ 6  dL Substitute for dA into the heat balance equation, separate the variables, and then integrate with appropriate limits. ðhÞð6 ft  dLÞðT w  T air Þ ¼ m_ air cp,air dT air ZL


 TZair,out m_ air cp,air dT air dL ¼ 6 ft  h T w  T air T air,in

0



 m_ air cp,air T w  T air,in L¼ ln 6 ft  h T w  T air,out

Calculate the mass flow rate of air using the continuity equation (Eq. 3.5). 

lbm ft3 60 min 0:0744 3  1200 h min ft ¼ 5357 lbm=h

m_ air ¼ ρair Qair ¼

10.5

Correlations Used in Calculating Convection Heat Transfer Coefficients

211

Calculate the velocity of air using the continuity equation (Eq. 3.4). 3

vair ¼

ft 1 min 1200 min  60 Qair sec ¼ ¼ 10 ft= sec Acs,duct 2 ft  1 ft

Calculation of the Reynolds number for flow through the noncircular, rectangular duct requires the use of the hydraulic diameter of the duct. Calculate the hydraulic diameter of the duct using Eq. 3.23. DH ¼ 4




  Acs 2 ft  1 ft ¼4 P 2ð2 ft þ 1 ftÞ ¼ 1:3333 ft

Calculate the Reynolds number for flow of air through the duct using Eq. 3.7.
 ft ð1:3333 ftÞ 10 sec D v Re ¼ H air ¼ νair ft2 1:66  104 sec ¼ 80, 319 Since Re > 10,000, calculate the Nusselt number using the Dittus-Boelter equation (Eq. 10.7) for heating of fluid. Nu ¼ 0:023 Re 0:8 Pr0:4 ¼ 0:023ð80319Þ0:8 ð0:713Þ0:4 ¼ 168:6 Calculate the heat transfer coefficient for air using the definition of Nusselt number (Table 10.1). hair DH k  ð168:6Þ 0:0148

Nu ¼

hair

Btu ðNuÞðk Þ hr  ft  F ¼ ¼ 1:3333 ft DH ¼ 1:871 Btu=hr  ft2  F



Calculate the length of the duct required by substituting the known values into the equation for length.

212

10

Convection Heat Transfer



 m_ air cp,air T w  T air,in ln L¼ 6 ft  h T w  T air,out
 0 1 lbm Btu 5357 0:2399  B hr lbm  F C
 C ¼B @ A Btu ð6 ftÞ 1:871 2  hr  ft  F
  90 F  50 F ln 90 F  70 F ¼ 79:4 ft

10.6

Typical Range of Convection Heat Transfer Coefficients

The value of convection heat transfer coefficient depends on the specific situation being considered. For example, heat transfer coefficients for free convection will be much smaller in value compared to heat transfer coefficients for forced convection. Further, in forced convection itself, the value of the heat transfer coefficient depends on the properties of the fluid involved and also the velocity of the fluid. However, typical ranges for convection heat transfer coefficients are observed for commonly encountered situations like flow of air in a duct or flow of water in a pipe/tube. The typical ranges of convection heat transfer coefficients [3] for different situations are presented in Table 10.3.

10.7

Overall Heat Transfer Coefficients in Conduction-Convection Systems

Many practical situations involve a combination of conduction and convection heat transfer [1, 3, 4]. As described in Sect. 8.5, heat transfer from a heated room to a cooler ambient air involves convection heat transfer from room air to inside wall, conduction heat transfer through the room wall, and finally convection heat transfer from the outside wall to ambient air. Thus, this situation involves three resistances to heat flow as shown in Fig. 8.2 and represented mathematically in Eq. 8.3. The overall heat transfer coefficient is the inverse of the overall resistance to heat flow in conduction-convection systems.

10.7

Overall Heat Transfer Coefficients in Conduction-Convection Systems

Table 10.3 Typical ranges for convection heat transfer coefficients

Fluid Free convection Air Gases/dry vapors Water Oil Other liquids Forced convection Air flow Gas/vapor flow Water flow in tubes Liquid flow in tubes Liquid metal flow in pipes Boiling Condensation

213

Range for h (W/m2.K) 5–80 5–35 50–800 30–300 30–1500 10–150 25–250 300–3000 500–5000 3000–30,000 2000–100,000 5000–100,000

Note: To obtain the heat transfer coefficient in Btu/hr – ft2 –  F, multiply the values in the table by 0.1761 since 1 W/m2.  K ¼ 0.1761 Btu/hr – ft2 – F

Consider the conduction-convection system shown in Fig. 8.2, reproduced here for convenient reference.

Combination of conduction and convection (Fig. 8.2)

The heat flow is represented by Eq. 8.3. ΔT T room  T air q ¼ Poverall ¼ R Rth conv,i þ Rcond þ Rconv,o

ðEq: 8:3 from Section 8:5Þ

214

10

Convection Heat Transfer

Under steady-state conditions, the heat flow is constant through the room air, through the wall, and through the ambient (outside) air. The cross-section area of the wall perpendicular to the heat flow direction is also constant. Equation 8.3 can be written in terms of heat flow per unit cross-section area and the corresponding resistances (due to conduction and convection) per unit area (Eqs. 9.4 and 10.5). q ΔT ΔT overall ¼ P overall ¼ A R þ Rcond,ua þ Rconv,o,ua Rth,ua conv,i,ua ΔT overall ¼ 1 ΔX 1 þ þ hi k ho

ð10:14Þ

In Eq. 10.14, hi and ho are the convection heat transfer coefficients for the inside and outside air, respectively. The three heat transfer resistances can be combined and represented by a single resistance, which is the reciprocal of the overall heat transfer coefficient commonly represented by U. q ΔT overall ΔT overall ¼ ¼ 1 1 A h1 þ ΔX þ k h U i

ð10:15Þ

o

The following equations can be written based on Eq. 10.15. 1 1 þ ΔX k þ ho

ð10:16Þ

q ¼ UAΔT overall

ð10:17Þ

U¼

1 hi

Overall heat transfer coefficients are extensively used in heat exchanger design equations. A heat exchanger facilitates heat transfer from a hot fluid to a cold fluid. In this situation, again there will be three resistances: convection resistance due to the outside fluid, conduction resistance of the tube wall, and convection resistance due to the inside fluid in the tubes. Overall heat transfer coefficients for heat exchangers are discussed in detail in Chap. 12. Example 10.4 A window has an effective cross-section area of 24 ft2, and thickness of 0.25 in. The thermal conductivity of the window material is 0.61 Btu/hr – ft –  F. The heat transfer coefficient of room air is 4.5 Btu/hr – ft2 –  F, and the heat transfer coefficient of ambient air is 6.2 Btu/hr – ft2 –  F. Calculate: A. The overall heat transfer coefficient for the situation described. B. The heat loss through each window if the average inside temperature is 70  F and the outside temperature is 45  F.

10.7

Overall Heat Transfer Coefficients in Conduction-Convection Systems

215

(Solution) A. Calculate the overall heat transfer coefficient by using Eq. 10.16. U¼

1 1 ΔX 1 þ þ hi k ho


1

 1 ft 12 in 1 1 þ þ Btu Btu Btu 0:61 4:5 6:2 hr  ft  F hr  ft2  F hr  ft2  F ¼ 2:394 Btu=hr  ft2  F

¼

0:25 in 

B. Calculate the heat loss through the window by using Eq. 10.17. q ¼ UAΔT overall
   Btu ¼ 2:394 24 ft2 ð70 F  45 FÞ hr  ft2  F ¼ 1436:4 Btu=hr

10.7.1 Order of Magnitude Analysis to Determine the Value of Overall Heat Transfer Coefficients

In conduction-convection systems, the value of the overall heat transfer coefficient will be close to the reciprocal value of the controlling resistance [1, 3]. This concept is illustrated in the following example involving a building wall. Inside heat transfer coefficient, hi ¼ 40 W/m2.K Thickness of board, ΔX ¼ 1.5 cm and thermal conductivity of board material, k ¼ 0.82 W/m.K Outside heat transfer coefficient, ho ¼ 4 W/m2.K Calculate the individual resistances per unit area. Inside convection resistance:

216

10

Ri ¼

Convection Heat Transfer

1 1 ¼ ¼ 0:025 m2  K=W hi 40 mW 2 K

Board conduction resistance: ΔX 1:5 cm  100 cm ¼ ¼ 0:0183 m2  K=W W k 0:82 mK 1m

Rw ¼

Outside convection resistance: Ro ¼

1 1 ¼ ¼ 0:25 m2  K=W ho 4 mW 2 K

Calculate the total resistance and then the percentage contribution of each resistance to the total resistance. Rtotal ¼ Ri þ Rw þ Ro m2  K m2  K m2  K þ 0:0183 þ 0:25 ¼ 0:025 W W W ¼ 0:2933 m2  K=W The percentage contribution of inside convection resistance is


Ri Rtotal

  100 ¼

m2 K W 2 0:2933 mWK

0:025

!  100 ¼ 8:52%

The percentage contribution of wall conduction resistance is


Rw Rtotal

  100 ¼

0:0183 0:2933

m2 K W m2 K W

!  100 ¼ 6:24%

The percentage contribution of outside convection resistance is


Ro Rtotal

  100 ¼

m2 K W 2 0:2933 mWK

0:25

!  100 ¼ 85:24%

Thus, the outside convection resistance is the controlling resistance with a contribution of approximately 85% to the total or overall resistance. In this example, the overall heat transfer coefficient is:

10.8

The Relationship Between Fluid Flow and Heat Transfer

U¼

1 1 ¼ Rtotal 0:2933

m2 K W

217

¼ 3:41 W=m2  K

Thus, the value of the overall heat transfer coefficient (3.41 W/m2.K) is close to the value of the outside heat transfer coefficient (4 W/m2.K). In any situation, if the order of magnitude of one of the heat transfer coefficients is significantly lower compared to the value of the other heat transfer coefficient, then the overall heat transfer coefficient will be approximately equal to the significantly lower heat transfer coefficient.

10.8

The Relationship Between Fluid Flow and Heat Transfer

The fact that the Prandtl number (Pr) represents the ratio of the thickness of the hydrodynamic boundary layer to the thickness of the thermal boundary layer indicates a strong relationship between fluid flow and heat transfer. In forced convection heat transfer, calculation of the heat transfer coefficient involves the use of Reynolds number. Since the heat and flow fields are related, it raises the possibility of deriving expressions that relate the heat transfer coefficient, h, to the fluid friction factor, f.

10.8.1 Colburn Analogy Between Fluid Friction Factor and Heat Transfer Coefficient For fully developed turbulent flow in smooth pipes (Re > 10,000), the Chilton – Colburn Analogy [2, 3] between fluid friction and the heat transfer coefficient can be written as shown in Eq. 10.18. f ¼ jH 8

ð10:18Þ

jH ¼ Nu  Re 1  Pr3 1

ð10:19Þ

jH is the Colburn j factor for heat transfer, and f is the Moody friction factor. Substitute the expressions for Nusselt, Reynolds, and Prandtl numbers from Table 10.1 into Eq. 10.19, simplify, and combine the result with Eq. 10.18.

218

10

f ¼ 8

Convection Heat Transfer




 2 h Pr3 ρcp v

ð10:20Þ

Equation 10.20 relates the heat transfer coefficient to the fluid flow friction factor for fully developed turbulent flow in smooth pipes. Example 10.5 The following data for flow of water in condenser tubes is provided in Practice Problem 10.2: tube ID ¼ 41 mm; density of water ¼ 997 kg/m3; and specific heat of water ¼ 4.180 kJ/kg.K. From the solution to Practice Problem 10.2, the velocity of water in the tubes is 3.80 m/s, the Reynolds number for the flow of water in the tubes is 1.75  105, and the Prandtl number is 6.12. Assuming a smooth tube surface, estimate the heat transfer coefficient for water using the Chilton-Colburn analogy (Eq. 10.20), and compare the result with that obtained in Practice Problem 10.2 using the conventional equation for Nusselt number (hPP 10.2 ¼ 11,002 W/m2.K). (Solution) Determine the friction factor from the Moody diagram (Fig. 3.4) using the Reynolds number (1.75  105) and the curve for smooth pipe as shown in the figure.

From the figure, f ¼ 0.016. Substitute the known values into Eq. 10.20, and solve for the heat transfer coefficient, h. f ¼ 8




 2 h Pr3 ρcp v

0 0:016 @ h ¼   kg 8 997 4:180  103 m3

1 

J kgK

Að6:12Þ3 m 2

3:80

s

Solve the preceding equation for the heat transfer coefficient to obtain h ¼ 9525 W/m2.K, which is slightly lower than the value of h obtained from the solution to Practice Problem 10.2, where the result was h ¼ 11,002 W/m2.K. The discrepancy is due to the fact that the Colburn analogy uses an exponent of

Practice Problems

219

(1/3) ¼ 0.3333 for the Prandtl number, whereas the Dittus-Boelter equation uses an exponent of 0.4 for the Prandtl number for heating of a fluid.

Practice Problems Practice Problem 10.1 Calculate the convection resistance for the situation described in Example 10.1. Practice Problem 10.2 Cooling water flows inside the tubes (ID ¼ 41 mm) of a condenser at a mass flow rate of 5 kg/s. The water enters the tubes at 20  C and leaves at 30  C. Determine the heat transfer coefficient for the water flowing in the condenser tubes. Water has the following properties at the average bulk temperature of 25  C: density ¼ 997 kg/m3; kinematic viscosity ¼ 8.92  107 m2/sec; specific heat ¼ 4.180 kJ/kg.K; and thermal conductivity ¼ 0.6071 W/m.K. Practice Problem 10.3 A heated vertical plate (10 ft high by 20 ft wide) at 150  F is exposed to air at 60  F. Calculate the heat transfer coefficient due to free convection. Properties of air at the film temperature of 105 F are density ¼ 0.0703 lbm/ft3; kinematic viscosity ¼ 1.83  104 ft2/sec; specific heat ¼ 0.2401 Btu/lbm –  F; thermal conductivity ¼ 0.0157 Btu/hr – ft –  F, Pr ¼ 0.710; and coefficient of thermal expansion, β ¼ 1.785  103/ F. Practice Problem 10.4 A metallurgical furnace consists of 2 cm thick insulating ceramic brick (k ¼ 0.48 W/ m.K) on the inside exposed to hot gases at a temperature of 400  C. The heat transfer coefficient of the hot gases is 57 W/m2.K. Adjacent to the ceramic layer is a 3 cm thick steel shell with relatively negligible conduction resistance. The steel shell is covered with insulation (k ¼ 0.198 W/m.K). The insulation layer is exposed to ambient air at 25  C with heat transfer coefficient of 33 W/m2.K. Calculate the thickness of the insulating layer required to limit the heat loss per unit area of the furnace wall to a maximum value of 1875 W/m2. Practice Problem 10.5 Water flows in a 7.98 in ID steel tube at a velocity 11 ft/sec. The water is heated from 50 to 70  F over a tube length of 180 ft by virtue of the walls of the tubes being maintained at 100  F. The relevant properties of water are density ¼ 62.2 lbm/ft3; kinematic viscosity ¼ 9.30  106 ft2/sec; specific heat ¼ 0.998 Btu/lbm –  F; thermal conductivity ¼ 0.3520 Btu/hr – ft –  F; and Prandtl number ¼ 5.96. Calculate the heat transfer coefficient required for heating water by using energy balance, and compare this calculated value with that predicted by the ChiltonColburn analogy (Eq. 10.20).

220

10

Convection Heat Transfer

Solutions to Practice Problems Practice Problem 10.1 (Solution) Calculate the convection resistance for the air flow in Example 10.1 using Eq. 10.3. Rconv ¼

1 hAs

¼
11:8

1

   Btu 5:18 ft2 2  hr  ft  F ¼ 0:0164 F=ðBtu=hrÞ

Practice Problem 10.2 (Solution) Calculate the velocity of water by using the continuity equation (Eq. 3.5). v¼

m_ m_  ¼  ρA ρ π D 2 i 4 5

kg s


 ¼
2  kg π 1m 997 3 41 mm  4 1000 mm m ¼ 3:80 m=s Calculate the Reynolds number for water using Eq. 3.7.   Di v ð0:041 mÞ 3:80 ms Re ¼ ¼ ¼ 1:75  105 2 ν 8:92  107 m s

Calculate the dynamic viscosity of water using Eq. 1.7. μ ¼ ρν ¼



kg m2 ¼ 8:893  104 kg=m  s 997 3 8:92  107 s m

Calculate the Prandtl number for water using the formula given in Table 10.1.

Solutions to Practice Problems

221





 J 4 kg 8:893  10 4:180  10 kg  K ms cp μ Pr ¼ ¼   W k 0:6071 mK ¼ 6:12 3

For flow of water through this pipe, since Re ¼ 1.75  105 > 10,000, calculate the Nusselt number using the Dittus-Boelter equation (Eq. 10.7). The exponent n ¼ 0.4 since water is being heated. Nu ¼ 0:023 Re 0:8 Pr0:4  0:8 ¼ ð0:023Þ 1:75  105 ð6:12Þ0:4 ¼ 743 Calculate the heat transfer coefficient for water by using the definition of Nusselt number (Table 10.1). The characteristic dimension, L, in the case of a pipe is the inside diameter, Di. hw D i k   W ð 743 Þ 0:6071 743  k mK ¼ hw ¼ 0:041 m Di ¼ 11002 W=m2  K Nu ¼ 743 ¼

Practice Problem 10.3 (Solution) The Nusselt number for free convection in a vertical immersed plate can be calculated by using Eq. 10.11. This involves the use of the Rayleigh number (Ra) and Prandtl number (Pr). Calculate the thermal diffusivity of air using the formula given in Table 10.1.
 Btu 1 hr 0:0157  hr  ft  F 3600 sec k  α¼ ¼
 ρcp lbm Btu 0:0703 3 0:2401 lbm  F ft ¼ 2:584  104 ft2 = sec Calculate the Rayleigh number by using the formula given in Table 10.1. The characteristic dimension, L, for a vertical plate is the height of the plate.

222

10

Convection Heat Transfer

gL3 βðT s  T 1 Þ αν 

3 ft 3 1:785  10 32:2 ð150 F  60 FÞ ð10 ftÞ F sec 2

 ¼ 2 2 4 ft 4 ft 2:584  10 1:83  10 sec sec

Ra ¼

¼ 1:094  1011 2

3 1 6

0:387RaL 6 7 NuL ¼ 40:825 þ  0:492169 278 5 1 þ Pr 2 3  1 11 6 0:387 1:094  10 6 7 ¼ 40:825 þ  0:492169 278 5 1 þ 0:710 ¼ 23:26 Calculate the heat transfer coefficient for air using the definition of Nusselt number (Table 10.1). hair L k  ð23:26Þ 0:0157 Nu ¼

hair

Btu ðNuÞðk Þ hr  ft  F ¼ ¼ L 10 ft ¼ 0:0365 Btu=hr  ft2  F



Practice Problem 10.4 (Solution) Calculate the overall heat transfer coefficient for this situation using Eq. 10.17. U¼

1875 mW2 ðq=AÞ ¼ 5:0 W=m2  C ¼ ΔT overall 400 C  25 C

Neglecting the conduction resistance of the steel shell, the heat transfer resistances involved in this situation are inside convection resistance of hot gases, conduction resistance of ceramic layer, conduction resistance of insulation, and outside convection resistance of ambient air. Subscript “c” represents the ceramic layer, and subscript “i” represents the insulation. For this situation, Eq. 10.16 can be written as follows, and then all the known values can be substituted.

Solutions to Practice Problems

223

1 1 ΔX c ΔX i 1 ¼ þ þ þ U hi ho kc ki 1m 2 cm  1 1 ΔX i 100 cm þ ¼ þ W W W W 0:48 5 2 57 2 0:198 mK mK m K m K 1 þ W 33 2 m K Solve the preceding equation for ΔXi, the thickness of insulation required. ΔX i ¼ 0:0219 m ð2:19 cm ’ 22 mmÞ Practice Problem 10.5 (Solution) Convert the inside diameter of the tube to feet.
D¼

7:98 in 

1 ft 12 in

 ¼ 0:665 ft

Determine the mass flow water by using the continuity equation (Eq. 3.5).   
lbm π ft 2 ð0:665 ftÞ ¼ ρwater Avwater ¼ 62:2 3 11 4 sec ft ¼ 237:64 lbm= sec


m_ water

Energy (heat) balance across a differential length, dL, of the tube results in the following equation: heat lost by the tube wall ¼ heat gained by water ðhÞðdAÞðT wall  T water Þ ¼ m_ water cp,water dT water The differential surface area corresponding to a differential length, dL, of the duct is the perimeter of the tube multiplied by the differential length. dA ¼ P  dL ¼ πDdL ¼ π ð0:665 ftÞdL ¼ 2:09 ft  dL Substitute for dA into the heat balance equation, separate the variables, and then integrate with appropriate limits.

224

10

Convection Heat Transfer

ðhÞð2:09 ft  dLÞðT wall  T water Þ ¼ m_ water cp,water dT water ZL 0

water,out
 TZ m_ water cp,water dT water dL ¼ 2:09 ft  h T wall  T water

T water,in



 m_ water cp,water T wall  T water,in L¼ ln 2:09 ft  h T wall  T water,out

Solve the preceding equation for h, and substitute the known values.

 m_ water cp,water T wall  T water,in h¼ ln 2:09 ft  L T wall  T water,out 0 1 lbm Btu 3600 sec
 237:64  0:998  F 100 F  50 F B C sec lbm‐ hr ¼@ A ln 100 F  70 F 2:09 ft  180 ft ¼ 1, 159 Btu=hr‐ft2 ‐ F To calculate the heat transfer coefficient by using the Chilton-Colburn analogy, determine the Reynolds number by using Eq. 3.7.   ft Dvwater ð0:665 ftÞ 11 sec Re ¼ ¼ ¼ 7:9  105 ft2 νwater 9:30  106 sec Using the Reynolds number and the curve for a smooth pipe, determine the Darcy friction factor from the Moody diagram as shown here.

From the Moody diagram, f ¼ 0.012. Using this value of f, and other known values, determine the heat transfer coefficient from Eq. 10.20.

References

f ¼ 8

225




 2 h Pr3 ρcp v 0

0:012 B ¼@ 8

1

2 h C Að5:96Þ3 lbm Btu ft 3600 sec   11 62:2 3  0:998 lbm‐ F sec hr ft

Solve the preceding equation for the heat transfer coefficient, h ¼ 1140 Btu/hr-ft2F. The heat transfer coefficient obtained from the heat balance method is h ¼ 1159 Btu/hr-ft2- F. The two values are fairly consistent with each other.



References 1. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P. (eds.): Fundamentals of Heat and Mass Transfer, 8th edn. Wiley (2019) 2. Coletti, F. (ed.): Heat Exchanger Design Handbook Multimedia Edition. Begell Digital Portal 3. Flynn, A.M., Akashige, T., Theodor, L.: Kern’s Process Heat Transfer, 2nd edn. Wiley (2019) 4. Holman, J.P.: Heat Transfer, 10th edn. McGraw Hill (2009) 5. Kirkpatrick, A.: Heat Transfer Resistance Modeling, Heat Transfer in Engines. Download from https://www.engr.colostate.edu/~allan/heat_trans/page5/page5.html 6. Nuclear Engineering, Newton’s Law of Cooling. Download from https://www.nuclear-power. net/nuclear-engineering/heat-transfer/convection-convective-heat-transfer/newtons-law-ofcooling/ 7. Whitaker, S.: Forced convection heat transfer correlations for flow in pipes, past flat plates, single cylinders, single spheres, and for flow in packed beds and tube bundles. AICHE J. 18(2), 361–371 (1972)

Chapter 11

Radiation Heat Transfer

11.1

Introduction

Every object emits electromagnetic radiation [2, 6] that depends on the absolute temperature (K or
R) of the object. The radiation energy emitted by a body is due to the vibrational and rotational motions of molecules, atoms, and electrons within the body. The flow of heat caused by the emission of radiant energy is called radiation heat transfer. The vibrational and rotational activities of the fundamental particles in a body are directly related to the temperature of the body. Therefore, the rate of radiation energy (heat) transfer from a body increases with increasing temperature. In reality, there is a constant exchange of radiant energy between bodies. Consider a red hot metal rod in a room. The rod emits considerable amount of radiant energy due to its high temperature, and this energy reaches the surrounding objects such as the walls of the room. However, the walls of the room also emit radiation energy by virtue of the fact that the walls are at a temperature higher than absolute zero. Some of the radiation energy from the walls could reach the red hot metal rod depending on the geometry, configuration, and positioning of the walls and the rod. Thus, radiation heat transfer involves net energy exchange between the hot and cold bodies [1, 5, 11]. Since the temperature of the red hot metal rod (designated as body “1”) is much higher than the temperature of the surrounding walls (designated as body “2”), there will be a net radiation heat flow from the red hot metal rod to the walls. The net heat transfer from body “1” to body “2” is symbolically represented as q12. The calculation of the net radiation heat transfer is described in detail in the following sections.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_11

227

228

11.2

11

Radiation Heat Transfer

Stefan-Boltzmann’s Law of Thermal Radiation

Stefan-Boltzmann’s law quantifies the radiation heat emitted by a body. It states that the radiation energy emitted per unit surface area of a perfect or ideal radiationemitting body (also known as a “black body”) is proportional to the fourth power of absolute temperature [2, 3, 5]. An ideal radiator or black body is one that can emit the maximum possible radiation by virtue of its absolute temperature. A black body also absorbs all the radiation that is incident upon it [9]. Black bodies are represented with subscript “b” in equations. Mathematically, Stefan-Boltzmann’s law can be written as: Eb / T 4

ð11:1Þ

In Eq. 11.1, Eb is energy emitted by a black body per unit surface area per unit time, which is the same as the heat flux due to radiation with units W/m2 (S I) and Btu / hr – ft2 (USCS). Also in Eq. 11.1, T is the absolute temperature of the body in Kelvin (S I) or degree Rankine (USCS), which can be obtained from the following equations: K ¼
C þ 273

and




R ¼
F þ 460

The proportionality relationship in Eq. 11.1 can be transformed into a regular equation by using a proportionality constant known as Stefan-Boltzmann’s constant [7], represented by the symbol σ. The values of Stefan-Boltzmann’s constant in both the S I and USCS units are given here. σ ¼ 5:669  108 W=m2  K4

ðS IÞ

σ ¼ 0:1714  108 Btu=hr  ft2 
R4

ðUSCSÞ

Based on Stefan-Boltzmann’s law, the rate of radiation heat flux (heat transfer per unit surface area) from a black body can be calculated by using Eq. 11.2. Eb ¼ σT 4

ð11:2Þ

Example 11.1 Calculate the radiation heat flux from a red hot metal object kept in a furnace at a temperature of 700
F. (Solution) Calculate the absolute temperature of the red hot metal object. T abs ¼ T ∘ F þ 460 ¼ 700∘ F þ 460 ¼ 1160∘ R

11.2

Stefan-Boltzmann’s Law of Thermal Radiation

229

Calculate the radiation heat flux from the object using Eq. 11.2. E b ¼ σT 4
¼ 0:1714  108

 Btu ð1160∘ RÞ4 hr  ft2  ∘ R4

¼ 3103Btu=hr  ft2 Hint for easy calculation in situations involving radiation: Absolute temperatures to the fourth power typically have very high numerical values. Calculations related to radiation heat transfer can avoid the problem of dealing with very high value numbers by dividing the absolute temperature by 100 and then taking the fourth power, which will result in the handling of reasonably smaller numbers. This approach works because dividing the absolute temperature by 100 should be accompanied by multiplying the absolute temperature by 100. Raising 100 to the fourth power will result in 108 which will cancel with 108 in the Stefan-Boltzmann’s constant.

11.2.1 Nonideal Radiators: Gray Bodies In most practical situations, objects emit only a fraction of the radiation energy that is predicted by Stefan-Boltzmann’s law. This is because the objects are not perfect radiators and do not emit the same amount of energy as a black body. Thus, most of the objects are gray bodies. Gray bodies are characterized by a factor known as emissivity represented by the symbol ε. The emissivity of a gray body is defined as the ratio of the actual energy, E, emitted by the body to the energy emitted by a black body, Eb, at the same temperature [2, 4, 5]. The emissivity, ε, of a body is the ratio of the emissive power of the body to that of a black body at the same temperature. Mathematically, emissivity is represented as shown in Eq. 11.3. ε¼

E Eb

ð11:3Þ

The radiation heat transfer per unit surface area of a gray body can be calculated by combining Eqs. 11.2 and 11.3. E ¼ εE b ¼ εσT 4

ð11:4Þ

The emissivity of materials depends to a great extent on the nature of the surface. Polished surfaces tend to have low values of emissivity since they reflect most of the incident radiation. Rough and weathered surfaces tend to have relatively higher

230 Table 11.1 Typical emissivity of surfaces

11

Radiation Heat Transfer

Surface description Cold rolled steel Type 301 stainless steel Weathered stainless steel Commercial aluminum sheet Polished metal surfaces Matte copper surface Fire brick Rough red brick Concrete Sand Graphite Smooth glass Plastics Oil paint Water Snow Ice

Emissivity, ε 0.80 0.60 0.85 0.09 0.02–0.06 0.22 0.75 0.92 0.90 0.90 0.75 0.93 0.90–0.97 0.94 0.96 0.97 0.98

values of emissivity. The typical values of emissivity of gray surfaces are given in Table 11.1.

11.2.2 Absorptivity, Transmissivity, and Reflectivity The incident radiation on a surface (also known as irradiation) can be either absorbed or transmitted or reflected. The fraction of irradiation absorbed is called absorptivity, α; the fraction of irradiation transmitted is known as transmissivity, τ; and the fraction of irradiation reflected is called reflectivity, ρ. The sum of these fractions should be one. αþτþρ¼1

ð11:5Þ

Kirchhoff’s law of radiation [2, 8] states that objects have identical capacities to emit as well as absorb radiation. For a black body, the absorptivity is equal to emissivity, and both are equal to unity since a black body absorbs all the irradiation and also emits the maximum possible radiation at a given temperature. For a black body, α¼ε¼1

ð11:6Þ

11.4

11.3

Calculation of Net Radiation Heat Transfer

231

Radiation View Factor

Radiation is due to electromagnetic waves emanating from a surface. Hence, it is a phenomenon based on the characteristics of the surface. As explained in Sect. 11.1, radiation heat transfer involves net heat exchange between two bodies, body “1” and body “2.” Since radiation heat transfer is a surface-based phenomenon, the net radiation heat transfer from body “1” to body “2” (represented as q12 and assuming T1 > T2) will depend on the geometric configuration of the two surfaces. To include the effects of geometric configuration in the calculation of net radiation heat transfer from body “1” to body “2,” a factor known as radiation view factor is used [1–3, 5]. The radiation view factor is represented by the symbol F12, and it is also known as shape factor or configuration factor. Depending on the geometric configuration of the two bodies, it is likely that some radiation from body “1” will bypass body “2.” The view factor, F12, represents the fraction of total radiation from body “1” that actually reaches body “2.”

11.3.1 View Factor Relationships The view factor for net radiation heat flow from body “i” to body “j” can be represented by Fij. Ai represents the surface area of body “i.” It can be shown that the following view factor relationships are valid in radiation heat transfer. Ai F ij ¼ A j F ji n X

F ij ¼ 1

ð11:7Þ ð11:8Þ

j¼1

Equation 11.7 is known as the reciprocity rule, and Eq. 11.8 is known as the summation rule.

11.4

Calculation of Net Radiation Heat Transfer

Consider two objects, “1” and “2,” that are exchanging heat by radiation with the following assumptions: 1. Both the objects are gray bodies, with emissivities ε1 and ε2. 2. The temperature of object “1” is greater than the temperature of object “2.”

232

11

Radiation Heat Transfer

The net radiation heat transfer from body 1 to body 2 can be conveniently modeled using the concept of driving force and resistance [2, 5, 11]. The driving force is based on the temperature difference between bodies 1 and 2.   Radiation heat transfer driving force ¼ Eb1  E b2 ¼ σ T 1 4  T 2 4 The net radiation heat transfer from body 1 to body 2 encounters the following resistances: Surface Resistance of body 1 ¼

1  ε1 ε1 A1

Space Resistance between bodies 1 and 2 ¼ Surface Resistance of body 2 ¼

1 A1 F 12

1  ε2 ε2 A2

Based on the concept of heat flow being equal to the driving force divided by the sum of resistances, the equation for net radiation heat transfer from body 1 to body 2 can be written by combining the preceding equations.   σ T 14  T 24

q12 ¼ 1ε1 ε1 A 1

2 þ A11F 12 þ 1ε ε2 A2

ð11:9Þ

Equation 11.9 is a general equation for net radiation heat transfer from body 1 to body 2, and it can be simplified for specific cases.

11.4.1 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Black Bodies Assuming that both the small object and the enclosure are black bodies: ε1 ¼ ε2 ¼ 1:0 All the radiation from the small object (body 1) will reach the enclosure (body 2). Therefore, F12 ¼ 1.0. After substituting the preceding results in Eq. 11.9, the equation for radiation from a small body to an enclosure is:   q12 ¼ σA1 T 1 4  T 2 4

ð11:10Þ

11.4

Calculation of Net Radiation Heat Transfer

233

11.4.2 Radiation Heat Transfer from a Small Object to an Enclosure, Both Being Gray Bodies All the radiation from the small object (body 1) will reach the enclosure (body 2). Therefore, F12 ¼ 1.0. Also, the surface area of the enclosure is much larger than the surface area of the pipe. Therefore: A2 >>>>>> A1 and

1  ε2 ¼0 ε2 A2

Substitute the preceding results into Eq. 11.9. q12 ¼

  σ T 14  T 24 1ε1 ε1 A1

þ

1 A1

  ¼ σε1 A1 T 1 4  T 2 4

ð11:11Þ

Example 11.2 A 100 W bulb can be considered as a black body-emitting radiation into an enclosure at 70
F, also considered as a black body. The filament of the bulb has a diameter of 0.005 in and total length of 2.5 in. Determine the filament temperature. (Solution) The 100 W bulb is body “1,” and the enclosure is body “2.” Convert the radiation heat transfer rate from watts to Btu/hr. q12


 3:412 Btu=hr ¼ ð100 WÞ ¼ 341:2 Btu=hr 1W

Calculate the surface area of the filament.

 1 ft 1 ft 2:5 in  A1 ¼ πDL ¼ ðπ Þ 0:005 in  12 in 12 in ¼ 2:73  104 ft2 Convert the temperature of the enclosure to degree Rankine. T 2 ¼ 70∘ F þ 460∘ ¼ 530∘ R All the radiation from the bulb reaches the enclosure. Hence, the view factor for radiation from the bulb to the enclosure is 1.0, that is, F12 ¼ 1.0. Since the filament and enclosure are both black bodies, their emissivities are 1.0.

234

11

Radiation Heat Transfer

ε1 ¼ ε2 ¼ 1:0 Substitute for the emissivities in Eq. 11.9 and simplify.   q12 ¼ σA1 F 12 T 1 4  T 2 4 Substitute the known values into the preceding equation, and solve for the temperature T1. 341:2

Btu ¼ hr


0:1714  108

 Btu ð2:73  104 ft2 Þð1:0Þ hr  ft2  ∘ R4  ðT 1 4  ð530∘ RÞ4

Solve for T1. T 1 ¼ 5197∘ R ð4737∘ FÞ Example 11.3 Two black parallel plates face each other as shown in the figure. The temperature of plate A is 300
C. Plate B is at 100
C, and it is centered parallel to plate A. Calculate the net radiation heat transfer from plate A to plate B.

(Solution) Since plate A is much bigger than plate B and since plate B is centered while facing plate A, it is reasonable to assume that all the radiation from plate B will reach plate A. Hence, the view factor from B to A is FBA ¼ 1.0. Since TA > TB, the net

11.5

Heat Transfer Equilibrium in Radiation Systems

235

radiation heat transfer will be from A to B. Calculate the view factor from A to B by using the reciprocity rule (Eq. 11.7). Ai F ij ¼ A j F ji ) F AB ¼


   AB 8 cm  8 cm ð1:0Þ ¼ 0:40 F BA ¼ 8 cm  20 cm AA

Calculate the absolute temperatures of plates A and B. T A ¼ 300∘ C þ 273∘ ¼ 573K

T B ¼ 100∘ C þ 273∘ ¼ 373K

Calculate the net radiation heat transfer from A to B using Eq. 11.9.   σ T A4  T B4

qAB ¼ 1εA εA AA

B þ AA F1 AB þ 1ε εB AB

Since both the plates are black, εA ¼ 1.0 and εB ¼ 1.0. The preceding equation can be simplified and used as shown.   qAB ¼ σAA F AB T A 4  T B 4
 W ¼ 5:669  108 2 4 ð0:08 m  0:20 m Þð0:4Þ m K   4  ð573 KÞ  ð373 KÞ4 ¼ 32:09 W

11.5

Heat Transfer Equilibrium in Radiation Systems

Radiation incident on a surface is often dissipated by conduction or convection, resulting in a combination of at least two modes of heat transfer. In such cases, an equilibrium temperature can be determined by using steady-state energy balance, which has the following general format: Radiation heat transfer from body 1 to body 2 ¼ Heat dissipated by body 2 either by conduction or convection The following examples illustrate the calculation of steady-state equilibrium temperatures. Example 11.4 A thin cylindrical, hot wire of diameter 0.5 in and length 6 in radiates heat to the surrounding wall enclosure with a total area of 4 square feet. The thickness of walls

236

11

Radiation Heat Transfer

in the enclosure is 3 in. The inside wall temperature is 200
F, the outside wall temperature is 70
F, and the thermal conductivity of the wall material is 0.6356 Btu/hr – ft –
F. The difference in surface areas of the inside and outside walls can be neglected. Determine the temperature of the wire. (Solution) Calculate the heat conducted by the wall using Eq. 9.2.
qcond ¼ kw Aw

ΔT ΔX w



0 1  ∘ ∘ Btu B200 F  70 FC ð4ft2 Þ@ A 1ft hr  ft  ∘ F 3in  12in ¼ 1322 Btu=hr

 ¼ 0:6356

Calculate the surface area of the wire to be used in radiation calculations.

 1 ft 1 ft A1 ¼ πDL ¼ ðπ Þ 0:5 in  6 in  ¼ 0:0654 ft2 12 in 12 in Under steady-state conditions, the heat conducted by the wall is equal to the heat transferred from the hot wire to the wall by radiation. qcond ¼ qrad ¼ 1322 Btu=hr Calculate the absolute temperature of the wall enclosure (body 2). T 2 ¼ 200∘ F þ 460∘ ¼ 660∘ R Calculate the temperature of the wire (body 1) using the equation for radiation heat transfer from a small object to an enclosure, with both being black bodies (Eq. 11.10).   q12 ¼ σA1 T 1 4  T 2 4 ) T1

4

Btu hr þ ð660∘ RÞ4 Btu 2 8 ð0:1714  10 Þð0:0654 ft Þ hr  ft2  ∘ R4 ¼ 1:198  1013 1322

q ¼ 12 þ T 2 4 ¼ σA1

1

T 1 ¼ ð1:198  1013 Þ4 ¼ 1860∘ R

ð1400∘ FÞ

11.5

Heat Transfer Equilibrium in Radiation Systems

237

11.5.1 Correction for Thermocouple Readings Thermocouples, used in measuring the temperature of hot gases, are usually situated inside a thermowell. In this situation, the temperature indicated by the thermocouple, TTC, is actually an equilibrium temperature rather than the actual temperature of the gas, TG [2, 3, 10]. The hot gas transfers heat to the thermowell by convection, and the thermowell in turn dissipates the heat to the surroundings (at temperature Ts), such as the pipe wall, by radiation. The steady-state energy balance is: Heat transferred by the gas to the thermocouple by convection ¼ heat dissipated by the thermocouple to the surroundings by radiation   hðT G  T TC Þ ¼ σεTC T TC 4  T S 4

ð11:12Þ

Example 11.5 The thermocouple used in measuring the temperature of a hot gas in a duct indicates a temperature of 300
C. The emissivity of the thermocouple material is 0.45, and the wall temperature of the duct is 70
C. The heat transfer coefficient for the gas is 355 W/m2.K. Determine the actual gas temperature and the percent error in the thermocouple measurement. (Solution) Calculate the absolute temperatures of the thermocouple reading and the surrounding walls. T TC ¼ 300∘ C þ 273∘ ¼ 573K

T S ¼ 70∘ C þ 273∘ ¼ 343K

Substitute the known values, and calculate the gas temperature, TG, using Eq. 11.12.   hðT G  T TC Þ ¼ σεTC T TC 4  T S 4 )    σεTC  T TC 4  T S 4 þ T TC TG ¼ h  0
1 W 8 5:669  10 ð0:45Þ B C m2  K4 Cðð573 KÞ4  ð343 KÞ4 Þ þ 300∘ C ¼B @ A W 355 2 m K ¼ 306:75∘ C Note: There is no need to convert the temperatures to absolute values for convection heat transfer (on the left-hand side of the equilibrium equation, Eq. 11.12) since ΔT
C ¼ ΔT K Calculate the percent error in the thermocouple measurement.

238

11

Radiation Heat Transfer



 T G  T TC 306:75∘ C  300∘ C  100 ¼ 2:20% % error ¼  100 ¼ 306:75∘ C TG Example 11.6 A metallic pipe 18 in long, 3 in outer diameter, has a surface emissivity of 0.68 and a surface temperature of 350
F. The pipe is exposed to an enclosure at an ambient temperature of 75
F. The convection heat transfer coefficient of the air in the enclosure is 1.25 Btu/hr – ft2 –
F. Calculate: A. The fraction of heat transfer from the pipe due to radiation B. The overall heat transfer coefficient for the total heat transfer from the pipe (Solution) Body 1 is the pipe, and the enclosure is body 2. A. Calculate the external surface area of the pipe.

 1 ft 1 ft A1 ¼ πDL ¼ ðπ Þ 18 in  3 in  ¼ 1:178 ft2 12 in 12 in All the radiation from the pipe surface will reach the enclosure. Therefore,  A1 F12 ¼ 1.0 and from Eq. 11.7 F 21 ¼ A2 F 12 ¼ 0 since A2>>>> A1 Calculate the absolute temperatures of the pipe surface and the enclosure. T 1 ¼ 350∘ F þ 460∘ ¼ 810∘ R

T 2 ¼ 75∘ F þ 460∘ ¼ 535∘ R

Substitute the preceding results into Eq. 11.9, and simplify to calculate the net radiation heat transfer from the pipe.

Practice Problems

239

Calculate the convection heat transfer from the pipe to the air in the enclosure using Eq. 10.1.  Btu ð1:178 ft2 Þð350∘ F  75∘ FÞ hr  ft2  ∘ F ¼ 404:94 Btu=hr

qconv ¼ hA1 ðT 1  T 1 Þ ¼


1:25

Calculate the total heat transfer from the pipe. qtotal ¼ qrad þ qconv ¼ 478:54

Btu Btu þ 404:94 ¼ 883:88 Btu=hr hr hr

Calculate the fraction of heat transfer from the pipe due to radiation. f rad ¼

478:54 Btu qrad hr ¼ ¼ 0:5414 qtotal 883:88 Btu hr

Calculate the overall heat transfer coefficient using Eq. 10.17. U¼

883:88 Btu qtotal hr ¼ ¼ 2:728 Btu=hr  ft2  ∘ F A1 ΔT overall ð1:178ft2 Þð350∘ F  75∘ FÞ

Practice Problems Practice Problem 11.1 Estimate the temperature in
C of a black body that emits a heat flux of 550 W/m2. Practice Problem 11.2 A steel pipe (ID ¼ 12 in, OD ¼ 12.75 in) carries steam at a temperature of 500oF. The pipe is insulated with calcium silicate insulation, and the surface temperature of insulation is 150
F. The pipe is located in a facility where the walls of the enclosure are at 80
F. The emissivity of the insulation surface is 0.75. Calculate the heat loss per foot length of the insulated pipe due to radiation. Practice Problem 11.3 Calculate the net radiation heat transfer from plate A to plate B in Example 11.3 if both the plates have an emissivity of 0.7. Practice Problem 11.4 Use the same data as in Example 11.3. If the wire temperature is 8000F, determine the temperature of the inside wall.

240

11

Radiation Heat Transfer

Practice Problem 11.5 A thin metal plate, 1.5 ft  1.2 ft, receives radiation flux of 1250 Btu/hr – ft2. Some of the incident heat is dissipated by convection, and the convection heat transfer coefficient is 17 Btu / hr – ft2 –
F. The plate attains an equilibrium temperature of 130
F, and the surrounding air temperature is 80
F. The plate is opaque, and it receives and dissipates heat only on one side. Determine: A. The emissivity of the plate material B. The reflectivity of the plate material Practice Problem 11.6 A heat pipe with an outer diameter of 219 mm and surface temperature of 190
C is placed in a room adjacent to a 75 mm thick, 3.25 m  3.25 m brick wall (k ¼ 1.25 W/m.K) with an inside surface temperature of 30
C. The emissivities of the pipe surface and the brick wall are 0.75 and 0.85, respectively. The geometric configuration is such that 65% of the radiation from the pipe reaches the brick wall. The outside air temperature is 18
C. At steady-state condition, determine the required outside heat transfer coefficient of the brick wall.

Solutions to Practice Problems Practice Problem 11.1 (Solution) Calculate the absolute temperature of the black body by using Eq. 11.2.  14 Eb E b ¼ σT 4 ) T ¼ ¼ σ

550 mW2 5:669  108 m2WK4

!14 ¼ 314 K

Convert the temperature to degree Celsius. ∘

C ¼ 314K  273∘ ¼ 41∘ C

Practice Problem 11.2 (Solution) The radiation occurs from the outside surface of the pipe. Calculate the outside surface area of the pipe in terms of L, the length of the pipe.
 1 ft Ao ¼ πDo L ¼ π 12:75 in  L ¼ 3:338L ft2 12 in

Solutions to Practice Problems

241

Determine the absolute temperatures of the two bodies. Body 1 is the insulated pipe, and the enclosure is body 2. T 1 ¼ 150∘ F þ 460∘ ¼ 610∘ R

T 2 ¼ 80∘ F þ 460∘ ¼ 540∘ R

All the radiation from the pipe will reach the walls of the enclosure. Therefore, the view factor is F12 ¼ 1.0. The surface area of the enclosure will be much greater than 2 the surface area of the pipe, that is, A2 >>> A1, and the term 1ε ε2 A2 can be neglected relative to the other terms in the denominator of Eq. 11.9. Substitute the preceding results and the known values into Eq. 11.9.

q12 ¼

σðT 1 4  T 2 4 Þ ¼ σA1 ε1 ðT 1 4  T 2 4 Þ 1  ε1 1 þ ε1 A1 A1
 Btu 8 ¼ 0:1714  10 ð3:338L ft2 Þð0:75Þ hr  ft2  ∘ R4  ðð610∘ RÞ4  ð540∘ RÞ4 Þ ¼ 229:26 L Btu=hr

The heat loss per unit length is qL12 ¼ 229:26 Btu=hr  ft Practice Problem 11.3 (Solution) Use the same data as in Example 11.3, and calculate the net radiation heat transfer from A to B with their emissivities using Eq. 11.9. The following data is available from the solution to Example 11.3: T A ¼ 300∘ C þ 273∘ ¼ 573 K,

T B ¼ 100∘ C þ 273∘ ¼ 373 K, F AB ¼ 0:40

Substitute all the known information into Eq. 11.9.

242

11

Radiation Heat Transfer

  σ T A4  T B4 qAB ¼ 1  εA 1 1  εB þ þ AA F AB εA AA εB A B
  W 5:669  108 2 4 ð573 KÞ4  ð373 KÞ4 m K ¼2 3 1  0:7 1 þ 6 ð0:7Þð0:08 m  0:20 m Þ ð0:08 m  0:20 m Þð0:4Þ 7 6 7 4 5 1  0:7 þ ð0:7Þð0:08 m  0:08 m Þ ¼ 20:06 W Practice Problem 11.4 (Solution) Use the following data from Example 11.4 and its solution. For the wall: kw ¼ 0:6356 Btu=hr  ft  ∘ F, Aw ¼ 4ft2 , T wo ¼ 70∘ F, ΔX w ¼ 3 in ¼ 0:25 ft For the wire, T 1 ¼ 800∘ F þ 460∘ ¼ 1260∘ RðGiven for this problemÞ:

A1 ¼ 0:0654 ft2 ,

Let Twi be the inside temperature of the wall. Under steady-state conditions, the heat transferred from the hot wire to the wall by radiation is equal to the heat conducted by the wall. qrad ¼ qcond Use Eq. 11.10 for radiation heat transfer from a small object to an enclosure and Eq. 9.2 for conduction heat transfer. 

σA1 T 1  T wi 4

4




 T wi  T wo ¼ k w Aw ΔX w

Substitute the known values into the preceding equation.


0:1714  108

 i  h Btu 0:0654 ft2 ð1260
RÞ4  T wi 4 2
4 hr  ft  R
    T wi  70
F Btu 2 ¼ 0:6356 4 ft hr  ft 
F 0:25 ft

Solutions to Practice Problems

243

Simplify the preceding equation, and solve for the inside wall temperature, Twi, by trial and error.
0:0112

   Btu Btu ∘ 4 4 R  T Þ ¼ 10:17 ðT wi  70∘ FÞ ð25205 wi ∘ 4 hr  F hr  ∘ R
 ∘ F 0:0011 ∘ 4 ð25205∘ R4  T wi 4 Þ ¼ ðT wi  70∘ FÞ R

Solve for Twiby trial and error, Twi¼ 97
F. Practice Problem 11.5 (Solution) A. Since the plate is opaque, the incident radiation is either absorbed or reflected. Let α be the absorptivity of the plate. Under steady-state conditions, the heat balance is: Radiation heat absorbed by the plate ¼ heat dissipated by the air due to convection αðAÞ

    q ¼ hA T eq:plate  T a A

Solve the preceding equation for α, and substitute the known values.

α¼

  h T eq:plate  T a q A

 ¼

 Btu 17 hrft ð130
F  80
FÞ 2
 F Btu 1250 hrft 2

¼ 0:68

From Kirchhoff’s law of radiation, the absorptivity and emissivities of materials are equal. ε ¼ α ¼ 0:68 Since the plate is opaque, the transmissivity of the plate, τ ¼ 0. Calculate the reflectivity of the plate material using Eq. 11.5. ρ ¼ 1  α ¼ 1  0:68 ¼ 0:32 Practice Problem 11.6 (Solution) At steady state, the heat is transferred from the pipe to the brick wall by radiation, through the brick wall by conduction and finally to the outside air by convection. Subscript 1 denotes the pipe, and subscript 2 denotes the brick wall. Assume one meter length of the pipe. Calculate the absolute temperatures of pipe and brick wall surfaces.

244

11

T 1 ¼ 190
C þ 273
¼ 463 K

Radiation Heat Transfer

T 2 ¼ 30
C þ 273
¼ 303 K

It is specified that 65% of the radiation from the pipe reaches the brick wall; therefore, F12 ¼ 0.65. Calculate the surface area of the pipe. A1 ¼ πDL ¼ ðπ Þð0:219 mÞð1:0 mÞ ¼ 0:6880 m2 Calculate the surface area of the wall. A2 ¼ ð3:25 mÞ2 ¼ 10:56 m2 Substitute the known values, and calculate the radiation heat transfer from the pipe using Eq. 11.9.   σ T 14  T 24 q12 ¼ 1  ε1 1 1  ε2 þ þ A1 F 12 ε1 A 1 ε2 A 2
  W 4 4 8 5:669  10  ð 303 K Þ ð 463 K Þ m2  K4 ¼ 1  0:75 1 1  0:85 þ þ 0:75  0:6880 m2 0:6880 m2  0:65 0:85  10:56 m2 ¼ 777:15 W At steady state, q12, rad ¼ q2, cond ¼ q3, conv ¼ 777.15 W Calculate the temperature drop across the brick wall using Eq. 9.2. ΔT wall

  q2,cond ðΔX wall Þ ð777:15 WÞð0:075 mÞ ¼ ¼   W A2 k wall ð10:56 m2 Þ 1:25 mK ¼ 4:42 K ð4:42
CÞ

Note: ΔT K ¼ ΔT
C ¼ 4.42




C

Calculate the outside temperature of the brick wall. T wo ¼ T wi  ΔT w ¼ 30
C  4:4
C ¼ 25:6
C Calculate the required outside heat transfer coefficient of the brick wall using Eq. 10.1. h¼

q3,conv 777:15 W ¼ A2 ðT wo  T air Þ ð10:56 m2 Þð25:6
C  18
CÞ ¼ 9:683 W=m2 
C

References

245

References 1. Amirault, S.B.: Radiation Shape Factor, SBA Invent – Mechanical Engineering Reference and Example Problems. Download from https://sbainvent.com/heat-transfer/radiation/radiationshape-factor/ 2. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P.: Fundamentals of Heat and Mass Transfer, 8th edn. Wiley, Hoboken (2019) 3. Flynn, A.M., Akashige, T., Theodor, L.: Kern’s Process Heat Transfer, 2nd edn. Wiley, Hoboken (2019) 4. Gedeon, M.: Thermal Emissivity and Radiation Heat Transfer, Materion Technical Tidbits (2018). Download from https://materion.com/-/media/files/alloy/newsletters/technical-tidbits/ issue-no-114-thermal-emissivity-and-radiative-heat-transfer.pdf 5. Holman, J.P.: Heat Transfer, 10th edn. McGraw Hill, New York (2009) 6. Lucas, J.: What is Electromagnetic Radiation?, Live Science (2015). Download from https:// www.livescience.com/38169-electromagnetism.html 7. Modest, M.F.: Fundamentals of Thermal Radiation – Stefan Boltzmann’s Constant, Science Direct (2013). Download from https://www.sciencedirect.com/topics/engineering/stefanboltzmann-constant 8. More, H.: Kirchhoff’s Law of Radiation, The Fact Factor (2020). Download from https:// thefactfactor.com/facts/pure_science/physics/kirchhoffs-law-of-radiation/8235/ 9. Nave, R.: Blackbody Radiation, Hyper Physics. Download from http://hyperphysics.phy-astr. gsu.edu/hbase/mod6.html 10. Shaddix, C.R.: A New Method to Compute the Proper Radiant Heat Transfer Correction of Bare Wire Thermocouple Measurements. In: Proc. of 10th U.S. National Combustion Meeting, (2017) 11. Srikanth, R.: Radiative Heat Transfer (2020). Download from https://www.jntua.ac.in/gateonline-classes/registration/downloads/material/a159101052285.pdf

Chapter 12

Heat Exchangers

12.1

Introduction

Heat exchangers [1, 7, 8] facilitate heat transfer from a hot fluid to a cold fluid. The two most widely used heat exchangers are shell and tube heat exchangers [5] and double pipe heat exchangers [4]. The flow arrangement can be either parallel (both fluids have the same flow direction) or counterflow (the two fluids flow in opposite directions). The counterflow arrangement is usually preferred since it results in a greater average temperature difference between the hot and cold fluids. Figures 12.1 and 12.2 illustrate the typical layouts of shell and tube and double pipe heat exchangers.

12.2

Heat Balance

The heat balance [1, 7] in a heat exchanger is an application of conservation of energy principle. Ideally, the exchanger is well insulated, resulting in negligible heat transfer to/from the surroundings. With this reasonable assumption, all the energy released by the hot fluid (subscript “h”) is absorbed by the cold fluid (subscript “c”). The rate of heat exchange between the hot and cold fluids is also known as the “heat duty” of the exchanger represented by the symbol q. The heat balance equation is: q ¼ m_ h cph ΔT h ¼ m_ c cpc ΔT c

ð12:1Þ

The heat balance equation is particularly important in solving heat exchanger problems. It is used in calculating any unknown parameter in the heat exchanger, as illustrated in the following example.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_12

247

248

12

Heat Exchangers

Fig. 12.1 Shell and tube heat exchanger schematic. (Source: https://commons.wikimedia.org/wiki/ File:Shell_heat_exchanger_LS.JPG,Public Domain)

Fig. 12.2 Schematic of a counterflow double pipe heat exchanger

Example 12.1 In a double pipe heat exchanger, 5000 kg/hr of water is heated from 15 to 75  C by using 15000 kg/hr of hot oil available at 140  C. The specific heats of water and oil are cp,water ¼ 4.187 kJ/kg. C and cp,oil ¼ 1.97 kJ/kg. C. Calculate the exit temperature of the oil and the heat duty of the exchanger. (Solution) Here, oil is the hot fluid, and water is the cold fluid. Use the heat balance equation (Eq. 12.1) to calculate the temperature decrease of the oil. q ¼ m_ h cph ΔT h ¼ m_ c cpc ΔT c ) 
 kg   kJ 4:187 5000  hr m_ c cpc ΔT c kg: C ð75 C  15 CÞ   ¼ ΔT h ¼ ¼ 43 C
 kg m_ h cph kJ 15000 hr 1:97 kg: C

12.2

Heat Balance

249

T exit,oil ¼ T inlet,oil  ΔT h ¼ 140 C  43 C ¼ 97 C The heat duty of the exchanger is either the heat lost by the oil or the heat gained by water. 

  kg kJ q ¼ m_ c cpc ΔT c ¼ 5000 ð75 C  15 CÞ 4:187 hr kg  C ¼ 1256100 kJ=hr Usually, the heat duty is expressed in watts (W) or kilowatts (kW) in S I units.    kJ 1 hr q ¼ 1256100 ¼ 348:92 kW ð349 kWÞ hr 3600 s

Standard values of specific heat of water, cp ¼ 1.0 Btu/lbm   F (USCS) and cp ¼ 4.187 kJ/kg.K (S I) Many industrial problems can be solved by simple heat balance as shown in Example 12.2. Example 12.2 The heat duty of a condenser in a refrigeration system is 30,000 Btu/hr. Cooling water enters the condenser at 65  F and leaves at 80  F. The power input to the compressor in the same system is 4.50 kW. Calculate the ratio gpm of water flow rate in the condenser per ton of refrigeration (12,000 Btu/hr) in the system. The schematic for a refrigeration system is shown in the figure.

250

12

Heat Exchangers

(Solution) Calculate the mass flow rate of water using the heat duty of the condenser and heat balance. heat rejected in the condenser ¼ heat absorbed by water: Therefore, qcond ¼ m_ w cpw ΔT w q m_ w ¼ cond ¼ cpw ΔT w

30000

Btu hr

Btu 1  15 F lbm F ¼ 2, 000 lbm=hr

The standard density of water is ρw ¼ 8.34 lbm/gal Calculate the volume flow rate of cooling water.

V_ water

   lbm 1 hr 2000 hr 60 min m_ ¼ w ¼ lbm ρw 8:34 gal ¼ 4:0 gpm

The refrigerant absorbs the heat from the refrigerated space in the evaporator of the refrigeration system. Calculate the heat absorbed in the evaporator by energy balance on the refrigeration system as per the schematic shown in the figure. energy added to the system ¼ energy removed from the system _ c,in ¼ Q_ out Q_ in þ W

_ c,in Q_ in ¼ Q_ out  W

0

1 Btu 3412 Btu B hr C  ð4:50 kWÞ@ ¼ 30000 A hr kW ¼ 14, 646 Btu=hr

Convert the heat transfer to the evaporator to tons of refrigeration. 0 1   Btu B1 ton refrigerationC qin ¼ 14646 @ A Btu hr 12000 hr ¼ 1:22 tons Calculate the required ratio.

12.3

Log Mean Temperature Difference (LMTD)

251

gpm water 4:0 gpm ¼ ton refrig: 1:22 tons ¼ 3:28 gpm water=ton refrig:

Standard Values for Density of Water: ρw ¼ 62.4 lbm/ft3 (USCS) ¼ 8.34 lbm/ gal (USCS) ¼ 1000 kg/m3 (S I) ¼ 1 kg/L (S I)

12.3

Log Mean Temperature Difference (LMTD)

The driving force for heat transfer in a heat exchanger is the temperature difference between the hot fluid and the cold fluid. This temperature difference varies along the length of the exchanger as shown in Figs. 12.3a and 12.3b. The log mean temperature difference (LMTD) represents an average temperature difference between the hot and cold fluids in the exchanger [1, 7, 8]. Fig. 12.3a Variation in temperature difference for parallel flow

Fig. 12.3b Variation in temperature difference for counterflow

252

12

Heat Exchangers

The log-mean temperature difference can be calculated by using the following formula written in terms of temperature differences at end “1” and at end “2” of the heat exchanger. ΔT LMTD ¼

ΔT 1  ΔT 2 ΔT 2  ΔT 1   ¼   1 2 ln ΔT ln ΔT ΔT 2 ΔT 1

ð12:2Þ

Example 12.3 Calculate the log-mean temperature difference for both parallel and counterflow for the heat exchanger in Example 12.1. (Solution) In Example 12.1, water is heated from 15 to 75  C, and oil is cooled from 140 to 97  C. Draw the schematic diagrams for parallel and counterflows as shown, and calculate the LMTD in each case by using Eq. 12.2. Parallel flow 1 2 Water 15∘ C ! 75∘ C Oil 140∘ C ! 97∘ C ∘ ΔT1 ¼ 125 C ΔT2 ¼ 22∘ C ΔT LMTD,PF ¼

ΔT 1  ΔT 2 125∘ C  22∘ C   ¼   ΔT 1 125∘ C ln ln ΔT 2 22∘ C

¼ 59:3∘ C

Counter flow 1 2 Water 15∘ C ! 75∘ C Oil 97∘ C   140∘ C ∘ ΔT1 ¼ 82 C ΔT2 ¼ 65∘ C ΔT LMTD,CF ¼

ΔT 1  ΔT 2 82∘ C  65∘ C   ¼  ∘  ΔT 1 82 C ln ln ΔT 2 65∘ C ¼ 73:2∘ C

Example 12.3 clearly illustrates the fact that the log mean temperature difference is always greater for counterflow, and hence the counterflow arrangement is preferred in heat exchangers.

12.3.1 LMTD Correction Factors In multi-pass heat exchangers, it is common for the tube side fluid to make several passes through the exchanger. In a “1–2 exchanger,” the shell side fluid makes a single pass through the exchanger, and the tube side fluid makes two passes through the exchanger. Hence the nomenclature is “1–2 exchanger.”

12.3

Log Mean Temperature Difference (LMTD)

253

Fig. 12.4a MTD correction factors for one shell pass and multiple tube passes (2, 4, 6, etc.). (Source: Basic Design Methods of Heat Exchangers (by Cuneyt Ezgi), Open Access Peer Reviewed Chapter, https://www.intechopen.com/chapters/54521)

Fig. 12.4b MTD correction factors for two shell passes and multiple tube passes (4, 8, 16, etc.). (Source: Basic Design Methods of Heat Exchangers (by Cuneyt Ezgi), Open Access Peer Reviewed Chapter, https://www.intechopen.com/chapters/54521)

In this exchanger, there is parallel flow during the first tube pass and counterflow during the second tube pass. Usually, LMTD is calculated based on counterflow since it results in a higher value. However, in multi-pass exchangers, the LMTD value will be lower than the calculated value due to the drop in LMTD during parallel flow. The LMTD calculated based on counterflow is adjusted by using a correction factor [6, 7] known as “MTD correction factor.” MTD correction factors are obtained from graphs generated by TEMA (Figs. 12.4a, and 12.4b). The MTD correction factor will be 1.0 for single-pass (1–1) exchangers and for condensers and evaporators. This is because condensation and evaporation are isothermal (constant temperature) processes resulting in phase change. The use of MTD correction factors will be illustrated later in this chapter.

254

12.4

12

Heat Exchangers

Overall Heat Transfer Coefficient

The overall heat transfer coefficient [1–3, 7] used in the design of heat exchangers is the reciprocal of the overall resistance to heat transfer. In the most general case, there are five resistances to heat transfer between the hot and cold fluids in a heat exchanger. Proceeding from the fluid inside the tubes to the outside atmosphere, the resistances are: 1. 2. 3. 4. 5.

Convection resistance due to the fluid inside the tube ¼ hi1Ai Fouling resistance due to the inside fluid ¼ Ri/Ai ðro =ri Þ Do =Di Þ Conduction resistance due to the inside pipe wall ¼ ln2πkL ¼ ln ð2πkL Fouling resistance [2, 3, 7] due to the fluid outside the tube ¼ Ro/Ao Convection resistance due to the outside fluid ¼ ho1Ao

The fouling factors, Ri, and Ro, have the reciprocal units of heat transfer coefficient (hr-ft2- F/Btu or m2. C/W). Fouling resistances are due to the formation of coatings on tube surfaces. The heat transfer areas are the inside-and-outside surface areas of the tube (not the cross-section area). Therefore, Ao ¼ πDoL and Ai ¼ πDiL, the outside and inside surface areas of the inner tube. L is the length of the exchanger, and Di and D0 are the ID and OD of the inner pipe. The difference between the inside and outside surface areas of the tube gives rise to two overall heat transfer coefficients. Ui is the overall heat transfer coefficient based on the inside surface area, Ai, of the inner tube, and Uo is the overall heat transfer coefficient based on the outside surface area, Ao, of the inner tube. The overall resistance is the sum of the individual resistances. Therefore, the following equation is obtained after adding up all the resistances mentioned earlier. 1 U i Ai

¼

ln ðDo =Di Þ Ro 1 1 R 1 þ þ ¼ þ iþ 2πkL U o Ao hi Ai Ai Ao ho Ao

ð12:3Þ

Since Ai ¼ πDiL and Ao ¼ πDoL, the overall heat transfer coefficient, Ui, can be calculated by using the following equation obtained by multiplying Eq. 12.3 throughout by Ai ¼ πDiL.     ðDi Þ ln ðDo =Di Þ 1 1 Di 1 Di þ Ro ¼ þ Ri þ þ 2k Ui hi ho D o Do

ð12:4Þ

Similarly, the overall heat transfer coefficient, Uo, can be calculated by using the following equation obtained by multiplying Eq. 12.3 throughout by Ao ¼ πDoL. 1 1 ¼ Uo hi

    ðD Þ ln ðDo =Di Þ Do Do 1 þ Ro þ þ Ri þ o 2k ho Di Di

ð12:5Þ

12.4

Overall Heat Transfer Coefficient

255

Invariably, the conduction resistance due to the tube wall can be neglected because the thermal conductivity of the tube wall material (usually copper or steel) is high, and the wall resistance will be much smaller than the other resistances. Further, by assuming Di  Do, the equation for the overall heat transfer coefficient takes the following simple form. 1 1 1 1 1 ¼ ¼ ¼ þ þ Ri þ Ro Uo Ui U hi ho

ð12:6Þ

Sometimes, the inside and outside fouling factors are combined into a single fouling factor, Rf. Following conservative design practices, heat exchangers are designed using a design overall heat transfer coefficient, UD. The overall heat transfer coefficient under clean conditions, UC, does not include the fouling factor. Equation 12.6 can be simplified to the expressions shown in Equations 12.7 and 12.8. 1 1 1 ¼ þ þ Rf UD hi ho

ð12:7Þ

1 1 1 ¼ þ UC hi ho

ð12:8Þ

Since UD includes an additional resistance due to fouling, UC > UD. Therefore, heat exchangers will be overdesigned when they are first put into service, and they will have more surface area than required. Combining Eqs. 12.7 and 12.8, 1 1 ¼ þ Rf UD UC

ð12:9Þ

Equation 12.9 is frequently used in the design of heat exchangers. Table 12.1 lists typical fouling resistances for different fluids.

Table 12.1 Typical fouling resistances for different fluids

Fluid Cooling water Organic liquids Refrigerants Salt solutions Steam Organic vapors Flue gases



 Fouling resistance, Rf hrft Btu 0.0014 0.0011 0.0014 0.0028 0.0009 0.0011 0.0019 2

F

Note: Multiply the values listed in the table by 0.1761 to obtain the values in m2. K/W

256

12

Heat Exchangers

Conversion Factor for Fouling Resistance: 1:0 hr  ft2  F=Btu ¼ 0:17612 m2 : C=W

Table 12.2 lists typical ranges of design overall heat transfer coefficients for different pairs of fluids with fouling included. Conversion Factor: 1.0 Btu/hrft2 F ¼ 5.678 W/m2 C

Example 12.4 The following data is available for the heat exchanger in Example 12.1, with water flowing in the inner tube and oil in the outer pipe. Heat transfer coefficient for water: 700 Btu/hrft2 F Heat transfer coefficient for oil: 75 Btu/hrft2 F Inner tube: 0.50 in OD, 0.049 in wall thickness (18 BWG) Thermal conductivity of tube material, k ¼ 29 Btu/hrft F Length: 8 ft Fouling factor for water ¼ 0.001 hrft2 F/Btu Fouling resistance for oil ¼ 0.002 hrft2 F/Btu Determine the thermal resistance of each component of the overall heat transfer coefficient based on the inside area of the inner tube, and calculate the percentage of each based on the overall resistance. Calculate the overall heat transfer coefficient and comment on the results. Table 12.2 Typical overall heat transfer coefficients (U ) for shell and tube heat exchangers (U is in Btu /hrft2-F)

Fluid combination Water-water Organic liquid-organic liquid Heavy oil-heavy oil Light oil-light oil Gas-gas Gas-water Organic liquid-water Organic vapor-water Steam-organic liquid Steam-water Light oil-steam/water Heavy oil-steam/water Steam-brine

Design U range 150–250 30–90 10–50 15–60 5–10 5–50 40–120 125–150 100–200 150–250 50–150 10–60 250–500

Note: Multiply the values listed in the table by 5.678 to obtain the values in W/m2. K

12.4

Overall Heat Transfer Coefficient

257

(Solution) Calculate the inside and outside surface areas of the inner tube. 

1 ft Do ¼ ð0:5 inÞ 12 in

 ¼ 0:0417 ft 

1 ft Di ¼ ð0:5 in  0:049 in  2Þ 12 in

 ¼ 0:0335 ft

Ao ¼ πDo L ¼ π ð0:0417 ftÞð8 ftÞ ¼ 1:048 ft2 Ai ¼ πDi L ¼ π ð0:0335 ftÞð8 ftÞ ¼ 0:8419 ft2 Calculate the individual resistances using Eq. 12.3. Inside convection resistance: 1 1  
¼  ¼ 0:0017 hr  F=Btu hi A i 2 Btu 700 hrft2  F 0:8419 ft Inside fouling resistance: 

 F Ri 0:001 hrft Btu ¼ ¼ 0:0012 hr  F=Btu 2 Ai 0:8419 ft 2

Conduction resistance due to the inside pipe wall: ln DDoi ln 0:0417 ftft
0:0335  ¼ 0:00015 hr  F=Btu ¼ Btu 2πkL ð2πÞ 29 hrft  F ð8 ftÞ Outside fouling resistance: 

 F Ro 0:002 hrft Btu ¼ ¼ 0:0019 hr  F=Btu 2 Ao 1:048 ft 2

Outside convection resistance: 1 1  
¼  ¼ 0:0127 hr  F=Btu ho A o 2 Btu 75 hrft 1:048 ft 2   F Calculate the total resistance by adding up the individual resistances.

258

12

Heat Exchangers

hr  F hr  F hr  F þ 0:0012 þ 0:00015 Btu  Btu  Btu hr  F hr  F þ0:0019 þ 0:0127 Btu Btu ¼ 0:01765 hr  F=Btu

Rtotal ¼ 0:0017

Calculate the percentage resistance of each component based on the total resistance.  F 0:0017 hr Btu Inside convection resistance ¼ 0:01765hr  F 100 ¼ 9:6%  Btu hr F 0:0012 Btu Inside fouling resistance ¼ 0:01765hr  F 100 ¼ 6:8% Btu    0:00015 hr F Conduction resistance due to the inside pipe wall ¼ 0:01765hrBtu F 100 ¼ 0:8% Btu  F 0:0019 hr Btu Outside fouling resistance ¼ 0:01765hr  F 100 ¼ 10:8%  Btu hr F  0:0127 Btu Outside convection resistance ¼ 0:01765hr  F 100 ¼ 72% Btu 

F From Eq. 12.3, Rtotal ¼ U1i Ai ¼ 0:01765 hr Btu

Ui ¼


1  ¼ 67:30 Btu=hr  ft2  F  F
2 0:8419 ft 0:01765 hr Btu

The following conclusions can be drawn from the results. 1. The tube wall resistance is extremely small and can be safely neglected. 2. The fouling resistances are significant (18%) and must be included for satisfactory design. 3. The outside convection resistance is an order of magnitude of nine times the inside convection resistance. In such cases, the lower heat transfer coefficient (75 Btu/hr-ft2- F) has the greatest influence on the overall heat transfer coefficient (67 Btu/hr-ft2- F). 4. The outside convection resistance and the fouling resistances constitute 91% of the total resistance.

12.5

Heat Exchanger Design Equation

The heat exchanger design equation [3, 6, 7] is used in determining the heat transfer surface area required for the exchanger. The general form of the heat exchanger design equation is: q ¼ UAFΔT LMTD

ð12:10Þ

The heat duty, q, is determined from heat balance (Eq. 12.1). The MTD correction factor, F, is used only for multi-pass heat exchangers, and it can be determined from

12.5

Heat Exchanger Design Equation

259

the graphs (Figs. 12.4a and 12.4b). The MTD correction factor is 1 for single-pass exchanger and also for multi-pass condensers and evaporators. Example 12.5 illustrates the use of the heat exchanger design equation. Example 12.5 10,000 lbm/hr of hot oil at 200  F is to be cooled to 140  F using cooling water available at 60  F in a 1–1 shell and tube exchanger with counterflow. The maximum allowable temperature rise of the cooling water, available at 60  F, is 20  F. The overall design heat transfer coefficient is 125 Btu/hr-ft2- F. The specific heats of water and oil are 1.0 Btu/lbm- F and 0.55 Btu/lbm- F, respectively. Determine: A. The heat exchanger surface area required B. The number of 3=8 in (ID ¼ 0.326 in) tubes required to achieve a water velocity of 9 ft/sec C. The length of each tube (Solution) A. Calculate the heat duty using Eq. 12.1. q ¼ m_ oil cp,oil ΔT oil   lbm Btu 0:55 ¼ 10000 ð200 F  140 FÞ  hr lbm  F ¼ 330, 000 Btu=hr Calculate the log mean temperature difference using Eq. 12.2. 1 Water 60 C ! 80 C 140 C

Oil

200 C



ΔT2 ¼ 120 C

ΔT1 ¼ 80 C ΔT LMTD ¼

2

ΔT 1  ΔT 2   ΔT 1 ln ΔT 2

120 C  80 C   120 C ln 80 C  ¼ 98:6 C

¼

For a 1–1 exchanger, the MTD correction factor, F, is 1. Rearrange the heat exchanger design equation (Eq. 12.10) to obtain a formula for the heat exchanger surface area required. As ¼

q UFΔT LMTD

ð12:11Þ

Substitute the known values into Eq. 12.11 to obtain the heat transfer surface area required.

260

12

Heat Exchangers

Btu 330000 hr Btu ð1:0Þð98:6 FÞ hr  ft2  F ¼ 26:77 ft2

q As ¼ ¼ UFΔT LMTD 125

B. Calculate the mass flow rate of water required by using the heat balance equation (Eq. 12.1). Btu 330000 q hr m_ w ¼ ¼  Btu cpw ΔT w 1:0 ð20 FÞ lbm  F ¼ 16, 500 lbm=hr
 Calculate the ID of the tubes, Di ¼ ð0:326 inÞ 121 ftin ¼ 0:0272 ft. Use the continuity equation from fluid mechanics, Eq. 3.5, to calculate the mass flow rate through each tube. The number of tubes required is the total mass flow rate of water divided by the mass flow rate of water in each tube.

n¼

m_ w ρvAcs

   lbm 1 hr 16500 hr 3600 sec    ¼ lbm ft π 62:4 3 ð0:0272 ftÞ2 9 sec 4 ft ¼ 14 tubes

C. Calculate the length of each tube by using the heat exchanger surface area required (As). L¼

As 26:77 ft2 ¼ nðπDi Þ ð14Þðπ Þð0:0272 ftÞ ¼ 22:38 ft

Example 12.6 Redesign the heat exchanger in Example 12.5 if the tube length is restricted to 12 ft with all the other parameters being the same, except the heat exchanger surface area. (Solution) Two tube bundles will be required to satisfy the tube length restriction and keep the parameters such as velocity, mass flow rate of water, and the number of tubes in

12.5

Heat Exchanger Design Equation

261

each bundle to be the same as in Example 12.5. With two tube bundles, this will be a 1–2 exchanger requiring the use of the MTD correction factor. Determine the MTD correction factor from Fig. 12.4(a). Use the nomenclature given in Fig. 12.4(a) (shown here for reference). R¼

T 1  T 2 200 F  140 F ¼ ¼ 3:0 80 F  60 F t2  t1

P¼

t2  t1 80 F  60 F ¼ T 1  t 1 200 F  60 F

¼ 0:1429 With R ¼ 3 and P ¼ 0.143 as parameters, obtain the MTD correction factor as shown in the figure. F ¼ 0.97

Calculate the revised heat exchanger area required by using the MTD correction factor. As,rev ¼

As 26:77 ft2 ¼ ¼ 27:6 ft2 F 0:97

Calculate the length of each tube in the two tube bundles. L¼

As,rev 27:6 ft2 ¼ 2nðπDi Þ 2ð14Þðπ Þð0:0272 ftÞ ¼ 11:54 ft

Therefore, the final design parameters for the heat exchanger will be: 1–2 shell and tube heat exchanger with 14 tubes per pass. 3=8 in OD, 18 BWG tubes (ID ¼ 0.326 in) and 12 ft in length resulting in a heat exchanger surface area of 28.2 ft2.

262

12.6

12

Heat Exchangers

Heat Exchanger Effectiveness

Heat exchanger effectiveness, ε, is defined as the ratio of actual heat transfer to the maximum possible heat transfer [3, 6, 7]. ε¼

qactual qmax

ð12:12Þ

The product of the mass flow rate and heat capacity of a fluid is known as the “heat capacity rate” and is represented by the letter C. The heat capacity rates of the hot and cold fluids can be written in terms of the following equations. Ch ¼ m_ h cph and C c ¼ m_ c cpc

ð12:13Þ

The lower of Ch and Cc values is designated as Cmin, and the higher value is designated as Cmax. The maximum possible temperature difference is Th, in  Tc, in, and therefore the maximum possible heat transfer can be calculated by using the following equation: qmax ¼ C min ðT h,in  T c,in Þ

ð12:14Þ

In terms of the heat capacity rates, the actual heat transfer is: qactual ¼ Cc ΔT c ¼ C h ΔT h

ð12:15Þ

Using Eqs. 12.12, 12.14, and 12.15, the heat exchanger effectiveness can be written as: ε¼

C ΔT ðor Ch ΔT h Þ qactual ¼ c c qmax Cmin ðT h,in  T c,in Þ

ð12:16Þ

Knowing the effectiveness of the heat exchanger, the actual heat transfer can be calculated by the following equation. qactual ¼ C c ΔT c ¼ C h ΔT c ¼ εC min ðT h,in  T c,in Þ

ðEq:12:16aÞ

12.6.1 Effectiveness-NTU Method The number of transfer units (NTU) required for heat transfer depends on the overall heat transfer coefficient, the heat exchanger surface area, and the heat capacity rate. NTU is a dimensionless parameter and is calculated using the following equation.

12.6

Heat Exchanger Effectiveness

263

NTU ¼

UAs Cmin

ð12:17Þ

In situations where the exit temperatures of both the fluids are not known, a trial and procedure will have to be used. The trial and procedure consists of assuming the exit temperature of one of the fluids and checking if the assumed temperature satisfies the heat balance and heat exchanger design equations. Such problems can be solved quickly and explicitly by using the effectiveness-NTU method [6, 7]. The ratio of Cmin to Cmax is defined as the heat capacity ratio, Cr. Cr ¼

Cmin Cmax

ð12:18Þ

The heat exchanger effectiveness can be obtained from formulas or from graphs. The formulas are given in terms of NTU and Cr. Double pipe heat exchanger (parallel flow): ε¼

1  exp ½NTUð1 þ Cr Þ 1 þ Cr

ð12:19Þ

Double pipe heat exchanger (counterflow): ε¼

1  exp ½NTUð1  Cr Þ 1  Cr exp ½NTUð1  C r Þ

ð12:20Þ

Shell and tube exchangers: h 8
 i911 2 1=2 = < 1 þ exp NTU 1 þ C r
1=2 h i A ε ¼ 2@ 1 þ C r þ 1 þ C r 2 :1  exp NTU
1 þ C 2 1=2 ; r 0

In graphs, the heat exchanger effectiveness is plotted vs. NTU with Cr as the parameter. The graphs for heat exchanger effectiveness are shown Figures 12.5(a) and 12.5(b). Example 12.7 illustrates the use of the effectiveness-NTU method. Example 12.7 The overall fouling factor used in the design of the exchanger in Example 12.6 was 0.003 hr-ft2- F/Btu. The mass flow rates and the entrance temperatures of oil and water remain the same. Calculate the exit temperatures of oil and water when the heat exchanger is first put into service. (Solution) Useful data from Example 12.6:

264

12

Heat Exchangers

Fig. 12.5a Heat exchanger effectiveness for 1  1 counterflow heat exchanger. (Source: Basic Design Methods of Heat Exchangers (by Cuneyt Ezgi), Open Access Peer Reviewed Chapter, https:// www.intechopen.com/ chapters/54521)

Fig. 12.5b Heat exchanger effectiveness for shell and tube heat exchanger (1 shell pass and 2, 4, 6, etc. tube passes). (Source: Basic Design Methods of Heat Exchangers (by Cu- neyt Ezgi), Open Access Peer Reviewed Chapter https:// www.intechopen.com/ chapters/54521)

1–2 shell and tube heat exchanger with a heat exchanger surface area of 28.2 ft2, mass flow rate of oil is 10,000 lbm/hr, and it is cooled from 200 F to 140  F. Mass flow rate of cooling water available at 60  F is 16,500 lbm/hr cooling water. The overall design heat transfer coefficient is 125 Btu/hr-ft2- F. The specific heats of water and oil are 1.0 Btu/lbm- F and 0.55 Btu/lbm- F, respectively. When the heat exchanger is first put into service, it operates under clean conditions with no fouling. However, the design overall coefficient, UD, includes the fouling

12.6

Heat Exchanger Effectiveness

265

Fig. 12.5c Heat exchanger effectiveness for shell and tube heat exchanger (2 shell passes and 4, 8, 12, etc. tube passes). (Source: Basic Design Methods of Heat Exchangers (by Cuneyt Ezgi), Open Access Peer Reviewed Chapter https:// www.intechopen.com/ chapters/54521)

factor. Since the overall heat transfer coefficient under clean conditions, UC, is higher than UD, there will be greater heat transfer under clean conditions resulting in different exit temperatures for both the fluids. Calculate the overall heat transfer coefficient under clean conditions using Eq. 12.10.

1 hr  ft2  F  0:003 Btu Btu 125 2  hr  ft  F ¼ 0:005 hr  ft2  F=Btu

1 1 ¼  Rf ¼ UC UD

UC ¼

1 2   F 0:005 hrft Btu

¼ 200 Btu=hr  ft2  F

Calculate the heat capacity rates of oil and water. 

  lbm Btu m_ oil cp,oil ¼ 10000 0:55 hr lbm  F ¼ 5, 500 Btu=hr  F    lbm Btu 1:0 m_ water cp,water ¼ 16500  hr lbm  F ¼ 16, 500 Btu=hr  F Therefore, C min ¼ 5500 Btu=hr  ∘ F,

C max ¼ 16500 Btu=hr  ∘ F

Calculate the NTU under clean conditions using Eq. 12.17.

266

12

NTU ¼

U C As ¼ Cmin

 200

¼ 1:025

Heat Exchangers


 Btu 28:2 ft2 2  hr  ft  F Btu 5500 hr  F

Calculate the heat capacity rate ratio using Eq. 12.18. Cr ¼

Btu 5500 hr C min F ¼ ¼ 0:3333 Btu Cmax 16500 hr F

Determine the heat exchanger effectiveness from Fig. 12.5b using NTU ¼ 1.03 and Cr ¼ 0.3333 as parameters as shown in the figure.

The heat exchanger effectiveness is ε ¼ 63 % ¼ 0.63. Water is the cold fluid and oil is the hot fluid. Calculate ΔTc and ΔTh using Eq. 12.16a. εC min ðT h,in  T c,in Þ  Cc  Btu   ð0:63Þ 5500  F ð200 F  60 FÞ hr  ¼ Btu 16500 hr  F ¼ 29:4 F

ΔT c ¼

12.6

Heat Exchanger Effectiveness

267

ε C min ðT h,in  T c,in Þ  Ch  Btu ð0:63Þ 5500 ð200 F  60 FÞ hr  F ¼ Btu 5500 hr  F ¼ 88:2 F

ΔT h ¼

Calculate the exit temperatures of water and oil using ΔTc and ΔTh. T e,water ¼ T i,water þ ΔT c ¼ 60∘ F þ 29:4∘ F ¼ 89:4∘ F T e,oil ¼ T i,oil  ΔT h ¼ 200∘ F  88:2∘ F ¼ 111:8∘ F Thus, when the heat exchanger is first put into service, the actual heat transfer is:   Btu qactual ¼ Cc ΔT c ¼ 16500 ð29:4 FÞ  hr  F ¼ 485, 100 Btu=hr This is much higher than the heat transfer rate of 330,000 Btu/hr under fouled conditions. Example 12.8 is a comprehensive problem based on a feedwater heater (FWH) in a power plant. Example 12.8 In a steam power plant, 60000 kg/hr of bleed steam from a turbine enters a feedwater heater (FWH) at 300 kPa and with 90% quality. The FWH heats 500000 kg/hr of feedwater entering at 20  C. The bleed steam leaves the FWH as a saturated liquid at 300 kPa. The FWH is a 1–2 shell and tube heat exchanger with the condensing steam on the shell side and with the feedwater flowing in the tubes (ID ¼ 20.98 mm). The design overall heat transfer coefficient for the exchanger is 4200 W/m2. C. Use the standard density of water, ρ ¼ 1000 kg/m3. The following properties can be obtained from steam tables or from online steam calculators (https://www.tlv.com/global/TI/ calculator/steam-table-temperature.html): At 0.30 MPa, Tsat ¼ 133  C, hf ¼ 561.4 kJ/kg, and hfg ¼ 2163.5 kJ/kg At 20  C, hf ¼ 83.91 kJ/kg At 75  C, hf ¼ 314.03 kJ/kg At 80  C, hf ¼ 355.01 kJ/kg A. B. C. D.

The exit temperature of the feedwater The heat exchanger surface area required The number of tubes per pass if the required water velocity is 3 m/s The length of each tube

268

12

Heat Exchangers

Due to decrease in demand for power, the feedwater flow rate is reduced to 300000 kg/hr. The inlet and exit conditions of the fluids remain the same as before. Determine: E. The mass flow rate of the bleed steam required. F. The velocity of the water in the tubes. G. the heat exchanger surface area required if the design “U” is controlled by the tube side heat transfer coefficient. (Solution) A. The exit temperature of the feedwater can be determined from energy balance. Subscript “s” represents steam and subscript “w” represents water. Further, subscript “i” represents inlet streams and subscript “e” represents outlet streams. Energy balance for the FWH results in the following equation. Heat lost by the bleed steam ¼ heat gained by the feedwater m_ s ðhsi  hse Þ ¼ m_ w ðhwe  hwi Þ Solve the preceding equation for hwe. hwe ¼ hwi þ

m_ s ðhsi  hse Þ m_ w

 
 kJ kJ hsi ¼ h f þ xhfg 300 kPa ¼ 561:4 þ ð0:90Þ 2163:5 kg kg ¼ 2508 kJ=kg hse ¼ h f at 300 kPa ¼ 561:4 kJ=kg hwi ¼ h f at 20 C ¼ 83:91 kJ=kg Substitute the known values into the equation for the exit enthalpy of water. hwe ¼ hwi þ ¼ 83:91

m_ s ðhsi  hse Þ m_w kJ þ kg

¼ 317:5 kJ=kg

kg 60000 hr

  kJ kJ 2508  561:4 kg kg kg 500000 hr

Interpolate for the preceding enthalpy between 75 and 80  C, to obtain, T ¼ 76  C Therefore, the exit temperature of the feedwater is 76  C.

12.6

Heat Exchanger Effectiveness

269

B. Calculate the heat duty of the FWH. _ s ðhsi  hseÞ q ¼m   kg kJ kJ ¼ 60000 2508  561:4 hr kg kg 8 ¼ 1:168  10 kJ=hr Convert the heat duty to Watts (W). 

kJ hr ¼ 3:244  107 W

q ¼

1:168  108



1000 J kJ



1 hr 3600 s



The saturation temperature at 300 kPa is 133  C, which is the temperature at which condensation takes place. Calculate the log mean temperature difference for counterflow. 1 133 C 76 C

2 Steam

133 C

Water

20 C

!

ΔT 1 ¼ 57 C ΔT LMTD,CF ¼

ΔT 2 ¼ 113 C

ΔT 2  ΔT 1   ΔT 2 ln ΔT 1

113 C  57 C   113 C ln 57 C  ¼ 81:8 C

¼

Since the temperature of the condensing steam remains constant, the MTD correction factor is F ¼ 1.0 The heat exchanger surface area required is determined from the heat exchanger design equation (Eq. 12.11). As ¼

q 3:244  107 W ¼  W UFΔT LMTD 4200 2  ð1:0Þð81:8 CÞ m  C ¼ 94:42 m2

C. Calculate the ID of the tube in meters (m).   1m D ¼ ð20:98 mmÞ 1000 mm ¼ 0:02098 m

270

12

Heat Exchangers

Use the equation for mass flow rate (Eq. 3.5 from fluid mechanics) to determine the number of tubes required per pass.

n¼

m_ w ρvAcs

   kg 1 hr 500000 hr 3600 sec  ¼   kg m π 1000 3 3 ð0:02098 mÞ2 sec 4 m ¼ 134 tubes

D. Calculate the length of the tubes by using the heat exchanger surface area required.

L¼

As 94:42 m2 ¼ 2nðπDÞ 2ð134Þðπ Þð0:02098 mÞ ¼ 5:35 m

E. Use the energy balance equation to determine the mass flow rate of bleed steam required for the reduced feedwater flow rate and the same inlet and exit conditions as before. m_ s ðhsi  hse Þ ¼ m_ w ðhwe  hwi Þ    kg kJ kJ 317:5  83:91 300000 hr kg kg m_ ðh  hwi Þ   ¼ m_ s ¼ w we ðhsi  hse Þ kJ kJ 2508  561:4 kg kg ¼ 36000 kg=hr F. Use the equation for mass flow rate (Eq. 3.5 from fluid mechanics) to determine the velocity of water in the tubes.

vw ¼

m_ w nρw Acs

   kg 1 hr 300000 hr 3600 sec    ¼ kg π ð134Þ 1000 3 ð0:02098 mÞ2 4 m ¼ 1:80 m=s

G. Calculate the new value of the overall heat transfer coefficient due to the decrease in tube side velocity. Subscript “1” refers to the original feedwater flow rate, and subscript “2” refers to the reduced feedwater flow rate. U is controlled by the tube side heat transfer coefficient, hi. From typical heat transfer correlations for

Practice Problems

271

turbulent flow, h / Re0.8 / v0.8. Therefore, the following proportionality relationship can be written.  0:8  m0:8 v2 1:30 ¼ 3:0ms s v1 ¼ 0:5122   W U 2 ¼ 0:5122U 1 ¼ ð0:5122Þ 4200 2  m  C ¼ 2151 W=m2  C U 2 hi2 ¼ ¼ U 1 hi1

Calculate the revised heat duty. q2 ¼  m_ w ðhwe  hwi Þ  kg kJ kJ ¼ 300000 317:5  83:91 hr kg kg 7 ¼ 7:01  10 kJ=hr Convert the revised heat duty to watts (W).     kJ 1000 J 1 hr 7:01  107 hr kJ 3600 s ¼ 1:95  107 W

q2 ¼

Substitute the known values into the heat exchanger design equation (Eq. 12.11) to obtain the revised heat exchanger surface area required. As2 ¼

1:95  107 W  W 2151 2  ð1:0Þð81:8 CÞ m  C ¼ 110:83 m2

q2 ¼ U 2 FΔT LMTD

Comment: The existing heat exchanger is undersized due to the reduced value of the overall heat transfer coefficient.

Practice Problems Practice Problem 12.1 A process liquid with specific heat 0.41 Btu/lbm F is to be cooled from 130 to 60  F using cooling water. The mass flow rate of the process liquid is 2300 lbm/hr, and the temperature rise of water is to be limited to 20  F. The specific heat of water

272

12

Heat Exchangers

is 1 Btu/lbm- F. Calculate the minimum flow rate of water required in gallons per minute. Practice Problem 12.2 A closed feedwater heater (FWH) in a power generation facility uses extraction steam at 2 MPa and 300  C to heat the boiler feedwater from 20 to 150  C. The steam leaves the FWH as a saturated liquid. The following properties of superheated steam and saturated water can be obtained from steam tables or online steam calculator: (https://www.tlv.com/global/TI/calculator/steam-table-temperature.html): h (at 2 MPa, 300  C) ¼ 3024.2 kJ/kg hf at 2 Mpa ¼ 908.5 kJ/kg hf at 20  C ¼ 83.91 kJ/kg hf at 150  C ¼ 632.18 kJ/kg Calculate the ratio, kg steam/kg feedwater. Practice Problem 12.3 Determine the overall heat transfer coefficient under clean conditions in Example 12.4. Practice Problem 12.4 Design a suitable shell and tube heat exchanger under the following conditions. • Water flowing in the tubes at a mass flow rate of 12500 kg/hr is to be heated from 40 to 60  C using 6000 kg/hr waste hot water available at 90  C on the shell side. • The design velocity to be maintained in the tubes is 2 m/s, and the tubes are 6.350 mm OD with a wall thickness of 0.711 mm. Maximum tube length specified is 2.50 m. • The overall heat transfer coefficient is 1300 W/m2. C, and the specific heat of water is 4.187 kJ/kg. C. Practice Problem 12.5 The mass flow rate of cold water in the exchanger in Practice Problem 12.4 is reduced from 12500 to 10000 kg/hr. The entrance temperature of the cold water remains the same at 40  C. The mass flow and the entrance temperature of the waste hot water remain the same at 6000 kg/hr and 90  C, respectively. The reduced flow rate of cold water has no effect on the overall heat transfer coefficient. Determine: A. B. C. D.

The number of transfer units The heat exchanger effectiveness The actual heat transfer The exit temperatures of the cold water and the hot water

Practice Problem 12.6 0.85 kg/s of steam at 0.70 MPa and 200  C is used in a closed feedwater heater (FWH) to heat feedwater from 110 to 180  C. The feedwater enters the FWH at 6 MPa and steam leaves the FWH as a saturated liquid. The FWH is a single-pass shell and tube heat exchanger with 40 tubes (OD ¼ 15.88 mm, ID ¼ 12.57 mm). The

Practice Problems

273

design overall heat transfer coefficient is 4500 W/m2. C. The pressure drop of the streams across the FWH is negligible. The density of feedwater is ρ ¼ 910 kg/m3. The following properties can be obtained from steam tables or from online steam calculators (https://www.tlv.com/global/TI/calculator/steam-table-temperature. html): h (at 0.70 MPa, 200  C) ¼ 2845.3 kJ/kg, Tsat at 0.70 Mpa ¼ 165  C hf at 0.70 Mpa ¼ 697.0 kJ/kg hf at 1100C ¼ 461.4 kJ/kg hf at 1800C ¼ 763.0 kJ/kg Determine: A. The mass flow of the feedwater that can be heated B. The length of the tubes required C. The velocity of water through the tubes Practice Problem 12.7 A 1–2 heat exchanger has been designed for heating 25 gpm of water on the tube side from 60 to 140  F, using 100 gpm of hot water entering the shell at 200  F. When the heat exchanger is first put into service, the exit temperature of water on the shell side is measured to be 170  F. The overall heat transfer coefficient used in the design of the exchanger is 300 Btu/hr-ft2- F. The density of water can be assumed to be constant at 8.34 lbm/gal. The following properties can be obtained from steam tables or from online steam calculators (https://www.tlv.com/global/TI/calculator/steamtable-temperature.html): hf at 200  F ¼ 168.13 Btu/lbm hf at 180  F ¼ 148.04 Btu/lbm hf at 170  F ¼ 138.01 Btu/lbm hf at 140  F ¼ 107.99 Btu/lbm hf at 60  F ¼ 28.08 Btu/lbm Determine the fouling factor used in the design of the exchanger. Practice Problem 12.8 An evaporator has a heat transfer area of 110 m2 and operates at a pressure of 101 kPa to produce steam by using feedwater at 30  C. The design overall heat transfer coefficient is 2200 W/m2. C. The heating medium is condensing steam available at 150 kPa (gage). The saturation temperature and the enthalpy of vaporization at 0.250 Mpa are127  C and 2182 kJ/kg, respectively. Determine the required steam flow rate in kg/min. Practice Problem 12.9 A counterflow condenser with a surface area of 6 ft2 condenses 155 lbm/hr of saturated R-134a vapor at 200 psia to saturated liquid at the same pressure. The inlet temperature of the cooling water is 60  F, and the temperature rise of the cooling water is 20  F. At 200 psia, the saturation temperature of R-134a is 125  F,

274

12

Heat Exchangers

and the enthalpy of condensation is 64.5 Btu/lbm. The increase in enthalpy of the cooling water is 20 Btu/lbm. Determine: A. The cooling water flow rate required (gpm) B. The overall heat-transfer coefficient used in designing the condenser Practice Problem 12.10 A cogeneration facility generates electricity and supplies steam to produce hot water for district heating. The hot water is produced in a 1–1 shell and tube heat exchanger using extraction steam at 400 psia and 500 F. The condensate leaves the heat exchanger as a saturated liquid at 400 psia. The normal demand for district heating is such that 50 gpm of return water from the district needs to be heated from 70 F to 150 F. The following properties can be obtained from steam tables or from online steam calculators (https://www.tlv.com/global/TI/calculator/steam-table-tempera ture.html): h (at 400 psia, 500  F) ¼ 1245.5 Btu/lbm, Tsat at 400 psia ¼445  F hf at 400 psia ¼ 424 Btu/lbm hf at 70  F ¼ 38Btu/lbm hf at 150  F ¼ 118 Btu/lbm Determine: A. The mass flow of the extraction steam required (lbm/hr) B. The heat transfer area required if the design overall heat transfer coefficient is 325 Btu/hr-ft2- F

Solutions to Practice Problems Practice Problem 12.1 (Solution) Subscript “l” represents the process liquid and subscript “w” represents water. From heat balance, Heat lost by process liquid ¼ heat gained by water m_ l cpl ΔT l ¼ m_ w cpw ΔT w   lbm Btu 0:41 2300 ð130 F  60 FÞ  hr lbm  F m_ l cpl ΔT l ¼ m_ w ¼   Btu cpw ΔT w ð20 FÞ 1:0 lbm  F ¼ 3300:5 lbm=hr 

Solutions to Practice Problems

275

Convert the mass flow rate to gpm.
gpmw ¼


 3300:5 lbm hr  ¼ 6:6 gpm 

60 min hr

8:34 lbm gal

Practice Problem 12.2 (Solution) Perform a heat balance for the FWH. Subscript “s” represents extraction steam, and subscript “w” represents feedwater. Further, subscript “i” represents inlet streams and subscript “e” represents exit streams. Heat given up by extraction steam ¼ heat absorbed by feedwater m_ s ðhsi  hse Þ ¼ m_ w ðhwe  hwi Þ Solve the preceding equation for the required ratio. ðh  hwi Þ kg steam m_ ¼ s ¼ we kg feed water m_ w ðhsi  hse Þ Use the given enthalpy data for steam. hsi ¼ h (2 MPa, 300  C, superheated steam) ¼ 3024.2 kJ/kg hse ¼ hf at 2 MPa ¼ 908.5 kJ/kg hwi ¼ hf at 20  C ¼ 83.91 kJ/kg hwe ¼ hf at 150  C ¼ 632.2 kJ/kg Substitute the known values into the modified heat balance equation to get the required ratio. 

 kJ kJ 632:2  83:91 kg kg ðh  hwi Þ kg steam m_  ¼ s ¼ we ¼ kg feed water m_ w ðhsi  hse Þ kJ kJ 3024:2  908:5 kg kg ¼ 0:2951 Practice Problem 12.3 (Solution) From the solution to Example 12.4, the overall heat transfer coefficient (based on the inside surface area of the tubes) with the fouling resistances is UiD ¼ 61.27 Btu/hrft2- F. Combine the inside and outside fouling factors to get the total fouling factor.

276

12

Heat Exchangers

Rf ¼ Ri þ Ro hr  ft2  F hr  ft2  F ¼ 0:001 þ 0:002 Btu Btu ¼ 0:003 hr  ft2  F=Btu Calculate the overall heat transfer coefficient under clean conditions by using Eq. 12.10. 1 1 ¼  Rf ¼ UC UD

1

 0:003

Btu hr  ft2  F ¼ 0:013321 hr  ft2  F=Btu

UC ¼

61:27

hr  ft2  F Btu

1 2  2  ¼ 75:07 Btu=hr‐ft ‐ F ‐ F 0:013321 hr‐ft Btu

Practice Problem 12.4 (Solution) Calculate the heat duty. 0

1 kg   12500 kJ B C hr q ¼ m_ c cpc ΔT c ¼ @ ð60 C  40 CÞ A 4:187 3600 s kg  C hr ¼ 290:76 kW Calculate the exit temperature of hot water by using the heat balance equation (Eq. 12.1). q ¼ m_ h cph ΔT h ¼ m_ c cpc ΔT c ) ΔT h ¼

q 290:76 kW 1 ¼0 m_ h cph kg   6000 kJ B C hr 4:187 @ A 3600 s kg  C hr ¼ 41:67 C ’ 42 C

T he ¼ T hi  ΔT h ¼ 90 C  42 C ¼ 48 C Calculate the LMTD for counterflow.

Solutions to Practice Problems

277

1 90 C

Hot water



Cold water

60 C

2 !



ΔT 1 ¼ 30 C ΔT LMTD ¼



48 C 40 C ΔT 2 ¼ 8 C

ΔT 1  ΔT 2 30 C  8 C   ¼    ΔT 1 30 C ln ln 8 C ΔT 2 ¼ 16:64 C

Calculate the heat transfer area required using the heat exchanger design equation (Eq. 12.11), assuming a 1–1 exchanger. As ¼

q 290:76  103 W ¼  W UFΔT LMTD 1300 2  ð1:0Þð16:64 CÞ m  C ¼ 13:44 m2

The standard density of water is 1000 kg/m3. The ID of each tube is: Di ¼

6:350 mm  2ð0:711 mmÞ 1000 mm m

¼ 0:00493 m

Calculate the number of tubes required to accommodate the mass flow rate of the cold water using the continuity equation (Eq. 3.5). m_ c ¼ nðρw Acs vÞ ) 0 1 kg 12500 B hr C @ A 3600 s m_ c   hr ¼ n¼   ρw Acs v kg π m 1000 3 ð0:00493 mÞ2 2 4 s m ¼ 91 tubes Calculate the heat transfer area available with 91 tubes and the maximum length allowed of 2.50 m.

278

12

0

Heat Exchangers

1

B6:350 mmC ð2:50 mÞ As ¼ nπDo L ¼ ð91ÞðπÞ@ 1000 mm A m ¼ 4:538 m2 Since the required surface area is 13.44 m2, try a 2–4 exchanger. For a multi-pass exchanger, the LMTD correction factor must be used. Determine the LMTD correction factor from Fig. 12.4b as shown. Calculate the parameters P and R using the nomenclature in Fig. 12.4b. R¼

T 1  T 2 90 C  48 C ¼  ¼ 2:1 60 C  40 C t2  t1

P¼

t2  t1 60 C  40 C ¼  ¼ 0:4 T 1  t 1 90 C  40 C

From the figure, the MTD correction factor is F ¼ 0.85. Therefore, the revised heat transfer surface area required is: As,rev ¼

As,orig 13:44 m2 ¼ ¼ 15:81 m2 F 0:85

Calculate the surface area available from four tube bundles. As4 ¼ 4ðAs1 Þ ¼ 4ð4:538 m2 Þ ¼ 18:152 m2 This is more than the required surface area. By trial and error, when the length is 2.20 m, the heat transfer surface area available with four tube passes is:

Solutions to Practice Problems

279

0

1

B6:350 mmC ð2:20 mÞ As4 ¼ 4πnDo L ¼ ð4πÞð91Þ@ 1000 mm A m ¼ 15:97 m2 Hence, the summary of the final design is as follows: 2–4 shell and tube heat exchanger, heat transfer surface area ¼ 15.97 m2, 91 tubes per pass, tube dimensions: L ¼ 2.20 m, OD ¼ 6.350 mm, ID ¼ 4.93 mm Practice Problem 12.5 (Solution) The following data is available from Practice Problem 12.4 and its solution: m_ h ¼ 6, 000Kg=hr, m_ c ¼ 10, 000kg=hr, cph ¼ cpc ¼ 4:187kJ=kg: C, U ¼ 1300W=m2 :K, As ¼ 15:97m2 : Calculate the heat capacity rates for both the hot water and for the cold water. Ch ¼ m_ h cph ¼

! 6000 kg hr 4:187 3600 s hr

C c ¼ m_ c cpc ¼

! 10000 kg hr 3600 s hr

4:187

 kJ ¼ 6:98 kW= C kg  C kJ kg  C



¼ 11:63 kW= C

Therefore, Cmax ¼ Cc ¼ 11.63 kW/ C and Cmin ¼ Ch ¼ 6.98 kW/ C A. Calculate the number of transfer units by using Eq. 12.17.   W 1300 ð15:97 m2 Þ 2  C UAs m   NTU ¼ ¼  C min kW 103 W 6:98  kW C ¼ 2:97 B. Calculate the heat capacity rate ratio.

Cr ¼

6:98 kW C min C ¼ ¼ 0:60 C max 11:63 kW C

Using NTU ¼ 2.97 and Cr ¼ 0.60 as parameters, determine the heat exchanger effectiveness from Fig. 12.5c as shown.

280

12

Heat Exchangers

From the graph, heat exchanger effectiveness, ε ¼ 79% ¼ 0.79 C. Calculate the actual heat transfer by using Eq. 12.16. qactual ¼ ðεÞðCmin  ÞðT h,in T c,in Þ kW ¼ ð0:79Þ 6:98  ð90 C  40 CÞ C ¼ 275:7 kW D. Calculate the temperature differences of hot water and cold water by using Eq. 12.16. ΔT c ¼

qactual 275:7 kW ¼ ¼ 39:5 C Cc 6:98 kW C

ΔT h ¼

qactual 275:7 kW ¼ ¼ 23:7 C Ch 11:63 kW C

Calculate the exit temperatures of hot water and cold water using the calculated temperature differences. T ce ¼ T ci þ ΔT c ¼ 40 C þ 39:5 C ¼ 79:5 C T he ¼ T hi  ΔT h ¼ 90 C  23:7 C ¼ 66:3 C

Solutions to Practice Problems

281

Practice Problem 12.6 (Solution) A. Use the heat balance equation to determine the mass flow of the feedwater that can be heated. Heat released by condensing steam ¼ heat absorbed by feedwater m_ s ðhsi  hse Þ ¼ m_ w ðhwe  hwi Þ Rearrange the preceding equation to obtain an equation for mass flow rate of water. m_ w ¼

m_ s ðhsi  hse Þ ðhwe  hwi Þ

Use the given enthalpies of water and steam. hsi ¼ h (0.7 MPa, 200  C, superheated steam) ¼ 2845.3 kJ/kg Steam leaves the FWH as a saturated liquid at 0.7 MPa hse ¼ hf at 0.7 MPa ¼ 697.0 kJ/kg hwi ¼ hf at 110  C ¼ 461.4 kJ/kg hwe ¼ hf at 180  C ¼ 763.0 kJ/kg Substitute the known values into the equation for water flow rate.    kg kJ kJ 0:85 2845:3  697 s kg kg m_ ðh  hse Þ   m_ w ¼ s si ¼ ðhwe  hwi Þ kJ kJ 763  461:4 kg kg ¼ 6:05 kg=s B. Calculate the heat duty for the exchanger.     kg kJ kJ 1000 W 0:85 2845:3  697 s kg kg kW ¼ 1826055 W

q ¼ m_ s ðhsi  hse Þ ¼

Calculate the LMTD for counterflow. From steam tables, the saturation temperature at 0.70 MPa is 165  C. This is the exit temperature of the steam condensate from the FWH.

282

12

1 200 C

2 Steam



!

Feed water



Heat Exchangers

180 C ΔT 1 ¼ 20 C ΔT LMTD ¼

165 C 110 C ΔT 2 ¼ 55 C

ΔT 2  ΔT 1 55 C  20 C      ¼ ΔT 2 55 C ln ln 20 C ΔT 1 ¼ 34:60 C

Calculate the heat transfer area required by using the heat exchanger design equation (Eq. 12.11). The MTD correction factor for a single-pass exchanger is 1.0. As ¼

q ¼ UFΔT LMTD 4500

1826055 W  W ð1:0Þð34:60 CÞ m2  C ¼ 11:73 m2

Convert the OD and ID of the tubes to meters. ! Do ¼

15:88 mm 1000 mm m

! Di ¼

¼ 0:0159 m

12:57 mm 1000 mm m

¼ 0:0126 m

Calculate the length of the tubes required by using the heat transfer area. L¼

As 11:73 m2 ¼ nπDo ð40ÞðπÞð0:0159 mÞ ¼ 5:87 m

C. Find specific volume of the feedwater at its average temperature of 145  C from the steam tables. v ¼ v f at 145 C ¼ 0:0011 m3 =kg The density of the feedwater is the inverse of the specific volume. ρ¼

1 1 ¼ ¼ 909 kg=m3 v 0:0011 m3 kg

Calculate the velocity in the tubes by using the equation for mass flow rate (Eq. 3.5 from fluid mechanics).

Solutions to Practice Problems

283

m_ w ¼ nðρAcs vÞ ) m_ w v¼ ¼ nðρAcs Þ

kg 6:05 m_ w  ¼   s 2 πDi 2 π ð 0:0126 m Þ kg ðnÞðρÞ ð40Þ 909 3 4 4 m ¼ 1:33 m=s

Practice Problem 12.7 (Solution) Determine the parameters for the heat exchanger under design (fouled) conditions. Calculate the exit enthalpy and hence the exit temperature of hot water from the heat balance equation. heat lost by hot water ¼ heat gained by cold water m_ h ðhhi  hhe Þ ¼ m_ c ðhce  hci Þ Solve the preceding equation for the exit enthalpy of hot water. hhe ¼ hhi 

m_ c ðhce  hci Þ m_ h

Use the given enthalpies. hhi ¼ hf at 200  F ¼ 168.13 Btu/lbm hf at 180  F ¼ 148.04 Btu/lbm hce ¼ hf at 140  F ¼ 107.99 Btu/lbm hci ¼ hf at 60  F ¼ 28.08 Btu/lbm Since the density of water is constant, the ratio of the mass flow rate of the cold water to the mass flow rate of hot water will be the same as the ratio of the volume flow rates. Therefore, 25 gpm m_ c V_ c ¼ 0:25 ¼ ¼ m_ h V_ h 100 gpm Substitute the known values into the equation for the exit enthalpy of hot water. m_ c ðhce  hci Þ m_ h   Btu Btu Btu  0:25 107:99  28:08 ¼ 168:13 lbm lbm lbm ¼ 148:15 Btu=lbm

hhe ¼ hhi 

284

12

Heat Exchangers

Find the saturation temperature of water corresponding to this enthalpy from the steam tables. At hf ¼ 148.04, The ¼ 180  F, which will be the exit temperature of hot water. Calculate the LMTD for counterflow. 1

2

200 F

Hot Water

140 F

Cold water

!

180 F 60 F

ΔT 1 ¼ 60 F ΔT LMTD ¼

ΔT 2 ¼ 120 F ΔT 2  ΔT 1 120 F  60 F     ¼ ΔT 2 120 F ln ln ΔT 1 60 F  ¼ 86:6 F

Calculate the heat duty of the exchanger under design conditions.     gal lbm Btu Btu 8:34 107:99  28:08 25 min gal lbm lbm ¼ 16, 661 Btu= min

qD ¼ m_ c ðhce  hci ÞD ¼

Since this is a 1–2 exchanger, the MTD correction factor will have to be used. Determine the MTD correction factor from Fig. 12.4a. Calculate the parameters P and R using the nomenclature in Fig. 12.4a. R¼

T 1  T 2 200 F  180 F ¼ 0:25 ¼ 140 F  60 F t2  t1

P¼

t2  t1 140 F  60 F ¼ 0:57 ¼ T 1  t 1 200 F  60 F

From Fig. 12.4a, for P ¼ 0.57, R ¼ 0.25, the MTD correction factor, F ¼ 0.96 as shown. Determine the designed heat transfer area of the exchanger using the design overall heat transfer coefficient, which includes the fouling factor. Calculate the heat transfer area by using the heat exchanger design equation (Eq. 12.11).

Solutions to Practice Problems

As ¼

285

   Btu 60 min 16661 min hr  ¼ Btu 300 ð0:96Þð86:6 FÞ hr  ft2  F ¼ 40:1 ft2

qD U D FΔT LMTD

Rework the calculations under clean conditions, that is, when the exchanger is first put into service. In this case, the exit temperature of the hot water is given to be 170  F. Calculate the exit enthalpy and hence the exit temperature of the cold water from the heat balance equation. heat lost by hot water ¼ heat gained by cold water m_ h ðhhi  hhe Þ ¼ m_ c ðhce  hci Þ Solve the preceding equation for the exit enthalpy of cold water. hce ¼ hci þ

m_ h ðhhi  hhe Þ m_ c

Use the given enthalpy of the hot water at 170  F. hhe ¼ hf at 170  F ¼ 138.01 Btu/lbm Substitute the known values into the equation for the exit temperature of cold water. hce ¼ hci þ

ðh  hhe Þ m_ h ðhhi  hhe Þ ¼ hci þ hi m_ c m_ c m_ h  168:13

Btu þ ¼ 28:08 lbm ¼ 148:60 Btu=lbm

Btu Btu  138:01 lbm lbm 0:25



From steam tables, the temperature corresponding to enthalpy of 148.6 Btu/lbm is 180oF. Hence, Tce ¼ 180oF. Calculate LMTD for counterflow under clean conditions. 1

2 

Hot Water



Cold water

!

200 F 180 F

ΔT 1 ¼ 20 F



170 F 60 F ΔT 2 ¼ 110 F

286

12

Heat Exchangers

ΔT 2  ΔT 1 110 F  20 F     ¼ ΔT 2 110 F ln ln 20 F ΔT 1  ¼ 52:8 F

ΔT LMTD ¼

Calculate the heat duty under clean conditions.     gal lbm Btu Btu 8:34 148:60  28:08 qC ¼ m_ c ðhce  hci ÞC ¼ 25 min gal lbm lbm ¼ 25, 128 Btu= min Since this is a 1–2 exchanger, the MTD correction factor will have to be used. Determine the MTD correction factor from Fig. 12.4a. Calculate the parameters P and R using the nomenclature in Fig. 12.4a. R¼

T1  T2 200 F  170 F ¼ ¼ 0:25 t2  t1 180 F  60 F

P¼

t2  t1 180 F  60 F ¼ ¼ 0:86 T1  t1 200 F  60 F

From the graph, the MTD correction factor, F ¼ 0.68, as shown. Since the same exchanger is used under clean conditions, the heat transfer surface area is the same as in the fouled conditions. Calculate the overall heat transfer coefficient under clean conditions by using the heat exchanger design equation.    Btu 60 min 25128 min hr qC  UC ¼ ¼
2 As ðFΔT LMTD ÞC 40:1 ft ð0:68Þð52:8 FÞ ¼ 1047:2 Btu=hr  ft2  F Calculate the fouling resistance used by using Eqn. 12.10.

Solutions to Practice Problems

Rf ¼

287

1 1  ¼ UD UC

1 1  Btu Btu 300 1047:2 hr  ft2  F hr  ft2  F ¼ 0:0024 hr  ft2  F=Btu

Practice Problem 12.8 (Solution) A. Calculate the heat duty of the evaporator by using the heat exchanger design equation (Eq. 12.10 shown here for reference).

q ¼ UAs FΔT LMTD For evaporators, the MTD correction factor is F ¼ 1.0. The exit temperature of steam produced from the feedwater is the saturation temperature at 101 kPa. The saturation temperature of water at 101 kPa is the normal boiling point of water, 100oC. Therefore, Tfwe ¼ Tsat at 101 kPa ¼ 100  C The absolute pressure of the condensing steam is: Pabs,steam ¼ Pgage,steam þ Patm ¼ 150 kPa þ 101 kPa ¼ 251 kPa The given saturation temperature at 250 kPa (0.250 MPa) is 127  C. Therefore, the temperature of condensing steam is: T se ¼ T si ¼ T sat at 0:251 MPa ¼ 127 C Calculate the LMTD for counterflow. 1

2



30 C 

127 C

Feed Water

!

Condensing Steam

ΔT 1 ¼ 97 C ΔT LMTD ¼



100 C 127 C ΔT 2 ¼ 27 C

ΔT 1  ΔT 2 97 C  27 C      ¼ ΔT 1 97 C ln ln 27 C ΔT 2 ¼ 54:74 C

Substitute the known values into the heat exchanger design equation.

288

12

Heat Exchangers

 q ¼ UAs FΔT LMTD ¼ 2200

 W ð110 m2 Þð1:0Þð54:74 CÞ : m2  C ¼ 1:325  107 W

The heat duty of the evaporator is supplied by the condensing steam at 0.251 MPa (abs). q ¼ m_ s hfg,0:251MPa The enthalpy of condensation is the same as the enthalpy of vaporization given as 2182 kJ/kg. Calculate the mass flow rate of steam required from the heat balance equation. m_ s ¼

q hfg,0:251MPa

1:325  107 W  ¼  kJ 1000 J 2182 kg kJ ¼ 6:07 kg=s

Convert the steam flow rate to kg/min.  m_ s ¼

6:07

kg s



60 s min

 ¼ 364:2 kg= min

Practice Problem 12.9 (Solution) A. Calculate the cooling water flow rate from the heat balance equation. Heat released by the condensing R  134a ¼ heat absorbed by the cooling water q ¼ m_ R hfg,200psia ¼ m_ w ðhwe  hwi Þ m_ w ¼

m_ R hfg,200psia ðhwe  hwi Þ

Substitute the known values into the equation for the mass flow rate of water. 

  Btu 64:5 lbm m_ R hfg,200psia m_ w ¼ ¼ Btu ðhwe  hwi Þ 20 lbm ¼ 500 lbm=hr Convert the water flow rate to gpm.

lbm 155 hr

Solutions to Practice Problems

289

m_ V_ w ¼ w ¼ ρw

1 hr 500 lbm hr  60 min ¼ 1 gpm 8:34 lbm gal

B. From R-134a tables, the saturation temperature at 200 psia is TR ¼ 125  F (refer to the solution of Part A). This will be the inlet as well as the exit temperature of R-134a in the condenser since condensation is an isothermal phase change from vapor to liquid. Calculate the LMTD for counterflow. 1

2 

125 F 

80 F

R‐314a



!

125 F

Cooling Water

60 F

ΔT1 ¼ 45 F ΔT LMTD ¼

ΔT2 ¼ 65 F ΔT 2  ΔT 1 65 F  45 F   ¼    ΔT 2 65 F ln ln ΔT 1 45 F ¼ 54:39 F

Calculate the heat duty of condenser. 

  lbm Btu 64:45 hr lbm ¼ 9, 990 Btu=hr

q ¼ m_ R hfg,250psia ¼

155

For a condenser, the MTD correction factor is F ¼ 1.0. Calculate the design overall heat transfer coefficient using the heat exchanger design equation (Eq. 12.11). Btu 9990 q hr U¼ ¼
2 As FΔT LMTD 6 ft ð1:0Þð54:39 FÞ ¼ 30:6 Btu=hr  ft2  F Practice Problem 12.10 (Solution) A. Obtain an equation for the mass flow of the extraction steam required by using heat balance.

290

12

Heat Exchangers

Heat lost by extraction steam ¼ heat gained by water q ¼ m_ s ðhsi  hse Þ ¼ m_ w ðhwe  hwi Þ m_ s ¼

m_ w ðhwe  hwi Þ ðhsi  hse Þ

h (at 400 psia, 500  F) ¼ 1245.5 Btu/lbm, Tsat at 400 psia ¼445  F hf at 400 psia ¼ 424 Btu/lbm hf at 70  F ¼ 38Btu/lbm hf at 150  F ¼ 118 Btu/lbm Substitute the given enthalpies into the preceding equation, and calculate the mass flow rate of extraction steam required. (Note: To convert gal / min of water to lbm / hr, multiply by the conversion factor 500 (lbm-min/gal-hr) and conversely to convert lbm / hr of water to gal / min (gpm) divide by the conversion factor 500 (lbm-min / gal-hr). m_ w ðhwe  hwi Þ  ðhsi  hse Þ   gal lbm min Btu Btu 50  8:34  60 118  38 min gal h lbm lbm ¼ Btu Btu 1245:5  424 lbm lbm ¼ 2437 lbm=hr

m_ s ¼

B. Calculate the heat duty of the exchanger. q ¼ m_ s ðhsi  hseÞ   lbm Btu Btu ¼ 2437 1245:5  424 hr lbm lbm ¼ 2:002  106 Btu=hr The extraction steam leaves as a saturated liquid at 400 psia, and the given saturation temperature at 400 psia is Ts ¼ 445  F. This will be the exit temperature of extraction steam in the heat exchanger. Calculate the LMTD for counterflow. 1

2 

500 F 

150 F

Extraction Steam

!

Water for district heating

ΔT 1 ¼ 350 F



445 F 70 F ΔT 2 ¼ 375 F

References

291

ΔT LMTD ¼

ΔT 2  ΔT 1 375 F  350 F   ¼   ΔT 2 375 F ln ln ΔT 1 350 F  ¼ 362 F

For a 1–1 exchanger, the MTD correction factor is F ¼ 1.0. Calculate the heat transfer surface area required using the heat exchanger design equation (Eq. 12.11). Btu 2:002  106  hr Btu 325 ð1:0Þð362 FÞ hr  ft2  F ¼ 17:02 ft2

q As ¼ ¼ UFΔT LMTD

References 1. Bergman, T.L., Lavine, A.S., Incropera, F.P., DeWitt David, P.: Fundamentals of Heat and Mass Transfer, 8th edn. Wiley, New York (2019) 2. Bildea, C.S., and Dimian, A.C.: PVC Manufacturing Process by Suspension Polymerization, Appendix B: Heat Exchanger Design, Wiley Online Library. Download from, https:// onlinelibrary.wiley.com/doi/pdf/10.1002/9783527621583.app2 (2008) 3. Coletti, F.: Editor, Heat Exchanger Design Handbook Multimedia Edition, Begell Digital Portal 4. Cavallo, C.: All About Double Pipe Heat Exchangers – What You Need To Know, Thomasnet. Download from. https://www.thomasnet.com/articles/process-equipment/double-pipe-heatexchangers/ 5. Cavallo, C.: All About Shell And Tube Heat Exchangers – What You Need To Know, Thomasnet. Download from. https://www.thomasnet.com/articles/process-equipment/shell-andtube-heat-exchangers/ 6. Ezgi, C.: Basic Design Methods of Heat Exchangers, Intech Open. Download from, https://www. intechopen.com/chapters/54521 (2017) 7. Flynn, A.: M, Akashige T, Theodor, L: Kern’s Process Heat Transfer, 2nd edn. Wiley, USA (2019) 8. Holman, J.P.: Heat Transfer, 10th edn. McGraw Hill, USA (2009)

Chapter 13

Thermodynamics Fundamentals

13.1

Introduction

In the term “thermodynamics,” thermo means heat and dynamics stands for changes or transformations. Thus, the subject matter of thermodynamics is concerned with primarily concerned with the generation of heat and eventual transformations of heat into useful forms of energy such as work and electricity. Thermodynamics has many practical applications in the engineering field, including automobile and aircraft engines, power generation, air conditioning, refrigeration, heat generation by combustion of fuels, and many more similar items. The first law of thermodynamics (Chap. 14) is essentially the principle of conservation of energy akin to the continuity equation (Eq. 3.4) in fluid mechanics, which is based on the principle of conservation of mass. Principles of mass and energy balances are extensively used in the analysis of thermodynamic systems.

13.2

Thermodynamic Properties and Variables

A thermodynamic system [1, 2, 4] is an entity that is being studied with respect to energy transformations or thermodynamic processes. A system is clearly defined by its boundary. Everything outside the system boundary constitutes the surrounding. Further, if the mass of the substance within the system remains constant, then the system is referred to as a closed system. In a closed system, the mass of any species will not cross the system boundary. However, energy can be added or removed from a closed system. A closed thermodynamic system is open to energy transfer across system boundaries, but it is closed for transfer of mass across the system boundary.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_13

293

294

13

Thermodynamics Fundamentals

Fig. 13.1 Closed thermodynamic system and thermodynamic properties

Thermodynamic properties [1–4] are defined with respect to the species within the system. A simple example of a closed thermodynamic system is air contained in a closed tank as shown Fig. 13.1. The mass of air in the tank is m, and it remains constant in this closed tank. The total thermodynamic properties of the air in the tank are represented by upper case variables as shown in Fig. 13.1. The thermodynamic properties are defined as follows. Pressure (P) The pressure of the system is the absolute pressure exerted by the substance within the system. Pressure is uniform within the system. As described in Chap. 2, the absolute pressure is the total pressure relative to zero pressure or perfect vacuum. It is the sum of the surrounding atmospheric pressure and the gage pressure. From Eq. 2.2, Pabs ¼ Patm þ Pgage

ð13:1Þ

It is customary to always use absolute pressure in thermodynamic calculations. This is in contrast to the customary usage of gage pressure in fluid mechanics. The units and conversion factors for pressure and also the standard values of atmospheric pressure are presented here. Conversion Factors: 1 psi ¼ 6.895 kPa ¼ 0.0689 bar ¼ 14.7 psi ¼ 101.3 kPa Standard Atmospheric Pressure at Sea Level: 1 atm ¼ 760 mm Hg ¼ 29.92 in Hg = 14.7 psi ¼ 101.3 kPa ¼ 1.013 bar

Note: In thermodynamics, the units of kgf/cm2 are also used for pressure. One kilogram force is defined as the force required to accelerate 1 kg mass at 9.81 m/s2. Thus, 1 kgf ¼ 9:81 kgm s2 ¼ 9:81 N and the relevant conversion factors are

13.2

Thermodynamic Properties and Variables

295

1 kgf/cm2 ¼ 98.1 kPa ¼ 14.2 psi, and therefore 1 kgf/cm2 is nearly equal to standard atmospheric pressure of 101 kPa or 14.7 psia.

Volume (V ) It is the total space occupied by the substance in the closed system. The typical units for volume are cubic feet, ft3, (USCS) and cubic meter, m3, (S I). Occasionally the unit of liter (L) is also used for volume. Conversion factors: 1 m3 ¼ 35.31 ft3 ¼ 1000 L

Temperature (T) The temperature of the substance within the system is indicative of the energy level of the system. It is customary to use the absolute temperature in thermodynamic calculations. The units for absolute temperature are degree Rankine,  R, (USCS), and Kelvin, K, (S I). The absolute temperature can be obtained by using the following equations. 

R ¼ 460 þ F 



K ¼ 273 þ C

ð13:2Þ ð13:3Þ

Conversion Factors:  R ¼ 0.5556 K,  F ¼ 1.8  C + 32 Note: Temperature difference will be identical in normal (common) and absolute scales since the same constant will be added and subtracted. Consider the temperature of water being increased from 20  C to 50  C. The temperature difference calculations are shown here.

ΔT  C ¼ 50 C  20 C ¼ 30  C Calculate the absolute temperatures in Kelvin by using Eq. 13.3. T 1 ¼ 273 þ 20 C ¼ 293 K and T 2 ¼ 273 þ 50 C ¼ 323 K ΔT K ¼ T 2  T 1 ¼ 323 K  293 K ¼ 30 K

Thus, ΔT K ¼ ΔT  C and similarly, ΔT R ¼ ΔT  F

296

13

Thermodynamics Fundamentals

Mass (m) This quantity represents the mass of the substance within the closed system. The units for mass are pound mass, lbm (USCS), and kilogram, kg (S I). Conversion Factor: 1 lbm ¼ 0.4536 kg

Internal Energy (U ) The molecules of the substance within the system have different types of energy such as kinetic energy, vibrational energy, and rotational energy. The internal energy of a system is sum total of the energies of all the molecules within the system. Since molecules have higher levels of energy at higher temperatures, the internal energy of a system is directly proportional to the temperature of the substance. The units of internal energy are British thermal unit, Btu (USCS), and Kilo Joules, kJ (S I). Conversion Factor: 1 Btu ¼ 1.055 kJ

Enthalpy (H ) The enthalpy of a substance is the sum of the internal energy of the substance and the product of its pressure and volume. The enthalpy of a substance represents the total heat content of the substance. Mathematically, H ¼ U þ PV

ð13:4Þ

Enthalpy has energy units (Btu and kJ). The product of pressure times volume of a substance will have units of energy as shown here. USCS:
PV 

 lbf  3  ft  ft  lbf ft2

Usually, the system pressure is specified in pound force per square inch (lbf/in2) or psi. 1 ft2 ¼ 144 in2 and 1 Btu ¼ 778 ft – lbf. Therefore, the equation to ensure consistent units (Btu) for enthalpy in USCS is:


 144 in2 P ðV Þ ft2 H¼Uþ 778 ft  lbf Btu

ð13:5Þ

In Eq. 13.5, the pressure P, is in psi and the volume V, is in ft3. Entropy (S) The entropy of a substance represents the degree of disorder of the molecules within the substance. Consider the same species, water (H2O) in three different phases—solid (ice), liquid (water), and gas or vapor (steam). In ice, the

13.2

Thermodynamic Properties and Variables

297

water molecules have an organized crystalline structure. After ice melts, the molecules in liquid water are more mobile enabling the water to flow under the application of an external force. Finally, when liquid water vaporizes, the molecules in steam have random motion resulting in greater level of disorder. Therefore, Ssteam > Sliq:water > Sice Entropy also represents the thermal energy per unit absolute temperature that cannot be converted to useful work, also known as Unavailable Energy. Entropy has the units of energy divided by absolute temperature (Btu/ R in USCS and kJ/K).

13.2.1 Specific Properties Thermodynamic tables and charts provide thermodynamic properties per unit mass of a substance. This makes it convenient for performing thermodynamic calculations in different situations. The properties per unit mass can be multiplied by the mass of the substance in thermodynamic calculations for closed systems, and they can be multiplied by mass flow rate of the species for situations involving open systems. The properties per unit mass are known as Specific Properties. Various specific properties [1–3] and their symbols and units are presented here. Specific properties are represented by the corresponding lower case letter of the upper case letter used in representing total properties. For example, specific volume is represented by v, and the total volume is represented by V.

Specific Volume (v) The specific volume of a substance is the volume per unit mass of the substance, and it has units of ft3/lbm (USCS) and m3/kg (S I). Since density is mass per unit volume, specific volume is the reciprocal of density. v¼

V 1 ¼ m ρ

ð13:6Þ

Specific Internal Energy (u) The specific internal energy of a substance is the internal energy per unit mass of the substance, and it has units of Btu/lbm (USCS) and kJ/kg (S I). u¼

U m

ð13:7Þ

Specific Enthalpy (h) The specific enthalpy of a substance is the enthalpy per unit mass of the substance, and it has units of Btu/lbm (USCS) and kJ/kg (SI).

298

13

h¼

H U þ PV ¼ ¼ u þ Pv m m

Thermodynamics Fundamentals

ð13:8Þ

Specific Entropy (s) The specific entropy of a substance is the entropy per unit mass of the substance, and it has units of Btu/lbm –  R (USCS) and kJ/kg.K (SI). s¼

S m

ð13:9Þ

13.2.2 Intensive and Extensive Properties The pressure and temperature of a system remain uniform throughout the system and are independent of the extent or size of the system. Hence, they are known as Intensive Properties. Properties such as total volume and total enthalpy of a system that depend on the extent or size of the system are known as extensive properties.

13.3

Ideal Gas Law

The thermodynamic state of a gaseous system is usually defined by the pressure, temperature, and volume of the system. The ideal gas law [1, 2] is an equation of state that relates the absolute pressure (P), volume (V ), and the absolute temperature (T ) of a gaseous (or vapor) system. Consider a gas in a piston-cylinder closed system. The piston is free to move back and forth depending on the state of the gas. When the gas in the cylinder is heated, it expands and pushes the piston. It is a well-known fact that a gas expands upon heating, that is, the volume of the gas increases upon heating. Conversely, the volume of a gas decreases upon cooling. This suggests a direct relationship between volume and temperature. Charles law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature. At constant pressure, V /T

ð13:10Þ

13.3

Ideal Gas Law

299

When a gas is compressed by pushing the piston, it occupies less volume. However, because of the compression process, the pressure is increasing. This suggests an inverse relationship between volume and pressure. Boyles law states that at constant temperature, the volume of a gas is inversely proportional to its absolute pressure. At constant temperature, V/

1 P

ð13:11Þ

The ideal gas law combines the effects of both Charles law and Boyles law, and it is mathematically represented by Eq. 13.12. PV ¼ mRT

ð13:12Þ

In Eq. 13.12, P is the absolute pressure, V is the volume, m is the mass, R is the individual gas constant [1, 2], and T is the absolute temperature. The value of R is different for different gases, and it depends on the molecular weight, M, of the gas. The molecular weight of a substance is also known as molar mass, and it has units of lbm/lbmol (USCS) and kg/kmol (S I). The molecular weight of a substance can be calculated by using the molecular formula and atomic weights. The molecular formula for water is H2O, and the atomic weights of hydrogen and oxygen are 1 lbm/lbmol and 16 lbm/lbmol, respectively. Therefore, the molecular weight of water is:


M H2 O

 lbm lbm þ 16 ¼ 18 lbm=lbmol ¼2 1 lbmol lbmol

Since the chemical or molecular formula of a substance remains the same, the numerical value of the molecular weight remains the same in different systems of units with corresponding changes for units of mass and moles. For example, the molecular weight of water is: M H2 O ¼ 18 lbm=lbmol ¼ 18 kg=kmol

Table 13.1 lists the molecular weights of common substances.

300

13

Thermodynamics Fundamentals

Table 13.1 Molecular weights of selected species Chemical species Air Oxygen Nitrogen Carbon dioxide Hydrogen Water Methane Ethane Propane

Molecular formula NA O2 N2 CO2 H2 H2O CH4 C2H6 C3H8

Molecular weight lbm/lbmol or kg/kmol 29 32 28 44 2 18 16 30 44

13.3.1 Concept of Mole A mole of a substance has a mass equivalent to its molecular weight, and it consists of Avagadro’s number (6.022  1023) of particles. Examples: The atomic weight of carbon is 12 lbm/lbmol. Therefore, 1 lbmol of carbon will have a mass 12 lbm and will have 6.022  1023 atoms of C. The molecular weight of water is 18 kg/kmol. Therefore, 1 kmol of water will have a mass of 18 kg and will have 6.022  1023 molecules of H2O. Mole is represented by the symbol N, and the moles of a substance can be obtained by dividing the mass of the substance by the molecular weight of the substance. N¼

m M

ð13:13Þ

Example for calculation of moles: Calculate the kilogram moles of water in 5 L of water. The density of water is 1000 kg/m3. Calculate the mass of water. mH2 O ¼ V H2 O  ρH2 O

 1 m3 kg ¼ 5L 1000 3 1000 L m ¼ 5 kg Calculate the mols of water using Eq. 13.13. N H2 O ¼

mH2 O 5 kg ¼ ¼ 0:2778 kmol M H2 O 18 kg kmol

13.3

Ideal Gas Law

301

13.3.2 Universal Gas Constant and Ideal Gas Equation in Molar Form The gas constant, R, in the ideal gas equation (Eq. 13.12) is known as the individual gas constant and has different values for different gases depending on the molecular formula of the gas. However, based on the concept of moles and Avagadro’s number, it is clear that equal moles of substances will have the same number of molecules of the substance. Based on this concept, a universal gas constant, represented by the symbol R can be used in the ideal gas equation based on moles as shown here. Manipulate the ideal gas equation by multiplying and dividing the right-hand side of Eq. 13.12 by the molecular weight of the gas. PV ¼

  m ðR  M ÞT M

ð13:14Þ

From Eq. 13.13, the mass of the gas divided by its molecular weight is the mols of the gas, represented by the symbol N. The product of the individual gas constant and the molecular weight is the universal gas constant [1, 2], R, which has the same value for all gases, and hence the term universal is used. Substitute the preceding results into Eq. 13.14. PV ¼ NRT

ð13:15Þ

Equation 13.15 is the ideal gas equation in molar form. The values of the universal gas constant in different units are presented here. Universal Gas Constant, 1545 ft  lbf=lbmol  R

R

10:73 psi  ft3 =lbmol  R 0:7302 atm  ft3 =lbmol  R 1:987 Btu=lbmol  R 8:314 kJ=kmol:K 0:08314 bar:m3 =kmol:K 83:14 bar:cm3 =kmol:K 0:0821 L:atm=mol:K Note : mol  g mole

1:987 cal=mol:K

The individual gas constant can be obtained by dividing the universal gas constant by the molecular weight of the gas. R¼

R M

ð13:16Þ

302

13

Thermodynamics Fundamentals

Example 13.1 20 lbm of steam is in a rigid container having a capacity of 2300 gallons. The temperature of steam is 300  F. The atmospheric pressure at the location is 14.7 psia. What will be the reading (in psi) of a pressure gage attached to the container? (Solution) Convert the volume of the container to cubic feet.
 0:1337 ft3 V ¼ ð2300 galÞ ¼ 307:5 ft3 gal Calculate the absolute temperature of steam by using Eq. 13.2. T  R ¼ 460 þ 300  F ¼ 760  R Calculate the individual gas constant of steam (water) using Eq. 13.16.

R¼

R ¼ M

psia  ft3 lbmol  R ¼ 0:5961 psia  ft3 =lbm  R lbm 18 lbmol

10:73

Calculate the absolute pressure of steam using the ideal gas law (Eq. 13.12).


 psia  ft3 ð20 lbmÞ 0:5961 ð760  RÞ lbm  R mRT P¼ ¼ ¼ 29:5 psia V 307:5 ft3 Calculate the gage pressure by using Eq. 13.1. Pgage ¼ Pabs  Patm ¼ 29.5 psia  14.7 psia ¼ 14.8 psig Note: It is always preferable to use steam tables (Sect. 13.9) for obtaining the properties of steam. In this problem, the temperature of steam is 300  F. Calculate the specific volume of steam by using Eq. 13.6. v¼

V 307:5 ft3 ¼ ¼ 15:37 ft3 =lbm m 20 lbm

Using temperature of 300  F and specific volume of 15.37 ft3/lbm as parameters, the absolute pressure from superheated steam tables is 28.94 psia.

13.3.3 STP, NTP, SCF, ACF, and Molar Volume of Ideal Gas The volume of a gas is very sensitive to changes in pressure and temperature. Different values of pressure and temperature are used in defining “standard” and “normal” conditions.

13.3

Ideal Gas Law

13.3.3.1

303

Standard Temperature and Pressure (STP) and Molar Volumes

The standard temperature and pressure are defined as 0  C/273 K, and 101.3 kPa (S I), 32  F/492  R, and 14.7 psia (USCS). The molar volumes of ideal gases at STP are commonly used in calculations for densities and volumes. 1 lbmol of any gas at STP occupies a volume of 359 ft3. 1 kmol of any gas at STP occupies a volume of 22.4 m3. The preceding results can be obtained by using Eq. 13.15 to calculate the volume of unit mole of ideal gas at STP.
 psia  ft3 ð1 lbmolÞ 10:73 ð492  R Þ lbmol  R NRT V¼ ¼ P 14:7 psia ¼ 359:1 ft3
 kJ ð273 K Þ ð1 kmolÞ 8:314 kmol  K NRT ¼ V¼ P 101:3 kPa ¼ 22:4 m3 Since the volume of a gas is inversely proportional to the pressure (Boyles Law) and directly proportional to the absolute temperature (Charles Law), the volume of “N” mols of an ideal gas at any condition can be calculated by using the following equations. The volume of N lbmol of any gas at any pressure P (psia) and any temperature T ( R) can be calculated by using Eq. 13.17. V¼



  359 ft3 14:7 psia T ðN lbmolÞ  P 492 R lbmol

ð13:17Þ

The volume of N kmol of any gas at any pressure P (kPa) and any temperature T (K) can be calculated by using Eq. 13.18.

  22:4 m3 101:3 kPa T ðN kmolÞ V¼ P 273 K kmol

ð13:18Þ

304

13.3.3.2

13

Thermodynamics Fundamentals

Normal Temperature Pressure (NTP)

The normal pressure is the same as the standard atmospheric pressure (14.7 psia and 101.3 kPa). However, there is considerable variation in the values used for normal temperature. The National Institute of Standards and Technology (NIST) in the USA defines normal temperature pressure as: • 68  F and 14.7 psia (USCS) • 20  C and 101.3 kPa

13.3.3.3

Standard Cubic Feet (SCF) and Actual Cubic Feet (ACF)

The preceding terms are widely used in the gas production and gas processing industries. The standard cubic feet is the volume of a gas measured at 14.7 psia pressure and 60  F, whereas the actual cubic feet is the volume of the gas measured at the actual conditions of pressure and temperature. Based on this concept, the volume flow rate of gases is specified in terms standard cubic feet per minute (SCFM). SCFM can be converted to actual cubic feet per minute (ACFM) by using the following equation, which is similar to Eq. 13.17 described earlier. ACFM ¼ SCFM 

13.4

14:7 psia T  actual Pactual 520  R

ð13:19Þ

Specific Heats of Gases

The specific heat of a substance is the heat required to raise the temperature of a unit mass of a substance by a unit degree of temperature. It is represented by the symbol “c” and has units of Btu/lbm –  R (USCS) and J/kg.K (S I). It is important to keep in mind that ΔT  R ¼ ΔT  F and ΔT K ¼ ΔT  C resulting in the following relationships: Btu/lbm- R ¼ Btu/lbm- F and kJ/kg.K ¼ kJ/kg C. Using the definition of specific heat, the heat required to heat any substance with mass m from temperature T1 to temperature T2 is: Q ¼ mcðT 2  T 1 Þ ¼ mcΔT

ð13:20Þ

When a gas is heated, it expands. If it is allowed to expand, the volume increases but the pressure remains constant. When a gas expands in this manner, it performs work by pushing the boundary. However, when a gas is heated in a rigid container, it occupies the volume of the container, which remains constant resulting an increase in pressure. Consequently, it is necessary to define two specific heats for gases— specific heat at constant pressure, represented by cp and specific heat at constant volume, represented by cv.

13.4

Specific Heats of Gases

305

The specific heat of an ideal gas at constant pressure [1, 2, 4] is defined as: cp ¼

dh dT

ð13:21Þ

The specific heat of an ideal gas at constant volume is defined as: cv ¼

du dT

ð13:22Þ

Also, for an ideal gas: cp  cv ¼ R

ð13:23Þ

The ratio of the specific heat at constant pressure to the specific heat at constant volume is represented by the symbol “k” (Table 13.2). cp ¼k cv

ð13:24Þ

The equation used in calculating the heat added to a gas when pressure is constant is: Q ¼ mcp ΔT

ð13:25aÞ

The equation used in calculating the heat added to a gas when volume is constant is: Q ¼ mcv ΔT

ð13:25bÞ

Table 13.2 cp, cv, and k for common gases Gas Air Oxygen Nitrogen Hydrogen Carbon dioxide Methane Ethane Propane

cp kJ/kg.K 1.0 0.92 1.04 14.3 0.85 2.25 1.77 1.68

cp Btu/lbm- R 0.24 0.22 0.25 3.43 0.20 0.53 0.43 0.41

cv kJ/kg.K 0.72 0.66 0.74 10.2 0.66 1.74 1.49 1.49

cv Btu/lbm- R 0.17 0.16 0.18 2.44 0.16 0.40 0.36 0.36

k 1.4 1.4 1.4 1.4 1.3 1.3 1.2 1.1

306

13.5

13

Thermodynamics Fundamentals

Nonideal Behavior of Gases

The ideal gas equation (Eq. 13.12) loses accuracy and breaks down at extremely high pressures and extremely low temperatures. The ideal gas law is valid only if the physical volume of the gas molecules is small compared to the macro volume occupied by the gas. This requirement is satisfied at moderate pressures and at moderate temperatures. However, the macro volume of the gas is very small at extremely high pressures and at very low temperatures. This results in the physical volume occupied by the molecules becoming comparable to the macro volume causing inaccuracies in calculations based on the ideal law. Gases that do not behave ideally are known as real gases. There are two options for obtaining accurate results for real gases: • Using real gas equations of state such as the Vander Waal’s equation • Using the generalized compressibility chart Real gas equations tend to be nonlinear requiring iterative solutions. However, using the generalized compressibility chart is the most commonly used option for calculations involving real gases.

13.5.1 Generalized Compressibility Chart The ideal gas equation (Eq. 13.12) is modified for use for real gases by introducing the “generalized compressibility factor” [1, 2, 6], which is represented by the symbol “Z.” The compressibility factor can be included in Eqs. 13.12 and 13.15 resulting in the following equations for real gases. PV ¼ ZmRT

ð13:26Þ

PV ¼ ZNRT

ð13:27Þ

The typical format of generalized compressibility chart is shown in Fig. 13.2. The generalized compressibility factor, Z, can be obtained by using the reduced pressure, Pr, and reduced temperature, Tr, as parameters. The reduced pressure and the reduced temperature can be calculated by the following equations. Pr ¼

P Pc

ð13:28Þ

Tr ¼

T Tc

ð13:29Þ

13.5

Nonideal Behavior of Gases

307

Fig. 13.2 Typical format of generalized compressibility chart

In Eq. 13.25, Pc is the critical pressure and in Eq. 13.26, Tc is the critical pressure, which are available from thermodynamic property tables. Example 13.2 3000 SCFM (measured at standard conditions of 14.7 psia and 60  F) of compressed natural gas enters a gas pipeline (ID ¼ 35.38 in) at 1500 psia and 70  F. The molecular weight of natural gas is 17 lbm/lbmol, and the critical properties are Pc ¼ 675 psi and Tc ¼ 340  R. Calculate: A. The mass flow rate of the gas in lbm/min. B. The volume flow rate of the gas at the pipeline entrance in cubic feet per minute (cfm) (Solution) A. Calculate the absolute value of standard temperature using Eq. 13.2. T  R ¼ 460 þ 60  F ¼ 520  R Calculate the individual gas constant for natural gas using Eq. 13.16.  R¼

R ¼ M NG

3

psiaft 10:73 lbmol R lbm 17 lbmol

 ¼ 0:6312 psia  ft3 =lbm  R

308

13

Thermodynamics Fundamentals

Calculate the mass flow rate using the ideal gas law (Eq. 13.12) at standard conditions.
 ft3 ð14:7 psiaÞ 3000 min PV_  m_ ¼ ¼
¼ 13:44 lbm= min 3 RT psia  ft  RÞ 0:6312 ð 520 lbm  R Calculate the absolute temperature of natural gas using Eq. 13.2. T  R ¼ 460 þ 70 F ¼ 530 R Since the pressure is quite high at 1500 psia, it is likely that the ideal gas law will not give accurate results. The compressibility factor, Z, will have to be used. Calculate the reduced pressure and the reduced temperature using Eqs.13.28 and 13.29. Pr ¼

P 1500 psia ¼ 2:22 ’ 2:2 ¼ Pc 675 psia

Tr ¼

T 530 ∘ R ¼ ¼ 1:56 ð’ 1:6Þ T c 340 ∘ R

Using Pr and Tr as parameters determine the generalized compressibility factor from the generalized compressibility chart (Fig. 13.2) as shown here.

From the graph, the compressibility factor, Z, is 0.87. Calculate the volume flow rate at the pipeline entrance using the real gas equation (Eq. 13.26).

13.6

Thermodynamic Processes Involving Ideal Gases

309



 lbm psia  ft3 0:6312 ð0:88Þ 13:44 ð530 RÞ min lbm  R _ Z mRT V_ ¼ ¼ P 1500 psia ¼ 2:64 ft3 = min

13.6

Thermodynamic Processes Involving Ideal Gases

A thermodynamic process is a procedure that causes a change in the state of the system. The different types of thermodynamic processes [1, 2, 4, 7] are described here. The ideal gas law (Eq. 3.12) can be written for any general thermodynamic process occurring between the initial state “1” and the final state “2.” The mass, m, remains constant in a closed system, and the individual gas constant, R, is constant for the gas undergoing the thermodynamic process. P1 V 1 ¼ mRT 1

ð13:30Þ

P2 V 2 ¼ mRT 2

ð13:31Þ

Divide Eq. 13.31 by Eq. 13.30. Note that m and R cancel out since they remain constant. P2 V 2 T 2 ¼ P1 V 1 T 1

ð13:32Þ

Equation 13.32 is very useful in the analysis of different thermodynamic processes.

13.6.1 Isothermal Process In an isothermal process, the temperature of the substance remains constant. Hence, T2 ¼ T1 From Eq. 13.32, the following equation can be written for an isothermal process.   1 P2 V 2 ¼ P1 V 1 , PV ¼ constant, V / P

ð13:33Þ

Equation 13.33 states that the product of pressure times volume is constant in an isothermal process, which implies that volume is inversely proportional to pressure

310

13

Thermodynamics Fundamentals

Fig. 13.3 P – V diagram for an isothermal expansion process

(Boyle’s law). An isothermal expansion process is shown on a P – V diagram in Fig. 13.3. Example 13.3 1000 ft3/min (cfm) of natural gas (methane, CH4) is compressed from 100 kPa to 300 kPa at a constant temperature of 30  C. Calculate the volume flow rate of the gas after the compression process. (Solution) From  Eq.13.33, for an isothermal process:    _V 2 ¼ P1 V_ 1 ¼ 100 kPa 1000 ft3 ¼ 333:33 ft3 = min P2

300 kPa

min

13.6.2 Isobaric Process In an isobaric process, the pressure of the system remains constant. Thus, in an isobaric process: P2 ¼ P1 The following equation is obtained by using the preceding result in Eq. 13.32. V2 T2 ¼ , ðV / T Þ V1 T1

ð13:34Þ

Equation 13.34 states that at constant pressure, the volume of an ideal gas is directly proportional to the absolute temperature of the gas (Charles law). A constant pressure process is shown in Fig. 13.4.

13.6

Thermodynamic Processes Involving Ideal Gases

311

Fig. 13.4 P – V diagram for an isobaric (constant pressure) process

Example 13.4 A piston-cylinder system consists of 1 lbm of air at 70  F and 20 psia. The air is heated at constant pressure such that the final volume increases by 60% of the initial volume. Calculate: A. The initial and final volumes during the process B. The final temperature (Solution) A. Calculate the individual gas constant for air using Eq. 13.16. psiaft3

R¼

10:73 lbmol R R ¼ ¼ 0:37 psia  ft3 =lbm  R lbm M Air 29 lbmol

Calculate the initial absolute temperature of air using Eq. 13.2. T 1 ¼ 460 þ 70 F ¼ 530 R Calculate the initial volume of air using the ideal gas law (Eq. 13.12).
 psia  ft3 ð1 lbmÞ 0:37 ð530 RÞ lbm  R mRT 1 V1 ¼ ¼ ¼ 9:81 ft3 20 psia P1 The final volume is 60% higher than the initial volume. V 2 ¼ 1:6V 1 ¼ 1:6  9:81 ft3 ¼ 15:70 ft3

312

13

Thermodynamics Fundamentals

B. Calculate the final temperature of air using Eq. 13.34. T2 ¼ T1


 V2 ¼ 530  R  1:6 ¼ 848  R V1

T 2  F ¼ 848  R  460 ¼ 388  F

13.6.3 Isochoric Process In an isochoric process, the volume of the substance remains constant. Thus, in an isochoric process: V2 ¼ V1 The following equation is obtained by using the preceding result in Eq. 13.32. P2 T 2 ¼ , ðP / T Þ P1 T 1

ð13:35Þ

Equation 13.35 states that at constant volume, the pressure of an ideal gas is directly proportional to the absolute temperature of the gas. A rigid container will have a constant volume. If a gas in a rigid container is heated, the pressure of the gas will increase since the gas is not free to expand when heated. A constant volume process is shown in Fig. 13.5.

13.6.4 Isentropic Process In an isentropic process, the entropy of the substance remains constant, that is, S ¼constant. Since the entropy does not change, the amount of unavailable energy remains constant during the process. This implies that all the available energy is converted to useful work in an isentropic process. Therefore, an isentropic process has the maximum possible efficiency in converting energy to useful work. The pressure-volume relationship for an isentropic process is: P2 V 2 k ¼ P1 V 1 k ¼ PV k ¼ C, a constant

ð13:36Þ

In Eq. 13.36, k is the ratio of specific heats, k ¼ cp/cv. Figure 13.6 shows a comparison of P – V diagrams for isothermal and isentropic expansion processes. The curve for isentropic expansion is steeper than the curve for isothermal expansion. This is because the exponent of V (in Eq. 13.36) for an isentropic process, k, is always greater than 1. In an isothermal process, the exponent of V (in Eq. 13.33) is 1.

13.6

Thermodynamic Processes Involving Ideal Gases

313

Fig. 13.5 P – V diagram for an isochoric (constant volume) process

Fig. 13.6 P – V diagram for isothermal and isentropic expansion processes

By combining the ideal gas equation (Eq. 13.12) and Eq. 13.36, the following additional P-V-T relationships can be written for an isentropic process. k1 V1 V2
k1 T2 P2 k ¼ T1 P1

T2 ¼ T1




ð13:37Þ ð13:38Þ

Note: An isentropic process is also known as a reversible adiabatic process. There is no heat transfer in an adiabatic process, that is, Q = 0 in an adiabatic process.

314

13

Thermodynamics Fundamentals

Example 13.5 Air is compressed isentropically to one-fourth its original volume. The initial pressure is 100 kPa, and the initial temperature is 25  C. Calculate the temperature and pressure after the compression process. (Solution) Calculate the initial absolute temperature of air using Eq. 13.3. T 1 ¼ 273 þ 25  C ¼ 298 K From Table 13.2, for air, k ¼ 1.4 Calculate the final temperature using Eq. 13.37.
1k  11:4 V2 1 T2 ¼ T1 ¼ ð298 KÞ ¼ 519 K 4 V1

ð59∘ CÞ

Calculate the final pressure by using Eq. 13.36.
P2 ¼ P1

V1 V2

k ¼ ð100 kPaÞð4Þ1:4 ¼ 696 kPa

13.6.5 Constant Enthalpy/Throttling Process During a throttling process, the enthalpy remains constant. A common example of a throttling process is flow of a fluid across a pressure-reducing throttling valve. The enthalpy remains constant because the throttling valve has no work or heat interaction with the surroundings. Other examples of throttling process are pressure and flow control valves, and pressure relief valves. The throttling process is also used in liquefaction of gases.

13.7

Calculation of Work, Internal Energy Changes, Enthalpy Changes, and Entropy Changes for Processes Involving Ideal Gas

13.7.1 Work Work is the product of force and distance. The pressure of a gas is the force exerted by the gas per unit cross-section area. Consider a gas expanding in a piston-cylinder system as shown in Fig. 13.7.

13.7

Calculation of Work, Internal Energy Changes, Enthalpy Changes. . .

315

Fig. 13.7 Work due to expansion of a gas

When the gas expands, it moves the piston thereby performing work. If Acs is the cross-section area of the piston, the force exerted by the gas on piston is F ¼ P  Acs The differential work performed by the gas in moving the piston through a distance dL is dW ¼ P  Acs  dL However, the product of the area of cross section and the differential length is the differential increase in the volume of the gas. dL  Acs ¼ dV Therefore, dW ¼ PdV

ð13:39Þ

To get the total work of expansion performed by the gas in moving the piston from the initial position (“1”) to the final position (“2”), the preceding equation is integrated with the corresponding limits for the volume. Z

ZV 2 dW ¼ W ¼

PdV

ð13:40Þ

V1

13.7.1.1

Work for a Constant Pressure (Isobaric) Process

Assuming the pressure to be constant, the work for a constant pressure can be obtained by integrating the preceding equation. W ¼ PðV 2  V 1 Þ ¼ PΔV

ð13:41Þ

316

13.7.1.2

13

Thermodynamics Fundamentals

Work for a Constant Volume (Isochoric) Process

When volume is constant, the change in volume, dV, is zero. V ¼ constant and dV ¼ 0 Therefore, W const:vol ¼ 0

13.7.1.3

ð13:42Þ

Work for a Constant Temperature (Isothermal) Process

From Eq. 13.33 for an isothermal process, PV ¼ C, a constant:Therefore, P ¼

C V

Substitute the preceding result into Eq. 13.40 and integrate. ZV 2 W¼

ZV 2
 dV PdV ¼ C V

V1

V1




 V2 V2 V2 W ¼ C ln ¼ P1 V 1 ln ¼ P2 V 2 ln V1 V1 V1

ð13:43Þ

From the ideal gas law (Eq. 13.12), for a constant temperature process, P1 V 1 ¼ P2 V 2 ¼ mRT

ð13:44Þ

From Eqs. 13.43 and 13.44, the equations for work in an isothermal process can be written as:


 V2 V2 V2 W ¼ P1 V 1 ln ¼ P2 V 2 ln ¼ mRT ln V1 V1 V1

ð13:45Þ

The volume ratio in Eq. 13.45 can be replaced by pressure ratio, VV 21 ¼ PP12, to obtain another set of equations for an isothermal process.


P1 W ¼ P1 V 1 ln P2






P1 ¼ P2 V 2 ln P2






P1 ¼ mRT ln P2

 ð13:46Þ

13.7

Calculation of Work, Internal Energy Changes, Enthalpy Changes. . .

13.7.1.4

317

Constant Entropy (Isentropic) Process

For an isentropic process, P2V2k ¼ P1V1k ¼ PVk ¼ C Solve the preceding equation for the pressure, P. P ¼ VCk Substitute for P in the equation for work. ZV 2 W¼

ZV 2 PdV ¼

V1

V1

 kþ1 V 2  1k C V V2  V 1 1k dV ¼ C ¼C k þ 1 V 1 1k Vk

Substitute for C from the P – V relationship.       P2 V 2 k V 2 1k  P1 V 1 k V 1 1k V 2 1k  V 1 1k W ¼C ¼ 1k 1k P2 V 2  P1 V 1 ¼ 1k 

Substitute for the product PV from the ideal gas law, PV ¼ mRT, to get the equations for work in an isentropic process. W¼

P2 V 2  P1 V 1 mRðT 2  T 1 Þ ¼ 1k 1k

ð13:47Þ

13.7.2 Internal Energy Change For an ideal gas with constant specific heats, the definition of specific heat at constant du volume is given by Eq. 13.22, cv ¼ dT Separate the variables, and integrate to get the following result. Δu ¼ cv ΔT

ð13:48Þ

Equation 13.48 can be used to calculate the internal change per unit mass for any process involving ideal gas. For an ideal gas with mass m, the equation for calculating the change in internal energy for any process is: ΔU ¼ mcv ΔT

ð13:49Þ

318

13

Thermodynamics Fundamentals

13.7.3 Enthalpy Change For an ideal gas with constant specific heats, the definition of specific heat at constant dh pressure is given by Eq. 13.21, cp ¼ dT Separate the variables, and integrate to get the following result. Δh ¼ cp ΔT

ð13:50Þ

Equation 13.50 can be used to calculate the enthalpy change per unit mass for any process involving ideal gas. For an ideal gas with mass m, the equation for calculating the change in enthalpy for any process is: ΔH ¼ mcp ΔT

ð13:51Þ

13.7.4 Entropy Change The entropy change for unit mass of an ideal gas for any process can be calculated by using the following equations.



 T2 P2  R ln T1 P1

 T2 V2 Δs ¼ cv ln þ R ln T1 V1 Δs ¼ cp ln

ð13:52Þ ð13:53Þ

For an ideal gas with mass m, the equations for calculating the change in entropy for any process are:


 T2 P2  R ln ΔS ¼ m cp ln T1 P1


 T2 V2 ΔS ¼ m cv ln þ R ln T1 V1

ð13:54Þ ð13:55Þ

Example 13.6 Methane at 327  C and 200 kPa is compressed isothermally from an initial volume of 3 L to a final volume of 2 L. Calculate: A. The work of compression required B. The change in entropy C. Comment on how the temperature can be maintained at 327  C.

13.7

Calculation of Work, Internal Energy Changes, Enthalpy Changes. . .

319

(Solution) Calculate the absolute temperature of methane by using Eq. 13.2. T 1 ¼ 273 þ 327  C ¼ 600 K The initial volume is V1 ¼ 3 L/(1000 L/m3) ¼ 0.003 m3. The final volume is V2 ¼ 2 L/(1000 L/m3) ¼ 0.002 m3. A. Calculate the isothermal compression work per unit mass by using Eq. 13.45.


   V2 0:002 m3 3 ln W ¼ P1 V 1 ln ¼ 200 kPa  0:003 m V1 0:003 m3 ¼ 0:2433 kJ ð243:3 JÞ The negative sign indicates work of compression being done on the gas. From Table 13.1, the molecular weight of methane is M ¼ 16 kg/kmol. B. Calculate the individual gas constant of methane using Eq. 13.16.

R¼

kJ R 8:314 kmolK ¼ ¼ 0:5196 kJ=kg  K M 16 kg kmol

Calculate the mass of methane by applying the ideal gas law (Eq. 13.12) at the initial condition (state1).  m¼

 kJ 0:5196 kgK ð600 KÞ

RT 1 ¼ ¼ 519:6 kg P1 V 1 ð200 kPaÞð0:003 m3 Þ

Calculate the entropy change using Eq. 13.55.

The negative sign indicates decrease in entropy, which is to be expected when a gas is compressed. The molecules of a compressed gas occupy less volume which results in lesser disorder of the molecules and hence lower entropy.

320

13

Thermodynamics Fundamentals

C. When a gas is compressed, the temperature increases since the absolute temperature of a gas is directly proportional to the pressure. The temperature can be maintained at 327  C (isothermal condition) by dissipating the heat generated during the compression process, which can be achieved by circulating a coolant around the compressor cylinder. Example 13.7 Carbon dioxide at 170  C occupies a volume of 1.9 m3 when the absolute pressure is 300 kPa. The gas is then heated to a final temperature of 300  C while maintaining the pressure constant at 300 kPa. Determine: A. B. C. D. E.

The mass of carbon dioxide The final volume The work done The heat added The changes in internal energy, enthalpy, and entropy

(Solution) Calculate the absolute temperature of carbon dioxide before and after heating by using Eq. 13.3. T 1 ¼ 273 þ 170  C ¼ 443 K

T 2 ¼ 273 þ 300  C ¼ 573 K

From Table 13.1, the molecular weight of carbon dioxide is 44 kg/kmol. Calculate the individual gas constant of carbon dioxide by using 13.16. R¼

kJ R 8:314 kmolK ¼ ¼ 0:1889 kJ=kg  K kg M 44 kmol

A. Calculate the mass of carbon dioxide by applying the ideal gas law (Eq.13.12) at state 1 (initial state). m¼

ð300 kPaÞð1:9 m3 Þ P1 V 1  ¼ ¼ 6:81 kg RT 1 kJ 0:1889 kgK ð443 KÞ

Note: kPa  kN/m2. Therefore, (kPa)(m3)  (kN/m2)( m3)  kN.m  kJ B. Calculate the final volume of carbon dioxide by using Eq. 13.34 (Charles law) for a constant pressure process.

   573 K T2 3 V2 ¼ V1 ¼ 2:46 m3 ¼ 1:9 m 443 K T1

13.7

Calculation of Work, Internal Energy Changes, Enthalpy Changes. . .

321

C. Calculate the work done due to expansion using Eq. 13.42 for a constant pressure process.   W ¼ PðV 2  V 1 Þ ¼ ð300 kPaÞ 2:46 m3  1:9 m3 ¼ 168 kJ D. From Table 13.2, the specific heat at constant pressure for carbon dioxide is cp ¼ 0.85 kJ/kg.K. Calculate the heat added during this constant pressure process by using Eq. 13.24.
 kJ ð573 K  443 KÞ ¼ 984:1 kJ Q ¼ mcp ΔT ¼ ð6:81 kgÞ 0:85 kg  K E. Change in Internal Energy From Table 13.2, the specific heat at constant volume for carbon dioxide is cv ¼ 0.66 kJ/kg.K. Calculate the change in internal energy by using Eq. 13.49, applicable for any process. Note that the internal energy of the gas will increase due to increase in temperature.


 kJ ΔU ¼ mcv ΔT ¼ ð6:81 kgÞ 0:66 ð573 K  443 KÞ ¼ 764:1 kJ kg  K Change in Enthalpy From Table 13.2, the specific heat at constant pressure for carbon dioxide is cp ¼ 0.85 kJ/kg.K. Calculate the change in enthalpy by using Eq. 13.51, applicable for any process.


 kJ ΔH ¼ mcp ΔT ¼ ð6:81 kgÞ 0:85 ð573 K  443 KÞ ¼ 984:1 kJ kg  K Change in Entropy Calculate the change in entropy by using Eq. 13.54, applicable for any process.

The positive sign for entropy change indicates increase in entropy, which is to be expected during expansion of a gas due to increase in the disorder of molecules.

322

13.8

13

Thermodynamics Fundamentals

Thermodynamic Phase Diagrams

Thermodynamic phase diagrams are graphs plotted with thermodynamic variables (typically, T, P, v, s, h) as parameters, and they illustrate the presence of different phases, the phase boundaries between two phases, and the values of the thermodynamic parameters. Only intensive and specific properties are used in the construction of thermodynamic phase diagrams.

13.8.1 Phase Diagram for Water The following points can be observed from the T – v phase diagram for water shown in Fig. 13.8. 1. Consider heating of liquid water at atmospheric pressure (14.7 psia) from an ambient temperature of 50  F, shown as reference point A in Fig. 13.8. The temperature will follow the path ABCD along the isobar for atmospheric pressure as shown in Fig. 13.8. Water will be in liquid phase up to reference point B, which is known as “saturated liquid” (commonly represented by the subscript f). At point B, the liquid has reached the saturation point (also known as the boiling point of the liquid) because any further heating will result in the formation of vapor (steam). The temperature at point B is the saturation temperature at atmospheric pressure, which is 212  F as shown in the figure. Therefore, at saturated liquid state, there is only one degree of freedom. Specifying a value of saturation pressure automatically fixes the value of saturation temperature. For example, if saturation pressure of 14.7 psia is specified, the saturation temperature Fig. 13.8 Temperaturespecific volume phase diagram for water

13.9

Properties of Steam

323

is automatically 212  F. Only one variable, either pressure or temperature, can be specified independently, and hence there is only one degree of freedom. Heating of the substance beyond B will result in the formation of a liquid-vapor mixture at a constant temperature of 212  F along the path BC, which represents the isothermal vaporization process. At reference point C, all the liquid has vaporized resulting in a homogeneous vapor phase. At reference point C, the substance has reached a saturated vapor state (commonly represented by the subscript g). Heating beyond reference point C results in superheated vapor state, reference point D, as shown in the figure. The isobar for 50 psia is also shown in the figure. The fixed pressure-temperature pair is Psat ¼ 50 psia and Tsat ¼ 281  F. Several isobars can be generated at different pressures. The locus of all the saturated liquid points will form the saturated liquid curve (phase boundary), and similarly the locus of all the saturated vapor points will form the saturated vapor curve. The two saturation curves will meet at the critical point, where there is no distinction between the liquid and vapor phases. 2. There are three different sections in the diagram based on the phase boundaries: compressed liquid to the left of the saturated liquid curve, liquid-vapor mixture between the saturated liquid and saturated vapor curves (inside the bell-shaped curve), and superheated vapor to the right of the saturated vapor curve. 3. At reference point A, the substance is a compressed liquid because at a temperature of 50  F, the corresponding saturation pressure is 0.18 psia. However, point A is on the 14.7 psia isobar. Since the pressure of the substance is much greater than the saturation pressure at the given temperature, the substance is in a compressed liquid state. This state is also referred to as subcooled liquid state because the temperature of the substance (50  F) is much lower than the saturation temperature (212  F) for the 14.7 psia isobar. 4. Reference point D lies on the 14.7 psia isobar, and the saturation temperature for this isobar is 212  F. However, the temperature of the substance at point D is 281  F as shown in the figure. Since the temperature of the substance is greater than the saturation temperature at the given pressure, the substance is in a superheated vapor state.

13.9

Properties of Steam

The properties of steam can be easily obtained from steam tables or from graphs based on different phase diagrams. Online calculator programs are also available to obtain the properties of steam when sufficient, correct inputs are provided by the user. Pressure and temperature are key variables required for obtaining the properties of steam. The data in the tables and charts are always listed on the basis of properties per unit mass (specific properties). Thus, one can obtain the specific volume, the specific internal energy, the specific enthalpy, and specific entropy from the steam tables. The steam tables are categorized as saturated tables, superheated tables, and compressed liquid tables.

324

13

Thermodynamics Fundamentals

13.9.1 Saturated Steam Tables The saturated steam tables can be used for situations between saturated liquid and saturated vapor, that is, path BC shown in Fig. 13.8. The saturated steam tables can be based either on temperature or pressure. Short excerpts from saturated steam tables in both USCS and S I units are shown here for illustrative purposes (Figs. 13.9 and 13.10).

13.9.1.1

Calculation of Properties of Liquid-Vapor Mixtures

In Fig. 13.8, the state of the substance between saturated liquid (reference point B) and saturated vapor (reference point C) is a liquid-vapor mixture. Liquid-vapor mixtures are characterized by the mass fraction of vapor in the mixture, which is known as the quality of mixture and is represented by the symbol x. The subscript f represents the liquid, and the subscript g represents the vapor. The mathematical representation of the quality of a liquid-vapor mixture is shown in Eq. 13.56. Quality, x ¼

mg mass of vapor ¼ mass of L  V mixture mf þ mg

ð13:56Þ

The quality of a liquid-vapor mixture represents the extent to which vaporization has occurred. The subscript “fg” is used in representing the change in any specific

Fig. 13.9 Saturated steam table excerpt (S I units)

Fig. 13.10 Saturated steam table excerpt (USCS units)

13.9

Properties of Steam

325

property due to complete vaporization. For example, vfg represents the increase in specific volume due to complete vaporization of the liquid, and vfg ¼ vg – vf. To find the value of any specific property of a liquid mixture, the change in property due to complete vaporization is weighted by the quality of the liquid-vapor mixture, and the result is added to the value of the specific property of saturated liquid. The following equations can be used in calculating the properties of liquid-vapor mixtures. v ¼ vf þ xvfg

ð13:57Þ

u ¼ uf þ xufg

ð13:58Þ

h ¼ hf þ xhfg

ð13:59Þ

s ¼ sf þ xsfg

ð13:60Þ

Example 13.8 A rigid container has a volume of 2 m3 and contains 5 kg of steam at a pressure of 200 kPa. Determine the quality of steam. (Solution) Calculate the specific volume of steam using Eq. 13.6. v¼

V 2 m3 ¼ ¼ 0:40 m3 =kg m 5 kg

From steam tables,

Since the specific volume of steam, v, is between vf and vg, steam is a liquid-vapor mixture. Solve Eq. 13.57 for the quality, x, and substitute the known values to obtain the quality. m3

m3

v  vf 0:40 kg  0:0011 kg x¼ ¼ ¼ 0:451 ð45:1%vaporÞ 3 vfg 0:8846 m kg

13.9.2 Superheated Steam Tables Steam becomes superheated upon further heating of saturated vapor. Superheated steam has two degrees of freedom. Both temperature and pressure have to be specified to get the properties of superheated steam. Short excerpts from superheated

326

13

Thermodynamics Fundamentals

Fig. 13.11 Superheated steam table excerpts (S I and USCS units)

steam tables in both USCS and S I units are shown here for illustrative purposes (Fig. 13.11). Example 13.9 Determine the enthalpy of steam at 50 psia and 350  F. (Solution) At the given conditions, steam is most likely superheated. This can be verified by examining the 50 psia superheated table.

From the table, steam is in a superheated state at 50 psia and 350  F since T (350  F) > Tsat(281  F) at 50 psia. By interpolating the values in the table, the enthalpy of steam at 50 psia and 350  F is h ¼ 1210 Btu/lbm.

13.9.3 Properties of Compressed Liquid The thermal properties of liquids are strongly dependent on temperature and not pressure. This is because liquids are essentially incompressible. Increasing the pressure of a liquid does not increase its enthalpy significantly. The increase in enthalpy of a liquid due to compression can be calculated by using Eq. 13.61. Δhcomprn: ¼ vðP2  P1 Þ

ð13:61Þ

13.9

Properties of Steam

327

Consider liquid water at ambient conditions of 30  C and 100 kPa. From saturated steam tables, the saturation pressure at 30  C is 0.0042 MPa (4.2 kPa). Since P > Psat, water is a compressed liquid at 30  C and 100 kPa. Also, from the saturated steam tables, the enthalpy of saturated liquid at 30  C is hf ¼ 125.73 kJ/kg, and the specific volume of saturated liquid at the same temperature is vf ¼ 0.001 m3/kg. Calculate the increase in enthalpy due to compression by substituting the known values into Eq. 13.61.
 m3 0:001 ð100 kPa  4:2 kPaÞ kg ¼ 0:0958 kJ=kg
 m3 kN m3 kN  m kJ Note : kPa      kg kg kg m2 kg Δhcomprn: ¼ vðP2  P1 Þ ¼

The increase in enthalpy due to compression is added to the enthalpy of saturated liquid at the given temperature of 30  C to obtain the enthalpy of the compressed liquid at 30  C and 100 kPa. hcomp:liq ¼ hf at 30

C

þ Δhcomprn ¼ 125:73

kJ kJ þ 0:096 ¼ 125:83 kJ=kg kg kg

The enthalpy of the compressed liquid is almost equal to the enthalpy of the saturated liquid at the given temperature. For compressed liquids, we can use the properties of the saturated liquid at the given temperature without any loss in accuracy. Example 13.10 Determine the state and enthalpy of steam at 50 psia and 200  F. (Solution) From the saturated steam table excerpt shown here, at 50 psia, Tsat ¼ 281  F. The temperature of steam T ¼ 200  F < Tsat ¼ 281  F. Hence, steam is a subcooled (compressed) liquid at the given conditions. For a compressed liquid, the enthalpy is the same as that of saturated liquid at the given temperature. Interpolating between the given values of saturated liquid enthalpies at 193  F and 212  F , the enthalpy of the saturated liquid at 200  F is h ¼ hf at 200  F ¼ 168.2 Btu/lbm.

328

13.10

13

Thermodynamics Fundamentals

Mollier Diagram

The Mollier diagram is a phase diagram for water plotted with specific enthalpy on the y-axis and specific entropy on the x-axis as shown in Fig. 13.12. The additional parameters used in the Mollier diagram are constant pressure lines, constant temperature curves, and constant moisture percent curves in the saturated region. Example 13.11 and Practice Problem 13.11 illustrate the use of the Mollier diagram. The Mollier diagram in S I units is shown in Fig. 13.12.

Fig. 13.12 Mollier diagram for steam (S I units). (Source: The Engineering Toolbox, reprinted with permission, Engineering ToolBox, (2008). Mollier Diagram for Water-Steam. [online] Available at: https://www.engineeringtoolbox.com/mollier-diagram-water-d_308.html)

13.11

Pressure-Enthalpy (P – h) Phase Diagram

329

Example 13.11 Steam at 200 psia and 500  F expands isentropically to a final pressure of 20 psia. Determine the enthalpy change for this process. (Solution)

Find the 200 psia constant pressure line and 500  F, and locate their intersection point as shown on the Mollier diagram illustration. This will be reference point “1” for the start of the isentropic expansion and h1 ¼ 1275 Btu/lbm. From reference point “1,” move vertically down along the constant entropy line to the final constant pressure line of 20 psia. This intersection point is reference point “2” and h2 ¼ 1090 Btu/lbm. Change is always the final value minus the initial value and the change in enthalpy is Δh ¼ h2  h1 ¼ 1090

Btu Btu  1275 ¼ 185 Btu=lbm lbm lbm

Note that in going from state 1 to state 2, the enthalpy decreases.

13.11

Pressure-Enthalpy (P – h) Phase Diagram

P – h phase diagrams are extensively used in analysis and calculations involving refrigeration and air-conditioning (HVAC) systems. The schematic layout of a typical P – h phase diagram is shown in Fig. 13.13. The P – h phase diagram has a dome-shaped phase boundary with saturated liquid on the left phase boundary and saturated vapor on the right phase boundary. The state of the refrigerant is a liquidvapor mixture between these boundaries inside the dome. To the left of the saturated liquid, the state of the refrigerant is compressed liquid (sub-cooled liquid), and the refrigerant is superheated to the right of the saturated vapor line.

330

13

Thermodynamics Fundamentals

Fig. 13.13 Typical P – h diagram for refrigerants

Example 13.12 Saturated R-134a liquid is throttled from 200 psia to 20 psia. Determine: A. The initial and final enthalpy B. The change in temperature C. The change in entropy (Solution)

Locate state point 1 (saturated liquid at 200 psia) of R-134a on the P – h diagram, (shown in the diagram for illustrative purposes). Determine the enthalpies, entropies, and temperatures at state points 1 and 2 using the enthalpy, entropy, and temperature lines as shown. A. The enthalpy lines are vertical, and therefore enthalpy remains constant during the throttling process. The enthalpy at state point 1 can be determined on the top or bottom horizontal enthalpy scale. From the diagram, the enthalpy is h ¼ 55 Btu/lbm ¼ const.

Practice Problems

331

State point 1 is located on the entropy s ¼ 0.10 line, and the entropy at state point 1 is s1 ¼ 0.10 Btu/lbm ‐  R State point 1 is located between the temperature lines of T ¼ 120  F and T ¼ 140  F. By interpolating in the figure for the temperature location of state point 1, temperature at state point 1 is T1 ¼ 128  F. Since enthalpy is constant during a throttling process, proceed vertically down along the constant enthalpy line to 20 psia constant pressure horizontal line. This intersection point is state point 2. B. State point 2 is located on the temperature line of T ¼ 0  F. The change in temperature is: ΔT ¼ T 2  T 1 ¼ 0 F  128  F ¼ 128  F C. State point 2 is located on the s ¼ 0.12 entropy line. The entropy at state point 2 is s2 ¼ 0.12 Btu/lbm -  R. The change in entropy is: Δs ¼ s2  s1 ¼ 0:12

Btu Btu  0:10 ¼ 0:02 Btu=lbm  R lbm  R lbm  R

Practice Problems Practice Problem 13.1 Find the volume (m3) occupied by 3 kmol of air at a pressure of 5 bar and temperature of 400 K. Practice Problem 13.2 Calculate the volume of 20 lbm of oxygen at 100 psia and 80  F using the standard molar volume as the basis. Practice Problem 13.3 Air is compressed isothermally to one-fourth of its original volume. If the initial pressure is 15 psia and the initial temperature is 70  F, calculate the final pressure of air in psia. Practice Problem 13.4 Methane coming from an arctic pipeline is at 10  C. 5000 L of this gas is stored in a rigid tank. A pressure gage attached to the tank shows a reading of 50 kPa. The atmospheric pressure at the location is 101 kPa. The temperature of methane is increased to 25  C. Calculate: A. The mass of methane in the tank B. The pressure gage reading after the heating process

332

13

Thermodynamics Fundamentals

Practice Problem 13.5 Calculate the work done during isothermal expansion of 10 kmol of carbon dioxide at an initial temperature of 30  C when the final volume is twice the initial volume. Practice Problem 13.6 Using the same data given in Example 13.6, calculate: A. The work of compression required in an isentropic process B. The temperature of methane after the compression process C. Comment on the results for the work required in isothermal and isentropic processes. Practice Problem 13.7 A rigid container with a volume 3.5 ft3 contains 1.8 lbm of oxygen at 80  F. The gas is heated to a final temperature 170  F. Calculate: A. The percent increase in the pressure B. The heat added C. The changes in internal energy and entropy Practice Problem 13.8 Steam exhausts from a turbine at 10 psia with 95% quality. Determine the enthalpy of the exhaust steam. Practice Problem 13.9 2 kg of steam at 200 kPa is in a rigid container with volume of 3 m3. Determine the state and temperature of steam. Practice Problem 13.10 Steam at a pressure of 200 kPa has an enthalpy of 455 kJ/kg. Determine the temperature of steam. Practice Problem 13.11 Steam at 300  C and 200 kPa expands isentropically such that the decrease in enthalpy is 500 kJ/kg. Determine the final pressure of steam after expansion.

Solutions to Practice Problems Practice Problem 13.1 (Solution) Calculate the volume of air using the molar form of ideal gas law (Eq. 13.15).


 bar  m3 ð3 kmolÞ 0:08314 ð400 KÞ kmol  K NRT V¼ ¼ P 5 bar ¼ 19:95 m3

Solutions to Practice Problems

333

Practice Problem 13.2 (Solution) Calculate the absolute temperature of oxygen by using Eq. 13.2. T  R ¼ 460 þ 80 F ¼ 540 R Calculate the lbmol of oxygen using Eq. 13.13. N¼

m 20 lbm ¼ ¼ 0:625 lbmol lbm M 32 lbmol

Calculate the volume of oxygen at the given conditions using Eq. 13.17

  359 ft3 14:7 psia T ðN lbmolÞ P 492∘ R lbmol


  359 ft3 14:7 psia 540∘ R ¼ ð0:625 lbmolÞ 100 psia 492∘ R lbmol

V¼

¼ 36:2 ft3 The preceding result can be compared with the result from the ideal gas law as follows. Calculate the individual gas constant for oxygen using Eq. 13.16.  R¼

R ¼ M O2

3

psiaft 10:73 lbmol R lbm 32 lbmol

 ¼ 0:3353 psia  ft3 =lbm  R




 psia  ft3 ð20 lbmÞ 0:3353 ð540 RÞ lbm  R mRT V¼ ¼ 100 psia P 3 ¼ 36:21 ft As one would expect, the results from the two methods are in perfect agreement. Practice Problem 13.3 (Solution) Since the final volume is one-fourth the initial volume, V 2 ¼ V41 , that is VV 12 ¼ 4 From Eq. 13.33, for an isothermal process,
 V1 P ¼ 4  15 psia ¼ 60 psia P2 ¼ V2 1 Note: Since pressure is inversely proportional to volume, a quick, intuitive solution to the problem is that the final pressure is four times the initial pressure.

334

13

Thermodynamics Fundamentals

Practice Problem 13.4 (Solution) A. Calculate the initial absolute temperature of methane using Eq. 13.3. T 1 K ¼ 273 þ 10 C ¼ 283 K Calculate the initial absolute pressure of methane using Eq. 13.1. P1,abs ¼ Patm þ P1,gage ¼ 101 kPa þ 50 kPa ¼ 151 kPa Calculate the individual gas constant of methane using Eq. 13.16. R¼

kJ 8:314 kmolK R ¼ ¼ 0:5196 kJ=kg  K M CH4 16 kg kmol

Calculate the mass of methane in the tank using the ideal gas law (Eq. 13.12).   1 m3 ð 151 kPa Þ 5000 L  1000 L PV   m¼ 1 1¼ ¼ 5:134 kg RT 1 kJ 0:5196 kgK ð283 KÞ Note: (kPa)(m3) ¼ (kN/m2)(m3) ¼ kN.m ¼ kJ B. The final temperature is T2 ¼ 273  + 25  C ¼ 298 K Since this is a rigid, closed tank, the volume and mass of methane in the tank will be constant. This is a constant volume process. Calculate the absolute pressure after the heating process by using Eq. 13.35 for a constant volume process.
   T 298 K P2 ¼ P1 2 ¼ ð151 kPaÞ ¼ 159 kPa 283 K T1

Calculate the gage pressure after the heating process by using Eq. 13.1. P2,gage ¼ P2,abs  Patm ¼ 159 kPa  101 kPa ¼ 58 kPa Practice Problem 13.5 (Solution) From Table 13.1, the molecular weight of carbon dioxide is MCO2 ¼ 44 kg/kmol. Calculate the mass of carbon dioxide by using Eq. 13.13.


kg m ¼ NM ¼ ð10 kmolÞ 44 kmol

 ¼ 440 kg

Solutions to Practice Problems

335

Convert the initial temperature of carbon dioxide to absolute value by using Eq. 13.3. T 1 ¼ 273 þ C ¼ 273 þ 30  C ¼ 303 K The final volume of carbon dioxide is twice the initial volume. Therefore,


V2 V1

 ¼2

Calculate the individual gas constant of carbon dioxide by using Eq. 13.16. R¼

kJ R 8:314 kmolK ¼ ¼ 0:1889 kJ=kg  K M 44 kg kmol

Substitute the known values into Eq. 13.45. W ¼ mRT ln



 V2 kJ ð303 KÞ ln ð2Þ ¼ ð440 kgÞ 0:1889 kg  K V1 ¼ 17456 kJ

Practice Problem 13.6 (Solution) The following data is available from Example 13.6 and its solution: From Table 13.1, the molecular weight of methane is M ¼ 16 kg/kmol. Initial absolute temperature of methane, T1 ¼ 600 K Individual gas constant of methane, R ¼ 0.5196 kJ/kg  K The initial volume is V1 ¼ 3 L/(1000 L/m3) ¼ 0.003 m3 The final volume is V2 ¼ 2 L/(1000 L/m3) ¼ 0.002 m3 A. From Table 13.2, the ratio of specific heats for methane is k ¼ 1.3. Rearrange Eq. 13.36 as shown, and calculate the final pressure of methane after isentropic compression. P2 ¼ P1


k
1:3 V1 0:003 m3 ¼ ð200 kPaÞ ¼ 339 kPa V2 0:002 m3

Calculate the isentropic compression work required using Eq. 13.47. W¼

P2 V 2  P1 V 1 339 kPa  0:002 m3  200 kPa  0:003 m3 ¼ 1k 1  1:3 ¼ 0:26 kJ ð260 JÞ

336

13

Thermodynamics Fundamentals

B. Calculate the temperature of methane after the compression process using Eq.13.37.
T2 ¼ T1

V2 V1

1k


11:3 0:002 m3 ¼ ð600 KÞ ¼ 678 K 0:003 m3

C. The work required for isentropic compression (260 J) is greater than the work required for isothermal compression (243.3 J) because part of the energy is used in compressing the gas against the heat generated during the compression process [5]. Since heat is not dissipated during isentropic compression, it takes more energy to compress a gas which is at a higher temperature. During isothermal compression, the temperature remains constant due to dissipation of heat, and it takes less energy to compress a gas at a lower temperature. Practice Problem 13.7 (Solution) Since oxygen is in a rigid container, the volume of oxygen will be constant. Hence, this is a constant volume process. Calculate the absolute values of the initial and final temperatures of oxygen by using Eq. 13.2. T 1 ¼ 460 þ 80  F ¼ 540  R

T 2 ¼ 460 þ 170  F ¼ 630  R

A. Calculate the ratio of final pressure to the initial pressure using Eq. 13.35 for a constant volume process. P2 T 2 630  R ¼ 1:167 ¼ ¼ P1 T 1 540  R

Calculate the percent increase in  as shown.    pressure % increase in pressure ¼

P2 P1 P1

100 ¼

P2 P1

 1 100 ¼ ð1:167  1Þ100 ¼ 16:7%

B. From Table 13.2, the specific heat at constant volume for oxygen is cv ¼ 0.16 Btu/lbm -  R. Calculate the heat added during this constant volume process by using Eq. 13.25.  Q ¼ mcv ΔT ¼ ð1:8 lbmÞ 0:16

 Btu ð630  R  540  RÞ ¼ 25:92 kJ  lbm  R

Solutions to Practice Problems

337

C. Change in Internal Energy Calculate the change in internal energy by using Eq. 13.49, applicable for any process. Note that the internal energy of the gas will increase due to increase in temperature.  ΔU ¼ mcv ΔT ¼ ð1:8 lbmÞ 0:16 ¼ 25:92 kJ

 Btu ð630  R  540  RÞ  lbm  R

Change in Entropy Calculate the change in entropy by using Eq. 13.55 applicable for any process.

Practice Problem 13.8 (Solution)

Find hf and hfg at 10 psia from the steam tables as shown, and substitute the known values into Eq. 13.59, and calculate the enthalpy. h ¼ hf þ xhfg ¼ 161:2

  Btu Btu þ ð0:95Þ 981:9 ¼ 1094 Btu=lbm lbm lbm

338

13

Thermodynamics Fundamentals

Practice Problem 13.9 (Solution) Calculate the specific volume of steam using Eq. 13.6. v¼

V 3 m3 ¼ ¼ 1:50 m3 =kg m 2 kg

From the saturated steam tables, the specific volume of saturated vapor (vg) at 200 kPa (0.20 MPa) is 0.8857 m3/kg, which is less than the specific volume of 1.50 m3/kg (v) of the given substance. Hence, steam in the container is in a superheated state.

From the 0.20 MPa superheated tables, when the specific volume is 1.5028 m3/kg, the temperature of steam is 380  C. Since the calculated specific volume of steam in the container is 1.50 m3/kg, the temperature of superheated steam in the container is 380  C.

Solutions to Practice Problems

339

Practice Problem 13.10 (Solution)

From the steam pressure tables, hf ¼ 505 kJ/kg at 200 kPa (0.2 MPa). The given enthalpy at 200 kPa is 455 kJ/kg < hf ¼ 505 kJ/kg. This point (0.2 MPa and 455 kJ/kg) is in the subcooled region as shown in the figure. Therefore, steam is a compressed liquid at the given conditions. From the saturated steam tables, when the enthalpy is 455 kJ/kg, the temperature is 108.5  C as shown (by interpolating the values given in the table). Practice Problem 13.11 (Solution)

340

13

Thermodynamics Fundamentals

Find the 200 kPa (2 bar) constant pressure line and 300  C, and locate their intersection point as shown on the Mollier diagram. This will be reference point “1” for the start of the isentropic expansion. From reference point “1,” move vertically down along the constant entropy line until the enthalpy decreases by 500 kJ/kg. This happens at the final constant pressure line of 19 kPa (0.19 bar). This intersection point is reference point “2,” and the final pressure is 19 kPa.

References 1. Burghardt, D.M., Harbach, J.A.: Engineering Thermodynamics, 4th edn. Tidewater Publications, Franklin (1999) 2. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill, New York (2019) 3. Conner, N.: Thermodynamic Properties, Thermal Engineering, USA (2019). Download from https://www.thermal-engineering.org/what-is-thermodynamic-property-definition/ 4. Dixon, J.R.: Thermodynamics I – An Introduction to Energy (Prentice Hall series in engi neering of the physical sciences). Prentice Hall, Englewood Cliffs (1975) 5. Duffy, A.: Comparing Isothermal and Adiabatic Processes, USA (2019). Download from http:// physics.bu.edu/~duffy/HTML5/PV_diagram_isothermal_adiabatic.html 6. Liley, P.E.: Compressibility Factor, Thermopaedia, USA (2011). Download from https://www. thermopedia.com/content/645/ 7. Ling, S.A.: Thermodynamic Processes, University Physics, Vol. 2, Canada (2016). Download from https://opentextbc.ca/universityphysicsv2openstax/chapter/thermodynamic-processes/

Chapter 14

Conservation of Energy and First Law of Thermodynamics

14.1

First Law of Thermodynamics

The first law of thermodynamics is based on the principle of conservation of energy [3, 4, 6]. The different forms of energy in a system are typically potential energy, kinetic energy, and internal energy. The potential energy of a system is based on its position relative to the datum, that is, the elevation of the system. The kinetic energy of a system is based on the velocity of the system. The internal energy of a system is based on the energy of the molecules within the system, which in turn is dependent on the temperature of the system. The temperature of the system can change due to addition of heat to the system or due to removal of heat from the system. The principle of conservation of energy states that energy is neither created nor destroyed. However, energy can be transformed from one form to another, without any net loss or net gain in the total energy of the system and the surroundings taken together [2, 3, 6].

14.1.1 First Law for a Closed System Consider a closed system that is clearly defined by its boundaries as shown in Fig. 14.1. The mass of the substance within the system remains constant since there is no transfer of mass across the boundaries of a closed system. However, energy can be added to the system, and energy can also be removed from the system. Thus, the system boundaries are closed for mass transfer but open for energy transfer. As shown in Fig. 14.1, energy is added to the system in the form of heat, Q, and energy is removed from the system in the form of work, W. As a result of the energy transfers, the total energy of the system changes from an initial value of Ei to a final value of Ef. Intuitively, the first law of thermodynamics for a closed system can be stated as: due to energy transformations, the final energy of the system will be © The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_14

341

342

14 Conservation of Energy and First Law of Thermodynamics

Fig. 14.1 Energy transformations for a closed thermodynamic system

the initial energy of the system plus the energy added to the system minus the energy removed from the system. In simple terms, the first law of thermodynamics for a closed system can be stated as: change in system energy will be equal to the net energy added to the system [1, 3, 4, 6]. The preceding statement can be mathematically represented by Eq. 14.1. ΔE ¼ E f  Ei ¼ Eadded  Eremoved

ð14:1Þ

Note that the change, delta, is always the final value minus the initial value. For the most general case, all forms of energy are included, and Eq. 14.1 can be written as follows: ΔE ¼ ðU þ PE þ KE Þ f  ðU þ PE þ KE Þi ¼ E added  E removed

ð14:2Þ

Most thermodynamic closed systems are stationary, and consequently there are no changes in potential and kinetic energies. Hence, only changes in internal energy of the system needs to be considered. For a stationary, closed, thermodynamic system as shown in Fig. 14.1, the equation for the first law simplifies to Eq. 14.3. ΔU ¼ U f  U i ¼ Eadded  Eremoved ¼ Q  W

ð14:3Þ

In Eq. 14.3, the heat transfer, Q, is positive if heat is added to the system, and the work, W, is positive if work is produced by the system. If heat is removed from the system, Q, will have negative sign and if work is performed on the system, then W, will have a negative sign. All confusion about the signs for Q and W can be overcome by drawing a schematic diagram as per the problem statement and then applying Eq. 14.3.

14.1

First Law of Thermodynamics

343

Example 14.1 2 lbm of air at 60  F and 15 psia is compressed to one-fifth of its original volume in a closed piston-cylinder arrangement. Determine the final temperature of air when A. The compression is isentropic. B. There is a 10% heat loss based on the isentropic work required. (Solution) A. Calculate the initial absolute temperature of air using Eq. 13.2. T 1 ¼ 460 þ 60 F ¼ 520 R: The ratio of final volume to the initial volume is 1:5. Use the temperature-volume relationship for an isentropic process (Eq. 13.37) to calculate the final temperature as shown. For air, the ratio of specific heats, k ¼ 1.4.
T2 ¼ T1

V2 V1

1k

¼ ð520 RÞ

 11:4 1 ¼ 990 R ð530 CÞ 5

B. Calculate the individual gas constant for air using Eq. 13.16. From Table 13.1, the molecular weight of air is 29 lbm/lbmole.

R¼

ftlbf R 1545 lbmol R ¼ ¼ 53:38 ft‐lbf=lbm  R lbm M 29 lbmol

Calculate the isentropic work required using Eq. 13.47.
 ft  lbf ð2 lbmÞ 53:28 ð990 R  520 RÞ lbm  R mRðT 2  T 1 Þ ¼ W¼ 1k 1  1:4  1 Btu ¼ ð125, 208 ft  lbf Þ 778 ft  lbf ¼ 160:94 Btu The negative sign for the work indicates that the compression work is done on the gas. Calculate the heat loss based on the isentropic compression work. Q ¼ ð0:10Þð160:94 BtuÞ ¼ 16:09 Btu Draw a schematic for the process with appropriate representation of arrows for work and heat transfer.

344

14 Conservation of Energy and First Law of Thermodynamics

Apply the first law, net energy added to the system ¼ change in system energy, to the schematic shown. ΔU ¼ W  Q ¼ 160:94  16:09 ¼ 144:85 Btu The internal energy of the system increases due to the work input to the system. From Table 13.3, for air, cv ¼ 0.17 Btu/lbm ‐  R. Calculate the final temperature of air using Eq. 13.49. ΔU ¼ mcv ΔT ) T 2 ¼ T 1 þ

ΔU 144:85 Btu   ¼ 946 R ¼ 520 R þ Btu mcv ð2 lbmÞ 0:17 lbm R

Due to heat loss, the final temperature is less than the value (990 R) for isentropic compression. Isentropic compression is a reversible-adiabatic (no heat transfer) process. Example 14.2 A closed system containing hot gases loses 20 J of heat to the surroundings during an expansion process. The work done during the expansion process is 30 J. Calculate the change in internal energy. (Solution) Draw a schematic for the process with appropriate representation of arrows for work and heat transfer.

Apply the first law, net energy added to the system ¼ change in system energy, to the schematic shown. 0  Q  W ¼ ΔU ) ΔU ¼ 0 J  20 J  30 J ¼ 50 J Intuitively, it can be stated that the total decrease in internal energy is due to work done by the system and in addition due to heat loss.

14.1

First Law of Thermodynamics

345

Fig. 14.2 Control volume and streams for an open thermodynamic system

14.1.2 I Law for Open Systems-Energy Balance In an open system, the system boundaries are open to both mass and energy flows. Consider the open system control volume (CV) shown in Fig. 14.2. Subscript “i” is used for inlet streams, and subscript “e” is used for exit streams. The species going in and out of the control volume are labeled 1,2,. . .. . . .i,. . .. . .n. The fundamental equation for the I Law applied to an open system at steady state is [2, 3, 5, 7]: Rate of energy into the CV ¼ Rate of energy out of the CV

ð14:4Þ

_ is the rate of energy (power) Q_ is the rate of heat transfer into the CV, and W extracted from the CV. The total energy of each stream consists of internal energy, kinetic energy, and potential energy. Since the streams flow in and out of the control volume of open systems, the flow energies need to be included for each stream. This implies that specific enthalpies (h ¼ u + Pv) should be used for open systems unlike specific internal energy used for closed systems. From Chap. 4 (Sect. 4.2), the kinetic and potential energies per unit mass are v2/2 and gz, respectively. v is the velocity of stream, and z is the elevation of the stream. Apply Eq. 14.4 to the control volume in Fig. 14.2. n X j¼1 n X j¼1

m_ je

m_ ji

v2ji hji þ þ gzji 2

v2je hje þ þ gzje 2

! þ Q_ ¼

n X j¼1

! 

n X j¼1

m_ ji

m_ je

v2je hje þ þ gzje 2

v2ji hji þ þ gzji 2

! _ þW

! _ ¼ Q_  W

ð14:5Þ

Equation 14.5 is known as the steady flow energy equation (SFEE) [2, 3, 5, 7]. Most thermodynamic systems consist of a single stream of steam, water, or air. In addition, in case the potential and kinetic energy changes are very small relative to enthalpy changes. At steady state, the mass flow rate remains constant, that is,

346

14 Conservation of Energy and First Law of Thermodynamics

_ Apply the preceding simplifications to Eq. 14.5 to obtain the followm_ i ¼ m_ e ¼ m. ing equation. _ m_ ðhe  hi Þ ¼ Q_  W

ð14:6Þ

Equation14.6 is called single stream steady flow energy equation (SSSFEE) [3, 5, 7] can be used on most occasions for devices like turbines and compressors. However, as in closed systems, it is always preferable to draw the schematic for the system device and represent all the mass and energy streams accurately and then apply the simple energy balance equation, Rate of energy in ¼ Rate of energy out (Eq. 14.4), to the schematic.

14.2

I Law Applied to Turbines and Compressors

For isentropic turbines and compressors, the changes in potential and kinetic energies are very small compared to the changes in enthalpy [3, 6]. Further, since an isentropic process is a reversible adiabatic process, there is no heat transfer from the turbine or compressor. Therefore, only the enthalpy changes and work need to be considered for turbines and compressors as shown in Fig. 14.3. A turbine produces work, whereas a compressor consumes work. Apply the I Law for open system (energy balance) to the turbine shown in Fig. 14.3. The variables associated with arrows pointing inward should be on the left-hand side of the following equation, and the variables associated with arrows pointing outward should be on the righthand side of the equation. Rate of energy in ¼ Rate of energy out _t)W _ t ¼ m_ ðh1  h2 Þ _ 1 ¼ mh _ 2þW mh The turbine work produced per unit mass of the fluid is:

Fig. 14.3 Energy balance schematic diagrams for turbine and compressor

ð14:7Þ

14.2

I Law Applied to Turbines and Compressors

_t W ¼ wt ¼ h1  h2 m_

347

ð14:8Þ



The preceding results can also be obtained by rearranging and simplifying 0 ðrev:adiabaticÞ _ )W _ ¼ m_ ðhi  he Þ ¼ Eq. 14.6 as shown: m_ ðhe  hi Þ ¼ Q_ W m_ ðh1  h2 Þ A similar approach can be used in obtaining the work consumed by a compressor. Apply the I Law for open system (energy balance) to the compressor shown in Fig. 14.3. Rate of energy in ¼ Rate of energy out _ c ¼ mh _ c ¼ m_ ðh2  h1 Þ _ 1þW _ 2)W mh

ð14:9Þ

The work consumed by the compressor per unit mass of the fluid is: _c W ¼ wc ¼ h2  h1 m_

ð14:10Þ

Compressors handle gases, and in most cases, ideal gas behavior can be assumed. For an ideal gas, the change in enthalpy can be calculated by using Eq. 13.50. Δh ¼ cp ΔT ) h2  h1 ¼ cp ðT 2  T 1 Þ Combine the preceding equation with Eqs. 14.9 and 14.10 to obtain the following equation for compressor power required. _ c ¼ mc _ p ðT 2  T 1 Þ W

ð14:11Þ

Note: The isentropic or ideal compressor power required can also be calculated by using the following equation.

_ c,ideal ¼ W

"
k1 #  k P2 k _ 1Þ ðmRT 1 k1 P1




ð14:11aÞ

From Eq. 14.11, the ideal compressor work required per unit mass is: _c W ¼ w c ¼ cp ð T 2  T 1 Þ m_

ð14:12Þ

348

14 Conservation of Energy and First Law of Thermodynamics

Fig. 14.4 Isentropic and Non-isentropic expansion through a turbine

14.2.1 Isentropic Efficiency of Turbines Turbines are extensively used in power generation systems. Figure 14.4 is a temperature-specific entropy phase diagram (T – s diagram) that illustrates and compares isentropic and non-isentropic expansion through a turbine. Isentropic expansion occurs along the path 1–2 since entropy remains constant along this vertical line. The maximum possible turbine work is obtained in isentropic expansion [2, 3]. However, isentropic (reversible, adiabatic) conditions are difficult to achieve in practice due to heat losses and friction losses. As a result, the actual work obtained from the turbine is always less than the isentropic or ideal work. Non-isentropic expansion takes place along the path 1–20 , which results in the increase in entropy of the fluid. This implies that the turbine is producing less than the available amount of work, which leads to the definition of isentropic efficiency of a turbine as shown in Eq. 14.13. Subscript “t” represents the turbine. ηt ¼

wt,actual h1  h20 ¼ wt,ideal h1  h2

ð14:13Þ

Since it is easy to obtain isentropic work, especially by using the Mollier diagram, isentropic turbine work is obtained first, and then the actual turbine work is obtained by using the isentropic efficiency. Example 14.3 Steam at 1200 psia and 800  F expands to a final pressure of 10 psia. The isentropic efficiency of the turbine is 80%. Determine: A. The actual work produced by the turbine B. The enthalpy of steam as it leaves the turbine

14.2

I Law Applied to Turbines and Compressors

349

(Solution) It is much easier and more efficient to use Mollier diagram for isentropic expansion in a steam turbine. However, the solution method that uses steam tables is also presented here for reference.

Finding the enthalpies from Mollier diagram: Locate the initial state of steam at 800  F, 1200 psia (state point 1). From state point 1, move vertically down along the constant entropy line to 10 psia. This is state point 2 after isentropic expansion along the path 1!2. As shown in the representative diagram, the enthalpies are h1 ¼ 1380 Btu/lbm and h2 ¼ 990 Btu/lbm. A. Determine the actual turbine work using Eq. 14.13. wt,actual ¼ ηt ðh1  h2 Þ

  Btu Btu  990 ¼ ð0:80Þ 1380 lbm lbm ¼ 312 Btu=lbm

350

14 Conservation of Energy and First Law of Thermodynamics

B. Determine the exit enthalpy of steam using Eq. 14.13. h20

¼ h1  η t ð h1  h2 Þ   Btu Btu Btu ¼ 1380  ð0:80Þ 1380  990 lbm lbm lbm ¼ 1068 Btu=lbm

Alternate method: Finding the enthalpies from the tables.

From the superheated steam table, h1 ¼ 1379 Btu/lbm @ 1200 psia, 800  F, and s1 ¼ 1.541 Btu/lbm- R. For isentropic expansion, s2 ¼ s1 ¼ 1.541 Btu/lbm- R. At state point 2, P2 ¼ 10 psia. From the saturated tables, at 10 psia, sg (1.7876) > s2 (¼ 1.541 Btu/lbm- R). Hence at state 2, steam is a liquid-vapor mixture. Determine the quality of steam at state point 2 using Eq. 13.60. x¼


 Btu Btu 1:541 lbm‐ s2  s f  R  0:2836 lbm‐ R ¼ ¼ 0:84 Btu sfg 1:5040 lbm‐ R 10psia

Calculate the enthalpy of steam after isentropic expansion using Eq. 13.59.   h2 ¼ h f þ xhfg 10psia

  Btu Btu þ ð0:84Þ 981:82 ¼ 161:24 lbm lbm ¼ 986 Btu=lbm

The enthalpies obtained from the Mollier diagram and from the steam tables are in close agreement with each other. However, by using Mollier diagram, we can avoid the additional step of determining the quality of the steam leaving the turbine.

14.2

I Law Applied to Turbines and Compressors

351

Fig. 14.5 Isentropic and non-isentropic compression

14.2.2 Isentropic Efficiency of Compressors Compressors are extensively used in gas compression systems and in gas turbines. Figure 14.5 is a temperature-specific entropy diagram (T – s diagram) that illustrates and compares isentropic and non-isentropic compression processes. Isentropic compression occurs along the path 1–2 since entropy remains constant along this vertical line. Isentropic compression requires the least amount of work input to the compressor [2, 3]. However, isentropic (reversible, adiabatic) conditions are difficult to achieve in practice due to heat losses and friction losses. As a result, the actual work used in the compression process is always greater than the isentropic or ideal work used. Non-isentropic compression takes place along the path 1–20 , which results in the increase in entropy of the fluid. This implies that a portion of the input work is used in increasing the temperature and entropy of the gas. The isentropic efficiency of a compressor is defined as shown in Eq. 14.8. Subscript “c” represents the compressor. ηc ¼

_ c,ideal W w h  h1 T  T1 ¼ c,ideal ¼ 2 ¼ 2 _ c,actual wc,actual h20  h1 T 20  T 1 W

ð14:14Þ

Since equations are available to obtain isentropic work, it is obtained first, and then the actual compressor work required is obtained by using the isentropic efficiency. Example 14.4 Air is compressed from 100 to 500 kPa. The inlet air temperature is 20  C, and the isentropic efficiency of the compressor is 80%. The volume flow rate at the compressor intake is 1500 m3/min. Calculate the power required in kW and the temperature of air after the compression process.

352

14 Conservation of Energy and First Law of Thermodynamics

(Solution) Calculate the individual gas constant for air using Eq. 13.16. From Table 13.1, the molecular weight of air is 29 kg/kmol. R¼

kJ R 8:314 kmolK ¼ 0:287 kJ=kg  K ¼ kg M 29 kmol

Calculate the absolute temperature of air at the compressor inlet by using Eq. 13.3. T 1 ¼ 273 þ 20 C ¼ 293 K Calculate the mass flow rate of air by using the ideal gas law (Eq. 13.12).   m3 1 min P1 V_ 1 ð100 kPaÞ 1500 min  60 s   ¼ m_ ¼ ¼ 29:73 kg=s RT 1 0:287 kJ ð293 KÞ kgK

Draw the schematic diagram for the compressor as shown.

Energy balance for the compressor results in the following equation. Rate of energy in ¼ Rate of energy out _ c ¼ mh _ c ¼ m_ ðh2  h1 Þ ¼ mc _ 1þW _ 2 )W _ p ðT 2  T 1 Þ mh From Table 13.2, the specific heat of air at constant pressure is cp ¼ 1.0 kJ/ kg  Kand the ratio of specific heats is k ¼ 1.4. Assuming isentropic compression, calculate the exit temperature of air using Eq. 13.38.
1:41
k1 P2 k 500 kPa 1:4 T2 ¼ T1 ¼ ð293 KÞ ¼ 464 K ð191 CÞ 100 kPa P1

14.3

I Law Applied to Heating and Cooling of Fluids

353

Calculate the ideal compressor power required by substituting the known values into the energy balance equation. _ c,ideal ¼ mc _ p ðT 2  T 1 Þ W




kg s ¼ 5084 kW

¼

29:73


1:0

 kJ ð464 K  293 KÞ kg  K

Calculate the actual power required by using Eq. 14.14. _ _ c,actual ¼ W c,ideal ¼ 5084 kW ¼ 6355 kW W 0:80 ηc Calculate the actual exit temperature of air using Eq. 14.14.
T 20 ¼

14.3

T2  T1 ηc




 þ T1

 464 K  293 K ¼ þ 293 K 0:8  ¼ 507 K ð234 CÞ

I Law Applied to Heating and Cooling of Fluids

The I Law and the energy balance equation have a very intuitive form in cases involving heating and cooling of fluids. Heat gained by the cold fluid ¼ heat lost by the hot fluid

ð14:15Þ

The transfer of heat from the hot fluid to the cold fluid takes place in a heat exchanger. The topic of heat exchangers has been extensively covered in Chap. 12 in the heat transfer section. Apply the energy balance equation (Eq. 14.15) to the heat exchanger shown in Fig. 14.6. Heat gained by the cold fluid ¼ heat lost by the hot fluid

Fig. 14.6 Heat exchanger

354

14 Conservation of Energy and First Law of Thermodynamics

m_ c ðhce  hci Þ ¼ m_ h ðhhi  hhe Þ

ð14:16Þ

Equation 14.16 is the basic heat balance equation [2, 3, 5] used in calculations related to heat exchangers. Since Δh ¼ cpΔT, Eq. 14.16 can be modified as shown. However, it is preferable to use steam tables to obtain the enthalpies for water and steam. m_ c cpc ðT ce  T ci Þ ¼ m_ h cph ðT hi  T he Þ

ð14:17Þ

Example 14.5 It is proposed to heat 2000 kg/hr of a process liquid from 10 to 70  C by using wastewater available at 90  C. The wastewater needs to be cooled to 40  C before discharging into a water body. The specific heat of the process liquid is 3.48 kJ/kg.K, and the specific heat of wastewater is 4.19 kJ/kg.K. Calculate the mass flow rate of the wastewater that can be cooled. (Solution) The process liquid is the cold fluid (subscript “c”), and the wastewater is the hot fluid (subscript “h”). Solve Eq. 14.17 for the mass flow rate of hot wastewater, m_ h , and substitute the known values.


m_ h ¼

14.4

m_ c cpc ðT ce  T ci Þ cph ðT hi  T he Þ


 kg kJ 3:48 2000 ð70 C  10 CÞ hr kg  K
 ¼ kJ ð90 C  40 CÞ 4:19 kg  K ¼ 1993 kg=hr

I Law Applied to Nozzles and Diffusers

Nozzles and diffusers have differing cross-section areas of flow resulting in velocity and kinetic energy changes [3, 5, 7]. Usually, the nozzles are well insulted to ensure minimal heat transfer, and they can be considered as adiabatic devices. There is no work output from nozzles nor is there any work input to nozzles. Hence, only enthalpy and kinetic energy terms need to be considered for nozzles. The schematic for a nozzle is shown in Fig. 14.7. Apply the I Law for open system (energy balance) to the flow nozzle shown in Fig. 14.7. Rate of energy in ¼ Rate of energy out

14.4

I Law Applied to Nozzles and Diffusers

355

Fig. 14.7 I Law schematic for a flow nozzle

_ 1þ mh

_ 12 _ 2 mv mv _ 2þ 2 ) ¼ mh 2 2

h2  h1 ¼ Δh ¼

v1 2  v2 2 2

ð14:12Þ

Example 14.6 Air enters an isentropic nozzle at 200  F and 200 psia. The diameter and velocity at the nozzle entrance are 20 in. and 50 ft/sec, respectively. The area ratio of the nozzle is 2:1. The velocity at the nozzle exit is 500 ft/sec. Determine the following: A. The mass flow rate of air through the nozzle B. The exit temperature C. The exit pressure (Solution) Draw the schematic diagram for the nozzle as shown.

A. Convert the inlet temperature to absolute value using Eq. 13.2. T 1 ¼ 460 þ 200 F ¼ 660 R Calculate the individual gas constant for air using Eq. 13.16. From Table 13.1, the molecular weight of air is 29 lbm/lbmol. psia‐ft3

R 10:73 lbmol R R¼ ¼ ¼ 0:37 psia  ft3 =lbm  R lbm M 29 lbmol Calculate the flow cross-section area at the nozzle entrance.

356

14 Conservation of Energy and First Law of Thermodynamics

A1 ¼

     π π 20 2 D1 2 ¼ ft ¼ 2:182 ft2 4 4 12

Calculate the density of air at the nozzle entrance using the ideal gas law (Eq. 13.12). ρ1 ¼

m P 200 psia  ¼ 0:8190 lbm=ft3 ¼ 1 ¼ psiaft3 V 1 RT 1  0:37 lbm R ð660 RÞ

Calculate the mass flow rate of air at the nozzle entrance using the continuity equation (Eq. 3.5). m_ ¼ ρ1 A1 v1 ¼



  lbm  ft ¼ 89:35 lbm= sec 0:8190 3 2:182 ft2 50 sec ft

B. At the given conditions of moderate temperatures and pressure, air can be considered as an ideal gas. For an ideal gas, h2  h1 ¼ Δh ¼ cp ΔT ¼ cp ðT 2  T 1 Þ Combine the preceding result with Eq. 14.12 to obtain a modified energy balance equation for a nozzle. cp ð T 2  T 1 Þ ¼

v1 2  v2 2 2

Reconciliation of units in the preceding equation: The left-hand side of the equation is enthalpy change (Btu/lbm). The right-hand side of the equation is velocity squared with units ft2/sec2, which can be converted to Btu/lbm by using the conversion constants gc (32.2 lbmft/lbfsec2) and j (778 ftlbf/Btu) as shown here. cp ð T 2  T 1 Þ ¼

ft2 = sec 2 v1 2  v2 2  Btu=lbm  lbmft 778 ftlbf 2gc j Btu lbf sec 2 

Hence, the final form of the energy balance equation (I Law) for a nozzle in USCS units is: cp ð T 2  T 1 Þ ¼

v1 2  v2 2 2gc j

From Table 13.3, the specific heat of air at constant pressure is:

14.5

Pumps

357

cp ¼ 0:24 Btu=lbm  R Solve the energy balance equation in USCS units for the exit temperature, and substitute all the known values. cp ð T 2  T 1 Þ ¼ T2

¼ T1 þ

v1 2  v2 2 ) 2gc j

v1 2  v2 2 2gc jcp

    ft 2 ft 2 50 sec  500 sec 

 lbm  ft 778 ft  lbf Btu ð2Þ 32:2 0:24 Btu lbm  R lbf  sec 2 ¼ 639 R ð179 FÞ

¼ 660 R þ

Calculate the exit pressure using the mass flow rate and the ideal gas law. Calculate the density at the nozzle exit using the continuity equation (Eq. 3.5). The mass flow rate is the same at the entrance and exit due to the principle of conservation of mass. ρ2 ¼

89:35 lbm m_ m_ sec    ¼ 0:1638 lbm=ft3 ¼ ¼ A2 v2 0:5A1 v2 ð0:5Þ 2:182 ft2 500 ft sec

Use the ideal gas law (Eq. 13.12) to calculate the exit pressure.
P2 ¼ ρ2 RT 2 ¼

14.5

0:1638

lbm ft3

 
psia  ft3 ð639 RÞ ¼ 38:73 psia 0:37 lbm  R

Pumps

Pumps increase the pressure of constant density/constant-specific volume incompressible liquids. The energy input to the pump is primarily used in increasing the pressure energy of the liquid [3, 5, 7]. From Sect. 4.2 in fluid mechanics, the pressure energy per unit mass is Pρ ¼ vP, where v is the specific volume of the liquid. The I Law schematic for a pump is shown in Fig. 14.8. Apply the I Law for open systems to the pump schematic shown in Fig. 14.8 assuming no heat losses (well insulated, adiabatic pump) Energy In ¼ Energy Out

358

14 Conservation of Energy and First Law of Thermodynamics

Fig. 14.8 I Law schematic for a pump

νP1 þ wpump ¼ νP2 The pump work is also equivalent to the enthalpy change of the liquid. Δh ¼ wpump ¼ νðP2  P1 Þ

ð14:13Þ

The power input required for the pump can be obtained by multiplying the ideal pump work per unit mass by the mass flow rate and then dividing the result by the efficiency of the pump. _ pump _ pump ¼ mw W ηpump

ð14:14Þ

Example 14.7 The feedwater requirement for a boiler is 8000 kg/hr at a pressure of 4.5 MPa. The inlet pressure of water to the feedwater pump is 90 kPa. The average density of water is 890 kg/m3. Determine the power input required to the pump if the efficiency of the pump is 75%. (Solution) Calculate the specific volume of the feedwater by taking the reciprocal of the density. υ¼

1 1 ¼ ¼ 0:0011 m3 =kg ρ 890 kg3 m

Substitute the known values into Eq. 14.13, and calculate pump work required per unit mass of feedwater. wpump ¼ νðP2  P1 Þ



m3 1000 kPa  90 kPa 0:0011 4:5 MPa  MPa kg ¼ 4:851 kJ=kg

¼

Note: (m3)(kPa)  (m3)(kN/m2)  kN.m  kJ Calculate the power input required for the pump using Eq. 14.14.

Practice Problems

_ pump _ pump ¼ mw W ¼ ηpump

359



1 hr 8000 kg hr  3600 s

  kJ 4:851 kg

0:75

¼ 14:37 kW

Note: kJ/s  kW

Practice Problems Practice Problem 14.1 Carbon dioxide in a piston-cylinder arrangement is heated from an initial temperature of 300 F to a final temperature of 500 F at a constant pressure of 45 psia. The initial volume of the gas is 10 ft3. Calculate: A. The work done B. The change in internal energy C. The heat added Practice Problem 14.2 2 lbm of methane is isentropically compressed to twice its initial pressure. The initial temperature is 150 F. Calculate: A. The compression work required B. The change in internal energy Practice Problem 14.3 10000 kg/hr of superheated steam at 1 MPa and 700  C expands in a turbine to a final pressure of 50 kPa. The turbine produces 2.5 MW power. Determine the isentropic efficiency of the turbine. Practice Problem 14.4 A compressor station in a natural gas pipeline compresses the gas from 570 to 650 psia. The mass flow rate of the gas, which is essentially methane, is 3000 lbm/ min. The power used by the compressor is 558 kW, and the inlet temperature of methane is 70  F. For methane, the ratio of specific heats, k ¼ 1.3 and the individual gas constant, R ¼ 96.56 ft  lbf/lbm- R. Determine the isentropic efficiency of the compressor. Practice Problem 14.5 Hot water at 180  F needs to be produced for heating purposes. 2000 lbm/hr of ambient water at 70  F will be mixed with steam at 300  F to produce the hot water required. All streams are at 14.7 psia. Calculate the mass flow rates of steam and the product hot water. Practice Problem 14.6 The velocity of natural gas is reduced to half its original value by using an adiabatic diffuser. The exit cross-section area is 50% greater than the cross-section area at the

360

14 Conservation of Energy and First Law of Thermodynamics

entrance. The parameters at the diffuser entrance are velocity ¼ 200 m/s, temperature = 20  C, and pressure ¼ 120 kPa. The properties of natural gas are approximately the same as the properties of methane. Determine the temperature and pressure at the diffuser exit. Practice Problem 14.7 A boiler feed pump takes in condensed water at 10 psia and delivers the water at 500 psia. In addition, the elevation difference between the pump suction line and the discharge line to the boiler is 25 ft and friction head loss is equivalent to 8 ft of water. Assume the specific volume of water to be constant at 0.0182 ft3/lbm. Calculate the horsepower required if the mass flow rate of water is 50 lbm/sec and the efficiency of the pump is 70%.

Solutions to Practice Problems Practice Problem 14.1 (Solution) Convert the temperatures to absolute values using Eq. 13.2. T 1 ¼ 460 þ 300 F ¼ 760 R

T 2 ¼ 460 þ 500 F ¼ 960 R

Calculate the final volume using Eq. 13.34 (Charles law) for a constant pressure process.

   960 R T2 3 V2 ¼ V1 ¼ 12:63 ft3 ¼ 10 ft T1 760 R A. Calculate the work done by using Eq. 13.41 for a constant pressure process.

W ¼ P ðV 2  V 1 Þ


  lbf 144 in2  45 2  12:63 ft3  10 ft3 2 in ft ¼ 17, 042 ft  lbf

¼

B. Calculate the individual gas constant for carbon dioxide using Eq. 13.16. From Table 13.1, the molecular weight of carbon dioxide is 44 lbm/lbmol. psiaft3

R¼

R 10:73 lbmol R ¼ ¼ 0:2439 psia  ft3 =lbm  R lbm M 44 lbmol

Calculate the mass of carbon dioxide by using the ideal gas law (Eq. 13.12) at state I.

Solutions to Practice Problems

361

  ð45 psiaÞ 10 ft3 P1 V 1  ¼ 2:43 lbm m¼ ¼ psiaft3 RT 1  0:2439 lbm  R ð760 RÞ From Table 13.2, the constant volume heat capacity for carbon dioxide is cv ¼ 0.16 Btu/lbm ‐  R. Calculate the change in internal energy using Eq. 13.49. ΔU ¼ mcv ΔT

  Btu ¼ ð2:43 lbmÞ 0:16 ð960 R  760 RÞ lbm  R ¼ 77:76 Btu

There is an increase in the internal energy of carbon dioxide due to increase in temperature. C. Draw a schematic for the process with appropriate representation of arrows for work and heat transfer.

Apply the first law, net energy added to the system ¼ change in system energy, to the schematic shown. Q  W ¼ ΔU ) Q ¼ ΔU þ W

1 Btu ¼ 77:76 Btu þ 17042 ft  lbf  778 ft  lbf ¼ 99:66 Btu

Part of the heat added is used in performing work, and the rest is used in increasing the temperature (internal energy) of carbon dioxide. Practice Problem 14.2 (Solution) A. Convert the initial temperature to its absolute value using Eq. 13.2. T 1 ¼ 460 þ 150 F ¼ 610 R Use the temperature-pressure relationship for an isentropic process (Eq. 13.38) to calculate the final temperature as shown. From Table 13.2, the ratio of specific heats for methane is k ¼ 1.3.

362

14 Conservation of Energy and First Law of Thermodynamics

T2 ¼ T1


k1 1:31 P2 k ¼ ð610 RÞð2Þ 1:3 ¼ 716 R P1

Calculate the individual gas constant for methane using Eq. 13.16. From Table 13.1, the molecular weight of methane is 16 lbm/lbmole. R¼

ftlbf R 1545 lbmol R ¼ ¼ 96:56 ft  lbf=lbm  R lbm M 16 lbmol

Calculate the work required for isentropic compression using Eq. 13.47.
 ft  lbf ð2 lbmÞ 96:56 ð716 R  610 RÞ lbm  R mRðT 2  T 1 Þ ¼ W¼ 1k 1  1:3  1 Btu ¼ ð68236 ft  lbf Þ 778 ft  lbf ¼ 87:71 Btu The negative sign for work indicates that the work is being done on the system (methane gas). B. Draw a schematic for the process with appropriate representation of arrows for work and heat transfer as shown. Isentropic process is reversible and adiabatic (no heat transfer).

Apply the first law, net energy added to the system ¼ change in system energy, to the schematic shown. 87:71 Btu  0 Btu ¼ ΔU ¼ 87:71 Btu Since there is no heat transfer, all the energy added in the form of work manifests an increase in internal energy of the gas. Practice Problem 14.3 (Solution) Draw a schematic diagram for the turbine as shown.

Solutions to Practice Problems

363

Energy balance around the turbine results in the following equations. Rate of energy in ¼ Rate of energy out _ _ t ) h1  h20 ¼ W t _ 1 ¼ mh _ 20 þ W mh m_ Determine the state points 1 and 2 for isentropic expansion on the excerpt from the Mollier diagram as shown in the figure. P1 ¼ 1 MPa ¼ 10 bar (Note: 1 bar ¼ 105Pa ¼ 100 kPa) and P2 ¼ 50 kPa ¼ 0.5 bar.

From the Mollier diagram, h1 ¼ 3920 kJ/kg and h2 ¼ 2920 kJ/kg for isentropic expansion. Substitute the known values into the energy balance equation.

364

14 Conservation of Energy and First Law of Thermodynamics

  1 kJ  _t 2:5  106 W 1000 W J h1  h20 ¼ ¼ 900 kJ=kg ¼ 1 hr m_  10000 kg 3600 s hr Note: W  J/s Calculate the isentropic efficiency of the turbine by substituting all the known values into Eq. 14.13. kJ

ηt ¼

900 kg h1  h20 ¼ ¼ 0:90 ð90%Þ kJ kJ h1  h2 3920 kg  2920 kg

Practice Problem 14.4 (Solution) Calculate the ideal or isentropic power required using Eq. 14.11a. _ c,ideal W




  k1 k P2 k _ ¼ ðmRT Þ P1 1 k1
1 
   1:3 lbm ft  lbf  ¼ 3000 96:56  R ð530 RÞ 1:3  1 min lbm   1:31 650 psia 1:3  1 570 psia ¼ 2:05  107 ft  lbf= min

Convert the ideal compressor power required to kW. _ c,ideal ¼ W




0:000023 kW 2:05  10 ft  lbf= min  ft  lbf= min 7

 ¼ 472 kW

Calculate the isentropic efficiency of the compressor using Eq. 14.14 and the given actual power used by the compressor (558 kW). ηc ¼

_ c,ideal W 472 kW ¼ 0:85 ð85%Þ ¼ _ c,actual 558 kW W

Practice Problem 14.5 (Solution) Draw a schematic diagram to represent the mixing process.

Solutions to Practice Problems

365

Mass balance for the mixer results in the following equation. Mass flow rate in ¼ Mass flow rate out m_ c þ m_ s ¼ m_ h

) 2000

lbm þ m_ s ¼ m_ h hr

Enthalpy (energy) balance for the mixer results in the following equation. Rate of energy in ¼ Rate of energy out m_ c hc þ m_ s hs ¼ m_ h hh Determine the enthalpies from the steam tables as shown. Note that the enthalpy of a compressed liquid is the same as the enthalpy of the saturated liquid at the given temperature.

hc ¼ h hh ¼ h

f ,70 F

f ,180 F

¼ 38:08 Btu=lbm ¼ 148:04 Btu=lbm

hs ð14:7 psia, 300 FÞ ¼ 1192:7 Btu=lbm Substitute the preceding results and the result from the mass balance equation into the energy balance equation.

366

14 Conservation of Energy and First Law of Thermodynamics


    lbm Btu Btu 38:08 þ m_ s 1192:7 2000 hr lbm lbm
  lbm Btu ¼ 2000 þ m_ s 148:04 hr lbm Solve the preceding equation for m_ s to get the mass flow rate of steam. Substitute the result into the mass balance equation to obtain the mass flow rate of hot water. m_ s ¼ 211 lbm=hr m_ h ¼ m_ c þ m_ s ¼ 2000

lbm lbm þ 211 ¼ 2211 lbm=hr hr hr

Practice Problem 14.6 (Solution) Draw the schematic for the diffuser as shown.

Convert the inlet temperature to absolute value using Eq. 13.3. T 1 ¼ 273 þ 20 C ¼ 293 K At the given conditions of moderate temperatures and pressure, natural gas can be considered as an ideal gas. For an ideal gas: h2  h1 ¼ Δh ¼ cp ΔT ¼ cp ðT 2  T 1 Þ Combine the preceding result with Eq. 14.12 to obtain a modified energy balance equation for a diffuser. cp ð T 2  T 1 Þ ¼

v1 2  v2 2 2

Reconciliation of units in the preceding equation: the left-hand side of the equation is specific enthalpy change (kJ/kg). The right-hand side of the equation is velocity squared with units m2/s2, which can be converted to kJ/kg by using the following technique.

Solutions to Practice Problems

cp ð T 2  T 1 Þ ¼

367

v1 2  v2 2 m2 kg kg  m m N  m     2  kg kg kg 2 s2 s J 1 kJ kJ    kg 1000 J kg

Alternately, the following conversion factor can be used: 1000 m2/s2 ¼ 1 kJ/kg Hence, the final form of the energy balance equation for a nozzle/diffuser in SI units is: v 2  v2 2  cp ðT 2  T 1 Þ ¼  1 m2 =s2 2 1000 1 kJ=kg Solve the preceding equation for the exit temperature, T2, and substitute the known values. From Table 13.2, the specific heat at constant pressure, cp, for methane is cP ¼ 2.25 kJ/kg  K. v 2  v2 2
1  þ T1 T2 ¼ 1000 m2 =s2 2cp 1 kJ=kg



2  2 200 ms  100 ms 
 þ 293 K ¼
1000 m2 =s2 kJ 2 2:25 kg  K 1 kJ=kg ¼ 300 K ð27 CÞ

The mass flow rate is the same at the entrance and exit due to the principle of conservation of mass. The mass flow rate can be calculated by using the continuity equation (Eq. 3.5). m_ ¼ ρ1 A1 v1 ¼ ρ2 A2 v2 The density of natural gas at the diffuser entrance and at the diffuser exit can be calculated using the ideal gas law (Eq. 13.12). ρ1 ¼

P1 P and ρ2 ¼ 2 RT 1 RT 2

Combine the preceding equations to get the following expression.



P1 A1 v1 P2 A2 v2 ¼ RT 1 RT 2 

m_ ¼

Solve the preceding equation for the exit pressure and substitute all the known values. 0 1


    200 m T2 A1 v1 300 K 1 @ sA P2 ¼ ðP1 Þ ¼ ð120 kPaÞ 293 K 1:5 100 m T1 A2 v2 s ¼ 164 kPa

368

14 Conservation of Energy and First Law of Thermodynamics

Practice Problem 14.7 (Solution) Draw the schematic for the pumping system as shown.

The energy input to the pump is used in increasing the pressure of water and also in overcoming the elevation difference and the friction head loss. From fluid mechanics (Chap. 4, Sect. 4.2), the potential energy per unit mass is gz and friction energy loss per unit mass is hf g. Therefore, the pump work required per unit mass of water is: wpump

¼
νðP2  P1 Þ þgz
þ hf g  ft3 lbf 144 in2 lbf 144 in2 ¼ 0:0182  10  500 2  lbm in
in2 ft2  ft2
 ft ft þ 32:2 ð25 ftÞ þ 32:2 ð8 ftÞ sec 2 sec 2 ¼ 2347 ft  lbf=lbm

Note: In fluid mechanics (Chap. 4, Sect. 4.2), it has been shown that ft2/sec2  ft– lbf/lbm. Calculate pump power required using Eq. 14.14.  lbm  50 sec 2347 ft‐lbf _ pump lbm  _ pump ¼ mw  W ¼ ¼ 305 hp sec ηpump ð0:70Þ 550 ft‐lbf= hp

References

369

References 1. Bahrami, M.: The First Law of Thermodynamics – Closed Systems, SFU Notes, Download from, https://www.sfu.ca/~mbahrami/ENSC%20388/Notes/First%20Law%20of%20Thermodynam ics_Closed%20Systems.pdf 2. Burghardt, D.M., Harbach, J.A.: Engineering Thermodynamics, 4th edn. Tidewater Publications, Franklin (1999) 3. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill, New York (2019) 4. Hall, N.: First Law of Thermodynamics, NASA Glenn Research Center. Download from https:// www.grc.nasa.gov/www/k-12/airplane/thermo1.html (2021) 5. Heany, G.: The First Law of Thermodynamics – Open System (Control Volume), Slide Serve. Download from, https://www.slideserve.com/garret/the-first-law-of-thermodynamics (2012) 6. Lucas, J.: What is the First Law of Thermodynamics?, Live Science. Download from https:// www.livescience.com/50881-first-law-thermodynamics.html (2015) 7. Mathew, S.: First Law of Thermodynamics for Open Systems, Lesics. Download from https:// www.lesics.com/first-law-of-thermodynamics-for-an-open-system.html (2018)

Chapter 15

Ideal Gas Mixtures and Psychrometrics

15.1

Ideal Gas Mixtures

An ideal gas mixture will have several different gases as components of the mixture. Each component in the mixture obeys the ideal gas law, and the mixture as a whole will also obey the ideal gas law [2, 3, 6]. The components are represented by subscript i and i can take on values ranging from 1 to n, where n is the number of components in the mixture. The mixture properties are represented by variables without subscripts, whereas the variables representing component properties will have subscripts representing the component. As an example, the variable P represents the pressure of the mixture as a whole, and P2 represents the pressure exerted by component “2” in the mixture. P2 is also known as partial pressure of component “2.” The basic variables used in ideal gas mixtures are illustrated in Fig. 15.1.

15.1.1 Key Definitions for Ideal Gas Mixtures Mass Fraction (xi) The mass fraction of component i is the mass of the component i divided by the total mass of the mixture. xi ¼

mi m

ð15:1Þ

Mole Fraction (yi) The mole fraction of component i is the moles of the component i divided by the total moles of the mixture.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_15

371

372

15 Ideal Gas Mixtures and Psychrometrics

Fig. 15.1 Mixture of ideal gases

yi ¼

Ni N

ð15:2Þ

Partial Pressure of Component i (Pi) The partial pressure of component i is defined as the pressure exerted by that component when it occupies the mixture volume, V, at the mixture temperature, T. From the ideal gas law (Eq. 13.12): Pi ¼

mi Ri T V

ð15:3Þ

Partial Volume of Component i (Vi) The partial volume of component i is defined as the volume occupied by that component at the mixture pressure, P, and at the mixture temperature, T. From the ideal gas law (Eq. 13.12): Vi ¼

mi Ri T P

ð15:4Þ

Since temperature is an intensive property, it is uniform throughout the mixture, and therefore the temperature of each component in the ideal gas mixture is the same as the temperature of the mixture itself.
, both pressure and volume are directly From the ideal gas law, PV ¼ N RT proportional to the moles of the gas. Hence, for an ideal gas mixture, the mole fraction the pressure fraction, volume fraction, and mole fraction of each component are identical. For any component, i, in an ideal gas mixture, mole fraction = pressure fraction = volume fraction [3, 6].

yi ¼

N i Pi V i ¼ ¼ N P V

ð15:5Þ

15.1

Ideal Gas Mixtures

373

15.1.2 Laws Related to Ideal Gas Mixtures 15.1.2.1

Dalton’s Law

Dalton’s law [2, 5] states that the sum of partial pressures of all the components in an ideal gas mixture will be equal to the total pressure of the mixture. n X

Pi ¼ P

ð15:6Þ

i¼1

15.1.2.2

Amagat’s Law

Amagat’s law [2, 5] states that the sum of partial volumes of all the components in an ideal gas mixture will be equal to the total volume of the mixture. n X

Vi ¼ V

ð15:7Þ

i¼1

Example 15.1 A tank consists of 64 kg of oxygen, 44 kg of carbon dioxide, and 84 kg of nitrogen. The temperature of the mixture is 30  C, and the pressure of the mixture is 300 kPa. Determine: A. B. C. D. E.

The mass fractions and mole fractions of each component The partial pressure of each component The molecular weight of the mixture The gas constant of the mixture The volume of the tank

(Solution) Draw the schematic diagram for this gas mixture as shown.

From Table 13.1, the molecular weights are O2 ¼ 32 kg/kmol, CO2 ¼ 44 kg/ kmol, and N2 ¼ 28 kg/kmol.

374

15 Ideal Gas Mixtures and Psychrometrics

It is best to generate and use the following table to solve this problem (Parts A and B).

Mass, Component mi, (kg) O2 64 CO2 44 N2 84 Sum m ¼ 192

Mol. Wt., Mi (kg/kmol) 32 44 28

Moles, (kmol), Ni ¼ mi /Mi 2 1 3 N¼6

Mole fraction, yi ¼ Ni / N 0.33 0.17 0.50 1.0

Mass fraction, xi ¼ m i / m 0.33 0.23 0.44 1.0

Partial Pr. (kPa), Pi ¼ yiP 99 69 132 300

C. Calculate the molecular weight of the mixture using Eq. 13.13. M¼

m 192 kg ¼ ¼ 32 kg=kmol N 6 kmol

Note: The average molecular weight of an ideal gas mixture is also the sum of the individual molecular weights weighted by the mole fractions of the components. n P Thus, M ¼ M i yi i¼1

D. Calculate the gas constant of the mixture using Eq. 13.16

R¼

kJ R 8:314 kmolK ¼ ¼ 0:26 kJ=kg  K kg M 32 kmol

E. The absolute temperature of the gas mixture is T ¼ 273  + 30  C ¼ 303 K. Calculate the volume of the tank by applying the ideal gas equation (Eq. 13.12) to the mixture.

V¼

mRT ¼ P

  kJ ð192 kgÞ 0:26 kgK ð303 KÞ 300 kPa

¼ 50:42 m3

Note: kJ/kPa  kN.m/(kN/m2)  m3.

15.2

Air-Water Vapor Mixture and Psychrometrics

Atmospheric air is an ideal gas mixture of dry air (subscript “a”) and water vapor (subscript “w”). This mixture is also referred to as moist air or humid air. The atmospheric pressure, P, is the total pressure due to the two components, dry air and

15.2

Air-Water Vapor Mixture and Psychrometrics

375

water vapor. From Dalton’s law, the total pressure of the gas mixture is the sum of the partial pressures of the components. Therefore: P ¼ Pa þ Pw

ð15:8Þ

In Eq. 15.8, P is the atmospheric pressure, Pa is the partial pressure of dry air, and Pw is the partial pressure of water vapor [1]. It is important to study and understand the properties of moist air [1, 3, 5] so that suitable methods can be used to manipulate the properties as desired. For example, atmospheric air can be hot and humid during summer months at certain locations. For human comfort, it is necessary to cool and remove condensed moisture from this air. This process is known as dehumidification and cooling, which is commonly known as air-conditioning.

15.2.1 Moist Air Properties and Definitions Dry Bulb Temperature (TDB) The dry bulb temperature is the ordinary temperature of air measured by a regular thermometer. Wet Bulb Temperature (TWB) The wet bulb temperature is the temperature measured by a wet bulb thermometer. As the name implies, in a wet bulb thermometer, the thermometer bulb is surrounded by a moist wick. Depending on the condition of the surrounding air, moisture from the wick will evaporate into air, using the internal energy of the thermometer itself. Dew Point Temperature (TDP) Some of the water vapor in air will condense when the air-water vapor mixture is cooled. The temperature at which condensation begins is called dew point temperature. Humidity Ratio (ω) Humidity ratio quantifies the moisture content of air on a mass basis. It is the ratio of the mass of water vapor to the mass of dry air. The mathematical equation for humidity ratio is: ω¼

lbm water vapor kg water vapor ¼ lbm dry air kg dry air

ð15:9Þ

Since specific humidity is dimensionless, it will have the same numerical value in all systems of units.

Relative Humidity (ϕ or RH) Relative humidity is a measure of how close the air is to the saturated state. Air can hold a limited amount of moisture at any given temperature. On nights when there is a rapid drop in temperature, the air tends to

376

15 Ideal Gas Mixtures and Psychrometrics

become supersaturated since the amount of moisture air can hold will decrease with decrease in temperature. Supersaturated air has excess moisture which condenses on automobile surfaces. The condensate can be noticed after a cool night. Fully saturated air has 100% relative humidity. Relative humidity is the ratio of the partial pressure of water vapor in the given mixture to the partial pressure of water vapor if the mixture were to be saturated at the mixture temperature. The saturation partial pressure can be obtained from steam tables, and it is the saturation pressure at the mixture temperature. The formula for relative humidity is ϕ¼

Pw Psat at T DB of mixture

ð15:10Þ

Example 15.2 Atmospheric air at 14.7 psia is at a temperature of 80  F and has a relative humidity of 70%. What is the partial pressure of dry air in the mixture? (Solution) From steam tables, at 80  F, the saturation pressure is Psat ¼ 0.51 psia. Calculate the partial pressure of water vapor in moist air by using the definition of relative humidity (Eq. 15.10). Pw ¼ ϕPsat ¼ 0:7  0:51 psia ¼ 0:357 psia Since atmospheric pressure is the sum of partial pressures of water vapor and dry air (Eq. 15.8), calculate the partial pressure of dry air as shown. Pa ¼ P  Pw ¼ 14:7 psia  0:36 psia ¼ 14:34 psia

15.2.2 Relationship Between Humidity Ratio and Relative Humidity Humidity ratio (ω) is a measure of moisture content in air, and relative humidity (ϕ) is an indication of how close the moisture content is to the saturation level. Hence, it follows that there should be a relationship between humidity ratio and relative humidity [1, 5], which is shown here. m N M ω¼ w¼ w w¼ ma NaMa

!     kg 18:0153 kmol Nw Nw ¼ 0:622 kg Na Na 28:9647 kmol

15.2

Air-Water Vapor Mixture and Psychrometrics

377

In an ideal gas mixture, the pressure ratio between components is the same as the mole ratio between components. Also, the partial pressure of dry air is Pa ¼ P  Pw. Therefore: ω ¼ 0:622

      Nw P Pw ¼ 0:622 w ¼ 0:622 Na Pa P  Pw

ð15:11Þ

From the definition of relative humidity (Eq. 15.10), Pw ¼ ϕPsat. Substitute this result into the preceding equation to obtain the relationship between humidity ratio and relative humidity. 

Pw ω ¼ 0:622 P  Pw





ϕPsat ¼ 0:622 P  ϕPsat

 ð15:12Þ

Example 15.3 At a certain location, the atmospheric pressure is 90 kPa, and air is at 30  C with 55% relative humidity. Calculate the humidity ratio of moist air at this location. (Solution) From steam tables, at 30  C, the saturation pressure is: Psat ¼ 0:0042 MPa ¼ 4:2 kPa Calculate the humidity ratio using Eq. 15.12.    ϕPsat 0:55  4:2 kPa ω ¼ 0:622 ¼ 0:622 90 kPa  0:55  4:2 kPa P  ϕPsat ¼ 0:0164 kg H2 O=kg dry air 

15.2.3 Use of Psychrometric Chart to Obtain Properties of Moist Air Psychrometrics is the study of moist air. The properties of moist air can be determined from the psychrometric chart [3, 4]. Figure 15.2 illustrates the general layout and the parameters used in the psychrometric chart. The state point of moist air can be located on the chart by using two parameters, for example, dry bulb temperature and relative humidity. Once the state point is located, all the properties of moist air can be determined [1, 3, 7] as shown in Fig. 15.2. The psychrometric chart for USCS units is shown in Fig. 15.3 and the psychrometric chart for SI units is shown in Fig. 15.4.

378

15 Ideal Gas Mixtures and Psychrometrics

Fig. 15.2 Determining properties of moist air from the psychrometric chart

Fig. 15.3 Psychrometric chart (USCS Units, Source: Engineering ToolBox, (2004). Air – Psychrometric Chart for Standard Atmospheric Conditions – Imperial Units. (Reproduced with permission. Available at: https://www.engineeringtoolbox.com/psychrometric-chart-d_816.html)

15.2

Air-Water Vapor Mixture and Psychrometrics

379

Fig. 15.4 Psychrometric Chart (SI Units)

The following example illustrates the use of psychrometric chart to find the properties of moist air. Example 15.4 Air is at 70  F and 60% relative humidity. Determine: A. B. C. D. E.

The humidity ratio The wet bulb temperature The dew point temperature The enthalpy per pound mass dry air The specific volume

(Solution) Locate the state point of moist air at the intersection of 70  F dry bulb vertical line and 60% relative humidity curve as shown in the figure. Using the state point as a reference, determine all the properties of moist air as shown.

380

A. B. C. D. E.

15 Ideal Gas Mixtures and Psychrometrics

ω ¼ 0.0097 lbm H2O/lbm dry air TWB ¼ 61  F TDP ¼ 55.5  F h ¼ 27.4 Btu/lbm dry air v ¼ 13.6 ft3/lbm dry air

15.3

Air-Conditioning Processes

An air-conditioning process is any process that alters the state of moist air [1, 5]. Air-conditioning goes beyond the most commonly perceived air-conditioning process of cooling and dehumidification (moisture removal) of hot, humid air. The different air-conditioning processes are illustrated in Fig. 15.5. This is followed by illustrative problems whose solutions involve representations on the psychrometric chart. The mass flow rate of dry air in any air-conditioning process remains constant. This is a key concept used in the solution of problems on air-conditioning. This is also the reasoning behind using unit mass of dry air as the basis for moist air properties. For example, the humidity ratio is lbm moisture/lbm dry air or kg moisture/kg dry air.

15.3

Air-Conditioning Processes

381

Fig. 15.5 Air-conditioning processes

15.3.1 Cooling and Dehumidification The most commonly perceived air-conditioning process is cooling and dehumidification of hot, moist air [1, 3, 5]. In this process, shown as process 1–2 in Fig. 15.6, the temperature of air decreases along with the moisture content. Example 15.5 8000 cfm of air at 80  F dry bulb and 80% relative humidity is conditioned to 70  F dry bulb and 60% relative humidity. Determine: A. The rate of condensate removal in gpm B. The tons of cooling achieved (1 ton of cooling ¼ 12,000 Btu/hr) (Solution) Obtain the properties of moist air at state point 1 (80  F DB and 80% RH) and at state point 2 (70  F DB and 60% RH) from the psychrometric chart as shown.

382

15 Ideal Gas Mixtures and Psychrometrics

Draw a schematic diagram for the air-conditioner as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 8000 min V_ m_ a ¼ ¼ ¼ 571:43 lbm d a= min ν1 14:0 ft3 lbm d a 3

The mass flow rate of dry air is not affected by any air-conditioning process, and hence it is constant. Mass balance for water results in the following equations.

15.3

Air-Conditioning Processes

383

Mass flow of water in ¼ mass flow of water out m_ a ω1 ¼ m_ a ω2 þ m_ w,condensed Solve the preceding equation for the rate of condensation of water, and substitute the known values. m_ w,condensed ¼ m_ a ðω1  ω2 Þ    lbm d a lbm water lbm water ¼ 571:43 0:018  0:0096 min lbm d a lbm d a ¼ 4:80 lbm water= min Convert the mass rate of condensation to gpm using the density of water, ρw ¼ 8.34 lbm/gal. Water condensed in gpm ¼

water m_ w,condensed 4:80 lbmmin ¼ ¼ 0:5755 gpm ρw 8:34 lbm gal

B. Calculate the rate of cooling achieved by performing an energy balance for the conditioner. Rate of energy in ¼ Rate of energy out m_ a ha1 ¼ m_ a ha2 þ m_ w hw þ Q_ out Solve the preceding equation for the rate of heat removal and substitute the known values. From steam tables, the enthalpy of the condensate is, hw ¼ hf at 70  F ¼ 38.08 Btu/lbm Q_ out ¼ m_ a ðha1  ha2 Þ  m_ w hw    lbm d a Btu Btu 39  27:4 ¼ 571:43 min lbm d a lbm d a    lbm water Btu  4:941 38:08 min lbm water ¼ 6446 Btu= min Convert the rate of cooling to tons. !   Btu 60 min 1 ton Q_ out ðtonsÞ ¼ 6446  ¼ 32:23 tons min hr 12000 Btu hr

384

15 Ideal Gas Mixtures and Psychrometrics

15.3.2 Heating and Humidification At some locations, the state of moist air could be cold and dry (low humidity) during the winter season. Cold, dry air can damage human skin and sensitive electronic devices. Cold, dry air needs to be conditioned by adding moisture and heat. This process, shown as process 7–8 in Fig. 15.5, is the heating and humidification process [1, 5]. In this process, both the moisture content and temperature will increase, and this is typically accomplished by spraying steam into the cold, dry air. Example 15.6 300 m3/min of air at 5  C and 40% relative humidity is to be humidified and heated a using steam spray. The desired conditions of air are 20  C and 60% relative humidity. Steam is supplied at 100 kPa and 150  C. Determine: (A) The rate at which steam must be supplied (B) The kW of additional heat input required, if any (Solution) Locate the state points on the psychrometric chart, and obtain the corresponding state properties as shown.

Draw the schematic diagram for the humidifier-heater as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1.

15.3

Air-Conditioning Processes

385

m 300 min V_ ¼ ¼ 380 kg= min ν1 0:79 m3 kg d a 3

m_ a ¼

Mass balance for water results in the following equations. Mass flow of water in ¼ mass flow of water out m_ a ω1 þ m_ s ¼ m_ a ω2 Solve the preceding equation for the mass flow rate of steam, and substitute the known values. m_ s ¼ m_ a ðω2  ω1 Þ    kg kg water kg water ¼ 380 0:009  0:0025 min kg d a kg d a ¼ 2:47 kg steam= min Rate of energy in ¼ Rate of energy out m_ a ha1 þ m_ s hs þ Q_ in ¼ m_ a ha2 Solve the preceding equation for the rate of heat input, and substitute the known values. From steam tables, the enthalpy of steam (100 kPa, 150  C, superheated steam) is hs(0.10 MPa, 150  C) ¼ 2776.6 kJ/kg Q_ in ¼ m_ a ðha2  ha1 Þ  m_ s hs    kg d a kJ kJ ¼ 380 43  12 min kg d a kg d a    kg steam kJ  2:47 2776:6 min kg steam ¼ 4923 kJ= min Convert the required heat input to kW. Q_ in ¼

kJ 4923 min 60 s min

¼ 82:05 kW

386

15 Ideal Gas Mixtures and Psychrometrics

15.3.3 Sensible Heating Sensible heating process, shown as process 3–4 in Fig. 15.5, increases the temperature of moist air without any changes to the moisture content [1, 5]. This process is used to heat up cold air. Example 15.7 3000 cfm of cold air at 50  F dry bulb and 40  F wet bulb is heated to 70  F dry bulb at constant moisture content. Determine: A. The change in relative humidity B. The moisture content of air C. The kW rating of the heating coil required (Solution) Obtain the properties of moist air at state point 1 (50  F DB and 40  F WB) and at state point 2 (70  F DB and ω2 ¼ ω1) from the psychrometric chart as shown.

A. From the preceding figure, RH1 ¼ 40% and RH2 ¼ 20%. ΔRH ¼ RH2  RH1 ¼ 20 %  40 % ¼  20%. There is a 20% decrease in relative humidity. B. From the figure, ω2 ¼ ω1 ¼ 0.003 lbm H2O/lbm d a Draw a schematic diagram for the air-conditioner as shown.

15.3

Air-Conditioning Processes

387

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. ft 3000 min V_ ¼ ¼ 232:56 lbm d a= min ν1 12:9 ft3 lbm d a 3

m_ a ¼

The mass flow rate of dry air is not affected by any air-conditioning process, and hence it is constant. C. Calculate the rate of heating required by performing an energy balance for the heater. Rate of energy in ¼ Rate of energy out m_ a ha1 þ Q_ in ¼ m_ a ha2 Solve the preceding equation for the rate of additional heat input, and substitute the known values. 

lbm d a 232:56 min ¼ 1116 Btu= min

Q_ in ¼ m_ a ðha2  ha1 Þ ¼

  Btu Btu 20:2  15:4 lbm d a lbm d a

Convert the heat input rate to kW. Q_ in ¼

Btu 1116 min Btu 57 min kW

¼ 19:58 kW

15.3.4 Other Air-Conditioning Processes Other, less common, air-conditioning processes shown in Fig. 15.6 are: 1–2: Humidifying only ω2 > ω1, T2 ¼ T1 3–4: Dehumidifying only ω4 < ω3, T4 ¼ T3 5–6: Evaporative cooling ω6 > ω5, T6 < T5 7–8: Chemical dehumidifying ω8 < ω7, T8 > T7

388

15 Ideal Gas Mixtures and Psychrometrics

Fig. 15.6 Other air-conditioning processes

15.4

Cooling Towers

In power plants and chemical processing plants, hot water from condensers and heat exchangers needs to be cooled before recirculating back to the exchangers in a closed loop. Typically, this is done in a cooling tower. A cooling tower operates adiabatically [2, 3] since it uses the heat from hot water to humidify the air flowing through the tower in a direction opposite to the hot water spray. Due to the supply of enthalpy of vaporization from the hot water to humidify the air, the hot water cools down. Since some hot water evaporates into the air stream, makeup water must be supplied to maintain constant circulation through the condensers and heat exchangers. The amount of air required to cool the hot water and the quantity of makeup water required can be determined by using mass and energy balances for the cooling tower as shown here. Figure 15.7 is a schematic of a cooling tower with all the relevant variables. Mass balance for water around the envelope (red, dashed lines) shown in Fig. 15.7 results in the following equations. Mass flow of water in ¼ mass flow of water out m_ w1 þ m_ a ω2 ¼ m_ w2 þ m_ a ω1 m_ w1  m_ w2 ¼ m_ a ðω1  ω2 Þ

ð15:13Þ

The left-hand side of the preceding equation also represents the quantity of makeup water required. m_ m ¼ m_ w1  m_ w2 ¼ m_ a ðω1  ω2 Þ Energy balance around the envelope results in the following equations.

ð15:14Þ

15.4

Cooling Towers

389

Fig. 15.7 Schematic for a cooling tower

Rate of energy in ¼ Rate of energy out m_ w1 hw1 þ m_ a ha2 ¼ m_ w2 hw2 þ m_ a ha1

ð15:15Þ

Equations 15.14 and 15.15 can be used simultaneously to solve problems on cooling towers as shown in Example 15.8. Example 15.8 8000 kg/hr of water at 40  C is to be cooled to a temperature of 25  C by using a cooling tower. Air enters the cooling tower at 20  C dry bulb and 50% relative humidity and leaves the tower at 32  C dry bulb and 70% relative humidity. Determine: A. The quantity of air required in terms of standard cubic meters per minute (SCMM). B. The amount of makeup water required in liters per minute. (Solution) A. Use the same nomenclature as in Fig. 15.7. Locate the state points (state point 1 is hot, moist air coming out of the top of the tower and state point 2 is cooler, drier air entering the bottom of the tower) of moist air on the psychrometric chart, and obtain the corresponding state properties as shown.

390

15 Ideal Gas Mixtures and Psychrometrics

Determine the enthalpies of water from the steam tables: hw1 ¼ hf at 40 C ¼ 167:53 kJ=kg water and hw2 ¼ hf at 25 C ¼ 104:83 kJ=kg water Express m_ w2 in terms of m_ w1 using Eq. 15.13, substitute the result into Eq. 15.15, and then solve for m_ a , the mass flow rate of dry air. Subsequently, substitute the known values to obtain the mass flow rate of dry air. m_ w1 ðhw1  hw2 Þ ha1  ha2  hw2 ðω1  ω2 Þ    kg water kJ kJ 167:53  104:83 8000 hr kg water kg water 0 1 ¼ kg water 0:022 C kg d a kJ kJ kJ B B C 88  39  104:83 @ kg d a kg d a kg water kg water A 0:0075 kg d a ¼ 10564 kg d a=hr

m_ a ¼

Calculate the volume flow rate of air at inlet conditions using the specific volume at inlet.

15.5

Mixing of Air Streams

V_ air,in ¼ m_ a ν2 ¼

391

   kg d a m3 0:84 ¼ 8874 m3 =hr 10564 hr kg d a

The air inlet temperature is 20  C (293 K), whereas the reference temperature for standard cubic meter is 15  C (288 K). Volume is directly proportional to absolute temperature. Calculate the standard cubic meter per minute (SCMM) of air required using the preceding concepts. V_ air req: ¼

   m3 1 hr 288 K ¼ 145:37 SCMM 8874  hr 60 min 293 K

B. Calculate the quantity of makeup water required by using Eq. 15.14 and by substituting the known values. m_ m ¼ m_ a ðω1  ω2 Þ    kg da kg water kg water ¼ 10564 0:022  0:0075 hr kg d a kg d a ¼ 153 kg water=hr Calculate the amount of makeup water required in liters per minute. Use the standard density of water, 1 kg/L. m_ m ¼

15.5

! kg water 1 hr 1   153 ¼ 2:55 L= min hr 60 min 1 kg L

Mixing of Air Streams

In buildings, it is a common practice to use a mix of return air and outside air for purposes of air quality, ventilation, and energy conservation. After mixing, the air is conditioned as required. For purposes of conditioning, it is essential to know the properties of mixed air. This can be determined by using mass and energy balances [1, 5] or by using a graphical method using the psychrometric chart. These methods are illustrated in the following example. Example 15.9 A facility needs 10,000 cfm of conditioned air. Prior to conditioning, outside air and return air are mixed in the proportion of 30% outside air and 70% return air by volume. The return air is at 70  F dry bulb and 60% relative humidity and outside air

392

15 Ideal Gas Mixtures and Psychrometrics

is at 85  F dry bulb and 80% relative humidity. Determine all the properties of the mixed air. (Solution) Locate the state points of return air (70  F DB and 60% rh – state point 1) and outside air (85  F DB and 80% rh – state point 2) on the psychrometric chart, and determine the properties of moist air as shown.

Draw a schematic for the mixing process as shown in the figure.

The volume flow rate of return air is V_ 1 ¼ 0:7  10000 ¼ 7000 cfm. The volume flow rate of outside air is V_ 2 ¼ 0:3  10000 ¼ 3000 cfm. Calculate the mass flow rates of return air and outside air (dry basis) using the specific volumes.

15.5

Mixing of Air Streams

393

ft 7000 min V_ 1 ¼ ¼ 514:71 lbm d a= min ν1 13:6 ft3 lbm d a 3

m_ a1 ¼

ft 3000 min V_ m_ a2 ¼ 2 ¼ ¼ 211:27 lbm d a= min ν2 14:2 ft3 lbm d a 3

Add up the mass flow rates of dry air in streams 1 and 2 to obtain the mass flow rate of dry air in the mixed stream (stream 3). lbm d a lbm d a þ 211:27 min min ¼ 725:98 lbm d a= min

m_ a3 ¼ m_ a1 þ m_ a2 ¼ 514:71

Mass balance for water at the junction, J, results in the following equations. Mass flow rate of water into the junction ¼ Mass flow rate of water out of the junction m_ a1 ω1 þ m_ a2 ω2 ¼ m_ a3 ω3 Solve the preceding equation for ω3, and substitute the known values to obtain ω3. m_ a1 ω1 þ m_ a2 ω2 m_ a3    lbm d a lbm water 514:71 0:0094 min lbm d a    lbm d a lbm water þ 211:27 0:0211 min lbm d a ¼ lbm d a 725:98 min ¼ 0:0128 lbm water=lbm d a

ω3 ¼

Energy balance at the junction, J, results in the following equations. Energy flow into the junction ¼ Energy flow out of the junction m_ a1 h1 þ m_ a2 h2 ¼ m_ a3 h3 Solve the preceding equation for h3, and substitute the known values to obtain h3.

394

15 Ideal Gas Mixtures and Psychrometrics

m_ a1 h1 þ m_ a2 h2 m_ a3    lbm d a Btu 27:4 514:71 min lbm d a    lbm d a Btu þ 211:27 43:4 min lbm d a ¼ lbm d a 725:98 min ¼ 32:06 Btu=lbm d a

h3 ¼

The intersection of ω3 ¼ 0.0128 and h3 ¼ 32.06 determines state point 3, which can be located on the psychrometric chart. At state point 3, the properties of moist air are ω3 ¼ 0.0128 lbm water/lbm d a, h3 ¼ 32.06Btu/lbm d a, TDB3 ¼ 74  F, v3 ¼ 13.53 ft3/lbm d a, rh3 ¼ 70%, TWB3 ¼ 67  F, and TDP3 ¼ 63.5  F. Alternate Method of Solution Using the lever rule, calculate the approximate value of the dry bulb temperature at state point 3 as shown. V_  T DB1 þ V_ 2  T DB2 T DB3 ¼ 1 V_  3   3 ft ft3 7000 ð70 FÞ þ 3000 ð85 FÞ min min ¼ ft3 10000 min ¼ 74:5 F Join state point 1 and state point 2 with a straight line and locate state point 3 on this line corresponding to the vertical dry bulb temperature line of 74.5  F. Determine all the properties of the mixed air at state point 3 as shown before. The results from the graphical method are consistent with the results obtained from the analytical method.

15.6

Use of Psychrometric Formulas

Commonly used psychrometric charts are valid only at sea level where the barometric pressure is 760 mm Hg (1 atmosphere). However, the barometric pressure decreases with increase in altitude. Hence, sea-level, 1 atmosphere barometric pressure psychrometric charts are not accurate at higher altitude locations. In such locations, the psychrometric formulas [1, 5] will have to be used. The following formulas were presented in the earlier sections of this chapter and are given here for

15.6

Use of Psychrometric Formulas

395

reference. Equation 15.10, shown here, can be used to calculate the relative humidity of moist air. ϕ¼

Pw Psat at T DB of mixture

Equations 15.11 and 15.12 can be used in calculating the humidity ratio.     Pw Pw ¼ 0:622 ω ¼ 0:622 Pa P  Pw   ϕPsat ω ¼ 0:622 P  ϕPsat Other useful psychrometric formulas include: Formula for calculating the specific volume (ft3/lbm dry air) v¼

Ra T Ra T ¼ 144ðP  Pw Þ 144ðP  ϕPsat Þ

ð15:16Þ

Formula to calculate the enthalpy of moist air (Btu/lbm d a). h ¼ 0:240T þ ωð1061 þ 0:444T Þ

ð15:17Þ

Example 15.10 At a given location, the atmospheric pressure is 13 psia, and the air temperature is 50  F. The relative humidity is 40%. For the moist air at this location, determine: A. The humidity ratio B. The enthalpy C. the specific volume. (Solution) A. Determine the saturation pressure at 50  F from the steam tables. Psat ¼ 0:18 psia Calculate the humidity ratio substituting all the known values into Eq. 15.12.    ϕPsat 0:40  0:18 psia ¼ 0:622 13 psia  0:40  0:18 psia P  ϕPsat ¼ 0:0035 lbm water=lbm d a

 ω ¼ 0:622

396

15 Ideal Gas Mixtures and Psychrometrics

B. Calculate the enthalpy of moist air using Eq. 15.17. h ¼ 0:240T þ ωð1061 þ 0:444T Þ   Btu ¼ 0:240 ð50 FÞ lbm d a‐ F 0 1   1061 Btu lbm water lbm water B C þ 0:0035 @ A lbm d a Btu  þ0:444 F  50 lbm steam‐ F ¼ 15:79 Btu=lbm d a C. Calculate the specific volume of moist air using Eq. 15.16. The absolute temperature of air is T ¼ 460  + 50  F ¼ 510  R 0

1 ft‐lbf 1545 B lbmol‐ RC @ Að510 RÞ lbm 29 Ra T   lbmol  v¼ ¼ 144ðP  ϕPsat Þ in2 lbf lbf 144 2 13 2  0:40  0:18 2 in in ft 3 ¼ 14:59 ft =lbm d a

Practice Problems Practice Problem 15.1 2 lbm of carbon dioxide escapes into a lab space with dimensions of 10 ft  20 ft  15 ft. The air in the lab space is at 14 psia and 70  F. Determine: A. B. C. D.

The mass of air in the lab space before contamination The mole fraction of carbon dioxide in the lab space The parts per million (ppm) of carbon dioxide in lab air on a volumetric basis The final pressure in the chamber assuming the temperature to be constant

Practice Problem 15.2 In atmospheric air at 100 kPa and 30  C, the partial pressure of water vapor is 1.3 kPa. Calculate: A. The partial pressure of dry air B. The relative humidity of atmospheric air

Solutions to Practice Problems

397

Practice Problem 15.3 An air-water vapor mixture at 14 psia and 65  F has 1% water vapor by mass. Calculate the relative humidity of this mixture. Practice Problem 15.4 Air is at 30  C, dry bulb, and 20  C wet bulb. Determine: A. B. C. D. E.

The humidity ratio The relative humidity The dew point temperature The enthalpy per kilogram mass of dry air The specific volume

Practice Problem 15.5 A cold surface at 45  F is in contact with air at 65  F dry bulb and 70% relative humidity. Will there be a condensate film on the cold surface? Justify your answer. Practice Problem 15.6 Outdoor air flowing at 150 m3/min has 50% relative humidity and is at 5  C dry bulb temperature. Determine the rate at which moisture needs to be added to outdoor air to make it compatible with air inside a facility where the conditions are 20  C and 70% relative humidity.

Solutions to Practice Problems Practice Problem 15.1 (Solution) A. Calculate the volume of the lab space, V ¼ 10 ft  20 ft  15 ft ¼ 3000 ft3 Calculate the individual gas constant for air (Mair ¼ 29 lbm/lbmol) using Eq. 13.16. psiaft3

Rair

10:73 lbmol R R ¼ ¼ ¼ 0:37 psia  ft3 =lbm  R lbm M air 29 lbmol

Calculate the absolute temperature of air by using Eq. 13.2. T ¼ 460 þ 70 F ¼ 530 R ¼ constant Calculate the mass of air in the lab space prior to contamination using the ideal gas law (Eq. 13.12). mair

  ð14 psiaÞ 3000 ft3 PV   ¼ 214:18 lbm ¼ ¼ psiaft3 Rair T  0:37 lbm  R ð530 RÞ

398

15 Ideal Gas Mixtures and Psychrometrics

B. Calculate the moles of carbon dioxide by using Eq. 13.13. From Table 13.1, the molecular weight of carbon dioxide is 44 lbm/lbmol. N CO2 ¼

mCO2 2 lbm ¼ ¼ 0:0454 lbmol lbm M CO2 44 lbmol

Calculate the mols of air by using Eq. 13.13. From Table 13.1, the molecular weight of air is 29 lbm/lbmol. N air ¼

mair 214:18 lbm ¼ ¼ 7:38 kmol lbm M air 29 lbmol

Calculate the mole fraction of carbon dioxide in the lab space using Eq. 15.2 yCO2 ¼

N CO2 N CO2 0:0454 lbmol ¼ ¼ ¼ 0:0061 N N CO2 þ N air 0:0454 lbmol þ 7:38 lbmol

C. Parts per million on a volumetric basis is one volume of carbon dioxide per million volumes of air, carbon dioxide mixture. For a mixture of ideal gases, volume fraction ¼ mole fraction (Eq. 15.5). From the results for part B, PPMCO2 ¼

0:00611 lbmol CO2 0:00611 vol CO2 106 ¼  6 1 lbmol CO2 , air mixture 1 vol CO2 , air mixture 10 ¼ 6110 ppm

Practice Problem 15.2 (Solution) A. Since atmospheric pressure is the sum of partial pressures of water vapor and dry air (Eq. 15.8), calculate the partial pressure of dry air as shown. Pa ¼ P  Pw ¼ 100 kPa  1:3 kPa ¼ 98:7 kPa B. From steam tables, at Psat ¼ 0.0042 MPa ¼ 4.2 kPa

30



C,

the

saturation

pressure

Calculate the relative humidity of atmospheric air using Eq. 15.10. ϕ¼

Pw 1:3 kPa ¼ 0:31 ð31%Þ ¼ Psat at T DB of mixture 4:2 kPa

is

Solutions to Practice Problems

399

Practice Problem 15.3 (Solution) Since the mass fraction of water vapor is 0.01, the mixture will have 0.01 lbm water/ lbm air-water vapor mixture. Calculate the humidity ratio using Eq. 15.9. ω¼

0:01 lbm H2 O 0:01 lbm H2 O ¼ ¼ 0:0111 lbm H2 O=lbm dry air lbm mixture 0:90 lbm dry air

Solve Eq. 15.11 for the partial pressure of water vapor, Pw, and substitute the known values.  Pw ¼

 lbm H2 O 0:0111 lbm dry air ð14 psiaÞ

ωP ¼ 0:622 þ ω 0:622

lbm H2 O lbm dry air

lbm H2 O þ 0:0111 lbm dry air

¼ 0:246 psia

From steam tables, at 65  F, the saturation pressure is Psat ¼ 0.315 psia. Calculate the relative humidity of the mixture using Eq. 15.10. ϕ¼

Pw 0:246 psia ¼ 0:78 ð78%Þ ¼ Psat at T DB of mixture 0:315 psia

Practice Problem 15.4 (Solution) Locate the state point of moist air at the intersection of 30  C dry bulb vertical line and 20  C wet bulb inclined line as shown in the figure. Using the state point as a reference, determine all the properties of moist air as shown.

400

A. B. C. D. E.

15 Ideal Gas Mixtures and Psychrometrics

ω ¼ 0.011 kg water/kg dry air RH ¼ 40% TDP ¼ 15  C h ¼ 58 kJ/kg dry air v ¼ 0.87 m3/kg dry air

Practice Problem 15.5 (Solution) The water vapor in the air will condense on the cold surface if the dew point of the moist air is below 45  F, which is the temperature of cold surface. Using the psychrometric chart, determine the dew point of the moist air as shown in the figure. First, locate the state point at the intersection of 65  F dry bulb vertical line and the 70% relative humidity curve. Move horizontally left from the state point to the saturation curve to determine the dew point.

From the figure, the dew point is 56  F. Since the dew point is greater than the surface temperature of 45  F, there will be no condensation.

Solutions to Practice Problems

401

Practice Problem 15.6 (Solution) Locate the state points of the outdoor air (5  C, 50% rh) and the inside air (20  C, 70% rh) as shown on the psychrometric chart. Determine the moisture content at the state points as shown.

Calculate the mass flow rate of dry air by using the specific volume of moist air at state point 1. m 150 min V_ ¼ 190 kg= min m_ a ¼ ¼ ν1 0:79 m3 kg d a 3

Mass balance for water results in the following equations. Mass flow of water in ¼ mass flow of water out m_ a ω1 þ m_ moisture ¼ m_ a ω2 Solve the preceding equation for the mass flow rate of moisture to be added, and substitute the known values. m_ moisture ¼ m_ a ðω2  ω1 Þ    kg kg water kg water ¼ 190 0:01  0:003 min kg d a kg d a ¼ 1:33 kg moisture= min

402

15 Ideal Gas Mixtures and Psychrometrics

References 1. Bahrami, M.: Gas Vapor Mixtures and HVAC, SFU Notes. Download from https://www.sfu.ca/ ~mbahrami/ENSC%20461/Notes/Gas%20Vapor%20Mixture_HVAC.pdf (2011) 2. Burghardt, D.M., Harbach, J.A.: Engineering Thermodynamics, 4th edn. Tidewater Publications, New York (1999) 3. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill, New York (2019) 4. Fly Carpet, Inc.: Free Online Psychrometric Chart. Download from http://www.flycarpet.net/en/ psyonline (2021) 5. Glicksman, L.R.: Air Water Vapor Mixtures – Psychrometrics. Open Course Ware, MIT. Download from https://ocw.mit.edu/courses/architecture/4-42j-fundamentals-of-energy-in-build ings-fall-2010/readings/MIT4_42JF10_water_vapor.pdf (2010) 6. Hachim, H.M.: Gas Mixtures Study Guide, Slide Share. Download from https://www.slideshare. net/HaiderAlumairy/gas-mixtures-61922378 (2016) 7. Rotronic Technical Note: How to Read a Psychrometric Chart, Rotronic Measurement Solutions, USA. Download from www.rotronic.com/media/productattachments/files/h/o/how_to_read_ psychchart_f_web.pdf (2014)

Chapter 16

Fuels and Combustion

16.1

Introduction

Combustion is burning of fuel in the presence of oxygen to generate heat [1, 2]. The heat generated can be used for generating steam, for powering vehicles and for many other similar purposes. Combustion processes produce emissions that can be harmful to the environment. The combustion process is basically a chemical reaction between the fuel and oxygen to produce combustion products and heat. Combustion of fuels is studied with two major objectives in mind: determining the amount of fuel and associated air required in any particular situation and finding the quantities of combustion products to assess environmental impact.

16.2

Fuels

Fuels used for combustion can be classified as solid, liquid, and gaseous fuels based on the state of the fuel. Coal is the most widely used solid fuel but also the most polluting. Gasoline (petrol) is the most widely used liquid fuel, and natural gas is commonly used in gas furnaces to generate heat, especially for home heating.

16.2.1 Heating Value of Fuels The heating value of a fuel is the amount of heat generated per unit mass or per unit mole of fuel [2, 7]. The typical units of heating value are Btu/lbm (or kJ/kg) and Btu/lbmol (or kJ/kmol).

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_16

403

404

16.2.1.1

16 Fuels and Combustion

Higher Heating Value of Fuels

The higher heating value (HHV) is the enthalpy of combustion per unit quantity of the fuel. Higher heating value is also known as standard heat of combustion. The definition of standard heat of combustion is it is the heat generated due to complete combustion of unit quantity of fuel, where the products of combustion are at a standard state of 78
F (25
C) and one atmosphere pressure. At these conditions, water will be in the liquid state. The higher heating value is also known as the gross heating value (GHV).

16.2.1.2

Lower Heating Value of Fuels

The lower heating value is the actual heating value of the fuel. In contrast to the higher heating value, the lower heating value does not assume the standard state of 78
F, 1 atm. or 25
C, 1 atm for combustion products. Water is always a product of combustion, and it will vaporize due to the heat generated during the combustion process. The enthalpy of vaporization is thus coming from the combustion of the fuel itself, and the enthalpy of vaporization of water needs to be subtracted from the total heat generated to get the actual heat of combustion. The lower heating value is also known as net heating value (NHV), or actual heating value (AHV), and it can be calculated using Eq. 16.1. LHV ¼ HHV  mwater hfg

ð16:1Þ

In Eq. 16.1, hfg is the enthalpy of vaporization of water at 25
C (78
F). Table 16.1 provides approximate net heating values (NHV) of common fuels [7].

16.3

Combustion Fundamentals and Definitions

Combustion is a chemical reaction between the fuel and oxygen. Oxygen oxidizes the combustible components of a fuel. During combustion of a hydrocarbon fuel, carbon is oxidized to form carbon dioxide, and hydrogen is oxidized to form water Table 16.1 Approximate net heating values of common fuels

Fuel Coal Diesel Gasoline Heating oil Bio-diesel Natural gas Methane

Heating value (USCS) 12,400 Btu/lbm 129,000 Btu/gal 114,000 Btu/gal 128,500 Btu/gal 126,000 Btu/gal 970 Btu/ft3 950 Btu/ft3

Heating value (S I) 28000 kJ/kg 36000 kJ/L 32000 kJ/L 35500 kJ/L 33000 kJ/L 36500 kJ/m3 35500 kJ/m3

16.3

Combustion Fundamentals and Definitions

405

[1, 2]. Water will be in the vapor phase due to the heat of the combustion reaction. Thus, the general format of combustion reaction of a hydrocarbon fuel will be: Hydrocarbon Fuel þ Oxygen ! Carbon dioxide þ water vapor Complete combustion occurs when all the carbon in the fuel is oxidized to carbon dioxide and all the hydrogen is oxidized to water. In partial combustion, some of the carbon is partially oxidized to carbon monoxide.

16.3.1 Stoichiometry of Combustion Reactions Stoichiometry is the process of generating and writing a balanced chemical reaction. In a balanced chemical reaction, the number of atoms of each species must be equal on both sides of the equation. The stoichiometric reactions for the combustion of carbon and hydrogen are [2, 8]: C þ O2 ! CO2 ðgÞ

ð16:1Þ

1 H2 þ O2 ! H2 O ) 2H2 þ O2 ! 2H2 OðgÞ 2

ð16:2Þ

In a stoichiometric reaction, the numbers preceding each element or molecule are known as stoichiometric coefficients, which are usually whole numbers. The stoichiometric coefficients of species represent the moles of the species. This is because chemical reactions occur between molecules of different species. Referring to the combustion reaction of hydrogen (Eq. 16.2), it can be stated that “two moles of hydrogen react with one mole of oxygen to form one mole of water.” Also, in Eq. 16.2, there are a total of 3 moles of reactants and 2 moles of product. The stoichiometric reaction shown in Eq. 16.2 can be written in terms of mass as shown. To obtain the mass of a species, multiply the moles of the species by the molecular weight of the species (from Eq. 13.13). 2H2 þ O2 ! 2H2 OðgÞ Mass ðlbm or kgÞ : 2  2 þ 1  32 ! 2  18 Thus, the total mass on the reactant side (4 lbm + 32 lbm ¼ 36 lbm) is equal to the total mass on the product side (36 lbm). An important conclusion is: In stoichiometric chemical reactions, mass is always conserved, while moles are not necessarily conserved between the reactant and product sides. The total mass of reactants will be equal to the total mass of products, whereas the total moles of reactants will not necessarily be equal to the total moles of products.

406

16 Fuels and Combustion

16.3.2 Combustion in Air Combustion reactions require oxygen. The source of oxygen is air. Only the oxygen in air reacts with the fuel components. The other species in air, which is mostly nitrogen, are inert and do not react with fuel components. The composition of air is 79 mole% nitrogen and 21 mole% oxygen. Based on this composition, the following mole ratios can be written [1, 2]. Moles N2 79 ¼ 3:76 ) Moles N2 ¼ 3:76  Moles O2 ¼ Moles O2 21

ð16:3Þ

Based on 1 mole of oxygen used, the moles of nitrogen present will be 3.76 moles. Therefore, the mole ratio between air and oxygen will be: Moles air 1 mole O2 þ 3:76 mole N2 ¼ ¼ 4:76 Moles O2 1 mole O2 ) Moles air ¼ 4:76  Moles O2

ð16:4Þ

However, since nitrogen is inert, it does not participate in the combustion reaction. The moles and mass of nitrogen will be the same on either side of the combustion equation. Example 16.1 Derive the theoretical combustion equation for stoichiometric combustion of methane (CH4) in air. (Solution) Methane is a hydrocarbon fuel, and complete combustion of methane will result in the formation of carbon dioxide and water vapor as products. In addition, the moles of nitrogen associated with oxygen in air must be included both on the reactant and product sides of the combustion equation. Using the preceding concepts, write an unbalanced equation with all the reactants and products as shown. Use algebraic symbols for unknown stoichiometric coefficients. CH4 þ aO2 þ bN2 ! cCO2 þ dH2 O þ b N2 Note that the stoichiometric coefficient of nitrogen must be the same on both sides of the equation since it is inert. Start balancing the equation by considering the elements in this order – first carbon, then hydrogen, and then oxygen. The coefficient of nitrogen will be 3.76 times the coefficient of oxygen. It is better to balance the elements rather than molecules. For example, for oxygen balance, balance O rather than O2.

16.3

Combustion Fundamentals and Definitions

407

C balance: c ¼ 1 H balance: 2d ¼ 4 ) d ¼ 2 O balance: 2a ¼ 2c + d ¼ 2(1) + 2 ) 2a ¼ 4 ) a ¼ 2 b ¼ 3.76  a ¼ 3.76  2 ¼ 7.52 Hence, the balanced equation for the stoichiometric combustion of methane is CH4 þ 2O2 þ 7:52N2 ! CO2 þ 2H2 O þ 7:52N2 The preceding equation is the theoretical combustion equation for methane.

16.3.3 Theoretical Air, Air-Fuel Ratio, and Excess Air 16.3.3.1

Theoretical Air

The amount of air required (mass or moles) as per the theoretical combustion equation is called theoretical air [2, 3, 6]. Example 16.2 Determine the amount (lbmol and lbm) of theoretical air required for combustion of 1 lbmol methane. (Solution) From the solution to Example 16.1, the theoretical combustion equation for 1 mol of methane is: CH4 þ 2O2 þ 7:52N2 ! CO2 þ 2H2 O þ 7:52N2 From the preceding equation, Theoretical moles of air required ¼ moles of O2 þ moles of N2 ¼ 2 lbmol O2 þ 7:52 lbmol N2 ¼ 9:52 lbmol of air: The theoretical mass of air required can be obtained by using the molecular weight of air (Mair ¼ 29 lbm/lbmol) as shown.   lbm ¼ 276:08 lbm mair ¼ N air M air ¼ ð9:52 lbmolÞ 29 lbmol

408

16.3.3.2

16 Fuels and Combustion

Air-Fuel Ratio

The air-fuel ratio can be represented either on a molar basis or on a mass basis as follows with associated symbols [2, 9]. RA=F




RA=F

molar


mass

moles air required mole fuel mass air required ¼ unit mass fuel

¼

ð16:6Þ ð16:7Þ

Example 16.3 Determine the air-fuel ratios on molar basis as well as on mass basis for the stoichiometric combustion of methane in air. (Solution) The stoichiometric equation for the combustion of 1 mole of methane can be obtained from the solution to Example 16.1. CH4 þ 2O2 þ 7:52N2 ! CO2 þ 2H2 O þ 7:52 N2 From the preceding stoichiometric equation, 9.52 lbmol of air are required per lbmol of fuel. Write the air-fuel ratio on molar basis using Eq. 16.6. RA=F


molar

¼

moles air required ¼ 9:52 lbmol air=lbmol fuel mole fuel

From the solution to Example 16.2, the mass of air required for complete combustion of 1 lbmol of methane is 276.08 lbm of air. From Table 13.1, the molecular weight of methane is 16 lbm / lbmol. Substitute the known values into Eq. 16.7 to obtain the air-fuel ratio on a mass basis. RA=F

16.3.3.3


mass

¼

276:08 lbm air ¼ 17:255 lbm air=lbm fuel 16 lbm fuel

Combustion Using Excess Air

To ensure adequate supply of air for complete combustion of fuel, it is a common practice to supply much more than the theoretical requirement of air. The amount of air supplied in excess of the theoretical requirement is called excess air [1, 2]. Typical amounts of excess air used are 10% for gaseous fuels, 20% for liquid fuels, and up to 60% for solid fuels [3, 6]. When excess air is used, only the theoretical amount of oxygen required for complete combustion of the fuel will be used. The remaining unused oxygen will emerge along with the products of combustion.

16.3

Combustion Fundamentals and Definitions

409

Combustion equations involving excess air must include excess oxygen on the product side of the combustion equation along with the corresponding moles of nitrogen associated with the oxygen supplied.

Example 16.4 Given the theoretical combustion equation for methane, derive the combustion equation for the combustion of methane using 10% excess air. CH4 þ 2O2 þ 7:52 N2 ! CO2 þ 2H2 O þ 7:52 N2 (Solution) Since 10% excess air is used, the stoichiometric coefficients of oxygen and nitrogen need to be multiplied by 1.1 (1 for theoretical air and 0.1 for the excess air part). Also, the excess (unused) oxygen must be included on the right-hand side of the equation. For balancing oxygen (O2), the coefficient of excess oxygen on the right side of the equation should be 2  0.1 ¼ 0.2 mols of unused oxygen. Hence, the combustion equation using 10% excess air will be as follows. CH4 þ ð1:1Þ2O2 þ ð1:1Þ7:52N2 ! CO2 þ 2H2 O þ 0:2O2 þ ð1:1Þ7:52N2 CH4 þ 2:2O2 þ 8:272N2 ! CO2 þ 2H2 O þ 0:2O2 þ 8:272N2

16.3.4 Analysis of Combustion Products-Flue Gas Analysis Combustion of hydrocarbon fuels typically results in the following products: carbon dioxide, water vapor, nitrogen associated with air, any excess oxygen, and occasionally small quantities of carbon monoxide in case there is partial combustion. If the fuel has sulfur, sulfur dioxide will also be part of the combustion products. Analysis of combustion products provides information needed for assessing the environmental impact due to the combustion products. The combustion products form a mixture of ideal gases. For an ideal gas mixture, mole fractions, volume fractions, and pressure fractions of components are equal to each other. In the combustion equation, the stoichiometric coefficients of the species represent the moles of the species. Thus, the individual moles of each product, the total moles of the combustion gases, and the mole fraction of each component can be obtained from the combustion equation. The products of combustion are also known as flue gases or stack gases.

410

16 Fuels and Combustion

Example 16.5 Combustion of propane (C3H8) takes place in the presence of 15% excess air. The products of combustion emerge at 100 kPa and 250
C. For the combustion of one kgmol of propane, determine: A. The total volume of the combustion products B. The volume fraction of carbon dioxide in the flue gas C. The partial pressure of water vapor in the stack gas (Solution) A. The volume of the product gas can be obtained from the ideal gas equation, which requires the total moles of the product gas after the actual combustion process using 15% excess air. First, obtain the theoretical combustion equation for propane using the same method as in the previous examples. C3 H8 þ aO2 þ bN2 ! cCO2 þ dH2 O þ bN2 C balance: c ¼ 3 H balance: 2d ¼ 8 ) d ¼ 4 O balance: 2a ¼ 2c + d ¼ 2(3) + 4 ) 2a ¼ 10 ) a ¼ 5 N2:O2 ratio: b ¼ 3.76  5 ¼ 3.76  5 ¼ 18.8 Hence, the balanced equation for the stoichiometric combustion of methane is: C3 H8 þ 5O2 þ 18:8N2 ! 3CO2 þ 4H2 O þ 18:8N2 Write the actual combustion equation with 15% excess air using the same method as before. C3 H8 þ ð1:15Þ5O2 þ ð1:15Þ18:8N2 ! 3CO2 þ 4H2 O þ ð0:15  5ÞO2 þ ð1:15Þ18:8N2 C3 H8 þ 5:75O2 þ 21:62N2 ! 3CO2 þ 4H2 O þ 0:75O2 þ 21:62N2 Calculate the total moles of the combustion product using the preceding equation. N prod ¼ 3 kmol þ 4 kmol þ 0:75 kmol þ 21:62 kmol ¼ 29:37 kmol Calculate the absolute temperature of the product gas using Eq. 13.3. T ¼ 273
+ 250
C ¼ 523 K. Calculate the volume of combustion product using the molar form of ideal gas law (Eq. 13.15).

16.3

Combustion Fundamentals and Definitions

411

 ð29:37 kmolÞ 8:314

V prod ¼

 kJ ð523 KÞ kmol  K 100 kPa

N prod RT ¼ P ¼ 1277 m3

Note: kJ/kPa  kN.m/(kN/m2) ¼ m3 B. For an ideal gas mixture, volume fraction ¼ mole fraction (Eq. 15.5). Therefore, Volume fraction CO2 ¼ yCO2 ¼

N CO2 3 kmol ¼ 0:1021 ð10:21%Þ ¼ N prod 29:37 kmol

C. Calculate the partial pressure of water vapor in the stack gas mixture using Eq. 15.5.

PH2 O ¼ yH2 O P ¼

16.3.4.1

    N H2 O 4 kmol ð100 kPaÞ ¼ 13:62 kPa P¼ 29:37 kmol N prod

Orsat Analysis of Flue Gases

Orsat analysis is a method used in determining the composition of combustion products on a dry basis [5]. The Orsat apparatus consists of tubes containing absorbents of components of combustion products such as carbon dioxide and excess oxygen. The volume of combustion products is measured subsequent to each absorption. This information can be used in obtaining the volume fraction of combustion products on a dry basis, that is, it does not include the fraction of water vapor. However, water vapor is a product of combustion and must be included in the combustion reaction equation. From the results of Orsat analysis, a skeleton combustion reaction is written using algebraic symbols for unknown stoichiometric coefficients and unknown molecular formula of the fuel. The values of the unknown variables are obtained using atomic balances for elements. Thus, useful, comprehensive information about the combustion processes can be obtained from Orsat analysis as illustrated in the following example. Example 16.6 The Orsat analysis of combustion of an unknown hydrocarbon provides the following result on a volume basis: carbon dioxide 11.33%, oxygen 4.12%, and the rest is nitrogen. Determine: A. The actual combustion equation B. The molar composition of the fuel C. The percent excess air used

412

16 Fuels and Combustion

(Solution) A. The mole fraction of a component is the same as volume fraction in a mixture of ideal gases. Based on the results of Orsat analysis, the volume percent and hence the mole percent of nitrogen in the combustion product are 100  11.33  4.12 ¼ 84.55%. Although Orsat analysis is on a dry basis, water vapor is a product of combustion, and it must be included in the combustion equation. Write the actual combustion equation based on 100 moles of combustion products using the data from the Orsat analysis and by using algebraic symbols for unknown entities. Ca Hb þ cO2 þ dN2 ! 11:33CO2 þ eH2 O þ 4:12O2 þ 84:55N2 Obtain the unknown stoichiometric coefficients by using balances for various species in the combustion equation. N2 balance: d ¼ 84.55 d N2 to O2 ratio in air: d ¼ 3.76c) c ¼ 3:76 ¼ 84:55 3:76 ¼ 22:49 C balance: a ¼ 11.33 O balance: 2c ¼ 2(11.33) + e + 2(4.12) )2(22.49) ¼ 2(11.33) + e + 2(4.12) ) e ¼ 14.08 H balance: b ¼ 2e ¼ 2(14.08) ¼ 28.16 Substitute the values for the unknowns, and write the actual combustion equation. C11:33 H28:16 þ 22:49O2 þ 84:55N2 ! 11:33CO2 þ 14:08H2 O þ 4:12O2 þ 84:55N2 B. The molecular formula for the fuel is C11.33H28.16. The total moles in the fuel is 11.33 + 28.16 ¼ 39.49 moles. Hence the molar composition of the fuel is: 

 11:33 moles 100 ¼ 28:69% 39:49 moles   28:16 moles Mole% H ¼ 100 ¼ 71:31% 39:49 moles Mole%

C¼

C. Unused oxygen appears on the product side of the actual combustion equation. This implies that excess air is used in the combustion of the fuel. Derive the theoretical combustion equation using the same procedure that was used in the solution of earlier problems.

16.4

Combustion of Coal: Use of Gravimetric Analysis of Coal

413

C11:33 H28:16 þ aO2 þ bN2 ! 11:33CO2 þ 14:08H2 O þ bN2 O balance: 2a ¼ 2(11.33) + 14.08 ¼ 36.74 ) a ¼ 18.37 N2 to O2 ratio in air: b ¼ 3.76 a)b ¼ 3.76a ¼ 3.76  18.37 ¼ 69.07 Therefore, the theoretical combustion equation is: C11:33 H28:16 þ 18:37O2 þ 69:07N2 ! 11:33CO2 þ 14:08H2 O þ 69:07N2 For reference, the actual combustion equation obtained from the solution to part A is: C11:33 H28:16 þ 22:49O2 þ 84:55N2 ! 11:33CO2 þ 14:08H2 O þ 4:12O2 þ 84:55N2     actual moles of air 22:49 þ 84:55 100 ¼ 100 % theoretical air ¼ theoretical moles of air 18:37 þ 69:07 ¼ 122:4% Hence % excess air used ¼ 122.4%  100% ¼ 22.4%

16.4

Combustion of Coal: Use of Gravimetric Analysis of Coal

The constituents of coal are reported on a gravimetric (mass) basis. Since the combustion reaction occurs on a molar basis, the constituents of coal must be converted to moles in order to determine the combustion equation. The free moisture associated with coal will also be part of the water vapor in the combustion product. The use of gravimetric data of coal is illustrated in the following example. Example 16.7 Gravimetric analysis of a type of coal provides the following information: carbon 70%, oxygen 4.2%, nitrogen 2.3%, hydrogen 6.7%, free moisture 8.8%, and the rest is ash. This coal is burnt with 30% excess air to generate steam in a power plant. On the basis of 1 pound of coal undergoing the combustion process, determine: A. The volume (ft3) of air to be supplied at 80
F and 14.5 psia B. The volume (ft3) of product gas at 450
F and 14.5 psia C. The lower heating value of coal if the higher heating value is 12,950 Btu/lbm

414

16 Fuels and Combustion

(Solution) A. It is best to work this problem using a table as shown. The basis is 1 lbm coal. Mass Species (lbm) C 0.7 O2,coal 0.042 N2,coal 0.023 H2 0.067 H2O, 0.088 free CO2 NA H2O, NA rxn O 2, exs* N2,air*

Mol.Wt. (lbm/lbmol) 12 32 28 2 18

Moles m/M (lbmol) 0.0583 0.0013 0.0008 0.0335 0.0049

O2 Reqd. Reaction (lbmol) C+O2!CO2 0.0583 NA NA NA NA H2+½O2!H2O 0.0168 NA NA

lbmol product NA NA 0.0008 NA 0.0049

44 18

NA NA

NA NA

0.0583 0.0335

NA NA

0.0221*

Total moles of product (lbmol)

0.3606* 0.4802

Please refer to the explanation below for the derivation of these figures

*

From the table, after subtracting the moles of oxygen already in coal, the net moles of oxygen required for the theoretical combustion of carbon and hydrogen in coal is: O2, required ¼ 0.0583 lbmol + 0.0168 lbmol ─ 0.0013 lbmol ¼ 0.0738 lbmol Since 30% excess air is used, the oxygen supplied is: O2, supplied ¼ (1.3)( 0.0738 lbmol) ¼ 0.0959 lbmol The excess oxygen which will be in the product stream is O2 , excess

¼ O2 , supplied  O2 , required ¼ 0:0959 lbmol  0:0738 lbmol ¼ 0:0221 lbmol

The nitrogen in air will also be part of the nitrogen in the product stream. The other part of nitrogen is the nitrogen in coal. Using the mole ratio of nitrogen to oxygen in air (3.76 : 1), determine the nitrogen in air and part of product stream. N2, air ¼ 3.76(O2, supplied) ¼ (3.76)(0.0959 lbmol) ¼ 0.3606 lbmol (also in product stream) Add the moles of oxygen supplied and the moles of nitrogen supplied to get the moles of air supplied,

16.4

Combustion of Coal: Use of Gravimetric Analysis of Coal

N air , supplied

415

¼ N O2, supplied þ N N2, supplied ¼ 0:0959 lbmol þ 0:3606 lbmol ¼ 0:4565 lbmol

Calculate the volume of air supplied using the ideal gas law (Eq. 13.15). 

 psia  ft3 ð0:4565 lbmolÞ 10:73  lbmol 
R V air ¼

N air RT air ¼ P ¼ 182:4 ft3

ðð80
F þ 460Þ
RÞ 14:5 Psia

B. Determine the total moles of product as shown in the table, Nprod ¼ 0.4802 lbmol. Calculate the volume of combustion product using the ideal gas law (Eq. 13.15).   psia  ft3 ð0:4802 lbmolÞ 10:73  lbmol 
R V prod ¼

N prod RT prod ¼ P ¼ 323:4 ft3

ðð450
F þ 460
Þ
RÞ 14:5 Psia

C. Calculate the mass of water in the product stream using the moles of water from the product column of the solution table.

mwater

  lbm ¼ Nwater Mwater ¼ ð0:0049 lbmol þ 0:0335 lbmolÞ 18 lbmol ¼ 0:6912 lbm

D. Calculate the lower heating value using Eq. 16.1. From steam tables, the enthalpy of vaporization of water at 78
F is hfg ¼ 1049.16 Btu / lbm. Substitute the known values into Eq. 16.1.   Btu Btu  ð0:6912 lbmÞ 1049:16 lbm lbm ¼ 12, 225 Btu=lbm

LHV ¼ HHV  mwater hfg ¼ 12950

416

16.5

16 Fuels and Combustion

Dew Point of Combustion Products

Water vapor is a product of combustion. The temperature in the stack system used in the discharge of combustion products must be high enough to prevent the condensation of this water vapor. Water condensate can combine with other product gases to form corrosive substances, resulting in damage to the stack system [4]. Water condensation in the stack system can be prevented by maintaining the temperature in the stack system well above the dew point of the combustion products. The partial pressure of water vapor in the combustion gases represents the saturation pressure of water vapor in the mixture. The dew point of the mixture of combustion gases is the saturation temperature at the saturation (partial) pressure of water vapor in the product gas mixture. Finding the dew point of combustion gases is illustrated in the following example. Example 16.8 The theoretical combustion equation for the combustion of methane (CH4) is: CH4 þ 2O2 þ 7:52N2 ! CO2 þ 2H2 O þ 7:52 N2 The products of combustion emerge at 101 kPa. Determine: A. The dew point of the combustion product when 10% excess air is used B. The impact of using 30% excess air on the dew point (Solution) A. Using the same technique as before, derive the combustion equation when 10% excess air is used. CH4 þ ð1:1Þ2O2 þ ð1:1Þ7:52N2 ! CO2 þ 2H2 O þ ð1:1Þ7:52 N2 þ ð0:1Þ2O2 CH4 þ 2:2O2 þ 8:27N2 ! CO2 þ 2H2 O þ 8:27N2 þ 0:2O2 Calculate the total moles of combustion products from the preceding equation. N prod ¼ 1 þ 2 þ 8:27 þ 0:2 ¼ 11:47 moles Calculate the mole fraction of water vapor in the product gas using Eq. 15.5. yH2 O ¼

N H2 O 2 moles ¼ 0:1744 ¼ 11:47 moles N

The combustion product is an ideal gas mixture, where mole fractions and pressure fractions are identical (Eq. 15.5).

16.5

Dew Point of Combustion Products

yH2 O ¼ 0:1744 ¼

417

PH2 O PH2 O ¼ P 101 kPa

Calculate the partial pressure of water vapor (which is the same as the saturation pressure) from the preceding equation. PH2 O ¼ ð0:1744Þð101 kPaÞ ¼ 17:61 kPa The dew point is the saturation temperature corresponding to the partial pressure of water vapor. From steam tables: T sat at 17:61 kPa ð¼ 0:0176 MPaÞ ¼ 57:2
C


The dew point of the combustion product is 57.2 C when 10% excess air is used. B. Determine the dew point when 30% excess air is used in a similar manner as shown here. Derive the actual combustion equation when 30% excess air is used. CH4 þ ð1:3Þ2O2 þ ð1:3Þ7:52N2 ! CO2 þ 2H2 O þ ð1:3Þ7:52 N2 þ ð0:3Þ2O2 CH4 þ 2:6O2 þ 9:78N2 ! CO2 þ 2H2 O þ 9:78N2 þ 0:6O2 Calculate the total moles of combustion products from the preceding equation. N prod ¼ 1 þ 2 þ 9:78 þ 0:6 ¼ 13:38 moles Calculate the mole fraction of water vapor in the product gas using Eq. 15.5. yH2 O ¼

N H2 O 2 moles ¼ 0:1495 ¼ 13:38 moles N

The combustion product is an ideal gas mixture, where mole fractions and pressure fractions are identical (Eq. 15.5). yH2 O ¼ 0:1495 ¼

PH2 O PH2 O ¼ P 101 kPa

Calculate the partial pressure of water vapor (which is the same as the saturation pressure) from the preceding equation. PH2 O ¼ ð0:1495Þð101 kPaÞ ¼ 15:1 kPa

418

16 Fuels and Combustion

The dew point is the saturation temperature corresponding to the partial pressure of water vapor. From steam tables: T sat at 15:1 kPa ð0:0151 MPaÞ ¼ 55
C The dew point of the combustion product is 55
C when 30% excess air is used. Conclusion: Increasing the amount of excess air will lower the dew point of the combustion product.

Practice Problems Practice Problem 16.1 A blend of gasoline fuel, which can be approximated by the chemical formula of octane (C8 H18), is burned in 20% excess air. Determine: A. The theoretical combustion equation B. The actual combustion equation C. The air-fuel ratio on a mass (kg) basis Practice Problem 16.2 The theoretical combustion equation for combustion of 1 mole of ethane (C2H6) is: C2 H6 þ 3:5O2 þ 13:16N2 ! 2CO2 þ 3H2 O þ 13:16N2 Flue gas analysis of combustion of 1 kmol of ethane with excess air reveals 2.8% oxygen by volume. Calculate the percent excess air used. Practice Problem 16.3 The gravimetric analysis of dry coal is as follows: carbon 80%, hydrogen 11%, and the rest being ash. The flue gases emerge at 14.7 psia and 500
F. Determine the volumes (ft3) of carbon dioxide and water vapor produced due to complete combustion of 1 lbm coal. Practice Problem 16.4 Theoretical combustion of 1 kmol of a hydrocarbon fuel results in the formation of water condensate when the stack temperature drops to 50
C. The volume percent of carbon dioxide in the combustion products is determined to be 6%, and the pressure in the stack is 105 kPa. Determine the empirical formula, that is, the ratio of carbon atoms to hydrogen atoms in the fuel.

Solutions to Practice Problems

419

Solutions to Practice Problems Practice Problem 16.1 (Solution) A. Write the skeletal theoretical combustion equation using variables for the unknown stoichiometric coefficients. C8 H18 þ aO2 þ bN2 ! cCO2 þ dH2 O þ bN2 Balance the elements in the preceding equation in the following order – C, H, and O. C balance: c ¼ 8 H balance: 2d ¼ 18 ) d ¼ 9 O balance: 2a ¼ 2c + d ¼ 2(8) + 9 ) a ¼ 12.5 N2:O2 ratio: b ¼ 3.76 a ¼ 3.76(12.5) ¼ 47 Substitute the values for the variables, and write the theoretical combustion equation. C8 H18 þ 12:5O2 þ 47 N2 ! 8CO2 þ 9H2 O þ 47N2 B. Since 20% excess air is used, multiply the coefficients of oxygen and nitrogen by 1.2, and account for excess oxygen on the right-hand side of the equation. C8 H18 þ ð1:2Þ12:5 O2 þ ð1:2Þ47 N2 ! 8CO2 þ 9H2 O þ ð1:2Þ47 N2 þ ð0:2Þ12:5 O2 C8 H18 þ 15O2 þ 56:4N2 ! 8CO2 þ 9H2 O þ 47N2 þ 2:5O2 C. Using information from the preceding actual combustion equation and Eq. 16.7, calculate air-fuel ratio on a mass basis as shown.

RA=F


mass

¼

¼

mass air required unit mass fuel ð15 þ 56:4Þkmol air  29

kg kmol

kg kg þ 18 kmol H  1 kmol kmol ¼ 18:16 kg air=kg fuel 8 kmol C  12

420

16 Fuels and Combustion

Practice Problem 16.2 (Solution) Let x be the fraction of excess air used. On the reactant side the stoichiometric coefficient of oxygen will be 3.5(1 + x) ¼ 3.5 + 3.5x. The stoichiometric coefficient of unused oxygen on the product side will be 3.5x. The stoichiometric coefficient of nitrogen on both sides of the equation will be 13.16(1 + x) ¼ 13.16 + 13.16x. Write the actual combustion equation using the preceding results. C2 H6 þ 3:5ð1 þ xÞO2 þ 13:16ð1 þ xÞN2 ! 2CO2 þ 3H2 O þ 13:16N2 þ ð13:16xÞN2 þ ð3:5xÞO2 The volume fraction of oxygen in the flue gas is 0.028 (equivalent to 2.8%). For a mixture of ideal gases, volume fraction ¼ mole fraction. The mole fraction of oxygen on the product side is: yO2 ¼ 0:028 ¼

N O2 3:5x ¼ N prod 2 þ 3 þ 13:16 þ 13:16x þ 3:5x

Simplify the preceding equation, and solve for x. 0:028 ¼

3:5x ) x ¼ 0:1676 18:16 þ 16:66x

Therefore, 16.76% excess air is used. Practice Problem 16.3 (Solution) Determine lbmol of carbon dioxide and lbmol of water vapor produced due to the combustion of 1 lbm of coal using the table shown here.

Species C H2 CO2 H 2O

Mass (lbm) 0.8 0.11 NA NA

Mol.Wt. (lbm/lbmol) 12 2 44 18

Moles m/M (lbmol) 0.0667 0.055 NA NA

Reaction C+O2!CO2 H2+½O2!H2O NA NA

lbmol product NA NA 0.0667 0.055

Calculate the volumes of the product gases using the ideal gas law in molar form (Eq. 13.15).

Solutions to Practice Problems

421

  psia  ft3 ð0:0667 lbmolÞ 10:73  lbmol 
R V CO2 ¼

N CO2 RT ¼ P ¼ 46:74 ft3

ðð500
F þ 460
Þ
RÞ 14:7 psia

N H2 O RT ¼ P ¼ 38:54 ft3

ðð500
F þ 460
Þ
RÞ 14:7 psia

  psia  ft3  ð0:055 lbmolÞ 10:73 lbmol 
R

V H2 O ¼

Practice Problem 16.4 (Solution) The dew point of water vapor is given as 50
C. Determine the saturation pressure at 50
C from the steam tables ) Psat ¼ 0:0124 MPa ¼ 0:0124 MPa 

1000 kPa ¼ 12:4 kPa MPa

Therefore, the partial pressure of water vapor in the combustion product is: PH2 O ¼ Psat ¼ 12:4 kPa For an ideal gas mixture such as the combustion product, the volume fraction, pressure fraction, and mole fraction are all equal to each other (Eq. 15.5). Determine the mole fractions of carbon dioxide and water vapor in the combustion product by using this concept. yCO2 ¼

V CO2 P 12:4 kPa ¼ 0:1181 ¼ 0:06 ð6%Þ and yH2 O ¼ H2 O ¼ 105 kPa V P

Using the preceding results and the molecular formulas of carbon dioxide and water vapor, determine the ratio of moles carbon (C) to moles hydrogen (H) in the combustion product. moles C ¼ moles CO2 ¼ 0:06 and moles H ¼ 2  moles H2 O ¼ 2  0:1181 ¼ 0:2362   moles C 0:06 1 1 ¼ ’ ¼ moles H comb:prod: 0:2362 3:94 4

422

16 Fuels and Combustion

The source for carbon and hydrogen in the combustion product is the fuel. Therefore, the mole ratio between carbon and hydrogen in the fuel will also be the same as the mole ratio between carbon and hydrogen in the combustion product. Therefore, the empirical formula of the fuel is CH4.

References 1. Burghardt, D.M., Harbach, J.A.: Engineering Thermodynamics, 4th edn. Tidewater Publications, Centreville (1999) 2. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill, New York (2019) 3. Chapman, R.: Understanding the Impact of Excess Air, Firebridge, Burlington (2019). Download from https://www.firebridgeinc.com/blog/understanding-the-impact-of-excess-air/ 4. Corrosionpedia: What does Dew Point Mean?, USA, (2017). Download from https://www. corrosionpedia.com/definition/1502/dew-point 5. Engineering Notes: Flue Gas Analysis by Orsat Apparatus. Download from https://www. engineeringenotes.com/thermal-engineering/fuels-and-combustion/flue-gas-analysis/flue-gasanalysis-by-orsat-apparatus-combustion-thermodynamics/49480 6. Engineering Tool Box: Combustion Efficiency and Excess Air, USA (2003). Download from https://www.engineeringtoolbox.com/boiler-combustion-efficiency-d_271.html 7. Engineering Tool Box: Fuels – Higher and Lower Calorific Values, USA (2003). Download from https://www.engineeringtoolbox.com/fuels-higher-calorific-values-d_169.html 8. Engineering Tool Box: Stoichiometric Combustion USA (2003). Download from https://www. engineeringtoolbox.com/stoichiometric-combustion-d_399.html 9. Steinberg, Bob: Combustion Efficiency and Air – Fuel Ratio, Sage Metering, USA (2013). Download from https://sagemetering.com/combustion-efficiency/air-fuel-ratio-effect-on-combus tion-efficiency/

Chapter 17

Thermodynamic Cycles

17.1

Introduction

Applications of thermodynamic cycles permeate through our daily lives as much as the roads and structures we use. For example, refrigerators and air-conditioners operate based on the vapor compression refrigeration cycle. The cars we drive operate based on the Otto cycle. Thermodynamic cycles are cyclic processes, as the name implies. After completing the cyclic process, the working fluid returns to its original state at the start of the cyclic process. Each thermodynamic cycle typically consists of four steps or subprocesses linked to each other. Each step is associated with a device such as turbine, pump, and heat exchangers. Since the working fluid flows through the device, each device will be an open thermodynamic system. Examples of working fluids that circulate through thermodynamic cycles are steam, air, combustion products, and refrigerants. Depending on the nature of the device in each step, there will be some work and/or heat interaction in each step. The preceding basic concepts of thermodynamic cycles are illustrated extensively in the subsequent sections through the use of practical examples.

17.2

Carnot Cycle and Reversed Carnot Cycle

Carnot cycle and reversed Carnot cycle are idealized cycles [3, 4, 9] that are used as stepping-stones for actual, real-world thermodynamic cycles. Since they achieve maximum possible performance levels (such as efficiency), they are also used as benchmarks for the assessment of actual cycles.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3_17

423

424

17 Thermodynamic Cycles

17.2.1 Carnot Cycle The Carnot cycle is an idealized power generation cycle. The Carnot cycle is based on the Carnot engine illustrated in Fig. 17.1 along with the associated temperatureentropy and pressure-volume phase diagrams. Carnot engine is a heat engine which essentially converts heat into useful work [3, 4, 9]. The Carnot engine takes heat from a high temperature heat reservoir and converts part of the heat (available energy) to useful work and rejects the remaining unavailable heat to a low temperature heat sink as shown in Fig. 17.1a. It is not feasible to convert all the heat absorbed into useful work because doing so violates the second law of thermodynamics (Sect. 17.2). Referring to Fig. 17.1a, b, the processes involved in the Carnot cycle are: 1–2: Isentropic compression of the working fluid (vertical, constant entropy line in the T – s diagram and increasing pressure, decreasing volume in the P – V diagram implies isentropic compression process) 2–3: Heat absorption at constant temperature and isothermal expansion of the working fluid (horizontal, constant temperature line in the T – s diagram and decreasing pressure, increasing volume in the P – V diagram implies isothermal expansion process) 3–4: Isentropic expansion of the working fluid, producing work (vertical, constant entropy line in the T – s diagram and decreasing pressure, increasing volume in the P – V diagram implies isentropic expansion process) 4–1: Heat rejection at constant temperature and isothermal compression of the working fluid (horizontal, constant temperature line in the T – s diagram and increasing pressure, decreasing volume in the P – V diagram implies isothermal compression process) Fig. 17.1 (a) Carnot engine, (b) Carnot cycle T – s diagram, (c) Carnot cycle P – V diagram

17.2

Carnot Cycle and Reversed Carnot Cycle

17.2.1.1

425

Thermal Efficiency of Carnot Cycle

The thermal efficiency of a thermodynamic cycle is a measure of the percentage of heat input converted to useful work as shown in Eq. 17.1. Using the nomenclature in Fig. 17.1a, the mathematical representation of Carnot thermal efficiency [3, 4] is: ηth,Carnot ¼

W QH

ð17:1Þ

Energy balance for the Carnot engine (Fig. 17.1a) results in the following equations. Energy In ¼ Energy Out QH ¼ QL þ W ) W ¼ QH  QL Substitute the preceding result into Eq. 17.1. Also, the heat energies can be replaced by the corresponding absolute temperatures. This results in a comprehensive set of equations for Carnot thermal efficiency. ηth,Carnot ¼

W Q  QL T H  T L ¼ H ¼ QH QH TH

ð17:2Þ

Example 17.1 A Carnot engine operates between temperature limits of 400  C and 100  C. It is desired to obtain useful work equivalent to 75 kJ from this engine. Determine: A. The thermal efficiency B. The heat input required C. The heat rejected (Solution) A. Calculate absolute temperatures using Eq. 13.3. T H ¼ 400∘ C þ 273∘ ¼ 673 K,

T L ¼ 100∘ C þ 273∘ ¼ 373 K

Calculate the thermal efficiency using the last part of Eq. 17.2. ηth,Carnot ¼

T H  T L 673 K  373 K ¼ 0:4458 ð’ 44:6%Þ ¼ 673 K TH

426

17 Thermodynamic Cycles

B. Calculate the heat input required by using the first part of Eq. 17.2. QH ¼

W 75 kJ ¼ ¼ 168:24 kJ ηth,Carnot 0:4458

C. Calculate the heat rejected using the energy balance equation. QL ¼ QH  W ¼ 168:24 kJ  75 kJ ¼ 93:24 kJ

17.2.1.2

Second Law of Thermodynamics and Carnot Efficiency

The first law of thermodynamics relates to the quantitative nature of energy based on the principle of energy conservation. The second law of thermodynamics [4, 14] looks at energy in a qualitative sense. The most practical interpretation of the second law is that it is impossible to convert heat completely into an equivalent amount of useful work. In the process of obtaining work from heat, some heat must be transferred to a heat sink. Thus, according to the second law of thermodynamics, heat is a superior quality of energy compared to work. The Kelvin-Planck statement of the second law is: It is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and deliver an equivalent amount of work. Carnot cycle converts all the energy available for conversion to work to useful work and minimizes the amount of heat rejection to the heat sink. Hence, a corollary of Kelvin-Planck statement of the second law of thermodynamics is: No heat engine can have a higher efficiency than a Carnot cycle operating between the same two temperature levels.

17.2

Carnot Cycle and Reversed Carnot Cycle

427

17.2.2 Reversed Carnot Cycle As the name implies, a reversed Carnot cycle is the exact opposite of the Carnot cycle. It is the same as the Carnot cycle operating in the reverse direction [3, 4, 12]. Similar to Carnot cycle being the basis for cycles producing net power, the reversed Carnot cycle forms the basis for power-consuming refrigeration cycles and heat pumps. The schematic diagram and the T – s diagram for the reversed Carnot cycle are shown in Fig. 17.2. A reversed Carnot engine takes heat from a low-temperature heat source, such as a refrigerated space, and transfers the heat to a high-temperature heat sink. Heat cannot spontaneously flow from low temperature to high temperature. Therefore, the reversed Carnot engine has to use external work to push the heat from low temperature to high temperature. As shown in Fig. 17.2b, the processes in the reversed Carnot cycle are: 1–2: Isentropic compression of the working fluid, which is usually a refrigerant. (vertical, constant entropy line in the T – s diagram and an increase in temperature of the working fluid implies an isentropic compression process). This compression process requires external work input. 2–3: Heat rejection at constant temperature (horizontal, constant temperature line in the T – s diagram and decreasing entropy implies isothermal compression process, that is, isothermal condensation). 3–4: Isentropic expansion of the working fluid, producing work (vertical, constant entropy line in the T – s diagram and decreasing temperature due to decrease in pressure implies an isentropic expansion process). 4–1: Heat absorption at constant temperature and isothermal expansion of the working fluid (horizontal, constant temperature line in the T – s diagram and increasing entropy due to evaporation of the working fluid implies an isothermal expansion process). Fig. 17.2 (a) Reversed Carnot engine, (b) reversed Carnot T – s diagram cycle

428

17.2.2.1

17 Thermodynamic Cycles

Performance Measure of Reversed Carnot Cycle: Coefficient of Performance (COP)

The performance measure of any device or process can be defined as a measure of desired effect achieved divided by the cost of achieving it. The desired effect in a reversed Carnot cycle is the heat absorbed from the low-temperature heat source (also known as the refrigerating effect), and cost of achieving the desired effect is the work input to the compressor. The preceding performance measure of reversed Carnot cycle is known as coefficient of performance (COP). Referring to Fig. 17.2a, the mathematical representation of COP for refrigeration or air-conditioning (heat removal) is COPrefg:=AC ¼

Q desired effect ¼ L cost of achieving it W

ð17:3Þ

Energy balance for the reversed Carnot engine (Fig. 17.2a) results in the following equations. Energy In ¼ Energy Out QL þ W ¼ QH ) W ¼ QH  QL Substitute the preceding result into Eq. 17.3. Also, the heat energies can be replaced by the corresponding absolute temperatures. This results in a comprehensive set of equations for coefficient of performance for refrigeration and air-conditioning. COPrefg:=AC ¼

QL QL TL ¼ ¼ W QH  QL T H  T L

ð17:4Þ

The reversed Carnot cycle can also be used for space heating in contrast to space cooling in refrigeration and air-conditioning processes. Space heating is achieved by capturing the heat rejected at the high-temperature heat sink (condenser). When the reversed Carnot cycle is used for space heating, it becomes a heat pump [7]. The desired effect in using a heat pump is the heat rejected at the heat sink, QH. Therefore, the COP for a heat pump will be: COPHeat Pump ¼

QH QH TH ¼ ¼ W QH  QL T H  T L

ð17:5Þ

The COP is always a number greater than one, with higher values indicating better performance.

17.2

Carnot Cycle and Reversed Carnot Cycle

429

The rate of cooling can be represented as tons of refrigeration (TOR) required. One ton of cooling is the rate of cooling achieved when 1 short ton (2000 lbm) of ice at 0  C melts in a 24-hour time period. By using the latent heat of melting (fusion) of ice, it can be shown that:

1 ton of refrigeration  12, 000 Btu=hr ¼ 3:517 kW

ð17:6Þ

Example 17.2 A Carnot refrigerator (freezer) operates between temperature limits of 18  C and 25  C. It is desired to remove heat equivalent to 75 kW from the cold space. Determine: A. The COPrefg./AC B. The power input required (kW) C. The rate of heat rejection (kJ/hr) (Solution) A. Calculate absolute temperatures using Eq. 13.3. T H ¼ 25 C þ 273 ¼ 298 K

T L ¼ 18 C þ 273 ¼ 255 K

Calculate the coefficient of performance for the freezer using the last part of Eq. 17.4, and substitute the known values. COPrefg:=AC ¼

TL 255 K ¼ 5:93 ¼ T H  T L 298 K  255 K

B. Calculate the power input required by using the first part of Eq. 17.4. The rate of heat removal from the cold space is 75 kW. _ ¼ W

Q_ L 75 kW ¼ 12:65 kW ¼ 5:93 COPrefg:=AC

C. Calculate the rate of heat rejection using the middle part of Eq. 17.4. Q_ H ¼


 Q_ L 1 þ COPrefg:=AC 75 kWð1 þ 5:93Þ ¼ 87:65 kW ¼ 5:93 COPrefg:=AC

430

17 Thermodynamic Cycles

Note: The same result can be obtained by using the energy balance equation for a reversed Carnot engine. _ ¼ 75 kW þ 12:65 kW ¼ 87:65 kW Q_ H ¼ Q_ L þ W Example 17.3 A Carnot heat pump is used in heating 1700 cfm of atmospheric air from 40 to 70  F. Calculate the COP if the power input is 5 hp. (Solution) Calculate the individual gas constant for air using Eq. 13.16. psia‐ft3

Rair ¼

10:73 lbmol‐ R R ¼ ¼ 0:37 psia‐ft3 =lbm‐ R lbm M air 29 lbmol

Calculate the average temperature of air and its absolute value using Eq. 13.2. T avg ¼ ð40 F þ 70 FÞ=2 ¼ 55 F

) T air ¼ 460 þ 55 F ¼ 515 R

Calculate the density of air at its average temperature using the ideal gas law (Eq. 13.12). ρ¼

m P 14:7 psia  ¼ ¼ 0:0771 lbm=ft3 ¼ psia‐ft3 V Rair T air  ð R Þ 0:37 lbm‐ 515 R

Calculate the mass flow rate of air using the continuity equation (Eq. 3.5). m_ air ¼ ρair V_ air ¼

 0:0771

lbm ft3

  ft3 ¼ 131:1 lbm= min 1700 min

From Table 13.2, the specific heat of air at constant pressure is 0.24 Btu/lbm- R. Energy balance at the high-temperature heat sink results in the following set of equations: Rate of heat rejected at the heat sink ¼ Rate of heat absorption by air    _QH ¼ m_ air cp,air ΔT air ¼ 131:1 lbm 0:24 Btu ð70 F  40 FÞ min lbm‐ R ¼ 943:9 Btu= min Calculate the coefficient of performance for the heat pump using Eq. 17.5.

17.3

Rankine Cycle

COPHeat Pump

17.3

431


0:0236 hp Btu 943:9 min Btu _QH 1 min ¼ 4:45 ¼ ¼ _ 5 hp W

Rankine Cycle

Rankine cycle is a commonly used cycle in steam power plants. Although Carnot cycle has the highest possible thermal efficiency for producing useful work, it is not feasible to implement it power plants. There are two major operational difficulties with Carnot cycle: 1. The pump handles a vapor-liquid mixture during the isentropic compression process (process 1–2 in the T – s diagram for Carnot cycle). This can lead to cavitation of pumps and frequent damage to the impellers. 2. The steam at the end of the isentropic expansion process (process 3–4 in the T – s diagram for Carnot cycle) is usually of low quality. This is because the expansion process, typically in a steam turbine, starts with saturated steam at state 3, and after generating work during the expansion process, the steam leaves the turbine with significant amounts of condensate. Again, the presence of a two -phase liquid-vapor mixture can damage the turbine blades. The preceding operational difficulties in the Carnot cycle are overcome by using superheated steam at the turbine inlet and completely condensing the steam after it leaves the turbine so that the pump is handling only liquid water. These modifications to Carnot cycle are illustrated in Fig. 17.3 with the Rankine cycle superimposed on the existing Carnot cycle. Clearly, the operational difficulties in Carnot cycle have been addressed in Rankine cycle operating with superheated steam.

Fig. 17.3 Comparison of Carnot and Rankine cycles

432

17 Thermodynamic Cycles

17.3.1 Analysis of Rankine Cycle The components of Rankine cycle and the T – s diagram for the Rankine cycle are shown in Fig. 17.4. Rankine cycle is used extensively for power generation. The working fluid is steam/water. The steam generator produces superheated steam by using heat generated by combustion of fuel. Superheated steam expands through the turbine producing work. After expansion through the turbine, steam is condensed in a surface condenser, which is a heat exchanger that uses cooling water as the cooling medium. The pressure of the condensed water is increased from condenser pressure to the boiler pressure by a pump. Pressurized water is conveyed to the boiler to continue the cycle. The following nomenclature is used in the analysis and calculations for Rankine cycle. All the energy terms are on the basis of unit mass of steam. For example, wt represents the work produced per unit mass of steam. qb ¼ heat input to the boiler wt ¼ work produced by the turbine qc ¼ heat rejected in the condenser wp ¼ work input to the pump P2 ¼ P3 ¼ boiler pressure P4 ¼ P1 ¼ condenser pressure Net energy ðworkÞ produced by the cycle ¼ wnet ¼ wt  wp

ð17:7Þ

m_ s ¼ mass flow rate of steam The power produced by the cycle can be obtained by multiplying the net energy (work) produced by the cycle by the mass flow rate of steam.

Fig. 17.4 (a) Rankine cycle components and schematic, (b) Rankine cycle T – s diagram

17.3

Rankine Cycle

433

_ net ¼ m_ s wnet W

ð17:8Þ

The rate of heat input to the boiler can be obtained using the following equation. Q_ b ¼ m_ s qb

ð17:9Þ

The efficiency of Rankine cycle is the net power produced (desired effect) divided by the rate of heat input to the boiler (cost of achieving the desired effect). ηth,Rankine ¼

_ net m_ s wnet wnet wt  wp W ¼ ¼ ¼ qb qb m_ s qb Q_ b

ð17:10Þ

The heat input of the boiler can be obtained by performing an energy balance around the boiler with Fig. 17.5 as a reference. Energy in ¼ Energy out h2 þ qb ¼ h3 ) qb ¼ h3  h2

ð17:11Þ

Similarly, the work produced by the turbine and the work input to pump can be obtained by performing energy balances around the turbine and pump with Fig. 17.6a, b as reference. Fig. 17.5 Energy balance around boiler

Fig. 17.6 (a) Energy balance around turbine, (b) energy balance around pump

434

17 Thermodynamic Cycles

Turbine energy balance: Energy in ¼ Energy out h3 ¼ wt þ h4 ) wt ¼ h3  h4

ð17:12Þ

Pump energy balance: Energy in ¼ Energy out h1 þ wp ¼ h2 ) wp ¼ h2  h1

ð17:13Þ

After determining the enthalpies at each state point using steam tables/Mollier diagram, the cycle efficiency and the required quantities can be calculated using the preceding equations. The enthalpy at state 2 can be calculated by using Eq. 17.13 after calculating the pump work using Eq. 14.13, reproduced here for reference. wpump ¼ ν1 ðP2  P1 Þ

ð14:13Þ

In the analysis of thermodynamic cycles and in calculations related to them, it is important to review, remember, and refer to the concepts and equations in Chap. 13: Thermodynamics Fundamentals and in Chap. 14: Conservation of Energy and First Law of Thermodynamics.

Example 17.4 Steam at a mass flow rate of 6500 lbm/hr, 500 psia pressure, and 600  F enters the turbine in a Rankine cycle. The steam expands to a final pressure of 14.7 psia. Determine: A. B. C. D. E. F.

The T – s diagram The enthalpies at each state point in the cycle The turbine and pump work per unit mass of steam The efficiency of the cycle The rate of heat input to the boiler The power produced by the cycle in MW

17.3

Rankine Cycle

435

(Solution) A.

B. It is best to determine the enthalpies at state points 3 and 4 using the Mollier diagram as shown here (the method for using tables and formula for determining the enthalpy after isentropic expansion is explained in detail in the solution to Example 14.3).

From the preceding illustration, h3 ¼ 1290 Btu/lbm and h4 ¼ 1015 Btu/lbm Determine the enthalpy at state point 1 from steam tables.

436

17 Thermodynamic Cycles

h1 ¼ hf at 14:7 psia ¼ 180:18 Btu=lbm Calculate the work input to the pump using Eq. 14.13 (similar to the solution shown in Practice Problem 14.7). wpump ¼ ν1 ðP2  P1 Þ ¼ νf,14:7psia ðP2  P1 Þ    ft3 lbf 144 in2 lbf 144 in2 0:0167  14:7  500 2  lbm in in2 ft2 ft2 ¼ 778 ft‐lbf Btu ¼ 1:50 Btu=lbm Note: Typically, the magnitude of pump work is very small compared to the turbine work, and it can be ignored without any loss of accuracy. Determine the enthalpy at state point 2 by using Eq. 17.13. h2 ¼ h1 þ wp ¼ 180:18

Btu Btu þ 1:50 ¼ 181:68 Btu=lbm lbm lbm

C. Calculate the turbine work per unit mass of steam using Eq. 17.12. wt ¼ h3  h4 ¼ 1290

Btu Btu  1015 ¼ 275 Btu=lbm lbm lbm

The pump work per unit mass of steam has already been determined, wpump ¼ 1.50 Btu/lbm D. Calculate the heat input to the boiler per unit mass of steam using Eq. 17.11. qb ¼ h3  h2 ¼ 1290

Btu Btu  181:68 ¼ 1108:32 Btu=lbm lbm lbm

Calculate the efficiency of the cycle using Eq. 17.10. ηth,Rankine ¼

Btu Btu  1:50 lbm wnet wt  wp 275 lbm ¼ ¼ ¼ 0:2468 ð’ 25%Þ Btu qb qb 1108:32 lbm

E. Calculate the rate of heat input to the boiler using Eq. 17.9. Q_ b ¼ m_ s qb ¼



lbm 65000 hr

 1108:32

 Btu ¼ 7:204  107 Btu=hr lbm

17.3

Rankine Cycle

437

F. Calculate the net power produced by the cycle using Eq. 17.8.     lbm Btu 2:9307  107 MW 273:5 65000 hr lbm Btu=hr ¼ 5:21 MW

_ net ¼ m_ s wnet ¼ W

Example 17.5 The boiler in Example 17.4 uses coal with a net heating value (NHV) of 12,200 Btu/ lbm as the fuel. The boiler efficiency is 85%. Determine: A. The required supply rate of fuel (lbm/hr) B. The hourly cost of fuel if the cost of coal is $57.79 per metric ton (Solution) A. The boiler efficiency denotes how much of the heat generated by the fuel is actually transferred to water to produce superheated steam.

m_ fuel ¼

7:204  107 Btu Q_ b hr ¼ ¼ 6947 lbm=hr Btu NHVfuel  ηb 12200 lbm  0:85

B. Fuel Cost=hr ¼ ð6947

lbm 57:79 Þð 1 MT hr MTÞð2204:6 lbmÞ¼182:10=hr

17.3.2 Rankine Cycle with Regenerative Feedwater Heating When the expansion of steam takes place in multiple stages, the expansion process approaches reversibility with minimal friction losses resulting in higher values of turbine work for the same heat input. This results in higher efficiency of the cycle. Rankine cycles with steam extraction between the turbine inlet and turbine exit are known as regenerative cycles. The thermal energy of the extracted steam is used in preheating the feedwater to the boiler. The entire process is called regenerative feedwater heating [4, 10]. Figure 17.7 illustrates the schematic diagram and the temperature-entropy diagram for a regenerative Rankine cycle with one stage of steam extraction and feedwater heating. As shown in Fig. 17.7, after the first stage of expansion from state point 3 to state point 4, a fraction of the steam, fe, is withdrawn for feedwater heating, while the remaining fraction of the steam, 1  fe, goes through the second stage of expansion from state point 4 to state point 5. Contrary to expectation, the fraction of stream be

438

17 Thermodynamic Cycles

Fig. 17.7 Rankine cycle with regenerative feedwater heating: (a) schematic, (b) T – s diagram (c) Energy balance for feedwater heater

Fig. 17.7 (continued)

extracted is not an arbitrary choice, rather it is determined from energy balance around the feedwater heater (FWH) as shown here. Energy in ¼ Energy out ð f e m_ s Þh4 þ ð1  f e Þm_ s h7 ¼ m_ s h1 ) f e h4 þ ð1  f e Þh7 ¼ h1 Solve the preceding equation for fe, the fraction of stream extracted for feedwater heating. fe ¼

h1  h7 h4  h7

ð17:14Þ

The turbine work occurs in two stages. From energy balance using the schematic diagram, the work produced by the high-pressure turbine T1 is wt1 ¼ h3  h4and the work produced by the low-pressure turbine T2 is wt2 ¼ (1  fe)(h4  h5)

17.3

Rankine Cycle

439

The total turbine work per unit mass of steam is: wt ¼ wt1 þ wt2 ¼ ðh3  h4 Þ þ ð1  f e Þðh4  h5 Þ

ð17:15Þ

The work input to the condensate pump is: wp1 ¼ (1  fe)(h7  h6) ¼ (1  fe)(v6) (P7  P6) Similarly, the work input to the feed pump is wp2 ¼ (h2  h1) ¼ (v1)(P2  P1) The total work input to the pumps is: wp ¼ wp1 þ wp2 ¼ ð1  f e Þðv6 ÞðP7  P6 Þ þ ðv1 ÞðP2  P1 Þ

ð17:16Þ

The heat input to the boiler per unit mass of steam is qb ¼ h3  h2

ð17:17Þ

The efficiency of a Rankine cycle with regenerative feedwater heating can be obtained by combining Eqs. 17.15, 17.16, and 17.17 and substituting the results into Eq. 17.10.

ηregen:Rankine

w ¼ net ¼ qb

ðh3  h4 Þ þ ð1  f e Þðh4  h5 Þ ½ ð 1  f e Þ ð h7  h6 Þ þ ð h2  h1 Þ  h3  h2

ð17:18Þ

Example 17.6 Use the data in the solutions to Examples 17.4 and 17.5 for a regenerative Rankine cycle which includes steam extraction for feedwater heating at a pressure of 100 psia. The power output remains the same as in Example 17.4. Determine: A. B. C. D.

The enthalpies at each state point in the cycle The net (energy) work produced per unit mass of steam The efficiency of the cycle The percentage savings in the hourly cost of fuel due to regenerative feedwater heating

(Solution) Draw the T – s diagram to help identify the state points and their enthalpies. Also, determine the enthalpies at state points 3, 4, and 5 using the Mollier diagram as illustrated in the diagram.

440

17 Thermodynamic Cycles

h3 ð500 psia, 600 FÞ ¼ 1290 Btu=lbm h4 ð100 psia, s4 ¼ s3 Þ ¼ 1150 Btu=lbm h5 ð14:7 psia, s5 ¼ s4 ¼ s3 Þ ¼ 1015 Btu=lbm h6 ¼ hf at 14:7 psia ¼ 180 Btu=lbm The work input to the pumps can be assumed to be 2 Btu/lbm based on the solution to Example 17.4.

17.3

Rankine Cycle

441

h7 ¼ h6 þ 2

Btu Btu Btu ¼ 180 þ2 ¼ 182 Btu=lbm lbm lbm lbm

h1 ¼ h f at 100 psia ¼ 298:5 Btu=lbm h2 ¼ h7 þ 2

Btu Btu Btu ¼ 298:5 þ2 ¼ 300:5 Btu=lbm lbm lbm lbm

B. Calculate the fraction of stream extracted for feedwater heating using Eq. 17.14.

fe ¼

Btu Btu  182 lbm h1  h7 298:5 lbm ¼ ¼ 0:1203 Btu Btu h4  h7 1150 lbm  182 lbm

Calculate the total turbine work using Eq. 17.15. wt ¼ wt1 þ wt2 ¼ ðh3  h4 Þ þ ð1  f e Þðh4  h5 Þ   Btu Btu ¼ 1290  1150 lbm lbm   Btu Btu þð1  0:1203Þ 1150  1015 lbm lbm ¼ 258:76 Btu=lbm Btu The work input to the two pumps is wp ¼ 2  2 lbm ¼ 4 Btu=lbm Calculate the net (energy) work by subtracting the total pump work from the total turbine work.

wnet ¼ wt  wp ¼ 258:76

Btu Btu 4 ¼ 254:76 Btu=lbm lbm lbm

C. Calculate the heat input to the boiler per unit mass of steam using Eq. 17.17 qb ¼ h3  h2 ¼ 1290

Btu Btu  300:5 ¼ 989:5 Btu=lbm lbm lbm

Calculate the efficiency of the cycle using Eq. 17.18. ηregen:Rankine ¼

Btu wnet 254:76 lbm ¼ Btu ¼ 0:2575 ð’ 26%Þ qb 989:5 lbm

D. Using Eq. 17.8, calculate the mass flow rate of steam required to generate the same power output as in Example 17.4.

442

17 Thermodynamic Cycles Btu

hr _ net 5:21 MW  3412142 W 1 MW ¼ ¼ 69, 780 lbm=hr m_ s ¼ Btu wnet 254:76 lbm

Calculate the rate of heat input to the boiler using Eq. 17.9. Q_ b ¼ m_ s qb ¼

 69780

lbm hr

 989:5

 Btu ¼ 6:905  107 Btu=hr lbm

Calculate the mass flow rate of fuel required and hence the hourly cost of fuel using the same method that was used in Example 17.4. 6:905  107 Btu Q_ b hr ¼ ¼ 6659 lbm=hr Btu NHVfuel  ηb 12200 lbm  0:85   lbm 57:79 Fuel Cost=hr ¼ 6659 1 MT MTÞð2204:6 hr lbmÞ¼174:55=hr

m_ fuel ¼

Calculate the percentage savings in the hourly cost of fuel due to regenerative feed water heating as shown.  %savings ¼ 182:10 

174:55

182:10Þ100 ¼ 0:0415 ð4:15%Þ

17.3.3 Rankine Cycle with Reheat In a Rankine cycle with reheat, the expansion of steam takes place in stages. If there are two stages of expansion, the steam is reheated after the first stage of expansion. The reheated steam then expands through the low-pressure turbine as shown in Fig. 17.8, which also includes the T – s diagram for a Rankine cycle with reheat. As shown in the schematic diagram of the Rankine cycle with reheat (Fig. 17.8a), the steam is withdrawn for reheating at state point 4, where the pressure is P4. After reheating, the steam enters the second-stage turbine at state point 5. From energy balance, the total turbine work per unit mass of steam is: wt ¼ wt1 þ wt2 ¼ ðh3  h4 Þ þ ðh5  h6 Þ

ð17:19Þ

The work input to the pump is: wp ¼ ðh2  h1 Þ ¼ ðv1 ÞðP2  P1 Þ

ð17:20Þ

17.3

Rankine Cycle

443

Fig. 17.8 Rankine cycle with reheat (a) Schematic (b) T – s Diagram

The heat input to the boiler per unit mass of steam consists of the usual heat input to produce superheated steam (from state 2 to state 3). However, there is additional heat input from state 4 to state 5 for reheating the steam. The total heat input to the boiler is: qb ¼ q1 þ q2 ¼ ðh3  h2 Þ þ ðh5  h4 Þ

ð17:21Þ

The efficiency of a Rankine cycle with reheat can be obtained by combining Eqs. 17.19, 17.20, and 17.21 and substituting the results into Eq. 17.10. ηReheat Rankine ¼

wnet ðh3  h4 Þ þ ðh5  h6 Þ  ðh2  h1 Þ ¼ qb ðh3  h2 Þ þ ðh5  h4 Þ

ð17:22Þ

Example 17.7 Use the data in Examples 17.4 and 17.5 for a reheat Rankine cycle which includes steam extraction for reheating at a pressure of 100 psia. The steam is reheated to a temperature of 500  F. The power output remains the same as in Example 17.4. Determine: A. The enthalpies at each state point in the cycle B. The net (energy) work produced per unit mass of steam C. The efficiency of the cycle (Solution) A. Draw the T – s diagram to help identify the state points and their enthalpies. Also, determine the enthalpies at state points 3, 4, 5, and 6 using the Mollier diagram as shown.

444

17 Thermodynamic Cycles

h3 ð500 psia, 600 FÞ ¼ 1290 Btu=lbm h4 ð100 psia, s4 ¼ s3 Þ ¼ 1150 Btu=lbm h5 ð100 psia, 500 FÞ ¼ 1280 Btu=lbm h6 ð14:7 psia, s6 ¼ s5 Þ ¼ 1110 Btu=lbm h1 ¼ hf at 14:7 psia ¼ 180:2 Btu=lbm The work input to the pump is 1.5 Btu/lbm based on the solution to Example 17.4. h2 ¼ h1 þ 1:5

Btu Btu Btu ¼ 180:2 þ 1:5 ¼ 181:7 Btu=lbm lbm lbm lbm

B. Calculate the turbine work per unit mass of steam using Eq. 17.19. wt ¼ wt1 þ wt2 ¼ ðh3  h4 Þ þ ðh5  h6 Þ   Btu Btu ¼ 1290  1150 lbm lbm   Btu Btu þ 1280  1110 lbm lbm ¼ 310 Btu=lbm Calculate the net (energy) work produced per unit mass of steam by subtracting the work input to the pump from the work produced by the turbine.

17.4

Brayton Cycle

445

wnet ¼ wt  wp ¼ 310

Btu Btu  1:50 ¼ 308:5 Btu=lbm lbm lbm

C. Calculate the total heat input to the boiler using Eq. 17.21.   Btu Btu qb ¼ ðh3  h2 Þ þ ðh5  h4 Þ ¼ 1290  181:70 lbm lbm   Btu Btu  1150 þ 1280 lbm lbm ¼ 1238 Btu=lbm Calculate the efficiency of Rankine cycle with reheat using Eq. 17.22. ηReheat Rankine ¼

Btu wnet 308:5 lbm ¼ ¼ 0:2492 ð’ 25%Þ Btu qb 1238 lbm

Comment: Reheat increased the efficiency only slightly [5]. This is because of the additional heat input required to reheat steam after the first stage of expansion.

17.4

Brayton Cycle

The Brayton cycle [3, 4] uses air as the working fluid. It forms the basis of gas turbines, which are extensively used in small-scale power generation in places like offshore platforms and ships. Gas turbines are also extensively used in powering aircraft engines. The schematic diagram, T – s diagram, and the P – V diagram for Brayton cycle are shown in Fig. 17.9. In Brayton cycle, compressed air undergoes combustion with fuel resulting in combustion gases at high temperature and high pressure, which expand through the gas turbine resulting in net (energy) work produced after subtracting the work required for compression. The compression and expansion processes are assumed to be isentropic initially, and then the isentropic efficiencies of the compressor and turbine are used in obtaining the actual work. The P-V-T relationships for isentropic processes (Eqs. 13.36, 13.37, and 13.38, reproduced here for reference) presented in Chap. 13 will be extensively used in the solution of Brayton cycle problems. P2 V 2 k ¼ P1 V 1 k  k1 T2 V1 ¼ T1 V2

ð13:36Þ ð13:37Þ

446

17 Thermodynamic Cycles

Fig. 17.9 Brayton cycle: (a) schematic diagram, (b) T – s diagram, (c) P – V diagram

T2 ¼ T1

 k1 P2 k P1

ð13:38Þ

17.4.1 Analysis of Brayton Cycle In Brayton cycle, air intake and exhaust of combustion gases take place at atmospheric pressure. Therefore, P1 ¼ P4. The compressor compresses air isentropically from pressure P1 to P2. The ratio of P2 to P1 is called pressure ratio, and it is represented by the symbol rP. Due to isentropic compression, the temperature of air increases. The temperature after the compression process can be determined by solving for temperature T2 in Eq. 13.38. The result is:  k1 P k T2 ¼ T1 2 P1

ð17:23Þ

In Eq. 17.23, k is the ratio of specific heats and for air, k ¼ 1.40. The compressed air reacts with the fuel to form combustion gases at high pressure (P3) and high

17.4

Brayton Cycle

447

temperature. The heat added in the combustion chamber can be calculated using the following equation based on energy balance for the combustion chamber: qin ¼ h3  h2 ¼ cp ðT 3  T 2 Þ

ð17:24Þ

In Eq. 17.24, cp is the specific heat of air at constant pressure available from Table 13.2 in Chap. 13. It is reasonable to assume negligible pressure drop in the combustion chamber and hence, P3 ¼ P2. The expansion of combustion gases in the turbine is an isentropic process, and using Eq. 13.38, the temperature of the exhaust gases from the turbine can be calculated by using Eq. 17.25. T4 ¼ T3

 k1 P4 k P3

ð17:25Þ

Using energy balance for the compressor will result in the following equation for the work input to the compressor per unit mass of air. wc ¼ h2  h1 ¼ cp ðT 2  T 1 Þ

ð17:26Þ

Similarly, the work output from the turbine per unit mass of air is: wt ¼ h3  h4 ¼ cp ðT 3  T 4 Þ

ð17:27Þ

There are two methods available to calculate the thermal efficiency of Brayton cycle. Similar to Rankine cycle, the efficiency of Brayton cycle will be net (energy) produced by the cycle divided by the heat input. ηth,Brayton ¼

wnet wt  wc ¼ qin qin

ð17:28Þ

The efficiency of Brayton cycle can also be calculated by using the following direct formula. 1 ηth,Brayton ¼ 1 
k1 rp k

ð17:29Þ

Example 17.8 Air enters the compressor in a Brayton cycle at 101 kPa and 15  C. The combustion gases enter the turbine at 700  C. The cycle has a pressure ratio of 8. Determine: A. The compressor work and the turbine work per unit mass of air B. The heat added in the combustion chamber C. The cycle efficiency using the results from the preceding questions and compare it with the efficiency obtained by using the direct formula

448

17 Thermodynamic Cycles

D. The volume flow rate of air (in cubic meters per second) required at the compressor inlet to produce 500 kW power (Solution) A. Calculate the absolute temperature of inlet air using Eq. 13.3. T 1 ¼ 273 þ 15 C ¼ 288 K From Table 13.2, the ratio of specific heats for air is k ¼ 1.4. Calculate the absolute temperature after the compression process using Eq. 17.23.  k1 1:41 P k T2 ¼ T1 2 ¼ ð288 KÞð8Þ 1:4 ¼ 522 K P1 From Table 13.2, the specific heat of air at constant pressure is cp ¼ 1.0 kJ/ kg-K. Calculate the work input to the compressor using Eq. 17.26.  w c ¼ cp ð T 2  T 1 Þ ¼

1:0

 kJ ð522 K  288 KÞ ¼ 234 kJ=kg kg‐K

Calculate the absolute temperature after combustion using Eq. 13.3. T 3 ¼ 273 þ 700 C ¼ 973 K Calculate the absolute temperature of the exhaust gases from the turbine using Eq. 17.25.  T4 ¼ T3

P4 P3

k1 k

 1:41 1 1:4 ¼ ð973 KÞ ¼ 537 K 8

Calculate the work output from the turbine using Eq. 17.27.   kJ ð973 K  537 KÞ ¼ 436 kJ=kg wt ¼ cp ðT 3  T 4 Þ ¼ 1:0 kg‐K B. Calculate the heat added in the combustion chamber using Eq. 17.24.   kJ ð973 K  522 KÞ ¼ 451 kJ=kg qin ¼ cp ðT 3  T 2 Þ ¼ 1:0 kg‐K

17.4

Brayton Cycle

449

C. Calculate the cycle efficiency using Eq. 17.28.

ηth,Brayton ¼

kJ kJ wt  wc 436 kg  234 kg ¼ ¼ 0:4479 ð’ 45%Þ kJ qin 451 kg

Calculate efficiency obtained by using the direct formula (Eq.17.29). 1 1 ηth,Brayton ¼ 1 
k1 ¼ 1  1:41 ¼ 0:4479 ð’ 45%Þ k ð8Þ 1:4 rp Comment: The efficiency obtained by using compressor work, turbine work, and heat input is in perfect agreement with the efficiency obtained by using the direct formula. D. The net (energy) work produced per unit mass of air is: wnet ¼ wt  wc ¼ 436

kJ kJ  234 ¼ 202 kJ=kg kg kg

From the net power produced, calculate the mass flow rate of air required. m_ a ¼

_ net 500 kJs W ¼ ¼ 2:4752 kg=s kJ wnet 202 kg

The pressure and temperature at the compressor inlet are P1 ¼ 101 kPa and T1 ¼ 288 K. Calculate the individual gas constant for air using Eq. 13.16. Rair ¼

kJ 8:314 kmolK R ¼ ¼ 0:2867 kJ=kg  K a1 M air 29 kg kmol

Calculate the volume flow rate of air required using the ideal gas law (Eq. 13.12).  2:4752

m_ R T V_ 1 ¼ air air 1 ¼ P1 ¼ 2:023 m3 =s Note: kJ/kPa ¼ kN.m/(kN/m2) ¼ m3

kg s



 kJ ð288 KÞ kg  K 101 kPa 0:2867

450

17 Thermodynamic Cycles

17.4.2 Brayton Cycle with Regeneration The exhaust gases leaving the turbine in Brayton cycle have significant amount of thermal energy. This heat is recovered in a regenerator, where the compressed air is preheated before combustion. The schematic diagram and the T – s diagram for a regenerative Brayton cycle are shown in Fig. 17.10. As shown in Fig. 17.10, the exhaust gases from the turbine exchange heat with the compressed air in the regenerator. As a result, the compressed air is preheated from state 2 to state 3 before undergoing combustion. Consequently, the heat to be supplied in the combustion chamber decreases, thereby reducing the fuel consumption and increasing the efficiency of the cycle. In the T – s diagram, the state point 30 represents the maximum possible or ideal regeneration. However, since T3' and T5 are equal, heat exchange is not possible due to zero temperature difference. Hence, T3, the actual temperature after preheating or regeneration will always be less than T30 , the ideal temperature after regeneration. The regenerator effectiveness, ε, is defined as the actual heat transfer in the regenerator divided by the maximum possible heat transfer [4]. ε¼

qact h  h2 T  T2 ¼ 3 ¼ 3 qmax h30  h2 T 30  T 2

ð17:30Þ

Since T30 and T5 are equal, the regenerator effectiveness can also be calculated using the following equation. ε¼

qact h  h2 T 3  T 2 ¼ 3 ¼ qmax h5  h2 T 5  T 2

Fig. 17.10 Brayton cycle with regeneration: (a) schematic, (b) T – s diagram

ð17:31Þ

17.4

Brayton Cycle

451

Example 17.9 A regenerator with an effectiveness of 80% is added to the Brayton cycle given in Example 17.8. Determine: A. The new heat input required in the combustion chamber B. The percentage increase in the cycle efficiency due to the use of the regenerator (Solution) A. From the solution to Example 17.8, the absolute temperature after compression (process 1–2) is T2 ¼ 522 K From the solution to Example 17.8, the absolute temperature after combustion is (note that the nomenclature has changed from T3 to T4 due to regeneration) T4 ¼ 273  + 700  C ¼ 973 K. From the solution to Example 17.8, the absolute temperature of the exhaust gases from the turbine is T5 ¼ 537 K. Using the regenerator effectiveness and Eq. 17.31, calculate the temperature (T3) after regeneration. T3  T2 T5  T2 ) T 3 ¼ T 2 þ 0:80ðT 5  T 2 Þ ¼ 522 K þ 0:80ð537 K  522 KÞ ¼ 534 K ε ¼ 0:80 ¼

With regeneration, the heating in the combustion chamber occurs from state point 3 to state point 4. Calculate the new heat input in the combustion chamber as shown. qin,regen

  kJ ð973 K  534 KÞ ¼ 439 kJ=kg ¼ cp ðT 4  T 3 Þ ¼ 1:0 kg‐K

B. The work input to the compressor and the work produced by the turbine will not be affected by the use of the regenerator. From the solution to Example 17.8, the work input to the compressor and the work produced by the turbine are: wc ¼ 234 kJ=kg and wt ¼ 436 kJ=kg: Calculate the new cycle efficiency using Eq. 17.28. ηth,Brayton,regen ¼

kJ kJ wt  wc 436 kg  234 kg ¼ ¼ 0:4601 ð’ 46%Þ kJ qin 439 kg

Calculate the percentage increase in the efficiency due to regeneration as shown.  %increase in efficiency ¼

 0:4601  0:4497 100 ¼ 2:31% 0:4497

452

17 Thermodynamic Cycles

Fig. 17.11 Combined cycle schematic

17.5

Combined Cycle

As the name suggests, the combined cycle is a combination of gas-powered Brayton cycle and steam-powered Rankine cycle [4, 6, 8]. The key component in the combined cycle is the heat recovery steam generator (HRSG), which recovers the waste heat from the exhaust gases of the Brayton cycle to produce steam for the Rankine cycle. The schematic diagram for a combined cycle is shown in Fig. 17.11. The mass flow rate of steam is m_ s and the mass flow rate of gas is m_ g . The ratio between the mass flow rate of steam and the mass flow rate of gas can be obtained by performing an energy balance around HRSG. Rate of energy in ¼ Rate of energy out m_ s h6 þ m_ g h4 ¼ m_ g h5 þ m_ s h7 ) m_ s ðh7  h6 Þ ¼ m_ g ðh4  h5 Þ m_ s h4  h5 ¼ m_ g h7  h6

ð17:32Þ

Example 17.10 The gas turbine in a combined gas-steam power cycle has a pressure ratio of 7. Air enters the compressor at 20  C and 100 kPa. The hot gases enter the gas turbine at

17.5

Combined Cycle

453

800  C, and the exhaust gases leave the HRSG at 130  C. The steam enters the steam turbine at 500  C and 10 MPa, and the condenser pressure is 50 kPa. Determine: A. B. C. D.

The heat added in the gas power cycle The net-work produced by the gas power cycle The net-work produced by the steam power cycle The ratio of mass flow rate of steam to the mass flow rate of the combustion gases E. The mass flow rate of steam required for the steam cycle to produce 3 MW F. The power produced by the gas turbine cycle G. The thermal efficiency of the combined cycle (Solution) The subscript “g” is used for the gas cycle, and the subscript “s” is used for the steam cycle. A. Calculate the absolute temperature of inlet air to the compressor using Eq. 13.3. T 1 ¼ 273 þ 20 C ¼ 293 K From Table 13.2, the ratio of specific heats for air is k ¼ 1.4. Calculate the absolute temperature after the compression process using Eq. 17.23.  k1 1:41 P k T2 ¼ T1 2 ¼ ð293 KÞð7Þ 1:4 ¼ 511 K P1 The hot gases leave the combustion chamber and enter the gas turbine at 800  C. Calculate the absolute temperature after combustion using Eq. 13.3. T 3 ¼ 273 þ 800 C ¼ 1073 K From Table 13.2, the specific heat of air at constant pressure is cp ¼ 1.0 kJ/kg.K. Calculate the heat added in the combustion chamber of the gas power cycle using Eq. 17.24.  qin,g ¼ cp ðT 3  T 2 Þ ¼

 kJ ð1073 K  511 KÞ ¼ 562 kJ=kg 1:0 kg‐K

B. Calculate the work input to the compressor using Eq. 17.26.  wc,g ¼ cp ðT 2  T 1 Þ ¼

1:0

 kJ ð511 K  293 KÞ ¼ 218 kJ=kg kg‐K

454

17 Thermodynamic Cycles

Calculate the absolute temperature of the exhaust gases from the gas turbine using Eq. 17.25.  T4 ¼ T3

P4 P3

k1 k

 1:41 1 1:4 ¼ ð1073 KÞ ¼ 615 K 7

Calculate the work output from the gas turbine using Eq.17.27.  wt,g ¼ cp ðT 3  T 4 Þ ¼

 kJ ð1073 K  615 KÞ ¼ 458 kJ=kg 1:0 kg‐K

Calculate the net (energy) work produced per unit mass of gas in the gas cycle by subtracting the compressor work from the work produced by the gas turbine. wnet,g ¼ wt,g  wc,g ¼ 458

kJ kJ  218 ¼ 240 kJ=kg kg kg

C. Determine the enthalpies of steam at state points 7 and 8 using the Mollier diagram as shown.

17.5

Combined Cycle

455

Note that in the preceding diagram, the pressures are given in bars and 1 bar ¼ 100 kPa, therefore, 10 MPa ¼ 10000 kPa ¼ 100 bar and 50 kPa ¼ 0.5 bar. h7 ð10 MPa, 500 CÞ ¼ 3390 kJ=kg, h8 ð50 kPa, s8 ¼ s7 Þ ¼ 2290 kJ=kg At state point 9, the steam has condensed, hence: h9 ¼ hf at 50 kPa ¼ 340 kJ=kg Calculate the enthalpy at state point 6 by first calculating the work input to the pump using Eq. 14.13 (similar to the solution shown in Practice Problem 14.7) and then adding the pump work to the enthalpy at state point 9. wpump,s ¼ ν9 ðP6  P9 Þ ¼ νf,50kPa ðP6  P9 Þ   m3 ð10000 kPa  50 kPaÞ ¼ 0:0011 kg ¼ 10:95 kJ=kg ð’ 11 kJ=kgÞ 3

3

kNm kJ Note: mkg  kPa  mkg  kN m2  kg  kg

h6 ¼ h9 þ wpump,s ¼ 340

kJ kJ þ 11 ¼ 351 kJ=kg kg kg

Calculate the work produced by the turbine in the steam cycle by applying Eq. 17.12 to state points 7 and 8. wt,s ¼ h7  h8 ¼ 3390

kJ kJ  2290 ¼ 1100 kJ=kg kg kg

Calculate the net (energy) work produced by the steam cycle by subtracting the pump work from the turbine work. wnet,s ¼ wt,s  wpump,s ¼ 1100

kJ kJ  11 ¼ 1089 kJ=kg kg kg

D. Note that mass flow rates are different in each cycle and the preceding equations per unit mass are based on the mass flow rates of the respective cycles. Calculate the absolute temperature (T5) of the exhaust gases leaving the HRSG using Eq. 13.3.

456

17 Thermodynamic Cycles

T 5 ¼ 273 þ 130 C ¼ 403 K Calculate the ratio of mass flow rate of steam to the mass flow rate of the combustion gases using Eq. 17.32. kJ Þð615 K  403 KÞ kg  K kJ kJ 3390  351 kg kg ¼ 0:0698 kg steam=kg gas

m_ s h4  h5 cpg ðT 4  T 5 Þ ¼ ¼ ¼ m_ g h7  h6 h7  h 6

ð1:0

E. Calculate the mass flow rate of steam required for the steam cycle to produce 3 MW using Eq. 17.8.

m_ s ¼

_ net,s 3000 kW W ¼ ¼ 2:775 kg steam=s kJ wnet,s 1089 kg

F. Calculate the mass flow rate of the gas in the gas cycle using the results from the preceding parts (D and E). kg steam m_ s ¼ 0:0698 kg gas m_ g kg steam 2:775 m_ s s ¼ ) m_ g ¼ kg steam kg steam 0:0698 0:0698 kg gas kg gas ¼ 39:76 kg gas=s Calculate the power produced by the gas turbine cycle by multiplying the net (energy) work produced per unit mass by the mass flow rate of the gas. _ net,g ¼ m_ g wnet,g ¼ ð39:76 kg gasÞð240 kJ Þ ¼ 9542 kW ð9:452 MWÞ W s kg G. The only heat input to the combined cycle is in the combustion chamber. Calculate the thermal efficiency of the combined cycle by dividing the total power produced by the rate of heat input to the combustion chamber of the gas cycle.

17.6

Cogeneration Power Plants: Combined Heat and Power

ηcombined ¼

17.6

457

_ gþW _s _ total W W 9542 kW þ 3000 kW ¼ ¼ _Qin,g kg kJ ðm_ g Þðqin,g Þ ð39:76 Þð562 Þ s kg ¼ 0:5613 ð56:13%Þ

Cogeneration Power Plants: Combined Heat and Power

Cogeneration power plants produce utilities such as hot water and steam in addition to generation of power, hence the name “cogeneration.” Cogeneration power plants are also called combined heat and power (CHP) units [1, 11, 13]. The utilities produced by cogeneration power plants can be used for process heating in refineries and in chemical processing facilities. They can also be used for district heating of residential and commercial buildings. Cogeneration power plants make better use of the heat input to the boiler compared to conventional power plants. For example, if a cogeneration facility uses all the steam produced for process heating, there is no heat loss due to condensation of steam that occurs in conventional power plants. In addition, cogeneration facilities have the flexibility to meet fluctuating demands of utilities and power. In a cogeneration cycle, typically, after one stage of expansion in the turbine, some of the steam is diverted to produce heating utilities. Fluctuating demands for heat and power can be easily fulfilled by diverting the required quantity of the steam produced in the boiler for heating utilities or power depending on the needs. Cogeneration also provides a decentralized approach to meet local needs of heat and power by locating the facility close to the place of use. Figure 17.12 illustrates the difference between convention power plants and cogeneration power plants. The performance of cogeneration units is measured by the utilization factor, εu. The utilization factor [1, 2, 4, 11, 13] is a measure of the fraction of heat input being used for useful purposes such as power generation and process heating, and it can be calculated using Eq. 17.33. εu ¼

_ net þ Q_ h W Q_ in

ð17:33Þ

The calculations associated with cogeneration plants are illustrated in Example 17.11. Example 17.11 The boiler in a cogeneration facility produces 40 lbm/sec of superheated steam at 2000 psia and 800  F. After the steam expands through the first stage of the turbine to a pressure of 200 psia, 60% of the steam is extracted for process heating, while the

458

17 Thermodynamic Cycles

Fig. 17.12 Comparison of cogeneration and conventional power plants

remaining 40% expands to 10 psia. The steam leaves the process heat generator as a saturated liquid at 200 psia. The work inputs to the pumps, being relatively small compared to other energy terms, can be ignored. Determine: A. B. C. D.

The schematic diagram with appropriate numbering of the state points The net power produced The rate of process heat produced The utilization factor

(Solution) A.

17.6

Cogeneration Power Plants: Combined Heat and Power

459

B. Determine the enthalpies at state points 3, 4, and 6 using the Mollier diagram as illustrated.

h3 ð2000 psia, 800 FÞ ¼ 1335 Btu=lbm h4 ð200 psia, s3 ¼ s4 Þ ¼ 1115 Btu=lbm h6 ð10 psia, s6 ¼ s4 Þ ¼ 915 Btu=lbm The fraction of steam extracted for process heating is 0.60, and the remaining fraction (0.40) goes through the second stage of expansion. The total rate of steam produced is 40 lbm/sec. Since the work input to the pumps can be neglected, only the work produced by the high-pressure (HP) and low-pressure (LP) turbines need to be considered. Multiply the work terms by the appropriate mass flow rates to obtain the net power produced. _ net ¼ m_ HP wHP þ m_ LP wLP ¼ m_ HP ðh3  h4 Þ þ m_ LP ðh4  h6 Þ W    lbm Btu Btu ¼ 40 1335  1115 sec lbm lbm    lbm Btu Btu þ 0:40  40 1115  915 sec lbm lbm ¼ 12, 000 Btu= sec

460

17 Thermodynamic Cycles

C. In the utility heat generator, the extraction steam at state point 4, with enthalpy h4 ¼ 1115 Btu/lbm, condenses to saturated liquid at 200 psia (state point 5) to provide the heat required for process heating. From steam tables, h5 ¼ hf at 200 psia ¼ 355 Btu=lbm: The mass flow rate of the extraction steam is 60% of the rate of generation. The rate of production of process heat is the mass flow rate of extraction steam multiplied by the difference in enthalpy as it condenses.    lbm Btu Btu 1115  355 0:6  40 sec lbm lbm ¼ 18, 240 Btu= sec

Q_ h ¼ ðm_ ext Þðh4  h5 Þ ¼

D. Since the work inputs to pumps are negligible, h8 ’ h7 ¼ hf at 10 psia ¼ 161 Btu=lbm The enthalpy at state 9 can be obtained by energy balance around the mixer. Note that the mass flow rate of the extract steam is: m_ ext ¼ 0:6  40

lbm ¼ 24 lbm= sec sec

The mass flow rate of the steam condensed and pumped to the mixer is the difference between the total steam flow rate and the extract steam flow rate. m_ cond ¼ m_ s  m_ ext ¼ 40

lbm lbm  24 ¼ 16 lbm= sec sec sec

From the schematic diagram generated as solution to part A, the streams coming into the mixer are stream 5 and stream 8, and the stream leaving the mixer is stream 9. Energy balance for the mixer results in the following set of equations. Rate of energy into the mixer ¼ Rate of energy out of the mixer m_ ext h5 þ m_ cond h8 ¼ m_ s h9 Solve the preceding equation for the enthalpy at state point 9, and substitute the known values to obtain h9.

17.7

Otto Cycle

461

      lbm Btu lbm Btu 355 þ 16 161 24 sec lbm sec lbm m_ h þ m_ cond h8   h9 ¼ ext 5 ¼ m_ s lbm 40 sec ¼ 277 Btu=lbm Since the work inputs to pumps are negligible: h1 ’ h9 ¼ 277 Btu=lbm The enthalpy of steam leaving the boiler is the same as the enthalpy of steam entering the turbine, that is, h2 ¼ h3 ¼ 1335 Btu/lbm Calculate the heat input to the boiler as shown.    lbm Btu Btu 40 1335  277 sec lbm lbm ¼ 42, 320 Btu= sec

Q_ in ¼ m_ s ðh2  h1 Þ ¼

Calculate the utilization factor using Eq. 17.33. εu ¼

Btu _ net þ Q_ h 12000 Btu W sec þ 18240 sec ¼ ¼ 0:7145 ð’ 71%Þ Btu 42320 Q_ in sec

17.7

Otto Cycle

The Otto cycle [4] is used in gasoline engines. In the first step, Otto cycle compresses intake air isentropically. Then, the compressed air undergoes combustion with the fuel at constant volume. At the end of the combustion process, the engine cylinder consists of high-pressure and high-temperature combustion gases with significant amount of energy. In the next step, the combustion gases expand isentropically producing work. In the final step, the cylinder cools back to the starting point to continue the cycle. The temperature-entropy (T -s) diagram and the pressure-volume (P – V) diagram for Otto cycle are shown in Fig. 17.13. A key parameter in Otto cycle is the compression ratio, which is the ratio of the volume of air at the beginning of the compression process to the volume of air at the end of the compression process. The symbol for compression ratio of Otto cycle is r, and it is represented as shown in Eq. 17.34.

462

17 Thermodynamic Cycles

Fig. 17.13 Otto cycle: (a) T – s diagram, (b) P – V diagram

r¼

V1 V2

ð17:34Þ

The efficiency of Otto cycle can be directly determined from the compression ratio using the following equation. ηOtto ¼ 1  r 1k

ð17:35Þ

The work and heat terms in Otto cycle can be obtained by applying the I Law of thermodynamics for closed systems. For example, the I Law for closed systems can be applied to the isentropic compression process (process 1–2) as shown here. Energy added to air ¼ increase in internal energy of air: Since an isentropic process is a reversible adiabatic process, there is no heat loss or gain. Hence, the only energy added to the air is the work input in the compression process. win ¼ ðΔuÞ12 ¼ cv ðΔT Þ12 ¼ cv ðT 2  T 1 Þ

ð17:36Þ

In the preceding equation, the intake air temperature, T1, is usually a known quantity. The temperature of air after the compression process, T2, can be determined by using the P V T relationships for isentropic processes (Eqs. 13.36, 13.37, and 13.38, reproduced here for reference). P2 V 2 k ¼ P1 V 1 k  k1 T2 V1 ¼ T1 V2

ð13:36Þ ð13:37Þ

17.7

Otto Cycle

463

T2 ¼ T1

 k1 P2 k P1

ð13:38Þ

In all the preceding equations, k ¼ 1.4 (ratio of specific heats for air). The heat input to the cycle can be determined by the following equation, which can be derived by using the same procedure that was used in deriving Eq. 17.36. qin ¼ ðΔuÞ23 ¼ cv ðΔT Þ23 ¼ cv ðT 3  T 2 Þ

ð17:37Þ

Similarly, the work output from the cycle can be determined by the following equation. wout ¼ ðΔuÞ34 ¼ cv ðΔT Þ34 ¼ cv ðT 3  T 4 Þ

ð17:38Þ

The efficiency for Otto cycle can also be calculated by using the following formula. ηOtto ¼

wnet wout  win ¼ qin qin

ð17:39Þ

Example 17.12 An Otto cycle has a compression ratio of 7. Air intake is at 20  C and 95 kPa, and the highest temperature achieved in the cycle is 1200  C. Determine A. B. C. D. E.

the pressure and temperature at each point in the cycle the work input to the compressor the heat added the work output the cycle efficiency using the results from parts B, C, and D and compare it with the efficiency calculated using only the compression ratio

(Solution) Since the compression ratio is 7, r p ¼ VV 12 ¼ 7 A. State point 1: Calculate the absolute temperature of intake air using Eq. 13.3. T 1 ¼ 273 þ 20 C ¼ 293 K, P1 ¼ 95 kPa

464

17 Thermodynamic Cycles

State point 2: Calculate the absolute temperature after the isentropic compression process using Eq. 13.37. T2 ¼ T1

 k1 V1 ¼ ðT 1 Þr p k1 ¼ ð293 KÞ7ð1:41Þ ¼ 638 K V2

Calculate the pressure after the isentropic compression process using Eq. 13.36. P2 ¼ P1 ð

V1 k Þ ¼ ð95 kPaÞð7Þ1:4 ¼ 1448 kPa ð’ 1:45 MPaÞ V2

State point 3: The maximum temperature in the cycle occurs at state point 3, which is at 1200  C. Calculate the absolute temperature after combustion using Eq. 13.3 and from the P – V diagram for Otto cycle (Fig. 17.13). T 3 ¼ 273 þ 1200 C ¼ 1473 K Since process 2–3 is a constant volume process, calculate the pressure at state point 3 using Eq. 13.35. P3 ¼ P2

    T3 1473 K ¼ 3343 kPa ð’ 3:34 MPaÞ ¼ ð1448 kPaÞ 638 K T2

State point 4: Apply Eq. 13.37 to the isentropic expansion process from state point 3 to state point 4. Simplify the result, and substitute the known quantities to obtain the absolute temperature at state point 4. Also note from the P – V diagram for Otto cycle (Fig. 17.13), V3 ¼ V2, V4 ¼ V1.  T4 ¼ T3

V3 V4

k1

 k1  ð1:41Þ V 1 ¼ T3 2 ¼ ð1473 KÞ ¼ 676 K 7 V1

Again, since process 4–1 is a constant volume process, calculate the pressure at state point 4 using Eq. 13.35.     T4 676 K ¼ 1534 kPa ¼ ð3343 kPaÞ P4 ¼ P3 1473 K T3

17.8

Vapor Compression Refrigeration Cycle

465

B. From Table 13.2, the specific heat at constant volume for air is cv ¼ 0:72 kJ=kg  K: Calculate the work input in the compression process by using Eq. 17.36.  win ¼ cv ðT 2  T 1 Þ ¼

 kJ ð638 K  293 KÞ ¼ 248:4 kJ=kg 0:72 kg  K

C. Calculate the heat input in the compression process by using Eq. 17.37.

qin ¼ cv ðT 3  T 2 Þ ¼

  kJ ð1473 K  638 KÞ ¼ 601:2 kJ=kg 0:72 kg  K

D. Calculate the work output in the expansion process by using Eq. 17.38.

wout

  kJ ¼ cv ðT 3  T 4 Þ ¼ 0:72 ð1473 K  676 KÞ ¼ 573:8 kJ=kg kg  K

E. Calculate the cycle efficiency using Eq. 17.39. wout  win 573:8 kg  248:4 kg ¼ ¼ 0:54 ð54%Þ kJ qin 601:2 kg kJ

ηOtto ¼

kJ

Calculate the cycle efficiency using only the compression ratio (Eq. 17.35). ηOtto ¼ 1  r 1k ¼ 1  711:4 ¼ 0:54 ð54%Þ Comment: The same efficiency is obtained either by using energy balance-based formulas or the direct calculation formula.

17.8

Vapor Compression Refrigeration Cycle

Vapor compression cycles are widely used in refrigeration and air-conditioning applications, where heat is transferred from a low-temperature environment to surroundings at higher temperature. The objective is to keep the low-temperature environment at a constant, cooler temperature. Vapor compression cycle [3, 4] is a modified version of the reversed Carnot cycle. The modifications result in making

466

17 Thermodynamic Cycles

Fig. 17.14 Comparison of reversed Carnot and vapor compression cycles

the reversed Carnot cycle practically feasible. The explanation for this will follow Fig. 17.14, which compares reversed Carnot cycle and vapor compression cycles. As shown in Fig. 17.14, the compressor in reversed Carnot cycle handles a liquidvapor mixture at state point 1. This can result in damage to compressor blades due to the high velocity impact and collapse of liquid-vapor bubbles on the blade surface. The vapor compression cycle overcomes this problem by compressing “dry” saturated vapor. The term “dry” here refers to the absence of liquid or condensate. The saturated vapor becomes superheated due to the addition of compressor work (energy). The other modification is the isentropic expansion process, 1–2, in the reversed Carnot cycle is replaced by constant enthalpy expansion through a throttling valve, which is a much simpler device compared to an isentropic turbine. Dry, saturated vapor is compressed, hence the terminology “vapor compression cycle.” Figure 17.15 illustrates the schematic, the temperature-entropy (T – s), and the pressure-enthalpy (P – h) diagrams for a typical vapor compression cycle.

17.8.1 Analysis of Vapor Compression Cycle In Fig. 17.15(a), m_ R is the mass flow rate of the refrigerant circulating in the cycle. Commonly used refrigerants are R-22, R-134a, and R-410. Usually, the tons of refrigeration or air-conditioning (cooling) required are specified. This can be converted to the rate of heat absorption from the space in terms of Btu/hr or kW using the conversion factors presented earlier in Eq. 17.6 and reproduced here for reference.

1 ton of refrigeration  12, 000 Btu=hr  3:517 kW

ð17:6Þ

17.8

Vapor Compression Refrigeration Cycle

467

Fig. 17.15 Vapor compression cycle: (a) schematic diagram, (b) T – s diagram, (c) P – h diagram

The heat absorbed in the evaporator per unit mass of refrigerant can be obtained by energy balance around the evaporator. Energy in ¼ Energy out h4 þ qin ¼ h1 ) qin ¼ h1  h4

ð17:40Þ

Rate of heat absorption ¼ Q_ in ¼ m_ R qin

ð17:41Þ

The heat absorbed in the evaporator is used in evaporating the refrigerant from state point 4 (liquid-vapor mixture) to state point 1 (saturated vapor). The work input to the compressor per unit mass of refrigerant can be obtained by energy balance around the compressor. Energy in ¼ Energy out h1 þ wc,ideal ¼ h2 ) wc,ideal ¼ h2  h1 _ c,ideal ¼ m_ R wc,ideal Isentropic power input to the compressor ¼ W

ð17:42Þ ð17:43Þ

The energy supplied by the compressor results in the refrigerant going from saturated vapor (state point 1) to superheated vapor (state point 2). The heat rejected in the condenser per unit mass of refrigerant can be obtained by energy balance around the condenser. Energy in ¼ Energy out h2 ¼ h3 þ qout ) qout ¼ h2  h3

ð17:44Þ

Rate of heat rejection ¼ Q_ out ¼ m_ R qout

ð17:45Þ

When the superheated refrigerant (state point 2) condenses to saturated liquid (state point 3), heat is rejected in the condenser. The enthalpies at each state point can be determined as follows:

468

17 Thermodynamic Cycles

h1 ¼ hg at evaporator pressure h2 ðs2 ¼ s1 Þ at condenser pressure h3 ¼ hf at evaporator pressure Process 3–4 is a constant enthalpy process )h4 ¼ h3

17.8.2 Performance Measures for Vapor Compression Cycles Vapor compression cycles can be used for cooling (refrigeration and air-conditioning) or for heating [4, 7] (heat pump). Depending on the purpose, the performance measures are defined as follows: COPref:=AC ¼ COPheat pump ¼

Q_ in qin h1  h4 ¼ ¼ _ c wc h2  h1 W

ð17:46Þ

Q_ out qout h2  h3 ¼ ¼ _c wc h2  h1 W

ð17:47Þ

Example 17.13 It is necessary to remove heat at the rate of 40 tons to maintain a cold storage at the desired temperature. This is accomplished by using a vapor compression cycle using R-134a as the refrigerant. The evaporator pressure is 300 kPa, and the condenser pressure is 1.5 MPa. Determine: A. The enthalpy of the refrigerant at each state point B. The power input required for the compressor (kW) if the compressor efficiency is 80% C. The COPrefg./AC (Solution) A. Draw the P – h diagram as a reference for state points and for heat/work flow. Also, determine the enthalpies at state points using the P – h diagram for R-134a as shown.

17.8

Vapor Compression Refrigeration Cycle

469

h1 ¼ hg,300 kPa ¼ 249 kJ=kg h2 ¼ h1500 kPa,s2 ¼s1 ¼ 287 kJ=kg h3 ¼ hf,1500 kPa ¼ 125 kJ=kg h4 ¼ h3 ¼ 125 kJ=kg B. Calculate the heat absorption per unit mass of refrigerant using Eq. 17.40. qin ¼ h1  h4 ¼ 249

kJ kJ  125 ¼ 124 kJ=kg kg kg

Using the information of 40 tons of cooling required and the equation for the rate of absorption of heat from the refrigerated space (Eq. 17.41), calculate the mass flow rate of refrigerant. m_ R ¼

kW Q_ in 40 tons  3:517 ton ¼ ¼ 1:134 kg=s kJ qin 124 kg

470

17 Thermodynamic Cycles

Calculate the isentropic power input required for the compressor by using Eq. 17.43, and then divide the result by the isentropic efficiency of the compressor to get the actual power required. _ c,actual ¼ m_ R ðh2  h1 Þ ¼ W ηc


 kJ kJ 1:134 kgs 287 kg  249 kg 0:80

¼ 53:86 kW

C. Calculate the coefficient of performance for refrigeration using Eq. 17.46.

COPrefg: ¼

Q_ in ð40  3:517Þ kW ¼ 2:61 ¼ _c 53:86 kW W

Practice Problems Practice Problem 17.1 A Carnot cycle has an efficiency of 62% and produces 15 hp power. Determine A. The rate of heat input (Btu/hr) required B. The temperature (in  F) of the heat source if the heat sink is at 50  F C. The rate at which heat should be removed from the heat sink Practice Problem 17.2 A Carnot air-conditioner has a COPref./AC of 3.72 and achieves 3 tons of cooling (1 ton of cooling  12,000 Btu/hr). Determine: A. The power input (hp) required B. The temperature of the heat sink ( F) if the temperature is to be maintained at 70  F in the air-conditioned space C. The rate at which heat is rejected to the heat sink in Btu/hr Practice Problem 17.3 The turbine and pump in Example 17.4 have efficiencies of 90% and 80%, respectively. The power produced will remain the same as in Example 17.4. Using the fuel and boiler efficiency data given in Example 17.5, calculate the percentage increase in coal required per hour. Hint: Refer to isentropic efficiencies for turbines and compressors in Chap. 14. The consequence of isentropic efficiency of pump is similar to the impact of isentropic efficiency in a compressor.

Solutions to Practice Problems

471

Practice Problem 17.4 An Otto cycle has an efficiency of 55% and takes in air at 14 psia, 70  F. The volume at the start of the compression process is 3 liters. Determine the volume, pressure, and temperature after compression process. Practice Problem 17.5 A vapor compression cycle is used in heating 1700 cfm of atmospheric air from 40 to 70  F. The refrigerant, R-134a, is compressed from 30 to 250 psia. Determine: A. The enthalpy of the refrigerant at each state point B. The isentropic horsepower required for the compressor C. The COPheat pump

Solutions to Practice Problems Practice Problem 17.1 (Solution) A. Calculate the rate of heat input required using the first part of Eq. 17.2. Q_ H ¼

_ W ηth,Carnot

  15 hp 2544 Btu=hr  ¼ ¼ 61, 548 Btu=hr 0:62 1 hp

B. Calculate the absolute temperature of the heat sink using Eq. 13.2. T L ¼ 50 F þ 460 ¼ 510 R Substitute the known quantities into the second part of Eq. 17.2, and solve for TH. ηth,Carnot ¼ 0:62 ¼

T H  T L T H  510 R ¼ ) T H ¼ 1342 R TH TH

Convert the absolute temperature to degree Fahrenheit using Eq. 13.2. T H ¼ 1342 R  460 ¼ 882 F C. Calculate the rate at which heat should be removed from the heat sink using the middle part of Eq. 13.2, and substitute the known values.

472

17 Thermodynamic Cycles

Btu _ Q_ H  Q_ L 61548 hr  QL ¼ Btu Q_ H 61548 hr ) Q_ L ¼ 23, 388 Btu=hr

ηth,Carnot ¼ 0:62 ¼

Practice Problem 17.2 (Solution) A. Calculate the power input required using the first part of Eq. 17.4.    12000 Btu=hr 0:000393 hp ð3 tonsÞ ton Btu=hr Q_ L _ ¼ W ¼ 3:72 COPrefg:=AC ¼ 3:80 hp B. Convert the temperature of the air-conditioned space to absolute value using Eq. 13.2. T L ¼ 70 F þ 460 ¼ 530 R Use the last part of Eq. 17.4, and substitute the known values to obtain the absolute temperature of the heat sink. COPrefg:=AC ¼ TH ¼

TL ) TH  TL

TL 530∘ R þ TL ¼ þ 530∘ R ¼ 672:5∘ R 3:72 COPrefg:=AC

Convert the absolute temperature of the heat sink to  F using Eq. 13.2. T H ¼ 672:5 R  460 ¼ 212:5 F C. Calculate the rate of heat rejection to the heat sink using the energy balance equation for the reversed Carnot engine. The rate of work input in Btu/hr from the solution to part(A) is:   ð3 tonsÞ 12000tonBtu=hr _QL _ ¼ W ¼ ¼ 9677 Btu=hr COPrefg:=AC 3:72 _ ¼ 36000 Btu þ 9677 Btu ¼ 45, 677 Btu=hr Q_ H ¼ Q_ L þ W hr hr

Solutions to Practice Problems

473

Practice Problem 17.3 (Solution) From the solution to Example 17.4, the following information is available: wp ¼ 1:50 Btu=lbm, wt ¼ 275 Btu=lbm, and _ net ¼ 5:21 MW W Calculate the actual turbine work produced using Eq. 14.13.   Btu wt,actual ¼ ηt  wt,ideal ¼ ð0:90Þ 275 ¼ 247:5 Btu=lbm lbm Calculate the actual work consumed by the pump using Eq. 14.14 (same equation can be applied to both pumps and compressors). wp,actual ¼

Btu wp,ideal 1:50 lbm ¼ ¼ 1:88 Btu=lbm ηp 0:80

Calculate the actual net (energy) work produced. wnet,actual ¼ wt,actual  wp,actual ¼ 247:5

Btu Btu  1:88 ¼ 245:62 Btu=lbm lbm lbm

Using Eq. 17.8, calculate the new mass flow rate of steam required due to turbine and pump efficiencies. The power output remains the same as before. m_ s,actual ¼

_ net W wnet,actual

3412142Btu

¼

5:21 MW  1 MW hr ¼ 72, 377 lbm=hr Btu 245:62 lbm

Using Eq. 14.14 (same equation can be applied to pumps and compressors), calculate the enthalpy of water leaving the pump since it would have changed due to the efficiency of the pump. From the solution to Example 17.4, the enthalpy of water entering the pump is h1 ¼ 180.18 Btu/lbm. The actual enthalpy of water leaving the pump (same as the enthalpy of water entering the boiler) can be obtained by adding actual pump work to the enthalpy of water entering the pump. h20 ¼ 180:18

Btu Btu þ 1:88 ¼ 182:06 Btu=lbm lbm lbm

From the solution to Example 17.4, the enthalpy of steam leaving the boiler is h3 ¼ 1290 Btu/lbm. Calculate the actual heat input to the boiler per unit mass of steam using Eq. 17.11.

474

17 Thermodynamic Cycles

qb ¼ h3  h20 ¼ 1290

Btu Btu  182:06 ¼ 1107:94 Btu=lbm lbm lbm

The preceding result is almost the same as that was obtained assuming ideal (or isentropic) pump. Hence, we can safely neglect the effect of pump efficiency on the heat input to the boiler. Calculate the actual rate of heat input to the boiler using Eq. 17.9. Q_ b,actual ¼ m_ s,actual qb,actual ¼



lbm 72377 hr

  Btu 1107:64 lbm

¼ 8:036  107 Btu=hr Calculate the actual mass rate of fuel required by using the net heating value of the fuel and boiler efficiency, same as in Example 17.4. m_ fuel,actual ¼

8:036  107 Btu Q_ b,actual hr ¼ ¼ 7749 lbm=hr Btu NHVfuel  ηb 12200 lbm  0:85

Calculate the percentage increase in the fuel requirement per hour. %increase ¼

! lbm 7749 lbm  6947 hr hr 100 ¼ 11:54% 6947 lbm hr

Practice Problem 17.4 (Solution) Calculate the compression ratio using Eq. 17.35. 1

1

ηOtto ¼ 1  r 1k ) r ¼ ð1  ηOtto Þ1k ¼ ð1  0:55Þ11:4 ¼ 7:36 Calculate the volume after the compression process using the definition of compression ratio (Eq. 17.34). r¼

V1 V 3L ) V2 ¼ 1 ¼ ¼ 0:41 L 7:36 V2 r

Calculate the pressure after the isentropic compression process using Eq. 17.36. P2 ¼ P1

 k V1 ¼ ð14 psiaÞð7:36Þ1:4 ¼ 229 psia V2

Calculate the absolute temperature of intake air using Eq. 13.2.

Solutions to Practice Problems

475

T 1 ¼ 460 þ 70 F ¼ 530 R Calculate the absolute temperature after the isentropic compression process using Eq. 13.37.  T2 ¼ T1

V1 V2

k1

¼ ðT 1 Þr p k1 ¼ ð530∘ RÞ7:361:41 ¼ 1178∘ R ð718∘ FÞ

Practice Problem 17.5 (Solution) A. Draw the P – h diagram as a reference for state points and for heat/work flow. Also, determine the enthalpies at state points using the P – h diagram for R-134a as shown.

From the preceding illustration,

476

17 Thermodynamic Cycles

h1 ¼ 104 Btu=lbm h2 ¼ 120 Btu=lbm h4 ¼ h3 ¼ 55 Btu=lbm B. Obtain the heating load of the heat pump by calculating the rate of heating required to heat the air. Calculate the individual gas constant for air using Eq. 13.16. psia‐ft3

Rair ¼

10:73 lbmol‐ R R ¼ ¼ 0:37 psia‐ft3 =lbm‐ R lbm M air 29 lbmol

Calculate the average temperature of air and its absolute value using Eq. 13.2. Tavg ¼ (40  F + 70  F)/2 ¼ 55  F ) Tair ¼ 460  + 55  F ¼ 515  R Calculate the density of air at its average temperature using the ideal gas law (Eq. 13.12). ρ¼

m P 14:7 psia  ¼ ¼ 0:0771 lbm=ft3 ¼ psia‐ft3 V Rair T air  0:37 lbm‐ ð R Þ 515 R

Calculate the mass flow rate of air using the continuity equation (Eq. 3.5). m_ air ¼ ρair V_ air ¼

 0:0771

lbm ft3

  ft3 ¼ 131:1 lbm= min 1700 min

From Table 13.2, the specific heat of air at constant pressure is 0.24 Btu/lbm- R Energy balance at the high-temperature heat sink results in the following set of equations: Rate of heat rejected at the heat sink ¼ Rate of heat absorption by air    _Qout ¼ m_ air cp,air ΔT air ¼ 131:1 lbm 0:24 Btu ð70 F  40 FÞ min lbm‐ R ¼ 943:9 Btu= min Calculate the heat rejection per unit mass of refrigerant using Eq. 17.44. qout ¼ h2  h3 ¼ 120

Btu Btu  55 ¼ 65 Btu=lbm lbm lbm

Using the information of heating load required and the equation for the rate of heat rejection from the condenser (Eq. 17.45), calculate the mass flow rate of refrigerant.

References

477

m_ R ¼

Btu Q_ out 943:9 min ¼ ¼ 14:52 lbm= min Btu qout 65 lbm

Calculate the isentropic horsepower input required for the compressor by using Eq. 17.43.    lbm Btu Btu 120  104  14:52 min lbm lbm   0:0236 hp Btu= min ¼ 5:48 hp

_ c,ideal ¼ m_ R ðh2  h1 Þ ¼ W

C. Calculate the coefficient of performance for the heat pump using Eq. 17.47.

COPheat pump ¼

Btu 943:9 min Q_ out 
 ¼ 4:06 ¼
lbm Btu Btu _c 14:52 min 120 lbm  104 lbm W

References 1. Aydin, D.: Cogeneration/Combined Heat and Power. Department of Mechanical Engineering, Eastern Mediterranean University, Cyprus. Download from https://staff.emu.edu.tr/ devrimaydin/documents/meng446/chapter%205%20combined%20heat%20and%20power.pdf (2017) 2. Bureau of Energy Efficiency: Cogeneration with Case Study, India. Download from https:// www.beeindia.gov.in/sites/default/files/2Ch7.pdf (2005) 3. Burghardt, D.M., Harbach, J.A.: Engineering Thermodynamics, 4th edn. Tidewater Publications, New York (1999) 4. Cengel, Y., Boles, M.: Thermodynamics – An Engineering Approach, 9th edn. McGraw Hill, New York (2019) 5. Davis, B.: What Are the Disadvantages of Reheating? MvOrganizing.org Knowledge Bank, USA. Download from https://www.mvorganizing.org/what-are-the-disadvantages-ofreheating/ (2021) 6. Energy Efficiency Solutions: Combined Cycle Gas Turbines, International Petroleum Indus try Environmental Conservation Association (IPIECA), London. Download from https://www. ipieca.org/resources/energy-efficiency-solutions/power-and-heat-generation/combined-cyclegas-turbines/ (2020) 7. Energy Saver: Heat Pump Systems. U.S. Department of Energy, USA. Download from https:// www.energy.gov/energysaver/heat-pump-systems (2021) 8. GE Education Resources: Combined Cycle Power Plants, USA. Download from https://www. ge.com/gas-power/resources/education/combined-cycle-power-plants (2021) 9. Hall, N., NASA Glenn Research Center, Ideal Carnot Cycle. Download from https://www.grc. nasa.gov/www/k-12/airplane/carnot.html (2021)

478

17 Thermodynamic Cycles

10. Kumar, D., Vijayakumar, C.: Performance analysis of regenerative feed heating in a steam power plant. IOSR J. Mech. Civil Eng. (IOSR-JMCE). 11(2), 01–08 (2014) 11. Office of Energy Efficiency and Renewable Energy: Combined Heat and Power Basics. Department of Energy, USA. Download from https://www.energy.gov/eere/amo/combinedheat-and-power-basics (2018) 12. Rudra, S.S.: Reversed Carnot Cycle, Process, COP, Limitations, and Applications, Mechanical Basics, USA. Download from https://mechanicalbasics.com/reversed-carnot-cycle-processcop-limitations/ (2021) 13. Scientific American Article: How Does Cogeneration Provide Heat and Power? Scientific American, USA. Download from https://www.scientificamerican.com/article/how-does-cogen eration-provide-heat-power/ (2008) 14. Urone, P: Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated. University of British Columbia, College Physics Open Text, Canada. Download from https:// opentextbc.ca/openstaxcollegephysics/chapter/carnots-perfect-heat-engine-the-second-law-ofthermodynamics-restated/ (2012)

Index

A Absolute pressure, 16, 17, 28 Absolute temperature, 236, 374, 391, 396, 397 Absorptivity, 230 Actual cubic feet (ACF), 304 Actual heating value (AHV), 404 Adiabatic saturation process, 388 Affinity laws, 121 Air-conditioning process, 381 cooling and dehumidification, 381, 383 heating and humidification, 384, 385, 387 sensible heating process, 386 Air-fuel ratio, 408 Air pressure, 17, 18, 28–30 Air-water vapor mixture air-conditioning, 375 dew point temperature, 375 dry bulb temperature, 375 humidity ratio, 375–377 psychrometrics, 377, 379 relative humidity, 375, 376 wet bulb temperature, 375 Amagat’s law, 373 Atmospheric pressure, 15–17

B Bernoulli’s equation, 112, 137, 139, 141 Black body, 228–230, 233, 240 Boundary layer theory, 69–71 Boyles Law, 303 Brake horse power (BHP), 107, 126 Brayton cycle, 445 analysis of, 446–449

with regeneration, 450, 451 Buckingham Pi theorem, 173–175, 179 Bulk temperature, 205

C Carnot cycle, 424 second law of, 426 thermal efficiency, 425, 426 Carnot engine, 424 Cavitation of pumps, 123 Center of pressure (CP), 24, 26, 27, 33, 34 Centrifugal pump, 111, 113, 127 Charles law, 303, 310, 360 Chemical dehumidification, 387 Chilton-Colburn analogy, 224 Closed system, 293 Coal, 403 Coefficient of contraction, 140 Coefficient of performance (COP) reversed Carnot cycle, 428–430 Coefficient of velocity, 140 Cogeneration power plants, 457–461 Colebrook equation, 50, 52 Combined cycle thermodynamic cycles, 452–456 Combined heat and power (CHP) thermodynamic cycles, 457, 459–461 Combustion actual, 410–412 air, 406 atomic balances, 411 coal, 413, 414 definition, 403, 404

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2022 N. S. Nandagopal, PE, Fluid and Thermal Sciences, https://doi.org/10.1007/978-3-030-93940-3

479

480 Combustion (cont.) dew point, 416, 417, 421 excess air, 408 excess oxygen, 409, 411, 414 methane, 406 Orsat analysis, 411, 412 percent excess air, 411 products-flue gas analysis, 409, 411 stoichiometric coefficients, 419, 420 stoichiometry, 405 Combustion of propane (C3H8), 410 Compressed liquid properties of, 326, 327 Compressibility factor, 308 Compressible flow adiabatic compressible flow with friction loss, 161, 163 calculation of density of gases, 150, 151 impact of high-speed winds, 149 isentropic gas flow stagnation-static relationships, 155, 156 steady flow energy equation, 154 with area changes, 157–159 Mach Number, 150–153 Compression ratio, 461 Compressors, 351 Condensation of moisture, 375, 383 Conduction-convection systems, 186, 187 Conduction heat transfer, 184 Configuration factor, 231 Conservation of mass, 35, 36 Constant enthalpy, 314, 468 Continuity equation, 35, 36, 86, 90, 91, 104, 109, 114, 119, 131, 132, 139 Control volume (CV), 102, 345 Convection heat transfer balance heat, 223 coefficients, 207–211 conduction cooler ambient, 212 heat flow, 213 order of magnitude analysis, 215, 216 transfer coefficients, 214 window, 214, 215 cooling water, 219 dimensionless parameters, 207, 208 fluid flow/heat transfer colburn analogy, 217, 218 Pr, 217 free convection and forced convection, 207 overall resistance, 216 resistance, 205, 206, 220, 222 typical range, 212

Index values, 224 Conversion factors, 1, 6–9 Cooling, 381, 383 Cooling towers, 388, 389, 391 Current flow, 183

D Dalton’s law, 373, 375 Darcy equation, 44, 45, 55, 96, 102, 105, 114, 128, 132 Darcy friction factor, 108, 161 Dehumidification, 381 Dehumidification and cooling, 375 Density, 3 Dew point, 375, 379, 397, 400 Differential manometers, 19, 21, 137, 139, 145 Differential pressure gage, 137, 139, 143, 145, 146 Diffusers, 354 Dimensionless parameter, 179, 181 benefits, 173 Buckingham Pi theorem, 173–175 Discharge coefficient, 140 Discharge pressure of pump, 104, 106 Dittus-Boelter equation, 221 Double pipe heat exchanger, 247, 248, 263 Drag coefficient, 180 Drag force, 71, 72 coefficient, 72–75 terminal velocity, 76, 77 Dry bulb temperature, 375, 377, 394, 397 Dynamic similarity, 176, 177, 180 Dynamic viscosity, 6, 7

E Electromagnetic radiation, 227 Emissivity, 229, 230 Energy balance for compressor, 346, 347, 364 Energy balance for diffuser, 366, 367 Energy balance for heat exchangers, 353, 354 Energy balance for nozzles, 354, 356, 357 Energy balance for open systems, 345 Energy balance for pumps, 358 Energy balance for turbines, 346, 347, 363 Energy equation affinity laws, 121, 122 Bernoulli’s equation, 112 cavitation of pumps, 123 forms, 101 mechanical energy equation, 102, 103 NPSH, 124

Index pump power, 106, 107, 109, 111, 127 (see also Pump power equation) specific weight, 129 Energy grade line (EGL), 102 Energy per unit mass, 102, 103 Energy per unit weight, 101, 102 Enthalpy change thermodynamics fundamentals, 318 Enthalpy of moist air, 395, 396 Entropy change thermodynamics fundamentals, 318–321 Equations of state, 306 Equilibrium in radiation systems, 235 Equivalent length method, 57–61 Ethylene glycol, 87, 88

F Fan laws, see Affinity laws Fanning friction factor, 45, 46 Feedwater heater (FWH), 267, 438 Film coefficient or film conductance, 205 First law for closed systems, 342 First Law for nozzles, 354, 356, 357 First Law for pumps, 358 First law of thermodynamics, 341 Flow through pipes in parallel, 66–69 through pipes in series, 64–66 Fluid dynamics, 35 boundary layer theory, 69–71 conservation of mass, 35, 36 drag force, 71, 72 drag coefficient, 72–75 terminal velocity, 76, 77 fluid flow, friction head loss and pressure drop for, 42–44 Darcy equation, 44, 45 Fanning friction factor, 45, 46 Hagen–Poiseuille equation, 46–48 Moody diagram, 48–52 impulse-momentum principle, 78–83 lift force, 77, 78 noncircular cross sections, flow through, 52–56 pipe fittings and valves, friction head loss across, 56 equivalent length method, 57–61 head loss at pipe entrance and at pipe exit, 61, 62 pipe cross section, head loss due to, 62, 63 velocity head method, 56, 57

481 pipes in parallel, flow through, 66–69 pipes in series, flow through, 64–66 solutions, to practice problems, 86–99 standard pipe sizes and nomenclature, 37–40 laminar and turbulent flow, 42 Reynolds number, 41 Fluid flow friction head loss and pressure drop for, 42–44 Darcy equation, 44, 45 Fanning friction factor, 45, 46 Hagen–Poiseuille equation, 46–48 Moody diagram, 48–52 Fluid flow measurements orifice meter, 139, 140, 144 permanent pressure loss in orifice meter, 144 pitot tube, 137, 138 venturi meter, 141–144 Fluid properties and units density, 3 specific gravity, 4, 5 specific weight, 3 systems of units, 1 units of force, 2 viscosity of fluid dynamic viscosity, 6, 7 kinematic viscosity, 7–9 Fluid statics differential manometers, 19 forces on submerged surfaces, 23 inclined manometers, 21 pressure due to a fluid column, 15, 16 Force, 171 Forced convection, 185 Fouling factor, 254–256, 263, 264, 273, 275, 284 Fouling resistance, 254–258, 275, 286 Fourier’s law, 189, 190 Free convection, 185 Friction factor, 95 Friction head loss across pipe fittings and valves, 56 equivalent length method, 57–61 head loss at pipe entrance and at pipe exit, 61, 62 pipe cross section, head loss due to, 62, 63 velocity head method, 56, 57 Fuels definition, 403 heating value, 403

482 Fuels (cont.) HHV, 404 Fundamental physical quantities and units, 2

G Gage pressure, 15, 29 Gases specific heats of, 304–306 generalized compressibility chart, 306–308 Gasoline (petrol), 403 Generalized compressibility chart, 151, 306–308 Geometric similarity, 175 Gravimetric analysis, 418 Gray bodies, 229 Gross heating value (GHV), 404

H Hagen–Poiseuille equation, 46–48, 89 Head loss due to change in pipe cross section, 62 gradual contraction, 63 gradual expansion, 63 sudden contraction, 63 sudden expansion, 62 at pipe entrance and at pipe exit, 61, 62 Heat balance, 247, 249 Heat balance equation, 247, 248 Heat capacity rate, 262 Heat conduction cylindrical wall, 194–197 Fourier’s law, 189 multilayer, 192–194 rectangular slab, 190–192 spherical shell, 197–199 thermal conductivity, 189 Heat duty, 247 Heat exchanger design equation, 258, 259, 261, 271 Heat exchanger effectiveness, 262, 263, 266, 272, 279, 280 Heat exchangers, 353 cooling water flow rate, 274, 288 double pipe heat exchangers, 247, 248 effectiveness, 262 NTU, 262, 263 exit enthalpy of water, 268 heat balance, 247, 249 heat capacity rates of oil and water, 265 heat duty, 269

Index heat exchanger design equation, 258, 261 LMTD, 251 correction factors, 252, 253 overall heat transfer coefficient, 254–256, 272 revised heat duty, 271 shell and tube heat exchanger, 248 steam flow rate, 273 Heat pump, 428 Heat recovery steam generator (HRSG), 452 Heat transfer coefficient, 205 Heat transfer mechanism, 205 Heat transfer principles conduction-convection systems, 186, 187 conduction heat transfer, 184 convection heat transfer forced convection, 185 free convection, 185 radiation heat transfer, 185 equation for, 183 thermal circuit to electrical circuit, 185 Heat transfer surface area, 258, 259, 278, 286, 291 Heating and cooling of fluids, 353, 354 Higher heating value (HHV), 404 Humid air, 374 Humidification, 384 Humidity ratio, 375–377, 379, 380, 395, 397, 399 Hydraulic diameter, 53 Hydraulic grade line (HGL), 102 Hydraulic horse power (HHP), 106, 111, 134 Hydrostatic force, 23–27, 29, 34 Hydrostatic static force, 23

I Ideal gas law thermodynamics fundamentals, 298–300 normal temperature pressure, 304 standard cubic feet and actual cubic feet, 304 standard temperature and pressure and molar volumes, 303 universal gas constant and ideal gas equation, 301, 302 Ideal gas mixtures Amagat’s law, 373, 374 Dalton’s law, 373 mass fraction, 371 mole fraction, 371 partial pressure of component, 372 partial volume of component, 372

Index variables, 372 Impulse-momentum principle, 78–83 Inclined manometer, 21 Independent variables, 173 Individual gas constant, 299, 301 Internal energy, 101, 337, 341, 344 thermodynamics fundamentals, 317 Isentropic Compressible Flow Functions, 166 Isentropic compression, 351, 362 Isentropic expansion, 348 Isentropic process, 312–314, 343 Isentropic turbine work, 348 Isobaric process, 310, 311 Isochoric process, 312 Isothermal process, 309, 310

K Kelvin, 295 Kelvin-Planck statement of the second law, 426 Kinematic similarity, 176 Kinematic viscosity (ν), 7, 9, 178 Kinetic energy, 101, 341 Kirchhoff’s law of radiation, 230

L Laminar and turbulent flow in pipes, 42 Laminar flow, 40 Law of thermodynamics, 101, 102 Lift Force, 77, 78 Liquid-vapor mixtures, 324–325 Log mean temperature difference (LMTD), 251–253, 276, 278, 281, 284, 285, 287, 289, 290

M Manometric fluid, 19, 30 Mass balance, 365, 366 Mass fraction, 371 Mechanical energy equation, 102, 103, 105 Mercury (Hg), 19 Mercury manometer, 21 Methane, 334 Mixing of air streams, 391, 393, 394 Moist air, 374 Moisture content of air, 375, 376, 381, 384, 386, 401 Molar volumes, 303, 304 Mole fraction, 371 Mole fraction of carbon dioxide, 398

483 Mollier diagram, 328, 329, 459 Moody diagram, 48–52, 92, 104 MTD correction factor, 253, 261, 269

N Net heating value (NHV), 404 Net positive suction head (NPSH), 123–125 Newton’s law of cooling, 205 Newton’s second law, 3 Newton’s third law, 79 Non ideal radiators, 229 Noncircular cross sections, flow through fluid dynamics, 52–56 Non-isentropic compression, 351 Non-isentropic expansion, 348 Normal temperature pressure (NTP), 304 Nozzles, 354 Number of transfer units (NTU), 262, 263, 265, 266

O Oil column, 17, 18, 28, 30 Orifice coefficient, 140 Orifice meters, 112, 139, 140 Orsat analysis, 411 Otto cycle, 461–465, 471 Outside air, 391, 392 Overall heat transfer coefficient, 212, 214, 254–256, 272 Overall resistance, 254, 256

P Partial pressure of water vapor, 375, 376, 396, 399 Partial volume, 372 Permanent pressure drop, 144 Pipe sizes and nomenclature, 37–40 laminar and turbulent flow, 42 Reynolds number, 41 Pipe cross section head loss due to change in, 62 gradual contraction, 63 gradual expansion, 63 sudden contraction, 63 sudden expansion, 62 Pipe diameters, 97 Piston-cylinder system, 311 Pitot tube, 137 Potential energy, 101, 341

484

Index

Prandtl number (Pr), 217, 221 Pressure, 171 Pressure energy, 101 Pressure-enthalpy (P – h) phase diagram, 329–331 Pressure equivalents, 17 Pressure fraction, 372 Pressure in water pipe, 28 Pressure intensity, 15 Psychrometric chart, 377–379, 381, 384, 386, 389, 391, 392, 394, 400 Psychrometric formulas, 394 Psychrometrics, 377, 379 Pump efficiency, 129 Pump laws, see Affinity laws Pump operating point, 115, 117, 118, 121, 133 Pump performance operating point, 115 pump performance curves, 113 pumps in parallel, 119–121 pumps in series, 117, 118 system curve, 114–116 Pump performance curves, 113 Pump power equation, 106–108, 111 Pumping system, 368 Pumps, 357 Pumps in parallel, 119–121 Pumps in series, 117, 118

transmissivity, 230 Radiation view factor definition, 231 view factor relationships, 231 Radiation-conduction systems, 235 Radiation-convection systems, 235 Rankine, 295 Rankine cycle, 431 analysis of, 432–437 with regenerative feedwater heating, 437–439, 441, 442 with reheat, 442–445 Ratio of orifice diameter to pipe diameter, 144 Ratio of throat diameter to pipe diameter, 141–143 Rayleigh number (Ra), 221 Reciprocity rule, 231 Reflectivity, 230 Refrigerating effect, 428 Regenerator effectiveness, 450 Relative humidity, 375–377, 379, 381, 384, 386, 389, 392, 395, 397, 398, 400 Return air, 391, 392 Reversed Carnot cycle, 427 coefficient of performance, 428–430 Reynolds number, 41, 42, 70, 90, 91, 104, 128 R-values, 194 building industry, 194

Q Quality of steam, 325

S Saturated steam tables, 324–326 Saturation pressure, 135, 376, 377, 395, 398, 399 Sensible heating process, 386 Sharp-edged orifice, 140, 145 Shape factor, 231 Shape factor reciprocity rule, 235 Shape factor summation rule, 231 Shell and tube exchangers, 263 Shell and tube heat exchangers, 247, 256, 264, 265, 279 Similitude dynamic similarity, 176, 177 geometric similarity, 175 kinematic similarity, 176 Single stream steady flow energy equation (SSSFEE), 346 Specific Enthalpy (h), 297 Specific Entropy (s), 298 Specific gravity (SG), 4, 105 Specific gravity of oil, 18 Specific heat at constant pressure, 304, 305

R R-134a, 466, 468 Radiation geometric factor, 231 Radiation heat transfer, 185 absorptivity, 230 correction for thermocouple readings, 237 definition, 227 equilibrium in radiation systems, 235 from pipe due to radiation, 238, 239 from small object to enclosure both being black bodies, 232 both being gray bodies, 233 Gray bodies, 229 net radiation heat transfer calculation, 231 surface resistance, 232 nonideal radiators, 229 overall heat transfer coefficient, 238, 239 reflectivity, 230 Stefan-Boltzmann’s law, 228

Index Specific Internal Energy (u), 297 Specific volume (v), 297 Specific volume of moist air, 382, 384, 387, 396, 401 Specific weight, 3, 30 Stagnation pressure, 137, 145 Standard atmospheric pressure at sea level, 16 Standard cubic feet (SCF), 304 Standard cubic feet per minute (SCFM), 304 Standard cubic meter per minute (SCMM) of air, 391 Standard heat of combustion, 404 Standard temperature and pressure (STP), 303 Static head, 19 Static pressure, 15, 18, 19, 22–24, 30–32, 137, 138, 145 Steady flow energy equation (SFEE), 345 Steam properties of, 323 compressed liquid, 326, 327 saturated steam tables, 324, 325 superheated steam tables, 325, 326 Steam flow rate, 273 Stefan-Boltzmann’s constant, 228, 229 Stefan-Boltzmann’s law, 228 Stoichiometric coefficients, 405 Stoichiometry, 405 Summation rule, 231 System curve, 114, 115, 117–121 Systeme Internationale (SI), 1–5, 7 Systems of units, 1

T Temperature (T ), 295 Temperature-pressure relationship, 361 Tensile force, 25, 27 Theoretical air, 407 Theoretical combustion, 406, 407 Thermal conductivity, 189 materials, 190 Thermal resistance, 183, 184 Thermocouples, 237 Thermodynamic cycles, 423 Brayton cycle, 445 analysis of, 446–449 with regeneration, 450, 451 Carnot cycle, 424 second law of, 426 thermal efficiency, 425, 426 combined cycle, 452–456 combined heat and power, 457, 459–461

485 Otto cycle, 461–465 practice problems, 470, 471 Rankine cycle, 431 analysis of, 432–437 with regenerative feedwater heating, 437–439, 441, 442 with reheat, 442–445 reversed Carnot cycle, 427 coefficient of performance, 428–430 solutions, to practice problems, 471–477 vapor compression refrigeration cycle, 465, 466 analysis of, 466, 467 performance measures for, 468, 470 Thermodynamic phase diagrams, 322 for water, 322, 323 Thermodynamic processes, involving ideal gases, 309 constant enthalpy/throttling process, 314 isobaric process, 310, 311 isochoric process, 312–314 isothermal process, 309, 310 Thermodynamics fundamentals, 293 enthalpy change, 318 entropy change, 318–321 gases, specific heats of, 304–306 generalized compressibility chart, 306–308 ideal gas law, 298–300 normal temperature pressure, 304 standard cubic feet and actual cubic feet, 304 standard temperature and pressure and molar volumes, 303 universal gas constant and ideal gas equation, 301, 302 internal energy change, 317 Mollier diagram, 328, 329 practice problems, 331, 332 pressure-enthalpy (P – h) phase diagram, 329 properties and variables, 293–296 intensive and extensive properties, 298 specific properties, 297, 298 solutions to, practice problems, 332–340 steam, properties of, 323 compressed liquid, 326, 327 saturated steam tables, 324, 325 superheated steam tables, 325, 326 thermodynamic phase diagrams, 322 for water, 322, 323 thermodynamic processes, involving ideal gases, 309

486 Thermodynamics fundamentals (cont.) constant enthalpy/throttling process, 314 isobaric process, 310, 311 isochoric process, 312–314 isothermal process, 309, 310 work, 314, 315 constant entropy (isentropic) process, 317 constant temperature (isothermal) process, 316 constant volume (isochoric) process, 316 Throttling process, 314 Tons of cooling, 381, 469, 470 Total dynamic head (TDH), 113 Transmissivity, 230 Turbines, 348

U United States Customary System (USCS), 1, 2, 4, 5 Units of force, 2 Universal gas constant, 301 U-tube differential manometer, 20 U-tube manometer, 28

V Vacuum gauge, 17 Vander Waal’s equation, 306 Vapor compression cycle, 471

Index Vapor compression refrigeration cycle, 465, 466 analysis of, 466, 467 performance measures for, 468, 470 Velocity, 171 Velocity head method, 56, 57 Vena contracta, 139, 140, 145 Venturi meter, 112, 141, 143 Viscosity absolute viscosities, 7 dynamic viscosity, 6 kinematic viscosity, 7, 9 Volume (V ), 295 Volume fraction, 372

W Water gage, 94 Water horse power, 106 Wet bulb temperature, 375 Work thermodynamics fundamentals, 314, 315 constant entropy (isentropic) process, 317 constant temperature (isothermal) process, 316 constant volume (isochoric) process, 316

Z Zero velocity, 6