POWER PLANT ENGINEERING 3

POWER PLANT
ENGINEERING 3

KUMARAN SUNDRAJ
MAIZUL AFZAIRIZAL MOHD ADNAN

JABATAN KEJURUTERAAN
MEKANIKAL
POLITEKNIK SEBERANG PERAI

POWER PLANT
ENGINEERING 3

Kumaran Sundraj
Maizul Afzairizal Mohd Adnan

2021
Mechanical Engineering Department

Politeknik Seberang Perai

©All rights reserved. No part of this publication may be translated or reproduced in any
retrieval system, or transmitted in any form or by any means, electronic, mechanical,
recording, or otherwise, without prior permission in writing from Politeknik Seberang

Perai.

ii PSP eBook | Power Plant Engineering 3

All rights reserved

No part of this publication may be translated or reproduced in any retrieval system,
or transmitted in any form or by any means, electronic, mechanical, recording, or

otherwise, without prior permission in writing from Politeknik Seberang Perai.

Published by

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Email: [email protected] Website : www.psp.edu.my
FB : politeknikseberangperai Ig : politeknikseberangperai

Perpustakaan Negara Malaysia Cataloguing-in-Publication Data

Kumaran Sundraj, 1971-
POWER PLANT ENGINEERING. 3 / KUMARAN A/L SUNDRAJ, MAIZUL

AFZAIRIZAL BIN MOHD ADNAN.

Mode of access: Internet
eISBN 978-967-0783-82-6

1. Power-plants.

2. Government publications--Malaysia.
3. Electronic books.

I. Maizul Afzairizal Mohd Adnan.

II. Title.

621.4

PSP eBook | Power Plant Engineering 3 iii

Acknowledgement

I would like to express my gratitude to the many people who saw me through this
book; to all those who provided their support, talked things over, read, wrote, offered
comments, allowed me to quote their remarks and assisted in the editing, proofreading
and design. I would like to thank Miss Ruhil Naznin binti Azamanfor her continuous
support, encouragement and generosity in allowing me to contribute to this book and
enabling me to publish this book. I also owe my thanks and gratitude to Mr. Maizul
Afzairizal bin Mohd Adnan for agreeing to be my co-author, and for updating and
refining the book’s contents and finishing the book. I wish to personally thank my
Head of Course, Mr. Ahmad Sabri bin Mohamed for his contributions and for sharing
with me his knowledge. I wish to thank my Head of Department, Mr. Muhammad Nasir
bin Marzuki for giving me the opportunity to write the book. Last but not least, I am
thankful to all those who have journeyed with me over the years, who have directly
or indirectly encouraged me in my work, and whose names are too many to mention.

Kumaran Sundraj
Maizul Afzairizal Mohd Adnan

iv PSP eBook | Power Plant Engineering 3

Preface

Power Plant Engineering provides exposure to students of as the entry-level personnel
into Power generation industry for both utility and manufacturing process plant.
Emphasis of this book is on general and basic operating principle includes plant fitting
and mountings, steam power cycle and prime movers used in power plant.

This Power Plant Engineering book is written for polytechnics’ students as a
reference in their study. The contents in this reference book are relevant to the syllabus
for Diploma in Mechanical Engineering specialized in Power Plant Engineering 3. The
main objective of this book is to provide a basic knowledge in Mechanical Engineering,
in a very simple, lucid and understandable language. Each chapter contains short
review explanation and example calculation to get complete knowledge of the
contents.

There are many illustrative solved examples and plenty of exercises are
provided in each chapter to stimulate interest in the students. SI units have been used
throughout the text. The entire book has been thoroughly revised; a large number of
solved example (questions having been selected from various universities and
competitive examinations) and ample additional text have been added.

We welcome constructive suggestion and comments from lecturer and
students. Good feedback is always given careful consideration and is helpful for future
improvements. We hope this book becomes a useful tool for students to tackle
important concepts and prepare themselves for their examination.

PSP eBook | Power Plant Engineering 3 v

Table of Content

Chapter Pages

CHAPTER 1: STEAM POWER PLANT 1
2
1.1 Rankine cycle 3
1.2 Difference between Carnot cycle and Rankine cycle 4
4
1.2.1 Rankine efficiency 5
1.2.2 Work Ratio 5
1.2.3 Specific Steam Consumption (S.S.C) 5
1.2.4 Isentropic Efficiency 10
1.2.5 Isentropic Efficiency for Turbine 14
1.3 Rankine Superheated Steam Cycle 16
1.4 Rankine With Double Reheat (Reheating Process)
1.5 Tutorial Steam Power Plant Cycle 20
20
CHAPTER 2: VAPOUR COMPRESSION SYSTEM 21
21
2.1 Vapour Compression System 22
2.2 Application of Refrigeration System 23
2.3 Vapour Compression Process 24
2.3.1 The Compressor 25
2.4 2.3.2 The Condenser 26
2.5 2.3.3 Metering Devices 28
2.6 2.3.4 The Evaporator 29
Basic Refrigeration Cycle 30
2.7 The Refrigerant Cycle: What Is It? 31
Process That Happen in Compressor 32
2.8 2.6.1 Process That Happen in Condenser 32
2.9 2.6.2 Process That Happen in Expansion Device 32
2.10 2.6.3 Process That Happen in Evaporator 33
2.11 Refrigerants 33
2.7.1 Definition 33
2.12 2.7.2 Classification of Refrigerants 34
35
2.7.2.1 Azeotrope Refrigerants 37
2.7.2.2 Halo-Carbon Refrigerants 38
2.7.2.3 Inorganic Refrigerants
Desirable Properties of An Ideal Refrigerant 38
Reversed Heat Engine System Operating on The Carnot Cycle 38
Vapour Compression Cycles 38
Vapor Compression Cycles Modification Made Are as Follows 45

2.11.1 Replacement of The Expansion Engine by A Throttle

Valve

2.11.2 Condition at The Compressor Inlet
2.11.3 Undercooling of The Condensed Vapor
Tutorial: Refrigerant System

vi PSP eBook | Power Plant Engineering 3 48
49
CHAPTER 3: VAPOUR COMPRESSION SYSTEM 49
49
3.0 Gas Turbine 49
3.1 Main Components of Gas Turbine 50
50
3.1.1 Compressor 52
3.1.2 Combustion Chamber 52
3.1.3 Turbine 53
3.2 Working of Open Cycle Gas Turbine 53
3.3 Brayton Cycle or Joule Cycle 54
3.4 Open Cycle Gas Turbine 56
3.4.1 T-S Diagram for Basic Turbine Gas 60
3.4.2 Power Turbine
3.5 Two Stage Turbine Gas 64
3.6 Reheat 64
3.7 Intercooler (Turbine Gas) 64
3.8 Tutorial Turbine Gas 65
65
CHAPTER 4: INTERNAL COMBUSTION ENGINE 65
67
4.0 Introduction Internal Combustion Engine 67
4.1 The Process of Four Stroke Engine Cycle 67
69
4.1.1 Suction/Intake Stroke 69
4.1.2 Compression Stroke 70
4.1.3 Power Stroke 70
4.1.4 Exhaust Stroke 71
4.2 Working of 2 Stroke Engine 72
4.2.1 Down Stroke 72
4.2.2 Up Stroke 72
4.3 Difference Between 4 Stroke & 2 Stroke Engine 73
4.3.1 Advantages of 4 Stroke Engine 73
4.3.2 Disadvantages of 4 Stroke Engine 73
4.3.3 Advantages of 2 Stroke Engine 74
4.3.4 Disadvantages of 2 Stroke Engine 74
4.4 Internal Combustion Engine Formula 74
4.4.1 Indicated Mean Effective Pressure (Imep) 74
4.4.2 Indicated Power (I.P) 75
4.4.3 Brake Power (Bp) 75
4.4.4 Brake Mean Effective Pressure (Bmep) 75
4.4.5 Friction Power 75
4.4.6 Efficiencies Of Internal Combustion Engine 76
4.4.7 Mechanical Efficiency (M) 76
4.4.8 Indicated Thermal Efficiency
4.4.9 Brake Thermal Efficiency
4.4.10 Specific Fuel Consumption
4.4.11 Volumetric Efficiency
4.4.12 Heat Balance Sheet
4.4.12a Heat Supplied By The Fuel
4.4.12b Heat Absorbed In Brake Power
4.4.12c Heat Rejected To The Cooling Water

PSP eBook | Power Plant Engineering 3 vii

4.4.12d Heat Carried Away by Exhaust Gases 76
4.4.12e Unaccounted Heat 77
4.5 Tutorial Internal Combustion Engine 85

CHAPTER 5: AIR STANDARD CYCLE 91
94
5.1 Constant Volume or Otto Cycle 99
5.2 Constant Pressure or Diesel Cycle
5.3 Tutorial Air Standard Cycle 103
105
CHAPTER 6: HEAT TRANSFER 105
108
6.0 Introduction to Heat 113
6.1 Fourier’s Law of Conduction 122
6.2 Newton’s Law of Cooling 123
6.3 The composite wall and the electrical analogy 124
6.4 Heat flow through a cylinder 125
6.5 Heat Exchanger 125
6.5 The basic types of Heat Exchanger 128
6.6 Counter Flow
6.7 Cross flow 130
6.8 Parallel flow and counter flow Recuperators
6.9 Tutorial Heat Transfer viii

SUMMARY

REFERENCES



1CHAPTER PSP eBook | Power Plant Engineering 3 1

STEAM POWER PLANT CYCLE

Objectives:
At the end of this chapter, students should be able to:

• Define Carnot cycle and Rankine cycle
• Sketch the block diagram and temperature-entropy (T-s) diagram
• Perform calculation related to Carnot Cycle, Rankine Cycle, Superheated Cycle

and Reheat Cycle.

2 PSP eBook | Power Plant Engineering 3

1.1 Rankine Cycle

Figure 1.1 The ideal Rankine cycle

Figure 1.2 T-s diagram for Rankine Cycle
The Rankine cycle is a practical cycle, and most steam power plants are based on it. The
problems with Carnot cycle are:
(i) Low work ratio
(ii) Complications during compression

PSP eBook | Power Plant Engineering 3 3

• It is impractical to compress wet steam because the water content separates out and fills
the rotary compressor.

• The difficulty in stopping the condensation at wet steam, so that subsequent compression
would bring the state point to saturated liquid.

• A very large compressor would be required.

To solve these problems, the wet steam is needed to fully condense to a saturated liquid. The
result of the modification of Carnot cycle is the Rankine cycle.

1.2 Difference Between Carnot Cycle and Rankine Cycle

Table 1.1 Comparison between Carnot Cycle and Rankine Cycle

No. Carnot Cycle Rankine Cycle

1. It is a theoretical cycle. It gives maximum This one is a practical cycle of steam engine

efficiency between two temperature and turbine.

difference

2. Heat is added and rejected at constant Heat is added and rejected at constant

temperature. pressure.

3. It has highest efficiency between two Rankine cycle has lower efficiency than Carnot

temperature differences. cycle.

4. Carnot cycle uses air as the working Rankine cycle uses water as working

substance. substance.

5. Carnot cycle is an ideal cycle for heat It is an ideal cycle for vapour power engine.

engine.

Processes in the cycle are:

4-5-1 (boiler) : Constant pressure heat addition in a boiler

• Liquid water enters the boiler and is heated to superheated

state in the boiler. (1)

( )Qboiler = h1 − h4

4 PSP eBook | Power Plant Engineering 3

1-2 (turbine) : Isentropic expansion in a turbine

• Steam from the boiler, which has an elevated temperature and

pressure, expands through the turbine to produce work, and

then is discharged to the condenser with relatively low

pressure. (2)

( )Wturbine = h1 − h2

2-3 (condenser) : Constant pressure heat rejection in a condenser

• Steam from the turbine is condensed to liquid water in the

condenser. (3)

( )Qcondenser = h3 − h2

3-4 (pump) : Isentropic compression in a pump

• Pump pressurized the liquid water from the condenser prior to

going back to the boiler.

( )Wpump = h4 − h3 or

( )Wpump = v f  105 (If pressure is in bar) (4)
P − Pboiler condenser

Where vf = 0.001 m3/kg

1.2.1 Rankine Efficiency

Rankine efficiency Rankine

= Wnet
Qsup lied

= Wturbine − W pump (5)
Qboiler

= (h1 − h2 ) − (h4 − h3 )
(h1 − h4 )

PSP eBook | Power Plant Engineering 3 5

1.2.2 Work Ratio

Work ratio,

w = Wnet
Wgross

= Wturbine − Wpump
Wturbine
(6)
(h1 − h2 ) − (h4 − h3 )
= (h1 − h2 )

1.2.3 Specific Steam Consumption (s.s.c)

It relates the power output to the steam flow necessary to produce steam.

The specific steam consumption is the steam flow in kg/h required to develop 1 kW.

= 3600  kg 
s.s.c Wnet  kWh 
3600 (7)

= (h1 − h2 ) − (h4 − h3 )

1.2.4 Isentropic Efficiency

Efficiency ratio is the ratio between the actual efficiency of the theoretical efficiency. In steam

power plant efficiency ratio is a comparison between the actual efficiency and the Rankine

cycle efficiency.

Isentropic Efficiency = Actual efficiency / Rankine efficiency (8)

1.2.5 Isentropic Efficiency for Turbine

For the turbine operating in steady state, the pressure at the input and output are fixed.
Figure below shows an isentropic process and the actual process through the turbine between
the two pressure values. The isentropic process shown by the lines directly and the actual
process by dotted lines.

6 PSP eBook | Power Plant Engineering 3

Isentropic Efficiency, isentropic turbine = Actual work of turbine / Isentropic work

 = WWisentropic (turbine)12'
12

= h1 − h2' (9)
h1 − h2

Example 1.1

A steam power plant works between boiler pressure of 40 bar and condenser pressure of
0.035 bar. Assume the work done on the pump is not neglected. Determine:

i. Work done by feed water pump
ii. Heat supplied to the system
iii. Work output from turbine
iv. Net work
v. Rankine efficiency
vi. Work ratio
viii. Specific steam consumption

i) Work done by feed water pump

( )Wpump = vf
P − PBOILER CONDENSER

Wpump = 0.001(40 − 0.035)105

Wpump = 4kJ / kg

ii) Heat supplied to the system
h1 = hg at 40 bar = 2801 kJ/kg

h3 = hf at 0.035 bar = 112 kJ/kg

W pump = h4 − h3
h4 = Wpump + h3 = 4 + 112 = 116kJ / kg

( )Qboiler = h1 − h4
Qboiler = (2801 − 116)

Qboiler = 2685kJ / kg

PSP eBook | Power Plant Engineering 3 7

iii) Work output from turbine

Since the turbine process is isentropic, s1=s2

s1 = sg at 40 bar = 6.070 kJ/kg K

s2 = s f + x2s fg at 0.035 bar

6.070 = 0.391+ x(8.130)

x2 = 6.070 − 0.391
8.130

x2 = 0.699

h2 = h f + x2h fg at 0.035 bar

h2 = 112 + (0.699)(2438)

h2 = 1816.2kJ / kg

iv) Net Work
Wnet = Wturbine − W pump
Wnet = 984.8 − 4
Wnet = 980.8kJ / kg

v) Rankine efficiency

 = QWRankine net
sup ply

 = W Q − WRankineturbine pump

boiler

 Rankine = 984.8 − 4 ( )Wturbine = h1 − h2
2685 Wturbine = (2801 −1816.2)

 Rankine = 0.365@ 36.5% Wturbine = 984.8kJ / kg

8 PSP eBook | Power Plant Engineering 3

vi) Work ratio

w = Wnet
Wgross

w = Wturbine − W pump
Wturbine

w = 984.8 − 4
984.8

 w = 0.996

vii) Specific steam consumption
s.s.c = 3600
Wnet
s.s.c = 3600
980.8
s.s.c = 3.67kg.steam / kWh

Example 1.2

A steam power plant operates with saturated steam at the boiler pressure of 40 bar and
0.035 bar condenser pressure. If the system works on the Rankine cycle with isentropic
efficiency of 80% during the expansion process and work pump is neglected, determine the
following:

i. Heat supplied to the system
ii. Work output from turbine
iii. Rankine efficiency
iv. Specific steam consumption

i) Heat supplied to the system

h1 = hg at 40 bar = 2801 kJ/kg

h3 = hf at 0.035 bar = 112 kJ/kg

( )Qboiler = h1 − h3
Qboiler = (2801 −116)

Qboiler = 2685kJ / kg

PSP eBook | Power Plant Engineering 3 9

ii) Work output from turbine

Since the turbine process is isentropic, s1=s2

s1 = s2 = sg at 40 bar = 6.070 kJ/kg K

s2 = s f + x2s fg at 0.035 bar

6.070 = 0.391+ x(8.130)

x2 = 6.070 − 0.391
8.130

x2 = 0.699

h2 = h f + x2h fg at 0.035 bar

h2 = 112 + (0.699)(2438)

h2 = 1816.2kJ / kg

Actual isentropic process

(( ))isentropic
= h1 − h2' = 0.80
h1 − h2

0.80 = 2801 − h2'

(2801 − 1816.2)

h2' = 2013.16kJ / kg

( )Wturbine = h1 − h2'

Wturbine = (2801 − 2013.16)

Wturbine = 787.84kJ / kg

iii) Rankine efficiency

 = QWRankine net
sup ply

 = W Q − WRankineturbine pump

boiler

 Rankine = 787.84 − 0
2685

 Rankine = 0.293@ 29.3%

iv) Specific steam consumption
s.s.c = 3600
Wnet
s.s.c = 3600
787.84
s.s.c = 4.57kg.steam / kWh

10 PSP eBook | Power Plant Engineering 3

1.3 Rankine Superheated Steam Cycle

Figure 1.3 Rankine Superheated Steam Cycle

Steam generated in the boiler is approaching the saturated steam and due to the limitation
of the material properties found in steam, boiler temperature will reach certain limits only. In
this condition, steam is not suitable for use in the turbine because the dryness fraction of the
steam during exit from the turbine is low. The presence of high moisture effects can lead to
rusting of turbine blades and so on. Reheating equipment is used to increase the steam
temperature without increasing existing pressure of the steam. The advantages of using
superheated steam are as follows:

(i) The superheat steam can increase the working capacity of the product due to higher heat
content. Therefore, the most economical use of steam can be achieved.

(ii) The use of superheated steam can increase the overall efficiency of the plant. Reheating
steam temperature can provide higher heat efficiency when used to drive the main
actuator.

(iii) The use of excessive steam can prevent corrosion during the final process of steam
expansion in the turbine. To prevent the corrosion process, the moisture content should
be controlled and the best-known steam dryness fraction when exiting the turbine is
approximately between 10% - 12% (x = 0.9 – 0.88).

PSP eBook | Power Plant Engineering 3 11

Example 1.3

A steam power plant operates on the Rankine cycle. The steam produced by the boiler is at
40 bar and 400C. The condenser is 0.035 bar. By neglecting the work pump, calculate the
Rankine cycle efficiency.

From steam table:
h1 = hg at 40 bar, 400C = 3214 kJ/kg
s1 = sg at 40 bar, 400C = 6.769 kJ/kg K

Since the 1-2 process is isentropic,

Then, s1 = s2

s2 = s f + x2s fg at 0.035 bar

6.769 = 0.391+ x(8.130)

x2 = 6.769 − 0.391
8.130

x2 = 0.785

h2 = h f + x2h fg at 0.035 bar

h2 = 112 + (0.785)(2438)

h2 = 2026kJ / kg

h3 = hf at 0.035 bar = 112 kJ/kg

Since the pump work is ignored,
Wpump = h4 − h3 = 0 40
h4 = h3 = 112kJ / kg

12 PSP eBook | Power Plant Engineering 3

Thus, Rankine efficiency

Rankine efficiency

 = QWRankine net
sup ply

 = W Q − WRankine turbine pump

boiler

 Rankine = (3214 − 2026) − 0
(3214 −112)

 Rankine = 0.383@ 38.3%

Example 1.4

A steam power plant operates with superheated steam at 450C and pressure of 40 bar with
a condenser of 0.045 bar. If the system works on the Rankine cycle with isentropic efficiency
of 86% during the expansion process and the work done on the pump is negligible. Define the
following:
i. Heat supplied to the system
ii. Work output of the turbine
iii. Rankine efficiency
iv. Specific steam consumption

From steam table,
Heat supplied to the system
Qsup plied = h1 − h4
h1 = hg at 40 bar, 450C = 3330 kJ/kg
h3 = hf at 0.035 bar = 130 kJ/kg
Work done by pump is neglected, h4 – h3 = 0
h4 = h3 = 130kJ / kg

PSP eBook | Power Plant Engineering 3 13

ii) Work output of the turbine

Since the 1-2 process is isentropic,

Then, s1 = s2

s1 = s2 = sg at 40 bar, 450C = 6.935 kJ/kg K

s2 = s f + x2s fg at 0.045 bar

6.935 = 0.451+ x(7.980)

x2 = 6.935 − 0.451
7.980

x2 = 0.813

h2 = h f + x2h fg at 0.045 bar

h2 = 130 + (0.813)(2428)

h2 = 2103kJ / kg

Actual isentropic process

(( )) =isentropic
h1 − h2' = 0.86
h1 − h2

0.86 = 3330 − h '
2

(3330 − 2103)

h2' = 2275kJ / kg

( )Wturbine = h1 − h2'

Wturbine = (3330 − 2275)

Wturbine = 1055kJ / kg

iii) Rankine efficiency

 = QWRankine net
sup ply

 = W Q − WRankine turbine pump

boiler

 Rankine = 1055 − 0
3200

 Rankine = 0.33@ 33%

Qsup plied = h1 − h4
Qsup plied = 3330 −130
Qsup plied = 3200kJ / kg

14 PSP eBook | Power Plant Engineering 3

iv) Specific steam consumption
s.s.c = 3600
Wnet
s.s.c = 3600
1055
s.s.c = 4.412kg.steam / kWh

1.4 Rankine with Double Reheat (Reheating Process)

The reheating process is a process whereby the steam is reheated and is usually done if the turbine
has more than one stage. The exhaust steam from the high-pressure turbine will be heated up to a
certain temperature, typically near the original superheat temperature before it is developed in the
next stage turbine.

Figure 1.4 Rankine with Double Reheat
Figure 1.4 shows the T-s diagram for reheat cycle.
At stage 1 high pressure, superheated steam is expanded in a high-pressure turbine to an intermediate
pressure at stage 2. The fluid then returned to a second stage boiler and superheated and reheated
to state 3. The reheated steam is then expanded in a low-pressure turbine to the final exhaust pressure
4. The moisture content of the working fluid is drastically reduced by use of reheat and this approach
is used in all fossil-fuelled and many nuclear power plants.

PSP eBook | Power Plant Engineering 3 15

Figure 1.5 T-s Diagram for Reheat Cycle
Heat Supplied to The System
Qsup ply = Qboiler + Qreheater

( ) ( )Qsup ply = h1 − h6 + h3 − h2

Actual heat supplied
Qsup ply = Qboiler + Qreheater

Qsup ply = (h1 − h6 ) + (h3 − h2 ')

Work of Turbine
Wturbine = WHPT + WLPT

( ) ( )Wturbine = h1 − h2 + h3 − h4

Actual work of turbine
Wturbine = WHPT + WLPT

( ) ( )Wturbine = h1 − h2 ' + h3 − h4 '

16 PSP eBook | Power Plant Engineering 3

Tutorial Steam Power Plant Cycle

Question 1
A steam power plant works between boiler pressure of 30 bar and condenser pressure of 0.04
bar. Assume the work done on the pump is not neglected. Determine:

(i) Work done by feed water pump
(ii) Heat supplied to the system
(iii) Work output from turbine
(iv) Net work
(v) Cycle efficiency
(vi) Work ratio
(vii) Specific steam consumption

Question 2
A steam power plant operates with saturated steam at the boiler pressure of 40 bar to 0.045
bar condenser pressure. If the system works on the Rankine cycle with isentropic efficiency
of 86% during the expansion process and work pump is neglected, determine the following:

(i) The heat supplied to the system
(ii) The work output of the turbine
(iii) The cycle efficiency
(iv) The specific steam consumption

Question 3
A steam power plant is based on the Rankine cycle. The steam produced by the boiler is at 40
bar and 400C. The condenser is 0.035 bar. By neglecting the work pump, determine the
following:

(i) The heat supplied to the system
(ii) The work output of the turbine
(iii) The cycle efficiency
(iv) The specific steam consumption

PSP eBook | Power Plant Engineering 3 17

Question 4
A steam power plant operates with superheated steam at 450C and pressure of 40 bar with
a condenser pressure of 0.045 bar. If the system works on the Rankine cycle with isentropic
efficiency of 86% during the expansion process and the work done on the pump is negligible.
Calculate the following:

(i) The heat supplied to the system
(ii) The work output of the turbine
(iii) The cycle efficiency
(iv) The specific steam consumption

Question 5
A reheat cycle operating between 30 and 0.04 bar has a superheat and reheat temperature
of 450C. The first expansion takes until the steam reaches the dry saturated stage and then
it reheats to reach boiler pressure. Neglecting feed pump work, determine:

(i) The heat supplied to the system
(ii) The work output of the turbine
(iii) The cycle efficiency
(iv) The specific steam consumption

Question 6
Steam at pressure of 40 bar and 400C is supplied to a high-pressure turbine. It comes out at
condition of saturated steam and then reheated up to 400C. The steam then entered the
low-pressure turbine and exit at condenser pressure of 0.045 bar. The system works on the
Rankine cycle with isentropic efficiency for high pressure turbine and low-pressure turbine
are 0.86 and 0.82 respectively. If the work done on the pump is negligible, determine:

(i) The steam pressure from the high-pressure turbine
(ii) Heat supplied to the system
(iii) The turbine output
(iv) The cycle efficiency
(v) The specific steam consumption

18 PSP eBook | Power Plant Engineering 3

Question 7
In a steam power plant, steam at 50 bar and 400℃ is expanded in the high-pressure turbine
to a pressure of 5 bar. The exhaust steam from the high-pressure turbine is reheated at
constant pressure until the temperature of 400℃ before expanding into the low-pressure
turbine. Steam exit to the condenser at the pressure of 0.09 bar. Assuming isentropic
efficiency for both turbines are 90% respectively, and the work done by feed water is not
negligible, determine:

(i) Total work output
(ii) The amount of heat supplied
(iii) Cycle efficiency
(iv) Specific steam consumption
(v) Draw the cycles on T-s diagram

Question 8
An ideal reheat Rankine cycle operates with steam as the working fluid. The reheat pressure
is set at 20 bar. Steam enters the high-pressure turbine at 130 bar, 600℃. The steam is
reheated to 600℃ before entering the low-pressure turbine and is condensed in a condenser
at 0.06 bar.

(i) Draw T-s diagram of the system
(ii) Identify work done by the pump
(iii) Determine heat supplied to the system
(iv) Calculate network of the system
(v) Calculate cycle efficiency of the system
(vi) Calculate the specific steam consumption

2CHAPTER PSP eBook | Power Plant Engineering 3 19

VAPOUR COMPRESSION SYSTEM

Objectives:
At the end of this chapter, students should be able to:

• Define Vapor Compression system and its main components
• Sketch the block diagram and temperature-entropy (T-s) diagram.
• Perform calculation related to Vapor Compression System.

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2.1 Vapour Compression System

• Refrigeration may be defined as the process of removing heat from a body or enclosed
space so that its temperature is first lowered and then maintained at a level below the
temperature of surroundings.

• The system maintained at the lower temperature is known as refrigerated system while
equipment used to maintain this lower temperature is known as refrigerating equipment.

• Work is required to transfer heat from lower temperature body to higher temperature
body.

• Amount heat removed by refrigerating equipment from refrigerated system is known as
refrigerating effect. Unit kJ/s.

• Effectiveness of refrigeration is given by Coefficient of Performance (COP)
• COP = Refrigerating effect / Work supplied.

2.2 Application of Refrigeration System

• Comfort air conditioning of auditoriums, hospitals, residents, offices, hotel.
• Manufacturing and preservation of medicine
• Storage and transportation of food stuffs such as dairy products, fruits, vegetables, meat,

and fish.
• Processing of textiles, printing work and photo graphics materials.
• Manufacturing of ice
• Cooling of concrete for dam
• Treatment of air for blast furnace
• Processing of petroleum and other chemical products.
• Production of rockets fuel.
• Computer functioning

PSP eBook | Power Plant Engineering 3 21

2.3 Vapour Compression Process

In simple vapor compression system processes are completed in one cycle. These are:
• Compression (compressor)
• Condensation (condenser)
• Expansion (expansion valve)
• Vaporization (evaporator)

2.3.1 The Compressor

• The compressor is the heart of the system. The compressor does just what its name is. It
compresses the low-pressure refrigerant vapor from the evaporator and compresses it
into a high pressure vapor.

• The inlet to the compressor is called the “Suction Line”. It brings the low pressure vapor
into the compressor.

• After the compressor compresses the refrigerant into a high pressure Vapor, it removes
it to the outlet called the “Discharge Line”.

Figure 2.1 The Compressor

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2.3.2 The Condenser

• The “Discharge Line” leaves the compressor and runs to the inlet of the condenser.
• Because the refrigerant was compressed, it is a hot high pressure vapor (as pressure goes

up – temperature goes up).
• The hot vapor enters the condenser and starts to flow through the tubes.
• Cool air is blown across the outside of the finned tubes of the condenser (usually by a fan

or water with a pump).
• Since the air is cooler than the refrigerant, heat jumps from the tubing to the cooler air

(energy goes from hot to cold – “latent heat”).
• As the heat is removed from the refrigerant, it reaches its “saturated temperature” and

starts to “flash” (change states), into a high-pressure liquid.
• The high-pressure liquid leaves the condenser through the “liquid line” and travels to the

“metering device”. Sometimes running through a filter dryer first, to remove any dirt or
foreign particles.

Figure 2.2 The Condenser

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2.3.3 Metering Devices

• Metering devices regulate how much liquid refrigerant enters the evaporator.
• Commonly used metering devices are, small thin copper tubes referred to as “cap

tubes”, thermally controller diaphragm valves called “TXV’s” (thermal expansion valves)
and single opening “orifices”.
• The metering device tries to maintain a present temperature difference or “super heat”,
between the inlet and outlet openings of the evaporator.
• As the metering devices regulates the amount of refrigerant going into the evaporator,
the device lets small amounts of refrigerant out into the line and loses the high pressure
it has behind it.
• Now we have a low pressure, cooler liquid refrigerant entering the evaporative coil
(pressure went down – so temperature goes down).

Figure 2.3 Metering Devices
Thermal Expansion Valves

• A very common type of metering device is called a TX Valve (Thermostatic Expansion
Valve). This valve has the capability of controlling the refrigerant flow. If the load on
the evaporator changes, the valve can respond to the change and increase or decrease
the flow accordingly.

• The TXV has a sensing bulb attached to the outlet of the evaporator. This bulb senses
the suction line temperature and sends a signal to the TXV allowing it to adjust the
flow rate. This is important because, if not all, the refrigerant in the evaporator

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changes state into a gas, there could be liquid refrigerant content returning to the
compressor. This can be fatal to the compressor. Liquid cannot be compressed and
when a compressor tries to compress a liquid, mechanical failing can happen. The
compressor can suffer mechanical damage in the valves and bearings. This is called”
liquid slugging”.

• Normally TXV's are set to maintain 10 degrees of superheat. That means that the gas
returning to the compressor is at least 10 degrees away from the risk of having any
liquid.

Figure 2.4 Thermal Expansion Valve

2.3.4 The Evaporator

• The evaporator is where the heat is removed from your house, business, or refrigeration
box.

• Low pressure liquid leaves the metering device and enters the evaporator.
• Usually, a fan will move warm air from the conditioned space across the evaporator

finned coils.
• The cooler refrigerant in the evaporator tubes, absorb the warm room air. The change of

temperature causes the refrigerant to “flash” or “boil”, and changes from a low-pressure
liquid to a low pressure cold vapor.
• The low pressure vapor is pulled into the compressor and the cycle starts over.

PSP eBook | Power Plant Engineering 3 25

• The amount of heat added to the liquid to make it saturated and change states is called
“Super Heat”.

• One way to charge a system with refrigerant is by super heat.

Figure 2.5 The Evaporator

2.4 Basic Refrigeration Cycle

• Starting at the compressor.
• Low pressure vapor refrigerant is compressed and discharged out of the compressor.
• The refrigerant at this point is a high temperature, high pressure, “superheated” vapor.
• The high-pressure refrigerant flows to the condenser by way of the "Discharge Line".
• The condenser changes the high-pressure refrigerant from a high temperature vapor to a low

temperature, high pressure liquid and leaves through the "Liquid Line".
• The high-pressure refrigerant then flows through a filter dryer to the Thermal Expansion valve

or TXV.
• The TXV meters the correct amount of liquid refrigerant into the evaporator.
• As the TXV meters the refrigerant, the high-pressure liquid changes to a low pressure, low

temperature, saturated liquid/vapor.
• This saturated liquid/vapor enters the evaporator and is changed to a low pressure, dry

vapor.
• The low pressure, dry vapor is then returned to the compressor in the "Suction line".
• The cycle then starts over.

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Figure: 2.6 Basic Refrigeration Cycle

2.5 The Refrigerant Cycle: What Is It?

An air conditioner works like a refrigerator. The refrigerant flows through the system, and
changes in state or condition. There are four processes in the 'refrigeration cycle'.
Process 1:
The compressor which pumps the refrigerant around the system, is the heart of the air
conditioner. Before the compressor, the refrigerant is a gas at low pressure. Because of the
compressor, the gas becomes high pressure, gets heated and flows towards the condenser.
Process 2:
At the condenser, the high temperature, high pressure gas releases its heat to the outdoor
air and becomes sub cooled high pressure liquid.

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Process 3:
The high-pressure liquid goes through the expansion valve, which reduces the pressure, and
thus temperature goes below the temperature of the refrigerated space. This results in cold,
low pressure refrigerant liquid.

Process 4:
The low-pressure refrigerant flows to the evaporator where it absorbs heat from the indoor
air through evaporation and becomes low pressure gas. The gas flows back to the compressor
where the cycle starts all over again.
In case of a heat pump the cycle can be reversed.

Condenser 1
2
High
Expansion Pressure
Device Compressor
Side

3 4 Low
Pressur

Evaporator

Figure 2.7 Schematic diagram of vapor compression system

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2.6 Process that Occurs in Compressor

The superheated vapor enters the

compressor where its pressure is 3

raised Condenser High
Pressure
4
Side

Expansion
Device Compressor

1 2 Low
Pressure
Side

Evaporator

Figure 2.8 Process that occurs in compressor

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2.6.1 Process that Occurs in Condenser

The high pressure superheated gas is
cooled in several stages in the condenser

4 Condenser 3
Compressor
Expansion High
Device Pressure

Side

12 Low
Evaporator Pressure

Figure 2.9 Process that occurs in condenser Side

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2.6.2 Process that Occurs in Expansion Device

Liquid passes through expansion
device, which reduces its pressure

and controls the flow into the
evaporator

Condenser 3
4
High
Pressure

Side

Expansion Compressor

Device

1 2 Low
Pressure

Evaporator Side

Figure 2.10 Process that occurs in expansion device

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2.6.3 Process that Occurs in Evaporator

Low pressure liquid 3
refrigerant in evaporator
absorbs heat and changes High
Pressure
to a gas
Side
Condenser

4

Expansion Compressor
Device

1 2 Low
Pressure

Evaporator Side

Figure 2.11 Process that happen in evaporator

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2.7 REFRIGERANTS
2.7.1 Definition

A refrigerant is defined as any substance that absorbs heat through expansion or vaporization
and loses it through condensation in a refrigeration system.

Figure 2.11 Types of Refrigerants

2.7.2 Classification of Refrigerants
2.7.2.1 Azeotrope Refrigerants

The term ‘azeotrope’ refers to a stable mixture of refrigerants whose vapor and liquid phase
retain identical compositions over a wide range of temperature.

• R-500
• R-502
• R-503
• R-504

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2.7.2.2 Halo-Carbon Refrigerants

• R-11
• R-12
• R-13
• R-14
• R-21
• R-22
• R-30
• R-40
• R-100
• R-113
• R-114
• R-134

2.7.2.3 Inorganic Refrigerants

The inorganic refrigerants were exclusively used before the introduction of halocarbon

refrigerants.

These refrigerants are still in due to their inherent thermodynamic and physical properties.

• R-717 (Ammonia)

• R-729 (Air)

• R-744 (Carbon dioxide)

• R-764 (Sulphur dioxide)

• R-118 (Water)

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Hydrocarbon

Most of the hydrocarbon refrigerants are successfully used in industrial and commercial

installations.

They possess satisfactory thermodynamic properties but are highly flammable and explosive.

• R-170 (Ethane)

• R-290 (Propane)

• R-600 (Butane)

• R-1120

• R-1130

• R-1150

• R-1270

2.8 Desirable Properties of An Ideal Refrigerant

• Low boiling point
• High critical temperature
• High latent heat of vaporization
• Low specific heat of liquid
• Low specific volume of vapor
• Non-corrosive to metal
• Non-flammable and non-explosive
• Non-toxic
• Low cost
• Easy to liquefy at moderate pressure and temperature
• Easy of locating leaks by odour or suitable indicator
• Mixes well with oil

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2.9 Reversed Heat Engine System Operating on The Carnot Cycle

The best COP will be given by a cycle which is a Carnot cycle operating between the given
temperature conditions.
Such a cycle using a wet vapor as the working substance is shown diagrammatically in Figure
below:
Wet vapor is used as the example, since the process of constant-pressure heat supply and
heat rejection are made at constant temperature, a necessary requirement of the Carnot
cycle and one which is not fulfilled by using a superheated vapor.

Figure 2.12 (a) Figure 2.12 (b)

The changes in the thermodynamic properties of the refrigerant throughout the cycle are
indicated on the T-s diagram of Fig. 2.12 (b). The cycle events are as follows:

1 – 2 Isentropic compression process (Work required)
Wet vapor at stage 1 enters the compressor and is compressed isentropic ally to state 2. The
work input for this process is represented by W1-2.

2 – 3 Condensation process (Heat rejected)
The vapor enters the condenser at stage 2 and is condensed at constant pressure and
temperature to state 3 when it is completely liquid. The heat rejected by the refrigerant is
Q2-3.

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3 – 4 Isentropic expansion process (work produced)
The saturated liquid expands isentropic ally where the temperature drops from T2 to T1 and
its state changes to wet vapor.

4 – 1 Evaporation process (Heat supplied)
At the lower pressure and temperature of state 4 the refrigerant enters the evaporator where
the heat necessary for evaporation, Q4-1, is supplied from the cold source.

The boundaries of the system are as shown in Figure 2.12 (b) and therefore

The Net Work input to the system W = W1-2 + W3-4
The Net Heat supplied to the system Q = Q2-3 + Q4-1
Refrigerating effect Q: (Maximum)
Q = T (s1 – s4) = T (s2 – s3)

Refrigerating effect Q: (Actual)
Q = h1 – h4

Coefficient of performance (COPref): (Actual)

.  

COPref = Q14 = h1 − h4
W21 h2 − h1

Coefficient of performance (COPref): (Maximum)

COPref = T1  (. − s4 )

s1 − s4

(T2 − T1 )(s1 )

COPref = T1
T2 − T1

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Example 2.1

A refrigerator has working temperatures in the evaporator and condenser coils of -30 and
32C respectively. What is the possible maximum COP? If the actual refrigerator has a COP of
0.75, calculate the required power input for a refrigerating effect of 5 kW.

Solution: and T2 = 32 + 273 = 305 K
T1 = -30 +273 = 243 K

COPref = T1
T2 − T1

COPref = 243 = 3.92
305 − 243

Actual COPref =0.75 x 3.92 =2.94

.  

COPref = Q14 = h1 − h4
W21 h2 − h1

W = Q14
COPref

W = 5 = 1.7kW
2.94

Required power input = 1.7 kW.

2.10 Vapour Compression Cycles

• The most widely used refrigerators and heat pumps are those which use a liquefiable
vapor as the refrigerant.

• The evaporation and condensation process take place when the fluid is receiving and
rejecting the specific enthalpy of vapor of vaporization, and these are constant –
temperature and constant-pressure process.

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• The cycle used in this process is reverse Carnot cycle but due to practical consideration
have led to several modifications to the ideal Carnot cycle which transform to Rankine
cycle.

2.11 Vapour Compression Cycles Modification
2.11.1 Expansion Engine by A Throttle Valve

• The plant is simplified by replacing the expansion cylinder with a simple throttle valve.
• The throttling process was shown to occur such that initial enthalpy equals the final

enthalpy.
• The process is highly irreversible so that the whole cycle becomes irreversible.
• The process is represented by the dotted line 3-4 on Figure 2.10 below.
• A comparison of Figure 1.3 and 1.2 shows that refrigerating effect Q1=(s1-s4), is reduced

by using a throttle valve instead of the expansion cylinder.

Figure 2.13 Reversed cycle using a throttle valve

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2.11.2 Condition at The Compressor Inlet

• In practice it is very difficult to determine condition 1 (figure 1.2) that is wet steam.
• In a practical unit this process is extended to give the vapor a definite amount of

superheated as it leaves the evaporator.
• This is undesirable, since the work to be done by the compressor is increased, as will be

shown later.
• It is a practical necessity to allow the refrigerant to become superheated in this way to

prevent the carry-over of liquid refrigerant into the compressor, where it interferes with
the lubrication.

Figure 2.14 T-s diagram for a reversed cycle with superheated vapor in the compressor

2.11.3 Undercooling of The Condensed Vapor

• The condensed vapor can be cooled at constant pressure to a temperature below that
of the saturation temperature corresponding to the condenser pressure.

• This effect is shown in Figure 1.5, in which the constant-pressure line is shown further
from the liquid line than it would appear, to illustrate the point.

• The effect of undercooling is to move the 3-4, representing the throttling process, to
the left on the diagram.

• The result of this is that the refrigerating effect in process 4-1 is increased.

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Figure 2.15 T-s diagram for a reversed cycle with undercooling in the condenser

Figure above includes all the modification of this section, and this can be taken as showing
the practical cycle.

For process 4-1:
h4 + Q1 = h1 + 0
Q1 = (h1 – h4)

Refrigerating effect Q1 = (h1 – h4)
For process 1-2:

h1 + 0 = h2 + W
W = (h2 – h1)

Work done on the refrigerant = (h2 – h1)

For process 2-3:
h2 + Q2 = h3 + 0
Q2 = (h2 – h3)

Heat rejected by the refrigerant = (h2 – h3)

For process 3-4:

h3 + 0 = h4 + 0
h3 = h4

PSP eBook | Power Plant Engineering 3 41

In throttling process = (h3 – h4)

The solution to numerical problems depends on the means of obtaining the enthalpies h1,
h2, and h3.

Example 2.2

A refrigeration machine operates between a condenser temperature of 45C and a refrigerant
temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant.

Solution:
Modification replaces the expansion device by using a throttle valve.

h1 = h f + xh fg pada -15C

From the diagram above s1 =s2 and s1 is in wet condition

S2 = sf1 + x1sfg pada -15C

0.681 = 0.0906 + x1 (0.6145)

= 0.6811− 0.00906
0.6145

= 0.96

h1 = hf + xhfg at -15C

h1 = 22.33 + 0.96 (180.97 -22.33)

h1 = 174.62 kJ/kg