Gas Turbines
Third Edition
About the Author
V Ganesan, currently working as Professor of
Mechanical Engineering, Indian Institute of
Technology Madras, is the recipient of Anna
University National Award for Outstanding Academic
for the Year 1997. He was the Head of the Department
of Mechanical Engineering, between October 2000 and
June 2002 and was the Dean (Academic Research)
between January 1998 and October 2000. He has so
far published more than 300 research papers in
national and international conferences and journals,
and has guided 20 MS and 40 PhDs.
Among other awards received by him are the Babcock Power Award for the
best fundamental scientific paper of Journal of Energy (1987), the Institution
of Engineers Merit Prize and Citation (1993), SVRCET Surat Prize (1995), Sri
Rajendra Nath Mookerjee Memorial Medal (1996), Automobile Engineer of the
Year by the Institution of Automobile Engineers (India) (2001), Institution of
Engineers (India), Tamil Nadu Scientist Award (TANSA) – 2003 by Tamil Nadu
State Council for Science and Technology, ISTE Periyar Award for Best
Engineering College Teacher (2004), N K Iyengar Memorial Prize (2004) by
Institution of Engineers (India) SVRCET Surat Prize (2004), Khosla National
Award (2004), Bharat Jyoti Award (2006) UWA Outstanding Intellectuals of
the 21st Century Award by United Writers Association, Chennai (2006), 2006
SAE Cliff Garrett Turbomachinery Engineering Award by SAE International,
USA, Sir Rajendra Nath Mookerjee Memorial Prize (2006) by Institution of
Engineers, Environmental Engineering Design Award 2006 by The Institution
of Engineers (India), 2006 SAE Cliff Garrett Turbomachinery Engineering
Award (2007) and Excellence in Engineering Education (Triple “E”) Award by
SAE International, USA (2007). He has received the Best Faculty Award from
Nehru Group of Institution in Coimbatore in 2009. He is the Fellow of Indian
National Academy of Engineering, National Environmental Science Academy,
Fellow of SAE International, USA, and Fellow of Institution of Engineers (India).
He has authored several books including the first edition of Internal Combustion Engines, McGraw Hill International Publications, Gas Turbines,
Computer Simulation of Four-Stroke Spark-Ignition Engines, Computer Simulation of Four-Stroke Compression-Ignition Engines, and Heat Transfer and has
edited several proceedings. He was formerly the Chairman of Combustion
Institute (Indian Section) and the Chairman of Engineering Education Board
of SAE (India), besides being a member of many other professional societies.
Dr Ganesan is actively engaged in a number of sponsored research projects
and is a consultant for various industries and R&D organizations.
Gas Turbines
Third Edition
V Ganesan
Professor
Department of Mechanical Engineering
Indian Institute of Technology
Madras
Tata McGraw Hill Education Private Limited
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Dedicated to Him
Preface
This text on gas turbines has been prepared keeping in mind the need of
a first-level textbook for the undergraduate, graduate and a professional
reference for practicing engineers. When I started writing the manuscript
a few years ago, to my knowledge, suitable text material was practically
non-existent, especially in SI units, and it seems to be true even today. It
is my earnest hope that this textbook will simplify the teaching of this
important subject.
Target Audience
This book is primarily intended for undergraduate/postgraduate students.
Hence, I have endeavoured to explain various topics right from the
fundamentals so that even a beginner can understand the exposition.
Besides, topics such as inlets, blades, environmental consideration, etc.,
are included to help the practising engineers as well. Professionals who
are engaged in the design and development of various power plants, who
are often called upon to undertake the work related to gas turbines, will
find this book extremely useful.
New to this Edition
It is gratifying to note that this book on Gas Turbines is being well received
by the academic community. Based on the suggestions and comments
received, all the chapters have been thoroughly revised. A set of multiplechoice questions has been introduced in each chapter. New sections and
subsections have been added in various chapters. A new chapter on
Transonic and Supersonic Compressors and Turbines has been included.
Further, minor errors brought to my notice have been rectified.
Organisation of the Book
SI units have been used consistently throughout the book. The book includes
a large number of typical worked-out examples and several illustrative
figures for an easier understanding of the subject. Exercises have been
given in various chapters so that the inquisitive students may solve these
problems and compare their answers with the answers provided.
Quite a lot of technical literature is available on this subject and owing
to the rate at which these technical papers now appear, a comprehensive
viii Preface
bibliography has not been included. However, a few references have been
selected and are appended to appropriate chapters. The text material is
divided into eight parts.
∑ Introduction and Fundamentals (Chapters 1 to 4)
∑ Cycle Analysis (Chapters 5 to 7)
∑ Compressors (Chapters 8 and 9)
∑ Combustion Systems (Chapter 10)
∑ Turbines (Chapter 11)
∑ Advanced Topics (Chapters 12 to 15)
∑ Environmental Considerations (Chapter 16)
∑ Rocket Propulsion (Chapter 17)
By dividing the chapters in the above manner, it may be seen that the
work has been organised to form a continuous logical narrative.
Web Supplements
The web supplements can be accessed at http://www.mhhe.com/ganesan/
gt3e and contain Solution Manual and PowerPoint Lecture Slides for
Instructors, and a Sample Chapter and Links to Reference Material for
students.
Acknowledgements
It would be impossible to refer in detail to all those who have been consulted
in the compilation of this work. A special note of appreciation is due to my
gas-turbines course students of the 1999 batch, for their help in checking
the worked out as well as exercise problems.
I am thankful for the help and support received from colleagues of the
department and sister departments of IIT Madras. I am particularly
grateful to Prof. P Srinivasa Rao and Prof. S Sampath who spared a great
amount of their valuable time in correcting the entire manuscript. Their
critical comments and suggestions were of immense help in improving
the presentation. In this context, I would also like to mention the names
of all the reviewers of the book.
Avinash Kumar Agarwal
Indian Institute of Technology (IIT)
Kanpur
Kanpur, Uttar Pradesh
V K Bajpai
National Institute of Technology (NIT)
Kurukshetra
Kurukshetra, Haryana
L D Garg
Punjab Engineering College (PEC)
University of Technology
Chandigarh, Punjab
Preface
ix
Souvik Bhattacharyya
Indian Institute of Technology (IIT)
Kharagpur
Kharagpur, West Bengal
Dipankar Chaterjee
Haldia Institute of Technology (HIT)
Haldia, West Bengal
Ranjan Basak
Sikkim Manipal Institute of Technology
Majitar, Sikkim
Mahesh Rathore
SNJB’s KBJ College of Engineering
Nashik, Maharashtra
Venkatraj Varadharajulu
Bharati Vidyapeeth College of
Engineering and Technology
Mumbai, Maharashtra
M Udayakumar
National Institute of Technolgy (NIT)
Tiruchirapalli
Tiruchirapalli, Tamil Nadu
P Chitambarnathan
Dr Sivanthi Adithanar College of
Engineering
Thiruchendur, Tamil Nadu
B Srinivas Reddy
G Pulla Reddy Engineering College
Kurnool, Andhra Pradesh
I wish to express my thanks to the Centre for Continuing Education for
encouraging me to write this book and helping me in the preparation of
this manuscript.
I will be failing in my duty if I do not acknowledge the help of my family
members, Ms P Rajalakshmi, Ms Vijayashree, Mr Venkatasubramanian
and Ms Aparna. Words are not sufficient to express my gratitude to Ms
Vijayashree who has put her heart and soul in the preparation of the book.
My sincere thanks are due to all of them.
Feedback
Although maximum care has been taken to minimize the errors, I will be
grateful for any constructive criticism for further improvement of the book.
I hope and wish the present edition also would be well received.
V GANESAN
x Preface
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Contents
Preface
Preface to the Second Edition
Preface to the First Edition
Nomenclature
1 Introduction
1.1 Prime Movers
1.2 Simple Gas Turbine
Review Questions
Multiple Choice Questions
ii
viii
ix
xix
1
1
2
4
5
2 Review of Basic Principles
2.1 Definitions and Laws
2.2 Energy Equation
2.3 Fluid Dynamics
2.4 Basic Definitions
2.5 Streamtube Area-Velocity Relation
2.6 Normal Shock Waves
2.7 Equations of Motion for a Normal Shock Wave
2.8 Oblique Shock and Expansion Waves
2.9 Flow with Friction and Heat Transfer
2.10 Flow in Constant-Area Duct with Friction
Review Questions
Multiple Choice Questions
7
7
15
19
19
25
28
28
30
32
32
36
37
3 Fundamentals of Rotating Machines
3.1 General Fluid Dynamics Analysis
3.2 The Physical Meaning of the Energy Equation
3.3 Classification of Machines
3.4 General Thermodynamic Analysis
39
39
42
44
45
xii
Contents
3.5 Efficiency of Rotating Machines
3.6 Dimensional Analysis of Rotating Machines
3.7 Elementary Airfoil Theory
Review Questions
Multiple Choice Questions
4 Cycle Arrangements
4.1 Open-Cycle Arrangements
4.2 The Closed-Cycle
4.3 Basic Requirements of the Working Medium
4.4 Properties of Various Working Media
4.5 Applications
4.6 Comparison of Gas Turbines with Reciprocating Engines
Review Questions
Multiple Choice Questions
47
48
54
59
60
63
63
68
71
72
72
73
75
76
5 Ideal Cycles and their Analysis
5.1 Assumptions in Ideal Cycle Analysis
5.2 The Simple Gas Turbine Cycle
5.3 The Heat Exchange Cycle
5.4 The Reheat Cycle
5.5 The Reheat and Heat Exchange Cycle
5.6 The Intercooled Cycle
5.7 The Intercooled Cycle with Heat Exchanger
5.8 The Intercooled and Reheat Cycle
5.9 Intercooled Cycle with Heat Exchange and Reheat
5.10 Comparison of Various Cycles
5.11 Ericsson Cycle
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
79
80
80
83
86
89
92
94
96
99
101
103
104
128
130
134
6 Practical Cycles and their Analysis
6.1 Assumptions
6.2 Stagnation Properties
6.3 Compressor and Turbine Efficiency
6.4 Pressure or Flow Losses
6.5 Heat Exchanger Effectiveness
6.6 Effect of Varying Mass Flow
6.7 Effect of Variable Specific Heat
6.8 Mechanical Losses
137
137
138
140
143
147
148
149
150
Contents
6.9 Loss due to Incomplete Combustion
6.10 Cycle Efficiency
6.11 Polytropic Efficiency
6.12 Performance of Actual Cycle
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
xiii
150
151
152
156
165
203
203
210
7 Jet Propulsion Cycles and Their Analysis
7.1 Reciprocating or Propeller Engines
7.2 Gas Turbine Engines
7.3 The Ramjet Engine
7.4 The Pulse Jet Engine
7.5 The Turboprop Engine
7.6 The Turbojet Engine
7.7 Thrust and Thrust Equation
7.8 Specific Thrust of the Turbojet Engine
7.9 Efficiencies
7.10 Parameters Affecting Flight Performance
7.11 Thrust Augmentation
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
213
214
215
216
220
224
227
235
237
238
246
246
252
271
272
276
8 Centrifugal Compressors
8.1 Essential Parts of a Centrifugal Compressor
8.2 Principle of Operation
8.3 Ideal Energy Transfer
8.4 Blade Shapes and Velocity Triangles
8.5 Analysis of Flow through the Compressor
8.6 Diffuser
8.7 Volute Casing
8.8 Performance Parameter
8.9 Losses in Centrifugal Compressors
8.10 Compressor Characteristics
8.11 Surging and Choking
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
279
280
282
286
287
290
302
306
307
310
310
311
313
326
327
331
xiv
Contents
9 Axial Flow Compressors
9.1 Historical Background
9.2 Geometry and Working Principle
9.3 Stage Velocity Triangles
9.4 Work Done Factor
9.5 Enthalpy–Entropy Diagram
9.6 Compressor Stage Efficiency
9.7 Performance Coefficients
9.8 Degree of Reaction
9.9 Flow through Blade Rows
9.10 Flow Losses
9.11 Stage Losses
9.12 Pressure Rise Calculation in a Blade Ring
9.13 Performance Characteristics
9.14 Comparison of Axial and Centrifugal Compressors
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
333
333
335
336
340
342
343
344
345
351
353
356
357
359
365
366
386
387
390
10 Combustion Systems
10.1 Combustion Theory Applied to Gas Turbine Combustor
10.2 Factors Affecting Combustion Chamber Design
10.3 Factors Affecting Combustion Chamber Performance
10.4 Form of Combustion System
10.5 Requirements of the Combustion Chamber
10.6 The Process of Combustion in a Gas Turbine
10.7 Combustion Chamber Geometry
10.8 Mixing and Dilution
10.9 Combustion Chamber Arrangements
10.10Some Practical Problems
Review Questions
Multiple Choice Questions
393
394
397
398
402
403
405
406
409
409
411
418
419
11 Impulse and Reaction Turbines
11.1 A Single Impulse Stage
11.2 A Single Reaction Stage
11.3 Multistage Machines
11.4 Velocity Triangles of a Single Stage Machine
11.5 Expression for Work Output
11.6 Blade loading and flow coefficients
11.7 Blade and Stage Efficiencies
421
422
422
423
424
426
427
427
Contents
xv
11.8 Maximum Utilization Factor for a Single Impulse Stage
11.9 Velocity-compounding of Multistage Impulse Turbine
11.10Pressure Compounding of Multistage Impulse Turbine
11.11The Reaction Turbine
11.12Multistage Reaction Turbines
11.13Blade-to-Gas Speed Ratio
11.14Losses and Efficiencies
11.15Performance Graphs
Worked out Examples
Review Questions
Exercise
Multiple Choice Questions
428
430
435
435
436
450
451
451
454
463
464
465
12 Transonic and Supersonic Compressors and Turbines
12.1 The Supersonic Compressor
12.2 Supersonic Axial Flow Compressors
12.3 Supersonic Radial Compressors
12.4 Supersonic Axial Flow Turbine Stages
Review Questions
Multiple Choice Questions
467
468
469
474
475
478
478
13 Inlets and Nozzles
13.1 Inlets
13.2 Subsonic Inlets
13.3 Diffuser
13.4 Supersonic Inlets
13.5 Exhaust Nozzles
Review Questions
Multiple Choice Questions
481
481
482
484
486
488
498
498
14 Blades
14.1 Blade Materials
14.2 Manufacturing Techniques
14.3 Blade Fixing
14.4 Problems of High Temperature Operation
14.5 Blade Cooling
14.6 Liquid Cooling
14.7 Air Cooling
14.8 Practical Air Cooled Blades
Review Questions
Multiple Choice Questions
501
501
503
508
510
513
515
515
518
521
521
xvi
Contents
15 Component Matching and Performance Evaluation
15.1 Performance Characteristics
15.2 Equilibrium Running Diagram
15.3 To Find the Equilibrium Points
15.4 Procedure to find Equilibrium Point
15.5 Performance Evaluation of Single-spool Turbojet Engine
15.6 Operating Line
15.7 General Matching Procedure
15.8 Transient Operation
Review Questions
Multiple Choice Questions
523
526
526
527
530
531
537
543
545
546
547
16 Environmental Considerations and Applications
16.1 Air Pollution
16.2 Aircraft Emission Standards
16.3 Stationary Engine Emission Standards
16.4 NOx Formation
16.5 NOx Reduction in Stationary Engines
16.6 Noise
16.7 Noise Standards
16.8 Noise Reduction
16.9 Assessment of the Gas Turbine
16.10Typical Applications of Gas Turbines
16.11The Small Gas Turbine Applications
16.12Electric Power Generation Applications
16.13Marine Application
16.14Gas Pumping Applications
16.15Locomotive Applications
16.16Automotive Applications
16.17Aircraft Applications
16.18Process Applications
16.19Additional Features of Gas Turbine Engines
16.20Trends in the Future Development
Review Questions
Multiple Choice Questions
549
549
551
555
557
559
560
562
565
567
569
569
570
570
571
572
572
573
573
574
578
578
579
17 Rocket Propulsion
17.1 Classification of Rockets
17.2 Principle of Rocket Propulsion
17.3 Analysis of an Ideal Chemical Rocket
17.4 Optimum Expansion Ratio for Rocket
17.5 The Chemical Rocket
581
582
583
584
587
589
Contents
xvii
17.6 Advantages of Liquid Propellant Rockets over Solid Propellant Rockets
17.7 Free Radical Propulsion
17.8 Nuclear Propulsion
17.9 Electro Dynamic Propulsion
17.10Photon Propulsion
17.11Comparison of various types of rockets
17.12Staging
17.13Multistage Rocket
17.14Comparison of Various Propulsion Systems
17.15Propulsive Efficiency
Review Questions
Multiple Choice Questions
600
601
601
602
606
606
607
608
609
609
611
613
Appendix
615
Index
619
xviii
Contents
1
INTRODUCTION
1.1
PRIME MOVERS
The distinctive feature of our civilization, one that makes it different from
all others, is the wide use of power from mechanical means. At one time
the primary source of power, or prime mover, for the work of peace and war
was chiefly man’s muscles. Even after animals had been trained to help and
after the wind and running streams had been harnessed, man was mainly
depending on his muscle power. But when he learned to convert the heat
of chemical reactions into mechanical energy, it revolutionized the world.
Machines which serve this purpose are known as heat engines.
The first heat engine to have a revolutionary effect was the gun. It is
perhaps hard to realize that the vital difference between the bow and the
gun was the substitution of gunpowder for the bowman’s muscles, because
the musket was much inferior to the bow in range and accuracy. In principle,
the gun is the ancestor of our internal combustion piston engines.
Another revolution began when Watt perfected the steam piston engine. Here, an intermediate working fluid was used, so that the products
of combustion did not act directly on the moving parts of the mechanism.
Later, the ancient principle of the turbine was adopted to steam, and the
piston engine took the back seat.
Among engines for the production of mechanical power there were, a
few years ago, three principal competitors in the field:
(i) the steam turbine plant,
(ii) the diesel piston engine, and
(iii) the gasoline piston engine.
The steam turbine, since the beginning of its career around the turn of
the nineteenth century, has become the most important prime mover for
power generation and a widely used power plant for marine application.
However, it has an inherent disadvantage of the need to produce highpressure, high-temperature steam. This involves the installation of a bulky
2
Gas Turbines
and expensive steam-generating equipment, a boiler or a nuclear reactor.
The hot gases produced in the boiler furnace or reactor core never reach the
turbine. They, merely are used to produce an intermediate working fluid,
namely steam.
Clearly, a much more compact power plant results when the water to
steam step is eliminated and the hot gases themselves are used to drive the
turbine. Diesel piston engines started replacing bulky steam power plants
for power generation. Gasoline engines were used in early days for aircraft
propulsion. A device known as the gas turbine came into existence to a
limited extent in certain types of superchargers used with piston engines.
The earliest patent on gas turbine was that of the Englishman, John
Barber, in 1879. Early designs were unsuccessful, largely due to two factors:
(i) the low efficiency of the compressors, and
(ii) the combustion temperature limitations imposed by the materials
then available.
Serious development of gas turbine began only after the second world war
with shaft power in mind, but attention was soon shifted to the turbojet
engine for aircraft propulsion. Since then, the gas turbine made a progressively greater impact in an increasing variety of applications. However, only
in the recent past much research effort has been focused on the design and
development of efficient gas turbine units. Of the various means of producing mechanical power available today the gas turbine, in many respects,
seems to be the most satisfactory power plant. It is mainly due to
(i) the absence of reciprocating and rubbing members which reduces the
vibration and balancing problems,
(ii) high reliability,
(iii) low lubricating-oil consumption, and
(iv) high power-to-weight ratio.
1.2
SIMPLE GAS TURBINE
In order to produce an expansion through a turbine a pressure ratio must
be provided. Hence, the first necessary step in the cycle of a gas turbine
plant must be compression of the working fluid. If, after the compression,
the working fluid is to be expanded directly in the turbine and there were
no losses in either component, the power developed by the turbine would
just equal that absorbed by the compressor. Thus, if the two were coupled
together, the combination would do no more than turn itself round.
The power developed by the turbine can be increased by the addition of
energy to raise the temperature of the fluid prior to expansion. When the
working fluid is air, a very suitable means of doing this is by the combustion
of fuel in the air which has been compressed. Expansion of hot working fluid
Introduction
3
then produces a greater power output from the turbine than the power necessary to drive the compressor. The three main components are, therefore,
a compressor, a combustion chamber and a turbine, connected together as
shown in Fig. 1.1.
Fuel
Combustion chamber
Products of combustion
Air
Power
output
Compressor
Turbine
Fig. 1.1 A simple gas turbine
In a practical cycle, losses do occur in both the compressor and the
turbine which increase the power absorbed by the compressor and decrease
the power output of the turbine. A certain minimum addition of energy in
the form of fuel to the fluid will therefore be required before one component
can drive the other. This fuel produces no useful power, due to component
losses and lowering of efficiency of the machine. Further addition of fuel will
result in a useful power output. However, there is a limit to the amount of
fuel that can be added per unit mass of air and therefore to the net power
output. The fuel-air ratio that may be used is governed by the working
temperature of the highly stressed turbine blades. This temperature is
limited by the creep strength of the materials used in the turbine blades
and the working life required.
The two main factors which affect the performance of gas turbines are
the efficiencies of various components and turbine working temperature.
The higher they are made, the better is the all-round performance of the
plant. It was, in fact, low efficiencies and poor turbine materials which
caused the failure of a number of early attempts. For example, in 1904 two
French engineers built a unit which did little more than turn itself over, with
compressor efficiency of about 60% and the maximum gas temperature of
about 740 K. The overall efficiency of the gas turbine cycle mainly depends
upon the pressure ratio of the compressor. The development of science of
aerodynamics and that of metallurgy made it possible to employ very high
pressure ratios (20:1) with an adequate compressor efficiency (85-90%) and
high turbine inlet temperatures, up to 1500 K.
Two possible combustion systems were proposed: one at constant-pressure
and the other at constant-volume. Theoretically, the thermal efficiency
of the constant-volume cycle is higher than that of constant-pressure cycle. However, constant-volume combustion involves mechanical difficulties
requiring valves to isolate the combustion chamber from the compressor and
turbine. Combustion is intermittent impairing the smooth running of the
4
Gas Turbines
machine. After certain initial attempts, constant-volume type combustion
was discontinued.
In the constant-pressure gas turbine, combustion is continuous and valves
are not necessary. It was soon accepted that the constant-pressure cycle
had the greater possibilities for future development.
It is important to realize that in the gas turbine the process of compression, combustion and expansion do not occur in a single component
as they do in a reciprocating engine. They occur in components which
are separate, in the sense that they can be designed, tested and developed
individually and these components can be linked together to form a gas
turbine unit in a variety of ways. The possible number of components is
not limited to those already mentioned. More compressors and turbines can
be added with intercoolers between the compressors, and reheat combustion chambers between turbines. Water coolers and heat exchangers can be
additionally fitted. These refinements are used to increase the power output and efficiency of the plant at the expense of added complexity, weight
and cost. The way these components are added not only affects the maximum overall thermal efficiency but also the variation of thermal efficiency.
Each arrangement is to be chosen depending on its suitability for a given
application. Thus, it is seen that a simple gas turbine consists mainly of
three components. Of the three, two are rotating machines and the third
one is a heat addition device. However, more complex system is possible
with the addition of auxiliary devices such as heat exchanger, intercooler
and reheater. In order to understand the working principle and cycle arrangements one should first have some fundamental knowledge of rotating
machines. This will be taken up in Chapter 3 after the review of some basic
principles in Chapter 2.
Review Questions
1.1 What is the difference between our civilization and the ancient ones
and what action of the man revolutionized the world?
1.2 What is a heat engine and what is its origin?
1.3 What is a steam engine and what are its characteristics?
1.4 What is the inherent disadvantage of a steam power plant?
1.5 Which were the three contemporary power plants for the production
of power?
1.6 When was the earliest gas turbine design attempted and what was the
result?
1.7 What are the latest research efforts in gas turbines and why is this
power plant considered to be satisfactory?
1.8 Explain with a neat sketch the details of a simple gas turbine power
plant.
Introduction
5
1.9 Compare the steam and gas turbine power plants.
1.10 Which parameter is most important to determine the overall efficiency
of a gas turbine cycle?
1.11 What are the two factors which affect the performance of a gas turbine?
1.12 What are the two types of combustion systems?
1.13 Why is constant-pressure heat addition is more advantageous in a gas
turbine?
1.14 What advantage does a gas turbine power plant have over a reciprocating engine from the point of view of component developments?
1.15 How many components are there in a simple gas turbine power plant?
Mention the additional components that can be added to improve the
power output and efficiency.
Multiple Choice Questions (choose the most appropriate answer)
1. Around the turn of nineteenth century the most important prime
mover was
(a) gas turbines
(b) diesel engines
(c) steam turbines
(d) gasoline engines
2. Early aircraft engines used
(a) gas turbine engines
(b) gasoline engines
(c) Diesel engines
(d) steam turbines
3. The gas turbine was invented by
(a) John Barber
(b) Brayton
(c) Otto
(d) Atkinson
4. Early designs of the gas turbine were unsuccessful due to
(a) low efficiency of the compressor
(b) low efficiency of the turbine
6
Gas Turbines
(c) combustion temperature limitation
(d) all of the above
5. Reciprocating engines are preferred over gas turbines because of
(a)
(b)
(c)
(d)
high reliability
high power to weight ratio
all of the above
none of the above
6. A simple open-cycle gas turbine power plant consists of
(a)
(b)
(c)
(d)
turbine,
turbine,
turbine,
turbine,
combustion chamber
combustion chamber
combustion chamber
compressor and heat
and heat exchanger
and charge cooler
and compressor
exchanger
7. The combustion in a gas turbine is at
(a)
(b)
(c)
(d)
constant-pressure
constant-volume
partly constant-pressure and partly constant-volume
constant temperature
8. The performance of a simple gas turbine depends on
(a)
(b)
(c)
(d)
efficiency of the compressor
efficiency of the turbine
efficiency of the compressor and turbine
none of the above
9. The pressure ratio of the modern gas turbine power plant is
(a)
(b)
(c)
(d)
5:1
10:1
15:1
20:1
10. The highest turbine inlet temperature is
(a)
(b)
(c)
(d)
1000
1200
1500
1800
K
K
K
K
Ans:
1. – (c)
6. – (c)
2. – (b)
7. – (a)
3. – (a)
8. – (c)
4. – (d)
9. – (d)
5. – (d)
10. – (c)
2
REVIEW OF BASIC
PRINCIPLES
INTRODUCTION
Turbines and compressors are usually analyzed using thermodynamic
and fluid dynamic equations. The thermodynamic equations relate temperature, pressure and volume whereas the fluid dynamic equations relate
force, mass and velocity. The following are the laws that are frequently
used in dealing with problems of and operation of these machines:
(i) energy equation in its various forms from the first law of thermodynamics,
(ii) temperature, entropy and gas relations from the second law of thermodynamics,
(iii) continuity relationships from the law of conservation of mass, and
(iv) momentum equation from Newton’s second law of motion.
2.1
DEFINITIONS AND LAWS
Before discussing the various aero-thermodynamic aspects of compressors
and turbines, let us review some important definitions used in the analysis
of compressible flow useful for rotating machines. However, the reader
is advised to refer to standard textbooks on thermodynamics and fluid
dynamics for more details.
2.1.1
System
A fixed identity with an arbitrary collection of matter is known as a system.
The boundary is an imaginary surface which separates the system from its
surroundings. Surroundings are those which are outside the system. System
can be classified as either an open system or a closed system.
8
Gas Turbines
Open System When there is a continuous flow of matter, it is called an open
system. Such a system is usually depicted by a control volume. It has a
fixed space but does not contain a fixed mass of matter; instead there is a
continuous flow of mass through it. The properties of the matter occupying
the control volume can vary with time. The surface which encloses a control
volume is called control surface.
Closed System When there is a fixed quantity of matter (fluid or gas),
it is called a closed system. There is no inflow or outflow of matter to
and from a closed system. However, a closed system can interact with its
surroundings through work and heat transfers. The boundaries of a closed
system containing the fixed mass of matter can change. Expanding gas in
a reciprocating internal combustion engine is one such example.
2.1.2
State
Condition of a system, defined by its properties, is known as the state of a
system.
2.1.3
Process
A change or a series of changes in the state of a system is known as a
process.
2.1.4
Cycle
If the initial and final states of a system experiencing a series of processes
are identical, it is said to have executed a cycle.
2.1.5
Pressure
Pressure at a point surrounded by an infinitesimal area, ΔA, is the force
per unit area. Pressure is usually designated by Pascal in SI units. It may
also be expressed as N/m2 or bar. In this book, we will follow ’bar’ as the
unit of pressure.
2.1.6
Density
The density of a medium is the mass of the matter (gas) per unit volume.
Density is expressed in kg/m3 .
2.1.7
Temperature
When two systems are in contact with each other and are in thermal equilibrium, the property common to both the systems having the same value is
called temperature. Thus temperature is a measure of the thermal potential
of a system.
Review of Basic Principles
2.1.8
9
Energy
Energy is the capacity to do work. The state of a system can be changed
by adding or removing energy. Heat and work are different forms of energy
in transit. They are not contained in any system.
Heat is the form of energy which transfers between two systems by
virtue of the temperature difference between them. Heat transfer to or
from a system changes its state.
Work is said to be done by a system on its surroundings when they
are moved through a distance by the action of a force; this is exerted by
the system on the surroundings in the direction of the displacement of the
surroundings. The magnitude of mechanical work is given by
Work done
=
Force × Distance in the direction of force
Both heat and work are path functions and they depend on the type of
process and therefore, are not properties of a system. In SI units, energy,
heat and work are all expressed in joules (J), kilojoules (kJ) or Newton
metres (Nm).
2.1.9
First Law of Thermodynamics
It states that when a system executes a cyclic process, the algebraic sum of
the work transfers is proportional to the algebraic sum of the heat transfers.
dW
∝
dQ
=
J
dQ
(2.1)
When heat and work are expressed in the same units, then
dQ −
dW
= 0
(2.2)
It can easily be shown that the quantity (dQ − dW ) is independent of
the path of the process and hence it represents a change in the property
of the system. This property is referred to as energy, denoted here by the
symbol E. Thus,
dE
=
dQ − dW
(2.3)
Equation 2.3 for the two states of a system can be written as
E2 − E1
=
Q−W
Q
=
W + (E2 − E1 )
Heat transfer =
2.1.10
(2.4)
Work done + Change in energy
Specific Heats of Gases
The specific heat of a gas is the heat carrying capacity in a process. It is
the amount of heat that is required to raise the temperature of a unit mass
of the gas by one degree.
10
Gas Turbines
In thermodynamic analysis two different types of specific heats are used:
(i) Specific heat at constant-volume, and
(ii) Specific heat at constant-pressure.
The specific heat at constant-volume (Cv ) is the amount of heat required
to raise the temperature of a unit mass of the gas by one degree at constantvolume. It is given by
Cv
∂q
∂T
=
=
v
∂U
∂T
(2.5)
v
The specific heat at constant-pressure (Cp ) is the amount of heat required
to raise the temperature of a unit mass of the gas by one degree at constantpressure . It is given by
Cp
∂q
∂T
=
=
p
∂h
∂T
(2.6)
p
The specific heats of actual gases are a function of temperature and vary
with temperature
Cp , Cv = f (T )
(2.7)
The ratio of the two specific heats, (γ), is an important parameter in compressible flow problems of turbomachines and is defined as
γ
2.1.11
=
Cp
Cv
(2.8)
Internal Energy
The internal energy of a gas is the energy stored in it by virtue of its
molecular motion. If it is assumed that the internal energy of a perfect gas
is zero at the absolute zero temperature, its value at a temperature T is
given by
U = Cv T
(2.9)
2.1.12
Enthalpy
The heat supplied to or rejected by a system at constant-pressure is the
change of enthalpy during the process. The value of enthalpy at a given
state is given by
h =
U + pv
h =
Cp T
=
U+
p
ρ
(2.10)
and for an ideal gas
(2.11)
Review of Basic Principles
2.1.13
11
Ideal Gas
An ideal gas obeys both Boyle’s and Charle’s laws, i.e.,
(pv)T
v
T
p
=
constant (Boyle’s law)
(2.12)
=
constant (Charle’s law)
(2.13)
Thus, an ideal gas obeys the simple equation of state
pv = RT
p =
ρRT
(2.14)
(2.15)
The two specific heats and the gas constant for an ideal gas are related by
the following equation:
Cp − Cv
=
R
(2.16)
Substituting Eq. 2.8 in 2.16, we get
1
R
Cv =
γ−1
(2.17)
γ
R
γ−1
(2.18)
Cp
2.1.14
=
Perfect Gas
A perfect gas is an ideal gas whose specific heats remain constant at all
temperatures.
d
(Cv )
dT
= 0
(2.19)
Cv = constant with temperature and
d
(Cp )
dT
= 0
(2.20)
Cp = constant with temperature. Unless otherwise stated, analyses of
compressible flow given in this book assume perfect gas relations.
2.1.15
Semi-perfect Gas
A semi-perfect gas is an ideal gas whose specific heats vary with temperature.
Cv = f (T ) and Cp = f (T )
(2.21)
2.1.16
Real Gas
The real gas behaviour deviates from that of an ideal gas. It does not obey
the equation of state (Eqs. 2.14, 2.15, etc.,). Different equations of state
are used for real gases.
12
Gas Turbines
2.1.17
Second Law of Thermodynamics
The second law of thermodynamics can be stated in a number of ways.
Some of them are as follows:
(i) Clausius statement Heat cannot, on its own, flow from a body at
lower temperature to a body at higher temperature.
(ii) Kelvin-Planck’s statement It is impossible to construct a heat engine
which performs a complete cycle and delivers work exchanging heat
from a single source.
The following relations can be derived from the second law of thermodynamics:
Definition of entropy
2
s2 − s1
Clausius inequality
dQ
T
=
1
dQR
T
≤ 0
(2.22)
(2.23)
In any irreversible process,
2
s2 − s1
>
1
dQ
T
In an irreversible adiabatic process,
s2 − s1 > 0
(2.24)
(2.25)
In any reversible cycle,
dQR
T
= 0
(2.26)
In an isentropic, i.e., a reversible adiabatic process,
s 2 − s1
2.1.18
= 0
(2.27)
Reversible Process
A process is reversible if the system and its surroundings can be restored to
their initial states by reversing the process. A reversible process in a flow
machine is possible only in the absence of fluid friction and heat transfer
with finite temperature difference. Since these conditions are impossible to
achieve in actual processes, all real flows in compressors and turbines are
irreversible. The reversible process is used only as an ideal reference process
for comparison with its equivalent actual process.
Review of Basic Principles
2.1.19
13
Irreversible Process
An irreversible process is one which does not satisfy the above conditions
of reversibility.
2.1.20
Adiabatic Process
During a process if there is no heat transfer between the system and the
surroundings, it is known as an adiabatic process. All the rotating machines
discussed in this book are assumed to follow only adiabatic processes.
2.1.21
Isentropic Process
An adiabatic process in which entropy remains constant is known as a
reversibly adiabatic or isentropic process. For unit mass, this is governed
by the following relations:
pv γ
T1
T2
=
=
T ds =
2.1.22
constant
p1
p2
(2.28)
γ−1
γ
=
dh − vdp
=
v2
v1
γ−1
ρ1
ρ2
=
1
dh − dp
ρ
=
0
γ−1
(2.29)
(2.30)
Non-flow Process
A process that occurs in a closed system is a non-flow process. One such
example is shown in Fig. 2.1. It represents the expansion of a fixed mass
of gas inside the cylinders of a reciprocating engine.
1
pV = constant
dV
p dV
Volume
2
14
Gas Turbines
2
Wnf p
=
pdv
(2.32)
1
For an adiabatic process, assuming perfect gas relations
Wnf p
2.1.23
=
1
(p1 v1 − p2 v2 )
γ−1
(2.33)
=
Cv (T1 − T2 ) = u1 − u2
(2.34)
Flow Process
A process that occurs in an open system or in a control volume is a flow
process. Processes occurring in all turbomachines are of this type. Figure
2.2 represents such a process.
Pressure
1
dp
γ
pV = constant
V dp
2
Volume
Fig. 2.2 Illustration of a flow process
The infinitesimal work done in a reversible process is given by
dWf p
=
−vdp
Wf p
=
−
For a finite process
(2.35)
2
vdp
(2.36)
1
For an adiabatic process in a perfect gas,
γ
Wf p =
(p1 v1 − p2 v2 )
γ−1
(2.37)
Wf p
=
Cp (T1 − T2 )
(2.38)
Wf p
=
h1 − h2
(2.39)
Review of Basic Principles
2.2
15
ENERGY EQUATION
The energy equation (Eq. 2.4) is basically derived from the first law of
thermodynamics as given in Section 2.1.9. It is written as
Q
=
W + (E2 − E1 )
(2.40)
For application in turbomachines, the energy terms will include internal
energy, gravitational potential energy and kinetic energy. Other forms of
energy which can be included but are not relevant here are strain energy,
magnetic energy, etc.,
1
E = U + m(gZ) + mc2
(2.41)
2
dE
=
dU + mg dZ + m d
1 2
c
2
(2.42)
where U is the internal energy and Z is the potential energy and c is the
velocity of the fluid.
The change in the energy in a finite process between two states is given by
1
E2 − E1 = (U2 − U1 ) + mg(Z2 − Z1 ) + m c22 − c21
(2.43)
2
Substituting Eq. 2.43 in Eq. 2.4, a general form of the energy equation can
be obtained.
1
Q = W + (U2 − U1 ) + mg(Z2 − Z1 ) + m c22 − c21
(2.44)
2
Dividing throughout by m
1
q = w + (u2 − u1 ) + g(Z2 − Z1 ) + c22 − c21
(2.45)
2
2.2.1
Steady-flow Energy Equation
For steady flow processes through turbomachines, the work term in Eqs. 2.44
and 2.45 contains shaft work and flow work. Thus,
W
=
Ws + (p2 V2 − p1 V1 )
(2.46)
Substituting Eq. 2.46 in Eq. 2.44 and rearranging, we get
1
Q = Ws + (U2 + p2 V2 ) − (U1 + p1 V1 ) + mg(Z2 − Z1 ) + m c22 − c21 (2.47)
2
Writing enthalpy H for the quantity U + pV ,
1
1
H1 + mgZ1 + mc21 + Q = H2 + mgZ2 + mc22 + Ws
(2.48)
2
2
In terms of specific quantities,
1
1
h1 + gZ1 + c21 + q = h2 + gZ2 + c22 + Ws
(2.49)
2
2
Equation 2.48 or 2.49 is the steady-flow energy equation for a control volume
or an open system. This will now be rewritten for processes in various
turbomachines and their components.
16
2.2.2
Gas Turbines
Hydro-turbomachines
In hydro-turbomachines
ρ
=
1
v
u1
≈
u2
q
≈
0
=
constant
Therefore, from Eq. 2.49, shaft work is given by
1
Ws = g(Z1 − Z2 ) + c21 − c22 + (p1 − p2 )V
2
(2.50)
In a stationary component, such as guide vanes or draught tubes, shaft
work is absent. Therefore, Eq. 2.50 gives
c22 − c21
2.2.3
=
2[g(Z1 − Z2 ) + (p1 − p2 )V ]
(2.51)
Compressible Flow Machines
Most of the compressible flow turbomachines such as turbines and compressors are considered as adiabatic machines, i.e., q ≈ 0. In these machines, change in potential energy (Z1 − Z2 ) is also negligible as compared
c2
c2
to changes in enthalpy (h1 − h2 ) and kinetic energy 21 − 22 . Therefore,
Eq. 2.49 yields
1
h1 + c21
2
=
1
h2 + c22 + Ws
2
(2.52)
The shaft work is given by
Ws
=
1
h1 + c21
2
1
− h2 + c22
2
(2.53)
If the entry and exit velocities are small or the difference between them
is negligible, then shaft work is given by the difference between the static
enthalpies at the two states
Ws
2.2.4
=
h1 − h2
(2.54)
Energy Transformation
It may be noted that energy transfer (shaft work input or output) in a
turbomachine stage is possible only in the rotor, whereas energy transformation can occur both in moving and fixed blades. An application of
the energy equation for stationary components of compressors and turbines
such as nozzle blade rings, diffusers and volute casings can be made. The
shaft work is absent in these components and the flow is almost adiabatic
(q ≈ 0). Therefore Eq. 2.52 gives
1
h1 + c21
2
=
1
h2 + c22
2
=
constant
(2.55)
Review of Basic Principles
2.2.5
17
Stagnation Enthalpy
In an adiabatic energy transformation process, if the initial state is represented by h, T, c, etc., and the final gas velocity is zero, the resulting value
of the enthalpy (h2 = h0 ) has a special significance. Under these conditions,
Eq. 2.55 yields
1
h 0 = h + c2
(2.56)
2
Since the gas is stagnant or stationary in the final state, the quantity
h0 in Eq. 2.56 is known as the stagnation enthalpy. Thus, the stagnation
enthalpy can be defined as the enthalpy of a gas or vapour when it is
adiabatically brought to rest. It may be observed that the definition of
stagnation enthalpy in Eq. 2.56 is only another form of the energy equation.
2.2.6
Stagnation Temperature
For a perfect gas, a stagnation temperature is defined through stagnation
enthalpy. From Eq. 2.56,
Cp T0
T0
1
= Cp T + c2
2
= T+
c2
2Cp
(2.57)
T0 is known as the stagnation temperature whereas T is the static temperature and c2 /2Cp is the equivalent of kinetic energy temperature (Tc ).
c2
2Cp
Tc
=
(2.58)
T0
= T + Tc
(2.59)
Equation 2.57 can be used to obtain an important relation for compressible
flow machines.
T0
T
= 1+
c2
2Cp T
(2.60)
Using Eq. 2.18,
T0
T
= 1 + c2
γ−1
2γRT
The velocity of sound in a gas at a local temperature T is given by
a
=
γRT
(2.61)
The Mach number of the flow is defined as the ratio of the local velocity of
the gas and of sound
18
Gas Turbines
c
a
=
M
=
T0
T
= 1+
c
√
γRT
Therefore,
2.2.7
γ−1
2
c2
a2
(2.62)
=
1+
γ−1
M2
2
(2.63)
Stagnation Velocity of Sound
Stagnation values of various flow parameters are used as reference values
in the analysis of compressible flow machines. Therefore, an expression for
the stagnation velocity of sound is derived here. By definition,
a0
=
γRT0
(2.64)
a0
=
(γ − 1)Cp T0
(2.65)
a0
=
(γ − 1)h0
(2.66)
Substituting for R,
Since Cp T0 = h0
2.2.8
Stagnation Pressure
The stagnation pressure is the pressure of the gas or fluid when it is brought
to rest adiabatically and reversibly (i.e., isentropically). The ratio of the
stagnation and static pressures can be obtained from Eq. 2.63
p0
p
=
T0
T
γ
γ−1
=
γ−1 2
M
1+
2
γ
γ−1
(2.67)
When the pressure changes are small, the process can be assumed to
be incompressible (ρ ≈ constant). Then the stagnation pressure can be
determined from
1
p0 = p + ρc2
(2.68)
2
2.2.9
Stagnation Density
The density of a stationary gas or vapour is the stagnation density. For an
ideal gas, its value at known values of stagnation temperature and pressure
is given by
ρ0
=
p0
RT0
(2.69)
For an isentropic process from Eq. 2.29,
ρ0
ρ
=
T0
T
1
γ−1
=
γ−1 2
M
1+
2
1
γ−1
(2.70)
Review of Basic Principles
2.2.10
19
Stagnation State
The concept of a reference state of the gas in a compressible flow machine
is very useful. The stagnation state of a gas is often used as a reference
state. A state defined by the stagnation temperature and pressure is the
stagnation state of the gas. This state is obtained by decelerating a gas
isentropically to zero velocity.
It should be observed that it is necessary here to qualify the deceleration
process as an isentropic process. This is not necessary in defining stagnation
enthalpy and temperature.
2.3
FLUID DYNAMICS
In this section we shall discuss some basic definitions used in fluid dynamics.
The analysis of flow in turbomachines requires the application of Newton’s
second law of motion along with the equations of continuity and energy
which will be discussed in this section.
The application of Newton’s second law of motion provides the equations
of motion which are also known as Euler’s momentum equations. These,
and the continuity equation will be used in the subsequent chapters. We
will summarize the various forms of these equations in the next section.
2.4
BASIC DEFINITIONS
First of all we will define various terms associated with fluid dynamics.
These definitions will help us in understanding the various equations used
in the fluid dynamics which will be applied to analyze the flow in turbomachines.
2.4.1
Fluid
A fluid is a substance which gets deformed continuously when shearing
forces are applied. Liquids, gases and vapours are all fluids. A non-viscous
or inviscid fluid is referred to as an ideal fluid.
2.4.2
Fluid Velocity
The instantaneous velocity of the fluid particle passing through a point is
known as the fluid velocity at that point.
2.4.3
Streamline
A curve in a flow field which is always tangential to the direction of flow is
referred to as a streamline. These are shown in Fig. 2.3.
20
Gas Turbines
Streamlines
Stream tube
Fig. 2.3 One-dimensional flow through a stream tube
2.4.4
Stream Tube
A stream tube (Fig. 2.3) is an infinitesimal portion of the flow field. It is
a collection of a number of streamlines forming an imaginary tube. There
is no flow through the walls of a stream tube. The properties of the flow
are constant across the section of a stream tube. Therefore, the flow in a
stream tube is one-dimensional.
2.4.5
Incompressible Flow
If the relative change in the density of a fluid in a process is negligible, it
is referred to as an incompressible process. In such a flow (or process) the
fluid velocity is much smaller than the local velocity of sound in it. The
flow of gases and vapours at Mach numbers less than 0.30 can be assumed
to be incompressible without much sacrifice in accuracy.
2.4.6
Compressible Flow
In compressible flows the relative changes in the fluid density are considerable and cannot be neglected. The fluid velocities in such flows are appreciable compared to the local velocity of sound. If the Mach number in
a flow is higher than 0.3, it is considered to be compressible.
2.4.7
Steady Flow
A flow is known to be steady if its properties do not change with time. The
shape of the stream tube does not change in a steady flow. For such a flow
∂c
∂t
2.4.8
=
∂p
∂t
=
∂T
∂t
=
∂ρ
∂t
=
∂m
∂t
=
0
(2.71)
Unsteady Flow
If one or more parameters (c, p, T, ρ, m etc) in a flow change with time, it
is known as unsteady flow.
Review of Basic Principles
2.4.9
21
Viscosity
Viscosity is the property which resists the shearing motion of two adjacent
layers of the fluid.
A fluid is known as a Newtonian fluid if the relation between the shear
stress and the angular deformation is linear. The shear stress is given by
τ
∝
dc
dy
=
μ
dc
dy
(2.72)
The constant of proportionality, μ, is known as the coefficient of viscosity
or dynamic viscosity.
The kinematic viscosity, ν, is the ratio of the dynamic viscosity and the
density of the fluid.
ν
=
μ
ρ
(2.73)
All real flows experience fluid viscosity. Therefore, their behaviour is influenced by the viscous forces.
2.4.10
Inviscid Flow
If the viscosity of the fluid is assumed to be absent, the flow is referred
to as inviscid flow. Such a flow glides freely over its boundaries without
experiencing viscous forces.
2.4.11
Reynolds Number
The Reynolds number is the ratio of inertia forces to viscous forces.
Inertia force =
Viscous force =
ρAc2
μcl
Therefore, the Reynolds number is given by
Re =
ρAc2
μcl
where l= characteristic length, A = l2 and μρ = ν. Therefore,
cl
ρcl
=
Re =
μ
ν
(2.74)
The value of the Reynolds number in a flow gives an idea about its nature.
For example, at higher Reynolds numbers the magnitude of viscous forces
is small compared to the inertia forces.
22
Gas Turbines
2.4.12
Mach Number
The Mach number is an index of the ratio of inertia and elastic forces. This
is defined by
M2
=
Inertia force
Elastic force
=
ρAc2
KA
where K is the bulk modulus of elasticity and A is the flow area. However,
K = ρa2 and therefore, M = ac .
This relation gives another important definition of the Mach number as
the ratio of the fluid velocity to the local velocity of sound. Thermodynamic
relations derived in sections 2.2.6, 2.2.7, 2.2.8 and 2.2.9 demonstrate its
application.
2.4.13
Mach Angle and Mach Cone
In a flow field if a small disturbance is introduced, it will be felt throughout
the field in the form of a wave traveling at the local velocity of sound relative
to the medium.
Consider the propagation of pressure disturbance shown in Fig.2.4. Assume that an object (say, an aircraft) is moving with velocity c = 0, c =
a/2, c = a and c > a as shown in Figs. 2.4(a – d). The moving object
creates disturbance waves. As in Figs. 2.4(a and b) the disturbance waves
will reach a stationary observer before the source of disturbance (moving
object) could reach him in subsonic flow.
In supersonic flows, Fig. 2.4(d) it takes considerable amount of time
for an observer to perceive the pressure disturbance, after the source has
passed him. This is one of the fundamental differences between subsonic
and supersonic flows.
Therefore, in a subsonic flow, the streamlines sense the presence of any
obstacle in the flow field and adjust themselves ahead of it and flow around
it smoothly. But in the supersonic flow, the streamlines feel the obstacle
only when they hit it. The obstacle acts as a source of disturbance and so
the streamlines deviate as the Mach cone as shown in Fig. 2.4(d). It should
be noted that the disturbance due to obstacle is sudden and therefore, the
flow behind the obstacle has to change abruptly.
In Fig. 2.4(d), it is shown that for supersonic motion of an object, there
is a well-defined conical zone in the flow field with the object located at
the nose of the cone, and the disturbance created by the moving object
is confined only to the field included inside the cone. The flow field zone
outside the cone does not even feel the disturbance.
For this reason, von Karman termed the region inside the cone as the
zone of action, and the region outside the cone as the zone of silence. The
lines at which the pressure disturbance is concentrated and which generate
the cone are called Mach waves or Mach lines. The angle between the Mach
line and the direction of motion of the body is called the Mach angle, μ.
Review of Basic Principles
(a) c = 0
23
(c) c = a
(b) c = a /2
Mac
h con
e
Zone of action
at
μ
ct
Zone of silence
(d) c > a
Fig. 2.4 Propagation of disturbance wave
From Fig. 2.4(d), it can be written as
sin μ =
at
ct
sin μ =
1
M
=
a
c
(2.75)
(2.76)
From the disturbance waves propagation shown in Fig. 2.4, we can infer
the following features of the flow regimes.
(i) When the medium is incompressible [M = 0, Fig. 2.4(a)] or when the
speed of the moving disturbance is negligibly small compared to the
local sound speed, the pressure pulse created by the disturbance will
spread uniformly in all directions.
(ii) When the disturbance source moves with a subsonic speed [M < 1,
Fig. 2.4(b)], the pressure disturbance will be felt in all directions.
At all points in space (neglecting viscous dissipation), the pressure
pattern will not be symmetrical.
(iii) For sonic velocity [M = 1, Fig. 2.4(c)] the pressure pulse will be
at the boundary between subsonic and supersonic flow and the wave
front will be in a plane.
(iv) For supersonic speeds [M > 1, Fig. 2.4(d)] the disturbance wave
phenomena are totally different from those at subsonic speeds. All the
24
Gas Turbines
pressure disturbances are included in a cone which has the disturbance
source at its apex. The effect of the disturbance is not felt upstream
of the disturbance source.
Sho
ck
Flow around a wedge shown in Figs. 2.5(a) and (b)can be seen with
smooth change and abrupt change in flow direction for subsonic and supersonic flow, respectively.
Moo< 1
M oo >1
(a) Subsonic flow
(b) Supersonic flow
Fig. 2.5 Flow around a wedge
For M∞ < 1, the flow changes its direction smoothly and pressure
decreases with acceleration; for M∞ > 1; there is sudden change in flow
direction at the body and pressure increases downstream of the shock.
2.4.14
Small Disturbance
When the apex angle of wedge, δ, is extremely small, the disturbances will
be small. Then, we can consider these to be identical to sound pulses.
In such a case, the deviation of streamlines will be small. There will be
infinitesimally small increase in pressure across the Mach cone as shown in
Fig. 2.6.
e
av
w
ch
a
M
μ
δ
Fig. 2.6 Mach cone
Review of Basic Principles
2.4.15
25
Finite Disturbance
Sh
oc
k
When the wedge angle, δ, is finite, the disturbances introduced are finite,
and then the wave is not called Mach wave but a shock or shock wave
(Fig. 2.7). The angle of shock, β, is always smaller than the Mach angle.
β
δ
Fig. 2.7 Shock wave
The deviation of streamline is finite and there is finite pressure increase
across the shock wave.
2.4.16
Transonic Flow
When a body is kept in transonic flow (Mach number between 0.8 and 1.2),
it experiences subsonic flow over some portions of its surface and supersonic
flow over other portions. There is also a possibility of shock formation on
the body. It is this mixed nature of the flow field which makes the study
of transonic flows complicated.
2.4.17
Hypersonic Flow
The temperature at stagnation point and over the surface of an object in
the hypersonic flow becomes very high and, therefore, it requires special
treatment. That is, we must consider the thermodynamic aspects of the
flow along with gas dynamic aspects. That is why hypersonic flow theory
is also called aero-thermodynamic theory. Besides, because of high temperature, the specific heats become functions of temperature and hence the
gas cannot be treated as perfect gas. If the temperature is quite high (of
the order of more than 2000 K), even dissociation of gas can take place.
The complexities due to high temperatures associated with hypersonic flow
makes its study complicated.
2.5
STREAMTUBE AREA-VELOCITY RELATION
In this section, let us consider quasi-one-dimensional flow, allowing the
streamtube area A to vary with distance x, as shown in Fig. 2.8.
26
Gas Turbines
A = A(x)
p = p(x)
y
ρ = ρ(x)
T = T(x)
V = V(x)
x
z
Fig. 2.8 Quasi-one-dimensional flow
Let us continue to assume that all flow properties are uniform across
any given cross-section of the streamtube, and hence are functions of x
only for steady flows. Such a flow, where A = A(x), p = p(x), ρ = ρ(x),
and c = c(x) for steady flow, is defined as quasi-one-dimensional flow.
Algebraic equations for steady quasi-one-dimensional flow can be obtained
by applying the integral form of the conservation equations.
For any streamtube of area A, the continuity equation is given by
ρAc
= constant
(2.77)
Differentiating with respect to c, we obtain
d(ρAc)
dA
d(ρc)
= ρc
+A
=0
dc
dc
dc
A
Since,
d(ρc)
dc
dp
dc
= A ρ+c
dρ dp
dp dc
= Aρ 1 − M 2
= −ρc
and from the Laplace equation, we have
dp
= a2
dρ
Therefore,
ρc
dA
+ Aρ 1 − M 2
dc
dA
dc
= 0
= −
A
1 − M2
c
(2.78)
Equation 2.78 is an important result. It is called the area-velocity relation.
The following information can be derived from the area-velocity relation:
(i) For incompressible flow limit, i.e., for M → 0, Eq. 2.78 shows that
Ac = constant. This is the famous volume-conservation equation or
continuity equation for incompressible flow.
Review of Basic Principles
27
(ii) For 0 ≤ M ≤ 1, a decrease in area results in increase of velocity, and
vice versa. Therefore, the velocity increases in a convergent duct and
decreases in a divergent duct. This result for compressible subsonic
flows is the same as that for incompressible flow.
(iii) For M > 1, an increase in area results in increase of velocity and vice
versa, i.e., the velocity increases in a divergent duct and decreases
in a convergent duct. This is directly opposite to the behaviour of
subsonic flow in convergent and divergent ducts.
(iv) For M = 1, by Eq. 2.78, dA/A = 0, which implies that the location
where the Mach number is unity, the area of the passage is either
minimum or maximum. We can easily show that the minimum in
area is the only physically realistic solution.
The above results are shown schematically in Fig. 2.9.
Velocity, c
c2 > c1
Increases
c2 < c1
Subsonic, M < 1
Supersonic, M > 1
c1
c2
c1
c1
c2
c1
c2
c2
Decreases
Fig. 2.9 Flow in convergent and divergent ducts
From the above discussions, it is c1ear that for a gas to expand isentropically from subsonic to supersonic speeds, it must flow through a convergentdivergent duct, as shown in Fig.2.10. The minimum area that divides the
convergent and divergent sections of the duct is called the throat. From the
point (iv) above, we know that the flow at the throat must be sonic with
M = 1. Conversely, for a gas to get compressed isentropically from supersonic to subsonic speeds, it must again flow through a convergent-divergent
duct, with a throat where sonic flow occurs.
28
Gas Turbines
V increasing
M<1
M>1
Throat�
M =1
Fig. 2.10 Flow in a convergent-divergent duct
2.6
NORMAL SHOCK WAVES
The shock may be described as a compression front in a supersonic flow
field and the flow process across the front results in an abrupt change in
fluid properties. The thickness of the shocks is comparable to the mean
free path of the gas molecules in the flow field. To have some physical feel
about the formation of such shock waves, consider a cylinder placed in a
flow as shown in Fig. 2.11. From the kinetic theory of gases it is known that
the flow consists of a large number of fluid molecules in unit volume. The
transport of mass, momentum and energy takes place through the motion
of these molecules. Also, the molecules carry the signals about the presence
of the cylinder around the flow field at a speed equal to the speed of sound.
In Fig. 2.11(a), the incoming stream is subsonic, c∞ < a∞ , and the
molecules far upstream of the cylinder get the information about the presence of the body through the signals which travel with speed, a∞ , well
in advance before reaching the cylinder. Therefore, the molecules orient
themselves in order to flow around the cylinder as shown in Fig. 2.11(a).
But when the incoming stream is supersonic, the molecules travel faster
than the signals. Therefore, there is no possibility that they will be informed of the presence of the body, before they reach the cylinder. Also,
the reflected signals from the face of the cylinder tend to coalesce a short
distance ahead of the body. Their coalescence forms a thin compression
front called shock wave, as shown in Fig. 2.11(b). Upstream of the shock,
the flow has no information about the presence of the body. However, the
streamlines behind the normal shock quickly compensate for the obstruction, since the flow is subsonic after a normal shock. Although the shock
formation discussed above is for a specific situation, the mechanism described is, in general, valid. However, we should realize that, when the flow
just starts, there is no shock. The formation of shock takes place after the
fluid molecules impinge on to the face of the cylinder and rebound.
2.7
EQUATIONS OF MOTION FOR A NORMAL SHOCK WAVE
For a quantitative analysis of changes across a normal shock wave, let us
consider an adiabatic, constant-area flow through a non-equilibrium region,
Review of Basic Principles
29
Voo < a oo
Moo< 1
(a) Subsonic flow
Voo > a oo
Moo> 1
Shoc
k
(b) Supersonic flow
Fig. 2.11 Streamlines around a blunt-faced cylinder in subsonic and supersonic flows
as shown in Fig. 2.12(a). Let we assume that sections 1 and 2 be sufficiently
away from the nonequilibrium regions so that we can define flow properties
at these stations, as shown in Fig. 2.12(a). Now we can write the equations
of motion for the flow considered, as follows: By continuity,
ρ1 c1
=
ρ2 c2
(2.79)
The momentum equation
p1 + ρ1 c21
=
p2 + ρ2 c22
(2.80)
=
1
h2 + c22
2
(2.81)
The energy equation
1
h1 + c21
2
Equations 2.79 to 2.81 are general – they apply to all gases. Also, there is no
restriction on the size or details of the nonequilibrium region as long as the
reference sections 1 and 2 are outside of it. The solution of these equations
will provide the relations that must exist between the flow parameters at
these two sections. Since, there is no restriction on the size or details of
the nonequilibrium region, let us idealize it by a vanishingly thin region, as
shown in Fig. 2.12(b), across which the flow parameters are said to jump.
The control sections 1 and 2 may also be brought arbitrarily close to it.
30
Gas Turbines
p
p
V1
V2
T1
T2
ρ
ρ
1
2
1
2
1
2
Non-equilibrium region
(a)
Shock
p
p
V1
V2
T1
T2
ρ
ρ
2
1
2
1
1
2
(b)
Fig. 2.12 Flow through a normal shock
Such a front or discontinuity across which there is sudden change in flow
properties is called a shock wave. There is no heat added or taken away
from the flow as it traverses the shock wave; hence the flow across the shock
wave is adiabatic.
At this stage, one question obviously arises. In a real fluid, is it possible to have a discontinuity in a continuum flow field? It is to be clearly
understood that the above consideration is only an idealization. Actually
very high gradients occur in a shock wave, in the transition from state 1
to state 2. These severe gradients produce viscous stress and heat transfer,
i.e., nonequilibrium conditions inside the shock. The processes taking place
inside the shock wave are extremely complex. It cannot be explained on the
basis of equilibrium thermodynamics. Temperature and velocity gradients
internal to the shock provide heat conduction and viscous dissipation. This
makes the process in the shock wave internally irreversible. In most of the
practical applications, the primary interest is not generally focused on the
internal mechanism of the shock wave. We are more interested on the net
changes in fluid properties taking place across the wave. However, there
are situations where the detailed information about the flow mechanism
inside the shock describing its structure is essential for studying practical
problems! However, these are beyond the scope of the book.
2.8
OBLIQUE SHOCK AND EXPANSION WAVES
It should be understood that in the normal shock, a compression wave
normal to the flow direction is generated. However, in a majority of physical
Review of Basic Principles
31
situations, a compression wave inclined at an angle to the flow occurs. Such
a wave is called an oblique shock.
Hence, the normal shock wave can be considered as a special case of
oblique shock wave. If we superimpose a uniform velocity normal to the
upstream flow of the normal shock, then, it will result in a flow filed of
oblique shock wave. This phenomenon can be used to get the oblique shock
relations. Oblique shocks usually occur when a supersonic flow is turned
into itself. The opposite of this, i.e., when a supersonic flow is turned away
from itself, results in the formation of an expansion fan. These two families
of waves play a dominant role in all flow fields involving supersonic velocities. Typical flows with oblique shock and expansion fan are illustrated in
Fig. 2.13.
Oblique shock
1
Expansion fan
2
M 2 < M1
M 2 > M1
1
M1
M1
2
M2
θ
θ
(a) Compressor corner
(b) Expansion corner
Fig. 2.13 Supersonic flow over corners
In Fig. 2.13(a) the flow is deflected into itself by the oblique shock. All
the streamlines are deflected to the same angle θ at the shock, resulting in
uniform parallel flow downstream of shock. The angle θ is referred to as flow
deflection angle. Across the shock wave, the Mach number decreases, and
the pressure, density, and temperature increase. The corner which turns
the flow into itself is called compression or concave corner. In contrast, in
an expansion or convex corner, the flow is turned away from itself through
an expansion fan. All the streamlines are deflected to the same angle θ
after the expansion fan, resulting in uniform parallel flow downstream of
the fan. Across the expansion wave, the Mach number increases, and the
pressure, density and temperature decrease. From Fig. 2.13, it is seen that
the flow turns suddenly across the shock and the turning is gradual across
the expansion fan, and hence all flow properties through the expansion fan
change smoothly, with the exception of the wall streamline which changes
suddenly.
Oblique shock and expansion waves prevail in two- and three-dimensional
supersonic flows, in contrast to normal shock waves, which are one-dimensional.
In this chapter, we shall focus our attention on steady, two-dimensional
(plane) supersonic flow.
32
Gas Turbines
2.9
FLOW WITH FRICTION AND HEAT TRANSFER
So far, we have discussed compressible flow of gases in ducts, where changes
in flow properties were brought about solely by area change, i.e., where
effects of viscosity are neglected. But, in a real flow situation like stationary power plants, aircraft propulsion engines, high-vacuum technology,
transport of natural gas in long pipelines, transport of fluids in chemical
process plants, and various types of flow systems, the high-speed flow travels through passages of sufficient length, the effects of viscosity (friction)
cannot be neglected. In many practical flow situations, friction may even
have a decisive effect on the resultant flow characteristics. The inclusion of
friction terms in the equations of motion makes the analysis of the problem
far more complex.
2.10
FLOW IN CONSTANT-AREA DUCT WITH FRICTION
Consider one-dimensional steady flow of a perfect gas, with constant specific heats, through a constant-area duct. Also, let there be neither external
heat exchange nor external shaft work and let differences in elevation produce negligible changes compared to frictional effects. The flow with the
above mentioned conditions, namely, adiabatic flow with no external work,
is called Fanno line flow.
Let the wall friction (due to viscosity) be the chief factor bringing about
changes in fluid properties, for the adiabatic compressible flow through
ducts of constant-area,under consideration.
The energy equation of steady flow under the above assumptions may
be written as
h+
c2
2
=
h0
(2.82)
where h and c are respectively the corresponding values of the enthalpy
and velocity at an arbitrary section of the duct, and h0 (the stagnation
enthalpy) has a constant value for all sections of the duct. By equation of
continuity,
ṁ
= ρc = G
(2.83)
A
where ρ is the density at the section where c and h are measured, and G is
called the mass velocity, which has a constant value for all sections of the
duct.
Combining Eqs. 2.82 and 2.83, we get the equation of the Fanno line in
terms of the enthalpy and density as
G2
h = h0 − 2
(2.84)
2ρ
where h and h0 are the static and stagnation enthalpy, ρ is density and
G is mass velocity. This equation shows that for a given initial condition,
Review of Basic Principles
33
the relation between the local density ρ and local enthalpy h is fixed. This
implies that the relation between any two properties of the flowing gas is
also fixed.
The friction coefficient f is defined as
f
=
wall shearing stress
dynamic head of the stream
The hydraulic diameter, D is defined as
4 (cross-sectional area)
D =
wetted perimeter
(2.85)
(2.86)
The advantage of using hydraulic diameter is that the equations, in terms of
hydraulic diameter, are valid even for ducts with non-circular cross-section.
The maximum length of the duct required for the flow to choke for a given
initial Mach number is given by
4f
Lmax
D
=
1 − M2
(γ + 1)M 2
γ+1
ln
+
2
γM 2
2γ
2 1 + γ−1
2 M
(2.87)
where f is the mean friction coefficient with respect to duct length, defined
by
f
=
Lmax
1
Lmax
f dx
(2.88)
0
The duct length required for the flow to pass from a given initial Mach
number M1 to a given final Mach number M2 can be obtained from the
expression
4f
L
D
=
4f
Lmax
D
M1
− 4f
Lmax
D
(2.89)
M2
It should be understood that Fanno flow friction always drives the Mach
number towards unity, decelerating a supersonic flow and accelerating a
subsonic flow. For any given initial Mach number, for a certain value of L
the flow becomes sonic. For this condition the flow is said to be choked,
since any further increase in L is not possible without causing a drastic
change of the inlet conditions. For instance, if the inlet conditions were
achieved by expansion through a supersonic nozzle, and if L were larger
than that allowed for attaining Mach 1 at the exit, then a normal shock
would form inside the supersonic nozzle and the duct inlet conditions would
suddenly become subsonic.
It is important to note that friction always causes the total pressure to
decrease whether the inlet flow is subsonic or supersonic. Further, unlike
the Rayleigh curve for flow with heating or cooling, the upper and lower portions of the Fanno curve cannot be traversed by the same one-dimensional
flow. In other words, it is not possible to first decelerate a supersonic flow
to sonic condition by friction, and then further retard it to subsonic speeds
34
Gas Turbines
also by friction, since such a subsonic deceleration violates the second law
of thermodynamics.
In summary, it can be stated that change of state in flow properties is
achieved by the following three means:
(i) with area change, treating the fluid to be inviscid and passage to be
frictionless,
(ii) with friction, considering the heat transfer between the surrounding
and system to be negligible, and
(iii) with heat transfer, assuming the fluid to be inviscid and passage to
be frictionless.
These three types of flows are called isentropic flow, frictional or Fannotype flow , and Rayleigh-type flow , respectively.
All gas dynamic problems encountered in practice can be classified under these three flow processes, of course with appropriate assumptions.
Although it is impossible to have a flow process which is purely isentropic
or Fanno-type or Rayleigh-type. In practice, it is justified in assuming so,
since the results obtained from these processes prove to be accurate enough
for most of the practical situations in gas dynamics.
Flows in which wall friction is the chief factor bringing about changes
in fluid properties, assuming that no heat is transferred to or from the fluid
stream are termed Fanno-type flow. When the ducts are short, the flow is
approximately adiabatic. However, when the ducts are extremely long, as
in the case of natural gas pipe lines, there is sufficient area for heat transfer
to make the flow non-adiabatic and approximately isothermal.
2.10.1
Laminar Flow
In laminar flow the fluid flows over a body in orderly parallel layers with
no components of fluctuations in any of the three directions (x, y and z
directions). In such a flow, the surface friction force predominates and
keeps the flow parallel to the surface. Other layers of flow slide on top of the
other. The values of the Reynolds number in such flows are comparatively
lower.
2.10.2
Turbulent Flow
At higher values of the Reynolds number, the inertia force becomes predominant and the fluid particles are no longer constrained to move in parallel
layers. Such a flow experiences small fluctuation components cx , cy and cz
in the three reference directions. These fluctuations cause continuous mixing of various layers of the flow leading to flow equalization in the major
part of the flow field.
On account of different flow patterns in laminar and turbulent flow, the
velocity profiles (shown in Fig. 2.14) in these are different. The nature of
flow in blade passages in a turbine can be identified to a great extent by
the velocity profiles and the values of Reynolds number.
Review of Basic Principles
Flow passage
Laminar
Turbulent
35
36
Gas Turbines
2.10.5
Friction Factor
Friction factor or the coefficient of skin friction is a measure of the frictional
resistance offered to the flow. This is defined by
f
=
τw
1
2
2 ρc
(2.91)
where f is the Fanning’s coefficient of skin friction. It may be noted that
Darcy’s friction factor is four times the Fanning’s coefficient.
2.10.6
Boundary Layer Separation
The boundary layer is the slow-moving or tired layer of the flow near a
solid surface. When the flow occurs in the direction of static pressure
rise (adverse pressure gradient), the boundary layer becomes thicker and
reverses if this static pressure gradient (or the pressure hill) is too high. The
leaving of the boundary layer from the surface and its reversal is known as
separation. This leads to chaotic flow, large drag and high energy losses.
In an accelerating flow, on account of the continuously decreasing static
pressure the thickening of the boundary layer is prevented; in fact, the
higher inertia forces make it thinner. The available pressure drop helps in
washing down any localized thickening of the boundary layer or its separation.
The separation of boundary layer and the point of separation depend on
the geometry and roughness of the surface, nature of the flow (turbulent or
laminar) and Reynolds number. The laminar boundary layer gets separated
earlier than the turbulent boundary layer.
In order to achieve high lift and good performance it is necessary to
prevent or delay the separation of the boundary layer. Two of the methods
to achieve this are (i) sucking away the decelerated layer, and (ii) energizing
it by injecting high energy fluid parallel to the surface. Separation can also
be delayed by achieving transition of the laminar flow into the turbulent
earlier. In all turbomachines boundary layer separation should be avoided.
Review Questions
2.1 State the various laws used in the design of turbomachines.
2.2 Explain (i) open system, and (ii) closed system.
2.3 State the relationship between enthalpy and internal energy?
2.4 State Clausius and Kelvin-Planck’s statement of second law of thermodynamic.
2.5 Explain the flow and non-flow process.
2.6 Derive steady-state energy equation.
2.7 Explain the terms ’stagnation pressure’ and ’stagnation temperature’.
Review of Basic Principles
37
2.8 What are streamlines and streamtubes? Explain with a sketch.
2.9 Explain the difference between laminar and turbulent flow.
2.10 What is a boundary layer? Give a brief account of boundary layer
separation.
Multiple Choice Questions (choose the most appropriate answer)
1. Pressure can be expressed in units
(a) Pascal
(b) N/m2
(c) bar
(d) none of the above
2. Cv can be expressed as
(a) Cv =
∂q
∂T
(b) Cv =
∂T
∂q
(c) Cv =
(d) Cv =
v
v
∂T
∂v v
∂h
∂P v
3. Cp can be expressed as
(a) Cp =
∂T
∂h v
(b) Cp =
∂T
∂q
(c) Cp =
∂q
∂T
(d) Cp =
v
v
∂h
∂T v
4. A process that occurs in a closed system is
(a) closed flow process
(b) non-flow process
(c) open flow process
(d) none of the above
5. A process that occurs in a open system is
(a) open flow process
(b) non-flow process
(c) flow process
(d) none of the above
38
Gas Turbines
6. Energy transformation can occur
(a) in moving blades
(b) in fixed blades
(c) both moving and fixed blades
(d) none of the above
7. If the relative change in the density of a fluid in a process is negligible
then it is called
(a) compressible flow
(b) incompressible flow
(c) steady flow
(d) unsteady flow
8. Reynolds number is the ratio of
(a) elastic force / viscous force
(b) inertia force / elastic force
(c) viscous force / inertia force
(d) inertia force / viscous force
9. Mach number is the ratio of
(a) inertia force / viscous force
(b) inertia force / elastic force
(c) viscous force / inertia force
(d) elastic force / viscous force
10. In a pipe flow if the Reynolds number is greater than 2000 then it is
considered as
(a) laminar
(b) steady
(c) turbulent
(d) unsteady
Ans:
1. – (d)
6. – (c)
2. – (a)
7. – (b)
3. – (d)
8. – (d)
4. – (b)
9. – (b)
5. – (c)
10. – (c)
3
FUNDAMENTALS OF
ROTATING MACHINES
INTRODUCTION
In the last chapter we reviewed some basic principles useful for the gas
turbine cycle analysis. In this chapter we will deal with the fundamental
principle of operation of rotating machines. Rotating machines are usually called turbomachines. These machines work on the principle of work
addition or extraction. When a fluid passes through a rotating machine
two things happen, viz., energy transfer and energy transformation. The
energy transfer means transfer of available energy from one part (rotor) to
the medium (fluid) or vice versa. Energy transformation means change of
one form of energy into another form, for example, change of kinetic energy
to pressure energy in a compressor. The energy transfer can occur only
in its moving or rotating elements whereas the energy transformation can
occur in both stationary and rotating elements.
As both the compressors and turbines are concerned with energy transfer, we will consider their basic performance together. In compressors the
energy is transferred from the rotor to the fluid while in turbines it is from
fluid to the rotor. The effectiveness of this transfer of energy in a fluid
machine is governed mainly by the fluid dynamics of the system.
3.1
GENERAL FLUID DYNAMICS ANALYSIS
Figure 3.1 shows the details of the passage of a fluid through a rotor of any
shape. The rotor has an axis A − A and rotates at a steady angular velocity
of ω radians per second.
Let us assume that at point 1 the fluid enters with a velocity c1 and
leaves at point 2 with velocity c2 . The radial distance of these points from
the axis A − A is r1 and r2 respectively. The velocity c1 can be represented
by three velocity components, viz.,
(i)
ca1
: axial velocity in a direction parallel to the axis A − A
40
Gas Turbines
c r2
ca2
c2
A
2
r2
ct2
w rad/s
r1
A
1
c t1
c a1
c r1 c
1
Fig. 3.1 Fluid flow through a rotor
(ii) cr1
(iii) ct1
: radial velocity in the direction normal to A − A
: tangential velocity in the direction normal to any radius
Similarly, at the exit point 2, the velocity c2 will have the three components,
ca2 , cr2 and ct2 respectively.
Figure 3.2 shows the velocity triangles at the entry as well as at the
exit of a general rotating machine. All velocity vectors shown are in the
same plane and assumed to remain constant over the entire entry and exit
section.
Exit
Blade row
Entry
c
1
c2
c
r2
w2
c t2
u2
c r1 w1
ct1
w
u1
r1
r2
Fig. 3.2 Energy transfer in a turbomachine
The angular speed of the rotor is ω radians per second.
ω
=
2πN
60
Fundamentals of Rotating Machines
41
The peripheral velocities of the blades at the entry and exit corresponding to diameters d1 and d2 are
u1
=
πN d1
60
u2
=
πN d2
60
The directions of the relative velocity vectors correspond to the rotor blade
angles. The absolute velocity vectors, c, are those which will be observed by
a person standing outside the rotor, whereas, the relative velocity vectors,
w are the ones which will be observed by an observer positioned on the
rotor.
The three velocity vectors c, w and u at a section (or station) are related
by the simple vector equation
c
=
u+w
The absolute velocity c at both the entry and exit has a tangential
component ct and a radial component cr .
The torque exerted by the rotor or by the fluid is obtained by employing
Newton’s second law of motion for the change of moment of momentum.
Torque
=
Rate of change in moment of momentum
Considering the interaction of fluid and rotor it will be seen that axial
and radial components do not contribute to the rotation. The axial component, ca , produces thrust and the radial component, cr , the radial force.
The tangential component, ct , produces rotational effects. Considering unit
mass of fluid, entering and leaving in unit time, we can write
The angular moment of momentum at inlet
= ct1 r1
(3.1)
The angular moment of momentum at exit = ct2 r2
(3.2)
As per Newton’s law, the rate of change of angular moment of momentum
is equal to the torque produced. Therefore,
T
= ct1 r1 − ct2 r2
(3.3)
and hence the rate of energy transfer for unit mass flow per unit time will
be given by
E
=
Tω
= ω(ct1 r1 − ct2 r2 )
=
ct1 r1 ω − ct2 r2 ω
(3.4)
But, ωr1 = u1 and ωr2 = u2 . Hence,
E
= ct1 u1 − ct2 u2
(3.5)
The above equation is called Euler’s energy equation. The equation can be
analyzed under two conditions, viz.,
42
Gas Turbines
(i) For ct1 u1 > ct2 u2 , in which the energy transfer is positive
(ii) For ct1 u1 < ct2 u2 , in which the energy transfer is negative
Positive energy transfer means that the energy is transferred from fluid
to rotor and therefore it is called turbine rotor. Negative energy transfer
means that the energy is transferred from rotor to the fluid and therefore
it is called compressor rotor. Therefore, the energy transfer equation can
be written now separately as
ET
= ct1 u1 − ct2 u2
(3.6)
EC
= ct2 u2 − ct1 u1
(3.7)
Equations 3.6 and 3.7 can be used freely for any type of turbine or compressor respectively which satisfy the following few conditions.
(i) The flow must be steady, i.e., there should not be any change in
angular velocity, flow rate, fluid properties and heat transfer rate
with respect to time.
(ii) The relationship applies strictly to every infinitesimal stream line,
i.e., if the velocity is not uniform over the inlet and exit areas then
an integration over each area must be done.
(iii) There should not be any discontinuity of pressure, i.e., a choked nozzle
at the rotor discharge.
In almost all applications, conditions (i), (ii) and (iii) are mostly satisfied.
3.2
THE PHYSICAL MEANING OF THE ENERGY EQUATION
Consider Fig. 3.3 which is the velocity diagram at the exit from a rotor.
From geometry of the velocity triangle we have
c2r2
=
c22 − c2t2
c2r2
=
w22 − (u2 − ct2 )
and also
(3.8)
2
(3.9)
where cr2 is called meridional component. Equating the values of c2r2 and
expanding
c22 − c2t2
=
w22 − u22 + 2ct2 u2 − c2t2
(3.10)
ct2 u2
=
1 2
c + u22 − w22
2 2
(3.11)
hence,
Fundamentals of Rotating Machines
43
A similar expression can be obtained for inlet, thus
ct1 u1
=
1 2
c + u21 − w12
2 1
(3.12)
Now the energy transfer as per Eq. 3.5 can be written as
E = ct1 u1 − ct2 u2
hence,
E
=
⎤
⎡
1⎢ 2
⎥
⎣ c1 − c22 + u21 − u22 + w22 − w12 ⎦
2
I
II
c2
(3.13)
III
w2
cr2
c t2
u2
Fig. 3.3 Velocity diagram at the exit of a rotor
Let us consider the three terms (I), (II) and (III) separately, in order
to understand their significance.
c2 −c2
The first term 1 2 2 represents the energy transfer due to change of absolute kinetic energy of the fluid between the entrance and the exit sections.
This effect is also known as impulse effect. Since it is concerned with the
absolute velocity, it represents a virtual pressure rise which can be accomplished if c2 is reduced in a diffuser so as to bring about a rise in pressure.
The rotor of a compressor will usually increase the absolute velocity of the
fluid but it does not necessarily increase the fluid pressure. Therefore, an
additional device is needed in order to slow down the absolute velocity to
some lower magnitude so that a static pressure rise may be accomplished.
All compressors have either a scroll or a diffuser or some type of vanes
which follow the rotating passage in order to reduce the exit velocity of the
fluid. As this absolute kinetic energy change may be used to accomplish a
rise in pressure, it can be called a virtual pressure rise or a pressure rise
which is theoretically possible to attain. If we can slow down the air velocity
by some means from c2 to c1 we can attain the ideal head provided that
the process is isentropic. Since this pressure rise comes from some device
c2 −c2
which is external to the rotor, the term, 2 2 1 is sometimes called external
effect.
44
Gas Turbines
The other two terms, viz., the second and the third in Eq. 3.13 are
factors which produce pressure rise within the rotor itself and therefore
they are called internal effect.
u2 −u2
The term 1 2 2 represents the energy transfer due to centrifugal effect.
It is well known from fluid mechanics that a rotating fluid in equilibrium
has a pressure at any radius proportional to the product of the square of
radius and the square of the angular velocity, ω 2 r2 , i.e., u2 .
w 2 −w 2
The third term 2 2 1 represents the energy transfer due to the change
of the relative kinetic energy of the fluid. This is nothing but a reaction
effect and gives rise to the change in static pressure.
From the above discussion, it is apparent that a turbomachine derives
its pressure change due to three distinct effects. The first term represents
energy transfer due to the change of kinetic energy. The last two terms
represent energy change involving change of static pressure. Thus, energy
transfer is dependent on a change of stagnation or total pressure, of which
dynamic component may be important. In some rotating machines it may
happen that only one of the three individual types of energy transfer is
present. In particular, the second term of the Eq. 3.13 shows that a radial
path of the fluid can be effective. That is to say that substantial effect may
be produced by allowing the fluid to enter and leave the rotor at different
radii. It is also possible for one of the components to be acting in the
opposite direction to the other two so that the net effect is as desired.
3.3
CLASSIFICATION OF MACHINES
The compressor and turbine can be classified based upon (i) the flow direction, and (ii) the degree of reaction.
The details of these two parameters are discussed in the following sections.
3.3.1
Classification Based upon Flow Direction
Based on the flow direction, machines can be divided into axial, radial or
mixed flow type.
(i) Axial Flow The machines which have no significant change of radius
between flow entry and exit, i.e., u1 ≈ u2 are called axial machines.
(ii) Radial Flow In this type there is a substantial change of energy due
to change of radius between the entry and the exit of flow. These
are often called as centrifugal machines in case of compressors and
centripetal machines in case of turbines.
(iii) Mixed Flow Although the majority of compressors and turbines in
present day use may be classified as axial or radial, it is possible to
have mixed-flow types in which neither effect clearly dominates.
Fundamentals of Rotating Machines
3.3.2
45
Classification Based upon the Degree of Reaction
Degree of reaction [R] is usually defined as the rate of energy transfer by
virtue of change of static pressure to the total energy transfer.
R =
1
2
u21 − u22 + w22 − w12
E
(3.14)
where E may be expressed either by Eq. 3.6 or by Eq. 3.7 depending on
whether it is for turbine or compressor respectively. Another way of defining
R is as follows:
Change of enthalpy in rotor
R =
(3.15)
Change of enthalpy in stage
where stage is the combination of stator and rotor.
Hence, for compressor
R =
Enthalpy increase in rotor
Enthalpy increase in stage
R =
Enthalpy drop in rotor
Enthalpy drop in stage
and for turbine
(3.16)
Based on the above definitions the machines can be classified as
(i) Impulse Machine In impulse machines there is no change of static
pressure in the rotor. Therefore these machines have a degree of
reaction (R) equal to zero. In the case of a turbine of this type the
energy transfer is wholly effected by a jet of fluid striking the blade.
A simple example of pure impulse machine is a paddle wheel or a
Pelton wheel.
(ii) Reaction Machine The reaction principle is best illustrated by rocket
propulsion or ordinary lawn sprinkler. In a pure reaction machine,
R = 1, all energy transfer occurs by virtue of change of static pressure
in the rotor. A reaction of unity for a compressor means that the fluid
enters and leaves with the same absolute velocity.
3.4
GENERAL THERMODYNAMIC ANALYSIS
So far we have talked about the energy transfer by the application of the
principle of fluid dynamics, i.e., in terms of rotor and fluid velocities. The
work output or the energy transfer can also be determined based on thermodynamic considerations. The two equations, viz., the fluid dynamics
and thermodynamics equations can be linked to determine the unknown
quantities.
The process of expansion and compression taking place in a gas turbine
engine will fall into the category of steady flow process. Considering unit
46
Gas Turbines
Q
W
1
2
c1
c2
.
.
.
.
.
.
.
.
Z1
.
Z2 = Z 1
Fig. 3.4 Thermodynamic analysis of fluid flow
mass of the working fluid and writing the general energy equation (Fig. 3.4),
we get
U1 + p1 V1 +
c21
c2
+ gZ1 + Q + W = U2 + p2 V2 + 2 + gZ2
2
2
(3.17)
where the symbols denote
U
pV
c
Z
Q&W
:
:
:
:
:
Internal energy
Flow work done on or by the fluid
Velocity of the fluid
Potential energy
Heat and work supplied from external source
In the case of gas turbine, the rate of flow of working fluid is quite large,
and the surface area available for transfer of heat is quite small. Therefore,
the process may be assumed to be adiabatic, i.e.,
Q
=
0
Equation 3.17 can now be written as
U1 + p1 V1 +
c21
+ gZ1 + W
2
=
U2 + p2 V2 +
c22
+ gZ2
2
If Z2 ≈ Z1 as in Fig. 3.4, Eq. 3.18 reduces to
W
= (U2 + p2 V2 ) − (U1 + p1 V1 ) −
c2
c21
− 2
2
2
= (Cv T2 + RT2 ) − (Cv T1 + RT1 ) −
c2
c21
+ 2
2
2
(3.18)
Fundamentals of Rotating Machines
= (T2 − T1 )(Cv + R) −
W
= Cp (T2 − T1 ) −
=
W
h2 +
c22
2
c2
c21
+ 2
2
2
47
(3.19)
c2
c2
c21
c2
+ 2 = (h2 − h1 ) − 1 + 2
2
2
2
2
− h1 +
c21
2
(3.20)
= h02 − h01
(3.21)
= Δh0
(3.22)
and if we neglect the change in kinetic energy then
W
= Δh
(3.23)
[Suffix 0 denotes total head or stagnation condition].
If in Eq. 3.22 or 3.23, change of enthalpy is positive then the work input
is considered positive and the case is that of a compressor. If it is negative,
that means the work input is negative, i.e., there is a work output and the
case is that of a turbine.
(3.24)
3.5
EFFICIENCY OF ROTATING MACHINES
An actual machine receiving gas at one particular condition and exhausting
at another condition will, because of losses, do less work than represented
by Eq. 3.23 for an expansion process and will absorb more amount of work
for a compression process. This action gives rise to the term efficiency
commonly called isentropic efficiency of the machine.
For a compression machine
Isentropic efficiency (ηC )
=
Isentropic work input
Actual work input
ηC
=
h2 − h1
h2 − h1
(3.25)
ηC
=
T2 − T1
T2 − T1
(3.26)
Isentropic efficiency (ηT )
=
Actual work output
Isentropic work output
ηT
=
h3 − h4
h3 − h4
If Cp is taken to be constant, then
For an expansion machine
If Cp is taken to be constant, then
(3.27)
48
Gas Turbines
ηT
=
T3 − T4
T3 − T4
(3.28)
ta
nt
The isentropic efficiencies of compressor and turbine can be evaluated from
the enthalpy-entropy (h-s) diagrams, the details of which are given in
Figs.3.5(a) and (b) respectively.
co
ns
3
p
=
nt
h
2’
ta
ns
2
2
=
co
p1
nt
h
a
nst
o
=c
p2
p 1=
ant
st
con
4
4’
1
s
s
(a) Compressor
(b) Turbine
Fig. 3.5 h-s diagram for compressor and turbine
3.6
DIMENSIONAL ANALYSIS OF ROTATING MACHINES
Basically, dimensional analysis is a method for reducing the number and
complexity of experimental variables which affect a given physical phenomenon, using a sort of compacting technique. If a phenomenon depends
upon n dimensional variables, dimensional analysis will reduce the problem to only k dimensionless variables, where the reduction n − k = 1, 2, 3
or 4 depending upon the problem complexity. Generally n − k equals the
number of different dimensions (sometimes called basic or primary or fundamental dimensions) which govern the problem. In fluid flow problems,
the four basic dimensions are usually taken to be mass M , length L, time
T and temperature Θ or an M LT Θ system for short. Sometimes one uses
an F LT Θ system, with force F replacing mass, M . In SI system of units
(System Internationale d’units) the units and dimensions used for fluid mechanics properties are given in Table 3.1.
Dimensional analysis of problems in turbomachines identifies the variables involved and groups them into non-dimensional quantities much lesser
in number than the variables themselves. In a design problem or performance test these non-dimensional quantities (or dimensional parameters or
numbers as they are sometimes called) are varied instead of the large number of parameters forming these groups. While a great convenience and
economy in test runs provided by employing this technique is obvious, the
design procedure uses these non-dimensional numbers to obtain optimum
performance. Some non-dimensional numbers give an idea of the type of
Fundamentals of Rotating Machines
Table 3.1 Units and Dimensions of Fluid-Mechanics Properties
Quantity
Units
Symbol
Dimensions
M LT Θ
F LT Θ
Length
m
L
L
L
Area
m2
A
L2
L2
Volume
m3
V
L3
L3
Velocity
m/s
c or u
LT −1
LT −1
Speed of sound
m/s
a
LT −1
LT −1
Volume flow
m3 /s
Q
L3 T −1
L3 T −1
Mass flow
kg/s
ṁ
M T −1
F T L−1
Pressure, stress
kg/ms2
p, σ
M L−1 T −2
F L−2
Strain rate
1/s
T −1
T −1
Angle
None
θ
None
None
Angular velocity
1/s
ω
T −1
T −1
Viscosity
kg/m s
μ
M L−1 T −1
F T L−2
Kinematic viscosity
m2 /s
ν
L2 T −1
L2 T −1
Surface tension
kg/s2
γ
M T −2
F L−1
Force
kg m/s2
F
M LT −2
F
Moment, torque
kg m2 /s2
M
M L2 T −2
FL
Power
kg m2 /s
P
M L2 T −1
F LT −1
Work, energy
kg m2 /s2
W, E
M L2 T −2
FL
Density
kg/m3
ρ
M L−3
F T 2 L−4
Temperature
K
T
Θ
Θ
Specific heat
m2 /s2 K
Cp , Cv
L2 T −2 Θ−1
L2 T −2 Θ−1
Thermal conductivity
kg m/s3 K
k
M LT −3 Θ−1
F T −1 Θ−1
Expansion coefficient
1/K
β
Θ−1
Θ−1
49
50
Gas Turbines
machine and its range of operation. The presentation of the performance
of a machine is also considerably simplified by adopting non-dimensional
numbers.
3.6.1
The pi Theorem
There are several methods of reducing a number of dimensional variables
into a smaller number of dimensionless groups. The scheme given here was
proposed in 1914 by Buckingham and is now called the Buckingham pi
theorem. The name pi comes from the mathematical notation Π, meaning
a product of variables. The dimensionless groups found from the theorem
are power products denoted by Π1 , Π2 , Π3 , etc., The method allows the pis
to be found in sequential order without resorting to free exponents.
The first part of the pi theorem explains what reduction in variables to
expect:
If a physical process satisfies the Principle of Dimensional Homogeneity
(PDH) and involves n dimensionless variables, it can be reduced to a relation between only k dimensionless variables or Π’s. The reduction j = n−k
equals the maximum number of variables which do not form a pi among
themselves and is always less than or equal to the number of dimensions
describing the variables.
The second part of the theorem shows how to find the pis one at a time:
Find the reduction j, then select j variables which do not form a pi
among themselves† . Each desired pi group will be a power product of these
j variables plus one additional variable which is assigned any convenient
nonzero exponent. Each pi group, thus, found is independent.
To be specific, suppose that the process involves n variables and assuming n = 5, we have,
v1 = f (v2 , v3 , v4 , v5 )
If the five variables can be expressed in terms of three (j) dimensions
(M LT ), then the number of dimensionless variables (k) or Πs will be
k = 5 − 3 = 2. That is, we can expect from the theorem, two and only two
pi groups.
Now let us pick out three convenient variables which do not form a pi
and suppose, these turn out to be v2 , v3 and v4 . Then the two pi groups
are formed by power products of these three plus one additional variable
Π1
=
M 0 L0 T 0
=
(v2 )a (v3 )b (v4 )c v1
Π2
=
M 0 L0 T 0
=
(v2 )a (v3 )b (v4 )c v5
Here, we have arbitrarily chosen the added variables, v1 and v5 , whose
exponents is taken as unity. Equating exponents of the various dimensions
is guaranteed by the theorem to give unique values of a, b and c for each
pi. And they are independent because only Π1 contains v1 and only Π2
contains v5 . It is a very neat system once you get used to the procedure.
† Make a clever choice here because all pis will contain these j variables in various
groupings
Fundamentals of Rotating Machines
51
Typically, there are six steps involved:
1. List and count the n variables involved in the problem. If any important variables are missing, dimensional analysis will fail.
2. List the dimensions of each variable according to M LT Θ or F LT Θ.
A list is given in Table 3.1.
3. Find j, initially guess j equal to the number of different dimensions
present and look for j variables which do not form a pi product. If
no luck, reduce j by 1 and look again. With practice, you will find j
rapidly.
4. Select j variables which do not form a pi product. Make sure they
please you and have some generality if possible, because they will then
appear in every one of your pi groups.
5. Add one additional variable to your j variables and form a power
product. Algebraically find the exponents which make the product
dimensionless. Try to arrange for your output or dependent variables
(force, pressure drop, torque, power) to appear in the numerator and
your plots will look better. Do this sequentially, adding one new
variable each time, and you will find all n − j = k desired pi products.
6. Write the final dimensionless function and check your work to make
sure all pi groups are dimensionless.
3.6.2
Non-dimensional Parameters for Rotating Machines
The technique of dimensional analysis is used to reduce the number of
variables into a few number of dimensionless groups. The dimensionless
groups involved in rotating machines for plotting the necessary performance
curves can be obtained in the following way.
Step 1 Write the function and count the variables. Let us express the
outlet pressure from the compressor, p2 , as function of other variables, viz.,
p2
where
p2
p1
D
ρ1
ṁ
N
:
:
:
:
:
:
=
f (p1 , ρ1 , D, ṁ, N )
(3.29)
outlet pressure from the compressor
inlet pressure to the compressor
characteristic linear dimension
density of the fluid
mass flow rate of the fluid
rotational speed
Note that in this problem there are six variables.
Step 2 Let us now write the dimensions of each of the variables (refer Table
3.1).
p : M L−1 T −2 ρ : M L−3 D : L ṁ : M T −1 N : T −1
Step 3 Find j. No variable contains the dimension Θ and so j is less or
equal to 3 (M LT ). On inspection of the list it can be seen that p1 , ρ and
52
Gas Turbines
D cannot form a pi group. Therefore, j equals 3, and n − j = 6 − 3 = 3.
The pi theorem guarantees for this problem that there will be exactly three
independent dimensionless groups.
Step 4 Select j variables. The group p1 , ρ, D will do the job.
Step 5 Combine p1 , ρ and D with one additional variable, in sequence to
find the three pi products.
First add p2 to find Π1 . You may select any exponent on this additional
term as you please, to place it in the numerator or denominator to any
power. Since p2 is the output, or dependent variable, we select it to appear
to the first power in the numerator.
Π1
=
pa1 ρb Dc p2
Π1
=
M 0 L0 T 0
M a L−a T −2a M b L−3b [Lc ] M L−1 T −2
=
=
(3.30)
M a+b+1 L−a−3b+c−1 T −2a−2
(3.31)
Equating exponents,
Mass
Length
Time
: a+b+1
: −a − 3b + c − 1
: −2a − 2
=
=
=
0
0
0
On solving the above three simultaneous equations, we get
a
=
−1;
b
=
0;
c
=
0
Π1
=
p2
p1
(3.32)
Π2
=
pa1 ρb Dc N
(3.33)
Π2
=
M 0 L0 T 0
Hence, from Eq. 3.30,
M a L−a T −2a M b L−3b [Lc ] T −1
=
M a+b L−a−3b+c T −2a−1
=
Equating exponents,
Mass : a + b
Length : −a − 3b + c
Time : −2a − 1
=
=
=
0
0
0
(3.34)
Fundamentals of Rotating Machines
53
On solving the above simultaneous equations, we have
a
1
= − ;
2
b
=
1
;
2
c
=
1
√
ND ρ
√
p1
Π2
=
Π3
= pa1 ρb Dc ṁ
Π3
= M 0 L0 T 0
(3.35)
(3.36)
M a L−a T −2a M b L−3b [Lc ] M T −1
=
= M a+b+1 L−a−3b+c T −2a−1
(3.37)
(3.38)
Equating exponents
Mass : a + b + 1
Length : −a − 3b + c
Time : −2a − 1
=
=
=
0
0
0
On solving the above simultaneous equations, we have
a
1
− ;
2
=
b
Π3
1
= − ;
2
=
c
=
ṁ
√
D 2 p1 ρ
−2
(3.39)
Step 6
Π1
= f (Π2 , Π3 )
p2
p1
= f
Therefore,
√
N D ρ1
ṁ
, 2√
√
p1
D p1 ρ
For a given compressor and working fluid, D and the gas constant, R, are
fixed. Writing p in terms of ρRT , the pressure ratio is given by
√
p2
N ṁ T1
(3.40)
= f √ ,
p1
p1
T1
The process may be repeated, letting the outlet temperature T2 be a function of the same variables, in which case TT21 will again be found to be the
√
function of √NT and ṁp1T1 . If the stagnation pressures and temperatures
1
are considered (as in the case of a gas turbine), then again Eq. 3.40 will result except that stagnation temperatures and pressures are substituted for
their static parameters in the equations. The experimentally determined
54
Gas Turbines
N
Surge line
= constant
T 01
p
02
p
01
m T 01
p
01
Fig. 3.6 Performance map of a compressor
performance map for a compressor based on the Eq. 3.40 will be as shown
in Fig. 3.6. In the above analysis, it may be noted that the viscosity was
not considered. Over the operating range of compressors and turbines the
variation of μ has a negligible effect on their performance and hence the
exclusion of viscosity from the analysis may be justified.
3.7
ELEMENTARY AIRFOIL THEORY
An airfoil may be defined as a streamlined body bounded principally by two
flattened curves and whose length and width are very large in comparison
with the thickness. It has a thick, rounded leading edge and a thin (sometimes sharp) trailing edge. Its maximum thickness occurs somewhere near
the midpoint of the chord. The backbone line lying midway between the
upper and lower surfaces is known as the camber line. When such a blade
is suitably shaped and properly oriented in the flow, the force acting on
it - normal to the direction of flow - is considerably larger than the force
resisting its motion. Aerofoil shapes are used for aircraft wing sections and
the blades of various turbomachines.
3.7.1
Nomenclature
A blade section of infinitesimal thickness is a curved line known as “camber
line” (Fig. 3.7). This forms the backbone line of a blade of finite thickness.
Its shape is specified by x-y coordinates which are conventionally presented
in the non-dimensional form x/ and y/ as shown in the figure. They are
applicable to a blade section of any chord length ( ). Maximum camber
(y/ )max and its position (a/ ) are important properties.
Fundamentals of Rotating Machines
55
y/l
0.1
Maximum camber
Camber line
y
0
0.5
x
x/l
1.0
a’
Fig. 3.7 Camber line profile
The shapes of profiles of the upper and lower surfaces of the blades are
specified by the thickness distribution along the chord or the camber line.
This may be symmetrical or unsymmetrical giving a symmetrical or an
unsymmetrical profile. Figure 3.8 shows the base profile of a symmetrical
blade. This may by used for an actual curved blade in a cascade of an axial
flow compressor.
The maximum thickness (tmax ) of the profile and its distance (a ) from
the leading edge, and the shapes of the leading and trailing edges (LE and
TE) are important parameters which govern the flow pattern and losses.
0.1
x
t/2 t max
LE 0
TE
a’
-0.1
Chord (l) or length of the camber line
0
0.5
1.0
Fig. 3.8 Thickness distribution of an aerofoil (base profile)
3.7.2
Nature of Flow
When a flat plate is moved through a fluid, it experiences certain resistance
to its motion on account of the fluid friction on its surface. If the flow
direction is parallel to its length, the force normal to the plate will be
zero. However, when the plate is inclined (at an angle i) to its direction
of motion, it will experience a resultant force R normal to its length. This
force has a drag force component (D) parallel to the flow and another lift
56
Gas Turbines
D
L
R
Flat plate
Free stream
+
i
+
-
-
-
+
+
+
Fig. 3.9 Lift force on an inclined flat plate
force component (L) perpendicular to it. The drag and lift forces are as
shown in Fig. 3.9.
As can be seen in Fig. 3.9 the lift force arises on account of the negative
and positive pressures prevailing on the upper and lower surfaces respectively. The figure shows only constant average values of pressures. In actual
practice the pressures on the two sides will vary from the leading to the
trailing edge.
When the angle of incidence (i) as shown in Fig. 3.10 is zero, the lift
force will be zero though CL = 0. The lift will increase up to an optimum
angle of attack and decrease as can be seen from Fig. 3.10. Along with lift
the drag also increases. Beyond the optimum angle of attack the drag force
starts dominating and increasing very rapidly followed by a decrease in the
lift force. This is obviously undesirable. The maximum drag occurs when
the plate is normal (i = 90◦ ) to the flow direction; the lift is zero for this
position.
In practical applications for aircraft wings and turbomachine blades, the
flat plate (if used) will have to be of finite thickness. In order to achieve
a high lift-to-drag ratio, the leading edge is rounded and the blade section
is tapered towards the thin trailing edge. To obtain further increase in the
value of L/D, the blade is slightly curved, thus giving a curved camber
line. It may be seen here that such a blade approaches a cambered aerofoil
shape.
Figure 3.11 shows (a) an uncambered aerofoil blade with zero angle of
attack, (b) an uncambered blade with angle of attack i, and (c) a cambered blade with an angle of attack i. Static pressure distribution around
a cambered aerofoil blade is shown in Fig. 3.12.
The centrifugal force on the fluid particles on the upper (convex) side
tries to move them away from the surface. This reduces the static pressure
on this side below the free-stream pressure. On account of this suction
effect, the convex surface on the blade is known as the suction side. In
contrast to this, the centrifugal force on the lower (concave) side presses
the fluid harder on the blade surface, thus increasing the static pressure
Fundamentals of Rotating Machines
57
0.7
0.6
CL
C L or C D
0.5
CD
0.4
0
-0.1
i
-0.2
-0.3
-4
-2
0
2
4
6
8
10
Angle of incidence
Fig. 3.10 Variation of lift and drag coefficients with incidence
above that of the free stream. Therefore, this side of the blade is known as
the pressure side.
Due to the above phenomenon, the flow on the suction side begins accelerating along the blade chord accompanied by a deceleration on the pressure
side. However, the common boundary conditions at the trailing edge require the flows on the two sides to equalize. Therefore, the accelerated flow
on the suction side experiences deceleration of the flow as it approaches
the trailing edge. The already decelerated flow in the pressure side starts
accelerating at some point along the chord. In spite of these processes the
flows coming from the two sides of the blade surface may fail to equalize.
The resultant upward force on the blade (Fig. 3.12) is the result of the
cumulative effect of the positive static pressure on the pressure side and
the negative pressure on the suction side.
Now the total upward force acting on the aerofoil is equal to the projected area times the pressure difference between the two sides. On an
aircraft wing there is a large area available for the production of lift force.
Therefore, only a small pressure difference over its aerofoil wing section
will provide the required lift. This requires only a slight deflection of the
approaching flow over the aircraft wings which is achieved by only slightly
cambered sections.
In contrast to this, the projected areas of turbomachine blades are much
smaller. Therefore, a considerable difference of static pressure between the
pressure and suction sides is required to provide the necessary lift or the
tangential force. This can only be achieved by providing highly cambered
blade sections.
58
Gas Turbines
Trailing edge
Leading edge
i=0
Chord = l
(a) Uncambered aerofoil with zero incidence
i
Chord line
(b) Uncambered aerofoil with incidence
Camber line
i
Camber
Chord line
(c) Cambered aerofoil with incidence
Fig. 3.11 Flow around aerofoil blades
Suction side
-
-
Aerofoil
-
+
+
+
Pressure side
Fig. 3.12 Pressure distribution around a cambered aerofoil blade
Fundamentals of Rotating Machines
59
The lift force is exerted by the fluid on the aircraft wings and on the
turbine rotor blades, whereas in power-absorbing machines like compressors, this lift force (exerted by their rotor blades on the fluid) is supplied
by the torque input.
3.7.3
Coefficients of Lift and Drag
The resultant force due to the flow around an aerofoil blade acts at its
centre of pressure. It has two components – lift force, normal to the flow
direction (or blade chord) and the drag force parallel to the flow. These
forces depend only on the density and velocity of the fluid and the blade
chord. Thus, L and D are functions of density, velocity and chord length.
L, D
=
f (ρ, c, )
The projected area per unit length of the blade is
A =
×1
The lift and drag coefficients based on this area relate the dynamic
pressure 12 ρc2 to the lift and drag forces.
L =
CL A
1
ρ c2
2
D
CD A
1
ρ c2
2
=
Conventionally, these coefficients are expressed as
CL
=
CD
=
L
1
2ρ
c2
D
c2
1
2ρ
(3.41)
(3.42)
As already stated, the lift and drag coefficients are functions of angle of
attack and the variation of CL and CD with i for an aerofoil blade will be
similar to Fig. 3.10.
Review Questions
3.1 What is the principle of operation of a rotating machine and what
factor governs the energy transfer?
3.2 What is the general energy transfer equation for a rotating machine?
Derive the same.
3.3 How do you differentiate between a turbine rotor and a compressor
rotor from the point of view of energy equation?
60
Gas Turbines
3.4 What are the conditions to be satisfied for the Euler’s energy equation
to be valid?
3.5 Discuss the physical meaning of the Euler’s energy equation.
3.6 What is meant by internal effect and external effect?
3.7 What are the two ways by which the compressor and turbine can be
classified? Discuss them briefly.
3.8 By means of a thermodynamic analysis develop an expression for the
energy transfer in a rotating machine.
3.9 Define the isentropic efficiency of a rotating machine (compressor and
turbine) and show the details on an h-s or a T -s diagram.
3.10 What is meant by dimensional analysis? Derive the dimensionless
parameters for a rotating machine.
3.11 Discuss briefly the theory of airfoil.
3.12 What is meant by lift and drag? Show the effect of angle of attack on
CL and CD on a graph.
Multiple Choice Questions (choose the most appropriate answer)
1. When the fluid passes through rotating machine, there is
(a) only energy transfer
(b) only energy transformation
(c) both energy transfer and transformation
(d) none of the above
2. The energy transfer in a rotating machine is given by
(a) steady flow energy equation
(b) unsteady flow energy equation
(c) Euler’s energy equation
(d) all of the above
3. The Euler’s equation is given by
(a) E = Ct1 u1 × Ct2 u2
(b) E = Ct1 u1 − Ct2 u2
(c) E = Ct1 u1 + Ct2 u2
(d) E = Ct1 u1 /Ct2 u2
Fundamentals of Rotating Machines
4. The equation for energy transfer in a rotating flow is given by
(a) 0.5 ×
c21 − c22 + u21 − u22 w12 − w22
(b) 0.5 ×
c21 − c22 + u21 − u22 w22 − w12
(d) 0.5 ×
c22 − c21 + u22 − u21 w22 − w12
(c) 0.5 ×
c21 − c22 + u22 − u21 w22 − w12
5. In the energy transfer equation, Δc = c21 − c22 represents
(a) impulse effect
(b) reaction effect
(c) centrifugal effect
(d) centripetal effect
6. In the energy transfer equation Δu = u21 − u22 represents
(a) impulse effect
(b) reaction effect
(c) centrifugal effect
(d) centripetal effect
7. In the energy transfer equation Δw = w22 − w12 represents
(a) impulse effect
(b) reaction effect
(c) centrifugal effect
(d) centripetal effect
8. The degree of reaction for a rotating machine is defined as
(a)
Enthalpy change in the rotor
Enthalpy change in the stator
(b)
Enthalpy change in the stator
Enthalpy change in the rotor
(c)
Enthalpy change in the stator
Enthalpy change in the stage
(d)
Enthalpy change in the rotor
Enthalpy change in the stage
9. Isentropic efficiency of a turbine is
(a)
Actual temperature drop
Isentropic temperature drop
(b)
Isentropic temperature drop
Actual temperature drop
61
62
Gas Turbines
Actual pressure drop
Isentropic pressure drop
Isentropic pressure drop
(d)
Actual pressure drop
(c)
10. The efficiency of the modern compressors is
(a) 70%
(b) 75%
(c) 80%
(d) 85%
Ans:
1. – (c)
6. – (c)
2. – (c)
7. – (b)
3. – (b)
8. – (d)
4. – (b)
9. – (a)
5. – (a)
10. – (d)
4
CYCLE
ARRANGEMENTS
INTRODUCTION
As already seen in Chapter 1, a simple gas turbine unit consists of
three components, viz., a compressor, a heat addition device and a turbine.
These three components can be arranged either in an open or a closed form.
Accordingly, a gas turbine cycle can be classified into two categories:
(i) open-cycle arrangement
(ii) closed-cycle arrangement
Of the two, open-cycle arrangements are much more common. In this
arrangement fresh atmospheric air is drawn into the system continuously
and energy is added by combustion of fuel in the working fluid itself. The
products of combustion are expanded through the turbine and exhausted
into the atmosphere. In the closed-cycle, the same working fluid, be it air
or some other gas, is repeatedly circulated through the system. It may
be noted that in this type of plant whether the working fluid is air or
some other gas, fuel cannot be burnt directly in the working fluid and the
necessary energy must be added in a heater or gas boiler.
4.1
OPEN-CYCLE ARRANGEMENTS
If a gas turbine is to be operated at a fixed speed and fixed load condition
such as peak-load power generation, a single shaft arrangement as shown
in Fig. 4.1 may be suitable. Flexibility of operation, i.e., the rapidity with
which the machine can accommodate itself to changes of load and speed,
and efficiency at part load, are in this case considered unimportant. A
heat exchanger can be added as shown in Fig. 4.2 to improve the thermal
efficiency, although for a given size of the plant, power output may be
reduced by 10% due to pressure losses in the heat exchanger.
64
Gas Turbines
Combustion
chamber
Product of
combustion
1
3
2
Air
4
Fuel
Power
output
Compressor
Turbine
Fig. 4.1 A simple gas turbine with single-shaft arrangement
Heat exchanger
6
Combustion chamber
5
4
1
3
2
Fuel
Power
output
Compressor
Turbine
Fig. 4.2 A simple gas turbine with a heat exchanger
The modified form, as shown in Fig. 4.3, is more suitable for fuels whose
products of combustion contain constituents which may corrode or erode
the turbine blades. It is much less efficient than the simple cycle power plant
because the heat exchanger, inevitably less than perfect, in transferring the
energy input because of the effectiveness of the heat exchanger. Such a cycle
may be considered only if inferior fuels are to be used. When flexibility of
operation is of paramount importance, such as in road, rail and marine
applications, a mechanically independent power turbine is used.
In this twin-shaft arrangement (Fig. 4.4), the compressor and highpressure turbine combination acts as a gas generator for the low pressure
turbine. Fuel flow to the combustion chamber is controlled to achieve variation of power. It should be noted that this will cause a decrease in cycle
pressure ratio and maximum temperature. At off-design conditions the
power output reduces with the result that the thermal efficiency deteriorates considerably at part loads.
Alternative arrangements to overcome the above disadvantages are the
series flow and parallel flow gas turbines (Figs. 4.5 and 4.6). In these
arrangements power output is controlled by the adjustment of fuel supply
to the combustion chamber in the power turbine line.
Cycle Arrangements
65
Combustion chamber
Heat exchanger
4
6
1
Fuel
3
5
2
Power
output
Compressor
Turbine
Fig. 4.3 A simple gas turbine with heat exchanger – an alternative
arrangement
Heat exchanger
Combustion
chamber
Air
Fuel
HPC
HPT
Compressor
Turbine
Power
LPT
output
Separate power turbine
Fig. 4.4 A simple twin-shaft arrangement
66
Gas Turbines
Heat exchanger
Combustion
chamber
Air
Fuel
HPC
HPT
Turbine
Compressor
Fuel
Combustion chamber
Power
LPT
output
Separate power turbine
Fig. 4.5 Series flow twin-shaft arrangement
Heat exchanger
Fuel
Power
output
Compressor
Turbine
Separate power turbine
Fig. 4.6 Parallel flow twin-shaft arrangement
The performance of a gas turbine may be improved substantially by
reducing the work of compression and/or increasing the work of expansion.
For any given pressure ratio, the power required for compression per kg of
working fluid is directly proportional to the inlet temperature. If, therefore,
the compression process is carried out in two or more stages with intercooling, the work of compression will be reduced. This arrangement is shown
in (Fig. 4.7).
Similarly, the turbine output can be increased by dividing the expansion
into two or more stages and reheating the gas to the maximum permissible
temperature between the stages (Fig. 4.8). By employing a heat exchanger,
the cost of additional fuel can be minimized. Complex cycles offer good part
load performance and high flexibility but it is to be noted that the inherent
simplicity and compactness of the power plant are lost.
To obtain higher thermal efficiencies without a heat exchanger, a high
pressure ratio is essential. Axial compressors are normally preferred, particularly for large units, as its efficiency is appreciably higher than that
Cycle Arrangements
Coolant
Out
Coolant
In
67
Heat exchanger
Combustion
chamber
Air
Intercooler
Fuel
HPT
HPC
LPC
Compressor
Compressor
Turbine
Fuel
Combustion chamber
Power
LPT
Separate power turbine
output
Fig. 4.7 Series flow with intercooling
Heat exchanger
Combustion chamber
Fuel
Compressor
Turbine
Reheat
combustion
chamber
Fuel
Power
Fuel
output
Turbine
Turbine
Fig. 4.8 Series flow with reheating
of the centrifugal compressor. Unfortunately, axial compressors are more
prone to instability when operating at off-design conditions. The unstable
operation, manifested by violent aerodynamic vibration, is likely when a
gas turbine is started up or operated at low power. The problem is particularly severe if an attempt is made to obtain a pressure ratio of more
than 8:1 in one compressor. One way of overcoming this difficulty is to
divide the compressor into two or more sections. This mechanical separation permits each section to run at different rotational speed. When the
compressors are mechanically independent, each will have its own turbine.
The LP compressor is driven by the LP turbine and the HP compressor by
the HP turbine. This arrangement is called straight compounding. Power
is normally taken either from the low pressure turbine shaft or from an
additional free power turbine. This configuration is generally referred to as
68
Gas Turbines
a twin-spool engine (Fig. 4.9). This arrangement is widely used both for
shaft power units and for the turbojet aircraft engines, employing pressure
ratios in the range of 10:1 to 20:1.
Heat exchanger
Combustion
chamber
Fuel
HPC
Compressor
HPT
Turbine
Power
LPC
Compressor
LPT
output
Turbine
Fig. 4.9 Straight compounded twin-spool arrangement
An arrangement, known as cross compounding wherein the LP compressor is driven by the HP turbine and the HP compressor by the LP turbine,
is claimed to give better efficiency at part load. Unfortunately, the effect
on stability of operation is the opposite of straight compounding, i.e., it
makes the problem worse instead of better.
4.2
THE CLOSED-CYCLE
In all the arrangements discussed so far, atmospheric pressure and temperature have been considered as the datum, and the exhaust gases were discharged at atmospheric pressure. The average exhaust temperature will be
around 700 K for an average maximum temperature of about 1000 K. It follows that 1 kg of gas will occupy a volume according to the law P V = mRT
which is about of 2.34 m3 . At low pressure-ratios the constant-pressure cycle requires large mass flow rate, thus the total volume flow rate for any
given power output will be quite large. In order to accommodate such
large flow, a large rotor diameter and long blades are required. This results
in excessive stress at the root of the blades due to large centrifugal force.
Hence, there is an upper limit on the size and power output of a turbine or
compressor.
It will be shown in the next chapter that the efficiency expression of
various constant-pressure cycles is a function of pressure-ratio and the gas
temperatures;, i.e., the efficiency depends not upon the magnitude of the
Cycle Arrangements
69
pressures, but only upon their ratio. However, the volume occupied by a
gas depends upon the magnitude of the pressure. When the volume to be
circulated becomes large, the pressure level can be raised so that the volume per kg of fluid, i.e., the specific volume, is reduced to the level desired.
However, the same pressure ratio (i.e., ratio of maximum to minimum pressures) should be maintained as before for attaining the same efficiency. It
is true that the maximum pressure will increase, but only in proportion to
the change of absolute pressures on the system.
Thus, when the system is closed from the atmosphere, the same fluid
will circulate again and again. It follows that if the fuel is burnt directly
in the circulating air, the oxygen will soon deplete and combustion will
fail. It is, thus necessary to supply a certain amount of fresh air to the
closed circuit. The normal overall air-fuel ratio of a gas turbine is between
60:1 to 100:1. It follows that only a small fraction of burnt oxygen is to
be supplied each time through the combustion chamber in order to have
continuous operation. Thus, majority of the exhaust gas leaving the turbine
would return to the compressor entrance;, i.e., the inlet temperature to the
compressor would be greatly increased. It may be noted that the work of
compression per kg can be shown to be
WC
Cp T1 γ−1
r γ −1
ηC
=
which is seen to depend upon the inlet temperature, T1 and the pressure
ratio, r. Thus, in a closed-cycle, the net output would be much reduced
unless a gas cooler is added between the turbine exhaust and the compressor
inlet to cool the gas which is being recirculated, down to approximately
normal inlet temperature. The details of such a system with internal firing
is shown in Fig. 4.10.
Compressor
Main turbine
Generator
p
p
p
1
Gas
cooler
p
2
1
Regenerator
In Out
2
2
p
p
1
Auxiliary
turbine
2
p
Combustion
chamber
2
Fuel
p
Compressor
atm
Make up air inlet
p
atm
Exhaust
Fig. 4.10 A closed-cycle arrangement with air as working medium
The compressor draws gas from turbine exhaust at point (1) through
a regenerator and gas cooler under pressure p1 , which is greater than atmospheric pressure. After compression, the gas at pressure p2 is delivered
70
Gas Turbines
to combustion chamber through regenerator. In the regenerator the compressed gas is heated to higher temperature by the main turbine exhaust. In
the combustion chamber, fuel is added to obtain the desired maximum gas
temperature. From combustion chamber the gases pass on to main turbine,
where they expand from pressure p2 to pressure p1 . The turbine provides
net work to the load after meeting the compression work of the cycle. A
small portion of the gases leaving the combustion chamber is by-passed to
auxiliary turbine, where expansion takes place from p2 to patm .
The auxiliary turbine develops enough work to run a compressor, which
draws air from atmosphere and delivers it at pressure p2 of the combustion
chamber. The amount of fresh air supplied by this compressor must be
equal to the amount of gases by-passed after combustion to atmosphere via
auxiliary turbine and also to the amount of air which is sufficient to burn
completely the fuel that is added to maintain the maximum temperature
of the cycle and thereby the maximum output of the main turbine.
Figure 4.11 illustrates another method on the closed-cycle concept,
where instead of supplying the fuel directly to, and burning it with, the
air inside the system, the fuel could be burned in a separate combustion
chamber built like a regenerator. The heat of combustion is transferred
through the containing walls of the furnace to the air flow in the turbines
as shown in Fig. 4.11. Thus, the air or gas contained in the closed system
can be used over and over again without the necessity for make-up air. This
results in that the products of combustion do not come in contact with the
moving parts and no deposit will accumulate on the turbine blades. In such
type of closed circuit every kind of solid fuels like coal, can also be burned
in the furnace. Further the working medium other than air having the desired properties can be used. The details of working of such a system is
shown in Fig. 4.11. Advantages and disadvantages of a closed-cycle system
are enumerated in the following sections.
Compressor
Main turbine
Generator
Regenerator
Gas
precooler
Gas heater
In
Out
Water
Air
preheater
Fuel
Air
Exhaust
Fig. 4.11 Another closed-cycle arrangement with working medium other
than air
Cycle Arrangements
4.2.1
71
Advantages
(i) Use of high pressure (and hence gas density) level throughout the
cycle would result in a reduced size of the plant for a given output.
(ii) Wide range of load variation is possible by varying the pressure levels
without altering the maximum cycle temperature. Hence, there will
be almost no variation of overall efficiency.
(iii) Erosion of the turbine blades due to the products of combustion is
eliminated.
(iv) Filtration of working medium is not required except charging for the
first time.
(v) High density of the working medium improves the effectiveness of the
heat exchanger.
(vi) Gases other than air having more desirable thermal properties, such
as helium etc., with γ = 1.66 can be used to increase the power output
and thermal efficiency.
(vii) Cheaper fuels can be used.
4.2.2
Disadvantages
(i) The heating system is quite bulky.
(ii) It is quite difficult to make the system absolutely leak proof.
(iii) Large capacity cooler is necessary.
(iv) Useful only for stationary power plants.
4.3
BASIC REQUIREMENTS OF THE WORKING MEDIUM
In a closed-cycle arrangement, the operating medium can be other than air
and it must satisfy the following requirements.
(i) Availability as well as cheapness of the working medium.
(ii) The circulating working medium must be stable, non-explosive and
non-corrosive.
(iii) It should be non-toxic and non-inflammable.
(iv) It should have high specific heat value Cp and high specific heat ratio,
γ.
(v) It must have a higher thermal conductivity, k.
72
Gas Turbines
Table 4.1 Properties of Various Gases
Gas
Air
Argon
Helium
Hydrogen
4.4
Cp
kJ/kg K
1.005
0.515
5.191
14.186
γ
1.400
1.660
1.660
1.408
k at 0◦ C
Watt/mK
0.0241
0.0163
0.1425
0.1743
R
kJ/kg K
0.287
0.151
10.050
59.934
PROPERTIES OF VARIOUS WORKING MEDIA
Table 4.1 shows the properties of various gases which can be used as a
working medium in the gas turbine.
An examination of the properties of some representative gases shown
in the table would indicate that when desirable values of Cp and γ are
available, they are accompanied by high values of R. The work done during
compression per kg of flow of any gas is given by
WC
= Cp (T2 − T1 )
and the temperatures after compression, T2 , is given by
T2
= T1 r
γ−1
γ
It is clear, therefore, that low values of γ and Cp are conducive to lower
compression work. On the other hand, a high value of γ and Cp are essential
for higher turbine output, assuming the same R. It may be noted that the
net output of the cycle will always increase with high values of Cp and
γ. Then the increased output per kg of working medium would permit a
reduction in mass flow rate. However, high values of thermal conductivity
would result in appreciable saving in heat exchangers, i.e., regenerators and
precoolers. The values given in the Table 4.1 for k are based on 0◦ C for
comparative purpose. Actually, these values would be considerably higher
at the temperatures encountered in regenerators. Therefore, it may be
concluded that any serious reduction in size of the gas turbine plant by
using a gas other than air would have to be realized in heat exchangers. Of
course, a gas other than air could be used in an externally-fired closed-cycle.
But it should be the aim to have a reduction in the size of heat exchanger
and is particularly important.
4.5
APPLICATIONS
Gas turbines can be classified into aircraft and industrial gas turbines, the
second term meaning all those gas turbine power plants which are not
included in the first category.
The aircraft gas turbines differ from the industrial gas turbines in three
main aspects.
Cycle Arrangements
73
(i) The life of the industrial gas turbine is expected to be of the order of
120,000 hours without major overhaul as against 600-1200 hours for
aircraft gas turbines.
(ii) Size and the weight of an aircraft power plant is very crucial compared
to industrial units.
(iii) The aircraft power plant can make use of kinetic energy of the gases
leaving the exhaust whereas it is wasted in other types and consequently, this energy loss must be kept as minimum as possible.
These differences in the requirements have considerable effect on design,
although fundamental theory is same for both the categories. Industrial gas
turbines are rugged in construction, with many auxiliary equipments. They
often employ a single, large cylindrical combustion chamber. They are also
designed for multifuel capability.
Apart from the aircraft market, the widest application of gas turbines
have been in pump sets for oil and gas transmission pipe lines and generation of electricity. So far gas turbines have made no inroads into the
world of merchant shipping but it is extensively used in naval operations.
A major disadvantage of the gas turbine in naval use is its poor part load
performance and higher specific fuel consumption. To overcome this problem, combined power plants consisting of gas turbines in conjunction with
steam turbines, diesel engines and other gas turbines have been considered.
To date, little impact has been made in the field of rail transport. Experimental trains have been operating in some countries. The high speed
passenger train with gas turbine power is an attractive concept for the future. Maybe in the near future, a long haul truck market will provide a
major application for the gas turbine. Major automobile industries are active in developing engines in the range of 200 – 300 kW. These vehicular
engines employ low pressure ratio, centrifugal compressor, free power turbine and a rotary heat exchanger. Concern with exhaust pollution will be
a critical factor in favour of gas turbine. The major problem is still with
high part-load fuel consumption.
Another concept of potentially great importance is the so-called Total
Energy Plant, where exhaust heat is used to provide building heating in
winter and refrigeration/air conditioning in the summer. Other uses for
energy in a gas turbine’s exhaust are found in process industries. The gas
turbine can also be used as a compact air compressor suitable for supplying
large quantities of air at moderate pressures.
4.6
COMPARISON OF GAS TURBINES WITH RECIPROCATING
ENGINES
Gas turbine is also an internal combustion engine. Its competitor in early
stages was the reciprocating internal combustion engines. Let us compare
them.
74
4.6.1
Gas Turbines
Advantages of Gas Turbines over Reciprocating Engines
(i) Mechanical efficiency Mechanical efficiency of the gas turbine is considerably higher than that of the best reciprocating engine. For simple gas turbine design mechanical efficiency of 90% to 95% has been
claimed while for reciprocating engine it is from 85 to 90% under full
load conditions. It is due to more frictional losses in reciprocating
engines.
(ii) Balancing Due to absence of any reciprocating mass in gas turbine
engine, balancing can be near perfect. Torsional vibrations are absent
because gas turbine is a steady flow machine.
(iii) Cost In case of larger output gas turbine units of 2500 kW, it can
be built at an appreciably lower cost and in a shorter time than the
corresponding multicylinder petrol or diesel engines.
(iv) Weight The fuel consumption per kW hour of best available aircraft
gas turbine is almost twice that of the normal petrol engine. However,
it has much lighter weight per kW so that the total weight of turbine
plus fuel does not compare unfavourably with reciprocating type of
engine and its fuel. To give quantitative example, the specific weight
of (a) steam turbine is about 53 kg/kW, (b) diesel engine is about
115 kg/kW and (c) gas turbine is about 20 kg/kW.
(v) External shape and size The basic cylindrical shape of turbine and
compressor unit renders the gas turbine more convenient to start,
especially in aircraft and locomotives.
(vi) Fuel The turbine can be designed to operate with cheaper and more
readily available fuels such as benzene, powdered coal, and heavy
graded hydrocarbons. Promising results have been obtained using
furnace oil and also pulverized coal as fuel.
(vii) Lubrication Compared with reciprocating engines the lubrication of
gas turbines is comparatively simpler. The requirement is chiefly to
lubricate the main bearing, compressor shaft and bearings of auxiliaries.
(viii) Maintenance The fact that the gas turbine consists of essentially a single turbine and compressor unit with a common or coupled shaft running in a relatively smaller number of main bearings, only minimum
maintenance is necessary as compared to the reciprocating internal
combustion engines.
(ix) Low operating pressures The gas turbine generally operates at relatively low pressures so that the parts exposed to these pressures can
be made light although the effects of thermal expansion and contraction must be taken into account. The maximum combustion pressure
is much lower than that in reciprocating engines so that the pressure
joints and piping do not pose any difficulty.
Cycle Arrangements
75
(x) Silent operation Since the exhaust from a gas turbine occurs under
practically constant-pressure conditions unlike the pulsating nature
of reciprocating engine exhaust, the turbine and compressor, if dynamically balanced, can run very smoothly. The usual vibrational
noises as in the case of reciprocating engine are almost absent.
(xi) Smokeless exhaust With the present tendency to use relatively large
surplus air for combustion in order to reduce temperature of gases,
the exhaust from the turbine is almost smokeless and generally free
from pungent odour associated with optimum and rich fuel mixture
which is characteristic of reciprocating engines.
(xii) High operational speed Turbine can be made lighter than the reciprocating engine of similar output. It can be run at much higher speed
than reciprocating engines. The output of any engine varies directly
as the product of the driving shaft torque and its rpm. Therefore,
for a given output and higher speed the torque will be lower. It may
be noted that the torque characteristics of the gas turbine is much
better than that of reciprocating engine, since the former gives a high
initial torque and its variation with speed is comparatively less.
4.6.2
Advantages of Reciprocating Engines over Gas Turbines
(i) Efficiency The overall efficiency of the turbine is much less than the
reciprocating engine since 70% of the output of the turbine is to be
fed to the compressor and other accessories and auxiliary parts.
(ii) Temperature limitation The maximum temperature in gas turbine
cannot exceed 1500 K because of the material consideration of the
blade while in reciprocating engines with complete combustion of the
fuel the maximum temperature can be raised to 2000 K. This high
temperature is permitted since the piston and cylinder head are subjected to this high temperature only for a fraction of a second.
(iii) Cooling We can achieve very good results by efficient cooling in reciprocating engine by which the heat of the cylinder walls is taken away,
which enables to keep the wall temperature only around 500 K but
in gas turbine, cooling is complicated, and, therefore, much higher
temperature cannot be allowed to reach.
(iv) Starting difficulties It is more difficult to start a gas turbine than a
reciprocating engine as it requires compressed air or some suitable
starter mechanism which are complicated.
Review Questions
4.1 What are the two types of cycle arrangements possible for a gas turbine
system and what is the basic difference between the two arrangements?
76
Gas Turbines
4.2 Explain the details of single-shaft arrangement for fixed load conditions.
4.3 What is the purpose of adding a heat exchanger? Explain the details
with suitable sketches.
4.4 With neat sketches explain the details of a gas turbine system when
flexibility of operation is of paramount importance.
4.5 Explain with neat sketches the cycle arrangements of a series flow
with intercooling and series flow with reheating.
4.6 With a neat sketch explain the straight compounded twin-spool arrangement.
4.7 With neat sketches explain the working of closed-cycle arrangements.
4.8 Discuss the advantages and disadvantages of closed-cycle system over
open-cycle system.
4.9 What are important factors to be considered if working media other
than air is to be used?
4.10 Briefly discuss the applications of gas turbines.
4.11 Discuss the advantages of gas turbines over reciprocating engines and
vice-versa.
4.12 Discuss the advantages of reciprocating engines over gas turbines.
Multiple Choice Questions (choose the most appropriate answer)
1. A gas turbine cycle can be operated
(a) only as an open-cycle arrangement
(b) only as a closed-cycle arrangement
(c) both as an open-cycle and closed-cycle arrangement
(d) none of the above
2. For the same pressure ratio and cycle peak temperature
(a) open-cycle is more efficient than closed-cycle
(b) open-cycle is less efficient than closed-cycle
(c) both open-cycle and closed-cycle will have the same efficiency
(d) none of the above
3. Adding heat exchanger to a simple ideal cycle
(a) improves work output
(b) reduces work output
Cycle Arrangements
77
(c) improves efficiency
(d) improves both work output and efficiency
4. When flexibility of operation is important
(a) single-shaft arrangement is better
(b) twin-shaft arrangement is better
(c) multiple compression is better
(d) none of the above
5. Major applications of gas turbines is for
(a) aircraft
(b) locomotive
(c) automotive
(d) all of the above
6. The specific weight of a gas turbine power plant is
(a) 50 kg/kW
(b) 20 kg/kW
(c) 100 kg/kW
(d) 80 kg/kW
7. The selection of the working medium of a closed-cycle gas turbine
power plant is based on
(a) the availability and cost
(b) non-explosive and non-corrosive properties
(c) high specific heat value and specific heat ratio
(d) all of the above
8. The life of an industrial gas turbine without major overhaul is
(a) 120,000 h
(b) 12,000 h
(c) 1200 h
(d) 120 h
9. The life of a aircraft gas turbine without major overhaul is
(a) 120,000 h
(b) 12,000 h
(c) 1200 h
(d) 120 h
78
Gas Turbines
10. The overall efficiency of a gas turbine engine compared to reciprocating engine for the same power output
(a) more
(b) less
(c) same as reciprocating engine
(d) none of the above
Ans:
1. – (c)
6. – (b)
2. – (c)
7. – (d)
3. – (c)
8. – (a)
4. – (b)
9. – (c)
5. – (a)
10. – (b)
5
IDEAL CYCLES AND
THEIR ANALYSIS
INTRODUCTION
In the previous chapter, we have seen, how large is the number of possible varieties of gas turbine cycle arrangements, when multistage compression and expansion, heat exchange, reheat and intercooling are incorporated. A comprehensive study of the performance of all such cycles allowing
for the various inefficiencies of the components, would result in a very large
number of performance curves. This chapter is mainly concerned with describing methods for calculating cycle performance under ideal conditions.
For convenience, we will treat cycles in two groups:
(i) shaft power cycle, and
(ii) aircraft propulsion cycle.
An important distinction between the two groups arises from the fact
that the performance of aircraft propulsion cycles depend very significantly
upon forward speed of the aircraft and altitude at which it is flying. However, these two variables do not enter into performance calculations of shaft
power cycle used in marine and land based power plants to which this chapter is confined.
Before proceeding with the main task, it will be useful to review the
performance of ideal gas turbine cycles in which perfection of the individual
components is assumed. The specific work output and cycle efficiency will
only be a function of pressure ratio and maximum cycle temperature. The
limited number of performance curves so obtained enables one to analyze
the major effects of various additions to the simple cycle. To be seen clearly,
such curves also show the upper limit of performance. In the limit this
is the maximum that can be expected of real cycles as the efficiency of
the gas turbine component is improved. Before going into the details of
performance analysis let us list the various assumptions we make in the
analysis of ideal cycle.
80
Gas Turbines
5.1
ASSUMPTIONS IN IDEAL CYCLE ANALYSIS
The assumption of ideal gas turbine cycle will be taken to imply the following:
(i) The change of kinetic energy of the working fluid between inlet and
outlet of each component is negligible.
(ii) Compression and expansion are reversible and adiabatic, i.e., isentropic.
(iii) There are no pressure losses in the inlet ducting, combustion chambers, heat exchangers, intercoolers, exhaust ducting and ducts connecting the components.
(iv) Heat transfer in a heat exchanger is assumed to be perfect (100%
effectiveness).
(v) The mass flow of gas is constant throughout the cycle.
(vi) Working fluid has the same composition throughout the cycle and is
a perfect gas with constant specific heats.
(vii) Bearing and windage friction, etc., are neglected.
5.2
THE SIMPLE GAS TURBINE CYCLE
The schematic details of a simple gas turbine are shown in Fig. 5.1. Figure
5.2 shows the various processes on a p-V diagram whereas Fig. 5.3 gives
the details on a T -s diagram. Figures 5.4 and 5.5 show the performance
curves of the cycle.
From the thermodynamic analysis the relevant steady flow energy equation has been shown to be (refer Eq. 2.54)
ws
=
Δh
=
h2 − h1
(5.1)
where ws is the work transfer per unit mass flow, and h stands for enthalpies.
Applying Eq. 5.1 to each component and assuming unit mass flow of
the working fluid, we can write the work input to the compressor (process
1→2) as
Compressor work [WC ]
W12
= (h2 − h1 )
=
Cp (T2 − T1 )
(5.2)
= (h3 − h2 )
=
Cp (T3 − T2 )
(5.3)
Heat addition [Q]
Q23
Turbine work [WT ]
Ideal Cycles and their Analysis
Combustion
chamber
Product of
combustion
1
3
2
Air
81
4
Fuel
Power
output
Compressor
Turbine
Fig. 5.1 Schematic arrangement of a simple gas turbine
2
10
1500
3
8
3
1000
T
p6
4
500
2
0
1
1
0
4
1
V
0
2
Fig. 5.2 p-V diagram
20
30
40
s
50
60
Fig. 5.3 T -s diagram
0.8
2
t=5
1.5
WN /CpT1
4
2
0.6
t=4
1
η 0.4
t=3
0.5
6
γ =1.6
0
γ =1.4
0.2
t=2
0
0
5
Fig. 5.4 r vs
10
r
WN
Cp T1
15
20
0
0
5
10
r
Fig. 5.5 r vs efficiency
15
20
82
Gas Turbines
W34
= (h3 − h4 )
=
Cp (T3 − T4 )
(5.4)
Net work output [WN ]
= WT − WC
= Cp (T3 − T4 ) − Cp (T2 − T1 )
T3
T4
T2
−
−
+1
T1
T1
T1
= Cp T1
Let
then
Let r
γ−1
γ
(5.5)
T3
T1
= t and
p2
p1
T2
T1
=
T3
T4
r
=
=
(5.6)
r
(5.7)
γ−1
γ
= c. From Eq. 5.6
WN
Cp T1
T3
T4 T3
T2
−
−
+1
T1
T3 T1
T1
=
= t−
WN
Cp T1
t
−c+1
c
= t 1−
1
c
(5.8)
− (c − 1)
(5.9)
N
Equation 5.9 shows that the specific work output CW
, upon which
p T1
the size of the plant depends is a function of not only the pressure ratio,
(r), but also of maximum cycle temperature T3 . Now,
η
=
Net work output
Heat input
=
Cp T1 t 1 − 1c − (c − 1)
Cp (T3 − T2 )
=
Cp T1 (t − c) −
Cp T1
η
=
1−
1
c
T3
T1
=
−
=
t−c
c
T2
T1
1−
1
r
γ−1
γ
WN
Q
(5.10)
(5.11)
=
(t − c) −
t−c
t−c
c
(5.12)
The efficiency of a simple gas turbine thus depends only on the pressure
ratio and the nature of the gas, (γ).
Figure 5.4 shows the plot of work output in non-dimensional form CWpN
T1
as a function of r and t. It may be noted that the value of t depends on the
Ideal Cycles and their Analysis
83
maximum cycle temperature, known as the metallurgical limit. The highly
stressed parts of the turbine should withstand this temperature during the
required working life. For an industrial plant it might be between 3.5 and
4.0, whereas a value of 5.0 to 5.5 is permissible for an aircraft engine with
cooled turbine blades.
A glance at the specific output curves (Fig. 5.4) show that a constant t
curve exhibits a maximum at a certain pressure ratio; W = 0 at r = 1 and
also at the value of the pressure ratio, r = tγ/(γ−1) (i.e., c = t) for which
the compression and expansion processes coincide (refer Eq. 5.9).
For any given value of t, there must be an optimum pressure ratio to
give a maximum specific output. This can be found by differentiating the
Eq. 5.9 with respect to c and equating it to zero; the result is,
√
(γ−1)/γ
ropt
=
t
(5.13)
Squaring both sides
t =
r
2
γ−1
γ
since,
r(γ−1)/γ
=
T2
T1
=
T3
T4
(5.14)
it can be written as
t =
T2 T3
T1 T4
=
T2
t
T4
(5.15)
If Eq. 5.15 is to be true then it follows that T2 must be equal to T4 , i.e.,
T2 = T4 .
This means that the specific work output is maximum when the pressure
ratio is such that the compressor and turbine outlet temperatures are equal.
For all values of r between 1 and tγ/2(γ−1) , T4 will be greater than T2 and
a heat exchanger can be incorporated to utilize the energy in the exhaust
gas to heat the air coming out of the compressor and thereby, the efficiency
improvement can be achieved.
Figure 5.5 shows the relation between η and r when the working fluid
is air (γ = 1.4), or a mono-atomic gas such as argon (γ = 1.66) is used. It
can be seen that the efficiency increases with pressure ratio but the rate of
increase reduces with the increase in pressure ratio.
5.3
THE HEAT EXCHANGE CYCLE
A schematic arrangement of a simple gas turbine cycle incorporating exhaust heat exchanger is shown in Fig. 5.6. The corresponding p-V and T -s
diagrams are shown in Fig. 5.7 and Fig. 5.8 respectively. Figures 5.9 and
5.10 show the performance curves of the cycle.
84
Gas Turbines
Heat exchanger
Combustion
chamber 4
6
1
5
2
3
Fuel
Power
output
Compressor
Turbine
Fig. 5.6 Schematic arrangement of a heat exchange cycle
12
2
10
5
3
1500
3
8
p6
1000
T
4
500
5
6
2
0
1
0
1
6
1
2
V
4
0
3
Fig. 5.7 p-V diagram
20
30
50
60
0.8
t=5
1.5
0.6
0.2
t=2
0
5
Fig. 5.9 r vs
10
r
WN
Cp T1
3
2
η 0.4
t=3
0.5
t=
t=5
t=4
t=
t=4
1
0
40
s
Fig. 5.8 T -s diagram
2
WN /CpT1
4
2
Simple cycle η
15
20
0
0
5
10
r
Fig. 5.10 r vs efficiency
15
20
Ideal Cycles and their Analysis
85
It may be noticed that the T -s diagram is unchanged in outline from
that of the simple gas turbine cycle, as can be seen from Figs. 5.3 and 5.8,
except for the presence of the heat exchanger as indicated by the two dotted
lines 4-5 and 2-6. It may be seen that the temperature of the compressed
air has been raised from T2 to T5 in the heat exchanger resulting in the fall
in temperature of the exhaust gases from T4 to T6 . Now, for a unit mass
flow rate
Compressor work [WC ]
W12
=
(h2 − h1 )
=
Cp (T2 − T1 )
(5.16)
=
(h3 − h5 )
=
Cp (T3 − T5 )
(5.17)
(h3 − h4 )
=
Cp (T3 − T4 )
(5.18)
Heat addition [Q]
Q53
Turbine work [WT ]
W34
=
Since, the expressions WN = WT −WC is identical compared to a simple
cycle (Eq.5.5), we have,
WN
Cp T1
=
t 1−
1
c
− (c − 1)
However, the expression for η will be different
WN
Q
Since, T5
=
=
η
Cp T1 t 1 − 1c − (c − 1)
Cp (T3 − T5 )
=
T4
η
=
Cp T1 t 1 −
Cp T1
η
=
t 1−
=
1−
c
t
1
c
T3
T1
1
c
−
− (c − 1)
T4
T3
− (c − 1)
t − ct
T3
T1
=
t−c
t
(5.19)
As can be seen from Eq. 5.19 the efficiency of the heat exchange cycle is
not independent of maximum cycle temperature and clearly increases as t
is increased. Further, it is evident that for a given value of t the efficiency
increases with decrease in pressure ratio and not with increase in pressure
ratio as for the simple cycle.
In Fig. 5.10 the solid line curves represent the above equation (Eq. 5.19).
Each curve for t starts at r = 1 with a value of η = 1− 1t , which is the Carnot
efficiency. This is to be expected because in this limiting case the Carnot
requirement of complex external heat reception and rejection at the upper
86
Gas Turbines
and lower cycle temperatures is satisfied. The curves fall with
√ increasing
pressure ratio until a value corresponding to the value of t equals c is
reached, and at this point the efficiency becomes that of the simple cycle.
At this pressure ratio T4 = T2 and the output is maximum. For higher
values of r a heat exchanger would cool the air leaving the compressor
reducing the efficiency.
The specific output is unchanged by the additions of a heat exchanger.
Compare the curves of work output of Fig. 5.4 and Fig. 5.9. From the
curves of efficiency for heat exchange cycle (Fig. 5.10) it can be concluded
that to obtain an appreciable improvement in efficiency by heat exchange
cycle we must have a value of r appreciably less than that for the optimum
specific work output. It may be noted that it is not necessary to use a
higher cycle pressure ratio as the maximum cycle temperature is increased.
The following observations can be made from the performance curves:
(i) With heat exchanger cycle, the cycle efficiency reduces as the pressure
ratio increases, which is opposite to that of a simple cycle. This is due
to the fact that as the pressure ratio increases the delivery temperature from the compressor increases and ultimately will exceed that of
the exhaust gas from the turbine. Then heat in the heat exchanger
will be lost from the air to the exhaust gases instead of desired gain.
The efficiency with lower temperatures, say at t = 2, is seen to become
negative soon after the pressure ratio 11.3 is exceeded (refer Fig. 5.10
WN < 0). The reason is that the temperature at compressor outlet
actually exceeds the assumed combustion temperature in this case.
(ii) In many cases, regeneration is not desirable. With high pressure
ratios, efficiencies are higher without regeneration, again because loss
of heat from the compressed air to the exhaust gases.
(iii) Efficiency with heat exchanger cycle rises very rapidly with increase
in maximum temperature of the cycle.
(iv) Lower pressure ratios and high cycle temperatures are favourable for
the regenerative cycle, since a large heat recovery is then possible.
(v) For a given temperature-ratio, the curve falls with increasing value of
pressure ratio until a value of c given by c2 = t is reached (Fig. 5.10).
After this, the efficiency is equal to the ideal cycle without regeneration. Any further increase of pressure ratio will yield an efficiency
which is lower than this value and is of no interest.
5.4
THE REHEAT CYCLE
A good improvement in specific work output can be obtained by splitting
the expansion and reheating the gas between the high pressure and low
pressure turbines. Figure 5.11 shows the schematic arrangement of the
reheat cycle along with the p-V (Fig. 5.12) and T -s (Fig. 5.13) diagrams.
Figures 5.14 and 5.15 show the performance curves of the cycle. The turbine
Ideal Cycles and their Analysis
87
work increase is obvious from the fact that the vertical distance between
any pair of constant-pressure lines increases as the entropy increases. Thus
(T3 − T4 ) + (T5 − T6 ) > (T3 − T4 )
Consider the T -s diagram in Fig. 5.13 in which the expansion is carried
out in two stages, reheating of the working fluid to the upper limit of
temperature T3 taking place between the stages. Let the pressure ratio in
compression be r and the pressure ratios of the expansion stages be r3 and
r4 so that
r
=
c =
r3 × r4
(5.20)
r(γ−1)/γ
=
T2
T1
(5.21)
(γ−1)/γ
=
T3
T4
(5.22)
(γ−1)/γ
=
T5
T6
(5.23)
c3
=
r3
c4
=
r4
Therefore,
c =
c3 × c4
(5.24)
For a unit quantity of fluid flow
Compressor work [WC ]
W12
=
(h2 − h1 )
=
Cp (T2 − T1 )
(5.25)
Heat addition [Q]
Q23
=
(h3 − h2 )
=
Cp (T3 − T2 )
(5.26)
Q45
=
(h5 − h4 )
=
Cp (T5 − T4 )
(5.27)
Turbine work [WT ]
W34
=
(h3 − h4 )
=
Cp (T3 − T4 )
(5.28)
W56
=
(h5 − h6 )
=
Cp (T5 − T6 )
(5.29)
Net work output [WN ]
WN
Cp T1
=
WT − WC
=
Cp (T3 − T4 ) + Cp (T5 − T6 ) − Cp (T2 − T1 )
=
Cp T1
=
T3
T4 T3
T5
T6 T5
T2
−
+
−
−
+1
T1
T3 T1
T1
T5 T1
T1
T3
T4
−
T1
T1
+
T5
T6
−
T1
T1
−
T2
−1
T1
(5.30)
88
Gas Turbines
Combustion chamber
1
2
Air
Fuel
Reheat combustion
chamber
5
4
Fuel
3
Product of
combustion
6 Power
output
Compressor
Turbine
Turbine
Fig. 5.11 Schematic arrangement of a reheat cycle
10
2
1600
3
8
1100
T
p6
4
4
5
4
6
4’
2
600
5
3
2
0
1
0
6
1
2
V
3
100
4
Fig. 5.12 p-V diagram
30
40
s
50
60
70
0.6
t=5
2
Wmax /CpT1
20
Fig. 5.13 T -s diagram
2.5
1.5
t=4
1
t=3
t=5
0.4
t=3
η
0.5
0
1
t=
Simple cycle
t=2
0
5
Fig. 5.14 r vs
10
r
WN
Cp T1
15
2
0.2
20
0
0
5
10
r
Fig. 5.15 r vs efficiency
15
20
Ideal Cycles and their Analysis
Since T5 = T3 and c3 · c4 = c, the above equation simplifies to
WN
t
t
= 2t −
−
−c+1
Cp T1
c3
c4
89
(5.31)
For maximum output
d
dc3
W
Cp T1
=
0
(5.32)
writing c4 in terms of c and c3 and treating c, T and t as constants with
respect to c3
t
t
= 0
−
c23
c
√
c3 =
c = c4
(5.33)
Equation 5.33 shows that to obtain maximum output the stage pressure ratios must be same and this should be the square root of the overall pressure
ratio for two stage expansion. Now,
Wmax
2t
(5.34)
= 2t − √ − c + 1
Cp T1
c
ηmax
ηmax
=
1
2t 1 − √
c
=
Wmax
Q
=
2t 1 −
=
2t 1 −
T3
T1
−
− (c − 1)
Cp T1 2t − c + 1 −
=
(5.35)
2t
√
c
Cp [T3 − T2 ] + Cp [T5 − T4 ]
√1
− (c − 1)
c
T2
T5
T4
T1 + T1 − T1
√1
c
− (c − 1)
2t − c −
√t
c
(5.36)
Comparisons of the CWp N
T1 curves (Fig. 5.14) with the simple cycle (Fig. 5.4)
indicates that reheat markedly increase the specific output, but the curves
for efficiency indicate that this is achieved at the expense of efficiency
(Fig. 5.15 ). This is to be expected because a less efficient cycle (4 456)
(refer Fig. 5.13) is added to the simple cycle – less efficient because it operates over a smaller temperature range. Note that the rate of reduction in
η becomes less as maximum cycle temperature is increased.
5.5
THE REHEAT AND HEAT EXCHANGE CYCLE
It is seen that improvement in specific power output is achieved in reheat
cycle at the expense of the efficiency. This can be overcome by adding a
90
Gas Turbines
heat exchanger to the reheat cycle. The schematic arrangements of the
reheat cycle with heat exchanger is given in Fig. 5.16. The corresponding
p-V and T -s diagrams are shown in Figs. 5.17 and 5.18 respectively. Figures
5.19 and 5.20 show the performance curves of the cycle. For a unit quantity
of mass flow,
Compressor work [WC ]
W12
=
Cp (T2 − T1 )
(5.37)
Cp (T3 − T7 ) + Cp (T5 − T4 )
(5.38)
Cp (T3 − T4 ) + Cp (T5 − T6 )
(5.39)
Heat addition [Q]
Q73 + Q45
=
Turbine work [WT ]
W34 + W56
=
Net work output [WN ]
=
WT − WC
=
Cp (T3 − T4 ) + Cp (T5 − T6 ) − Cp (T2 − T1 )
=
Cp T1
T3
T4
T5
T6
T2
−
+
−
−
+1
T1
T1
T1
T1
T1
(5.40)
The maximum work output expression will be same as Eq. 5.34 since
heat exchanger will improve only the efficiency and not the work output.
Hence,
Wmax
Cp T1
ηmax
=
2t
2t − c + 1 − √
c
=
Wmax
Q
=
Cp T1 2t − c + 1 −
=
Cp [T3 − T7 ] + Cp [T5 − T4 ]
2t − c + 1 −
T3
T1
−
T7
T1
+
2t
√
c
T5
T1
(5.41)
2t
√
c
− TT41
Since T7 = T4 and T5 = T3 , we have
ηmax
=
=
ηmax
=
2t − c + 1 −
T3
T1
−
T4
T1
+
T3
T1
2t − c + 1 −
2t −
1−
2t
√
c
− TT41
=
2t − c + 1 −
2
T3
T1
−
2t
√
c
T4
T1
2t
√
c
2t
√
c
c−1
2t
2t − √
c
(5.42)
Ideal Cycles and their Analysis
91
Heat exchanger
Combustion
chamber
8
7
1
Air
2
Compressor
6
5
4
3
Fuel
Reheat combustion
chamber
Fuel
Turbine
Power
output
Turbine
Fig. 5.16 Schematic arrangement of a reheat cycle with heat exchanger
10
2
1600
3
5
3
7
8
7
1100
T
p6
4
4
2
600
5
6
4
8
2
0
1
0
6
8
1
2
V
3
100
4
Fig. 5.17 p-V diagram
0.8
30
40
s
50
60
0.6
1.5
t=4
1
t=3
t=
η 0.4
t=5
2
t=3
t=
2
0.2
0.5
t=2
0
5
Fig. 5.19 r vs
10
r
Wmax
Cp T1
70
t=5
t=4
t=3
t=5
2
Wmax /Cp T1
20
Fig. 5.18 T -s diagram
2.5
0
1
15
Reheat cycle
20
0
0
5
10
r
Fig. 5.20 r vs efficiency
15
20
92
Gas Turbines
The higher exhaust gas temperature is now fully utilized in the heat exchanger. In fact, when a heat exchanger is employed, the efficiency is higher
with reheat than without. The family of constant t lines exhibit the same
features as those for simple cycle with heat exchanger. Each curve starts
with the Carnot values at r = 1 and falls with increasing r to meet the
corresponding curve of the reheat cycle without heat exchanger at a value
of r corresponding to maximum specific output.
5.6
THE INTERCOOLED CYCLE
We have seen in Section 5.4 that specific output of the cycle can be improved
by increasing the turbine work output incorporating the reheat cycle. Another way of achieving the same is reducing the work of compression, i.e.,
compression in more than one stage and using an intercooler in between.
That is, the compression of the working fluid is cut off at some intermediate pressure and the fluid is cooled by passing it through a heat exchanger
supplied with coolant from some external source before being compressed
in a second stage to the required pressure ratio, a certain improvement
in overall output can be achieved. The details of such an arrangement
are shown in Fig. 5.21. The corresponding p-V and T -s diagrams are as
shown in Figs. 5.22 and 5.23 respectively. Figures 5.24 and 5.25 show the
performance curves of the cycle. Now, for a unit quantity of mass flow
Compressor work [WC ]
W12 + W34
=
Cp (T2 − T1 ) + Cp (T4 − T3 )
(5.43)
Cp (T5 − T4 )
(5.44)
Cp (T5 − T6 )
(5.45)
Heat addition [Q]
Q45
=
Turbine work [WT ]
W56
=
Net work output [WN ]
Let,
=
WT − WC
=
Cp (T5 − T6 ) − Cp (T2 − T1 ) − Cp (T4 − T3 )
=
Cp T1
T5
T6
T2
T4
T3
−
−
+1−
+
T1
T1
T1
T1
T1
T5
T1
=
t;
T5
T6
T2
T1
=
c1 ;
T4
T3
=
=
(5.46)
c
c2
(5.47)
Ideal Cycles and their Analysis
93
Combustion chamber
1
2
Air
Intercooler
Compressor
4
3
Fuel
Compressor
5
Product of
combustion
6 Power
output
Turbine
Fig. 5.21 Schematic arrangement of a intercooled cycle
4
10
1600
5
8
5
1100
T
p6
6
4
3
600
2
0
0
1
3
6
1
V
2
100
3
Fig. 5.22 p-V diagram
8
1
18
28
s
38
48
58
Fig. 5.23 T -s diagram
0.6
2
t=5
t=5
t=4
1.5
Wmax /CpT1
2
4
2
t=4
0.4
t=3
η
1
t=3
t=
0.2
0.5
2
t=2
0
0
5
Fig. 5.24 r vs
10
r
Wmax
Cp T1
15
20
0
0
5
10
r
Fig. 5.25 r vs efficiency
15
20
94
Gas Turbines
It can be shown that for maximum power output and perfect intercooling, (following the similar procedure as we have done for reheat cycle)
√
c1 =
c = c2
(5.48)
=
T5
T6 T5
T2
T4
−
−
+1−
+1
T1
T5 T1
T1
T3
=
t−
√
t √
− c+1− c+1
c
Wmax
Cp T1
=
t−
√
t
−2 c+2
c
ηmax
=
Wmax
Q
=
t−
Wmax
Cp T1
Since T3 = T1
ηmax
=
=
t
c
(5.49)
√
Cp T1 t − ct − 2( c − 1)
Cp (T5 − T4 )
=
√
− 2( c − 1)
T5
T1
−
T4
T3
√
√
c − ct − c + 2
√
t− c
√
t
c−2
c +
√
1−
t− c
t−
(5.50)
(5.51)
The specific work output and the efficiency curves are as shown in Figs. 5.24
and 5.25 respectively. Intercooling will help to increase the net work output
of the cycle. Because of the lower compressor outlet temperature, the fuel
flow rate to obtain a given turbine inlet temperature will increase. Therefore, the thermal efficiency of the intercooled cycle will be less than that
for a simple cycle.
Intercooling is useful when the pressure ratios are high and the efficiency
of the compressor is low. Due to lower compressor outlet temperature there
will be more scope for adding a heat exchanger. Due to regeneration, a
substantial amount of heat from exhaust gases can be recovered. This will
be seen in the next section.
5.7
THE INTERCOOLED CYCLE WITH HEAT EXCHANGER
As seen in the previous section, by intercooling we can improve the work
output of the cycle. Suppose we have to improve the efficiency also, then
we can add a heat exchanger to the intercooled cycle. The basic schematic
arrangements along with the p-V and T -s diagrams are shown in Figs. 5.26,
5.27 and 5.28 respectively. Figures 5.29 and 5.30 show the performance
curves of the cycle. Now, for an unit quantity of mass flow
Compressor work [WC ]
Ideal Cycles and their Analysis
Products of
combustion
1
Air
8
2
95
Heat
exchanger
3
Intercooler
Compressor
7
Combustion
chamber
5
4 Fuel
6
Power
output
Compressor Turbine
Fig. 5.26 Schematic arrangement of an intercooled cycle with heat exchanger
4
10
1600
5
5
7
8
1100
T
p6
4
3
7
600
2
0
1
0
1
6
8
V
2
8
3
100
3
Fig. 5.27 p–V diagram
8
1
18
28
t=5
1.5
s
38
48
t=
0.6
t=4
t=
η 0.4
1
t=5
t=4
3
t=5
2
t=3
t=
2
t=3
0.5
0.2
t=2
0
0
5
Fig. 5.29 r vs
10
r
Wmax
Cp T1
58
Fig. 5.28 T -s diagram
0.8
2
Wmax /CpT1
2
4
2
6
15
Intercooled cycle
20
0
0
5
10
r
Fig. 5.30 r vs efficiency
15
20
96
Gas Turbines
W12 + W34
=
Cp (T4 − T3 ) + Cp (T2 − T1 )
(5.52)
Cp (T5 − T7 )
(5.53)
Cp (T5 − T6 )
(5.54)
Heat addition [Q]
Q75
=
Turbine work [WT ]
W56
=
Net work output [WN ]
=
WT − WC
=
Cp (T5 − T6 ) − Cp (T2 − T1 ) − Cp (T4 − T3 )
The above expression is identical to equation in a simple cycle with intercooling, Hence, the maximum specific work output is given by
Wmax
Cp T1
ηmax
√
t
−2 c+2
c
=
t−
=
Wmax
Q
=
t−
=
t−
t
c
ηmax
=
1−
√
2( c − 1)
t 1 − 1c
=
√
t
c −2 c
T5
T7
T1 − T1
Since T7 = T6
T5
T1
√
Cp T1 t − ct − 2 c + 2
Cp (T5 − T7 )
+2
√
−2 c+2
−
T6
T5
(5.55)
T5
T1
=
t−
t
c
√
−2 c+2
t − ct
(5.56)
Figures 5.29 and 5.30 give the variation of specific power output and
efficiency respectively with r for various t. As already discussed the specific
power output increases over a simple cycle because of intercooling. Because
of the addition of heat exchanger, substantial recovery of heat from the
exhaust gas is possible which increases the efficiency also. It may be noted
that the improvement in efficiency is substantial at lower pressure ratios
and higher turbine inlet temperatures.
5.8
THE INTERCOOLED AND REHEAT CYCLE
We have seen – in the previous section – how to improve the efficiency of
an intercooled cycle by adding an heat exchanger. We can further improve
the specific work output of the intercooled cycle by adding reheat. The
schematic diagram of the cycle is as shown in Fig. 5.31. The corresponding
Ideal Cycles and their Analysis
97
p-V and T -s diagrams are as shown in Figs. 5.32 and 5.33 respectively.
Figures 5.34 and 5.35 show the performance curves of the cycle. For a unit
quantity of mass flow,
Compressor work [WC ]
W34 + W12
= Cp (T4 − T3 ) + Cp (T2 − T1 )
(5.57)
Heat addition [Q]
Q45 + Q67
= Cp (T5 − T4 ) + Cp (T7 − T6 )
(5.58)
Turbine work [WT ]
W56 + W78
= Cp (T5 − T6 ) + Cp (T7 − T8 )
(5.59)
Net work output [WN ]
= WT − WC
= Cp (T5 − T6 + T7 − T8 ) −
Cp (T4 − T3 + T2 − T1 )
= Cp T1
WN
Cp T1
=
(5.60)
T5
T6
T7
T8
T4
T3
T2
−
+
−
−
+
−
+1
T1
T1
T1
T1
T1
T1
T1
T5
T6 T5
T5
T8 T5
T4
T1
T2
−
+
−
−
+
−
+1
T1
T5 T1
T1
T7 T1
T3
T1
T1
By following the usual procedure to get the maximum power output, we
can show that
√
√
Wmax
t
t
= t− √ +t− √ − c+1− c+1
Cp T1
c
c
√
2t
= 2t − √ − 2 c + 2
c
Wmax
Cp T1
ηmax
√
t
= 2 t− √ − c+1
c
=
=
=
(5.61)
Wmax
Q
2Cp T1 t −
√t
c
−
√
c+1
Cp (T5 − T4 ) + Cp (T7 − T6 )
2Cp T1 t −
Cp T1
T5
T1
−
√t
c
−
T4
T1
+
√
c+1
T7
T1
−
T6
T1
(5.62)
98
Gas Turbines
Products of
combustion
Reheat combustion
Combustion chamber
1 2
3
4
Intercooler
Air LP
HP
Compressor
Fuel
chamber
5
6
Fuel
8
Power
output
LP
HP
Compressor
7
Turbine
Turbine
Fig. 5.31 Schematic arrangement of an intercooled and reheat cycle
4
10
5
7
5
1500
8
1000
T
p6
4
3
2
0
2
7
500
6
8
1
0
1
2
V
0
3
Fig. 5.32 p–V diagram
2.5
Wmax /Cp T1
2
3
1
10 20 30 40 50 60 70
s
Fig. 5.33 T -s diagram
t=5
0.4
t=4
t=3
1.5
η
t=3
1
t=2
0.2
t=2
0.5
0
4
t=5
2
8
6
0
5
Fig. 5.34 r vs
10
r
Wmax
Cp T1
15
20
0
0
5
10
r
Fig. 5.35 r vs efficiency
15
20
Ideal Cycles and their Analysis
99
√
− c+1
√
2t − c − √tc
=
2 t−
=
2(t −
√t
c
√
c) − 2 √tc − 1
√
2t − c − √tc
√
+ c−2
= 1−
√
2t − c − √tc
√t
c
(5.63)
(5.64)
The specific work output and efficiency curves are shown in Figs. 5.34 and
5.35 respectively.
5.9
INTERCOOLED CYCLE WITH HEAT EXCHANGE AND REHEAT
This is the most complex cycle arrangement so far discussed. Even though
the specific work output and the efficiency of this cycle will be the maximum, it is at the expense of the simplicity. The schematic arrangement
and the corresponding p-V and T -s diagrams are given in Figs. 5.36, 5.37
and 5.38 respectively. Figures 5.39 and 5.40 show the performance curves
of the cycle.
For unit quantity of mass flow
Compressor work [WC ]
W34 + W12
=
Cp (T4 − T3 ) + Cp (T2 − T1 )
(5.65)
Cp (T5 − T9 ) + Cp (T7 − T8 )
(5.66)
Heat addition [Q]
Q95 + Q67
=
Turbine work, [WT ]
W56 + W78
=
Cp (T5 − T6 ) + Cp (T7 − T8 )
(5.67)
Net work output [WN ]
=
WT − WC
=
[Cp (T5 − T6 ) + Cp (T7 − T8 )] −
[Cp (T4 − T3 ) + Cp (T2 − T1 )]
(5.68)
This expression is exactly identical to the expression for the simple cycle
with intercooling and reheat (Eq. 5.60). Hence,
Wmax
Cp T1
=
√
t
2 t− √ − c+1
c
100
Gas Turbines
Heat
exchanger
Products of
combustion
Combustion Reheat Combustion
chamber
chamber
10
1
Air
9
2
Intercooler
3 4
LP
HP
Compressor
6
5
Fuel
Fuel
Power
output
LP
HP
Compressor
8
7
Turbine
Turbine
Fig. 5.36 Schematic arrangement of intercooled cycle with heat exchanger
and reheat
4
10
5
8
4
3
2
0
2
7
10
8
1
0
1
2
V
0
3
2.5
0
η
t=3
t=2
0
5
Fig. 5.39 r vs
10
r
Wmax
Cp T1
3
1
10
10 20 30 40 50 60 70
s
15
t=5
t=
0.4
t=5
2
t=3
t=2
0.2
20
0
t=4
t=3
0.6
1.5
0.5
2
0.8
t=4
1
4
Fig. 5.38 T -s diagram
t=5
2
8
6
500
6
Fig. 5.37 p–V diagram
Wmax /Cp T1
9
1000
T
p6
7
5
1500
9
Intercooled with
reheat
0
5
10
r
Fig. 5.40 r vs efficiency
15
20
Ideal Cycles and their Analysis
ηmax
=
=
101
Wmax
Q
2Cp T1 t −
√t
c
−
√
c+1
(5.69)
Cp (T5 − T9 ) + Cp (T7 − T6 )
Since T6 = T8
ηmax
ηmax
=
2 t−
=
2t −
=
T5
T1
−
√
√t −
c+1
c
T8
T7
T8
T1 + T1 − T1
2t
√
c
√
−2 c+2
2t −
2t
√
c
=
2t −
t−
2t
√
c
√t
c
√
−2 c+2
+t−
√t
c
√
c−1
1− t √
√ ( c − 1)
c
=
√
c
1−
t
(5.70)
The specific power output and the efficiency curves are as shown in Figs. 5.39
and 5.40 respectively.
5.10
COMPARISON OF VARIOUS CYCLES
However, the purpose of the foregoing sections has been to illustrate the
individual effects of the most important modifications which can be made
to a simple gas turbine cycle. Using the various expressions developed in
previous sections, Table 5.1 has been compiled for r = 4 and t = 3 from
which the percentage gains or losses over the performance of a typical simple
cycle, can be ascertained.
Table 5.1 Comparison of Various Cycles
S.No.
1.
2.
3.
4.
5.
6.
7.
Addition to simple cycle
Heat exchange
Intercooling
Reheat
Reheat and heat exchange
Intercooling and
heat exchange
Reheat and intercooling
Reheat, intercooling and
heat exchange
Effect on
W
η
Cp T1
+50.0 No change
− 6.50
+10.2
−10.4
+24.5
+66.7
+24.5
+68.0
+10.2
−18.2
+80.0
+34.7
+34.7
These values of pressure ratio and the maximum temperature ratio are
used in all the cycles, and equal division of the overall pressure ratio in
compression or expansion – corresponding to maximum work output, is
assumed when intercooling and/or reheat are used.
102
Gas Turbines
2
(a) Simple cycle
(b) Reheat
(c) Intercooling
(d) Intercooling &
reheat
1.5
d
c
W/CpT1
b
d
a
1
c
b
0.5
t=4
t=3
0
0
5
a
10
15
20
r
Fig. 5.41 Comparison of various cycles
It should be noted that this is not an exhaustive comparison, because the
value of the overall pressure ratio for maximum work output is also affected
by the various additions to the simple cycle. The effect may be seen from
the curves of Fig. 5.41, which show the variation of work output with r
for some of the cycles under two conditions of turbine inlet temperature.
The chain-dotted curves correspond to t = 3 so that, if the compressor inlet
temperature is taken as 288 K, the turbine inlet temperature will be around
865 K, a value suitable for a long life unit. The full line curves are for t = 4,
which gives T3 ≈ 1150 K a figure somewhat reasonable for the power plants
used at present in short life units such as jet engines for fighter aircraft.
No mention is made of heat exchange in Fig. 5.41, as we have seen that the
addition of this process to an ideal cycle makes no difference to the work
output.
It would of course be possible to make a comparison of the performance
of cycles without heat exchange using, for each cycle, the overall pressure
ratio giving maximum work output. This might, however, give a misleading impression because, as will be shown in the next Chapter, the efficiency
of compression is not only less than 100% in practice, but also decrease
with increase of pressure ratio. The gain in cycle efficiency with increase of
pressure ratio is not, therefore, so great as would appear from calculations
based on ideal cycles. It is, of course, possible to incorporate in a cycle
multiple arrangements of the above modification, such as several compression stages with intercooling, several expansion stages with reheating, and
heat exchanger. A study of any such cycle will reveal that improvements
in performance will follow the form of those realized with each individual
Ideal Cycles and their Analysis
103
addition as shown above. Appendix 1 provides in a nutshell the complete
details of the 10 cycles we have discussed in the previous sections. It is
intended as a ready reckoner for the reader.
5.11
ERICSSON CYCLE
A discussion of ideal cycles for the gas turbine would not be complete
without some reference to the theoretical cycle of Ericsson. In introducing
multi-stage compression with intercooling, and multistage expansion with
reheating, it is immediately seen from the T -s diagram that an approach
is being made towards isothermal processes. This would, of course, be
achieved only if the number of stages were infinite. In this case, the cycle
would follow the outline 1234 given in Fig. 5.42. With a perfect heat ex1000
4
3
800
T
600
400
1
2
200
0
10
20
30
s
40
50
60
Fig. 5.42 Ericsson cycle
changer, the heat supplied from 2 to 3 would be completely furnished by
the heat rejected from 4 to 1, so that the only heat supplied is that during
the isothermal expansion from 3 to 4. Similarly, the only heat rejected is
that during the isothermal compression from 1 to 2. These heat quantities
P2
2
are easily shown to be equal to RT3 loge P
P1 and RT1 loge P1 respectively,
so that the work done is equal to their difference, which can be written as
2
R(T3 − T1 ) loge P
P1 . The efficiency is therefore given by
η
=
T3 − T1
T3
since the pressure ratio is the same in compression and expansion. Here R
refers to gas constant. The cycle is therefore of optimum efficiency like the
Carnot cycle, all heat received at the upper temperature and rejected at
the lower temperature. However, this cycle is really an ideal one and is of
theoretical interest only.
104
Gas Turbines
Worked out Examples
5.1 Compute the indicated mean effective pressure and efficiency of a
Joule cycle if the temperature at the end of combustion is 2000 K
and the temperature and pressure before compression is 350 K and 1
bar. The pressure ratio is 1.3. Assume Cp = 1.005 kJ/kg K.
Solution
2
10
3
8
p6
4
2
0
1
0
1
4
2
V
Fig. 5.43
Consider the isentropic compression process 1→ 2,
γ−1
p2 γ
=
T1
= 350 × (1.3)0.286 = 377.27 K
T2
p1
Consider the isentropic expansion process 3→ 4
T4
=
T3
p4
p3
γ−1
γ
= 2000 ×
1
1.3
0.286
= 1855.42 K
Work done during compression, (WC )
WC
=
Cp (T2 − T1 ) = 1.005 × (377.27 − 350) = 27.40 kJ/kg
Work done during expansion, (WT )
WT
WN
=
Cp (T3 − T4 )
=
1.005 × (2000 − 1855.42)
=
WT − WC
Volume at various state points:
=
=
145.30 kJ/kg
145.30 − 27.40 = 117.90 kJ/kg
Ideal Cycles and their Analysis
0.287 × 103 × 350
= 1.0045 m3 /kg
1 × 105
V1
=
RT1
p1
V2
=
0.287 × 103 × 377.27
1.3 × 105
V3
=
0.287 × 103 × 2000
1.3 × 105
V4
=
0.287 × 103 × 1855.42
1.0 × 105
imep
=
net work done
change in volume
=
117.90 × 103
× 10−5
5.325 − 0.833
=
Cp (T3 − T2 ) = 1.005 × (2000 − 377.27)
=
1630.84 kJ/kg
=
WN
q
q
η
=
=
105
0.833 m3 /kg
=
4.415 m3 /kg
=
=
=
5.325 m3 /kg
WN
V4 − V2
=
117.9
× 100
1630.84
0.2625 bar
=
7.23%
Ans
⇐=
Ans
⇐=
Since the pressure ratio is quite small in spite of higher peak temperature, the efficiency is very low. This example illustrates that the
efficiency of a simple cycle depends mainly on the pressure ratio and
not on the peak temperature.
5.2 Calculate the improvement in the efficiency when a heat exchanger is
added to the simple cycle given in the previous example.
Solution
The T -s diagram of the cycle is shown in Fig. 5.44. As can be seen the heat
input is
q
=
Cp (T3 − T5 )
For a perfect heat exchange,
T5
=
T4
q
=
1.005 × (2000 − 1855.42) = 145.30 kJ/kg
η
=
117.90
WN
=
q
145.30
=
0.81
=
81%
Ans
⇐=
Note the tremendous improvement in the efficiency when a heat exchanger
is added to a simple cycle with a lower pressure ratio. This example amply
106
Gas Turbines
3
1500
1000
T
5
4
2
500
6
1
0
20
30
40
s
50
60
Fig. 5.44
illustrates that for a heat exchange cycle lower pressure ratio and higher
peak temperature is advantageous.
η2
η1
81
7.23
=
=
Ans
11.20
⇐=
More than 11 times efficiency improvement is achieved because of the addition of heat exchanger.
5.3 A gas turbine operates on a pressure ratio of 6. The inlet air temperature to the compressor is 300 K and the air entering the turbine
is at a temperature of 577◦ C. If the volume rate of air entering the
compressor is 240 m3 /s. Calculate the net power output of the cycle
in MW. Also compute its efficiency. Assume that the cycle operates
under ideal conditions.
Solution
Consider the isentropic compression process 1→ 2
T2
=
T1
p2
p1
γ−1
γ
0.286
= 500.81 K
=
509.18 K
= 300 × (6)
Consider the isentropic expansion process 3→ 4
T3
850
T4
=
=
γ−1
0.286
6
γ
p3
p4
Work input to the compressor, (WC ),
WC = Cp (T2 − T1 )
=
1.005 × (500.81 − 300) = 201.81 kJ/kg
Ideal Cycles and their Analysis
1500
107
3
1000
T
4
2
500
1
0
20
30
40
s
50
60
Fig. 5.45
Work done by the turbine, (WT ),
=
Cp (T3 − T4 )
=
1.005 × (850 − 509.18) = 342.52 kJ/kg
=
WT − WC
=
140.72 kJ/kg
=
Cp (T3 − T2 )
=
1.005 × (850 − 500.81) = 350.94 kJ/kg
ρ1
=
p1
1 × 105
3
= 1.161 kg/m
=
RT1
287 × 300
Power output
=
140.72 × 240 × 1.161 = 39210.22 kW
=
39.2102 MW
=
WN
q
WT
WN
q
η
=
=
342.52 − 201.80
140.72
× 100 = 40.16%
350.94
Ans
⇐=
Ans
⇐=
5.4 A gas turbine cycle has a perfect heat exchanger. Air enters the
compressor at a temperature and pressure of 300 K and 1 bar and
discharges at 475 K and 5 bar. After passing through the heat exchanger the air temperature increases to 655 K. The temperature of
air entering and leaving the turbine are 870◦C and 450◦ C. Assuming
no pressure drop through the heat exchanger, compute
(i) the output per kg of air,
(ii) the efficiency of the cycle, and
(iii) the work required to drive the compressor.
108
Gas Turbines
Solution
12
10
2
5
3
1500
3
8
p6
1000
T
5
4
2
4
500
6
2
0
1
0
6
1
V
2
1
4
0
3
20
30
40
s
50
60
Fig. 5.46
Work input to the compressor, (WC ),
WC
=
Cp (T2 − T1 )
=
175.9 kJ/kg
=
1.005 × (475 − 300)
=
1.005 × (1143 − 723)
=
1.005 × (1143 − 655)
Work done by the turbine, (WT ),
WT
Heat added, q
=
Cp (T3 − T4 )
=
422.1 kJ/kg
=
Cp (T3 − T5 )
=
490.44 kJ/kg
The output per kg of air
WN
=
WT − WC
=
246.2 kJ/kg
=
422.1 − 175.9
Ans
⇐=
The efficiency of the cycle,
η
=
Work output
Heat supplied
=
246.2
× 100
490.44
=
50.2%
Ans
⇐=
The work required to drive the compressor
=
175.9 kJ/kg
Ans
⇐=
Ideal Cycles and their Analysis
109
5.5 A closed-cycle regenerative gas turbine operating with air as the working medium. Assume the following data: p1 = 1.4 bar, T1 = 310 K,
p2 /p1 = 5, Tmax = 1050 K, effectiveness of the regenerator is 100%,
net output = 3000 kW. Assuming the compression and expansion to
be isentropic, calculate
(i) thermal efficiency, and
(ii) mass flow rate of air per minute.
Solution
3
1500
1000
T
5
4
2
500
6
1
0
20
30
40
s
50
60
Fig. 5.47
T2
=
T4
=
p2
p1
T1
γ−1
γ
T3
(rp )
310 × 50.286
=
1050
50.286
=
γ−1
γ
=
=
491.2 K
662.65 K
As regenerator effectiveness is 100%,
T4
=
T5
WN
=
ṁCp (T3 − T4 ) − ṁCp (T2 − T1 )
3000
=
ṁ × 1.005 × [(1050 − 662.65) − (491.2 − 310)]
3000
=
ṁ × 207.18
ṁ
=
14.48 kg/s
ηth
=
WN
Heat added
=
662.65 K
=
868.8 kg/min
Ans
⇐=
110
Gas Turbines
=
ṁCp [(T3 − T4 ) − (T2 − T1 )]
ṁCp (T3 − T5 )
=
(1050 − 662.65) − (491.2 − 310)
× 100
(1050 − 662.65)
=
53.22%
Ans
⇐=
5.6 A gas turbine with a regenerator has got the following data:
Compressor inlet temperature
:
290 K
Compressor outlet temperature
:
460 K
Inlet temperature of the turbine
:
900◦C
Outlet temperature to the turbine
:
467◦C
Assuming no pressure drop in the heat exchanger, calculate
(i) the pressure ratio of the compressor and turbine,
(ii) the specific power output,
(iii) the overall efficiency of the cycle, and
(iv) the work required to drive the compressor.
Assume 100% mechanical efficiency.
Solution
3
1500
1000
T
5
4
2
500
6
1
0
20
30
40
s
50
60
Fig. 5.48
T2
T1
=
c
=
460
290
=
1.586
Compressor pressure ratio,
r
=
γ
c γ−1
=
1.5863.5
Specific power output (per kg of air)
=
5
Ans
⇐=
Ideal Cycles and their Analysis
WN
=
Cp (T3 − T4 ) − Cp (T2 − T1 )
=
1.005 × (1173 − 740) − 1.005 × (460 − 290)
=
264.3 kJ/kg
Overall efficiency of the cycle =
η
111
Ans
⇐=
Specific power output
Heat supplied per kg of air
264.3
Cp (T3 − T5 )
=
Assuming regenerator effectiveness to be 100%, T5 = T4 .
264.3
× 100
1.005 × (1173 − 740)
=
=
60.7%
Ans
⇐=
Work required to drive the compressor
T3
T4
=
Cp (T2 − T1 )
=
170.85 kJ/kg
=
c
=
=
1173
740
1.005 × (460 − 290)
Ans
⇐=
=
1.585
Turbine pressure ratio,
r
γ
=
c γ−1
=
1.5853.5
=
Ans
5
⇐=
5.7 The ratio of net work to turbine work of an ideal gas turbine plant
is 0.563. Take the inlet temperature to the compressor as 300 K.
Calculate the temperature drop across the turbine if the thermal efficiency of the unit is 35%. Assume a mass flow rate of 10 kg/s, Cp
= 1 kJ/kg K and γ = 1.4.
Solution
For ideal simple cycle,
1
c
η
=
1−
c
=
1.5384
T2
=
T1 × c
=
300 × 1.5384
=
Cp (T2 − T1 )
=
1 × (461.52 − 300)
WC
=
0.35
=
461.52 K
=
161.52 kJ/kg
112
Gas Turbines
1500
3
1000
T
4
2
500
1
0
20
30
40
s
50
60
Fig. 5.49
Net work
Turbine work
=
0.563
WC
WT
=
0.563
WT
=
WC
1 − 0.563
=
369.61 kJ/kg of air
=
WT − WC
=
208.09 kJ/kg of air
1−
WN
161.52
0.437
=
=
369.61 − 161.52
Net heat supplied per kg of air
=
1
× WN
ηth
=
594.54 kJ/kg of air
q
=
Cp (T3 − T2 )
594.54
=
1 × (T3 − 461.52)
T3
=
1056.06 K
c
=
T3
T4
T4
=
T3
c
=
1056.06
1.5384
q
=
208.09
0.35
=
1.5384
=
686.47 K
Ideal Cycles and their Analysis
113
Hence, temperature drop across the turbine
T3 − T4
=
1056.06 − 686.47 = 369.59 K
Ans
⇐=
5.8 The specific power output of a turbine is 336.5 kW/kg and the exhaust
gases leave the turbine at 700 K. Calculate the turbine pressure ratio if
the value of Cp and Cv are 1 kJ/kg K and 0.717 kJ/kg K respectively.
Round off the pressure ratio to the nearest integer.
Solution
γ
=
1
Cp
=
Cv
0.717
Cp (T3 − T4 )
=
336.5
T3 − 700
=
336.5
T3
=
1036.5 K
p3
p4
γ
γ−1
T3
T4
=
=
=
=
1.3947
1036.5
700
1.3947
0.3947
Ans
4
⇐=
5.9 A gas turbine plant works between temperature limits of 300 K and
900 K. The pressure limits are 1 bar and 4 bar. Estimate the thermal
efficiency of the plant and the shaft power available for the external
load in kW. Assume mass rate of flow of air to the compressor as 1600
kg/min.
Solution
1500
3
1000
T
4
2
500
1
0
20
30
40
s
Fig. 5.50
50
60
114
Gas Turbines
For simple cycle gas turbine plant
η
1
c
=
1−
c
=
p2
p1
η
=
1−
W
Cp T1
=
t 1−
1
c
t
=
900
300
=
W
1.005 × 300
=
3× 1−
=
0.4952 kJ/kg of air
=
0.4952 × 1.005 × 300
=
1600
60
W
Mass flow of air,
Shaft power available
=
γ−1
γ
40.286
=
1
1.486
× 100
=
=
1.486
Ans
32.7%
⇐=
− (c − 1)
3
1
1.486
=
− (1.486 − 1)
=
149.3 kJ/kg of air
26.66 kg/s
149.3 × 26.66
=
3980.34 kW
Ans
⇐=
5.10 In a gas turbine, the pressure ratio to which air at 15◦ C is compressed is 6. The same air is then heated to a maximum permissible
temperature of 750◦ C. First in a heat exchanger and then combustion
chamber. It is then expanded in two stages such that the expansion
work is maximum. The air is reheated to 750◦ C after the first stage.
N
Determine the cycle thermal efficiency, the work ratio W
and net
WT
shaft work per kg of air.
Solution
c
=
60.286
T2
=
T1 × 1.669
For maximum expansion work
p3
p5
=
=
p4
p6
=
1.669
=
√
rc
288 × 1.669
=
√
6
=
=
2.45
480.67 K
Ideal Cycles and their Analysis
1600
7
6
4
2
600
100
5
3
1100
T
115
8
1
20
30
40
s
50
60
70
Fig. 5.51
T4
=
T3
p3
p4
γ−1
γ
1023
=
=
(2.45)0.286
791.7 K
As pressure ratio is same, T6 = T4 = 791.7 K.
WC
WT
q
=
Cp (T2 − T1 )
=
1.005 × (480.67 − 288)
=
193.63 kJ/kg of air
=
Cp (T3 − T4 ) + Cp (T5 − T6 )
=
2 × 1.005 × (1023 − 791.7) = 464.9 kJ/kg of air
=
Cp (T3 − T7 ) + Cp (T5 − T4 )
Because of 100% regeneration,
q
=
2 × 1.005 × (1023 − 791.7)
=
464.9 kJ/kg of air
=
WT − WC
=
271.34 kW
η
=
WN
q
=
271.34
× 100
464.9
Wratio
=
WN
WT
=
271.34
464.9
WN
=
Ans
⇐=
464.9 − 193.56
Ans
⇐=
=
=
58.36%
0.5836
Ans
⇐=
Ans
⇐=
116
Gas Turbines
5.11 A gas turbine plant operates between 5◦ C and 839◦ C. Find
(i) pressure ratio at which cycle efficiency equals Carnot cycle efficiency,
(ii) pressure ratio at which maximum work is obtained, and
(iii) efficiency under conditions giving maximum work.
Solution
1500
3
1000
T
4
2
500
1
0
20
30
40
s
50
60
Fig. 5.52
ηcarnot
=
1−
Tmin
Tmax
=
1−
1
c
=
1−
278
1112
=
0.75
For a simple cycle
η
c
=
4
p2
p1
=
43.5
=
0.75
p2
p1
=
=
γ−1
γ
128
Therefore, pressure ratio at which cycle efficiency equals Carnot efficiency
Ans
=
128
⇐=
For a simple cycle,
WN
Cp T1
=
t 1−
1
c
− (c − 1)
Pressure ratio for maximum work is obtained when
Ideal Cycles and their Analysis
dWN
dc
=
0
0
=
t 0+
t
−1
c2
=
0
c
=
√
t
=
2
=
23.5
p2
p1
1
c2
117
− (1 − 0)
√
4
=
=
2
γ−1
γ
p2
p1
=
11.31
Pressure ratio for maximum work
=
Ans
11.31
⇐=
Efficiency at maximum work output
1
=
=
1−
c
1−
1
2
=
50%
Ans
⇐=
5.12 An ideal open-cycle gas turbine plant using air operates in an overall
pressure ratio of 4 and between temperature limits of 300 K and 1000
K. Assuming the constant value of specific heat Cp = 1 kJ/kg K and
Cv = 0.717 kJ/kg K, evaluate the specific work output and thermal
efficiency for each of the modification below and state the percentage
change from the basic cycle. Assuming optimum stage pressure ratios,
perfect intercooling and perfect regeneration, find
(i) basic cycle,
(ii) basic cycle with heat exchanger,
(iii) basic cycle with two stage inter cooled compressor, and
(iv) basic cycle with heat exchanger and two-stage intercooled compressor.
Solution
Basic cycle: (refer Fig. 5.3)
k
=
Cp
Cv
c
=
(rp )
t
=
T3
T1
=
γ−1
γ
=
1
0.717
=
=
40.283
1000
300
=
1.3947
=
1.4804
3.333
118
Gas Turbines
WN
Cp T1
WN
η
1
c
− (c − 1)
=
t 1−
=
3.333 × 1 −
=
0.6012
=
Cp × T1 × 0.6012
=
1 × 300 × 0.6012
=
1−
1
c
1
1.4804
=
1−
− (1.4804 − 1)
=
180.36 kJ/kg
1
1.4804
× 100
=
32.45%
=
55.58%
Basic cycle with heat exchanger: (refer Fig. 5.8)
WN
=
180.36 kW
η
=
1−
c
t
=
1−
1.4804
3.333
× 100
Basic cycle with intercooled compressor: (refer Fig. 5.23)
WN
Cp T1
1
c
−2
√
c−1
=
t 1−
=
3.333 × 1 −
=
0.64815
WN
=
η
=
1 × 300 × 0.64815
√
t
c−2
c + √
1−
t− c
=
1−
1
1.4804
−2×
=
194.44 kJ/kg
√
1.4804 − 2
√
3.333 − 1.4804
3.333
1.4804
+
√
1.4804 − 1
× 100
=
30.63%
Basic cycle with heat exchanger and intercooled compressor: (refer Fig. 5.28)
WN
η
=
194.47 kJ/kg
=
√
√
2×
1.4804 − 1
2( c − 1)
= 1−
1−
1
1
t 1− c
3.333 × 1 − 1.4804
=
59.92%
× 100
Ans
⇐=
Ideal Cycles and their Analysis
Cycle
WN
(kJ/kg)
180.36
180.36
194.47
194.47
Basic cycle
With heat exchanger
With intercooling
With heat exchanger
& intercooling
η
(%)
32.45
55.58
30.63
59.92
% change
in WN
–
0
7.8
7.8
119
% change
in η
–
71.28
−5.61
84.65
5.13 Prove that the efficiency corresponding to the maximum work done
in a Brayton cycle is given by the relation
1
ηwmax = 1 − √
t
where t is the ratio of the maximum and minimum temperatures of
the cycle.
A gas turbine operating on Brayton cycle between 27◦ C and 827◦C.
Determine the maximum net work per kg and cycle efficiency. Also
compare Carnot efficiency with Brayton efficiency for these temperature limits.
Solution
1500
3
1000
T
4
2
500
1
0
20
30
40
s
50
60
Fig. 5.53
WT
=
Cp (T3 − T4 )
WC
=
Cp (T2 − T1 )
WN
=
WT − WC
=
Cp T3 1 −
=
T4
T3
Cp [(T3 − T4 ) − (T2 − T1 )]
− T1
T2
−1
T1
120
Gas Turbines
Let rp be the pressure ratio,
!
WN
Let c = (rp )
WN
=
γ−1
γ
Cp T3 1 −
(rp )
− T1 (rp )
γ−1
γ
γ−1
γ
−1
"
T3
T1
and t =
=
1
1
c
Cp T3 1 −
− T1 (c − 1)
For maximum work dWN /dc = 0.
0
=
Cp T3
T1
=
T3
c2
=
c
=
η
=
=
1
− T1
c2
1
c2
T3
T1
√
t
=
t
(T3 − T4 ) − (T2 − T1 )
(T3 − T2 )
T3 1 −
T4
T3
T3
T1
T1
T3
T1
=
=
=
=
1−
t
(rp
−1
T2
T1
− (rp )
γ−1
) γ
− (rp )
γ−1
γ
−1
γ−1
γ
1
c
− (c − 1)
(t − c)
c−1
c
− (c − 1)
(t − c)
(c − 1) t−c
c
(t − c)
For maximum work c =
−
1
T3
T1
t 1−
T2
T1
− T1
=
=
(c − 1) ct − 1
(t − c)
c−1
c
=
1−
1
c
√
t
ηwmax
=
1
1− √
t
T1
=
27 + 273
Ans
⇐=
=
300 K
Ideal Cycles and their Analysis
T2
=
1100 K
t
=
1100
300
Wmax
=
1
Cp T3 1 − √
t
=
1.005 ×
=
3.666
− T1
1
1100 × 1 − √
3.666
=
121
√
t−1
− 300 ×
√
3.666 − 1
Ans
⇐=
252.34 kJ/kg of air
ηwmax
=
1
1− √
3.666
ηcarnot
=
Tmax − Tmin
Tmax
=
0.7272
=
× 100
=
47.77%
Ans
⇐=
1100 − 300
1100
=
Ans
72.72%
⇐=
5.14 In an ideal gas turbine cycle with reheat, air at state (p1 , T1 ) is compressed to pressure rp1 and heated to T3 . The air is then expanded in
two stages, each turbine having the same pressure ratio, with reheat
to T3 between the stages. Assuming the working fluid to be a perfect gas with constant specific heats, and that the compression and
expansion are isentropic, show that the specific work output will be
a maximum when r is given by
r
where t =
T3
T1
and a =
=
t(2/3a)
=
c
γ−1
γ .
Solution
T3
T4
=
T2
T1
=
T5
T6
=
ra
√
r
a
=
√
c
Assume unit mass flow of air and given T5 = T3 .
Total turbine work output,
WT
=
Cp (T3 − T4 ) + Cp (T5 − T6 )
=
Cp (T3 − T4 ) + Cp (T3 − T6 )
122
Gas Turbines
1600
1100
T
5
4
6
4’
2
600
100
3
1
20
30
40
s
50
60
70
Fig. 5.54
=
1
2Cp (T3 − T4 ) = 2Cp T3 1 − √
c
=
Cp (T2 − T1 )
=
Cp T1 (c − 1)
=
1
2Cp T3 1 − √
c
=
2
=
2t 1 − c− 2 − (c − 1)
Compressor work input,
WC
=
Cp T1
T2
−1
T1
Net work output,
WN
− Cp T1 (c − 1)
Specific work output,
WN
Cp T1
1
T3
1− √
T1
c
1
Specific work output is maximum when
entiating Eq.(1) and equating it to zero,
2t −1 × −
1
2
3
× c− 2 − 1
r(
γ−1
γ
WN
Cp T1
= 0. Therefore, differ-
0
=
1
)
=
ra
r
=
t(2/3a)
3
=
d
dc
(1)
=
t × c− 2
c
− (c − 1)
=
2
t3
Ans
⇐=
Ideal Cycles and their Analysis
123
5.15 A Brayton cycle works between 1 bar, 300 K and 5 bar, 1250 K. There
are two stages of compression with perfect intercooling and two stages
of expansion. The work out of first expansion stage being used to drive
the two compressors, where the interstage pressure is optimized for
the compressor. The air from the first stage turbine is again heated
to 1250 K and expanded. Calculate the power output of free power
turbine and cycle efficiency without and with a perfect heat exchanger
and compare them. Also calculate the percentage improvement in the
efficiency because of the addition of heat exchangers.
Solution
5
1500
1000
T
500
0
7
4
2
3
1
9
8
6
5
7
6
8
T
2
4
10
1
3
10 20 30 40 50 60 70
s
s
Fig. 5.55
Because of perfect intercooling,
√
p2
=
r =
√
5
=
2.236 bar
Further, T4 = T2 and T3 = T1 .
c1
=
2.2360.286
T2
=
T 1 × c1
=
T4
=
T2
377.7 K
WC
=
Wc1 + Wc2
=
2 × 1.005 × (377.7 − 300)
WT 1
=
WC
Cp (T5 − T6 )
=
156.2
T6
=
1250 −
=
T5
T6
p5
p6
=
=
156.2
1.005
γ
γ−1
=
1.259
300 × 1.259
=
2Wc1
=
=
=
=
377.7 K
2Cp (T2 − T1 )
156.2 kJ/kg
1094.6 K
1250
1094.6
3.5
= 1.591
124
Gas Turbines
=
5
1.591
T7
T8
=
p7
p8
T8
=
1250 ×
WT 2
=
Cp (T7 − T8 )
=
1.005 × (1250 − 900.9)
η
=
WN
q
q
=
Cp (T5 − T4 ) + Cp (T7 − T6 )
=
1.005 × [(1250 − 377.7) + (1250 − 1094.6)]
=
1032.84 kJ/kg
=
350.8
× 100
1032.84
=
Cp × [(T5 − T9 ) + (T7 − T6 )]
=
1.005 × [(1250 − 900.9) + (1250 − 1094.6)]
=
507.02 kJ/kg
=
350.8
× 100
507.02
p6
η
=
3.143 bar
γ−1
γ
1
3.143
0.286
=
900.9 K
=
350.8 kJ/kg
WT 2
q
=
=
33.96%
With regenerator
q
η
=
69.2%
=
50.9%
Percentage improvement in efficiency
=
69.2 − 33.96
69.2
Ans
⇐=
5.16 The p-V diagrams of two gas turbine power plants operating on identical inlet conditions of 1 bar and 27◦ C are shown in the Fig. 5.56. The
maximum turbine inlet temperature is 1200 K for both the plants.
Assuming that the pressure ratio and temperature ratio are the same
for both the plants, calculate the efficiency ratio of the power plants.
Also find the pressure ratio of LPC, HPC, HPT and LPT.
Solution
η1
=
1−
c
(Eq. 5.19)
t
Ideal Cycles and their Analysis
12
10
2
5
5
9
3
8
8
p6
p6
4
4
3
2
7
6
2
2
0
4
10
125
1
0
1
4
6
V
2
0
3
10
8
1
0
1
2
V
3
Fig. 5.56
r1
=
√
c
(Eq. 5.70)
t
√
1 − c/t
=
1 − c/t
η2
=
1−
η2
η1
=
x
t
=
r
c
=
r
=
40.286
√
c
=
1.219
x
=
1 − (1.219/4)
1 − (1.487/4)
η2
η1
=
1.107
r2
=
r3
=
2
=
4
γ−1
γ
=
=
1.487
=
1.107
Ans
⇐=
r4
=
√
r
=
√
4
Ans
⇐=
5.17 An ideal gas turbine operates with m number of compressor stages
and n number of turbine stages with a overall pressure ratio, r. The
maximum temperature is Tmax and the minimum temperature is
Tmin . Assume pressure ratios in compressor is same in all m stages
and perfect intercooling and reheating. Also assume that pressure
ratios in all n stages in turbine are same. Show that
t
=
texit × (ra )
m+n
mn
T4
t = TTmax
, a = γ−1
γ , and texit = T2 . where T4 and T2 are the exit
min
temperatures of the turbine and compressor respectively.
126
Gas Turbines
Solution
Stage n
Stage 3
Tmin
4
Stage 1
Stage 2
Stage 1
Stage 3
2
Stage m
T
Stage 2
3
Tmax
1
s
Fig. 5.57
Let ric and rit be the stage pressure of the compressor and turbine respectively.
m
n
r
=
ric
and r = rit
Now referring to the Fig. 5.57,
T2
a
=
(r)
Tmin
(i)
and
Tmax
a
= (r)
T4
(i) × (ii) gives
Tmax
T2
×
Tmin
T4
=
(ric ) × (rit ) = r m × r n
t
=
texit × (ra )
a
a
1
m+n
mn
1
a
(ii)
= r
m+n
mn
a
Ans
⇐=
5.18 In the above problem, if the gas turbine is working on optimum pressure ratio to give the maximum specific work output, establish the
relation between t and c.
Solution
Let ri be the pressure ratio of each stage, where i = 1, 2, . . . m for compressor and i = 1, 2, . . . n for turbine.
For compressor
r
=
r1 × r2 × r3 × . . . × rm
=
rim
The exit temperature of the compressor is T2 and that of turbine is T4 .
Since, the stage pressure ratios are equal, the temperature ratio will also
be equal.
Ideal Cycles and their Analysis
127
Tmin
Stage n
Stage 3
4
Stage 1
Stage 1
Stage 2
Stage 3
Stage m
2
T
Stage 2
3
Tmax
1
s
Fig. 5.58
r
r
γ−1
γ
=
rim
T2
Tmin
=
T2
Tmin
=
T2m
m
Tmin
=
mn
Tmin
=
T2mn
cn
n
Tmax
=
T4n r
mn
Tmax
=
T4mn cm
=
T4
T2
r
γ−1
γ
=
m
=
γ
γ−1
Tmax
T4
m
n
T2m
c
(i)
Similarly,
γ−1
γ
From (i) and (ii),
tmn
(ii)
mn
cm+n
Since, for the given condition
T4
=
T2
t
=
c
m+n
mn
Ans
⇐=
5.19 A gas turbine unit operates at a mass flow of 30 kg/s. Air enters
the compressor at a pressure of 1 bar and temperature 15◦ C and is
discharged from the compressor at a pressure of 10.5 bar. Combustion
occurs at constant-pressure and results in a temperature rise of 420 K.
If the flow leaves the turbine at a pressure of 1.2 bar, determine the
net power output from the unit and also the thermal efficiency. Take
Cp = 1.005 kJ/kg K and γ = 1.4.
128
Gas Turbines
Solution
p2
p1
γ−1
γ
T2
=
T1
T3
=
564.22 + 420
T4
=
T3
p3
p4
γ−1
γ
= 288 × 10.50.286 = 564.22 K
=
984.22 K
984.22
=
10.5 0.286
1.2
= 529.27 K
=
ṁCp (T2 − T1 )
=
30 × 1.005 × (564.22 − 288) = 8328 kW
=
ṁCp (T3 − T4 )
=
30 × 1.005 × (984.22 − 529.27)
=
13716.74 kW
WN
=
W T − WC
Q
=
ṁCp (T3 − T2 )
=
12663 kW
=
Work done
Heat supplied
=
42.55%
⇐=
=
5388.74 kW
⇐=
WC
WT
ηth
Power output
=
5388.74 kW
=
30 × 1.005 × 420
Ans
⇐=
=
5388.74
× 100
12663
Ans
Ans
Review Questions
5.1 In what way the gas turbine cycles can be grouped and what are the
important distinctions between them?
5.2 State the assumptions made in an ideal cycle analysis of gas turbines.
5.3 Draw the schematic diagram of a simple cycle and explain briefly the
working of the cycle. Draw the p-V and T -s diagrams of the cycle.
5.4 Derive the expression for specific work output and the efficiency of a
simple cycle. Draw their trends as a function of pressure ratio.
5.5 Show that the specific work output is maximum when the pressure ratio
is such that the compressor outlet and turbine outlet temperatures are
equal.
Ideal Cycles and their Analysis
129
5.6 Find a suitable equation for the two pressure ratios where the specific
output becomes zero for a constant temperature ratio.
5.7 Explain why the specific power output becomes zero at these two pressure ratios for the above cycle.
5.8 Draw the schematic diagram of a simple cycle with a heat exchanger
and explain briefly the working principle. Draw also the p-V and T -s
diagrams of the cycle.
5.9 Derive the expression for specific work output and the efficiency of a
simple cycle with a heat exchanger. Draw their trends as a function
of pressure ratio.
5.10 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.11 Draw the schematic diagram of a simple cycle with reheat and explain
briefly the working principle. Draw also the p-V and T -s diagrams of
the cycle.
5.12 Derive the expression for specific work output and the efficiency of a
simple cycle with reheat. Draw their trends as a function of pressure
ratio.
5.13 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.14 Draw the schematic diagram of a simple cycle with reheat and heat
exchange and explain briefly the working principle. Draw also the p-V
and T -s diagrams of the cycle.
5.15 Derive the expression for specific work output and the efficiency of a
simple cycle with reheat and heat exchange. Draw their trends as a
function of pressure ratio.
5.16 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.17 Draw the schematic diagram of a simple cycle with intercooler and
explain briefly the working principle. Draw also the p-V and T -s
diagrams of the cycle.
5.18 Derive the expression for specific work output and the efficiency of
a simple cycle with intercooler. Draw their trends as a function of
pressure ratio.
5.19 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.20 Draw the schematic diagram of a simple cycle with intercooled and
heat exchanger and explain briefly the working principle. Draw also
the p-V and T -s diagrams of the cycle.
130
Gas Turbines
5.21 Derive the expression for specific work output and the efficiency of a
simple cycle with intercooled and heat exchanger. Draw their trends
as a function of pressure ratio.
5.22 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.23 Draw the schematic diagram of a simple cycle with intercooled and
reheat and explain briefly the working principle. Draw also the p-V
and T -s diagrams of the cycle.
5.24 Derive the expression for specific work output and the efficiency of
a simple cycle with intercooled and reheat. Draw their trends as a
function of pressure ratio.
5.25 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.26 Draw the schematic diagram of a simple cycle with intercooled, heat
exchange and reheat and explain briefly the working principle. Draw
also the p-V and T -s diagrams of the cycle.
5.27 Derive the expression for specific work output and the efficiency of a
simple cycle with intercooled, heat exchange and reheat. Draw their
trends as a function of pressure ratio.
5.28 Explain the important observations from the specific work output and
efficiency variation as a function of pressure ratio for the above cycle.
5.29 By means of suitable graphs compare the specific work output of various cycles without heat exchanger.
5.30 With a T -s diagram briefly explain the Ericsson cycle.
Exercise
[Note: Take γ−1
γ = 0.286 and Cp = 1.005 kJ/kg K for all problems,
unless stated otherwise.]
5.1 In a gas turbine plant, air enters the compressor at 1 bar and 27◦ C.
The pressure ratio is 6. The temperature at turbine inlet is 1000 K.
The mass flow rate of air is 10 kg/s. Determine
(i) power required to drive the compressor and the turbine power
output,
(ii) the ratio of the turbine to compressor work,
(iii) net power developed by the plant, and
(iv) the thermal efficiency.
Ans: (i) 2018.14 kW; 4029.75 kW (ii) 1.997
(iii) 2011.61 kW (iv) 40.10%
Ideal Cycles and their Analysis
131
5.2 A gas turbine is supplied with 60 kg/s of gas at 5 bar and 800◦ C and
expands it isentropically to 1 bar. Take the mean specific heats of the
gas at constant-pressure and constant-volume to be 1 kJ/kg K and
0.717 kJ/kg K respectively. Calculate the exhaust gas temperature
and the power developed in MW.
Ans: (i) 680.44 K (ii) 23.55 MW
5.3 An open-cycle gas turbine receives air at 0.98 bar and 23◦ C. The air is
compressed to 5.25 bar and reaches a maximum temperature of 650◦C
before entering into the turbine. The hot air expands back to 0.98 bar.
Assuming air-standard cycle and for unit mass flow rate, compute
the thermal efficiency of the plant if the compression and expansion
processes are isentropic. What is the ratio of the work required to
drive the compressor to the work developed by the turbine.
Ans: (i) 38.13% (ii) 0.5183
5.4 In a gas turbine plant the air enters the compressor at 1 bar and 300
K. The pressure ratio is 5. The temperature at the turbine inlet is
1200 K. The mass rate of flow is 12 kg/s.
Sketch the cycle on p-V and T-s planes and indicate the area representing the heat supply, heat rejection and net work of the cycle.
Determine,
(i)
(ii)
(iii)
(iv)
compressor and turbine work,
net work developed,
the ratio of turbine work to compressor work, and
the thermal efficiency.
Ans: (i) 2116.53 kW; 5341.37 kW (ii) 3224.84 kW
(iii) 2.5236 (iv) 36.9%
5.5 In an air-standard cycle heat supply is at constant-volume and the
heat rejection is at constant-pressure. The compression and expansion
are isentropic and the air at the start of the compression is at 30◦ C
and 1 bar. The pressure ratio is 6. The heat supply is 860 kJ/kg
of air and air flow is 2.0 kg/s. Assume Cp = 1.005 kJ/kg K and
Cv = 0.717 kJ/kg K. Calculate
(i) temperature at the end of each processes,
(ii) the power developed, and
(iii) the thermal efficiency.
Ans: (i) 505.82 K; 1705.26 K; 721.56 K
(ii) 1569.6 kW (iii) 91.25%
5.6 The inlet pressure and temperature before compression in a Joule
cycle are 1 atm and 300 K respectively. What will be the imep of the
cycle when the pressure ratio is 2.0 and peak temperature at the end
of combustion is 2200 K.
Ans: 0.7218 bar
132
Gas Turbines
5.7 If an ideal regenerator is added to the above cycle, for the same temperature and pressure ranges as for the previous problem, calculate
the work done and efficiency.
Ans: (i) 331.8 kJ/kg (ii) 83.4%
5.8 A simple ideal gas turbine works with a pressure ratio of 8. The
compressor and turbine inlet temperatures are 300 K and 800 K respectively. If the volume flow rate is 250 m3 /s, compute the net power
output and cycle efficiency.
Ans: (i) 33.557 MW (ii) 44.75%
5.9 A gas turbine power plant, working on an air-standard cycle, the
heat supply is at constant-volume and heat rejection is at constantpressure. The compression and expansion are isentropic. The atmospheric temperature and pressure are 27◦ C and 1 atm respectively.
The pressure ratio is 9. The heat supply is 600 kJ/kg of air. For an
air flow of 3 kg/s calculate, (i) temperature at the end of each process,
(ii) the net power developed and (iii) the thermal efficiency. Draw the
p-V and T -s diagrams.
Ans: (i) 562 K; 1398.8 K; 574.9 K
(ii) 1694.12 kW (iii) 94.1%
5.10 A turbine supplied with gas at 5.15 bar and 800◦ C and expands it
isentropically to 1.03 bar. If the mean specific heat of the gas at
constant-pressure and constant-volume are 1 kJ/kg K and 0.717 kJ/kg
K respectively. The air inlet temperature to compressor is at 30◦ C.
Calculate (i) the exhaust temperature and (ii)the power developed in
kJ per kg of gas per minute.
Ans: (i) 680.4 K (ii) 13068 kJ/kg
5.11 An open-cycle gas turbine plant receives air at 1 bar and 23◦ C. The
air is compressed to 5.5 bar and reaches a maximum temperature
of 700◦C in the cycle. The hot air expands back to 1 bar. Assuming air-standard cycle, compute thermal efficiency of the plant if the
compression and expansion are isentropic. What is the ratio of work
required to drive the compressor to the work developed by the turbine.
Ans: (i) 38.6% (ii) 0.495
5.12 In a gas turbine plant the air at 10◦ C and 1 bar is compressed to 4 bar
with compressor efficiency of 100%. The air is heated in the regenerator having 100 per cent effectiveness and the combustion chamber
till its temperature is raised to 700◦C and has a pressure drop of 0.14
bar. Determine the thermal efficiency of the plant. Ans: (i) 55.9%
5.13 A gas turbine operating between pressure limits of 1.5 bar and 5.5
bar. The inlet air temperature of the compressor is 20◦ C and the
air entering the turbine is at a temperature of 560◦ C. If the volume
rate of air entering the compressor is 1600 m3 /min, determine the
available power output for the cycle. Assume that the cycle operates
under ideal conditions.
Ans: (i) 6057.57 kW
Ideal Cycles and their Analysis
133
5.14 In a regenerator gas turbine cycle, air enters the compressor at a temperature and pressure of 30◦ C and 1.5 bar and discharges at 220◦C
and 5.2 bar. After passing through the regenerator the air temperature is 395◦ C. The temperature of air entering and leaving the gas
turbine are 900◦ C and 510◦C. Assuming no pressure drop through the
regenerator, determine (i) the output per kg of air, (ii) the efficiency
of the cycle and (iii) the work required to drive the compressor.
Ans: (i) 201 kJ/kg (ii) 39.6% (iii) 190.95 kJ/kg
5.15 Compare the maximum work delivered by an aircraft gas turbine
which works in the following two atmospheric conditions (two-stage
compression with perfect intercooling, but without reheat and regeneration). Compressor pressure ratio is 4 and metallurgical temperature limit is 1000 K. At ambient conditions: pressure = 1 atm and
temperature = 28◦ C and at 6000 m altitude : pressure = 0.5 atm
and temperature = −25◦ C. Find the percentage change in Net work
output, efficiency and exhaust temperature if the volume flow rate of
air is 2.5 m3 /s.
Ans: (i) 32.11% decrease (ii) 1.34% increase (iii) Nil
5.16 A closed-cycle gas turbine (with reheat) power plant operates using
helium as the working medium. The pressure ratio is 10. The maximum permitted temperature is 1000 K. Assuming the work output to
be maximum, calculate the efficiency. If air is used instead of helium,
calculate the efficiency and difference in heat added. Assume ideal
Brayton cycle. Temperature at the inlet of compressor = 27◦ C, Cp
of helium = 5.204 kJ/kg K and γ of helium = 1.67.
Ans: (i) 40.16%; 46.23% (ii) 2490.22 kJ/kg
5.17 A Brayton cycle operates with ideal air between 1 bar, 300 K and 5
bar, 1000 K. The air is compressed in two stages with perfect intercooling. Similarly in the turbine expansion occurs in two stages with
perfect reheating. Calculate the optimum pressure in bar, net work
output and the fraction of turbine output that has to be put back to
compressor (WC /WT )
Ans: (i) 2.236 bar (ii) 257.3 kJ/kg (iii) 0.378
5.18 A gas turbine unit operates at a mass flow of 30 kg/s. Air enters
the compressor at a pressure of 1 bar and temperature 15◦ C and is
discharged from the compressor at a pressure of 10.5 bar. Combustion
occurs at constant-pressure and results in a temperature rise of 420 K.
If the flow leaves the turbine at a pressure of 1.2 bar, determine the
net power output from the unit and also the thermal efficiency.
Ans: (i) 5388.74 kW (ii) 42.55%
5.19 An ideal open-cycle gas turbine plant using air operates on an overall
pressure ratio of 4 and between the temperature limits of 300 K and
1000 K. Assuming constant specific heats, Cp = 1.005 kJ/kg K and
134
Gas Turbines
Cv = 0.717 kJ/kg K, evaluate the specific work output and thermal
efficiency for each of the modifications below and state the percentage
change from the basic cycle. Assuming optimum stage pressure ratios,
perfect intercooling and perfect regeneration, find
(i) basic cycle,
(ii) basic cycle with heat exchanger,
(iii) basic cycle with two-stage intercooled compressor, and
(iv) basic cycle with heat exchanger and intercooled compressor.
ηth
% change
%
32.7
55.4
30.9
59.8
in
WN
Cp T01
Cycle
Basic
Heat exchange
Intercooled
Heat exchange & intercooled
0.6
0.6
0.653
0.653
W
Cp T01
–
0
8.8
8.8
% change
in ηth
–
69.4
−5.5
82.9
Multiple Choice Questions (choose the most appropriate answer)
1. Performance of an ideal cycle power plant pertains to
(a) work output
(b) efficiency
(c) specific fuel consumption
(d) all of the above
2. A simple ideal cycle consists of
(a) two adiabatic and two isentropic
(b) two adiabatic and two isothermal
(c) two isothermal and two constant-pressure
(d) two isentropic, one constant-pressure, one constant-volume
3. For the maximum specific output, for any given value of t, the optimum pressure ratio is given by
(a) c = t
(b) c = t2
√
(c) c = t
(d) c =
1
t
Ideal Cycles and their Analysis
135
4. Power output of simple cycle is a function of
(a) only pressure ratio
(b) only temperature ratio
(c) both pressure ratio and temperature ratio
(d) all of the above
5. The efficiency of the simple ideal cycle is a function of
(a) only the γ of the working fluid
(b) only the pressure ratio
(c) both pressure ratio and γ
(d) inlet temperature of the turbine
6. Maximum power output is achieved for a simple ideal cycle with respect to pressure ratio when
(a) outlet temperature of the compressor is equal to outlet temperature of the turbine
(b) outlet temperature of the compressor is lower than the outlet
temperature of the turbine
(c) outlet temperature of the compressor is higher than the outlet
temperature of the turbine
(d) It has nothing to do with outlet temperatures
7. When a heat exchanger is added to an ideal simple cycle,
(a) power output decreases but the efficiency increases
(b) power output increases but the efficiency decreases
(c) both remain the same
(d) power output remains the same but the efficiency increases
8. For better performance, an ideal heat exchange cycle should be operated with
(a) lower t and higher r
(b) higher t and higher r
(c) higher t and lower r
(d) r and t should be the same
9. Multistage compression with intercooling and multistage expansion
with reheating improves
(a) efficiency
(b) power output
136
Gas Turbines
(c) both power output and efficiency
(d) none of the above
10. An ideal cycle with reheat, intercooling and heat exchange will increase
(a) efficiency
(b) work output
(c) both efficiency and work output
(d) none of the above
Ans:
1. – (d)
6. – (a)
2. – (c)
7. – (d)
3. – (c)
8. – (c)
4. – (c)
9. – (b)
5. – (c)
10. – (c)
6
PRACTICAL CYCLES
AND THEIR ANALYSIS
INTRODUCTION
In the last chapter, ideal gas turbine cycles have been analyzed. In practice, it is difficult to achieve those ideal conditions. Therefore, in this chapter, we will discuss the performance calculations of practical gas turbine
cycles by taking into account the various losses in different components.
Before going into the details, various assumptions made are enumerated
below.
6.1
ASSUMPTIONS
The practical gas turbine cycles differ from ideal cycles in the following
main respects:
(i) As the fluid velocities are high in turbo machinery, the change in
kinetic energy between inlet and outlet of each component will be
taken into account.
(ii) The compression and expansion processes are irreversible adiabatic
involving increase in entropy.
(iii) Fluid friction results in pressure losses in combustion chamber and
heat exchangers and also in the inlet and exhaust ductings.
(iv) Complete heat exchange is not possible in a heat exchanger.
(v) The mass flow is assumed to be the same in spite of the addition of the
fuel. This is justified because the bleeding of air from the compressor
(which is around 1 to 2%) for cooling the turbine discs and blade
roots is compensated by the addition of fuel.
138
Gas Turbines
(vi) The values of Cp and γ of the working fluid vary throughout the
cycle due to change of temperature and due to changes in chemical
composition of the working medium.
(vii) Slightly more work than that required for the compression process
will be necessary to overcome bearing and windage friction in the
transmission and to drive ancillary components.
The implication of the above assumptions are discussed in the following
sections.
6.2
STAGNATION PROPERTIES
The first assumption brings out the effect of fluid velocity. The kinetic
energy terms in the steady flow energy equation can be accounted for,
implicitly by making use of stagnation (or total) enthalpy. The energy
equation is
1 2
h0 − h =
c −0
(6.1)
2
where, the stagnation enthalpy h0 is the enthalpy which a gas stream of
enthalpy h and velocity c would possess when brought to rest adiabatically
and without work transfer.
From Eq. 6.1,
c2
h0 = h +
(6.2)
2
when the fluid is a perfect gas, Cp T can be substituted for h, and the
corresponding concept of stagnation temperature T0 is defined by
T0
=
T+
c2
2Cp
(6.3)
In the above expression, c2 /2Cp is called the dynamic temperature and,
when it is necessary to emphasize the distinctions, T is referred to as the
static temperature.
An idea of the order of magnitude of the difference between T0 and T
can be obtained by considering air at atmospheric temperature for which
Cp = 1.005 kJ/kg K, flowing at 100 m/s, then
T0 − T
=
1002
2 × 1.005 × 103
≈
5K
It follows from the energy equation that, if there is no heat or work
transfer, T0 will remain constant. If the duct is varying in cross-sectional
area or friction is degrading the directed kinetic energy into random molecular energy, the static temperature will change, but T0 will not. Applying
Practical Cycles and their Analysis
139
the concept of an adiabatic compression, the energy equation becomes
⎤
⎡
W
=
=
⎢
⎢
(T2 − T1 )
+
− Cp ⎢
⎢
⎣
static component
c22
2Cp
−
c21
2Cp
⎥
⎥
⎥
⎥
⎦
dynamic component
− Cp (T02 − T01 )
(6.4)
stagnation
Similarly, for a heating process without work transfer
Q
= Cp (T02 − T01 )
(6.5)
Thus, if stagnation temperatures are employed there is no need to refer
explicitly to the kinetic energy term. A practical advantage is that it is
easier to measure the stagnation temperature of a high velocity stream
than the static temperature.
When a gas is brought to rest, not only adiabatically but also reversibly,
i.e., isentropically, the stagnation pressure p0 is defined by
p0
p
γ/(γ−1)
T0
T
=
(6.6)
p0 and T0 can be used in the same way as static values. The stagnation
and static states at the inlet and outlet of a compression process are shown
in Fig. 6.1.
p
02
02
p
2
02’
c 22 / 2 C p
2
2’
p
01
T
p
1
01
c21 / 2C p
1
s
Fig. 6.1 Stagnation and static properties
140
Gas Turbines
Thus the implication of the first assumption, viz., the effect of the inclusion of velocity in the calculation is that we shall refer to stagnation
conditions and not static conditions as we did in the ideal cycle analysis.
6.3
COMPRESSOR AND TURBINE EFFICIENCY
The effect of compressor and turbine efficiency is discussed in this section.
The actual simple gas turbine cycle is shown in Fig. 6.2, on a T -s diagram.
During compression process a considerable amount of energy supplied to
the compressor is wasted in churning up the working fluid. This energy does
not contribute to the pressure rise but is converted into heat by friction.
The outcome is that the temperature of the working fluid is higher at the
end of the compression than it would have been had the process been fully
and truly isentropic. The compression process on the T -s diagram will,
therefore be represented by the line 1 – 2, the temperature at 2 being
higher than at 2 , which is the temperature that would have been reached
by isentropic compression over the same pressure ratio. Because of this,
more work input to the compressor is required and thereby the efficiency
of the compressor comes into picture.
03
04
T
04’
02
02’
01
01-02’-03-04’ Ideal cycle
01-02-03-04 Actual cycle
s
Fig. 6.2 Actual gas turbine cycle on a T -s diagram
The compressor efficiency may be defined as the ratio of the work required for isentropic compression to the actual work input. In ideal cycle
analysis, it was assumed that the kinetic energy is the same before and after
the process. But in actual gas turbine cycles the changes in kinetic energy
are appreciable and therefore this assumption is not valid. Thus, under actual conditions we must consider the stagnation properties for calculating
work done in various components.
The compressor efficiency is given by
ηC
=
Isentropic compression work
Actual compression work
Practical Cycles and their Analysis
h02 − h01
h02 − h01
=
T02 − T01
T02 − T01
=
141
(6.7)
for the constant value of Cp . This ratio is also known as total-head isentropic
efficiency of the compression process.
Let rc be the total-head pressure ratio (or stagnation pressure ratio)
during compression, then in the isentropic process (1 − 2 )
T02
T01
= rc(γ−1)/γ
and
ηC
=
T02
T01
T01
(6.8)
(γ−1)/γ
−1
=
T02 − T01
T01 rc
−1
T02 − T01
(6.9)
or in the most suitable form for cycle calculations
T02 − T01
=
T01
ηC
rc(γ−1)/γ − 1
(6.10)
The above equation gives the actual stagnation temperature rise during
compression process.
Now referring to the expansion process of the gas turbine cycle, (Fig. 6.2)
a similar effect of wasteful heating of the working fluid will occur, as in the
case of compressor. This will result in the expansion following the line 3–4
in the Fig. 6.2, instead of ideal isentropic line 3–4 . The effect is therefore
to reduce the work output of the turbine.
The efficiency of the turbine can be written as
ηT
=
Actual turbine work output
Isentropic turbine work output
=
h03 − h04
h03 − h04
=
(T03 − T04 )
(T03 − T04 )
(6.11)
for the constant value of Cp .
If rt is the total head pressure ratio during the expansion process (3−4 )
T03
T04
ηT
(γ−1)/γ
= rt
=
(6.12)
T03 − T04
T03 1 −
(6.13)
1
(γ−1)/γ
rt
or
T03 − T04
= ηT T03 1 −
1
(γ−1)/γ
rt
(6.14)
When the turbine is exhausting direct to the atmosphere, the kinetic
energy of the exhaust is wasted. Then, the efficiency will be given by the
142
Gas Turbines
ratio of work output to the isentropic enthalpy drop from stagnation inlet
to static outlet conditions. This efficiency may be denoted by ηT and is
given by
T03 − T04
ηT =
(6.15)
T03 1 − (P /P 1)(γ−1)/γ
03
4
It should be noted that the extra work supplied to the compressor, because of its inefficiency, is not entirely wasteful. The additional heating
effect, resulting from non-isentropic compression, means that the heat to
be supplied in the combustion chamber is reduced. The gain in overall
efficiency so derived is not sufficient, however, to overcome the loss in efficiency due to the increased energy required to drive the compressor. The
higher exhaust temperature resulting from turbine inefficiency can be, to
some extent, utilized if an exhaust heat exchanger is included in the cycle.
The actual work of compression for unit mass flow rate is given by
Wcact
Cp T01
ηC
=
rc(γ−1)/γ − 1
(6.16)
and similarly, the actual work of expansion for unit mass flow rate is given
by
Wtact
ηT Cp T03 1 −
=
1
(γ−1)/γ
rt
(6.17)
Net work done is given by
WN = ηT Cp T03 1 −
1
(γ−1)/γ
rt
Cp T01
ηC
−
rc(γ−1)/γ − 1
(6.18)
If the works of compression and expansion are equated, the minimum temperature ratio TT03
= tmin can be calculated for which no net work output
01
is possible. Thus
ηT Cp T03 1 −
1
=
(γ−1)/γ
rt
Cp T01
ηC
rc(γ−1)/γ − 1
(6.19)
Assuming the specific heats in both processes are same and also rc = rt = r
T03
T01
=
r(γ−1)/γ
ηT ηC
tmin
=
T03
T01
=
(6.20)
r(γ−1)/γ
ηT ηC
(6.21)
For r = 5 and ηC = ηT = 0.85, with γ = 1.4, with minimum temperature
ratio about 2.19. Thus with T01 = 300 K, T03 must be about 657 K. The
Eq. 6.21 explains why the gas turbine has not been developed into a satisfactory prime mover, even many years after it had been first demonstrated.
Practical Cycles and their Analysis
143
Neither materials suitable for high temperature nor compressors of high
efficiency were available until about the middle 1930’s. Thus for r = 5 and
ηT = ηC = 0.7, T03 must be 970 K before the unit is self-driving and must
be considerably higher than this, for developing power to take any extra
load on the shaft.
The difference between the efficiencies of steam turbine plants and gas
turbine plants operating at the same temperature can be comprehended by
consideration of the net work expression. For steam turbines, the compression work is very small, since the working fluid is in liquid state, and thus
the compression efficiency is a negligible factor. For the gas turbine, the
negative work of compression is a considerable fraction of the total turbine
work and so the compression efficiency is very significant. Small variations
in the compressor and turbine efficiencies can thus have a considerable effect
on the overall performance. The magnitude of this effect can be explained
by a parameter known as the ‘work ratio’, defined as the ratio of net work
to the total turbine work.
Higher the value of work ratio, less is the effect on the variation of
compressor efficiency and the performance.
Work ratio =
WN
WT
=
WN
WT − WC
WC
=
=1−
WT
WT
WT
1−
Cp T01
ηC
(γ−1)/γ
rc
ηT Cp T03 1 −
−1
(6.22)
1
(γ−1)/γ
rt
If rc = rt = r
Work ratio =
1−
r(γ−1)/γ T01
ηC ηT T03
=
1−
c 1
t ηC ηT
(6.23)
.
where c = r(γ−1)/γ and t = TT03
01
From this expression, it will be noted that the work ratio is increased
by high temperature ratio, t, and by low pressure ratio, r. In general, the
value of work ratio is not used as any major criterion but it may be useful
as a deciding factor. Thus, the second assumption brings into light how
important the component efficiencies are to achieve the best performance
from the power plant. It may be noted that the higher values of efficiencies
culminate in better performance.
6.4
PRESSURE OR FLOW LOSSES
In this section, effect of pressure losses will be considered. Although the
irreversibility occurring in the compression and turbine has the major effect
on reducing the cycle performance from the ideal values, loss due to friction
144
Gas Turbines
and turbulence occurs throughout the whole plant. It is due to the fact that
no fluid flow process can be completely reversible. This overall loss may
conveniently be divided into the following losses.
(i) air-side intercooler loss,
(ii) air-side heat exchanger loss,
(iii) combustion chamber loss (both main and reheat),
(iv) gas-side heat exchanger loss, and
(v) duct losses between components and at intake and exhaust.
These losses are measured as differences of pressure from the ideal value.
All the losses up to the turbine inlet may be considered as being equivalent
to a reduction of pressure ratio during the compression and all those following the turbine being equivalent to a reduction of pressure ratio during
the expansion. The various losses are indicated in Fig. 6.3.
p
03
04
04’
cc
05
Δp
heg
Δp
06
hea
p
p
04
Δp
T
03
02’ 02
p
02
01
01
s
Fig. 6.3 T –s diagram showing various losses
It can be seen that because of the losses, the pressure ratios for compression and expansion are not equal. These fluid flow losses sometimes
called parasitic losses, are usually small individually compared with those
of compressor and turbine but are controllable to a large extent by suitably
designing the components and ducts. The loss in pressure due to flow is
proportional to ρc2 . From the continuity equation
ṁ =
therefore,
Acρ
(6.24)
Practical Cycles and their Analysis
c =
ṁ
Aρ
∝
ρc2
(6.25)
and
Δp
145
ṁ2
A2 ρ
∝
(6.26)
Thus, for a given mass flow rate ṁ and density ρ, the pressure loss is
inversely proportional to the square of flow area, or assuming circular duct
of diameter d,
1
Δp ∝
(6.27)
d4
Losses, thus can be minimized by increasing cross-sectional area. But the
problem is not that simple, since for components like diffusers and heat
exchangers, the length must be comparatively increased if low velocities
are to be achieved without higher flow losses. Thus, the overall bulk may
increase. The greatest difficulty is often on the exhaust side, i.e., following
the turbine, because the gases have a very low pressure and high temperature thereby the specific volume will be large. Hence, the gas-side of the
heat exchanger and the final exhaust duct is quite critical as far as design
is concerned.
It may be noted that the individual flow losses are small compared with
those due to irreversibility in compressor and turbine. Nevertheless, they
are extremely important because of their cumulative effect. Further, there
is little point in making attempts to get the best component efficiencies,
if their effect is nullified by parasitic losses. Because, flow losses are so
dependent on size and shape, the actual layout of a gas turbine plant has
a great influence on its overall performance.
The effect of varying flow losses on cycle performance can be generalized
in a manner convenient for calculation. The analysis consists of finding the
difference in performance between the cycle without losses and the cycle
with losses.
Without pressure losses p03 = p02 and p04 = p01 , hence, the turbine
work is
⎡
⎤
(γ−1)/γ
p02
−
1
⎢ p01
⎥
WT = ηT Cp T03 ⎣
(6.28)
⎦
(γ−1)/γ
p02
p01
Let
γ−1
γ
= a, then
⎡
p02
p01
a
−1
⎤
WT
=
ηT Cp T03 ⎣
with pressure losses
p03
=
p02 − Δpc
(6.30)
p04
=
p01 + Δpt
(6.31)
p02
p01
a
⎦
(6.29)
146
Gas Turbines
Wt
⎡
=
a
p02 −Δpc
p01 +Δpt
ηT Cp T03 ⎣
−1
a
p02 −Δpc
p01 +Δpt
⎤
(6.32)
⎦
where Δpc is the pressure loss between the compressor delivery, (state point
2) and turbine inlet, (state point 3) and Δpt between turbine discharge,
(state point 4) and the atmosphere
⎛ a
⎞
a
Δp
p02 1−
Wt
c
p02
⎜ pa 1+ Δpt a − 1 ⎟
p01
⎜ 01
⎟
ηT Cp T03 ⎜
⎟
a
c
⎝ pa02 1− Δp
⎠
p02
=
Δpt
pa
01 1+ p
The term 1 ±
Δp
p0
a
01
can be expanded as a series
1±a
Δp
± ...
p0
By omitting higher order terms of
work with losses becomes
Wt
Δp
p0
⎛
which by itself is small, the turbine
− 1⎟
⎟
⎟
c
1−a Δp
⎠
p02
02
Δp
1+a p t
01
⎜ pa
⎜ 01
ηT Cp T03 ⎜
⎝ pa02
=
=
⎡
ηT Cp T03 ⎣
p02
p01
a
Wloss
⎡
=
(6.34)
Δpt
01
c
t
1 − a Δp
− 1 + a Δp
p02
p01
p02
p01
Wt − Wt gives Wloss
⎞
Δpc
pa
02 1−a p
pa
01 1+a p
Wt
(6.33)
a
ηT Cp T03 ⎣
a
c
1 − a Δp
p02
Δpc
t
a Δp
p01 + a p02
p02
p01
a
1−
c
a Δp
p02
⎤
⎦
⎤
⎦
(6.35)
(6.36)
Δpc
c
In the term 1 − a Δp
p02 , the value of a p02 will be close to zero since
a
=
γ−1
γ
=
0.286
c
for exhaust gas and Δp
p02 should not exceed 0.10.
Thus, the loss of output can be simplified to
Wloss
=
ηT Cp T03
p02
p01
a
a
Δpt
Δpc
+
p01
p02
(6.37)
Practical Cycles and their Analysis
147
In the demonstration above, only two pressure losses were used, but we
can generalize on the basis of effect of a loss being inversely proportional
to the pressure level at which it occurs so that
) Δp
p
ΔpA
pA
=
+
ΔpB
pB
+
ΔpC
pC
+ ...
(6.38)
The above equation is not exact but is good enough for initial calculation
and particularly for estimating the effect of varying amount of losses in the
whole cycle.
6.5
HEAT EXCHANGER EFFECTIVENESS
The implication of the fourth assumptions, viz., the effectiveness of heat
exchanger is discussed in this section. For ideal cycle calculations the regeneration was taken as 100%, i.e., the air from the compressor was assumed
to be heated up to the turbine discharge temperature. Thus in Fig. 6.4,
T05 = T04 and as the fluid has constant specific heat, mass flow rate is the
same everywhere. Therefore,
T04 − T06
=
T05 − T02
T02
=
T06
(6.39)
i.e.,
1620
03
1420
1220
1020
05
T
820
04
02
620
06
420
20
1.0
0.8
0.6
01
220
20
30
40
s
50
Fig. 6.4 Effect of heat exchanger effectiveness
60
148
Gas Turbines
It is a well-known fact that heat transfer requires a temperature difference and when this difference is very small, as it is in the ideal heat
exchanger, then, infinite area is required. For actual heat exchanger the
required surface area increases very rapidly as the temperature difference
available decreases and thus for economic reasons the amount of regeneration is limited (Fig. 6.4) Thus, T05 is less than T04 and correspondingly
T06 > T02 . The mass flow rate of expanded gas in an actual gas turbine is
not quite the same as the mass flow rate of compressed air. It is usually
slightly greater by the amount of the fuel injected in the combustion chamber provided there is no bleeding of air from compressor outlet to cool the
turbine blades. Normally the mass of fuel injected and the amount of bleed
air is almost same.
The specific heat of the hot gases, Cpg , is greater than that of the
comparatively cool air by virtue of their higher average temperature. Thus,
for a heat balance
ṁa Cpa (T05 − T02 )
or
T05 − T02
=
=
ṁg Cpg (T04 − T06 )
ṁg Cpg
(T04 − T06 ) > T04 − T06
ṁa Cpa
(6.40)
(6.41)
The effectiveness of the heat exchanger is defined as the ratio of the temperature rise of the air (T05 − T02 ) to the maximum temperature difference
available (T04 − T02 ). This ratio is called the effectiveness or thermal ratio
and is denoted by . Thus
=
T05 − T02
T04 − T02
(6.42)
For ideal cycle with 100% effectiveness of the heat exchanger, the cycle
efficiency increases with decrease of pressure ratio and this holds true for the
actual cycle with < 1.0. However, it is modified by the effect of component
efficiency to produce an optimum pressure ratio for given values of , ηC and
ηT . Thus the implication of the fourth assumption is that effectiveness of
the heat exchanger has a say in the performance of the cycle.
6.6
EFFECT OF VARYING MASS FLOW
The compressor handles a mass flow rate, ṁa , taken from the atmosphere
and sends it to the combustion chamber where a fuel flow rate of ṁf is
added. Because of this the turbine gas flow rate, ṁg , is greater than the
compressor air flow rate, ṁa , by (1 + f ), where f is the fuel-air ratio by
mass, i.e.,
and
ṁg
=
ṁa + ṁf
ṁg
ṁa
=
1+
ṁf
ṁa
(6.43)
=
1+f
(6.44)
Practical Cycles and their Analysis
149
Although, the fuel-air ratio is convenient to work with the calculations,
it is usually more meaningful to think in terms of its reciprocal, the airfuel ratio, AF in gas turbine calculations. For most hydrocarbon fuels,
the stoichiometric air-fuel ratio is about 15:1 in whole numbers so that the
200% and 400% of theoretical air represent air-fuel ratios of about 30:1 and
60:1 respectively.
However, there is always nearly a compensating effect due to the bleeding of air from the compressor, either at discharge or intermediately, for a
variety of ancillary purposes, such as cooling air for bearing, turbine wheel,
cabin cooling etc., The quantity of bleed air is not known exactly when
design cycle calculations are being made. It is normally of the order of 1
to 2% of the total air. It is reasonable to assume that the air and gas mass
flows are equal. For final calculations, it is necessary to know or to estimate
as exactly as possible, the amount of bleed air because it can be of such
proportions as to reduce the nominal performance significantly. Because of
the higher speeds of military aircraft and use of higher combustion temperature, with correspondingly greater demand of cooling air, the demand for
compressor bleed air increases.
6.7
EFFECT OF VARIABLE SPECIFIC HEAT
The effect of variable specific heat is discussed in this section. The specific
heat of air is independent of pressure within the operating limits of gas
turbine. However, it varies considerably with temperature. At 300 K the
constant-pressure specific heat of air is 1.005 kJ/kg K but at 1000 K, it is
1.140 kJ/kg K – an increase of about 13.4%.
Further, the internal combustion of fuel causes the expanding exhaust
gas which contains products of combustion, principally CO2 and H2 O vapour,
both having higher value of specific heat than that of pure air. At 1000 K,
a typical value of Cp for gases after combustion is 1.147. In ideal cycle
calculations the specific heat has been taken to be constant throughout the
cycle, with a value of 1.005 kJ/kg K.
Thus, this assumption would seem to introduce considerable error, because the difference between the cold air and hot gas values given above
is about 15%. Although there is an error, it is much less than this value,
because of the compensating effect of the decrease in value of γ with temperature. As γ decreases, the exponent (γ−1)
decreases and thus the change
γ
of temperature ΔT for a given pressure ratio and initial temperature decreases. Because in the calculation of work input or output the change
of enthalpy, Cp ΔT , is involved. The effect of increased Cp is neutralized
to some extent. Actually the specific heat for air and gases, changes continuously during compression and expansion. For precise calculation, an
integration process is required. Keenan and Kaye∗ have provided tables of
air and gas properties at various air-fuel ratios in their ‘Gas Tables’ which
can be used for precise calculations.
∗ Keenan,
J.H. and Kaye, J. Gas Tables, Wiley, 1948
150
Gas Turbines
It has been shown that use of average value of specific heat over typical
ranges of conditions existing in gas turbine is very suitable and convenient
method. Adoption of a fixed value of Cp for the compression process and
another value for heating and expansion process is most appropriate. Corresponding values of γ can also be chosen. Numerical values commonly
used for gas turbines with air as working fluid are
Cpa = 1.005 kJ/kg K and γ = 1.40 during compression
Cpg = 1.147 kJ/kg K and γ = 1.33 during heating and expansion
The point to note is that even though Cp and γ values vary during a
cycle, it is recommended that a constant value for compressor (Cpa = 1.005
kJ/kg K) and another constant value for heating and expansion (Cpg =
1.147 kJ/kg K) will be more appropriate as this does not introduce any
significant error in the calculations.
6.8
MECHANICAL LOSSES
Let us see the last assumption in the practical cycle analysis. In most of the
gas turbine plants the power required to drive the compressor is transmitted
directly from the turbine without any intermediate gearing. Any loss that
occurs is therefore, only due to bearing friction, and windage. The loss is
very small and is usual to assume that it amounts to about 10% of the
power necessary to drive the compressor.
If any power is used to drive ancillary components such as oil pump,
fuel pump, etc., then this is accounted for by simply subtracting it from
the net output of the turbine unit.
If all these are taken into account then it is the practice to club it under
ηmech so that the work required to drive the compressor is calculated using
the relation
WC
6.9
=
1
Cpa (T02 − T01 ) kJ/kg of air
ηmech
(6.45)
LOSS DUE TO INCOMPLETE COMBUSTION
At the design operating conditions combustion efficiency in a gas turbine
limit is quite high, close to 100%. It is defined as the ratio of enthalpy
released to available enthalpy of fuel. It may be noted that the combustion
efficiency is extremely difficult to measure accurately and it is the conservative practice to assume it as 98% for purpose of cycle calculations. It is
taken into account in the performance calculation by dividing the theoretical amount of fuel required by 0.98 and it affects only cycle efficiency and
not work output.
It is extremely difficult to maintain combustion efficiency at this value
in very lean mixtures especially at low loads or very high altitude in a
turbo jet engine. It may be necessary to use lower values if operational
Practical Cycles and their Analysis
151
requirements dictate considerable periods at this condition. However, since
both part load and high altitude imply that the total fuel consumption is
relatively small, even a lower value of combustion efficiency may not be of
very great importance.
6.10
CYCLE EFFICIENCY
Air-fuel ratio has considerable effect on cycle efficiency since gas turbine
operate on an overall air-fuel ratio of the order of 60:1 to 100:1. In order
to calculate the cycle efficiency an accurate determination of air-fuel ratio is essential. Assume ṁf and ṁa are the mass flow rates of fuel and
air respectively. Equating heat supplied to the heat required to raise the
temperature of the gas to T03 we have
ṁf ηb ΔHrp
=
(ṁf + ṁa )Cpg (T03 − T02 )
(6.46)
where ηb is the combustion efficiency. ΔHrp is the enthalpy of reaction for
unit mass at 25◦ C (lower calorific value).
By dividing the above equation with ṁa
f ηb ΔHrp
= (1 + f )Cpg (T03 − T02 )
(6.47)
= Cpg (T03 − T02 )
(6.48)
f is the fuel-air ratio
f ηb ΔHrp
since f is small compared to 1.0.
Consider Eq. 6.48. There are two unknowns, viz., f and Cpg . Note
that Cpg is a function of temperature and the species concentration in
the exhaust. The species concentration depends upon the air-fuel ratio. In
order to solve the Eq. 6.48, start with a mean value of Cpg = 1.147 kJ/kg K;
ηb = 98% and the lower calorific value of the fuel as 43 MJ/kg. Find the
value of f . For this value of f , we can find the species concentration and
more accurate value of Cpg . With this new value of Cpg find f and the
process is repeated until the value of Cpg and f do not vary by a percentage,
say, 0.1%.
A set of standard curves, connecting the ideal temperature rise with
air-fuel ratio for a range of initial temperatures are also available in the
literature to find the value of fuel-air ratio.
Specific fuel consumption can be found directly from
sf c
=
3600 ṁf
WN ṁa
=
3600 f
kg/kW h
WN
(6.49)
where WN is the net work output in kW/kg/s of air flow.
The cycle η may be defined as
η
=
WN
f CV
=
3600
sf c CV
(6.50)
where the heating value, CV = 43 MJ/kg. Thus knowing the accurate
air-fuel ratio actual cycle efficiency can be determined.
152
Gas Turbines
6.11
POLYTROPIC EFFICIENCY
It should be noted that isentropic efficiencies of compression and expansion
process vary with pressure. Therefore, it is necessary to take this variation
into account while calculating the performance of actual cycles. This variation can be taken into account by defining a new parameter called polytropic
or small-stage efficiency. Let us consider an infinitesimal compressor stage
(small stage with pressure and temperature rise of dp and dT respectively)
in which the isentropic efficiency remains constant. Then, the small stage
efficiency is defined as the isentropic efficiency of an elemental stage of the
compression which is constant throughout the whole process.
To understand the effect of overall pressure ratio on the relationship between small-stage efficiency and overall efficiency, let us imagine the process
on a T -s diagram (Fig. 6.5).
1020
05
04’
820
04
05’’
T
620
03’
02’
420
02
01
220
20
03
0
10
20
30 s 40
50
60
70
Fig. 6.5 Multi-stage compressor process in a T –s diagram
Since the process is not isentropic, the starting point for any elemental
stage will be at higher level than the starting point of the whole process.
Thus, since the vertical distance between two constant-pressure lines increases with entropy, the isentropic temperature rise for the elemental or
small stage is greater than that for the corresponding stage in an isentropic
compression. Hence, the sum of isentropic temperature
* rises for all the
elemental stages making up the complete compression,
ΔT s, is greater
than the single overall temperature rise which would result from an isentropic compression ΔT and this effect is magnified as the overall pressure
ratio increases. Now, the actual temperature-rise is found by dividing each
ΔT by the polytropic efficiency. It follows that the overall efficiency is less
than the polytropic efficiency, and that it decreases with the increase in
overall pressure ratio.
A similar argument holds good for an expansion process also. Here
the overall efficiency is always greater than the polytropic efficiency and
increases with increase in pressure ratio.
Practical Cycles and their Analysis
153
Now, a relationship between overall efficiency and polytropic efficiency
for varying pressure ratio can be obtained by assuming a polytropic law for
the process, viz.,
pV n
=
constant
(6.51)
pV
=
RT
(6.52)
From Eqs. 6.51 and 6.52, we can obtain
p1−n T n
=
constant
(6.53)
=
log c
(6.54)
dp
dT
+n
p
T
=
0
(6.55)
1 − n dp
n p
=
n − 1 dp
n p
(6.56)
Also, for a perfect gas,
Taking the logarithm on both sides, we get
(1 − n) log p + n log T
Differentiating this equation, we get
(1 − n)
Therefore,
dT
T
=
−
Thus the temperature rise in an elemental stage is given by
dT
=
n − 1 dp
T
n p
(6.57)
Had the process been isentropic, temperature rise in the elemental stage
would have been
γ − 1 dp
dT =
T
(6.58)
γ p
Hence, the efficiency of small stage, which is defined as the ratio of the
isentropic temperature rise to the actual temperature rise, i.e.,
ηpc
=
dT
dT
=
γ−1 n
γ n−1
(6.59)
This expression has been developed on the basis of static temperatures and
pressures. The same result will apply to stagnation conditions also. The
index n refers to an assumed polytropic law connecting stagnation pressure
and temperature. ηpc is the small-stage total head isentropic efficiency or
it is called polytropic efficiency.
Similarly, the polytropic efficiency in expansion process can be shown
to be
γ n−1
ηpe =
(6.60)
γ−1 n
ηpc and ηpe enable us to calculate overall efficiency in terms of the polytropic
efficiency.
154
Gas Turbines
Let rc be total head pressure ratio in compression and (T02 − T01 ) be
the total head temperature rise, then
But,
ηC
=
T02 − T01
T02 − T01
T02
T01
=
rc γ
T02
T01
T02
T01
=
−1
−1
(6.61)
γ−1
Since T02 − T01 , is the temperature rise during isentropic process, and
T02
T01
γ−1 1
γ ηpc
n−1
=
rc n
=
rc
(6.62)
Therefore,
γ−1
ηC
=
rc γ − 1
γ−1
γ
1
ηpc
rc
(6.63)
−1
and similarly
ηT
=
γ−1
γ
1
rt
1−
1−
1
rt
ηpe
(6.64)
γ−1
γ
For any given value of the polytropic efficiency, curves can be drawn showing
the variation of overall efficiency with pressure ratio. Figure 6.6 gives curves
for both compression and expansion for the polytropic efficiencies of 80, 85
and 90%.
For polytropic process, the temperature rise across the compressor is
given by
T02 − T01
where
n−1
n
⎡
= T01 ⎣
=
p02
p01
(n−1)
n
⎤
− 1⎦
1 γ−1
ηpc γ
(6.65)
(6.66)
and the temperature drop across the turbine will be given by
⎡
(n−1) ⎤
n
1
⎦
T03 − T04 = T03 ⎣1 − p03
(6.67)
p04
where
n−1
n
=
γ−1
ηpe
γ
(6.68)
Practical Cycles and their Analysis
155
0.95
ηpe = 0.9
ηpe = 0.85
Isentropic efficiency
0.9
ηpc = 0
.9
0.85
η pe = 0.8
ηpc = 0.8
5
0.8
0.75
ηpc = 0
.8
0.7
0.65
0
5
10
15
20
25
r
Fig. 6.6 Variation of isentropic η with pressure ratio for different polytropic
efficiency
Another formula for polytropic efficiency can be found in terms of pressure and temperature. For a small stage compression
ηpc
ηpc
=
dT
dT
dT
T
=
γ − 1 dp
γ p
=
γ − 1 dp T
γ p dT
(6.69)
(6.70)
Since, ηpc is constant, the equation can be integrated. On integration, we
will get
ηpc ln
T2
T1
ηpc
= ln
p2
p1
ln
p2
p1
=
ln
γ−1
γ
(6.71)
γ−1
γ
(6.72)
T2
T1
when stagnation conditions are considered
ln
ηpc
=
ln
p02
p01
T02
T01
γ−1
γ
(6.73)
This expression can be used to calculate polytropic efficiency from measured
values of stagnation pressures and temperatures.
156
Gas Turbines
The corresponding expression for expansion process is
T03
T04
ln
ηpe
=
p03
p04
ln
6.12
(6.74)
γ−1
γ
PERFORMANCE OF ACTUAL CYCLE
As per the simple cycle on T -s diagram, the temperature rise during the
compression is given by
T02 − T01
=
T01 ( γ−1
)
rc γ a − 1
ηC
(6.75)
and during expansion,
T03 − T04
⎡
= ηT T03 ⎣1 −
1
( γ−1
γ )g
rt
Now the net work output is given by
WN
=
⎤
⎦
n−1
1 ( n )e
1
( n−1 )
−
Cpa T01 re n c − 1
rt
ηmech
Cpg T03 1 −
(6.76)
(6.77)
where rc and rt are the pressure ratios of compression and expansion respectively. Then overall cycle efficiency
η=
1
Cpg (T03 − T04 ) − ηmech
Cpa (T02 − T01 )
Net work
=
Heat supplied
Cpg (T03 − T02 )
Cpg ηT T03
1−
=
Cpg
( γ−1 )
( r1t ) γ g
+
T03 −T01
1
ηC
−
T01
1
C
ηmech pa ηC
( γ−1
γ )a
−1
rc
( γ−1
γ )a
−1
rc
,
(6.78)
+1
where ηmech is the mechanical transmission efficiency and Cpa and Cpg are
the mean specific heats for air and gas respectively.
Also, the formula for overall efficiency can be expressed in terms of
index of polytropic law for both compression and expansion which may be
determined from equations already derived for known values of polytropic
efficiencies. Then,
η=
Cpg T03 1 −
1
rt
( n−1
n )e
−
1
ηmech Cpa T01
( n−1 )
Cpg T03 − T01 rc n c
( n−1 )
rc n c − 1
(6.79)
The suffixes e and c on the n−1
n powers distinguish between expansion
and compression processes respectively.
Practical Cycles and their Analysis
6.12.1
157
The Simple Cycle
The above general expressions (Eqs. 6.78 and 6.79) enable us to draw curves
showing the variation of work output and cycle efficiency with pressure ratio
and turbine inlet temperature and for a given compressor inlet temperature
T01 . The various input values are given in Table 6.1. A set of such curves
using the values in Table 6.1 are shown in Fig. 6.7 for work output and
Fig. 6.8 for efficiency for a simple cycle.
Table 6.1 Input Values
ηpc
ηpe
ηmech
γair
γgas
Cpa
Cpg
γa −1
γa
γg −1
γg
T01
85%
87%
98%
1.4
1.33
1.005
1.147
0.286
0.248
300 K
The specific output curves for the conditions stated above are shown in
Fig. 6.7 for four values of t, where t is the ratio of turbine inlet temperature
to the compressor inlet temperature. For comparison, with the ideal cycles,
the work output is divided by Cp T01 . Both WN and η show a peak at
certain values of pressure ratio, r, the value of r being lower for the WN
than for η for a given temperature ratio. The effect of increasing t is
to increase the pressure ratio at which the peak value occur. Increase of
turbine inlet temperature has a comparatively greater effect on work output
than on efficiency. The importance of a higher value of t in order to obtain
reasonable performance is clearly demonstrated in these figures. The simple
cycle gas turbine, therefore, must use cheap fuels and have a high working
temperature in order to compete in fuel cost with reciprocating combustion
engines. It should be noted that, gas turbine power plants offer considerable
advantages in size and weight.
The effect of different values of compressor and turbine isentropic efficiencies on the performance of the gas turbine can now be easily studied.
This will show that, effect of reducing either or both, will reduce W and η.
On the other hand, the increased component efficiency increases markedly
both W and η and moves the maximum values to a higher pressure ratios.
It will also be noted that if low component efficiencies are expected then it
will not be worthwhile to use higher pressure ratio, since the maximum to
both W and η occur relatively at lower values of pressure ratio.
While plotting curves of Figs. 6.7 and 6.8, inlet air temperature was
assumed to be constant at 300 K. Since, the compressor work is directly
proportional to T01 , the specific work output and cycle efficiency will decrease as inlet air temperature increases. This effect is usually of little
importance in industrial applications. However, it has considerable importance in aircraft applications and results in improved performance at higher
altitudes due to reduced ambient temperatures. From the study of these
curves, it can be concluded that a comparatively lower turbine temperature may be advantageous in conjunction with high component efficiencies
in order to achieve better performance. The use of lower temperatures allow
Gas Turbines
1.5
t=5
1
W/CpT01
158
t=4
0.5
t=3
t=2
0
0
5
10
15
20
r
W
Cp T01
Fig. 6.7
vs r for a simple cycle
0.6
Practical
Ideal
0.4
t=5
t=4
η
0.2
t=
3
t=
2
0
0
4
8
12
16
r
Fig. 6.8 η vs r for a simple cycle
20
Practical Cycles and their Analysis
159
less expensive materials to be used and avoids possible fouling of turbine
blades.
6.12.2
The Heat Exchange Cycle
For better specific fuel consumption, it is necessary to add a heat exchanger
to the simple gas turbine cycle in order to approach the regenerative cycle. Figures 6.9 and 6.10 show the effect on performance of adding heat
exchanger of 0.80 effectiveness to the simple cycle. It can be seen that
addition of heat exchanger has almost negligible effect on net work output
since we have not assumed any pressure loss in the heat exchanger. If one
assumes pressure losses, the reduction in work output can be attributed to
the increased pressure losses and hence the reduced pressure ratio across
the turbine. However, the efficiency increase is quite notable. It may be
noted that with a heat exchanger effectiveness of 0.80 and t = 3, a peak
efficiency of about 29.0% is achieved. An almost equally important effect
is that regeneration lowers the pressure ratio at which the peak efficiency
occurs. This holds the promise for the possibility of simple and less expensive compressor so as to obtain high efficiency with regeneration. Further,
it is seen that the optimum pressure ratio for efficiency is lower than that
for specific work output. The range of possible heat exchanger effectiveness is very wide and higher values can be achieved practically only by the
regenerative type heat exchanger. This, however, represents a more difficult problem of design and construction. It should however be noted that
overall efficiency of over 30% is possible with standard values for the cycle,
which makes the gas turbine competitive in fuel consumption with other
combustion engines.
6.12.3
The Reheat Cycle
The next variation to the simple cycle is the addition of reheat. Reheating
is undertaken in order to achieve higher outputs. It may be noted that
reheat will not drastically alter the efficiency.
The variation of specific power output and efficiency with respect to
the pressure ratio is shown in Figs. 6.11 and 6.12. It is seen that there is
a marked improvement in the work output with increase in t whereas the
efficiency slightly decreases for t = 4 and t = 5. This is because of the less
efficient cycle is included due to reheat with lower expansion ratio.
6.12.4
The Reheat and Heat Exchange Cycle
Next variation to the simple cycle is to add both reheat and heat exchange
in order to improve the work output as well as the efficiency. The effect of
such addition is illustrated in Figs. 6.13 and 6.14.
With heat exchange, addition of reheat improves the specific output
considerably without loss of efficiency (Figs. 6.13 and 6.14). The curves
of Fig. 6.13 are based on the assumption that the gas is reheated to the
maximum cycle temperature at the point in the expansion giving equal
Gas Turbines
1.5
t=5
1
W/CpT01
160
t=4
0.5
t=3
t=2
0
0
4
8
12
16
20
r
Fig. 6.9
0.6
W
Cp T01
vs r for a heat exchange cycle
Without heat exchange
With heat exchange
t=5
t=4
0.4
t=5
t=
η
t=4
3
0.2
t=
3
t=
2
0
0
4
8
12
16
r
Fig. 6.10 η vs r for a heat exchange cycle
20
Practical Cycles and their Analysis
1.5
With reheat
Without reheat
t=5
t=5
1
W/CpT01
t=4
t=4
0.5
t=3
t=3
t=2
0
0
4
8
12
16
20
r
W
Cp T01
Fig. 6.11
vs r for a reheat cycle
0.4
t=5
With reheat
Without reheat
t=5
0.3
t=4
t=4
η 0.2
t=
3
t=
3
0.1
0
t=
0
2
4
8
12
16
r
Fig. 6.12 η vs r for a reheat cycle
20
161
Gas Turbines
2
With reheat and heat exchange
Without reheat and heat exchange
1.5
W/CpT01
162
t=
5
t=5
1
t=
4
t=4
0.5
0
t=3
t=3
t=2
0
4
8
12
16
20
r
Fig. 6.13
W
Cp T01
0.6
vs r for a reheat and heat exchange cycle
Without reheat and heat exchange
With reheat and heat exchange
t=5
t=4
0.4
t=5
t=3
η
t=4
0.2
t=3
t=
2
0
0
4
8
12
16
20
r
Fig. 6.14 η vs r for a reheat and heat exchange cycle
Practical Cycles and their Analysis
163
pressure ratios for the two turbines. The gain in efficiency due to reheat
obtained with the ideal cycle is not realized in practice, partly because of
the additional pressure loss in the reheat chamber and the inefficiency of the
expansion process. However, it may be pointed out that the improvement
is not realized mainly because, the effectiveness of the heat exchanger is
well short of unity and the additional energy in the exhaust gas is not fully
recovered. It is important to use a pressure ratio not less than the optimum
value for maximum efficiency, because at lower pressure ratios the addition
of reheat can actually reduce the efficiency as indicated by the curves.
It is worthwhile to note that reheat has not been widely used in practice
because of the additional combustion chamber, and the associated control
problems. It can offset the advantage gained from the decrease in size of the
main components consequent upon the increase in specific output. With
some possible exception of the application, reheat would be considered only
(i) if the expansion had to be split between two turbines for other reasons
and
(ii) if the additional flexibility of control provided by the reheat fuel supply was thought to be desirable.
With regard to (i), it must be noted that the natural division of
expansion between a compressor turbine and power turbine may not
be at the optimum point. If the expansion is not done optimally then
the full advantage of reheat will not be realized.
Finally, readers familiar with steam turbine design may appreciate
that reheat introduces additional mechanical problems arising from
the decrease in gas density. Hence, the need for longer blading, in the
low pressure stages will become necessary.
6.12.5
The Intercooled Cycle
The next variation to the simple cycle is the introduction of intercooling.
The performance of such a cycle is as shown in Fig. 6.15 and 6.16. From
cycle analysis it will be known that intercooling improves the work output
but decreases the efficiency at higher temperature ratios provided there
are no pressure losses. However, with pressure losses the efficiency will
decrease irrespective of temperature ratios. The analysis of actual cycle
will show that the work output increases considerably and peaks at much
higher pressure ratio. The efficiency, however, is but little changed at low
pressure ratios and increases as the pressure ratio increases. This is because
with irreversible compression the saving in negative work with intercooling
outweighs the additional fuel necessary for heating the air delivered by the
compressor at a lower temperature. The change in efficiency is not large
and the main use of intercooling is to achieve a greater work output per kg
of air. For high pressure ratios, intercooling is almost a necessity, because as
the air temperature increases, the increment of pressure for given amount of
work decreases. However, it is worthwhile to point out here that intercoolers
Gas Turbines
2
With intercooling
Without intercooling
1.5
W/CpT01
164
t=5
t=5
1
t=4
t=4
0.5
t=3
t=3
t=2
0
0
4
8
12
16
20
r
Fig. 6.15
W
Cp T01
vs r for an intercooled cycle
0.4
t=5
With intercooling
Without intercooling
t=4
0.3
t=4
t=3
η 0.2
t=
3
0.1
t=
t=2
0
0
4
2
8
12
16
r
Fig. 6.16 η vs r for an intercooled cycle
20
Practical Cycles and their Analysis
165
tend to be bulky and if they require cooling water, the self-contained nature
of the gas turbine is lost.
Figures 6.17 and 6.18 show the performance characteristics of intercooling combined with heat exchange and this provides one of the most
effective gas turbine cycles. Intercooling provides increased specific output,
while both intercooling and regeneration improve the efficiency. The heat
exchange modifies the effect of intercooling moving the pressure ratio up
for peak efficiency.
6.12.6
Intercooled Cycle with Reheat
Figures 6.19 and 6.20 show the results of intercooling with reheat which
are somewhat similar to these for heat exchange in increasing power and
efficiency. Reheat has been used in practice only on experimental units,
as the introduction of second combustion chamber leads to more complex
installations and control problems.
6.12.7
Intercooled Cycle with Heat Exchange and Reheat
Figures 6.21 and 6.22 show the complex cycle, i.e., the cycle with intercooling, reheat and heat exchange, representing the closest practical approach
to the ideal Ericsson cycle. It results in high efficiencies over a wide range
of pressure ratios and quite a high power outputs. It has been used only in
experimental plants, because in spite of its good performance, it requires
an elaborate plant which except for its more moderate use of cooling water,
offers little advantage over a conventional steam plant.
From the above analysis it may be concluded that in practice gas turbines utilize either higher pressure ratio simple cycle or a lower pressure
ratio heat exchange cycle. The other additions mentioned do not normally
show sufficient advantage to the offset increased complexity and capital
cost.
Worked out Examples
6.1 An oil gas turbine installation consists of a compressor, a combustion chamber and turbine. The air taken in at a pressure of 1 bar
and temperature of 30◦ C is compressed to 6 bar, with an isentropic
efficiency of 87%. Heat is added by the combustion of fuel in combustion chamber to raise the temperature to 700◦ C. The efficiency of
the turbine is 85%. The calorific value of the oil used is 43.1 MJ/kg.
Calculate for an air flow of 80 kg/min. Neglect the effect of fuel in
the mass flow rate. Calculate
(i) the air/fuel ratio of the turbine gases,
(ii) the final temperature of exhaust gases,
(iii) the net power of installation, and
(iv) the overall thermal efficiency of the installation.
Gas Turbines
2
With intercooling and heat exchange
Without intercooling and heat exchange
1.5
W/CpT01
166
t=
5
t=5
1
t=
4
t=4
0.5
t=3
t=3
t=2
0
0
4
8
12
16
20
r
Fig. 6.17
W
Cp T01
0.6
vs r for an intercooled with heat exchangers
With intercooling and heat exchange
Without intercooling and heat exchange
t=5
0.4
t=4
η
t=5
t=4
t=3
0.2
t=3
t=
2
0
0
4
8
12
16
20
r
Fig. 6.18 η vs r for an intercooled with heat exchangers
Practical Cycles and their Analysis
2
With intercooling and reheat
Without intercooling and reheat
t=
1.5
5
t=5
W/CpT01
t=4
1
t=4
t=3
0.5
t=3
t=2
0
0
4
8
12
16
20
r
Fig. 6.19
0.5
W
Cp T01
vs r for an intercooled with reheat cycle
With intercooling and reheat
Without intercooling and reheat
0.4
t=5
t=4
t=4
t=3
0.3
η
t=3
0.2
t=2
0.1
t=
0
0
4
2
8
12
16
20
r
Fig. 6.20 η vs r for an intercooled with reheat cycle
167
Gas Turbines
2.5
t=5
With pressure losses
Without pressure losses
Ideal cycle
2
t=5
t=4
W/CpT01
168
1.5
t=5
1
t=4
t=3
t=4
t=3
0.5
t=3
0
0
4
8
12
16
20
r
Fig. 6.21
W
Cp T01
vs r for a complex cycle
0.8
t=5
t=
4
0.6
t=
3
t=5
t=4
t=5
η 0.4
t=4
t=3
t=3
0.2
With pressure losses
Without pressure losses
Ideal cycle
0
0
4
8
12
16
r
Fig. 6.22 η vs r for a complex cycle
20
Practical Cycles and their Analysis
169
Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg = 1.147 kJ/kg K, γg =
1.33.
Solution
03
04
04’
02
T
02’
01
s
Fig. 6.23
The T -s diagram is as shown in the figure above.
p02
p01
T02
=
T01
T02
=
T01 +
γ−1
γ
303 × 60.2857
=
=
505.5 K
T02 − T01
505.5 − 303
= 535.8 K
= 303 +
ηC
0.87
Neglecting the addition of fuel in the combustion chamber, we have ṁf +
ṁa ≈ ṁa .
=
ṁa Cpg (T03 − T02 )
CV
=
80 1.147 × (973 − 535.8)
×
= 0.0155 kg/s
60
43100
ṁa
ṁf
=
1
80
×
60 0.0155
A/F
=
86
T04
=
T03
T04
=
T03 − ηT (T03 − T04 )
=
973 − 0.85 × (973 − 623.8)
ṁf
=
86
Ans
⇐=
p04
p03
γ−1
γ
= 973 ×
1
6
The net power of installation, WN (neglect fuel)
0.2481
= 623.8 K
=
676.2 K
Ans
⇐=
170
Gas Turbines
WN
=
ṁa Cpg (T03 − T04 ) − ṁa Cpa (T02 − T01 )
=
1.3333 × 1.147 × (973 − 676.2) −
1.3333 × 1.005 × (535.8 − 303)
=
Ans
142 kW
⇐=
The overall thermal efficiency, ηth
ηth
=
Net power
Heat input
=
21.25%
142
× 100
0.0155 × 43100
=
Ans
⇐=
6.2 In a gas turbine plant air enters the compressor at 1 bar and 7◦ C.
It is compressed to 4 bar with an isentropic efficiency of 82%. The
maximum temperature at the inlet to the turbine is 800◦ C. The isentropic efficiency of the turbine is 85%. The calorific value of the fuel
used is 43.1 MJ/kg. The heat losses are 15% of the calorific value.
Calculate the following:
(i) Compressor work in KJ/kg
(ii) Heat supplied in KJ/kg
(iii) Turbine work in KJ/kg
(iv) Net work in KJ/kg
(v) Thermal efficiency
(vi) Air/fuel ratio
(vii) Specific fuel consumption in kg/kW h
(viii) Ratio of compressor work to turbine work
Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg = 1.147 kJ/kg K, γg =
1.33.
Solution
T02
T02
WC
p02
p01
γ−1
γ
=
T01
=
280 × 40.2857
=
T01 +
T02 − T01
ηC
=
280 +
416.1 − 280
0.82
=
Cpa (T02 − T01 )
=
416.1 K
=
446 K
Practical Cycles and their Analysis
171
03
T
04
04’
02
02’
01
s
Fig. 6.24
Q
T04
=
1.005 × (446 − 280)
=
166.83 kJ/kg of air
=
Cpg (T03 − T02 ) = 1.147 × (1073 − 446)
=
719.17 kJ/kg of air
p04
p03
γ−1
γ
Ans
⇐=
1
4
0.2481
=
T03
=
760.7 K
=
T03 − ηT (T03 − T04 )
=
1073 − 0.85 × (1073 − 760.7) = 807.5 K
=
Cpg (T03 − T04 ) = 1.147 × (1073 − 807.5)
=
304.53 kJ/kg of air
=
WT − WC
=
137.7 kJ/kg of air
=
WN
Q/0.85
=
16.3%
ṁf CV (1 − 0.15)
=
(ṁa + ṁf )Q
ṁa
ṁf
=
0.85 CV
−1
Q
T04
WT
Net work, WN
ηth
= 1073 ×
Ans
⇐=
=
=
Ans
⇐=
304.53 − 166.83
Ans
⇐=
137.7
× 100
719.19/0.85
Ans
⇐=
172
Gas Turbines
0.85 × 43100
−1
719.17
=
sf c
WC
WT
=
49.94
Ans
⇐=
Fuel consumed/hour
Net power
=
=
ṁf × 3600
ṁa WN
=
0.524 kg/kW h
=
166.83 × ṁa
304.53 × (ṁa + ṁf )
=
166.83 × 49.94
304.53 × 50.94
=
3600
49.94 × 137.7
Ans
⇐=
=
0.537
Ans
⇐=
6.3 The following data refer to a gas turbine set employing a regenerator:
Isentropic efficiency of the compressor
Isentropic efficiency of the turbine
Mechanical transmission efficiency
Pressure ratio
Maximum cycle temperature
Combustion efficiency
Calorific value of the fuel
Air mass flow
Regenerator effectiveness
Regenerator gas-side pressure loss
Ambient temperature and pressure
:
:
:
:
:
:
:
:
:
:
:
82%
85%
99%
7:1
1000 K
97%
43.1 MJ/kg
20 kg/s
75%
0.1 bar
327 K; 1 bar
Calculate the output, specific fuel consumption and overall thermal
efficiency. Assume that the pressure losses in the air-side of the regenerator and combustion chamber are accounted for in the compressor
efficiency. Compare these results with those obtained for the same
plant without the regenerator, and with regenerator but without pressure loss and also comment on the results. Neglect the effect of fuel
mass addition in the heat balance in the combustion chamber but
include in the turbine calculation.
Solution
(i) With regeneration
p02
p01
T02
=
T01
T02
=
T01 +
γ−1
γ
= 327 × 70.2857 = 570.2 K
T02 − T01
ηC
Practical Cycles and their Analysis
173
03
04
04’
05
02
T
02’
01
s
Fig. 6.25
p04
570.2 − 327
0.82
=
327 +
=
p01 + Δp
p04
p03
=
=
1 + 0.1
γ−1
γ
=
1.1 bar
1.1
7
0.2481
T04
=
T03
T04
=
T03 − ηT (T03 − T04 )
=
1000 − 0.85 × (1000 − 631.8)
=
T02 + (T04 − T02 ) = 623.6 + 0.75 × (687 − 623.6)
=
671.2 K
T05
= 1000 ×
623.6 K
= 631.8 K
=
687 K
Neglecting the effect of change in mass flow rate due to ṁf in combustion
chamber, the heat balance is given by
ṁf CV ηcomb
=
ṁa Cpg (T03 − T05 )
ṁf
=
20 × 1.147 × (1000 − 671.2)
= 0.18 kg/s
43100 × 0.97
WT − WC
=
(ṁa + ṁf )Cpg (T03 − T04 ) −
ṁa Cpa
(T02 − T01 )
ηm
WN
=
20.18 × 1.147 × (1000 − 687) −
20 × 1.005 × (623.6 − 327)
0.99
=
7244.84 − 6021.88
=
1222.96 kW
Ans
⇐=
174
Gas Turbines
sf c
ηth
=
ṁf × 3600
WN
=
0.53 kg/kW h
=
1222.96
× 100
0.18 × 43100
0.18 × 3600
1222.96
=
=
Ans
⇐=
Ans
15.8%
⇐=
(ii) Without regenerator
p04
T04
=
1 bar
=
p04
p03
p01
γ−1
γ
=
617.1 K
=
T03 − ηT (T03 − T04 )
=
1000 − 0.85 × (1000 − 617.1)
=
ṁa Cpg (T03 − T02 )
CV ηcomb
=
20 × 1.147 × (1000 − 623.6)
= 0.207 kg/s
43100 × 0.97
=
WT − WC
=
20.207 × 1.147 × (1000 − 674.5) − 6021.88
=
1522.37 kW
⇐=
sf c
=
0.207 × 3600
= 0.49 kg/kW h
1522.37
⇐=
ηth
=
1522.37
× 100
0.207 × 43100
⇐=
ṁf
WN
1000 ×
0.2481
T03
T04
=
1
7
=
=
674.5 K
Ans
=
17.06%
Ans
Ans
Similarly, we can work out the results with regenerator but without pressure
loss. The following table gives the details.
Quantities
p03
p04
T04 (K)
ṁf (kg/s)
WT (kW)
sf c (kg/kW h)
ηth (%)
Regenerator
With Δp
Without Δp
6.364
7
687
674.5
0.18
0.186
7244.84
7536.41
0.53
0.44
15.8
18.9
Without
regenerator & Δp
7
674.5
0.207
7544.25
0.49
17.06
Practical Cycles and their Analysis
175
Remarks : As can be seen from the table that pressure loss plays a
major role in the efficiency than the regenerator. Hence, more care
should be taken in the design to have minimum pressure loss.
6.4 The following data apply to gas turbine set employing a separate
power turbine, regenerator, and intercooler between two-stage compression:
Isentropic efficiency of compression each stage
:
80%
Isentropic efficiency of compressor turbine
:
88%
Isentropic efficiency of power turbine
:
88%
Turbine to compressor transmission efficiency
:
98%
Pressure ratio in each stage of compression
:
3:1
Temperature after intercooler
:
297 K
Air mass flow
:
15 kg/s
Regenerator effectiveness
:
80%
Regenerator gas-side pressure loss
:
0.1 bar
Maximum turbine temperature
:
1000 K
Ambient temperature
:
327 K
Ambient pressure
:
1 bar
Calorific value of the fuel
:
43.1 MJ/kg
Calculate the net power output, specific fuel consumption, and overall
thermal efficiency. Assume that the pressure losses in the air-side of
the regenerator and combustion chamber are accounted for in the
compressor efficiency. Assume Cpa = 1.005 kJ/kg K, γa = 1.4, Cpg =
1.147 kJ/kg K, γg = 1.33.
Solution
03
09’
05
T
02
06
07
02’ 07’
08
09
04
04’
01
s
Fig. 6.26
p07
p01
T07
=
T01
T07
=
T01 +
=
477.8 K
γ−1
γ
T07 − T01
ηC
327 × 30.2857
=
=
327 +
=
447.6 K
447.6 − 327
0.8
176
Gas Turbines
WLP C
=
15 × 1.005 × (477.8 − 327)
p02
p08
γ−1
γ
=
2273.31 kW
297 × 30.2857
T02
=
T08
=
T02
=
T08 +
=
433.9 K
WHP C
=
15 × 1.005 × (433.9 − 297) = 2063.77 kW
WC
=
2063.77 + 2273.31
T02 − T08
ηC
=
297 +
=
Actual work input to compressor, Wca
WC
4337.08
WCa
=
=
ηtrans
0.98
=
406.5 K
406.5 − 297
0.8
4337.08 kW
=
4425.59 kW
Neglecting the effect of ṁf ,
WCT
=
ṁa Cpg (T03 − T09 )
T09
=
1000 −
T09
=
T03 −
=
707.7 K
p09
T04
T03 − T09
ηT
p03
=
T03
T09
T04
4425.59
15 × 1.147
γ
γ−1
p04
p09
=
=
=
742.8 K
1000 −
9
1000 4.0303
707.7
= 2.234 bar
γ−1
γ
=
1000 − 742.8
0.88
=
T09
742.8 ×
=
623.1 K
=
T09 − ηT (T09 − T04 )
=
742.8 − 0.88 × (742.8 − 623.1)
1.1
2.234
0.2481
=
637.5 K
Work output of the power turbine, WT P
WT P
=
15 × 1.147 × (742.8 − 637.5) = 1811.69 kW
(ii) Specific fuel consumption, sf c
T05
=
T02 + (T04 − T02 )
=
433.9 + 0.8 × (637.5 − 433.9) = 596.8 K
Practical Cycles and their Analysis
177
=
ṁa Cpg (T03 − T05 )
CV
=
15 × 1.147 × (1000 − 596.8)
= 0.161 kg/s
43100
sf c
=
0.161 × 3600
= 0.32 kg of fuel/kW h
1811.7
⇐=
ηth
=
1811.7 × 100
0.161 × 43100
⇐=
ṁf
=
26.1%
Ans
Ans
6.5 A 1850 kW open-cycle stationary plant is to have one intercooler, one
reheater and a regenerator. On one shaft the high pressure turbine
drives the low pressure compressor. On another shaft, the low pressure turbine drives the high pressure compressor and the load. The
following data may be assumed (Neglect the weight of the fuel and
the mechanical efficiencies).
Ambient pressure
:
1 bar
Ambient temperature
:
27◦ C
Maximum cycle temperature
:
720◦C
Pressure ratio in each stage of compression
:
2.5
Turbine and compressor efficiencies
:
80%
Pressure drop in each heater and
:
3%
each side of regenerator
Regenerator effectiveness
:
75%
Assume intercooling to ambient temperature and reheating to maximum cycle temperature.
Sketch the T -s diagram and index all state points. Calculate these
state points and the necessary ideal states. Calculate the thermal
efficiency and air rate. Find the power output of each turbine and
compressor. If the fuel used has 43.1 MJ/kg heating value and that
the combustion efficiency is 98%, calculate the fuel used per hour at
rated load and specific fuel consumption.
Solution
03 10
09’
09
05
T
06
02
02’
08
07’
07
01
s
Fig. 6.27
04
04’
178
Gas Turbines
T07
γ−1
γ
p07
p01
300 × 2.50.2857
=
T01
=
389.8 K
=
T01 +
=
412.3 K
WLP C
=
WHP T = 1.005 × (412.3 − 300) = 112.86 kJ/kg of air
WHP T
=
WLP C
T09
=
T03 −
WHP T
112.86
= 894.6 K
= 993 −
Cpg
1.147
⇐=
T09
=
T03 −
T03 − T09
993 − 894.6
= 870 K
= 993 −
ηT
0.8
⇐=
p03
=
0.97 × p02 × (1 − 0.03)
=
0.97 × 6.25 × 0.97
T07
p09
=
=
=
=
300 +
389.8 − 300
0.8
Ans
T02
⇐=
Cpg (T03 − T09 )
=
5.88
=
= 3.45 bar
1.03
3.347
Ans
0.2481
=
T10
T04
=
T10 − ηT (T10 − T04 )
=
993 − 0.8 × (993 − 741.2)
=
Cpg (T10 − T04 )
=
231 kJ/kg of air
WN
=
231 − 112.86
=
ṁa
=
Wplant
WN
1850
118.14
T05
=
T02 + (T04 − T02 ) = 412.3 + 0.75 × (791.6 − 412.3)
=
696.8 K
WLP T
=
=
=
Ans
⇐=
⇐=
T04
= 993 ×
Ans
⇐=
3.347 bar
γ−1
γ
Ans
Ans
5.88 bar
993 4.0303
870
3.45 × (1 − 0.03)
p04
p10
⇐=
=
=
γ
γ−1
Ans
T02
T07 − T01
ηC
p03
=
T03
T9
p10
=
Ans
= 741.2 K ⇐=
791.6 K
Ans
⇐=
1.147 × (993 − 791.6)
118.14 kJ/kg of air
=
15.7 kg/s
Ans
⇐=
Ans
⇐=
Practical Cycles and their Analysis
Q
=
Cpg T03 − T05 + T10 − T09
=
1.147 × (993 − 696.8 + 993 − 894.6)
=
452.61 kJ/kg of air
ηth
=
118.14
× 100
452.61
WLP C
=
WHP C
=
15.7 × 112.86
WLP T
=
15.7 × 231
ṁf
=
15.7 × 452.61 × 3600
ṁa × Heat input
=
ηcomb × CV
0.98 × 43100
=
605.65 kg of fuel/h
=
605.65
1850
sf c
=
=
26.1%
179
Ans
⇐=
WHP T
=
=
=
1771.9 kW
3626.7 kW
0.327 kg of fuel/kW h
Ans
⇐=
6.6 A gas turbine operating at a pressure ratio of 11.3137 produces zero
net work output when 476.354 kJ of heat is added per kg of air mass.
If the inlet air total temperature is 300 K and the turbine efficiency
is 71%. Find the compressor efficiency and the temperature ratio.
Assume γ = 1.4 and Cp = 1.005 kJ/kg K for the whole cycle.
Solution
03
T
04
04’
02
02’
01
s
Fig. 6.28
Since WN = 0, ΔTC = ΔTT and assuming Cp to be constant,
T02 − T01
From the above,
=
T03 − T04
180
Gas Turbines
T03 − T02
=
T04 − T01
(1)
γ−1
γ
T02
T01
=
T02
=
300 × (11.3137)0.4/1.4
ηT
=
T03 − T04
T03 − T04
T03
T04
=
T03
T04
=
p02
p01
0.4/1.4
=
(11.3137)
=
(2)
γ−1
γ
p03
p04
600 K
0.4/1.4
=
(11.3137)
2
(3)
Given that heat added = 476.354 kJ/kg of air,
Cp (T03 − T02 )
=
476.354
T03 − T02
=
476.354
1.005
=
474 K
(4)
From (2),
0.71
=
T03 1 −
T04
T03
T03
T04
T04
−1
T04
T03
1−
0.71
=
2×
T04
T03
=
1−
T04
=
T01 + (T03 − T02 ) = 300 + 474 = 774 K
T03
=
774
0.645
=
1200 K
t
=
1200
300
=
4
T02
=
1200 − 474
=
ηC
=
T02 − T01
T02 − T01
=
=
70.4%
2−1
0.71
2
=
Ans
0.645
⇐=
From (1) and (4)
Ans
⇐=
726 K
600 − 300
726 − 300
× 100
Ans
⇐=
Practical Cycles and their Analysis
181
6.7 The efficiencies of the compressor and turbine of a gas turbine are
70.42% and 71% respectively. The heat added in the combustion
chamber per kg of air is 476.354 kJ/kg. Find a suitable pressure
ratio such that the work ratio is 0.0544. Also find the corresponding
temperature ratio. The inlet total temperature of air 300 K.
Solution
03
T
04
04’
02
02’
01
s
Fig. 6.29
Work ratio
=
1−
c 1
t ηT ηC
0.0544
=
1−
c
t
c
1
×
t 0.7042 × 0.71
=
0.4728
Q
=
476.334 kJ/kg
Cpg (T03 − T02 )
=
476.334
T02
=
T01 +
476.334
=
1.147 × T03 − T01 +
476.334
1.147
=
T01
415.286
=
300 × [t − (1 + 0.6714t − 1.42)]
1.384
=
0.3286t + 0.42
0.3286t
=
0.964
t
=
2.934
T01
(c − 1)
ηC
T01
T01
c−
ηC
ηC
T03
1
1
− 1+
× 0.4728 × t −
T01
ηC
ηC
Ans
⇐=
182
Gas Turbines
(rp )
T03
T01
=
2.934
c
t
=
0.4728
c
=
2.934 × 0.4728
=
1.3872
=
(1.3872)3.5
γ−1
γ
rp
=
=
1.3872
Ans
3.144
⇐=
6.8 A practical gas turbine cycle having the isentropic compressor efficiency of ηC and isentropic turbine efficiency of ηT with a work ratio
of 0.30. If the pressure ratio r and temperature ratio t are 12 and
4 respectively, calculate the minimum temperature ratio required to
drive the compressor. If the above unit works at ideal condition what
will be its efficiency?
Solution
03
04
04’
02
T
02’
01
s
Fig. 6.30
Wratio
=
1−
c
1
·
t ηC ηT
0.3
=
1−
1
120.286
·
4
ηC ηT
ηC ηT
=
0.509
(1 − 0.3)
tmin
=
c
ηC ηT
=
120.286
= 2.79 = 2.8
0.73
η
=
1−
1
c
=
1−
=
=
1−
0.509
ηC ηT
0.73
1
120.286
=
0.508
Ans
⇐=
Ans
⇐=
Practical Cycles and their Analysis
183
6.9 The polytropic efficiency of the compressor is 85%. If the ideal outlet
temperature of the compressor is twice that of the inlet, calculate
the isentropic efficiency of the compressor. If the polytropic efficiency
of the turbine is same as the compressor, calculate the isentropic
efficiency of the turbine assuming that there is no pressure loss. Take
γa = 1.4 and γg = 1.33.
Solution
03
04
04’
02
T
02’
01
s
Fig. 6.31
γ−1
ηC
=
rc γ − 1
γ−1
rc γ ·
rc
=
ηC
=
ηT
=
−1
γ
γ−1
T02
T01
=
23.5
=
11.31
1
2−1
=
= 0.794 = 79.4%
1.26
11.310.286/0.85 − 1
1−
1−
=
1
ηpc
0.400
0.452
ηpe · γ−1
γ
1
rt
1
rt
=
γ−1
γ
=
1−
0.885
1
=
Ans
⇐=
0.85×0.248
1
11.31
0.248
1
− 11.31
Ans
88.5%
⇐=
6.10 A compressor has an isentropic efficiency of 0.85 at a pressure ratio
of 4.0. Calculate the corresponding polytropic efficiency, and thence
plot the variation of isentropic efficiency over a range of pressure ratio
from 2.0 to 10.0.
Solution
ηC
=
( γ−1 )
rc γ − 1
rc
γ−1 1
γ ηpc
−1
184
Gas Turbines
0.85
40.286 − 1
=
4
rc
ηc
−1
=
1.572
1
× ln 4
ηpc
=
ln 1.572
ηpc
=
0.877
⇐=
ηC
=
−1
r0.286
c
r0.326
−1
c
⇐=
4
0.286 ×
1
0.286× ηpc
1
0.286× ηpc
2
0.863
4
0.850
6
0.842
4
6
8
0.836
Ans
Ans
10
0.832
0.9
ηc
0.85
0.8
2
8
10
r
Fig. 6.32
6.11 A closed-cycle gas turbine is to be used in conjunction with a gas
cooled nuclear reactor. The working fluid is helium (Cp = 5.19
kJ/kg K and γ = 1.66).
The layout of the plant consists of two-stage compression with intercooling followed by a heat-exchanger; after leaving the cold side of
the heat-exchanger the helium passes through the reactor channels
and on to the turbine; from the turbine it passes through the hot-side
of the heat-exchanger and then a pre-cooler before returning to the
compressor inlet. The following data are applicable:
Compressor and turbine polytropic efficiencies
:
0.88
Temperature at LP compressor inlet
:
310 K
Pressure at LP compressor inlet
:
14.0 bar
Compressor pressure ratios (LP and HP)
:
2.0
Temperature at HP compressor inlet
:
300 K
Mass flow of helium
:
180 kg/s
Reactor thermal output (heat input to gas turbine)
:
500 MW
Pressure loss in pre-cooler and intercooler (each)
:
0.34 bar
Pressure loss in heat-exchanger (each side)
:
0.27 bar
Practical Cycles and their Analysis
Pressure loss in reactor channels
Helium temperature at entry to reactor channels
:
:
185
1.03 bar
700 K
Calculate the power output and thermal efficiency, and the heatexchanger effectiveness implied by the data.
Solution
05
06
07
T
06’
04
02
04’ 02’
01
03
s
Fig. 6.33
ηpc
=
γ−1 n
γ n−1
0.88
=
0.66
n
×
1.66 n − 1
n−1
n
=
0.4518 for compressor
T02
T01
=
(rc )
T02
=
310 × 20.4518
T04
=
T03 (rc )
Q
=
ṁCp (T05 − T07 )
500 × 103
=
180 × 5.19 × (T05 − 700)
T05
=
1235.217 K
Total pressure loss,
=
n−1
n
n−1
n
=
=
424 K
300 × 20.4518
0.34 + 0.34 + 0.27 + 1.03
p05
=
56.00 − 1.98
p06
=
14.00 + 0.27 + 0.34
ηpt
=
γ n−1
γ−1 n
=
=
=
1.98 bar
54.02 bar
=
410.32 K
14.61 bar
186
Gas Turbines
n−1
n
=
T06
=
T05
p05
p06
WC
WT
WN
η
0.66
1.66
0.88 ×
n−1
n
=
0.3498
1235.217
=
54.02 0.3498
14.61
= 781.790 K
=
ṁCp [(T02 − T01 ) + (T04 − T03 )]
=
180 × 5.19 × [(424 − 310) + (410.32 − 300)]
=
209559 kW
=
ṁCp (T05 − T06 ) = 180 × 5.19 × (1235.217 − 781.79)
=
423591 kW
=
WT − WC
=
214.032 MW
⇐=
=
214.032
WN
× 100 =
× 100 = 42.80%
q
500
⇐=
=
T07 − T04
T06 − T04
=
289.68
× 100
371.47
=
209.559 MW
=
=
=
423.591 MW
423.591 − 209.559
Ans
Ans
700 − 410.32
781.79 − 410.32
=
78%
Ans
⇐=
6.12 An open-cycle gas turbine plant consists of a compressor, combustion
chamber and turbine. The isentropic efficiencies of compressor and
turbine are ηC and ηT , the maximum and minimum cycle temperatures are T03 and T01 respectively, and rp is the pressure ratio for
both the compression and expansion. Neglecting pressure losses and
assuming that the working substance is a perfect gas, show that the
necessary condition for positive power output is
n−1
ηC ηT T03 > T01 rp n
Such a plant delivers 1500 kW and operates such that inlet pressure
and temperature at the compressor is 1 bar and 25◦ C and that at
turbine is 4 bar and 700◦ C. Calculate the isentropic efficiency of the
turbine and the requisite mass flow of air in kg/s if the compressor
efficiency is 85% and overall thermal efficiency is 21%. Assume Cp to
be 1.005 kJ/kg K throughout.
Solution
n−1
WC
=
Cp T01 rp n − 1
ηC
Practical Cycles and their Analysis
187
03
04
04’
02
T
02’
01
s
Fig. 6.34
WT
=
Cp ηT 1 −
1
T03
n−1
n
rp
If WT = WC , then
n−1
Cp ηT
n−1
rp n − 1
rp n − 1
T03
=
Cp T01
ηC ηT T03
=
T01 rp n
n−1
ηC
rp n
n−1
This means there will not be any power output. Hence for a positive power
output,
n−1
ηC ηT T03
T02
Given η0 = 0.21 =
>
T01 rp n
T01 r
γ−1
γ
−1
=
T01 +
=
298 +
Q
=
WN
1500
= 7142.86 kW
=
η0
0.21
ṁ
=
Q
Cp (T03 − T02 )
=
7142.86
1.005 × (973 − 468.59)
=
14.09 kg/s
ηC
298 × 40.286 − 1
= 468.59 K
0.85
1500
Q
Ans
⇐=
188
Gas Turbines
WN
=
1500
14.09
WN
=
Cp (T03 − T04 ) − Cp (T02 − T01 )
=
1.005 × (973 − T04 − 468.59 + 298)
T04
=
696.48 K
T04
=
T03
973
= 0.286 = 654.52
r0.286
4
ηT
=
T03 − T04
T03 − T04
=
=
0.868
86.8%
=
=
106.46 kJ/kg
973 − 696.48
973 − 654.52
Ans
⇐=
6.13 In a compound gas turbine the air from the compressor passes through
a heat exchanger heated by the exhaust gases from the low-pressure
turbine, and then into the high pressure combustion chamber. The
high-pressure turbine drives the compressor only. The exhaust gases
from the high pressure turbine pass through the low-pressure combustion chamber to the low pressure turbine which is coupled to an
external load. The following data refer to the plant:
Pressure ratio of the compressor : 4:1
Isentropic efficiency of compressor : 0.86
Isentropic efficiency of HP turbine : 0.84
Isentropic efficiency of LP turbine : 80.0
Mechanical efficiency of drive to compressor : 0.92
Temperature of gases entering HP turbine
Temperature of gases entering LP turbine
Atmospheric temperature
Atmospheric pressure
:
:
:
:
660◦ C
625◦ C
15◦ C
1 bar
In the heat exchanger 75% of the available heat is transferred to the
air. Assuming that the specific heat Cp of air and gas is 1.005, determine (i) the pressure of the gases entering the low pressure turbine,
and (ii) the overall efficiency.
Solution
p02
p01
γ−1
γ
T02
=
T01
ηC
=
T02 − T01
T02 − T01
0.86
=
428.13 − 288
T02 − 288
T02
=
450.94 K
= 288 × 40.286 = 428.13 K
Practical Cycles and their Analysis
T05 < T03
03 05
07 04’
T
189
02
06
06’
08
04
02’
01
s
Fig. 6.35
Work done in compressor = ηm × Work done in turbine
Cp (T02 − T01 )
=
0.92 × Cp (T03 − T04 )
450.94 − 288
=
0.92 × (933 − T04 )
T04
=
755.9 K
ηT
=
T03 − T04
T03 − T04
0.84
=
933 − 755.9
933 − T04
T04
=
722.16 K
T03
T04
=
p04
=
In HP turbine,
p03
p04
γ−1
γ
4
3.5
933
722.16
In LP turbine,
γ−1
γ
T05
T06
=
p05
p06
898
T06
=
1.63
1
T06
=
780.90 K
ηT
=
T05 − T06
T05 − T06
0.286
=
1.63 bar
Ans
⇐=
190
Gas Turbines
0.8
=
898 − T06
898 − 780.9
T06
=
804.3 K
=
T07 − T02
T06 − T02
0.75
=
T07 − 450.94
804.3 − 450.94
T07
=
716 K
Q
=
Cp (T03 − T07 + T05 − T04 )
=
1.005 × (933 − 716 + 898 − 755.9)
=
360.9 kJ/kg
=
Cp (T02 − T01 )
=
163.75 kJ/kg
=
Cp (T03 − T04 ) + Cp (T05 − T06 )
=
1.005 × [(933 − 755.9) + (898 − 804.3)]
=
272.15 kJ/kg
=
272.15 − 163.75
WT − WC
=
= 30%
Q
360.89
WC
WT
ηth
=
1.005 × (450.94 − 288)
Ans
⇐=
6.14 In a closed-cycle turbine plant the working fluid at 38◦ C is compressed
with an adiabatic efficiency of 82%. It is then heated at constantpressure to 650◦ C. The fluid then expands down to initial pressure
in a turbine with an adiabatic efficiency of 80%. The fluid after
expansion is cooled to 38◦ C.
The pressure ratio is such that work done per kg is maximum. For the
working fluid take Cpa = 1.005 kJ/kg K and Cpg = 1.147. Calculate
the pressure ratio and cycle efficiency.
Solution
t
=
T03
T01
=
923
311
=
2.96
For maximum specific work we know that
γ−1
√
√
γ
ropt
=
tηC ηT =
2.96 × 0.82 × 0.8
ropt
=
3.194
=
1.3935
Ans
⇐=
Practical Cycles and their Analysis
191
03
04
04’
02
T
02’
01
s
Fig. 6.36
p02
p01
γ−1
γ
= 311 × 3.1940.286 = 433.51 K
T02
=
T01
T02
=
T01 +
T04
=
T03
3.1940.248
ηT
=
T03 − T04
T03 − T04
0.80
=
923 − T04
923 − 692.03
T04
=
738.22 K
ηth
=
1.147 × (923 − 738.22) − 1.005 × (460.40 − 311)
1.147 × (923 − 460.40)
=
211.43 − 150.15
× 100
530.60
T02 − T01
433.51 − 311
= 460.40 K
= 311 +
ηC
0.82
=
923
3.1940.248
=
=
692.03 K
11.55%
Ans
⇐=
6.15 The following data relate to a test on simple gas turbine plant.
Stagnation pressure at entry is 1 bar, static pressure and temperature
at compressor entry are 0.93 bar and 10◦ C respectively; compressor
delivery total head pressure and temperature, 6 bar and 230◦ C respectively; turbine exhaust pipe temperature, 460◦ C, turbine horsepower
5100 kW.
Calculate the total head isentropic efficiency of the compressor, and
taking the compressor entry area as 0.10 m2 , calculate the air mass
flow and estimate the temperature of the gases at entry to the turbine.
It may be assumed that there are no losses at compressor entry, that
the entry velocity distribution is uniform and that increase of mass
flow due to fuel addition is negligible.
192
Gas Turbines
Take Cp = 1.005 kJ/kg K and γ = 1.4 for compression and Cp = 1.147
kJ/kg K, γ = 1.333 for expansion.
Solution
03
04
02’ 02
T
04’
01’ 01
s
Fig. 6.37
γ
γ−1
p01
pa
=
1+
γ−1 2
M
2
1
0.93
=
1+
1.4 − 1 2
M
2
-
1
3.5
1
0.93
−1
3.5
=
0.324
M
=
T01
T1
=
1 + 0.2 × 0.3242
T01
=
283 × 1 + 0.2 × 0.3242
0.2
p02
p01
γ−1
γ
=
288.94 K
= 288.94 × 60.286 =
T02
=
T01
ηC
=
T02 − T01
482.35 − 288.94
= 90.35%
=
T02 − T01
503 − 288.94
= 482.35 K
Ans
⇐=
We know ṁ = ρAV . By equation of state
p
=
ρRT
ρs
=
ps
RTs
a
=
V
=
=
γRT =
0.93 × 105
287 × 283
=
3
1.145 kg/m
√
1.4 × 0.287 × 283 × 1000 = 337.21 m/s
M × a = 0.324 × 337.21 = 109.26 m/s
Practical Cycles and their Analysis
ṁ
=
ρAV = 1.145 × 0.1 × 109.26 = 12.51 kg/s
193
Ans
⇐=
We know turbine output power = 5100 kW
5100
=
ṁCp (T03 − T04 )
5100
=
12.51 × 1.147 × (T03 − 733)
Turbine inlet stagnation temperature
T03
=
Ans
1088.43 K
⇐=
6.16 In the last seventy years, the gas turbine, inlet temperature has increased from 700◦C to 1000◦C, the turbine efficiencies have increased
from 70 to 90 per cent, and the compressor efficiencies from 65 to 85
per cent.
For a pressure ratio of 3 calculate:
(i) the efficiency and work ratio of turbines of 70 years ago and
(ii) the efficiency and work ratio of modern turbines.
Assume an inlet pressure of 1 bar and temperature of 27◦ .
Solution
03
04
04’
02
T
02’
01
s
Fig. 6.38
T02
=
T01 r
γ−1
γ
=
300 × 30.286
=
410.75 K
Turbines 70 years ago
T02
!
1
1+
ηC
=
T01
=
300 × 1 +
p02
p01
γ−1
γ
−1
1
× 30.286 − 1
0.65
"
=
470.40 K
194
Gas Turbines
T04
ηth
1
=
T03 1 − ηT 1 −
=
973 × 1 − 0.7 × 1 −
=
1.147 × (973 − 810.56) − 1.005 × (470.40 − 300)
1.147 × (973 − 470.4)
=
186.315 − 171.24
× 100 = 2.62%
576.48
=
186.32 − 171.24
186.32
Work ratio
Modern turbines
T02
=
T01 +
r
γ−1
γ
1
30.248
=
=
810.56 K
Ans
⇐=
0.0809
T02 − T01
410.75 − 300
= 430.29 K
= 300 +
ηC
0.85
T04
=
1273
30.248
T04
=
T03 − ηT (T03 − T04 )
=
1273 − 0.9 × (1273 − 969.398)
WC
=
1.005 × (430.29 − 300)
WT
=
1.147 × (1273 − 999.758)
ηth
=
313.4 − 130.941
= 18.88%
1.147 × (1273 − 430.29)
⇐=
=
WN
WT
⇐=
Work ratio
=
=
969.398 K
182.46
313.4
=
=
=
999.758 K
130.941 kJ/kg
=
0.58
313.4 kJ/kg
Ans
Ans
6.17 A gas turbine operates at a pressure ratio of 7 and maximum temperature is limited to 1000 K. The isentropic efficiency of the compressor
is 85 per cent and that of the turbine is 90 per cent. If the air enters
the compressor at a temperature of 288 K, calculate the specific output and efficiency of the plant and compare these values with those
achieved by the ideal Joule cycle. If the unit under actual condition
is required to produce a power output of 750 kW, determine the necessary mass flow rate. Take Cp equal to 1.005 for ideal and actual
cycles.
Solution
T02
=
T01 1 +
γ−1
1
r γ −1
ηC
Practical Cycles and their Analysis
195
03
04
04’
02
T
02’
01
s
Fig. 6.39
T04
WN
ηth
1
70.286 − 1
0.85
=
540.29 K
=
288 × 1 +
=
T03 1 − ηT 1 −
=
1000 × 1 − 0.9 1 −
=
Cpg (T03 − T04 ) − Cpa (T02 − T01 )
=
1.005 × (1000 − 615.87) − 1.005 × (540.29 − 288)
=
132.499 kJ/kg
=
WN
Cpg (T03 − T02 )
=
0.2867
=
1
r
γ−1
γ
1
70.286
=
615.87 K
Ans
⇐=
132.499
1.005 × (1000 − 540.29)
=
Ans
⇐=
28.68%
Ideal Joule cycle
WN
1
=
Cp T03 1 −
=
1.005 × 1000 × 1 −
=
213.42 kJ/kg
ηth
=
1−
ṁa
=
pN
WN
1
r
γ−1
γ
=
r
γ−1
γ
=1−
− T01 r
1
70.286
γ−1
γ
−1
− 288 × 70.286 − 1
Ans
⇐=
1
= 0.427 = 42.7%
70.286
750
132.499
=
5.60 kg/s
Ans
⇐=
Ans
⇐=
196
Gas Turbines
6.18 The layout of a gas turbine is shown in the diagram. The compressor
is driven by the HP stage of a two-stage turbine and compresses 5 kg
of air per second from 1 bar to 5 bar with an isentropic efficiency of
85%. The H.P. stage has an isentropic efficiency of 87% and its inlet
temperature is 675◦ C. The L.P. stage, which is mechanically independent has an isentropic efficiency of 82%. The expansion pressure
ratios of the two turbines are not equal. The exhaust gases from the
L.P. stage pass to a heat exchanger which transfers 70% of the heat
available in cooling the exhaust to raise the compressor temperature
at delivery.
Assuming the working fluid to be air throughout, of constant specific
heat, and neglecting pressure losses, estimate the intermediate pressure p04 and the temperature T04 between the two turbine stages,
the power output of the L.P. stage and the overall plant efficiency.
Assume an inlet pressure of 1 bar and temperature of 15◦ .
Solution
Heat exchanger
Combustion
chamber
Air
Fuel
HPC
HPT
Compressor
Turbine
Power
LPT
output
Separate power turbine
Fig. 6.40
p02
p01
γ−1
γ
T02
=
T01
ηC
=
T02 − T01
T02 − T01
0.85
=
456.35 − 288
T02 − 288
= 288 × 50.286 = 456.35 K
Practical Cycles and their Analysis
197
03
04
04’
0x
T
02
05
05’
02’
01
s
Fig. 6.41
T02
=
486.06 K
As LPT mechanically independent implies HPT power = Compressor power.
Now,
Cp (T02 − T01 )
=
Cp (T03 − T04 )
T04
=
T01 − T02 + T03 = 288 − 486.06 + 948
T04
=
749.94 K
ηHP T
=
T03 − T04
T03 − T04
0.87
=
948 − 749.94
948 − T04
T04
=
720.34 K
γ
γ−1
p03
p04
=
T03
T04
5
p04
=
948
720.34
p04
=
1.91 bar
T04
T05
=
T05
=
ηLP T
=
p04
p05
Ans
⇐=
3.5
Ans
⇐=
γ−1
γ
749.94
1.91 0.286
1
T04 − T05
T04 − T05
=
623.23 K
198
Gas Turbines
0.82
=
749.94 − T05
749.94 − 623.23
T05
=
646.04 K
Heat exchanger effectiveness
0.7
=
T0x − 486.06
T0x − T02
=
T05 − T02
646.04 − 486.06
T0x
=
598.046 K
WLP T
=
Cp (T04 − T05 )
=
1.005 × (749.94 − 646.04) = 104.42 kJ/kg
PLP T
=
104.42 × 5 = 522.10 kW
Q
=
Cp (T03 − T0x ) = 1.005 × (948 − 598.04)
=
351.71 kJ/kg
=
WLP T
Q
ηth
=
Ans
⇐=
Ans
⇐=
104.42
= 29.69%
351.71
Ans
⇐=
6.19 In the gas turbine plant shown, each compressor operates on a pressure ratio of 3 and an isentropic efficiency of 82%. After the low
pressure compressor, some of the air is extracted and passed to a
combustion chamber from which the products leave at a temperature
of 650◦ C and expand in power turbine. The remainder of the air
passes through the high pressure compressor and into a combustion
chamber from which it leaves at a temperature of 540◦C and expands
in a turbine which drives both the compressors. The isentropic efficiency of each turbine is 87%. If the temperature of the air at inlet
to the low pressure compressor is 15◦ C, determine the percentage of
the total air intake that passes to the power turbine and the thermal
efficiency of the plant.
For compression assume Cp = 1.005 kJ/kg K and γ = 1.4 and for
heating and expansion Cp = 1.147 kJ/kg K and γ = 1.33.
Solution
T02
T03
γ−1
1
rLPγ − 1
ηC
=
T01 1 +
=
288 × 1 +
=
T02 1 +
1
30.286 − 1
0.82
γ−1
1
γ
rHP
−1
ηC
= 417.66 K
Practical Cycles and their Analysis
199
04
05
07
07'
06
03
06'
03'
02'
02
01
Fig. 6.42
CC
07
04
PT
x
1-x
CC
01
03
02
HPC
LPC
05
06
CT
Fig. 6.43
1
30.286 − 1
0.82
=
417.66 × 1 +
T06
=
T05
90.248
T06
=
T05 − ηT (T05 − T06 )
=
813 − 0.87 × (813 − 471.45) = 515.85 K
=
= 605.7 K
813
= 471.45 K
90.248
Let x be the mass of air passing through the power turbine, then
ṁa Cpa (T02 − T01 ) + ṁa (1 − x)Cpa (T03 − T02 ) = ṁa (1 − x)Cpg (T05 − T06 )
(1 − x)
Cpg
(T05 − T06 ) − (T03 − T02 ) = T02 − T01
Cpa
200
Gas Turbines
x
=
1−
=
1−
=
0.142
Cpg
Cp (T05
1.147
1.005
T02 − T01
− T06 ) − (T03 − T02 )
417.66 − 288
× (813 − 515.85) − (605.7 − 417.66)
Percentage of the total air intake that passes to the power turbine
T07
ηth
Ans
=
14.2%
=
T04 1 − ηT 1 −
⇐=
1
3
γg −1
γg
1
= 731.50 K
=
923 × 1 − 0.87 × 1 −
=
xCpg (T04 − T07 )
(1 − x)Cpg (T05 − T03 ) + xCpg (T04 − T02 )
3
1.33−1
1.33
=
0.142×1.147×(923−731.50)
(1−0.142)×1.147×(813−605.7)+0.142×1.147×(923−417.66)
=
0.109
=
10.9%
Ans
⇐=
6.20 In an open-cycle gas turbine plant the air is compressed in a two-stage
compressor with complete intercooling to the initial temperature. After passing through an exhaust heat exchanger and combustion chamber, the gases are expanded in a two-stage turbine with reheating in a
second combustion chamber between the stages. Both the compressor
stages are driven by high pressure turbine stage and the power output
of the plant is taken from the mechanically independent low pressure
turbine stage. After expansion in the low pressure turbine, the gases
pass through the heat exchanger to atmosphere.
The pressure ratio of each compressor is 2:1, the air inlet pressure
and temperature are 1 bar and 15◦ C and the gas inlet temperature to
both turbines is 700◦ C. If the isentropic efficiency in each compressor
and turbine stage is 85% and the thermal ratio of heat exchanger is
50% determine the output per kg of air per second and the thermal efficiency of the plant, neglecting pressure losses in the heat exchanger,
intercooler and combustion chambers and any variation in the mass
flow of the working fluid due to addition of fuel. For gases during
heating and expansion take Cp = 1.147 and γ = 1.333.
Solution
T01
=
T03
Practical Cycles and their Analysis
05
08
08�
06� 011
011�
09
T
07
06
04
02
04� 02�
010
01
03
s
Fig. 6.44
p02
p01
=
2
p04
p01
=
p04
p02
×
p03
p01
p04
p01
=
4
p04
p03
=
=
2×2
=
4
p05
p08
=
γ−1
γ
p02
p01
= 288 × 20.286 = 351.15 K
T02
=
T01
T02
=
T01 +
=
362.3 K
T05
T06
=
20.248
T06
=
973
1.1876
T06
=
T05 − ηT (T05 − T06 ) = 973 − 0.85 × (973 − 819.3)
=
842.36 K
T02 − T01
351.15 − 288
= 288 +
ηC
0.85
=
=
T04
1.1876
=
819.3 K
=
T08
Heat exchanger effectiveness
T09
WN
=
T09 − T04
T08 − T04
=
T04 + (T08 − T04 )
=
362.3 + 0.5 × (842.36 − 362.3)
=
WLP T
=
Cp (T07 − T08 )
=
602.33 K
201
202
Gas Turbines
=
1.147 × (973 − 842.36) = 149.84 kJ/kg
Ans
⇐=
Heat addition
Q
ηth
=
Cpg (T05 − T09 ) + Cpg (T07 − T06 )
=
Cpg (2T05 − T09 − T06 )
=
1.147 × (2 × 973 − 602.33 − 842.36) = 575 kJ/kg
=
WN
Qs
149.84
575
=
=
26.06%
Ans
⇐=
6.21 The following data refer to a closed-cycle gas turbine plant using helium as working fluid and incorporating two-stage compression with
intercooling and two-stage expansion with reheating; temperature at
entry to each compression stage is 270◦C; pressure at entry to first
compression stage and exit from the second turbine stage is 1 bar;
first compression stage pressure ratio is 6; each compressor stage isentropic efficiency is 0.85; temperature at inlet to each expansion stage
is 1150◦C; isentropic efficiency of each expansion stage is 0.9; reheat
pressure is 6 bar; for helium polytropic index n is 1.24 and R is 10.05
kJ/kg K. Calculate the cycle thermal efficiency.
05 07
06’
T
06
08
08’
04
02
04’ 02’
01
03
s
Fig. 6.45
Solution
p02
p01
T02
=
T01
T02
=
T01 +
Cp
Cv
=
1.24
Cp − Cv
=
10.05
γ−1
γ
= 543 × 60.1935 = 768 K
T02 − T01
768 − 543
= 807.81 K
= 543 +
ηC
0.85
Practical Cycles and their Analysis
Cp
=
51.925 kJ/kg K
Cv
=
41.875 kJ/kg K
WC
=
2 × (807.81 − 543) × 51.925 = 27500.5 kJ/kg
T05
T06
=
60.24/1.24 = 1.4144
T06
=
1423
1.4144
T06
=
1423 − 0.9 × (1423 − 1006)
WT
=
2 × 51.925 × (1423 − 1047.7) = 38974.905 kJ/kg
Q
=
q1 + q2
=
51.925 × (1423 − 807.81) +
=
203
1006 K
=
1047.7 K
51.925 × (1423 − 1047.7) = 51431.193 kJ/kg
WN
=
WT − WC = 38974.905 − 27500.5 = 11474.405 kJ/kg
η
=
11474.405
51431.193
=
22.31%
Ans
⇐=
Review Questions
6.1 What are the various assumptions made in practical cycle analysis?
6.2 Bring out clearly the implications of the various assumptions.
6.3 Define polytropic efficiency. Derive suitable expressions for polytropic
efficiency and bring out the relation between the polytropic efficiency
and isentropic efficiency.
6.4 Show the variation of isentropic efficiency for various polytropic efficiency as a function of pressure ratio.
6.5 With suitable graphs, compare the performance of various practical
cycles with respect to a simple cycle.
Exercise
[Note: Take γaγ−1
= 0.286 and
a
stated otherwise]
γg −1
γg
= 0.248 for all problems unless
6.1 In a gas turbine the compressor takes in air at a temperature of 27◦ ,
and compresses it to five times the initial pressure with an isentropic
efficiency of 85%. The air is then passed through a regenerator heated
204
Gas Turbines
by the turbine exhaust before reaching the combustion chamber. The
effectiveness of the regenerator is 80%. The maximum temperature
after constant-pressure combustion is 677◦ C and the efficiency of the
turbine is 80%. Neglecting all losses except mentioned, and assuming
the working fluid throughout the cycle to have the characteristics of
air
(i) Sketch the cycle on the T -s diagram.
(ii) Calculate the efficiency of the cycle.
Ans: (i) 23.7%
6.2 A simple gas turbine takes in air at 1.0 bar and 27◦ C and compresses
to a pressure of 6 bar with the isentropic efficiency of compression
being 85%. The air passes to the combustion chamber, and after
combustion the gases enter the turbine at a temperature of 560◦C
and expand to 1.00 bar, the turbine efficiency being 80%. Neglecting
the change of mass flow rate due to fuel, calculate the flow of air
in kg per second for a net output of 1500 kW making the following
assumptions: Loss of pressure in combustion chamber = 0.08 bar.
Ans: (i) 42.67 kg/s
6.3 The axial compressor of a gas turbine delivers 20 kg/s of air at a pressure ratio of 5.0 with an isentropic efficiency of 80%. Inlet temperature and pressure are 22◦ C and 1 bar. Calculate the power required
by the compressor.
After heating at constant-pressure to 870◦C the gas is expanded to 1
bar through a turbine with efficiency of 85%. The turbine is direct
coupled to the compressor and to an airscrew reduction gear which
has an efficiency of 95%. What power is available for the airscrew?
Ans: (i) 4332.56 kW (ii) 2852.17 kW
6.4 For a gas turbine operating at a pressure ratio of 8.7 the maximum
temperature ratio to be maintained such that the turbine just supports the compressor, which is given by tmin = 3.0. If the compressor
inlet total temperature and the turbine efficiency are respectively 300
K and 0.75, find the following:
(i) the compressor efficiency
(ii) the temperature ratio at which the compressor work is 80% of
the power produced by turbine. Also find the corresponding
heat addition and Net work output per unit mass flow in the gas
turbine. Take Cpa = 1.005 kJ/kg K and Cpg = 1.147 kJ/kg K.
(iii) for the temperature ratio found at (ii) find the required compressor pressure ratio at which the compressor work and turbine
work are equal. Take γg = γa and Cpg = Cpa .
Ans: (i) 0.8 (ii) (a) 3 (b) 396.06 kW (c) 64.29 kW (iii) 12.33
Practical Cycles and their Analysis
205
6.5 In a gas turbine the pressure ratio of r is achieved by two stage compression with intercooling. If η1 is the efficiency of LP compressor,
η2 is the efficiency of the HP compressor and is the effectiveness of
the intercooler. Derive an expression for the above for stage pressure
ratios in terms of overall pressure ratio between first stage and second
stage such that work of compressor is minimum.
If the efficiencies of the HP and LP compressors are respectively 0.75
and 0.85, and if the effectiveness of the intercooler is 80%, find the
work of compression assuming the inlet temperature at LP compressor
as 300 K. The total pressure ratio for overall compression ratio is
11.3137.
−1
Ans: (i) c01 = c ηη12 +
+ −1
(iii) 327.368 kJ/kg
1
2
(ii) c02 = c
η2 + −1
η1 + −1
1
2
6.6 A simple gas turbine with heat-exchanger has a compressor and turbine having respective isentropic efficiencies ηC and ηT . Show that
the combined effect of small pressure drops Δphg (in gas-side of heatexchanger) and Δp (total in combustion chamber and air-side of heatexchanger) is to reduce the specific work output by an amount given
by
γ−1
Cp T03 ηT
Δp
×
Δphg +
γ−1
γ
r
γ
r
p01
where T03 is turbine inlet temperature, p01 is compressor inlet pressure and r, the compressor pressure ratio. Assume that Cp and γ are
constant throughout the cycle.
6.7 In a gas turbine plant, air is compressed from state (p01 , T01 ) to a
pressure rp01 and heated to temperature T03 . The air is then expanded in two stages with reheat to T03 between the turbines. The
isentropic efficiencies of the compressor and each turbine are ηC and
ηT . If xp01 is the intermediate pressure between the turbines, show
that, for given values p01 , T01 , √
T03 , ηC , ηT and r, the specific work output is a maximum when x = r.
If this division of the expansion between the turbines is maintained,
show that:
(i) when r is varied, the specific work output is a maximum with r
given by
γ/(γ−1)
ηC ηT T03
r3/2 =
T01
(ii) when a perfect heat-exchanger is added, the cycle efficiency is
given by
T01 r(γ−1)/2γ r(γ−1)/2γ + 1
η =1−
2ηC ηT T03
Assume that the working fluid is a perfect gas with constant specific heats, and that pressure losses in the heater, reheater, and heatexchanger are negligible.
206
Gas Turbines
6.8 A peak load generator is to be powered by a simple gas turbine with
free power turbine delivering 20 MW of shaft power. The following
data are applicable:
Compressor pressure ratio
Compressor isentropic efficiency
Combustion chamber pressure loss
Combustion efficiency
Turbine inlet temperature
Power turbine isentropic efficiency
Mechanical efficiency (each shaft)
Ambient conditions pa , Ta
:
:
:
:
:
:
:
:
11.0
0.82
0.4 bar
0.99
1150 K
0.89
0.98
1 bar, 300 K
Assume gas generator turbine isentropic efficiency = 0.87, Draw the
schematic diagram of the cycle. If the calorific value of the fuel =
43 MJ/kg, calculate the air mass flow required and the specific fuel
consumption.
Ans: (i) 130.51 kg/s (ii) 0.3726 kg/kW h
6.9 A gas turbine is to consist of a compressor, combustion chamber, turbine and heat-exchanger. It is proposed to examine the advantage of
bleeding off a fraction Δm
of the air delivered by the compressor
m
and using it to cool the turbine blades. By so doing the maximum
permissible cycle temperature may be increased from T to (T + ΔT ).
The gain in efficiency due to the increase of temperature will be offset
by a loss due to the decrease in effective air flow through the turbine.
Show that, on the following assumptions, there is no net gain in efficiency when
ΔT
Δm
T
=
m
1 + ΔT
T
and that this result is independent of the compressor and turbine
efficiencies. Assumptions:
(i) No pressure loss in combustion chamber or heat exchanger.
(ii) Working fluid is air throughout and the specific heats are constant.
(iii) Air bled for cooling purposes does no work in the turbine.
(iv) Temperature of the air entering the combustion chamber is equal
to that of the turbine exhaust.
A plant of this kind operates with an inlet temperature of 288 K,
a pressure ratio of 6.0, a turbine isentropic efficiency of 90 per cent
and a compressor isentropic efficiency of 87 per cent. Heat transfer
calculations indicate that if 5 per cent of the compressor delivery is
bled off for cooling purposes, the maximum temperature of the cycle
can be raised from 1000 to 1250 K. Find the percentage increase in
(a) efficiency, and (b) specific work output, which is achieved by the
combined bleeding and cooling process. Make the same assumptions
as before and take γ = 1.4 throughout. Ans: (i)25.1% (ii) 48.55%
Practical Cycles and their Analysis
207
6.10 The efficiencies of the compressor and turbine of a gas turbine are ηC = 0.80 and ηT = 0.75 respectively.
If the
temperature ratio is 4 and the inlet total temperature of
air 300 K, find the optimum pressure ratio, and corresponding Net work output, work ratio, heat addition and efficiency.
Ans: (i) 4.626 (ii) 158.96 kJ/kg (iii) 0.43 (iv) 796.16 kJ/kg (v) 19.96%
6.11 An auxiliary gas turbine for use on a large airliner uses a single-shaft
configuration with air bled from the compressor discharge for aircraft
services. The unit must provide 1.5 kg/s bleed air and a shaft power
of 200 kW. Calculate (a) the total compressor air mass flow and (b)
the power available with no bleed flow, assuming the following:
Compressor pressure ratio
Compressor isentropic efficiency
Combustion pressure loss
Turbine inlet temperature
Turbine isentropic efficiency
Mechanical efficiency (compressor rotor)
Mechanical efficiency (driven load)
Ambient conditions
:
:
:
:
:
:
:
:
3.80
0.85
0.12 bar
1050 K
0.88
0.99
0.98
1 bar, 288 K
Ans: (i) 4.97 kg/s (ii) 630.9 kW
6.12 In a simple gas turbine plant air enters the compressor at 1 bar and
15◦ C and leaves at 6 bar. It is then heated in combustion chamber
to 700◦ C and then enters the turbine and expands to atmospheric
pressure. The isentropic efficiency of compressor and turbine are 0.80
and 0.85 respectively and the combustion efficiency is 0.90. The fall
in pressure in the combustion chamber is 0.1 bar. Determine
(i) air-fuel ratio,
(ii) work ratio,
(iii) thermal efficiency,
(iv) air rate in kg per shaft kilowatt power, and
(v) specific fuel consumption.
Take calorific value of the fuel equal 42000 kJ/kg.
Ans: (i) 73.219:1 (ii) 0.2831 (iii) 16.89% (iv) 0.0105 kg/kW
(v) 0.5158 kg/kW h
6.13 An industrial gas turbine takes in air at 1 bar and 27◦ C and compresses it to 5.5 times the original pressure. The temperatures at the
salient points are compressor outlet, 251◦ C, turbine inlet 760◦ C and
turbine outlet 447◦C. Calculate (i) the compressor and (ii) turbine
efficiencies.
Compare the ideal cycle and the practical cycle considering component efficiencies for the following
(iii) work ratio,
208
Gas Turbines
(iv) thermal efficiency,
(v) optimum pressure ratio for maximum output.
Ans: (i) 0.842 (ii) 0.88 (iii) 0.47; 0.37 (iv) 0.308; 0.229 (v) 8.7; 5.7
6.14 Deduce an expression for the specific output (kJ/kg of working fluid)
of a simple constant-pressure gas turbine in terms of temperatures
at the beginning of compression and at the beginning of expansion,
the isentropic efficiencies of the compressor and turbine, the pressure
ratio and the isentropic index. The specific heat at constant-pressure
may be assumed constant, and the weight of the fuel added may be
neglected.
Hence determine the pressure ratio at which the specific output is
maximum for the following operating conditions. Temperature at
compressor inlet is 15◦ C, temperature at turbine inlet is 630◦C and
the isentropic efficiency for compressor and turbine is 0.85 and 0.90
respectively.
Neglect the effect of fuel flow and assume constant Cp of air. Determine in kg/kW-h the air flow and the specific fuel consumption, if
1 kg/s of fuel is flowing with the calorific value of 42000 kJ/kg.
⎡
⎤
⎢
Ans: (i) WN = Cp ηT T03 ⎢
⎣1 −
1
γ−1
rp γ
⎥
⎥−
⎦
Cp
ηC T01
γ−1
rp γ − 1
(ii) 4.62 (iii) 35.1 kg/kW h (iv) 0.362 kg/kW h
6.15 In a gas turbine plant, comprising a single stage compressor, combustion chamber and turbine, the compressor takes in air at 15◦ C
and compresses it to 4 times the initial pressure with an isentropic
efficiency of 85 per cent. The fuel-air ratio is 0.0125 and the calorific
value of the fuel is 42000 kJ/kg. If the isentropic efficiency of the
turbine is 82 per cent, find the amount of air intake for a power output of 260 kW and also the overall thermal efficiency. Take the mass
of the fuel into account. The turbine inlet temperature is 1000 K.
Ans: (i) 2.33 kg/s (ii) 21.3%
6.16 At design speed the following data apply to a gas turbine set employing a separate power turbine, heat exchanger, reheater and intercooler
between two-stage compression.
Efficiency of compression in each stage : 80%
Isentropic efficiency of compressor turbine : 87%
Isentropic efficiency of power turbine : 80%
Transmission efficiency : 99%
Pressure ratio in each stage of compression : 2:1
Pressure loss in intercooler : 0.07 bar
Temperature after intercooling : 300 K
Thermal ratio of heat exchanger : 0.75
Pressure loss in combustion chamber : 0.15 bar
Combustion efficiency of reheater : 98%
Practical Cycles and their Analysis
Maximum cycle temperature
Temperature after reheating
Air mass flow
Ambient air temperature
Ambient air pressure
:
:
:
:
:
209
1000 K
1000 K
25 kg/s
15◦ C
1 bar
Take the calorific value of fuel as 42 MJ/kg and pressure loss in each
side of heat exchanger as 0.1 bar.
Find the net power output, overall thermal efficiency, specific fuel
consumption. Neglect the kinetic energy of the gases leaving the
system.
Ans: (i) 2.457 MW (ii) 23.72% (iii) 0.3735 kg/kW h
6.17 The following diagram refers to an automotive gas turbine and the
data refer to this turbine.
Rated power (kW) : 220
Pressure ratio : 16:1
Turbine inlet temperature,◦ C : 925
Reheat turbine inlet temperature,◦ C : 925
bsf c, kg/kW h : 0.3465
Exhaust temperature,◦ C : 316
Regenerator efficiency, % : 75
Burner efficiency, % : 96
Compressor efficiency, % : 80
Turbine efficiency, % : 85
Ambient temperature,◦ C : 27
The schematic diagram is as shown below.
07
C2
03
T2
04
06
CC
Air cooler
Air
08
01
02
011
010
C
CC
05
012
T1
T4
T3
09
Fig. 6.46
Calculate the
(i) outlet temperature of all turbines,
(ii) net work output,
(iii) thermal efficiency,
Transmissio
210
Gas Turbines
(iv) work ratio, and
(v) mass flow rate.
Ans: (i) T07 = 1038 K; T09 = 1057.5 K; T010 = 953.1 K;
T011 = 793.1 K; (ii) 280.9 kJ/kg (iii) 36.6%
(iv) 0.434 (v) 0.783 kg/s
6.18 The maximum and minimum temperatures in a simple gas turbine
plant working on the Joule cycle are 1000 K and 288 K respectively.
The pressure ratio is 6, and the isentropic efficiencies of the compressor and turbine are 85 and 90 per cent respectively. Calculate work
output per kg and the efficiency of the plant.
Ans: (i) 143.041 kJ/kg (ii) 25.7%
6.19 In a gas turbine plant the air enters the compressor at 1 bar and
27◦ C. The pressure leaving the compressor is 5 bar and the maximum
temperature in the cycle is 850◦C. Compare the compressor work,
turbine work, work ratio and cycle efficiency for the following two
cases:
(i) the cycle is ideal
(ii) the cycle is actual with compressor efficiency of 80%, turbine efficiency of 85% and the pressure loss in the combustion chamber
is 0.15 bar.
If in the above cycle the air expands in the turbine to such a pressure
that the turbine work is just equal to compressor work, and further
expansion of air upto atmospheric pressure takes place in a nozzle;
calculate the velocity of air leaving the nozzle. Assume a nozzle efficiency of 90%.
If an ideal regenerator is incorporated in the ideal cycle, what is the
thermal efficiency of the cycle?
Ans: For an ideal cycle:
(i) 176.24 kJ/kg (ii) 416.35 kJ/kg (iii) 0.5767 (iv) 36.89%
For an actual cycle:
(i) 220.3 kJ/kg (ii) 354.76 kJ/kg (iii) 0.379 (iv) 19.41%
Ve = 543.52 m/s; ηth = 57.67%
Multiple Choice Questions (choose the most appropriate answer)
1. The important assumptions in practical cycle analysis are
(a) the compressor and expansion process are irreversible adiabatic
(b) complete heat exchange is possible in a heat exchanger
(c) the pressure losses are negligibly small
(d) none of the above
Practical Cycles and their Analysis
211
2. The isentropic efficiency of the modern compressor is
(a) 65%
(b) 75%
(c) 85%
(d) 90%
3. Isentropic efficiency of a compressor is defined as the ratio of
Isentropic work input
Stoichiometric work input
Isentropic work input
(b)
Actual work input
Actual work input
(c)
Isentropic work input
Stoichiometric work input
(d)
Isentropic work input
(a)
4. The isentropic efficiency of a turbine is defined as the ratio of
Actual work output
Isentropic work output
Isentropic work output
(b)
Actual work output
Actual work input
(c)
Isentropic work input
Isentropic work input
(d)
Actual work input
(a)
5. The work ratio is defined as
(a)
WT −WC
WT
(b) 1 −
(c)
WC
WT
WN
WT
(d) all of the above
6. The effectiveness of the modern heat exchanger is
(a)
= 55%
(b)
= 65%
(c)
= 75%
(d)
= 90%
7. The assumption of mass flow rate to be the same in spite of the fuel
addition is due to
(a) the gas turbine operates at a very lean air-fuel ratio of the order
of 60 to 100
212
Gas Turbines
(b) fuel flow rate compared to air flow rate is very small
(c) the bleed air from the compressor to cool the turbine blades
which compensates for the fuel addition
(d) none of the above
8. The mechanical losses are accounted by ηmech on
(a) compressor side
(b) combustion chamber side
(c) heat exchanger side
(d) turbine side
9. Cycle efficiency can be defined as
(a)
(b)
(c)
WT −WC
f ×CV
WN
f ×CV
3600
sf c×CV
(d) any of the above equation
10. For a constant value of ηpc (say 85%) ηc
(a) increases with increase in pressure ratio
(b) decreases with increase in pressure ratio
(c) remains constant with increase in pressure ratio
(d) none of the above
Ans:
1. – (d)
6. – (c)
2. – (c)
7. – (c)
3. – (b)
8. – (a)
4. – (a)
9. – (d)
5. – (d)
10. – (b)
7
JET PROPULSION
CYCLES AND THEIR
ANALYSIS
INTRODUCTION
The last two chapters were dealt with the ideal and practical shaft power
cycles respectively. In this chapter, the analysis of jet propulsion cycles will
be taken up. Gas turbine cycles for jet propulsion differ from shaft power
cycles because of the fact that the useful power output for jet propulsion is
produced, wholly or partially, as a result of expansion of gas in a propelling
nozzle; wholly in turbojet engines and partially in turboprop engines. A
second distinguishing feature is the need to consider the effect of forward
speed and altitude on the performance of propulsion engines. The beneficial
effect of those parameters, together with an inherently high power/weight
ratio have enabled the gas turbine to replace the reciprocating engine for
aircraft propulsion except for the very low power levels.
The principle of jet propulsion is obtained from the application of Newton’s laws of motion.∗ We know that when a fluid is to be accelerated,
a force is required to produce this acceleration in the fluid. At the same
time, there is an equal and opposite reaction force of the fluid on the engine
which is known as the thrust. Hence, it may be stated that the working
of jet propulsion is based on the reaction principle. Thus all devices that
move through fluids must follow this basic principle.
In principle, any fluid can be used to achieve the jet propulsion. Thus
water, steam or combustion gases can be used to propel a body in a fluid.
But there are limitations in the choice of the fluid when the bodies are to
be propelled in the atmosphere. Experience shows that only two types of
fluids are particularly suitable for jet propulsion.
∗ Newton’s Second Law: Rate of change of momentum in any direction is proportional
to the force acting in that direction.
Newton’s Third Law : For every action there is equal and opposite reaction.
214
Gas Turbines
(i) A heated and compressed atmospheric air – admixed with the products of combustion produced by burning fuel in that air can be used
for jet propulsion. The thermochemical energy of the fuel is utilized
for increasing the temperature of the air to the desired value. The jet
of this character is called a thermal jet and the jet propulsion engine
using atmospheric air is called air breathing engines.
(ii) Another class of jet-propulsion engines use a jet of gas produced by the
chemical reactions of fuel and oxidizer. Each of them is carried with
the system itself. The fuel-oxidant mixture is called the propellant.
No atmospheric air is used for the formation of the jet. But the
oxidant in the propellant is used for generating the thermal jet. A
jet produced in this way is known as rocket jet and the equipment
wherein the chemical reaction takes place is called a rocket motor.
The complete unit including the propellant is called a rocket engine.
From the above discussion it is clear that jet-propulsion engines may be
classified broadly into two groups.
(i) air breathing engines and
(ii) rocket engines
In this chapter, we will discuss the air breathing engines. Air breathing
engines can be further classified as follows:
(i) reciprocating or propeller engines
(ii) gas turbine engines.
7.1
RECIPROCATING OR PROPELLER ENGINES
In early days, the source of power for an air breathing engine was a reciprocating internal combustion engine which used to drive a propeller connected
to it. The propeller displaces rearwards, a large mass of air, accelerating
it in the process. Due to this acceleration of the fluid a propulsive force is
produced which drives the aircraft.
The extensive use of aircraft for military purposes led to a very rapid
development of reciprocating internal combustion engines during the two
world wars and it is now a highly developed piece of equipment as compared to its industrial counterpart. Further development resulted in the
use of highly supercharged and turbocharged engines. All these engines
were gasoline engines. Diesel engines in spite of their good fuel economy
and reliability was not used due to higher weight.
For small aircraft flying at velocities less than about 500 km/h reciprocating engine is in an enviable position due to its excellent fuel economy and
good take-off characteristics. However, due to comparatively large drop in
power with altitude operation and the need of using high octane fuels, along
Jet Propulsion Cycles and Their Analysis
215
with the difficult cooling and lubrication problems, high weight/power ratio, and larger frontal area these are being replaced by turbojets in higher
speed ranges.
Rapid developments in design of turbojet and turboprop engines have
started exploding the best fuel economy myth of the reciprocating engines.
They are nearing the specific fuel consumption value of reciprocating engines. Still the reciprocating engine is likely to be used for small aircraft
needing only a few hundred kilowatts both because of good take off characteristics and due to difficulties in the development of smaller gas turbine
engines giving reasonable fuel economy and cost.
However, the use of reciprocating engines is continuously on the decline
because its development has reached almost a saturation stage as far as
maximum power is concerned. The demand of present day aircraft, in
terms of high flight speeds, long distance travels and high load carrying
capacities, is soaring to new heights.
A power output more than 4000 kW is difficult to obtain without modifications in the present reciprocating engine power plant. The output can be
increased by increasing the cylinder sizes, installing large number of cylinders or by running the engine at higher speeds. Unfortunately all these
methods of raising the output of the engine increase the engine size, frontal
area of the aircraft, complexity and cost of the plant. The drag of the plane
will also increase to critical values with increase in engine size. Hence, for
aircraft propulsion, gas turbine engines are the ideal power plant.
7.2
GAS TURBINE ENGINES
World War II was the turning point for the development of gas turbine
technology. All modern aircrafts are fitted with gas turbines. Gas turbine
engines can be classified into
(i) ramjet engines,
(ii) pulse jet engines,
(iii) turbojet engines,
(iv) turboprop engines, and
(v) turbofan engines.
Taken in the above order they provide propulsive jets of increasing mass
flow and decreasing jet velocity. Therefore, in that order, it will be seen that
ramjet can be used for highest cruising speed whereas the turboprop engine
will be useful for the lower cruising speed at low altitudes. In practice, the
choice of the power plant will depend on the required cruising speed, desired
range of the aircraft and maximum rate of climb.
The details of various gas turbine engines mentioned above are discussed
under two categories: (i) pilotless operation, and (ii) piloted operation. The
ramjet and pulse jet engines come under the category of pilotless operation
whereas the turboprop and turbojet engines are used for piloted operation.
216
7.3
Gas Turbines
THE RAMJET ENGINE
The fact of obtaining very high pressure ratios of about 8 to 10 by ram compression has made it possible to design a jet engine without a mechanical
compressor. A deceleration of the air from Mach number 3 at diffuser inlet
to Mach number 0.3 in combustion chamber would cause pressure ratio of
more than 30. Due to shock and other losses inevitable at such velocities
all of this pressure rise is not available; still whatever we get is more than
sufficient for raising the air pressure to the required combustion pressure.
This principle of ram pressure rise is used in the ramjet engines. The ram
pressure rise can be achieved in diffusers.
It may be noted that the simplest types of air breathing engine is the
ramjet engine and a simplified sketch of the engine is illustrated in Fig. 7.1.
The engine consists of
(i) supersonic diffuser (1–2),
(ii) subsonic diffuser section (2–3),
(iii) combustion chamber (3–4), and
(iv) discharge nozzle section (4–5).
Air inlet and
supersonic diffuser Subsonic
diffuser Combustor Nozzle
F
A
ma
B
m +m
A
P
c
A
0 1
Oblique shock
2
3
mf
Normal shock
Central body housing accessories
4
c
B
5
Jet Propulsion Cycles and Their Analysis
217
the fuel is injected by suitable injectors and mixed with the unburnt air.
The air is heated to a temperature of the order of 1500 – 2000 K by the
continuous combustion of fuel. The fresh supply of air to the diffuser builds
up pressure at the diffuser end so that these gases cannot expand towards
the diffuser. Instead, the gases are made to expand in the combustion
chamber towards the tail pipe. Further, they are allowed to expand in the
exhaust nozzle section. The products will leave the engine with a speed
exceeding that of the entering air. Because of the rate of increase in the
momentum of the working fluid, a thrust, F , is developed in the direction
of flight.
Normally, the air enters the engine with a supersonic speed which must
be reduced to a subsonic value. This is necessary to prevent the blow out of
the flame in the combustion chamber. The velocity must be small enough to
make it possible to add the required quantity of fuel for stable combustion.
Both theory and experiment indicate that the speed of the air entering the
combustion chamber should not be higher than that corresponding to a
local Mach number of 0.2 approximately.
The cycle pressure ratio of a ramjet engine depends upon its flight velocity. The higher the flight velocity the larger is the ram pressure, and
consequently larger will be the thrust. This is true until a condition is
reached where the discharge nozzle becomes chocked. Thereafter, the nozzle operates with a constant Mach number of 1 at its throat. Therefore, a
ramjet having fixed geometry is designed for a specific Mach number and
altitude, and at the design point, will give the best performance.
Since the ramjet engine cannot operate under static conditions, as there
will be no pressure rise in the diffuser, it is not self-propelling at zero flight
velocity. To initiate its operation, the ramjet must be either launched from
an airplane in flight or be given an initial velocity by some auxiliary means,
such as launching rockets. Since the ramjet is an air breathing engine, its
maximum altitude is limited. Its field of operations is inherently in speed
ranges above those of the other air breathing engines. However, it has a
limited use in the high subsonic speed range. Its best performance capabilities, however, are in supersonic speed range of Mach numbers between
2 and 5. The upper speed is limited by the problem of cooling of the outer
skin of the engine body at the high flight Mach numbers.
7.3.1
Thermodynamic Cycle
The ramjet works on the thermodynamic cycle, viz., the Brayton cycle.
Figure 7.2 shows this cycle on a T -s diagram. In an ideal cycle process,
1 → 2 is isentropic ram compression in both the diffusers and process
3 → 4 is the isentropic expansion in the nozzle. In actual practice there
will be losses due to shock, friction and mixing at the diffuser and losses in
the nozzle. The actual compression and expansion is shown by the processes
1 → 2 and 3 → 4 respectively. Combustion is represented by the process
2 → 3. Since the expansion and compression is assumed isentropic, in an
ideal ramjet cycle the stagnation pressure must remain constant and the
pressure ratio of ram compression and nozzle expansion ratio must be same.
218
Gas Turbines
3
Nozzle
4
T
4’
2
2’
Ram
1
s
Fig. 7.2 Ramjet cycle on T -s diagram
p02
p1
=
1+
γ−1 2
Mi
2
p03
p4
=
1+
γ−1 2
Mj
2
and
γ
γ−1
γ
γ−1
where Mi and Mj are inlet and exhaust Mach number. We have,
p02
p03
=
p1
p4
which gives
or
Mi
=
Mj
c
√ i
γRT1
=
c
√ j
γRT4
cj
=
ci
or
T4
T1
(7.1)
Knowing cj and ci , thrust can be calculated.
7.3.2
Performance
Figure 7.3 shows the net thrust and thrust specific fuel consumption per
unit weight of a ramjet engine. It can be seen that at low flight speeds
Jet Propulsion Cycles and Their Analysis
219
l
leve
Sea
0m
700
00 m
130
00 m
200
00 m
300
Mach number
Net specific fuel consumption
(kg/N thrust)
Net thrust (kN)
the thrust fuel consumption is quite high, while at high speeds it reduces
considerably. This is because at low speeds the compression is poor.
As can be seen from the Fig.7.3 the thrust is maximum at sea level
and decreases as the altitude increases, due to reduction in air density
with altitude. However, the velocity of the ramjet increases due to lower
resistance at high altitude, thereby, improving specific fuel consumption. If
the increase in thrust due to increased jet velocity is more than the decrease
in thrust due to reduced density; the specific fuel consumption will improve.
Otherwise, the specific fuel consumption will be independent of altitude.
The thrust, at a given altitude, rises rapidly with an increase in the
flight speed due to increased pressure ratio but at too high Mach numbers
the losses in inlet diffuser reduce the thrust. Ramjets have highest thrust
per unit weight amongst air breathing engines and is only next to rockets
in this respect.
Though a ramjet can operate at subsonic velocities just below the sonic
velocity, it is most efficient at high velocities of about 2400–6000 km/h and
at very high altitudes.
Max. thrust
Cruise thrust
Mach number
Fig. 7.3 Performance of a ramjet engine
7.3.3
Advantages, Disadvantages and Characteristics
In this section, the various advantages and disadvantages of ramjet are
enumerated.
Advantages of ramjet
(i) Ramjet is very simple and does not have any moving part. It is very
cheap to produce and requires almost no maintenance.
(ii) Due to the fact that a turbine is not used to drive the mechanical compressor, the maximum temperature which can be allowed in ramjet
is very high, about 2000◦ C as compared to about 900◦ C in turbojets.
This allows a greater thrust to be obtained by burning fuel at air-fuel
ratio of about 13:1, which gives higher temperatures.
220
Gas Turbines
(iii) The specific fuel consumption is better than other gas turbine power
plants at high speed and high altitudes.
(iv) Theoretically there seems to be no upper limit to the flight speed of
the ramjet.
Disadvantages of ramjet
(i) Since the compression of air is obtained by virtue of its speed relative
to the engine, the take-off thrust is zero and it is not possible to start
a ramjet without an external launching device.
(ii) The engine heavily relies on the diffuser and it is very difficult to
design a diffuser which will give good pressure recovery over a wide
range of speeds.
(iii) Due to high air speed, the combustion chamber requires flame holder
to stabilize the combustion.
(iv) At very high temperatures of about 2000◦C dissociation of products
of combustion occurs which will reduce the efficiency of the plant if
not recovered in nozzle during expansion.
7.3.4
Basic Characteristics and Applications
The basic characteristics of the ramjet engine can be summarized as follows:
(i) It is a simple engine and should be adaptable for mass production at
relatively low cost.
(ii) It is independent of fuel technology and a wide range of liquid, and
even solid fuels can be used.
(iii) Its fuel consumption is comparatively very large for its application in
aircraft propulsion or in missiles at low and moderate speeds.
(iv) Its fuel consumption decreases with flight speed and approaches reasonable values when the flight Mach number is between 2 and 5 and
therefore, it is suitable for propelling supersonic missiles.
Due to its high thrust at high operational speed, it is widely used in
high-speed military aircrafts and missiles. Subsonic ramjets are used in
target weapons, in conjunction with turbojets or rockets for getting the
starting torque.
7.4
THE PULSE JET ENGINE
Pulse jet engine (Fig. 7.4) is very similar to ramjet engine in construction
except that in addition to the diffuser at intake, combustion chamber and
exhaust nozzle, it has mechanically operated flapper valve grids which can
allow or stop air flow in the combustion chamber. Thus pulse jet is an
Jet Propulsion Cycles and Their Analysis
221
intermittent flow, compressorless type of device with minimum number of
moving parts. Pulse jet was the power plant of German V-1 bomb popularly
known as ‘Buzz Bomb’ first used in World War II in 1944.
Inlet diffuser
Valve grid
Combustion chamber
Tail piece
c
Fuel
Spark plug
Fig. 7.4 The pulse jet engine
The basic features of the pulse jet engine are illustrated in Fig. 7.4. It
consists essentially of the following parts:
(i) a diffuser,
(ii) a valve grid which contains springs that close on their own spring
pressure,
(iii) a combustion chamber,
(iv) a spark plug, and
(v) a tail pipe or discharge nozzle.
The theoretical and actual p-V diagrams of the pulse jet engine are
shown in Fig. 7.5 and Fig. 7.6 respectively.
The operation of the pulse jet is as follows: During starting compressed
air is forced into the inlet which opens the spring loaded flapper valve grid;
the air enters combustion chamber into which fuel is injected and burnt
with the help of a spark plug. Combustion occurs with a sudden explosion
process 2–3 in Fig. 7.5, i.e., the combustion is at constant-volume instead
of at constant-pressure as in other propulsive devices. The pulse jet cycle
is more near to Otto cycle. Ram action can also be used to increase the
pressure of the cycle (Fig. 7.5).
The function of the diffuser is to convert the kinetic energy of the entering air into static pressure rise by slowing down the air velocity. When
a certain pressure difference builds up across the valve grid, the valves will
open. This makes the fresh air to enter the combustion chamber, where
fuel is mixed with the air and combustion starts. To start the combustion
initially the spark plug is used. Once the combustion starts it proceeds
at constant-volume. Thereby, there is a rapid increase in pressure, which
causes the valve to close rapidly. The products of combustion surges towards the nozzle. They expand in the nozzle and escape into the atmosphere
222
Gas Turbines
3
3’
p
Isentropic
2
4
1
V
Fig. 7.5 Theoretical pulse jet cycle on p-V diagram
3
p
1
Expansion and
acceleration of gases
2
4
Pulsating gas column
leaving the system
V
Fig. 7.6 Actual pulse jet cycle on p-V diagram
with a higher velocity so that the exit velocity is much higher than the inlet
velocity.
Thus, the rate of momentum of the working fluid is changed so as to
cause a propulsive thrust. Since, the combustion process causes the pressure
to increase, the engine can operate even at static conditions once it gets
started. When the combustion products accelerate from the chamber, they
leave a slight vacuum in the combustion chamber. This, in turn, produces
sufficient pressure drop across the valve grid, allowing the valves to open
again.
A new charge of air enters the combustion chamber which is mixed with
fuel that flows continuously. The fresh fuel-air mixture is ignited by the
charge leaving and/or by residual charge. New charge need not be ignited
with a spark plug again. Proper design allows the duct to fire at a given
pulse rate when the fuel flows continuously. The frequency of pulsation
is determined by the duct shape and working temperature and may be as
high as 500 cycles per second in very small units. The thrust of the pulse
jet engine is proportional to the average mass flow rate of gases through
the engine multiplied by its increase in velocity.
Like ramjet engines, the maximum operating altitude of the pulsejet
is also limited by air density consideration. Unlike the ramjet, the pulse
jet engine develops thrust at zero speed. A high initial launching velocity,
Jet Propulsion Cycles and Their Analysis
223
however, improves its performance. The thrust of the engine, of course,
decreases with altitude and does not continue to increase with increasing
flight speeds up to supersonic range as is true of ramjet. The maximum
flight speed of the pulse jet engine is limited by aerodynamic consideration
to below 800 km/h. The pulse jet engine is simple and cheap for subsonic
flight and well adapted to pilotless aircraft. The use of the pulse jet engine is
restricted to pilotless aircraft due to its severe vibration and high intensity
noise.
The pulse jet has low thermal efficiency and limited speed range. In
early designs the efficiency obtained was about 2 to 3 percent with a total
flight life of 30 to 60 minutes. The maximum operating speed of the pulse
jet is seriously limited by two factors:
(i) It is not possible to design a good diffuser at high speeds.
(ii) The flapper valves, the only mechanical part in the pulse jet, also have
certain natural frequency and if it coincides with the cycle frequency
resonance occurs and the valve may remain open and no compression
will take place.
Also, as the speed increases it is difficult for the air to flow back. This
reduces the total compression pressure as well as the mass flow of air which
results in inefficient combustion and lower thrust. The reduction in thrust
and efficiency is quite sharp as the speed increases. At subsonic speeds it
might not operate as the speed is not sufficient to raise the air pressure to
the required combustion pressure.
7.4.1
Advantages and Disadvantages
Advantages of pulse jet
(i) This is a very simple device next to ramjet and is light in weight. It
requires very small and occasional maintenance.
(ii) Unlike ramjet, it has static thrust because of the compressed air starting ; thus it does not need a device for initial propulsion. The static
thrust is even more than the cruise thrust.
(iii) It can run on almost any types of liquid fuels without much effect
on the performance. It can also operate on gaseous fuel with a little
modifications.
(iv) Pulsejet engine is relatively cheap.
Disadvantages of pulse jet
(i) The biggest disadvantage is very short life of flapper valves and high
rates of fuel consumption. The specific fuel consumption is as high
as that of ramjet.
224
Gas Turbines
(ii) The speed of the pulse jet is limited to a very narrow range of about
650–800 km/h because of the limitations in the aerodynamic design
of an efficient diffuser suitable for a wide speed range.
(iii) The operational range of the pulse jet is also limited in altitude range.
(iv) The high degree of vibrations due to intermittent nature of the cycle
and the buzzing noise has made it suitable for pilotless crafts only.
(v) It has lower propulsive efficiency than turbojet engines.
7.4.2
Applications
Pulse jet is highly suited for bombers like the German V-1 and it has also
been used in some helicopters, target aircrafts missiles, etc.
7.5
THE TURBOPROP ENGINE
It is a known fact that an higher thrust per unit mass flow of fuel can
be obtained by increasing the mass flow of air which results in better fuel
economy. This fact is utilized in a turboprop engine which is an intermediate between a pure jet engine and a propeller engine. Turboprop engine
attempts to increase the air flow by using a propeller driven by the turbine
in addition to a small thrust produced by the exhaust nozzle.
The details of this engine is illustrated in Fig. 7.7. In this engine the
turbine is designed so as to develop enough shaft power for driving a propeller which provides most of the propulsive thrust. It may be noted that
the thrust produced due to jet action is quite small.
Diffuser Compressor
Fuel Turbine Nozzle
G
Reduction gear
Propeller
Shaft
Fuel
Combustion chamber
Fig. 7.7 The turboprop engine
The engine consists of the following components:
(i) a diffuser,
(ii) a compressor,
(iii) a combustion chamber,
(iv) a turbine,
Jet Propulsion Cycles and Their Analysis
225
(v) an exhaust nozzle,
(vi) a reduction gear, and
(vii) a propeller.
In the turboprop engine, the turbine extracts much more power because
the turbine is to provide power for both the compressor and the propeller.
When all of this energy is extracted from the high temperature gases, there
is still a little energy left for producing jet thrust. Thus the turboprop
engine derives most of its propulsive thrust from the propeller and derives
only a small portion (10 to 20% depending upon the flight velocity) from
the exhaust nozzle. Since the shaft speed of gas turbine engine is very much
higher than that of a propeller, a reduction gear must be placed between the
turbine shaft and the propeller to enable the propeller to operate efficiently.
Still, as flight speed increases, the ratio of nozzle power to propeller power
for maximum thrust tends to increase. It may be noted that the propulsive
thrust developed is due to the following:
(i) the propeller increases the air momentum, and
(ii) the overall engine – from diffuser to nozzle – provides an internal
momentum increase.
The sum of these two thrusts is the total thrust developed by the engine.
7.5.1
Thermodynamic Cycle
The thermodynamic cycle of turboprop is similar to that of a pure jet engine
except that now more energy is used in the turbine as seen from the T -s
diagram of Fig. 7.8. About 80 to 90 per cent of the total energy is used
in the turbine and only about 10 to 20 per cent is used in exhaust nozzle.
The propeller produces its own thrust and thus the turboprop engine is
essentially a two fluid stream engine.
The optimum power ratio for turboprop engine can be determined as
follows: Let the total expansion be divided into parts such that
Total thrust = Nozzle thrust + Propeller thrust
= ṁa (cj − ci ) + ṁa ci
where
Δhnoz
ΔhT
ηtr
ηT
ηnoz
ṁa
=
=
=
=
=
=
= ṁa
2Δhnoz ηnoz − ci + ṁa
= ṁa
2Δhnoz ηnoz − ci +
ηT ΔhT ηtr
ηT ΔhT ηtr
enthalpy drop in the nozzle
enthalpy drop in the turbine
transmission efficiency of the propeller and gears
turbine efficiency
nozzle efficiency
mass flow rate in turbine
(7.2)
226
Gas Turbines
3
4
T
4’
2
2’
1
s
Fig. 7.8 T -s diagram for a turboprop engine
By assigning suitable values to ηT , ηnoz and ηtr this equation can be
optimized for maximum thrust.
7.5.2
Performance
Turboprop engines combine in them the high take-off thrust and good propeller efficiency of the propeller engines at speeds lower than 800 km/h and
the small weight, lower frontal area, and reduced vibration and noise of the
pure jet engine. Its operational range is between that of propeller engines
and turbojets though it can operate in any speed upto 800 km/h.
The power developed by the turboprop remains almost same at high
altitudes and high speeds as that under sea-level and take-off conditions
because as speed increases ram effect also increases. The specific fuel consumption increases with increase in speed and altitude. The thrust developed is high at take-off and reduces at increased speed.
7.5.3
Advantages and Disadvantages
Advantages of Turboprop
(i) Turboprop engines have a higher thrust at take-off and better fuel
economy.
(ii) The frontal area is less than propeller engines so that the drag is
reduced.
(iii) The turboprop can operate economically over a wide range of speeds
ranging from low speeds where pure jet engine is uneconomical to
high speeds of about 800 km/h where the propeller engine efficiency
is low.
Jet Propulsion Cycles and Their Analysis
227
(iv) It is easy to maintain and has lower vibrations and noise.
(v) The power output is not limited as in the case of propeller engines.
(vi) The multishaft arrangement allows a great flexibility of operation over
a wide range of speeds.
Disadvantages of Turboprop
(i) The main disadvantage is that at high speeds, due to shocks and flow
separation, the propeller efficiency decreases rapidly, thereby, putting
up a maximum speed limit on the engine.
(ii) It requires a reduction gear which increases the cost and also consumes
certain amount of energy developed by the turbine in addition to
requiring more space.
7.5.4
Applications
(i) The turboprop engine is widely used in commercial and military aircraft due to its high flexibility of operation and good fuel economy.
It is likely to eliminate propeller engines from moderate power and
speed aircraft.
7.6
THE TURBOJET ENGINE
The two pilotless air breathing engines described, viz., ramjet and pulsejet
are simple in construction. However, their application is limited and, to
date, they have not been used very extensively. The most common type
of air breathing engine apart from turboprop is the turbojet engine. The
important features are shown in Fig. 7.9.
Diffuser Compressor
0
1
Fuel
Turbine
2
Nozzle
4
3
Fuel
Shaft
Combustion chamber
Fig. 7.9 The turbojet engine
This engine consists of the following components:
(i) a diffuser,
(ii) a mechanical compressor,
(iii) a combustion chamber,
5
228
Gas Turbines
(iv) a mechanical turbine, and
(v) an exhaust nozzle.
The function of the diffuser is to convert the kinetic energy of the entering air into a static pressure rise which is achieved by the ram effect. After
this air enters the mechanical compressor.
The compressor used in a turbojet can be either centrifugal type or
axial flow type. The use of a particular type of compressor gives the turbojet typical characteristics. The centrifugal compressor produces a high
pressure ratio of about 4:1 to 5:1 in a single stage and usually a doublesided rotor is used to reduce the engine diameter. The turbojet using a
centrifugal compressor has a short and sturdy appearance. The advantages
of centrifugal compressor are high durability, ease of manufacture and low
cost, and good operation under adverse circumstances such as icing and
when sand and small foreign particles are inhaled in inlet duct.
The axial flow compressor is more efficient than the centrifugal type
and gives the turbojet a long, slim, streamlined appearance. The engine
diameter is reduced which results in low aircraft drag. A multistage axial
flow compressor can develop a pressure ratio as high as 6:1 or more. The
air handled by it is more than that handled by a centrifugal compressor of
the same diameter.
The demand for increased power has led to the use of split or twospool axial flow compressor. A very high pressure ratio of about 9:1 to
13:1 is obtained by using a high pressure and a low pressure rotor driven
by separate shafts. The use of high pressure ratio gives very good specific
fuel consumption (0.75 kg/kg thrust per hour) and use of two rotors allows
greater efficiency because firstly, the high pressure rotor can be governed
for speed and secondly, the low pressure rotor can be allowed to run at a
speed giving maximum efficiency.
The turbojets having centrifugal compressor have about 20 per cent
weight advantage over the axial flow turbojets. Thrust per unit weight is
more for the first type while thrust per unit diameter is more the second
type. The axial flow turbojets have about 6 to 8 per cent less specific fuel
consumption. After the compressor air enters to the combustion chamber,
the fuel nozzles feed fuel continuously and continuous combustion takes
place at constant-pressure. The high pressure, high temperature gases then
enter the turbine, where they expand to provide enough power output from
the turbine.
The turbine is directly connected to the compressor and all the power
developed by the turbine is absorbed by the compressor and the auxiliaries.
The main function of the turbine is to provide power, to drive the compressor. After the gases leave the turbine they expand further in the exhaust
nozzle, and are ejected into the atmosphere with a velocity greater than
the flight velocity thereby producing thrust for propulsion.
Current turbojet engines operate with compressor pressure ratios between 6 and 16, and with turbine inlet temperatures of the order of 1200 K.
Jet Propulsion Cycles and Their Analysis
229
The corresponding speed of the exhaust jet when propelling an aircraft at
900 km per hour (250 m/s) is of the order of 500 m/s.
Like the ramjet engine, the turbojet engine is a continuous flow engine.
Here a compressor run by a turbine is used to provide additional pressure
rise which is not available in a ramjet engine. Since this engine has a
separate mechanical compressor, it is capable of operating even under static
conditions. However, increase in flight velocity improves its performance
because of the benefit of ram pressure achieved by the diffuser.
Because of turbine material limitations, only a limited amount of fuel
can be burnt in the combustion chamber. The exhaust products downstream of the turbine still contain a considerable amount of excess oxygen.
Additional thrust augmentation can be achieved from the turbojet engine
by providing an afterburner in which additional fuel can be burnt. Properly
designed afterburner can greatly increase the temperature and hence the
velocity of exhaust gases – providing thrust augmentation.
Of the four air-breathing engines, discussed so far only the turbojet
and turboprop engines have been applied as the propulsion device for piloted aircraft. Turbojet engine is eminently suited for propelling aircraft
at speeds above 800 km/h. As the flight speed is increased, the ram pressure increases rapidly and at supersonic speeds, (Mach number = 3) the
characteristics of turbojet engine tend to merge with those of the ramjet
engine.
7.6.1
Thermodynamic Cycle
Figures 7.10 and 7.11 show the basic thermodynamic cycle of a turbojet
engine of p-V and T -s diagrams. This is Joule or Brayton cycle. In the
analysis of turbojet cycle following assumptions are made:
(i) There is no loss of pressure in the combustion chamber.
(ii) The specific heat is constant.
(iii) Power developed by the turbine is just sufficient to drive the compressor.
At the inlet to the diffuser air enters with a velocity equal to the forward
speed of the aircraft. In diffuser air velocity is decreased and pressure is
increased. In the ideal case the pressure will rise such that the velocity at
the exit of the diffuser is zero. However, in actual practice the air will have
a velocity of about 60–90 m/s at diffuser exit. If ηd is the efficiency of the
diffuser then the total pressure at the end of diffusion process is given by
p01
p0
=
γ−1 2
M
1 + ηd
2
γ
γ−1
From the diffuser air goes into a compressor. If ηC is the compressor efficiency and pp02
the pressure ratio, we get
01
Gas Turbines
Combustion
2
3
Pressure loss in
combustion chamber
ine
r
esso
rb
Tu
pr
Com
230
1
4
No
Diffuser
zzl
e
5
0
V
Fig. 7.10 p-V diagram of a turbojet engine
03
n
bustio
02’ 02
T
Com
04
Compressor
p
Turbine
Nozzle
04’
01’
01
5
5’
Diffuser
0
s
Fig. 7.11 T -s diagram of a turbojet engine
Jet Propulsion Cycles and Their Analysis
ηC
=
h02 − h01
h02 − h01
h02 − h01
=
1
(h02 − h01 )
ηC
=
Cp T01
ηC
or
=
T02
−1
T01
231
Cp
(T02 − T01 )
ηC
=
Cp T01
ηC
p02
p01
γ−1
γ
−1
The heat input per kg of air in the combustion chamber is given by
q
=
h03 − h02
We assume that there is no pressure loss in the combustion chamber
so that the pressure remains constant and full pressure is available for
expansion in the turbine and nozzle. Since we have assumed that turbine
and compressor work are same.
h02 − h01
=
h03 − h04
If ηT is the turbine efficiency and
h03 − h04
=
p04
p03
p04
p03
ηT Cp T03 1 −
is the turbine pressure ratio,
γ−1
γ
By equating the turbine and compressor work, we get
Cp T01
ηC
p02
p01
γ−1
γ
− 1 = ηT Cp T03 1 −
p04
p03
γ−1
γ
(7.3)
From Eq. 7.3, knowing the ram compression and the compressor pressure
ratio, the required turbine pressure ratio to produce a power equal to that
absorbed by the compressor can be obtained and from the turbine pressure
ratio the nozzle pressure can be obtained.
If ηnoz is the nozzle efficiency then,
ηnoz
=
h04 − h5
h04 − h5
(7.4)
It is assumed that there is no loss in passing the gas from turbine exhaust
to the nozzle. It should be noted that in Eq. 7.4, h5 is used instead of total
enthalpy h05 because the exhaust nozzle efficiency is an indication of the
percentage of total energy converted into velocity energy. Therefore,
h04 − h5
ηnoz Cp T04 1 −
=
p5
p04
γ−1
γ
The exit velocity at the nozzle end can be obtained by writing energy
balance equation.
h04 − h5
=
c2j
2
232
Gas Turbines
c2j
2
cj
γ−1
γ
p5
p04
=
ηnoz Cp T04 1 −
=
.
/
/
02η C T 1 −
noz p 04
p5
p04
γ−1
γ
Thrust developed
=
=
ṁa [(1 + f )cj − ci ]
.
⎧
/
⎨
/
ṁa (1 + f )02ηnoz Cp T04 1 −
⎩
p5
p04
γ−1
γ
− ci
⎫
⎬
(7.5)
⎭
From the above, thrust specific fuel consumption and also the thermal
efficiency can be calculated.
7.6.2
Performance of a Turbojet Engine
With the help of the above analysis it is possible to estimate the performance of a turbojet engine taking into account the component efficiencies
and other parameters. Figure 7.12 shows the thrust specific fuel consumption for various compressor pressure ratios at two different Mach numbers.
It is evident that for a given Mach number there is only one compressor
pressure ratio which gives best fuel economy for given values of component
efficiencies and maximum allowable temperature. As the pressure ratio increases with a given maximum temperature fuel consumption decreases to a
minimum. After that further increase in pressure ratio will not improve the
fuel economy until maximum temperature is not raised. For a given pressure
ratio higher maximum temperature will result in more thrust. Maximum
thrust per unit mass of fuel is achieved at a lower pressure ratio than the
one which produces minimum specific fuel consumption at the same turbine inlet temperature. In addition to the above parameters, three more
variables – flight speed, altitude (inlet temperature and pressure), and fuel
flow rate – greatly affect the performance of a jet engine.
The turbojet is almost a constant thrust engine. The specific fuel consumption based on thrust power reduces because with almost constant
thrust, the thrust power increases as shown in Fig. 7.13. Therefore, the
maximum speed is the most efficient operational point for the turbojet.
The static thrust of such engines is very low as compared to propeller
engine aircraft for which cruise thrust is about 60 per cent of the takeoff thrust. The thrust at first decreases with increase in speed because
the velocity cj in the thrust equation increases. After a minimum value,
the thrust starts increasing (see Figs. 7.14 and 7.15) due to increased ram
compression at higher speeds.
As the altitude increases, the thrust decreases due to decrease in density,
pressure and temperature of the air (see Fig. 7.15). However, the rate of
Jet Propulsion Cycles and Their Analysis
233
TSFC
M = 0.66
M = 0.33
r
Fig. 7.12 Thrust specific fuel consumption vs compressor pressure ratio for
a turbojet engine
TP
TP
TPSFC
TPSFC
Flight speed
Fig. 7.13 Thrust specific fuel consumption and thrust power vs flight speed
decrease of thrust is less than the rate of decrease of density with altitude
because some loss due to reduced density is compensated by lesser drag.
The thrust is maximum at sea level. Due to considerable reduction in drag
(at an altitude of 8000 m the drag is reduced to less than 25 per cent of
sea level drag), the turbojet is most efficient when flown at high altitudes
and at relatively high speeds. The fuel consumption on fuel mass per km
of travel increases with speed as power output also increases with speed.
The operational range of turbojet engine is about 800 to 1100 km/h and
the specific fuel consumption is about 1.0 to 1.5 kg/thrust h at cruising
speeds and are still greater at lower speeds. The altitude limit is about
10000 m.
234
Gas Turbines
r
lle
pe
ro
hp
itc
ep
bl
Relative thrust
sta
ju
Ad
Jet engine
Flight speed (km/h)
Fig. 7.14 Thrust vs speed for propeller driven and turbojet engine
Thrust
Sea level
3000 m
6000 m
9000 m
12000 m
Flight speed
Fig. 7.15 Effect of altitude on thrust at maximum rpm
7.6.3
Advantages and Disadvantages of Turbojet
Advantages of Turbojet
(i) The power to weight ratio of a turbojet is about 4 times that of a
propeller system having reciprocating engine.
(ii) It is simple, easy to maintain and requires lower lubricating oil consumption. Furthermore, complete absence of liquid cooling results in
reduced frontal area.
(iii) There is no limit to the power output which can be obtained from
a turbojet while the piston engines have reached almost their peak
power and further increase will be at the cost of complexity and
greater engine weight and frontal area of the aircraft.
Jet Propulsion Cycles and Their Analysis
235
(iv) The speed of a turbojet is not limited by the propeller and it can
attain higher flight speeds than engine propeller aircrafts.
Disadvantages of Turbojet
(i) The fuel economy at low operational speeds is extremely poor.
(ii) It has low take-off thrust and hence poor starting characteristics.
7.7
THRUST AND THRUST EQUATION
Having gone through the details of various propelling devices, their thermodynamic cycles and performances, it is worthwhile to discuss some of the
basic laws of thrust production and the factors which affect the performance
of the engine.
Let us consider the control volume of a schematic propulsive device
shown in Fig. 7.16. A mass ṁi of air enters the control volume with a
velocity ci and pressure pi and the products of combustion of mass ṁj
leaves the control volume with a velocity cj and pressure pj . The flow is
assumed to be steady and reversible outside the control volume, the pressure
and velocity being constant over the entire control volume except that at
the exhaust area Aj . Force F is the force necessary to balance the thrust
produced due to change in momentum of the fluid as it passes through the
control volume.
Fuel
Aj
p
j
cj
mj
Ai
p
i
ci
mi
Fig. 7.16 General schematic diagram of a propulsive device
If pa is the atmospheric pressure, then writing the momentum equation,
we get
ṁj cj − ṁi ci
= F + (pi − pa )Ai − (pj − pa )Aj
or thrust,
F
= (ṁj cj − ṁi ci ) + (pj − pa )Aj − (pi − pa )Ai
We have by mass balance,
(7.6)
236
Gas Turbines
ṁj
= ṁi + ṁf
(7.7)
where ṁj , ṁi and ṁf are the mass flow rates of exhaust gases, air, and fuel
respectively. If
f
ṁj
= Fuel air ratio
=
ṁf
ṁi
= ṁi (1 + f )
Thrust = ṁi [(1 + f )cj − ci ] + (pj − pa )Aj − (pi − pa )Ai
Momentum thrust
(7.8)
Pressure thrust
From Eq. 7.8 it is clear that the net thrust produced is made up two
parts, viz., momentum thrust and the pressure thrust . If the exhaust
velocity cj from the control volume is subsonic, then pj ≈ pa and also
pi ≈ pa so that the pressure thrust is quite small. Similar is the case for
propeller engines. For supersonic exhaust velocity the pressure pj may differ
from pa . However, the pressure thrust developed is so small as compared to
the momentum thrust that it can safely be neglected for simple calculations
and the net thrust is given by
Thrust
=
ṁi [(1 + f )cj − ci ]
(7.9)
The thrust, given by Eq. 7.9 can be increased by increasing the mass
flow or increasing the velocity of the exhaust jet for a given ci . Equation
7.9 has been derived for a simple control volume shown in Fig. 7.16, but is
equally applicable to an aircraft flying at a forward speed ci . In the latter
case the velocities are considered relative to the aircraft. The air has a
velocity ci equal to aircraft forward speed, relative to the engine. Thus,
we see that a large amount of thrust can be obtained either by propelling
a large mass of air and increasing its velocity by a small amount, or by
increasing the velocity of a small mass of air to a high value. In the case of
aircraft power plants the fuel-air ratio f is very small (about 0.01 to 0.02)
and hence the mass of fuel can be neglected safely without causing much
error in the performance calculations.
The thrust is, then, given by
F
=
ṁi (cj − ci )
the effective speed ratio, α, is given by
ci
α ≡
cj
and
F
=
ṁi ci
(7.10)
(7.11)
1
−1
α
(7.12)
Jet Propulsion Cycles and Their Analysis
237
The product ṁi cj is called the gross thrust and ṁi ci is called the inlet
drag or inlet momentum. Equation 7.12 is also applicable to turboprop
engines.
The thrust developed by the engine overcomes the drag on the aircraft
and in doing so it develops power, called the thrust power which is given
by
PT
=
F ci
=
ṁi (cj − ci )ci
(7.13)
Note that turboprop engine is rated in kilowatts while turbojet and
ramjet are rated on the basis of thrust developed.
7.8
SPECIFIC THRUST OF THE TURBOJET ENGINE
The thrust per kg of air flow is known as specific thrust or specific impulse.
Assuming ṁa is the mass flow rate at inlet, the specific impulse, Isp , is
given by
Isp
=
F
ṁa
=
(cj − ci )
In terms of the effective speed ratio, α =
Isp
=
ci
cj ,
cj (1 − α)
(7.14)
Eq. 7.14 becomes
(7.15)
The specific thrust is a criterion of the size of engine required for producing a given total thrust. It is apparent from Eq. 7.15 that, to achieve a
large specific thrust, the speed ratio α must be kept small. Also this equation shows that if the effective jet velocity cj is maintained constant, the
specific thrust, Isp , decreases with increase in the flight speed ci . The maximum value of Isp under the above mentioned conditions is realized when
the flight speed is zero. The magnitude of the specific thrust at ci = 0 is of
particular significance because it is the measure of the thrust available for
take off.
The specific static thrust, (Isp )st is given by
(Isp )st
=
cj
(7.16)
Hence to develop a large specific thrust at take off the effective jet speed
should be as large as possible.
Let Δhnoz be the enthalpy drop in the nozzle corresponding to the
velocity cj , then
c2j
2
= Δhnoz
cj
=
2Cp ΔT
(7.17)
238
7.9
Gas Turbines
EFFICIENCIES
To evaluate the performance of a turbojet engine the efficiencies of its
components must be known. Here, we shall define the various efficiencies
associated with jet propulsion plants which in turn will indicate the state
of development of the components of the plant.
7.9.1
Inlet Diffuser or Ram Efficiency
h
p
h01= h 01’
01
p
01
’
=
=
co
n
co
ns
tan
t
sta
n
t
In all jet propulsion plants, except turboprop and rocket engines, a diffuser
is used at the inlet to convert kinetic energy into static pressure rise. This
compression process is essentially adiabatic but it cannot be considered as
reversible since fluid friction is present. The most widely used efficiency
of the inlet jet process is based upon the pressure rise that actually takes
place; compared to the pressure rise which would have taken place had the
process been isentropic. By definition, this efficiency is (refer Fig. 7.17)
c 21
s
on
2
pa
Ideal
=c
Actual
t
tan
s
Fig. 7.17 h–s diagram for inlet diffuser
ηram
=
p01 − pa
p01 − pa
=
p01
pa
p01
pa
−1
−1
(7.18)
In an ideal diffuser, the process will be isentropic and the final pressure
will be given by
p01
pa
But
=
T01
Ta
γ
γ−1
(7.19)
Jet Propulsion Cycles and Their Analysis
T01
= Ta +
c2i
2Cp
239
(7.20)
where ci is the forward speed of the engine.
γ
⎡
⎤ γ−1
c2i
T
+
a
p01
2Cp
⎦
= ⎣
=
pa
Ta
c2i
1+
2Cp Ta
γ
γ−1
(7.21)
or
p01
= pa
c2i
1+
2Cp Ta
γ
γ−1
(7.22)
But due to losses in friction and shock, the actual pressure will not rise
to the value given by Eq. 7.22. From Eq. 7.18 actual pressure rise is
p01 − pa
=
ηram (p01 − pa )
(7.23)
or
p01
=
pa + ηram pa
γ
γ−1
c2i
1+
2Cp Ta
− pa
or
p01
=
pa 1 + ηram
!
γ
γ−1
c2i
1+
2Cp Ta
−1
"
(7.24)
Efficiencies of the order of 85 to 90 percent are obtained with well designed diffusers at subsonic speeds. The diffusion process is dependent more
on the type of aircraft and ducting than it is on the engine itself. Hence,
it should not be considered as an engine characteristic, although it affects
the overall engine cycle.
Equation 7.18 has one serious disadvantage. At low flight speeds p01 −pa
might actually become negative as a result of inlet duct pressure drop,
whereas if the airplane is standing still or if the engine is on test bed,
p01 − pa is equal to zero causing Eq. 7.18 to become negatively infinite. A
more consistent method of expressing diffuser friction losses is in the form
of a percentage pressure drop or pressure coefficient. Referring to Fig. 7.17,
the diffuser pressure drop coefficient is expressed by
pdc
7.9.2
=
p01 − p01
p01
=
1−
p01
p01
(7.25)
Thermal Efficiency of the Turbojet Engine
Thermal efficiency of a propulsive device is an indication of the degree of
utilization of energy in fuel in accelerating the fluid flow and is defined
as the ratio of propulsive power furnished to exhaust nozzle to the heat
240
Gas Turbines
supplied and is given by
Thermal efficiency =
Propulsive power
Fuel flow rate × Calorific value of fuel
ηth
=
P
ṁa Qi
(7.26)
where Qi = f CV = heat supplied to the engine per kg of air and f =
ṁf /ṁa is the fuel-air ratio and CV is the calorific value of the fuel.
The propulsive power, P , is given by the net increase in the energy of the
working fluid between inlet and exit, viz., stations 0 and 5 (refer Fig. 7.9).
Hence,
P
=
1
(ṁa + ṁf )c2j − ṁa c2i
2
(7.27)
Assuming ṁ = ṁa + ṁf ≈ ṁa , the Eq. 7.27 reduces to
P
=
ṁa 2
c − c2i
2 j
(7.28)
Introducing α = ci /cj and substituting P in Eq. 7.26, we get
ηth
=
ṁa c2j (1 − α2 )
2ṁf CV
=
c2j 1 − α2
2f CV
(7.29)
If h02 and h03 are specific stagnation enthalpies of the working fluid
entering and leaving the combustion system, and ηcomb denotes the combustion efficiency, then by referring Fig. 7.18,
Qi
= f CV
=
h03 − h02
ηcomb
(7.30)
Substituting the value of Qi in Eq. 7.26 we get
ηth
=
ηcomb c2j 1 − α2
2(h03 − h02 )
(7.31)
If we assume that the mean specific heat remains constant during process 2-3, then
ηth
=
ηcomb c2j 1 − α2
2Cp (T03 − T02 )
(7.32)
For constant values of altitude and turbine inlet temperature T03 , the
enthalpy added by the combustion process (h03 − h02 ) is also constant. It
is seen from the Eq. 7.32, that the thermal efficiency, ηth , reduces to zero
Jet Propulsion Cycles and Their Analysis
Turbine
03
h03
WT
04
h
Burner
q
Compressor
241
p 04
c 2j / 2
04’
5’
5
Nozzle (jet)
02
p
a
p=
’
p
01
p 01
01
c2i /2
01’
5
02’
WC
0
Diffuser (ram)
s
Fig. 7.18 h–s diagram for turbojet engine
when α = 1, i.e., when cj = ci . That conclusion is logical because when
α = 1, the engine develops no thrust and no energy is added to the fluid
flowing through the engine. On the other hand when α = 0, i.e., when
ci = 0 the thermal efficiency of the engine depends entirely on the jet
velocity cj .
It should be noted that Eq. 7.32 applies only to turbojet, ramjet and
pulse jet engines and does not apply to propeller engines and rocket engines.
For propeller engines thermal efficiency is defined as
ηth =
Brake power
Fuel rate × CV of fuel
(7.33)
In the case of turboprop engine shaft power as well as thrust power
both are used for propulsion. However, the shaft power is considerably
larger than the thrust power and it is usual to define thermal efficiency on
the equivalent shaft power basis such that the thrust power due to exhaust
jet is also included. For calculating thrust power some suitable velocity for
the aircraft is selected.
7.9.3
Propeller Efficiency
The propeller produces thrust power by accelerating the air. The propeller
itself is driven by the engine. The efficiency of the propeller is defined as
the ratio of the thrust power to the shaft power.
Propeller efficiency
=
Thrust power
Shaft power
=
F × ci
sp
(7.34)
In the case of turboprop engine the thrust power developed by the exhaust is also considered.
242
7.9.4
Gas Turbines
Transmission Efficiency
In many cases the engine or the turbine output cannot be directly applied to
the propeller some form of transmission is involved between the engine and
the propeller in the form of a reduction gear. The main reason for providing
reduction gear in the case of a turboprop engine is high rotational speed of
the turbine at which the propeller cannot be rotated efficiently. In addition
to this some layout problems always occur. Due to friction and other losses
the output from the transmission system is always less than input to it and
the transmission efficiency is defined as
Efficiency of transmission =
Output of the transmission
Input to the transmission
(7.35)
Measures of the thermal efficiency of an engine for a given air speed
and jet speed are; the thrust specific fuel consumption (T SF C) for cycle
efficiency and the thrust power specific fuel consumption for overall thermal
efficiency (T P SF C) by definition, the thrust specific fuel consumption is
T SF C
=
3600 × ṁf
kg of fuel/N thrust-hour
F
(7.36)
where ṁf is in kg/s. Dividing Eq. 7.36 by ṁa , we get
T SF C
=
3600 × f
Isp
(7.37)
Similarly thrust power specific fuel consumption is
T P SF C
=
3600 × ṁf
kg/thrust kW h
TP
(7.38)
=
3600 × f
(T P )s
(7.39)
The cycle efficiency of a turbojet engine remains relatively constant with
an increase in air speed, whereas the propulsive efficiency increases. Thus,
it may be expected that the thrust power specific fuel consumption would
improve with air speed. Typical curves showing the trend of thrust, thrust
power and fuel consumption are shown in Fig. 7.19.
7.9.5
Propulsive Efficiency
During the forward motion, the specific thrust, cj , is reduced by the inlet
drag ci . The net thrust thus, is dependent not only on the power plant
but also on the flight speed. The utilization of the gross thrust may be
considered in terms of the propulsive efficiency.
Thus, propulsive efficiency is the measure of the effectiveness with which
the kinetic energy imparted to the fluid is transferred into useful work.
The useful work done by the system is the product of the thrust and flight
velocity, i.e., F × ci which is called the thrust power. The kinetic energy
243
Th
rus
t po
we
r
Jet Propulsion Cycles and Their Analysis
Thrust
TSFC
TP
SF
C
Air speed or flight speed (km/h)
Fig. 7.19 Typical performance curves for a jet engine
imparted to the fluid is the difference between the kinetic energy at exit and
the kinetic energy at inlet and is called propulsive power. The difference
between propulsive power and thrust power is called the leaving losses.
Now,
Kinetic energy of air at inlet
c2i
2
=
ṁa
=
ṁa (1 + f )
c2j
2
(7.41)
Propulsive power =
ṁa (1 + f )
c2j
c2
− ṁa i
2
2
(7.42)
(7.40)
Kinetic energy of gases at exit
Thrust power =
Propulsive efficiency =
=
(7.43)
F × ci
Thrust power
Propulsive power
ṁa [(1 + f )cj − ci ]ci
c2
ṁa (1 + f ) 2j −
=
[(1 + f )cj − ci ]ci
c2
(1 + f ) 2j −
c2i
2
c2i
2
(7.44)
As the turbojet engines operate at very low fuel-air ratios, f , (i.e., very
high air-fuel ratios), f may be neglected, then
ηp
=
2(cj − ci )ci
c2j − c2i
244
Gas Turbines
=
ηp =
2ci
c j + ci
2(cj − ci )ci
(cj − ci )(cj + ci )
=
2α
1+α
(7.45)
(7.46)
ci
.
cj
Differentiating 7.46 with respect to the speed ratio α, and equating it
to zero, the conditions for the best propulsive efficiency can be obtained,
which is α = 1. This means that the velocity of the gas jet, cj , is equal to
the velocity of the aircraft. At this point, the propulsive efficiency becomes
100 per cent. However, from Eq.7.12, it is clear that at the condition of
maximum propulsive efficiency, the thrust developed is zero and no useful
work output can ever be obtained from the power plant. Hence, it is not
practical to maximize propulsive efficiency of a jet engine. Other parameter
such as thrust etc., must be considered for performance evaluation.
Now from Eq.7.13, thrust power
where α =
PT
=
ṁa c2j α(1 − α)
(7.47)
By differentiating the Eq. 7.47 with respect to α and equating to zero, it
can be easily seen that the maximum thrust occurs at α = 0.5, i.e., when
the speed of the gas jet at the exhaust is twice that of the air at inlet. The
corresponding value of the propulsive efficiency is given by
ηp
=
2α
1+α
=
2 × 0.5
1 + 0.5
=
0.667
(7.48)
Thus it is evident from the above analysis that for a turbojet, ramjet and
pulsejet, the points of maximum propulsive efficiency and maximum thrust
are different. Some compromise must be made to get reasonable thrust
with a good propulsive efficiency.
Turboprop engines are essentially two-fluid stream engines and this analysis cannot be applied to them. For each stream a separate equation must
be written.
In the case of propeller driven aircraft engine, the propulsive efficiency
is based on the brake power (bp) of the engine
ηp
=
Thrust power
Brake power
(7.49)
It can be noted that ηp , of an engine increases with the aircraft speed.
To put it in a nutshell large thrust per unit flow rate of working fluid
cannot be obtained from a small light weight propulsion system unless the
jet velocities are very large. To realize reasonable propulsive efficiencies the
flight speed must be high. These are met usually in turbojet engines.
Jet Propulsion Cycles and Their Analysis
7.9.6
245
Overall Efficiency of a Propulsive System
The performance of the propulsion system is normally evaluated in terms of
overall efficiency, η0 is defined as the ratio of rate at which useful propulsion
work is done to the rate at which energy is supplied to the system. In other
words the overall efficiency ηo of a propulsive device is the ratio of the useful
work done to the chemical energy supplied in the form of fuel.
ηo
=
Useful propulsive work
Chemical energy supplied
=
Transmission output
Engine output
×
×
Energy input
Transmission input (engine output)
Useful propulsive work
Input to the propulsive device (transmission output)
ηo
=
Thermal efficiency × Transmission efficiency ×
Propulsive efficiency
ηo
=
ηth × ηtr × ηp
(7.50)
It follows from the above that the overall effect of the propulsive device
is made of a power plant converting some percentage of the chemical energy supplied to it into a form useful for propulsion (shaft power in case
of propeller devices and acceleration of mass flow for jet devices). This includes the losses in the transmission, if used, (the transmission converting
the energy in a more useful form, i.e., in acceleration of the fluid); the final
propulsive device converting this energy into useful work output. This is
shown in Fig. 7.20.
Fuel
Engine
Engine thermal
efficiency
Transmission
system
Propulsive
device
Output of engine
= Input to
transmission
Output of transmission
= Input to propulsion
Transmission
efficiency
Propulsion
efficiency
Overall efficiency =
Useful propulsion work
Energy in fuel
Fig. 7.20 Block diagram for aircraft engine efficiencies
Useful
propulsive
work
246
Gas Turbines
7.10
PARAMETERS AFFECTING FLIGHT PERFORMANCE
As has been already mentioned that two parameters which affect the performance of the jet propulsion cycle are – the forward speed of the aircraft
and the altitude at which the aircraft flies. These two do not enter into
the analysis of shaft power cycle. The effect of the two parameters are
discussed in the following sections.
7.10.1
Effect of Forward Speed
The forward speed affects the inlet pressure and temperature of the compressor. The inlet duct to the compressor acts as a diffuser. The air which
enters the diffuser at flight speed is slowed down to the speed acceptable
to compressor, and at the same time raising its pressure and temperature.
This increase of pressure and temperature due to aircraft speed is known
as ram effect, or simply ram. It becomes more and more prominent as the
flight speed increases. For a given aircraft speed and ram efficiency, the
ram pressure ratio increases as the ambient temperature decreases at high
altitudes.
The second effect of aircraft forward speed is in relation to propulsive
efficiency. As flight velocity increases, the inlet drag also increases. If there
were no ram effect, the net specific thrust, Isp , would decrease. This is
because the jet velocity remains the same. With ram, the increase of inlet
temperature reduces the gross-thrust to some extent. However, the increase
of inlet pressure more than compensates for this. Because of this, the cycle
pressure ratio increases without shaft work being necessary. The overall
effect of forward speed on inlet drag and ram is to reduce somewhat the
net specific thrust.
7.10.2
Effect of Altitude
The effect of altitude on a turbojet is by virtue of reduction of ambient
pressure and temperature. The temperature of atmosphere varies considerably and continuously with location and time, so that the standard atmosphere is used for calculating the performance at various altitudes. The
most general data used are those for International Standard Atmosphere
or ICON atmosphere (International Commission On Navigation). It corresponds approximately to average values found in the middle latitudes. It is
normalized by using a linear decrease of temperature or lapse rate of 1.98◦C
per 300 m of altitude, starting with a ground level temperature of 15◦ C.
With the temperature fixed, the pressure can be calculated according to
the principles of hydrostatics using 1.03 bar as the ground level pressure.
Table 7.1 provides values of International Standard Atmosphere.
7.11
THRUST AUGMENTATION
The poor take-off characteristic of the turbojet engine can be improved by
augmenting the thrust. The thrust from a turbojet is given by
Jet Propulsion Cycles and Their Analysis
247
Table 7.1 International Standard Atmosphere.
Altitude [m]
0
300
1500
3000
4500
6000
7500
9000
12000
15000
18000
21000
F
Temperature [◦ C]
15.0
13.0
5.1
− 6.8
− 16.7
− 26.6
− 36.5
− 46.6
− 56.3
− 56.3
− 56.3
− 56.3
=
Pressure [bar]
1.030
0.992
0.856
0.708
0.579
0.473
0.382
0.306
0.190
0.118
0.073
0.066
ṁa [(1 + f )cj − ci ]
in which the exhaust cj , is the function of the maximum temperature in the
cycle. Higher the maximum temperature higher is the value of cj . Another
method of increasing thrust is to increase the mass flow rate. Improved
thrust results in shorter take-off distances, high climb rate and good manoeuvrability at high altitudes. The thrust augmentation can be effected
by the following methods:
(i) Burning of additional fuel in the tail pipe between the turbine exhaust
section and entrance section of the exhaust nozzle. This method
of thrust augmentation increases the jet velocity and is known as
afterburning. The device used is called the afterburner.
(ii) Injecting refrigerants, water or water-alcohol mixture at some point
between inlet and exit sections of the air compressor. This method of
thrust augmentation increases the mass flow rate and decreases the
work of compression.
(iii) Bleeding off air in excess of that required for stoichiometric combustion in the main combustion chamber – at the entrance section of the
combustion chamber, and burning it with the stoichiometric fuel-air
ratio in a separate one. The combustion products from the latter
combustor are expanded in a separate auxiliary exhaust nozzle. The
bled air is replaced by water which is injected in the main combustors. This method of thrust augmentation is known as the bleed burn
cycle.
7.11.1
The Afterburner
This method of thrust augmentation is being widely applied for obtaining
high thrust for short duration. It is known that turbine blade material
248
Gas Turbines
considerations limit the combustion chamber temperature rise. This in
turn, limits the basic engine fuel- air ratio to values of about 0.017. As
a result, the products of combustion leaving the turbine contain enough
unutilized oxygen to support further combustion. Thus if a suitable burner
is installed between the turbine and exhaust nozzle, a considerable amount
of fuel can be burned in this section to produce temperatures entering the
nozzle as high as 2000◦C. This increases the gas velocity, and hence provides
a thrust increase. A boost of about 30 per cent can be obtained in this
manner. However, the fuel consumption increases rapidly. For about 20 per
cent thrust increase by use of reheat the overall fuel consumption may be
increased by more than 100 per cent and this additional mass of fuel has to
be carried by the turbojet. Therefore, it is used only for take-off or for high
climbing rates and for a very short duration. Because of the temperature
rise in the afterburner, there is a large increase in specific volume of gases,
and to keep the pressure drop as small as possible, the tail pipe of the
afterburning area is more than that of the normal engine. Furthermore,
the afterburning engine must be equipped with a variable exit area exhaust
nozzle so that by varying its area with the afterburner operating, the normal
conditions at inlet to the afterburner will be unaffected.
Another way of seeing the need for an increase of nozzle exit area is to
examine the compressor performance chart. Figure 7.21 shows the sketch
of the typical thrust variation with speed for an afterburner engine.
On
Off
Thrust
After burner
Engine speed
100%
Fig. 7.21 Variation of thrust with speed
As shown by the thrust versus rpm plot, the augmented thrust occurs
at a constant 100% rpm; therefore, on the compressor performance chart
(Fig. 7.22) the augmented thrust should lie on 100% rpm line. Point X
represents operating point for 100% rpm operation without afterburner.
Now, when the afterburner is tuned on without increasing the nozzle area,
the flow resistance felt in the compressor will increase and the mass flow
rate will decrease along the constant rpm line to produce operating point
Y , which is in the stalled region. This type of operation of course, cannot
be tolerated. For optimum conditions, the added flow resistance, caused by
the specific volume increase in the afterburner, must be exactly balanced by
Jet Propulsion Cycles and Their Analysis
249
Operating line
Surge line
p
Y
03
X
100%
p
02
95%
N = 90%
ma
Fig. 7.22 Performance chart for an afterburner engine
the decrease in flow resistance brought about by opening the nozzle by the
proper amount. Under these conditions, the operating point for afterburner
operation will coincide with point X.
Figure 7.23 shows the thermodynamics cycle of a turbojet engine fitted
with afterburner on h-s plane. The process lines up to the turbine section
are the same as for a basic engine. The afterburner process 5–6 is ideally
a constant-pressure process, but the internal drag and momentum pressure
loss produce a total pressure drop of about 5%. It is apparent that more
thrust can be realized per kg of air by examining the relative magnitude of
enthalpy drops across the normal nozzle and across the afterburner nozzle.
About 1900 o C
6
2
0
a
p
p
e
p
7’
4
5
3
05
p
03
or T
About 850 o C
7
Nozzle process without afterburne
p 02
Atmospheric pressure
s
Fig. 7.23 h–s diagram of a turbojet engine with afterburner
250
7.11.2
Gas Turbines
Injection of Water-Alcohol Mixture
This method of thrust augmentation is probably the simplest one to achieve.
Mixture of water and alcohol or just water is injected at the combustion
chamber or compressor inlet section through a series of suitable spray nozzles to produce an increase in thrust.
The effect which produces the greater gain in thrust is the cooling effect
within the compressor through water evaporation, which brings about rise
in compressor discharge pressure. The rise in pressure can be seen from the
compressor h-s diagram shown in Fig. 7.24. Process 1–2 represents normal
dry compression process which is achieved by the turbine work ΔhT , being
delivered to the compressor. Now, when the water alcohol injection system
is turned on (usually at 100% rpm) the mixture as it evaporates within the
compressor, cools the air to produce the compression process line 1–2 . The
amount of work delivered by the turbine, with or without water injection,
remains essentially the same, therefore, points 2 and 2 must line on the
same enthalpy line. It is apparent that water alcohol injection process
produces a higher compressor discharge pressure which in turn produces an
increase in thrust.
2"
p 02
p 02"
h or T
2
2’
Δ hT = Δ h C
1
s
Fig. 7.24 Compressor h–s diagram
The factors which contribute to thrust augmentation by water-alcohol
mixture injection can be summarized such as:
(i) evaporative cooling which produces higher pressure and higher mass
flow;
(ii) additional mass of injected fluid; and
(iii) possibility of burning of alcohol.
Jet Propulsion Cycles and Their Analysis
251
Mass flow rate
(a) Axial compressor
e
in
e
gl
lin
tin
era
N/ T
0
all
0
Δp
N/ T
Op
Δp
X
St
Pressure ratio
St
all
lin
Op
e
era
tin
gl
in
e
X
Pressure ratio
The first factor determines whether a water injection system is practical
on a given engine installation. The first factor also provides the key to
determine where water injection produces the higher augmentation thrust
ratios. Cooling will be accomplished by the injected fluid until the air at
compressor discharge is saturated. Thus to get this saturation point, the
amount of cooling increases as the thrust ratio is greater on a hot day than
on a cold day and the augmented thrust ratio decreases with altitude and
increases the flight Mach number.
The water injected into an axial flow compressor tends to be centrifugally separated from the air. To eliminate this problem, water injection
into the combustion chamber has been developed. The principle of operation of this method is illustrated on the compressor-performance diagram
(Fig. 7.25).
Mass flow rate
(b) Centrifugal compressor
Fig. 7.25 Compressor performance charts for typical axial and centrifugal
flow machines
The
√N
T0
lines for an axial flow compressor are steeper near the surge
line than the √NT lines of the centrifugal compressor. Point 0 in the figures
0
represents the normal operating point at 100% speed. At this operating
conditions the gas-flow is normally chocked, thus the only way to increase
the thrust output of the engine is to raise the pressure upstream to turbine
or nozzle.
Now, when the water is injected into the combustion chamber, the flow
resistance on the compressor increases, and since the engine speed is fixed,
the compressor operating point shifts upward of constant √NT line until a
0
stabilized operating point X is reached. This point occurs when the pressure required by the new flow system is achieved. It is clear that if too
much flow resistance is added, compressor stall will occur, therefore, it is
necessary to maintain close control over the rate of water injection. The
increase in flow resistance causes the pressure level of the engine, and consequently the thrust, to increase, but at the same time, the added compressor
flow resistance produces a tendency to decrease the thrust by reducing the
mass flow through the compressor. However, since the pressure is higher
252
Gas Turbines
at the turbine and nozzle, these components will handle more mass, the
additional mass being the water injected into the burner. Effectively water
injection into the combustion chamber produces a thrust increase by
(i) increasing the compressor pressure ratio due to reduced compressor
air flow, and
(ii) increasing the total mass flow through turbine and exhaust nozzle.
The magnitude of thrust increase is entirely dependent upon compressor operating characteristics.
7.11.3
Bleed Burn Cycle
Since in a turbine excess air is also present, a small percentage of highpressure air from the compressor is bled to an auxiliary combustion chamber by by-passing the turbine. In auxiliary combustion chamber the bled
air is heated by an additional fuel supply to a higher temperature than
would be permissible in the main engine on account of the limiting temperature at the turbine blades. The hot gases are then discharged forming
an additional jet. A shut off valve is used to bring the engine to normal
position. Water is injected into main combustion chamber to replace the
mass of the extracted air, thus maintaining the discharge of main jet at
the same level. This method is usually used for take-offs only due to high
rate of liquid consumption which cannot be carried with engine during its
flight. The augmented thrust ratio is highest for this method among the
three methods.
Of the three methods discussed, afterburning seems to be the only practical method for thrust augmentation during flight. The air bleed-off system
gives maximum thrust augmentation but at the expense of large fuel consumption. This is used only when a large take-off thrust is needed. For
smaller thrust augmentation ratio water injection is used because of simplicity and light weight. Afterburner is used for medium thrust augmentation
ratio. Afterburner combined with water injection can also be used.
Worked out Examples
[Note : Take
γa −1
γa
= 0.286;
γg −1
γg
= 0.248]
7.1 A turbojet power plant uses aviation kerosene having a calorific value
of 43 MJ/kg. The fuel consumption is 0.18 kg per hour per N of
thrust, when the thrust is 9 kN. The aircraft velocity is 500 m/s the
mass of air passing through the compressor is 27 kg/s. Calculate the
air-fuel ratio and overall efficiency.
Solution
ṁf
=
0.18
× 9000
3600
=
0.45 kg/s
Jet Propulsion Cycles and Their Analysis
Air-fuel ratio
=
27
0.45
Thrust power, PT
=
F × ci
=
9 × 500
Heat input, Q
=
0.45 × 43000
η
=
PT
Q
=
4500
× 100
19350
=
23.26%
=
Ans
60 : 1
=
253
⇐=
4500 kW
=
19350 kW
Ans
⇐=
7.2 A simple turbojet unit operates with a maximum turbine inlet temperature of 1200 K, a pressure ratio of 4.25:1 and a mass flow of
25 kg/s under design conditions, the following component efficiencies
may be assumed.
Isentropic efficiency of compressor
Isentropic efficiency of turbine
Propelling nozzle efficiency
Transmission efficiency
Combustion chamber pressure loss
:
:
:
:
:
87%
91.5%
96.5%
98.5%
0.21 bar
Assume Cpa = 1.005 kJ/kg K and γ = 1.4, Cpg = 1.147 kJ/kg K and
γ = 1.33.
Calculate the total design thrust. Also calculate the total thrust and
specific fuel consumption taking into consideration the nozzle choking
condition. Assume that the unit is stationary and at sea level, where
the ambient conditions may be taken as 1 bar and 293 K. Assume
air-fuel ratio of 50.
Solution
03
04
04’
T
02
02’
01
s
Fig. 7.26
5
5’
254
Gas Turbines
Actual temperature rise in the compressor
T02 − T01
T02
=
T01 (γa −1)/γa
r
−1
ηC C
=
293
× 4.250.286 − 1 = 172.63 ◦ C
0.87
=
172.63 + 293
=
465.63 K
Since the work output of the turbine must be just sufficient to drive the
compressor we have,
ηtr Cpg (T03 − T04 )
=
Cpa (T02 − T01 )
T03 − T04
=
1.005 × 172.63
0.985 × 1.147
T04
=
1200 − 153.56
=
153.56 K
=
1046.44 K
Thus, the total head pressure ratio, rt across the turbine is,
1
T03 − T04
=
ηT T03 1 −
153.56
=
0.915 × 1200 × 1 −
=
1−
1
rt0.248
rt
=
(γ −1)/γg
rt g
153.56
0.915 × 1200
1
0.8602
1
rt0.248
=
0.8602
=
1.835
1/0.248
The total head pressure at exit from the turbine is,
p04
p02 − Δpcomb
rt
=
p03
rt
=
4.25 − 0.21
1.835
=
=
2.20 bar
For isentropic flow through the nozzle,
T04
T5
T5
γg −1
γg
=
p04
p5
=
2.20
1.0
=
1046.47
1.216
0.248
= 1.216
=
860.56 K
Jet Propulsion Cycles and Their Analysis
255
Now, the propelling nozzle efficiency is given by
ηn
=
T04 − T5
T04 − T5
T04 − T5
=
ηn (T04 − T5 )
=
0.965 × (1046.47 − 860.56) = 179.38 K
T5
=
1046.47 − 179.38 = 867.07 K
c2j
2Cp
=
T04 − T5 = 179.38
c5
=
√
2 × 1.147 × 1000 × 179.38
=
641.46 m/s
=
25 × 641.46 = 16036.6 N
Total design thrust/s
Ans
⇐=
Now, we should compare the operating pressure ratio, pp045 , of nozzle with
the corresponding critical pressure ratio, pp04
. If pp045 > pp04c , then the nozzle
c
will choke and a convergent-divergent nozzle will be required. If pp04
< pp04c ,
5
then the nozzle will not choke and only a simple convergent nozzle will be
sufficient.
In practice, a convergent nozzle is used, in which expansion takes place to
the critical pressure, pc , which is greater than the ambient pressure, pa = p5 .
Then in addition to momentum thrust there will be a pressure thrust, which
is given by
A(pc − pa )
where A is the cross-sectional area of the nozzle outlet.
Thus, in this problem,
p04
p5
=
p04
pa
=
2.2
Since the nozzle efficiency is given, critical pressure ratio is given by
p04
pc
1
=
1−
1
ηn
γg −1
γg +1
γg
γg −1
1
=
1−
1
0.965
1.33−1
1.33+1
4.03
= 1.895
p
Therefore, in this problem, p04
is greater than the critical pressure ratio of
a
the propelling nozzle and as such the complete expansion within nozzle is
not possible. Then, the static pressure at the nozzle exit will be
256
Gas Turbines
p5 = pc
=
p04
1
p04 /pc
pc
p04
γg −1
γg
T5 = Tc
=
T04
Tc
=
892.86 K
R
=
Cp (γ − 1)
γ
=
284.6 J/kg K
cj
ρc
= 2.2 ×
1
= 1.160 bar
1.895
1
1.895
= 1046.44 ×
0.33
1.33
1147 × 0.33
1.33
=
√
1.33 × 284.6 × 892.86
=
γRTc
=
=
581.34 m/s
=
1.160 × 105
284.6 × 892.86
3
=
0.456 kg/m
Let A be the required area of the propelling nozzle, then,
25
ṁ
= 0.0943 m2
=
A
=
ρ c cj
0.456 × 581.34
If the nozzle is circular, then
π 2
d
=
0.0943
4
d
=
4
× 0.0943
π
=
0.3466
=
34.66 cm
Actual diameter would have to be slightly greater than this owing to the
effect of the boundary layer at the nozzle outlet. Now,
Pressure thrust
=
(pc − pa )A
=
(1.16 − 1) × 105 × 0.0943
=
25 × 583.11
=
1508.8 N
Momentum thrust
=
14577.75 N
Therefore, total thrust per second,
=
sf c
=
=
14577.75 + 1508.8
=
16086.55 N
Ans
⇐=
ṁa
A/F
Total thrust
25 × 3600
50
= 0.111 kg/N thrust h
16086.55
Ans
⇐=
Jet Propulsion Cycles and Their Analysis
257
7.3 In an aircraft power plant, the gas expands through a turbine to
an intermediate pressure and on leaving the turbine it expands from
intermediate pressure to the back pressure, generating kinetic energy
for jet. All the power of turbine is absorbed in driving the associated
compressor.
In such a unit gas enters the turbine at 4.5 bar and 800◦ C and expands
therein to 1.75 bar. Turbine absorbs 75% of the available enthalpy
drop. Expansion occurs through the jet from exhaust condition to
1.03 bar. Assume that the velocity of gas entering the turbine and jet
(nozzle) is negligible. There is no heat loss and conversion of kinetic
energy is 100% of the available adiabatic enthalpy drop. Calculate (i)
temperature of the gas entering the jet (nozzle), and (ii) velocity of
the gas leaving the jet. Assume Cp = 1.05 kJ/kg K and γ = 1.38.
Solution
03
04
04’
T
02
5
5’
02’
01
s
Fig. 7.27
Temperature of gas entering the nozzle,
T04
=
T03 1 − ηT 1 −
=
1073 ×
!
=
888.92 K
1
r
γ−1
γ
1 − 0.75 1 −
"
1
4.5 0.275
1.75
Ans
⇐=
Velocity leaving the nozzle,
cj
=
2Cp (T04 − T5 )
=
.
/
/
02 × 1.05 × 888.92 × 1 −
=
503.2 m/s
1
1.75 0.2754
1.03
Ans
⇐=
258
Gas Turbines
7.4 The effective jet exit velocity from a jet engine is 2700 m/s. The
forward flight velocity is 1350 m/s and the air flow rate is 78.6 kg/s.
Calculate
(i) thrust,
(ii) thrust power, and
(iii) propulsive efficiency.
Solution
α
=
ci
cj
Thrust, F
=
ṁa (cj − ci )
=
78.6 × (2700 − 1350)
=
106110 N
=
F × ci
=
106110 × 1350
=
143.25 × 106 W
=
2α
α+1
=
2 × 0.5
1.5
=
66.67%
Thrust power
ηp
=
1350
2700
=
=
0.5
Ans
⇐=
0.6667
Ans
⇐=
7.5 The following data apply to a turbojet aircraft flying at an altitude
of 6.1 km where the ambient conditions are 0.458 bar and 248 K.
Speed of aircraft : 805 km/h
Pressure ratio of compressor : 4:1
Combustion chamber pressure loss : 0.21 bar
Turbine inlet temperature : 1100 K
Intake duct efficiency : 95%
Isentropic efficiency of compressor : 0.85
Isentropic efficiency of turbine : 0.90
Mechanical efficiency of transmission : 99%
Nozzle efficiency : 95%
Nozzle outlet area : 0.0935 m2
L.C.V. of fuel : 43 MJ/kg
Find the thrust and specific fuel consumption in kg/Nh of thrust.
Assume convergent nozzle. Take Cpa = 1.005 kJ/kg K and γ =
1.4, Cpg = 1.147 kJ/kg K and γ = 1.33.
Jet Propulsion Cycles and Their Analysis
03
04’
T
02
02’
01’
04
5
5’
01
a
s
Fig. 7.28
Solution
Speed of aircraft
=
p01
805 × 1000
3600
!
=
223.6 m/s
γ
γ−1
Ci2
1+
2Cp Ta
=
pa 1 + ηram
=
0.458× 1+0.95×
=
0.631 bar
p02
=
r × p01
T01
=
Ta +
=
272.87 K
T02
=
T01 +
429.08
=
272.87 +
T03 − T04
=
Cpa (T02 − T01 )
Cpg ηm
=
1.005 × (429.08 − 272.87)
1.147 × 0.99
=
138.25 K
T04
=
1100 − 138.25
p03
=
p02 − 0.21 = 2.52 − 0.21 = 2.31 bar
7
Ci2
2Cp
=
−1
2
223.6
1+ 2×1.005×248×1000
4 × 0.631
=
248 +
T01 γ−1
r γ −1
ηC
=
3.5
−1
223.62
2 × 1005
=
8
2.524 bar
429.08 K
272.87
× 40.286 − 1
0.85
=
"
961.75 K
259
260
Gas Turbines
T04
p04
T03 − T04
ηT
=
T03 −
=
946.389 K
=
p03
=
1.26 bar
=
1100 −
γ
γ−1
T04
T03
= 2.31 ×
138.25
0.90
946.389
1100
4.03
The nozzle pressure ratio is
p04
p04
=
p5
pa
=
1.26
0.458
=
2.751
The critical pressure ratio
1
=
1
ηn
1−
1
=
1
0.95
1−
p04
pc
=
γg
γg −1
γg −1
γg +1
4.03
1.33−1
1.33+1
1.917
Thus nozzle is choking.
γg −1
γg
pc
p04
T5 = Tc
=
T04
Tc
=
818.35 K
p5 = pc
=
1.26
p04
=
= 0.657 bar
1.917
1.917
ρ5
=
ρc
=
0.282 kg/m
=
γg RTc
cj
ṁ
=
pc
RTc
= 961.75 ×
=
1
1.917
0.33
1.33
0.657 × 105
284.6 × 818.35
3
=
√
1.33 × 0.2846 × 1000 × 818.35
=
556.56 m/s
=
ρc Ac cj
=
14.67 kg/s
=
0.282 × 0.0935 × 556.56
Jet Propulsion Cycles and Their Analysis
261
Total thrust
F
=
ṁa (cj − ci ) + A5 (pc − pa )
=
14.67 × (556.56 − 223.6) +
0.0935 × (0.657 − 0.458) × 105
ṁf
sf c
Ans
=
6745.17 N
=
ṁCp (T03 − T02 )
CV
=
14.67 × 1.147 × (1100 − 429.08)
43000
=
0.263 kg/s
=
ṁf
F
=
0.140 kg/N h
=
⇐=
0.263 × 3600
6745.17
Ans
⇐=
7.6 A turboprop aircraft is flying at 600 km/h at an altitude where the
ambient conditions are 0.458 bar and −15◦C. Compressor pressure ratio 9:1. Maximum gas temperature 1200 K. The intake duct efficiency
is 0.9 and total head isentropic efficiency of compressor and turbine
is 0.89 and 0.93 respectively. Calculate the specific power output in
kJ/kg, thermal efficiency of the unit taking mechanical efficiency of
transmission as 98% and neglecting the losses other than specified.
Assume that exhaust gases leave the aircraft at 600 km/h relative to
the aircraft. (Hint : No thrust as the unit is turboprop)
Solution
03
04’
T
02
02’
01’
04
5
5’
01
a
s
Fig. 7.29
ci
=
600 ×
5
18
=
166.67 m/s
262
Gas Turbines
5
18
=
166.67 m/s
c2i
2Cp
=
258 +
cj
=
600 ×
T01
=
Ta +
=
271.82 K
p01
=
166.672
2 × 1005
T01
γ
γ−1
T01
Ta
pa
=
0.5497 bar
=
0.9 × (0.5497 − 0.458) + 0.458
=
0.5405 bar
p02
=
9 × 0.5405
T02
=
T01 (r)
=
271.82 × 90.286
=
271.82 +
T04
=
T03
p04
p03
T04
=
1200 − 0.93 × (1200 − 695.87)
WN
=
WT −
=
Cpg (T03 − T04 ) −
=
1.147 × (1200 − 731.16) −
=
263.76 kJ/kg
=
263.76
× 100
1.147 × (1200 − 539)
p01
T02
ηth
=
271.82
258
=
=
0.458 ×
3.5
4.865 bar
γ−1
γ
=
509.60 K
509.60 − 271.82
0.89
γ−1
γ
= 1200 ×
=
1
9
539 K
0.248
= 695.87 K
=
731.16 K
WC
ηm
Cpa (T02 − T01 )
ηm
1.005 × (539 − 271.82)
0.98
Ans
⇐=
=
34.79%
Ans
⇐=
7.7 In a turbojet unit with forward facing ram intake, the jet velocity
relative to the propelling nozzle at exit is twice the flight velocity.
Determine the rate of fuel consumption in kg/s, when developing a
thrust of 25000 N under the following conditions.
Jet Propulsion Cycles and Their Analysis
Ambient pressure and temperature
Compression total head pressure ratio
Flight speed
CV of fuel
Ram efficiency
Isentropic efficiency of compressor
Isentropic efficiency of turbine
Isentropic efficiency of nozzle
Combustion efficiency
Turbine pressure ratio
:
:
:
:
:
:
:
:
:
:
263
0.7 bar; 1◦ C
5:1
800 km/h
42000 kJ/kg
100%
85%
90%
95%
98%
2.23
Assume the mass flow of fuel is small compared with the mass flow
of air and that the working fluid throughout has the properties of air
at low temperature. Neglect the extraneous pressure drop. Assume
Cpg = Cpa = 1.005 kJ/kg K.
Solution
03
04’
T
02
02’
01’
04
5
5’
01
a
s
Fig. 7.30
ci
=
800 km/h
cj
=
2 × ci = 2 × 222.22 = 444.44 m/s
T01
=
Ta +
=
298.55 K
T02 − T01
=
T01 (γ−1)/γ
r
−1
ηC c
T02 − 298.55
=
298.55
× 50.286 − 1
0.85
T02
=
503.9 K
p01
=
pa
c2i
2Cp
=
222.22 m/s
=
=
c2i
1+
2Cp Ta
274 +
222.222
2 × 1005
T01
γ
γ−1
=
205.31
264
Gas Turbines
222.222
2 × 274 × 1005
3.5
=
0.7 × 1 +
ηr
=
p01 − pa
p01 − pa
p01
=
1 × (0.945 − 0.7) + 0.7
=
0.945 bar
p02
=
5 × 0.945
T03 − T04
=
T02 − T01 = 503.9 − 298.6 = 205.3 K
T03 − T04
=
ηT T03 1 −
=
0.9 × T03 × 1 −
205.3
=
0.1845 × T03
T03
=
1112.73 K
F
=
ṁ(cj − ci )
25000
=
ṁ × (444.44 − 222.22)
ṁa
=
112.5 kg/s
=
= 0.945 bar
4.725 bar
1
rt0.286
1
2.230.286
Ans
⇐=
Neglecting the effect of mass addition of fuel on the right hand side.
ṁf CV
=
ṁa Cp (T03 − T02 )
ṁf
=
112.5 × 1.005 × (1112.73 − 503.9)
42000
=
1.64 kg/s
Ans
⇐=
7.8 Under take-off conditions when the ambient pressure and temperature
are 1.01 bar and 288 K, the stagnation pressure and temperature in
the jet pipe of a turbojet engine are 2.4 bar and 1000 K, and the
mass flow is 23 kg/s. Assuming that the expansion in the converging
propelling nozzle is isentropic, calculate the exit area required and
the thrust produced.
For a new version of the engine the thrust is to be increased by the
addition of an aft fan which provides a separate cold exhaust stream.
The fan has a by-pass ratio of 2.0 and a pressure ratio of 1.75, the
isentropic efficiencies of the fan and fan-turbine sections being 0.88
and 0.90 respectively. Calculate the take-off thrust assuming that the
Jet Propulsion Cycles and Their Analysis
265
expansion in the cold nozzle is also isentropic, and that the hot nozzle
area is adjusted so that the hot mass flow remains at 23 kg/s.
Solution
As expansion in the nozzle is isentropic, the nozzle efficiency is 100%. Then,
pc
γ
γ−1
2
γ+1
=
p04
=
1.2969 bar
=
2.4 ×
2
2.33
4.03
Since pc > pa the nozzle will choke. Therefore, all the variables are calculated with respect to throat conditions.
Tc
=
T04
2
γ+1
ρc
=
pc
RTc
=
=
0.531 kg/m3
cj
=
1000 ×
2
= 858.37 K
2.33
1.2969
284.6 × 858.37
√
1.33 × 284.6 × 858.37
=
γRT
=
=
570 m/s
=
ṁ
ρ c cj
=
0.076 m2
=
ṁcj + Ac (pc − p1 )
=
23 × 570 + 0.076 × 1.2969 × 105 − 1.01 × 105
=
15290.47 N
T02
=
T01
p02
p1
T02
=
T01 +
T02 − T01
ηf
=
288 +
338 − 288
0.88
At nozzle exit,
A
Total thrust
=
23
0.9824 × 570
Ans
⇐=
Ans
⇐=
Aft fan engine
Cold nozzle
γ−1
γ
= 288 × 1.750.286
=
344.8 K
=
338 K
266
Gas Turbines
Addition of an aft fan does not affect the main flow. Cold nozzle thrust
adds to the hot nozzle thrust.
Flow through cold nozzle
=
2 × 23
=
46 kg/s
Assuming nozzle efficiency to be 100%,
pc
Thrust (cold)
2
γ+1
γ
γ−1
=
p02
=
(1.01 × 1.75) ×
=
0.93374 bar
=
46 ×
=
15542.8 N
2
1.4 + 1
3.5
(below ambient)
2 × 1005 × (344.8 − 288)
Fan turbine
23 × 1.147 × (1000 − T05 )
=
46 × 1.005 × (344.8 − 288)
T05
=
900.5 K
T05
=
1000 −
=
889.44 K
p05
=
2.4 ×
889.44
1000
pc
=
1.5 ×
2
1.33 + 1
=
0.811 bar (subcritical)
=
ṁ
=
23 ×
=
27263.25 N
Fcold
=
15542.8 N
Take-off thrust
=
15542.8 + 27263.25
=
42806.05 N
Fhot
1000 − 900.5
0.9
4.03
=
1.5 bar
4.03
2Cp (T05 − T06 )
2 × 1147 × (900.5 − 288)
Ans
⇐=
Jet Propulsion Cycles and Their Analysis
267
7.9 A jet propelled plane consuming air at the rate of 18.2 kg/s is to fly at
Mach number 0.6 at an altitude of 4500 m (pa = 0.55 bar, Ta = 255
K). The diffuser which has a pressure coefficient of 0.9, decreases the
flow to a negligible velocity. The compressor pressure ratio is 5 and
maximum temperature in the combustion chamber is 1273 K. After
expanding in the turbine, the gases continue to expand in the nozzle
to a pressure of 0.69 bar. The isentropic efficiencies of compressor,
turbine and nozzle are 0.81, 0.85 and 0.915 respectively. The heating
value of the fuel is 45870 kJ/kg. Assuming Cp = 1.005 kJ/kg K, Cpg
= 1.147 kJ/kg K. γair = 1.4, γgas = 1.33, calculate
(i) power input to the compressor
(ii) power output of the turbine
(iii) the fuel-air ratio
(iv) the thrust provided by the engine and
(v) the thrust power developed.
Solution
03
04’
T
02
02’
01’
04
5
5’
01
a
s
Fig. 7.31
=
Mi
=
192.05 m/s
=
T01
=
Ta +
c2i
2Cp
=
255 +
192.052
2 × 1005
=
p01
pa
=
T01
Ta
p01
=
0.55 ×
ci
T01
γRTa
=
0.6 ×
√
1.4 × 287 × 255
273.35 K
γ
γ−1
273.35
255
3.5
=
0.701 bar
268
Gas Turbines
ηram
=
p01 − pa
p01 − pa
p01
=
0.9 × (p01 − pa ) + pa
=
0.9 × (0.701 − 0.55) + 0.55 = 0.686 bar
p02
p01
=
rc
p02
=
5 × 0.686
T02
=
T01 1 +
=
273.35 × 1 +
=
ṁa Cp (T02 − T01 )
=
18.2 × 1.005 × (470.62 − 273.35)
=
3608.26 kW
WC
=
=
0.9
5
=
3.43 bar
γ−1
1
rc γ − 1
ηC
1
50.286 − 1
0.81
= 470.62 K
Ans
⇐=
Power output of the turbine = Power input to the compressor
=
WC
Fuel-air ratio
=
(ṁa + ṁf )Cpg (T03 − T02 )
=
(18.2 + ṁf ) × 1.147 × (1273 − 470.62)
=
ṁf × 45870
ṁf
=
0.373 kg/s
f
=
0.373
18.2
WT
=
(ṁa + ṁf )Cpg (T03 − T04 )
3608.26
=
(18.2 + 0.373) × 1.147 × (1273 − T04 )
T04
=
1103.62 K
⎡
⎛
T04
=
=
Ans
WT
3608.26 kW
=
⇐=
=
ṁf CV
Ans
0.0205
T03 ⎣1 − ηT ⎝1 −
⎞⎤
1 ⎠⎦
γ−1
γ
rt
⇐=
1103.62
=
1273 × 1 − 0.85 × 1 −
rt
=
1.99
1
rt0.248
Jet Propulsion Cycles and Their Analysis
p03
p04
=
1.99
p04
=
1.724 bar
p04
pa
=
1.724
0.55
=
269
3.13
As nozzle efficiency is given, the critical pressure ratio is given by
p04
pc
1
=
=
γ
γ−1
γ−1
γ+1
1−
1
ηn
1−
1
0.915
1
×
4.03
0.33
2.33
=
1.97
Critical pressure, pc
=
1.724
1.97
=
0.875 bar
The nozzle will be choking
Tc
cj
pc
p04
=
T04
=
932.73 K
γg −1
γg
0.33
1.33
√
1.33 × 284.6 × 932.73
=
γRTC
=
594.18 m/s
=
pc
RTC
=
0.875 × 105
284.6 × 932.73
ṁa + ṁf
=
ρc An cj
18.2 + 0.373
=
0.33 × An × 594.18
An
=
0.0947 m2
Thrust, F
=
[(ṁa + ṁf )cj − ṁa ci ] + [An (pc − pa )]
=
[(18.2 + 0.373) × 594.18 − 18.2 × 192.05] +
ρc
=
= 1103.62 ×
1
1.97
=
3
0.33 kg/m
0.0947 × (0.875 − 0.55) × 105
=
10618 N
Ans
⇐=
270
Gas Turbines
Thrust power, Fp
=
F × ci
=
2060 kW
10726.7 × 192.1 × 10−3
=
Ans
⇐=
7.10 Air enters a turbojet engine at a rate of 12×104 kg/h at 15◦ C and 1.03
bar and is compressed adiabatically to 182◦ C and four times the pressure. Products of combustion enter the turbine at 815◦ C and leave
it at 650◦C to enter the nozzle. Calculate the isentropic efficiency of
the compressor, the power required to drive the compressor, the exit
speed of gases and thrust developed when flying at 800 km/h. Assume
the isentropic efficiency of turbine is same as that of the compressor
and the nozzle efficiency 90%.
Solution
03
04
04’
T
5
5’
02
02’
01
s
Fig. 7.32
T02 − T01
=
T01 ( γ−1
)
rc γ − 1
ηC
455 − 288
=
288
× 40.286 − 1
ηC
ηC
=
84%
ṁa
=
12 × 104
3600
WC
=
ṁa Cp (T02 − T01 ) = 33.33 × 1.005 × (455 − 288)
=
5594 kW
T03 − T04
=
ηT T03 1 −
1088 − 923
=
1088 × 0.84 × 1 −
Compressor work
=
Ans
ηT
=
⇐=
33.33 kg/s
Ans
⇐=
1
rt0.248
1
rt0.248
Jet Propulsion Cycles and Their Analysis
rt
=
2.23
p04
=
p03
rt
p04
pa
=
1.8475
= 1.79
1.03
p04
pc
=
=
4.12
2.23
=
=
271
1.8475 bar
1
1−
1
ηn
1−
1
0.9
γ
γ−1
γ−1
γ+1
1
×
0.33
2.33
4.03
=
1.99
Nozzle will not choke.
T5
T04
=
p5
p04
γ−1
γ
1.03
1.8478
0.33
1.33
T5
=
923 ×
ηn
=
T04 − T5
T04 − T5
T5
=
T04 − ηn (T04 − T5 )
=
923 − 0.9 × (923 − 798.46)
cj
F
=
=
798.46 K
=
810.91 K
2 × 1147 × (923 − 810.91)
=
507.1 m/s
=
ṁ(cj − ci )
=
9495.72 N
Ans
⇐=
=
33.33 × (507.1 − 222.2)
Ans
⇐=
Review Questions
7.1 What is meant by jet propulsion? What are the basic differences between jet propulsion cycle and shaft power cycle.
7.2 Explain the principle of jet propulsion and mention how the jet propulsion engines are classified.
7.3 What do you understand by the term ’air-breathing engines’? How
are they classified?
7.4 With the aid of a neat diagram, explain the working principle of a
ramjet engine.
272
Gas Turbines
7.5 Draw the thermodynamic cycle of the ramjet engine and derive the
equation for thrust.
7.6 What are the advantages and disadvantages of a ramjet engine and
what are its applications?
7.7 With the aid of a schematic diagram, explain the working principle of
pulse jet engine and also draw the ideal and actual p-V diagram.
7.8 Mention the various advantages and disadvantages of the pulse jet
engine.
7.9 With the aid of the schematic diagram and thermodynamic cycle, explain the working of a turboprop engine.
7.10 Mention the various advantages and disadvantages of a turboprop engine and also bring out the applications.
7.11 With a neat sketch and T -s diagram, explain the working of turbojet
engine and also derive the expression for the thrust developed.
7.12 Explain with suitable graphs the performance of a turbojet engine.
What are the advantages and disadvantages of a turbojet engine?
7.13 What is meant by thrust? Derive the thrust equation for a general
propulsion system.
7.14 Explain the various efficiencies associated with a propulsion device.
7.15 Explain clearly the various factors affecting the performance of a
propulsion device.
7.16 What is meant by thrust augmentation and explain how it is effected.
7.17 Explain the principle of afterburner in thrust augmentation.
7.18 Draw the performance chart of an afterburner engine and explain.
7.19 How does water-alcohol mixture injection help thrust augmentation?
Explain.
7.20 Explain the principle of bleed burn cycle for thrust augmentation.
Exercise
[Note: Take γaγ−1
= 0.286 and
a
stated otherwise]
γg −1
γg
= 0.248 for all problems unless
7.1 The exit velocity from a jet unit is 650 m/s for an air flow of 40
kg/s through the unit. The aircraft is flying at 250 km/h. Calculate
the thrust developed, the thrust power and the propulsion efficiency.
Neglect the effect of fuel.
Ans: (i) 23222.4 N (ii) 1612.56 kW (iii) 19.3%
Jet Propulsion Cycles and Their Analysis
273
7.2 A simple jet engine has compressor directly coupled to the turbine
mounted in an aircraft with forward facing intake and rearward convergent propelling nozzle. Calculate the total thrust when the aircraft
flies at true air speed of 300 m/s in the ambient total conditions of
−10◦ C and 0.58 bar.
Air mass flow rate
Compressor stagnation pressure ratio
Turbine inlet stagnation temperature
Combustion chamber loss in stagnation
Compressor stage efficiency
Turbine stage efficiency
Combustion efficiency
Ram efficiency
Nozzle efficiency
Mechanical efficiency
:
:
:
:
:
:
:
:
:
:
39 kg/s
7.5 : 1
650◦ C
4%
82%
85%
100%
90%
100%
100%
Ans: 9683.96 N
7.3 In a turbojet unit with forward facing ram intake, the jet velocity
relative to the propelling nozzle at exit is twice the flight velocity.
Determine the rate of fuel consumption in kg/s, when developing a
thrust of 25000 N under the following conditions.
Ambient pressure
Ambient temperature
Compression total head pressure ratio
Pressure at exit
Flight speed
L.C.V. of fuel
Ram efficiency
Isentropic efficiency of compressor
Isentropic efficiency of turbine
Isentropic efficiency of nozzle
Combustion efficiency
:
:
:
:
:
:
:
:
:
:
:
0.7 bar
1◦ C
5:1
0.7 bar
233 m/s
43 MJ/kg
100%
85%
90%
95%
98%
Neglect the effect of fuel in total mass flow rate.
Ans: 0.794 kg/s
7.4 A naval aircraft is powered by a turbojet engine, with provision for
flap blowing. When landing at 55 m/s, 15 % of the compressor delivery is bled off for flap blowing and it can be assumed to be discharged
perpendicularly to the direction of flight. If a propelling nozzle area
of 0.13 m2 is used, calculate the net thrust during landing given that
the engine operating conditions are as follows
274
Gas Turbines
Compressor pressure ratio
Compressor isentropic efficiency
Turbine inlet temperature
Turbine isentropic efficiency
Combustion pressure loss
Nozzle isentropic efficiency
Mechanical efficiency
Ambient conditions
:
:
:
:
:
:
:
:
9.0
0.82
1275 K
0.87
0.45 bar
0.95
0.98
1 bar, 288 K
The ram pressure and temperature rise can be regarded as negligible.
Ans: 18489.6 N
7.5 The following data apply to a twin-spool turbofan engine, with the
fan driven by the LP turbine and the compressor by the HP turbine.
Separate cold and hot nozzles are used.
Overall pressure ratio
Fan pressure ratio
By-pass ratio mc /mh
Turbine inlet temperature
Mechanical efficiency of each spool
Combustion pressure loss
Total air mass flow
:
:
:
:
:
:
:
19.0
1.65
3.0
1300 K
0.99
1.25 bar
115 kg/s
It is required to find the thrust under sea level static conditions where
the ambient pressure and temperature are 1.0 bar and 288 K. Assume
the fan, compressor and turbine polytropic efficiency as 0.90 and the
isentropic efficiency of each propelling nozzle as 0.95. The values of
n−1
n for the polytropic compression and expansion are
For compression,
n−1
n
=
1
ηpc
γ−1
γ
a
ηpt
γ −1
γ
g
=
0.3175
=
0.225
For expansion,
n−1
n
=
Draw the schematic as well as T -s diagrams.
Ans: 38892.3 N
7.6 Extend the above example on the turbofan, with the additional information that the combustion efficiency is 0.99, determine the sf c.
Also, calculate the thrust and sf c when a combustion chamber is
incorporated in the by-pass duct and the ‘cold’ stream is heated to
1000 K. The combustion efficiency and pressure loss for this process
may be assumed to be 0.97 and 0.05 bar respectively. Assume CV =
42 MJ/kg.
Ans: (i) 1.135 ×10−5 kg/N s (ii) 55823.17 N
(iii) 3.672 ×10−5 kg/N s
Jet Propulsion Cycles and Their Analysis
275
7.7 A turbojet engine is traveling at 920 km/h at standard sea-level conditions. The ram efficiency is 87%, the compressor pressure ratio is
4.3:1, the compressor efficiency is 82%, the burner pressure loss is
2%, the fuel-air ratio is 0.0119, the turbine inlet total temperature is
688◦ C. The turbine efficiency is 83.5%. Calculate
(i) the specific gross thrust and
(ii) the thrust specific fuel consumption.
Ans: (i) 255.05 N (ii) 0.168 kg/N h
7.8 A turbojet engine inducts 51 kg of air per second and propels an
aircraft with an uniform flight speed of 912 km/h. The isentropic
enthalpy change for the nozzle is 200 kJ/kg and its velocity coefficient
is 0.96. The fuel-air ratio is 0.0119, the combustion efficiency is 0.96
and the lower heating value of the fuel is 42 MJ/kg. Calculate
(i) the thermal efficiency of the engine,
(ii) the fuel flow rate in kg/h and T SF C,
(iii) the propulsive power in kW
(iv) the thrust power, and
(v) the propulsive efficiency.
Ans: (i) 30.9% (ii) 2184.84 kg/h; 0.1186 kg/N h (iii) 7876.16 kW
(iv) 4664.6 kW (v) 59.22%
7.9 A turbojet engine is being used to propel an aeroplane. The drag is
3900 N. The coefficient of drag is 0.01835. The wing area is 21.25 m2 .
The air consumption per second of the engine is 14.5 kg/s and the
thrust developed is 8900 N. Calculate the flight velocity and effective
jet velocity. Also calculate the specific thrust where the Cp = 1.005
kJ/kg K. What is the density ratio at this altitude of 10000 m. Take
ρ = 0.5 kg/m3 at this altitude.
Ans: (i) 200 m/s (ii) 813.8 m/s (iii) 613.8 N (iv) 0.41
7.10 In a theoretical cycle for jet propulsion unit both the compression
and expansion are considered isentropic, and the heat is supplied at
constant-pressure. Show that the thrust developed per kg of air per
second when the velocity of approach is neglected is
γ−1
γ
2Cp T (q − 1) rp
1
2
−1
where q is the ratio of absolute temperatures after and before combustion, rp is the compressor pressure ratio and T is the absolute
atmospheric temperature.
276
Gas Turbines
7.11 A simple turbojet is operating with a compressor pressure ratio of
8.0, a turbine inlet temperature of 1200 K, polytropic efficiencies of
compressor and turbine of 0.87 and a mass flow rate of 15 kg/s, when
the aircraft is flying at 260 m/s at an altitude of 7000 m. Assuming
the following component efficiencies, and I.S.A. conditions, calculate
the propelling nozzle area required, the net thrust developed and the
thrust specific fuel consumption.
Isentropic efficiency of intake
Isentropic efficiency of propelling nozzle
Mechanical efficiency
Combustion chamber pressure loss
Calorific value of the fuel
Combustion efficiency
:
:
:
:
:
:
0.95
0.95
0.99
6 % of p02
43 MJ/kg
0.97
Ans: (i) 0.069 m2 (ii) 7893.4 N (iii) 0.124 kg/h N
7.12 A turboprop aircraft is flying at 800 km/h at an altitude where the
ambient conditions are 0.567 bar and −20◦C. Compressor pressure ratio 8:1. Maximum gas temperature 1100 K. The intake duct efficiency
is 0.95 and total head isentropic efficiency of compressor and turbine
is 0.92 and 0.95 respectively. Calculate the specific power output in
kJ/kg of air and the thermal efficiency of the unit taking mechanical
efficiency of transmission as 0.96 and neglecting the losses other than
specified. Assume that exhaust gases leave the aircraft at 800 km/h
relative to the aircraft. (Hint : No thrust as the unit is turboprop)
Ans: (i) 226.23 kJ/kg (ii) 34.17%
Multiple Choice Questions (choose the most appropriate answer)
1. The useful power output for the propulsion cycle is from
(a) inlet diffuser
(b) outlet nozzle
(c) turbine
(d) compressor
2. The working of jet propulsion is based on
(a) impulse principle
(b) centrifugal principle
(c) reaction principle
(d) centripetal principle
3. The difference between air breathing engine and rocket engine are
(a) there is altitude limit
(b) there is temperature limit
Jet Propulsion Cycles and Their Analysis
277
(c) atmospheric air is used for formation of the jet
(d) all of the above
4. Reciprocating engines are not used in modern aircrafts due to
(a) high specific weight
(b) high speed requirement
(c) large drop in power with altitude
(d) all of the above
5. Ramjet engines are highly suited for
(a) subsonic application
(b) supersonic application
(c) hypersonic application
(d) none of the above
6. Pressure rise in a ramjet is achieved by a
(a) diffuser
(b) centrifugal compressor
(c) axial flow compressor
(d) none of the above
7. Compared to a ramjet engine, a pulse jet engine has
(a) higher propulsive efficiency
(b) no altitude limitation
(c) larger life
(d) none of the above
8. Turbo prop engines have
(a) lower thrust at take-off
(b) less frontal area
(c) difficult to maintain
(d) none of the above
9. In a turbojet engine compared to momentum thrust, pressure thrust
is
(a) quite low
(b) quite high
(c) almost equal
(d) none of the above
278
Gas Turbines
10. The propulsive efficiency of a turbo jet engine as a function of speed
ration, α is given by
(a)
(b)
(c)
(d)
1+α
2α
1−α
2α
2α
1+α
2α
1−α
Ans:
1. – (b)
6. – (a)
2. – (c)
7. – (d)
3. – (d)
8. – (b)
4. – (d)
9. – (a)
5. – (b)
10. – (c)
8
CENTRIFUGAL
COMPRESSORS
INTRODUCTION
The second World War was the turning point for the development of
gas turbine technology. The most rapid progress in the development of
gas turbines was made during this period with the use of the centrifugal compressors. Attention was focused on the simple turbojet unit and
the power to weight ratio was one of the most important considerations.
Centrifugal compressor was the best possible type available at that time.
Development time was another factor in the perfection of gas turbine design. Much expertise was derived from the experience gained on the design
of small high-speed centrifugal compressors for supercharging reciprocating
engines. Since the war, however, the axial compressor has been developed
to the point where it has an appreciably higher isentropic efficiency. We
have already seen in the analysis of the practical cycles (Chapter 6) the importance of efficiency of each component from the point of view of overall
performance of the power plant.
A centrifugal compressor is one of its class of machines in producing
pressure rise and is known as turbo-compressors. In this type, energy is
transferred by dynamic means from a rotating member (or impeller) to
the continuously flowing working fluid. The main feature of the centrifugal
compressor is that, the angular momentum of the fluid flowing through the
impeller is increased partly by virtue of the impeller’s outlet diameter being
significantly larger than its inlet diameter. The centrifugal compressor may
be known as a fan, blower, supercharger, booster, exhauster or compressor;
the distinction between these types being very vague. Broadly speaking,
fans are classified as low-pressure compressors and blowers as mediumpressure compressors. Boosters, exhausters and superchargers are named
from their point of view of applications.
Although the centrifugal compressor is unlikely to be used in gas turbine
power plants where specific fuel consumption is the main criterion, it still
has certain advantages for some applications.
280
Gas Turbines
(i) It occupies a smaller length than the equivalent axial flow compressor.
(ii) It is not so liable to loss of performance by build up of deposits on
the surfaces of the air channels.
(iii) It can work reasonably well in a contaminated atmosphere compared
to axial flow machine.
(iv) It is able to operate efficiently over a wider range of mass flow rate
at any particular rotational speed.
This latter feature, viz., a wider range of mass flow rate matching a wide
range of operating conditions with turbine, makes centrifugal compressors
attractive compared to axial flow machines.
A pressure ratio of the order of 4:1 can be obtained from a single-stage,
manufactured using conventional materials. It has already been shown in
chapters 5 and 6 that this is adequate for a heat-exchange cycle when the
turbine inlet temperature is in the region of 1000-1200 K. Certainly, it can
find an application in small power units. It is mainly because the higher
isentropic efficiency of axial compressors cannot be maintained for very
small sizes of machines. Most current proposals for vehicular gas turbines
utilize a centrifugal compressor in a heat-exchange cycle. Materials such as
titanium now enable pressure ratios of over 6:1 to be used.
There is a renewed interest in the centrifugal stage, used in conjunction
with one or more axial stages, for small turbofan and turboprop aircraft
engines. The centrifugal compressor is less suitable when the cycle pressure
ratio requires the use of more than one stage in series because of the tortuous
path the air must follow between the stages. Nevertheless, a single stage
centrifugal compressor has been used successfully in turboprop engines.
8.1
ESSENTIAL PARTS OF A CENTRIFUGAL COMPRESSOR
The compression process is carried out in a centrifugal compressor, which
comprises mainly of four elements (Fig. 8.1):
(i) The inlet casing with converging nozzle, whose function is to accelerate
the fluid to the impeller inlet. The outlet of the inlet casing is known
as the eye.
(ii) The impeller, in which the energy transfer takes place, resulting in a
rise of fluid kinetic energy and static pressure.
(iii) The diffuser, whose function is to transform the high kinetic energy
of the fluid at the impeller outlet into static pressure.
(iv) The outlet casing, which comprises a fluid collector known as a volute
or scroll.
Further definitions are required to describe the impeller (Fig. 8.2) and
diffuser (Fig. 8.3). The various impeller components shown in Fig. 8.2 are
the following:
Centrifugal Compressors
281
Volute
3
2
Diffuser
Inlet casing with
converging nozzle
Impeller
0
1
Fig. 8.1 A centrifugal compressor stage
(i) The impeller vanes, help to transfer the energy from the impeller to
the fluid.
(ii) The hub, which is surface AB. [Fig. 8.2(a)].
(iii) The shroud, which is surface CD [Fig. 8.2(a)]. Impellers enclosed
on the surface CD are known as shrouded impellers, but the surface
CD is referred to as the shroud in descriptions of impeller geometry
whether the impeller is enclosed or not.
(iv) The inducer, the section EF in impellers of the form shown in Fig. 8.2(b)
whose function is to increase the angular momentum of the fluid without increasing its radius of rotation.
The diffuser may consist of any annular space [Fig. 8.3(a)] known as a
vaneless diffuser or may be in the form of a set of guide vanes, when it is
known as a vaned diffuser [Fig. 8.3(b)]. The main aim of providing diffusers
is to increase the static pressure by reducing the kinetic energy. Diffusers
will be discussed in greater detail in a later section.
To put it in a nutshell, a centrifugal compressor has essential two parts
of energy transformation:
(i) Rotating impeller which imparts a high velocity to the fluid and at the
same time increases the static pressure. Impellers are housed inside
a stationary casing.
(ii) A number of fixed diverging passages in which the air is decelerated
increasing the static pressure.
282
Gas Turbines
Impeller vanes
Shroud D B
C
Inducer
hub
E
F
A
A
C
hub
Shroud D B
(a)
(b)
Fig. 8.2 The impeller
Outlet diffuser
Vaneless diffuser
Volute
Impeller
Diffuser vanes
(a)
(b)
Fig. 8.3 Diffusers
Figure 8.4 is a schematic diagram of a centrifugal compressor. The
impeller may be designed as single-sided or double-sided as in Fig. 8.5 (a)
or 8.5 (b), but both operates on the same principle. The double-sided
impeller was required in early aero-engines because of the relatively small
flow capacity of the centrifugal compressor for a given overall diameter.
8.2
PRINCIPLE OF OPERATION
Air is sucked into the impeller eye through an accelerating nozzle and
whirled round at high speed by the vanes on the impeller disc (refer Fig. 8.4).
At any point, in the impeller, the flow experiences a centripetal acceleration
due to a pressure head. Hence, the static pressure of the air increases from
the eye to the tip of the impeller. The remainder of the static pressure rise
is obtained in the diffuser.
It may be noted that air enters the impeller eye with a very high velocity.
The friction in the diffuser will cause some loss in stagnation pressure. It
Centrifugal Compressors
o
90 bend taking air to
combustion chambers
283
284
Gas Turbines
Depth of diffuser
Centrifugal Compressors
3
2
p
0
Inlet
casing
1
Impeller
Diffuser
Channel
Fig. 8.6 Pressure rise across centrifugal compressor
p
02
02
p
03
03
p
3
03’
2
C2
h03 , h02
2
C3
2
3
3’
2
p
h
2
2’
2
p
01
h00 , h 01
00
p
01
00
2
C1
p
1
2
1’ 1
s
Fig. 8.7 Enthalpy–entropy diagram of a centrifugal compressor
285
286
Gas Turbines
converted to pressure (isentropically), the delivery pressure could be p02 .
Since the diffusion process is not accomplished isentropically, the process
(2–3) results, and some kinetic energy remains at diffuser exit (velocity c3 ),
the static delivery pressure at point 3 is p3 . The final state, in the collector,
has static pressure p3 , low kinetic energy c23 /2 , and a stagnation pressure
p03 which is less than p02 since the diffusion process is incomplete as well
as irreversible.
8.3
IDEAL ENERGY TRANSFER
Let us first consider the case of an ideal compressor with the following
assumptions for a radial vaned impeller:
(i) Losses due to friction are negligible.
(ii) Energy loss or gain due to heat transfer to or from the gas is considered very small.
(iii) The gas leaves the impeller with a tangential velocity equal to the
impeller velocity (i.e., ct2 = u2 ), no slip condition is assumed.
(iv) The air enters the rotor directly from the atmosphere without any
tangential component, i.e., ct1 = 0.
Applying these assumptions the Euler’s energy equation (Eq. 8.1) under
ideal conditions becomes
E
=
u22
(8.2)
This is the maximum energy transfer that is possible. Therefore, the work
done by the impeller on unit quantity of air is given by
W
=
E
=
u22
(8.3)
Equation 8.3 represents the maximum work capacity of a radial vaned
impeller for ideal operation. It may be noted that air cannot leave the impeller with a tangential velocity greater than u2 , and in practical machines
the work or energy absorbed by the fluid will be less than that given by
Eq. 8.3. Also from thermodynamic analysis we have the equation (refer
Eq. 3.21) for the energy transfer, E, as
W
=
E
=
h02 − h01
If rc is the pressure ratio based on total pressure
(8.4)
p02
p01
; we shall have
γ−1
W
=
Cp T01 rc γ − 1
(8.5)
If c1 ∼
= c2 , i.e., when there is no kinetic energy change between inlet and
exit we can refer to static conditions so that
Centrifugal Compressors
287
γ−1
W
Cp T1 rc γ − 1
=
p2max
p1
Here, rc is the static pressure ratio
u22
(8.6)
. From Eq. 8.3 and 8.5 we have
γ−1
Cp T01 rc γ − 1
=
(8.7)
where Cp is in kJ/kg K.
8.4
BLADE SHAPES AND VELOCITY TRIANGLES
In the previous section, we dealt with ideal energy transfer. In order to
understand the actual energy transfer and flow through the compressor,
we will use two velocity triangles, viz., entry velocity triangle and exit
velocity triangle. The notations used here correspond to the x, r and θ
coordinate system. As per the convention for radial machines, at a given
point the angles are measured from the tangential direction. The absolute
and relative air angles at the entry and exit of the impeller are denoted by
α1 , α2 and β1 , β2 respectively. Based on the value of β2 the blade shapes
are given the name as forward curved blades (β2 > 90◦ ), radial blades
(β2 = 90◦ ) and backward curved blades (β2 < 90◦ ) (refer Fig. 8.8). Note
that angle β2 is measured with respect to negative of u (i.e., −u). The
impeller linear velocities at inlet and exit are given by
β > 90
u1
=
2πr1 N
60
u2
=
2πr2 N
60
o
β = 90
2
o
2
Forward
Radial
β < 90
o
2
Backward
Fig. 8.8 Blade shapes
8.4.1
Entry Velocity Triangle
Figure 8.9(a) shows the flow at the entry of the inducer section of the
impeller without Inlet Guide Vanes (IGV). The absolute velocity (c1 ) of
288
Gas Turbines
the flow is axial (α1 = 90◦ ) and the relative velocity (w1 ) is at an angle β1
from the tangential direction. Thus the swirl or whirl component ct1 = 0.
c1
ca1
=
(8.8)
tan β1 =
u1
u1
Flow
Inducer section
of the impeller
u1
ca1= c1
w1
α1
ct1 = 0
u1
β1
(a) Without inlet guide vanes
Inducer
c1
α1
u1 = ct1
β1
w 1 = c a1
Inlet guide vanes
Entry
(b) With inlet guide vanes
Fig. 8.9 Flow through the inducer section
Figure 8.9(b) shows the flow through axially straight inducer blades
in the presence of IGVs. The air angle (α1 ) at the exit of the IGVs is
such that it gives the direction of the relative velocity vector (w1 ) as axial,
i.e., β1 = 90◦ . This configuration seems to offer some manufacturing and
aerodynamic advantages, viz.,
(i) centrifugal impellers with straight blades are much easier and cheaper
to manufacture and
(ii) the relative velocity (w1 ) approaching the impeller is considerably
reduced. In this case β1 = 90◦ and the positive swirl component is
ct1
tan α1
= u1
=
w1
u1
(8.9)
=
ca1
u1
(8.10)
Centrifugal Compressors
289
Figure 8.10 shows the entry and exit velocity triangles for impeller blades
located only in the radial section. For the sake of generality, the absolute
velocity vector c1 is shown to have a swirl component ct1 . However, if there
are no guide vanes, c1 will be radial (c1 = cr1 ) and α1 = 90◦ , ct1 = 0. This
particular condition is expressed by zero whirl or swirl at the entry and
would be assumed in this chapter unless otherwise mentioned.
c2
α2
c t2
c r2
w2
β2
u2
c1
α1
Impeller blade ring
c t1
cr
1
u1
β1
w1
Fig. 8.10 Velocity triangles for backward swept impeller blades (β2 < 90◦ )
8.4.2
Exit Velocity Triangle
The impeller shown in Fig. 8.10 are having backward swept blades, i.e.,
β2 < 90◦ . The exit velocity triangle is as shown in the figure. The flow
leaves the blades with a relative velocity w2 and at an air angle, β2 . The
absolute velocity of flow leaving the impeller is c2 at an air angle α2 . Its
tangential (swirl or whirl) component is ct2 and the radial component cr2 .
The following relations can be obtained from the velocity triangles as shown
in Fig. 8.10 at the entry and exit.
cr1
=
c1 sin α1
= w1 sin β1
ct1
=
c1 cos α1
= cr1 cot α1
cr2
=
c2 sin α2
= w2 sin β2
ct2
=
c2 cos α2
= cr2 cot α2
(8.11)
=
u1 − cr1 cot β1
(8.12)
(8.13)
=
u2 − cr2 cot β2
(8.14)
Figure 8.11 shows the velocity triangles at the entry and exit of a radialtipped blade extending into the inducer section. The velocity triangle at the
entry is similar to that in Fig. 8.10; here ca1 replaces the velocity component
cr1 . The exit velocity triangle here is only a special case of the triangle in
Fig. 8.10 with β2 = 90◦ . This condition when applied in Eqs. 8.13 and 8.14,
gives
290
Gas Turbines
c2
(at exit)
w 2 = c r2
β2
α2
c t2 = u 2
c1
α1
ca1 w1
β1
c t1
u1
(at inlet)
Impeller blade ring
Fig. 8.11 Velocity triangles for radial-tipped impeller with inducer blades,
(β2 = 90◦ )
cr2
ct2
=
=
w2
c2 cos α2
= c2 sin α2
= cr2 cot α2
(8.15)
=
u2
(8.16)
Figure 8.12 shows the velocity triangles for forward swept blades, (β2 > 90◦ )
with zero swirl at the entry. It may be observed that such blades have large
fluid deflection and give ct2 > u2 . This increases the work capacity of the
impeller and the pressure rise across it. But in practice, this configuration
is unsuitable for higher speeds and leads to higher losses.
8.5
ANALYSIS OF FLOW THROUGH THE COMPRESSOR
In the previous section we have seen various blade shapes and the corresponding inlet and exit velocity triangles to understand flow conditions. In
this section we will analyze the flow through the compressor from inlet to
exit. Ideal conditions cannot be achieved in practice and various losses do
occur in an actual compressor. The losses are the major factors responsible
for the decrease of efficiency in the centrifugal compressors. In order to understand the various losses it is necessary to understand the flow through
the compressor first.
In the early days of radial flow compressor development, the design
and theoretical analysis was mainly based on the assumption that the air
flow through the compressor may be considered one-dimensional. However, there are complications due to flow separation and the appearance
of shocks. Still, one-dimensional analysis provides the base for the initial
design of the components. The technique has been further refined over the
years by the introduction of design parameters or design criteria gained
from experimental results. Finally, the design can be modified with the
Centrifugal Compressors
291
c2
α2
u
2
β2
c
w2
cr2
t2
c1
Impeller blade ring
w1
u1
β1
Fig. 8.12 Velocity triangles for forward swept blades, with zero swirl at
entry (β2 > 90◦ )
help of three-dimensional flow analysis to obtain optimum compressor performance.
With the aid of the momentum equation, steady-flow energy equation,
continuity equation and thermodynamic equation of state, it is possible to
estimate the representative state of the air (that is, pressure, temperature,
velocity, Mach number etc.,) at any point in the compressor channel. The
assumptions in the estimation are that the fluid is a perfect gas and the flow
is isentropic and one-dimensional. This will help to determine the shape
of the compressor components (channel geometry) to produce desirable
velocity or Mach number profiles from inlet to exit. When air flows from
inlet to exit, it passes through the following, viz., (1) the inlet casing,
with the accelerating nozzle, (2) the impeller, (3) the diffuser and (4) the
volute. The details of various elements through which air flows with the
representative dimensions are shown in Fig. 8.13.
8.5.1
The Inlet Casing
The inlet casing consists of an accelerating nozzle with or without inlet
guide vanes. The function of the inlet casing is to deliver air to the impeller
eye with minimum loss and to provide a uniform velocity profile at the eye.
The inlet flange is axisymmetric and the inlet duct takes the form of a simple
accelerating convergent nozzle. Since there is no energy transfer, stagnation
enthalpy remains constant. Hence, from the h-s diagram (Fig. 8.7),
T1
=
T01 −
c21
2Cp
(8.17)
292
Gas Turbines
Volute (4)
Diffuser (3)
Casing
Shroud
Impeller eye
d2
Impeller (2)
2
Hub
dt
2
d1
2
Driving shaft
dh
Inducer section
2
IGV
Inlet casing with
accelerating nozzle (1)
Fig. 8.13 Details of airflow passage of a centrifugal compressor stage
With an air filter fitted to the compressor inlet casing, a pressure drop will
occur which must be kept to a minimum. However, it should be accounted
for, in the flow analysis and prediction of compressor performance.
Significant losses can occur in inlet casings fitted with silencing baffles.
Under this condition an inlet casing efficiency (ηic ) may be used. Similar
to the nozzle efficiency, ηic , can be defined as (Fig. 8.7)
ηic
=
h00 − h1
h00 − h1
(8.18)
where h00 − h1 is the enthalpy change in isentropic expansion from p00
to p1 .
For a perfect gas assumption (Cp = constant) Eq. 8.18 becomes
ηic
=
T00 − T1
T00 − T1
=
T00 − T1
T00
=
1
1−
⎡
T00 − T1 ⎢
⎣
T00
T1
T00
1
1−
p1
p00
T00 = T01 since h00 = h01 (refer Fig. 8.7). Now,
T01 − T1
=
c21
2Cp
Rearranging and substituting in Eq. 8.19 gives
(γ−1)/γ
⎤
⎥
⎦
(8.19)
Centrifugal Compressors
p1
p00
=
1−
c21
2Cp ηic T00
293
γ/(γ−1)
From values of p1 and T1 = T00 − c21 /2Cp , the density at the impeller eye
can be calculated.
p1
ρ1 =
RT1
and hence the mass flow rate entering the impeller eye can be estimated
π 2
ṁ =
d − d2h1 ca1 ρ1
(8.20)
4 t1
where dt1 and dh1 are the impeller eye tip and eye hub diameters respectively, and ca1 is the axial component of the absolute velocity of the air
entering the impeller eye. In the case of a swirl-free intake
ca1
=
c1
(8.21)
For compressor intakes, where the air must be turned from a radial to an
axial direction, the air incidence angle at the impeller eye (inducer) has
to be chosen very carefully. This is mainly because of the variation in the
axial velocity distribution caused by the free vortex flow effect in the bend
of the inlet ducting.
8.5.2
The Inducer
The inducer, or entry section of the impeller has a very pronounced effect
on the impeller performance and hence on over-all compressor efficiency.
The eye hub diameter (dh1 ) of the inducer is determined by a number of
design considerations such as inducer stress/vibration, impeller manufacturing techniques and the number of blades.
The tip of the inducer eye is the point where the highest inlet relative
Mach number occurs, and therefore careful attention should be paid to the
choice of the inducer eye tip diameter (dt1 ). For a given design mass flow
rate and impeller speed, dt1 should be chosen so as to obtain a minimum
relative Mach number (Fig. 8.14).
To further reduce the relative Mach number (Mt1 ), pre-whirl (inlet guide
vanes) may be used. However, the penalty will be a reduction in energy
transfer in the impeller. Thus the technique is usually used only on highpressure ratio compressors, where the inlet Mach number exceeds unity
and shock waves reduce impeller efficiency. The use of inlet guide vanes
has to be carefully evaluated for each application. Since there is a trade-off
between improved stage performance from the impeller and reduced stage
performance from the diffuser.
8.5.3
The Impeller
Energy transfer occurs in the impeller of the compressor, hence there is an
increase in stagnation enthalpy and pressure. To attain peak compressor
Gas Turbines
Mach number at eye tip
294
Diameter of eye tip
Fig. 8.14 Effect of eye tip diameter
efficiency great care must be taken so that very efficient diffusion processes
is achieved in the impeller and the diffuser. Since the diffusion processes
are related to the flow, a reasonable relative Mach number at the impeller
inlet, and a minimum absolute Mach number at the impeller outlet are to
be ensured.
8.5.4
The Effect of Impeller Blade Shape on Performance
As already mentioned in section 8.4 the various blade shapes utilized in
impellers of centrifugal compressors can be classified as
(i)
forward-curved blades
(β2 > 90◦ ),
(ii) radial-curved blades
(β2 = 90◦ ),
(iii) backward-curved blades (β2 < 90◦ ).
Figure 8.15 represents the variation of pressure-ratio that can be obtained with respect to mass flow rate for the above mentioned various shapes
for operation at a given rpm. The basic velocity diagram at exit is shown
for each type of blade is Fig. 8.16. For example, the Euler energy equation
without inlet swirl (ct1 = 0) is written as
E
=
u2 ct2
=
u2 (u2 − cr2 cot β2 )
(8.22)
For any particular impeller running at a constant speed, energy equation
can be written as
E
=
K1 − K2 Q
(8.23)
where Q is the volume flow rate which is proportional to flow velocity for
given impeller, and K1 and K2 are constants as given by
K1
=
u22
(8.24)
K2
=
u2 cot β2
πd2 b2
(8.25)
and
Centrifugal Compressors
295
Pressure ratio or head
Forward
Radial
Backward
Mass flow rate
Fig. 8.15 The effect of impeller blade shape on performance
c2
c2
w2
β2
u2
c t2
Forward
w2 = cr2
c2
cr2
cr2
β2
c t2 = u2
Radial
w2
β2
c t2
u2
Backward
Fig. 8.16 Exit velocity diagrams for different shapes of blades
The comparison of performance for the three types of vanes are made
for the same volume flow rate, each blade having unit depth, and for the
same vector value of cr2 . Figure 8.16 shows the exit velocity triangles for
three types of impellers from which the relative performance of the blades
can be evaluated.
Centrifugal effects of the curved blades create a bending moment and
produce increased stresses which reduce the maximum speed at which the
impeller can be run. Good performance can be obtained with radial impeller blades. Backward-curved blades are slightly better in efficiency and
are stable over a wider range of flows than either radial or forward-curved
blades. The forward-curved impeller can produce the highest pressure ratio
for a given blade tip speed; but is inherently less stable and has a narrow
operating range. Its efficiencies are lower than that are possible with the
backward-curved or radial-curved blades.
Although all the three types can be used in compressors, the radial
blade is used almost exclusively in turbojet engine applications.
296
8.5.5
Gas Turbines
The Impeller Channel
The relative velocity of the air in the impeller channels undergoes a rapid
deceleration from the impeller eye to the impeller tip. In addition to the
function of transferring energy to the air, the impeller should act as an
efficient diffuser. A badly shaped channel will interfere with the diffusion
process, causing flow separation at the impeller walls; leading to higher
losses. To obtain gradual deceleration of the flow in the channel, the geometry of the impeller channels should be such that it can provide a smooth
change in the relative Mach number along the mean flow path.
The flow in an impeller is not completely guided by the impeller vanes
and hence the effective fluid outlet angle does not equal to the impeller
vane outlet angle. To account for this deviation, a factor known as the
slip factor is used to correct the energy transfer calculated from simple
one-dimensional theory. It may be kept in mind that impeller flow is not
one-dimensional. Many analyses have been made of the velocity distribution within an impeller assuming the flow to be isentropic. Although such
solutions may depart from experimental velocity distributions, they are
useful in predicting regions in an impeller channel where losses may occur.
No quantitative criteria for boundary layer separation in impellers are at
present available and much more work is necessary to take these factors
into account.
It is a known fact that the vorticity of frictionless fluid does not change
with time. Hence, if the flow at the inlet to an impeller is irrotational,
the absolute flow must remain irrotational throughout the impeller. As the
impeller has an angular velocity, ω, the fluid must also have an angular
velocity −ω, relative to the impeller which is known as the relative eddy.
Thus if there were no flow through the impeller the fluid in the impeller
channels would rotate with an angular velocity equal and opposite to the
impeller’s angular velocity [Fig. 8.17(a)].
(a) Relative eddy
(b) Relative flow through the impeller
Fig. 8.17 Relative eddy and flow through the impeller
The flow through a rotating impeller can be considered as the vector sum
of the flow through the impeller passages – when the impeller is stationary –
and the flow produced in the fluid by the rotation of the impeller. It may be
Centrifugal Compressors
297
noted that the large amount of the mass of air flowing through the impeller
has certain inertia and due to the formation relative eddies, the velocity
of air at the tip is always less than the blade speed. Further, the relative
flow through the impeller is not perfectly guided by the impeller vanes but
is deflected away from the direction of rotation of the impeller; because of
which air leaves at an angle smaller than the vane angle [Fig. 8.17(b)].
The resultant velocity triangle at impeller outlet will be as shown in
Fig. 8.18. It may be seen that the actual tangential component of absolute velocity (ct2 ) of the fluid is less than that of the perfectly guided value
(ct2 ). The perfectly guided value will be obtained if the fluid leaves the impeller at the vane angle (β2 ). The difference between the value of perfectly
guided and actual tangential component (ct2 − ct2 ) is called slip velocity
(cs ). Hence, the energy transfer can be written as
E
=
u2 ct2
(8.26)
It follows that the actual energy transfer in the impeller is less than the
perfectly guided value as the fluid does not come out at an angle β2 , the
exit blade angle.
Actual velocity triangle
cs
Perfectly guided
velocity triangle
c2
w2
w2'
c 2’
cr
β2
β2’
ct2
u2
ct2’
Fig. 8.18 Effect of relative eddy on impeller outlet velocity
8.5.6
Slip Factor
The actual velocity profiles at the impeller tip due to real flow behaviour are
shown in Figs. 8.19 and 8.20. Figure 8.19 corresponds to the velocity profile
at exit in the meridional plane whereas Fig. 8.20 depicts the details between
two vanes. The energy transfer occurring in the impeller corresponding to
these velocity profiles is less than the one that would have been obtained
with one-dimensional flow.
The relative eddy [Fig. 8.17(a)] mentioned earlier causes the flow in the
impeller passages to deviate (Fig. 8.21) from the blade angle (β2 ) at the
exit to an angle (β2 ), the difference being larger for a larger blade pitch or
smaller number of impeller blades.
298
Gas Turbines
cr= w
Exit velocity profile
Hub
Shroud
Meridional plane*
* Meridional plane is a plane
which contains the axis of
rotation and is swept around
that axis
Fig. 8.19 Velocity profile at the impeller tip in the meridional plane
Exit velocity profile
w
Vane to vane plane
ω
Fig. 8.20 Vane to vane velocity profile at the exit of the impeller
Centrifugal Compressors
Actual
Ideal
299
cs
cr
2
2
w
2
cr
2’
c
w 2’
c2
c 2’
ct
β2’ β
t2’
2
u
2
ω
Fig. 8.21 Exit velocity triangles with and without slip
On account of the aforementioned effects, the apex of the actual velocity
triangle at the impeller exit is shifted away (opposite to the direction of
rotation) from the apex of the perfectly guided velocity triangle as shown
in Fig. 8.21. This phenomenon is known as slip and the shift of the apex is
the slip velocity (cs ). It may be noted that on account of the slip, the whirl
component is reduced which in turn decreases the energy transfer and the
pressure developed. The ratio of the actual and perfectly guided values of
the whirl components at the exit is known as slip factor (μ)
μ =
ct2
ct2
(8.27)
(1 − μ)ct2
(8.28)
The slip velocity is given by
cs
8.5.7
=
ct2 − ct2
=
Determination of Slip Factor
Several theoretical and empirical equations have been derived for the slip
factor to allow calculation of energy transfer during the design of a new
impeller. The most widely used of these are theoretically based equations
300
Gas Turbines
derived by Stodola∗ , Stanitz† and Balje‡ , all of which assume the flow of
an inviscid fluid through the impeller.
8.5.8
Stodola’s Formula
Stodola∗ assumed that the relative eddy was equivalent to the rotation
of cylinder with an angular velocity equal in magnitude and opposite in
direction to that of the impeller.
Figure 8.22 depicts the model of flow with slip as suggested by Stodola.
The relative eddy is assumed to fill almost the entire exit section of the
impeller passage. It is considered equivalent to the rotation of a cylinder
of diameter d = 2r at an angular velocity ω which is equal and opposite to
that of the impeller as shown in the figure. The diameter, and hence, the
tangential velocity of the cylinder is approximately determined as follows:
s = 2 π r2 / z
β
2’
ω
2r
ω
Fig. 8.22 Stodola’s model of flow with slip
The blade pitch at the outer radius (r2 ) of the impeller with z number
of blades is
s =
2πr2
z
The diameter of the cylinder is
2r
≈ s sin β2
=
2πr2
sin β2
z
(8.29)
∗ Stodola, A., Steam and Gas Turbines, McGraw Hill Book Co., New York, Vol.I &
II, 1927, Reprint Peter Smith, New York, 1945.
† Stanitz, J.D., ’Some theoretical aerodynamic investigations of impellers in radial and
mixed flow centrifugal compressors’, Trans ASME, 74, 4, 1952
‡ Balje, O.E., ’A study of design criteria and matching of turbomachines. Pt. B –
Compressor and pump performance and matching of turbo components’, ASME paper
No.60–W A–231, 1960
Centrifugal Compressors
301
The slip velocity, is assumed to be due to the rotation of the cylinder and
can therefore be taken as,
cs
= ωr
Substituting in the above equation for r from Eq. 8.29, we have
ωπr2
sin β2
cs =
z
(8.30)
However, ωr2 = u2 . Therefore,
cs
π
u2 sin β2
z
=
(8.31)
Substituting Eq. 8.31 in Eq. 8.28, we have
π
u2 sin β2
(1 − μ)ct2 =
z
μ = 1−
π u2
sin β2
z ct2
From Fig. 8.21, we have
ct2
= u2 − cr2 cot β2 ,
and
μ = 1−
π
z
sin β2
1 − φ2 cot β2
(8.32)
where the flow coefficient, φ2 = cr2 /u2
For a radial-tipped impeller (β2 = 90◦ )
π
μ = 1−
z
(8.33)
The above expressions for μ show that for a given geometry of flow, (blade
exit angle) the slip factor increases with the number of impeller vanes.
When the slip factor approaches 1, the slip velocity tends to be zero (i.e.,
ct2 = ct2 ). Hence, larger the number of blades lower will be the slip
velocity. However, there is a limitation from the manufacturing point of
view to decide about the number of impeller vanes. Thus, the number of
impeller vanes is one of the governing parameters in the calculation of losses
and this should be given due consideration. It should be noted that because
of the phenomenon of the slip the actual energy transfer is given by
E
8.5.9
=
μu2
(8.34)
Stanitz’s Formula
A method based on the solution of potential flow in the impeller passages
is suggested by Stanitz for β2 between 45◦ and 90◦ . The slip velocity is
302
Gas Turbines
found to be independent of the blade exit angle and the compressibility.
This is given by
cs
=
1.98
u2
z
(1 − μ)ct2
=
1.98
u2
z
(8.35)
μ =
1−
1.98 u2
z ct2
μ =
1−
1.98
z
For β2 = 90◦
=
1−
1.98
z(1 − φ2 cot β2 )
(8.36)
(8.37)
Equations 8.33 and 8.37 are of identical form.
8.5.10
Balje’s Formula
Balje suggests an equation for radial-tipped (β2 = 90◦ ) blade impellers:
μ =
where
n =
8.6
1+
6.2
zn2/3
−1
(8.38)
Impeller tip diameter
Eye tip diameter
DIFFUSER
Centrifugal compressors are usually fitted with either a vaneless or a vaned
diffuser, although in some low-speed applications a volute is fitted directly
around the impeller. The influence of the diffuser upon compressor performance cannot be over emphasized: a considerable proportion of the fluid
energy at the impeller tip is kinetic energy (especially in radial-vaned impellers) and its efficient transformation into static pressure is important.
Losses may be high in the diffuser as the fluid is flowing against an adverse
pressure gradient. Hence, careful design of diffuser is a must.
8.6.1
Vaneless Diffuser
As the name indicates, the gas in a vaneless diffuser is diffused in the
vaneless space around the impeller before it leaves the stage through a
volute casing. In some applications the volute casing is omitted.
The gas in the vaneless diffuser gains static pressure rise simply due to
the diffusion process from a smaller diameter (d2 ) to a larger diameter (d3 ).
The corresponding area of cross-section in the radial direction are
A2
=
πd2 b2
=
2πr2 b2
(8.39)
Centrifugal Compressors
A3
=
πd3 b3
=
2πr3 b3
303
(8.40)
Such a flow in the vaneless space is a free-vortex flow in which the angular
momentum remains constant.
It can be shown that for a parallel walled (constant width) diffuser, the
ratio of tangential velocity at the exit to that of inlet is given by
ct3
cr3
c3
r2
=
=
=
(8.41)
ct2
cr2
c2
r3
Further, it can be shown that for a constant width diffuser with compressible flow
cr2
cr3
α2 = α3 = tan−1
= tan−1
(8.42)
ct2
ct3
Equation 8.41 clearly shows that the diffusion is directly proportional to
the diameter ratio (d3 /d2 ). This leads to a relatively large-sized diffuser
which is a serious disadvantage of the vaneless type. In some cases the
overall diameter of the compressor may be impractically large. This is a
serious limitation which prohibits the use of vaneless diffusers in aeronautical applications. Besides this, the vaneless diffuser has a lower efficiency
and can be used only for a small pressure rise. However, for industrial applications where large-sized compressors are needed, the vaneless diffuser is
economical and provides a wider range of operation. Besides this, it does
not suffer from blade stalling and shock waves.
8.6.2
Vaned Diffuser
For a higher pressure ratio across the radial diffuser, the diffusion process
has to be achieved across a relatively shorter radial distance. This requires
the application of vanes which provide greater guidance to the flow in the
diffusing passage. Diffuser blade rings can be fabricated from sheet metal
or casting can be made as cambered and uncambered shapes of uniform
thickness (see Figs. 8.23 and 8.24). Figure 8.25 shows a diffuser ring made
up of cambered aerofoil blades.
To avoid flow separation, the divergence of the diffuser blade passages
in the vaned diffuser ring can be kept small by employing a large number
of vanes which will lead to higher friction losses. Thus an optimum number
of diffuser vanes must be employed. It is the normal practice to have the
divergence of the flow passages not more than 12◦ .
The flow leaving the impeller has jets and wakes. When such a flow enter
a large number of diffuser passages, the quality of flow entering different
diffuser blade passages differs widely and some of the blades may experience
flow separation leading to rotating stall and poor performance. To avoid
such a possibility it is safer to provide a smaller number of diffuser blades
than that of the impeller. In some designs the number of diffuser blades is
kept one-third of the number of impeller blades. This arrangement provides
a diffuser passage with flows from a number of impeller blade channels.
304
Gas Turbines
Flow
Fig. 8.23 Diffuser ring with cambered blades
α3
(90- α
3)
α2
Straight flat blade
r3
r2
Fig. 8.24 Diffuser ring with straight (uncambered) flat blades
il
ofo
Aer
es
blad
Flow
Fig. 8.25 Diffuser ring with cambered aerofoil blades
Centrifugal Compressors
305
Thus the nature of flow entering various diffuser passages does not differ
significantly.
Another method to prevent steep velocity gradients at the diffuser entry
is to provide a small (0.05d2 − 0.1d2 ) vaneless space between the impeller
exit and the diffuser entry as shown in Fig. 8.26 and Fig.8.27. This allows
the non-uniform impeller flow to mix out and enter the diffuser with less
steep velocity profiles. Besides this the absolute velocity (Mach number)
of the flow is reduced at the diffuser entry. This is a great advantage,
especially, if the absolute Mach number at the impeller exit is greater than
unity. The supersonic flow at the impeller exit is decelerated in this vaneless
space at constant angular momentum without shock.
Diffuser blades
Flow
ω
d2
2
Impeller
blades
dh
2
Vaneless space
d1
2
Fig. 8.26 Vaneless space between impeller exit and diffuser entry
b3
Diffuser
θ
b2
r3
Vaneless space
Impeller
r2
Fig. 8.27 Radial diffuser passage with diverging walls
Every diffuser blade ring is designed for given flow conditions at the
entry at which optimum performance is obtained. Therefore, at off-design
306
Gas Turbines
operations the diffuser will give poor performance on account of mismatching of the flow. In this respect a vaneless diffuser or a vaned diffuser with
aerofoil blades (Fig. 8.25) is better. For some applications it is possible to
provide movable diffuser blades whose directions can be adjusted to suit
the changed conditions at the entry.
In some designs for industrial applications, a vaneless diffuser supplies
the air or gas direct to the scroll casing, whereas for aeronautical applications, various sectors of the vaned diffuser are connected to separate
combustion chambers placed around the main shaft.
8.7
VOLUTE CASING
The volute or scroll casing collects and guides the flow from the diffuser or
the impeller (in the absence of a diffuser). The flow is finally discharged
from the volute through the delivery pipe. For high pressure centrifugal
compressors or blowers, the gas from the impeller is discharged through a
vaned diffuser whereas for low pressure fans and blowers, the impeller flow
is invariably collected directly by the volute; since a diffuser is not required
owing to the relatively low pressures.
Figure 8.28 shows a volute casing along with impeller, diffuser and vaneless spaces. The volute base circle radius (r) is a little larger (0.05 to 0.10
times the diffuser or impeller radius) than the impeller or diffuser exit radius. The vaneless space before volute decreases the non-uniformities and
turbulence of flow entering the volute as well as noise level. Some degree
of diffusion in the volute passage is also achieved in some designs, while
others operate at constant static pressure.
Delivery pipe
Exit
Throat
Vaneless spaces
Tongue
Flow
Impeller
r
Di
Volute passage
r
ffu
s
er
Volute
section
dr
Fig. 8.28 Flow through volute casing of a centrifugal compressor
Centrifugal Compressors
307
Different cross-sections are employed for the volute passage as shown
in Fig. 8.29. The rectangular section is simple and convenient when the
volute casing is fabricated from sheet metal by welding the curved wall to
the two parallel side walls. While the rectangular section is very common
in centrifugal blowers, the circular section is widely used in compressor
practice.
(a)
(b)
(c)
r4
r4
r4
Circular
Trapozoidal
Rectangular
axis
Fig. 8.29 Different cross-section of the volute passage
While, on the one hand, the volute performance is dependent on the
quality of flow passed on to it from the impeller or diffuser, the performance
of the impeller or the diffuser also depends on the environment created by
the volute around them. The non-uniform pressure distribution around the
impeller provided by its volute gives rise to the undesirable radial thrust
and bearing pressures.
8.8
PERFORMANCE PARAMETER
As ideal energy transfer given by Eq. 8.2 will not hold good for an actual
compressor, in order to assess the performance, certain parameters are used.
In this section we will describe them briefly.
8.8.1
Power Input Factor
The ideal energy transfer is given by Eq. 8.2, which is E = u22 . In practice,
the actual energy transfer to the air from the impeller is lower than that
given by Eq. 8.34, which is μu22 . Further, some energy is lost in friction
between the casing and the air carried round by the vanes, and in disc
friction or windage. In order to take this into account, power input factor,
Pif is introduced, so that the work input to the compressor becomes
Wc
=
Pif μ u22
(8.43)
The normal value used for Pif is between 1.035 to 1.04.
Let ΔTc = T02 − T01 be the total head temperature rise across the
compressor. As there is no energy addition in the diffuser this temperature
rise is also the temperature rise across the impeller. Thus,
ΔTc
=
T02 − T01
=
Pif μu22
Cp
(8.44)
308
8.8.2
Gas Turbines
Pressure Coefficients
This is a performance parameter which is useful in comparing various centrifugal compressors.
Each impeller has a definite maximum work capacity limited by the
maximum tangential velocity at the exit. If this maximum work is utilized
to the maximum advantage, an isentropic compressor will result, and a
delivery pressure of p02 max will be obtained. Now referring to Fig. 8.30 we
have,
02’ max
T02’ max
T02
02
02’
T 02’
T
p
p
02
02max
p
01
04 03
T01
01
s
Fig. 8.30 T –s diagram of the compression process
p02max
p01
γ−1
γ
T02 max
=
T01
(8.45)
Wisen
=
Wmax = Cp T01
p02max
p01
γ−1
γ
−1
(8.46)
The actual compressor produces only p02 pressure after an expenditure
of adiabatic work given by
Wadia
=
Cp T01
p02
p01
γ−1
γ
Now pressure coefficient ψp is defined as
−1
(8.47)
Centrifugal Compressors
ψp
=
p02
p01
Cp T01
Wadia
Wisen
γ−1
γ
=
p02max
p01
Cp T01
−1
γ−1
γ
309
(8.48)
−1
and from Eq. 8.7, the denominator is u22 and therefore
ψp
8.8.3
=
Cp T01 (rc )
γ−1
γ
u22
−1
(8.49)
Compressor Efficiency
We have already dealt with the compressor efficiency. However, it is better
to recall it once again here. The overall stagnation isentropic efficiency
(Fig. 8.31) in terms of pressure ratio can be written as
p
02
2
p
01
h
2’
Δ h0
Δ h 0’
1
s
Fig. 8.31 h–s diagram of the compressor
ηc
=
Δh0
Δh0
In terms of temperature and pressure ratios
γ−1
ηc
From Eq. 8.44,
=
T01 rc γ − 1
(T02 − T01 )
(8.50)
310
Gas Turbines
T02 − T01
=
Pif μu22
Cp
(8.51)
=
ψp u22
Cp
(8.52)
=
ψp
Pif μ
(8.53)
From Eq. 8.49,
γ−1
T01 rc γ − 1
Substituting in the Eq.8.50
ηc
8.9
LOSSES IN CENTRIFUGAL COMPRESSORS
Total losses in a centrifugal compressor may be divided into two groups:
(i) Frictional losses These are proportional to c2 and hence proportional
to ṁ2 .
(ii) Incidence losses These loses in terms of drag coefficient CD are proportional to CD c2 .
Figure 8.32 shows the variation of the above two losses with respect to the
mass flow rate.
Frictional losses
Losses
Total losses
Incidence losses
Mass flow rate
Fig. 8.32 Variation of losses with respect to the mass flow rate
8.10
COMPRESSOR CHARACTERISTICS
The ideal performance characteristics as shown in Fig. 8.15 for a radialvane compressor are to be modified because of the losses mentioned above.
If these losses are subtracted from the ideal energy transfer for a radialvaned impeller, then the constant-pressure ratio straight line characteristics
becomes curved, with a maximum value of energy at some particular value
of the mass flow rate as shown in Fig. 8.33. Radial-vaned impellers are
Centrifugal Compressors
311
mostly used with gas turbine compressors. Therefore, a characteristic which
will have a point of maximum pressure ratio with a positive and a negative
slope is shown in Fig. 8.33. This particular variation decides upon the range
of compressor operation.
Ideal
Actual
For
= 90
Mass flow rate
Loss
Gas Turbines
6
N
Pressure ratio
Relative to design value
T 01
5
4
Locus of points of
maximum efficiency
3
Surge line
2
1
0
0.2
0.4
0.6
0.8
1.0
1.2
m T 01 Relative to design value
p
01
Fig. 8.35 Actual characteristics of a centrifugal compressor
100
Total head efficiency (%)
312
80
60
0.6 0.7 0.8 0.9 1.0
40
N
T
Relative to design value
01
20
0
0.2
m
0.4
T
01
0.6
0.8
1.0
1.2
Relative to design value
p01
Fig. 8.36 Total head efficiency of a centrifugal compressor
Centrifugal Compressors
313
pressure ratio starts reducing. At this moment, there is a higher pressure in
the downstream of the system (near exit) than at compressor delivery and
the flow stops momentarily and may even reverse its direction. This reduces
the pressure downstream. After a short interval of time, compressor again
starts to deliver fluid and the operating point shifts to C again. Again
the pressure starts increasing and the operating point moves from right to
left. If the downstream conditions are unchanged then once again the flow
will break down after point A and the cycle will be repeated with a high
frequency. This phenomenon is known as surging or pumping. This will be
again explained in greater detail with respect to axial flow compressor in
the next chapter.
This instability will be severe in compressors producing high pressure
ratios, which may ultimately lead to physical damage due to impact loads
and high-frequency vibration. Because of this particular phenomenon of
surging or pumping at low-mass flow rates, the compressor cannot be operated at any point to the left of the maximum pressure ratio point A, i.e.,
it cannot be operated on the positive slope of the characteristic.
At higher mass flow rate points on the characteristic a different situation
occurs. At a constant rotor speed the tangential velocity component at the
impeller tip remains constant. As can be seen from Fig. 8.34, with the
increase in mass flow rate the pressure ratio decreases (ABCD) and hence
the density also decreases. These effects result in a considerably increased
velocity which increases the absolute velocity and the incidence angle at
the diffuser vane top. Thus there is a rapid progress towards a choking
state. The slope of the characteristic therefore steepens and finally after
point D mass flows cannot be increased any further. The characteristic
finally becomes vertical. The point D on the characteristic curve is called
a choking point.
The actual characteristics and the total head efficiency of a centrifugal
compressor are shown in Figs. 8.35 and 8.36 respectively. Part of the curve
which is on the left of the maximum pressure ratio point is inoperable due
to surge, and the line joining these points is called the surge line. On the
higher mass flow part the range is limited due to the choke point. The
peak efficiencies at each speed are quite close to the surge line. When the
compressor is used in the gas turbine power plant then the characteristics
of compressor and turbine must be matched properly. Otherwise problems
will be experienced either due to surging or due to low efficiency.
Worked out Examples
8.1 A centrifugal compressor under test gave the following data:
Speed
Inlet total head temperature
Outlet and inlet total head pressure
Impeller dia
:
:
:
:
11,500 rev/min
21◦ C
4 bar, 1 bar
75 cm
If the slip factor is 0.92, what is the compressor efficiency?
314
Gas Turbines
Solution
02
T
02’
01
s
Fig. 8.37
11500
60
u
=
π × 0.75 ×
WC
=
μu2 =
W
=
Cp (T02 − T01 )
T02
=
187.63
+ 294
1.005
p01
=
1 bar
p02
p01
=
0.92 × 451.62
1000
=
451.6 m/s
=
480.7 K
γ−1
γ
T02
=
T01
=
ηc
=
437 − 294
× 100
480.7 − 294
187.63 kJ/kg
294 ×
=
4
1
0.286
=
437 K
Ans
76.6%
⇐=
8.2 A centrifugal compressor has to deliver 35 kg of air per sec. The
impeller is 76 cm diameter revolving at 11,500 rpm with an adiabatic
efficiency of 80%. If the pressure ratio is 4.2:1, estimate the probable
axial width of the impeller at the impeller tip if the radial velocity is
120 m/s. The inlet conditions are 1 bar and 47◦ C.
Solution
T02
=
320 × 4.20.286
T02
=
T01 +
T02 − T01
ηc
=
320 +
482.4 − 320
= 523 K
0.8
=
482.4 K
Centrifugal Compressors
315
02
T
02’
01
s
Fig. 8.38
Ignoring the effects of velocity of flow
ρ2
=
p02
4.2 × 105
3
= 2.8 kg/m
=
RT02
287 × 523
Atip
=
35
2.8 × 120
Axial width
=
0.1042
= 0.0436 m = 4.36 cm
π × 0.76
=
0.1042 m2
Ans
⇐=
8.3 A centrifugal compressor has an inlet eye 15 cm diameter. The impeller revolves at 20,000 rpm and the inlet air has an axial velocity of
107 m/s, inlet stagnation temperature 294 K and inlet pressure 1.03
kg/cm2 . Determine
(i) theoretical angle of the blade at this point and
(ii) Mach number of the flow at the tip of the eye.
Solution
Peripheral speed of eye tip,
20000
= 157.08 m/s
60
u1
=
π × 0.15 ×
β1
=
tan−1
axial velocity
ca1
= tan−1
peripheral velocity
u1
=
tan−1
107
157.08
=
Ans
34.26◦
⇐=
Relative velocity at eye tip,
=
u1
cos β1
=
157.08
cos 34.26
=
190 m/s
316
Gas Turbines
w1
ca1 = c1
α1
β
1
u1
Fig. 8.39
Velocity of sound,
-
=
γR T01 −
c2a
2Cp
1072
1.4 × 287 × 294 −
2 × 1005
=
=
1
2
340.35 m/s
Mach number at the tip
=
190
Relative velocity at eye tip
=
Velocity of sound
340.35
=
0.558
Ans
⇐=
8.4 A centrifugal compressor takes in gas at 0◦ C and 0.7 bar and delivers
at 1.05 bar. The efficiency of the process compared with the adiabatic
compression is 83%. The specific heat of the gas at constant-pressure
and constant-volume are 1.005 and 0.717 respectively. Calculate the
final temperature of the gas and work done per kg of gas.
If the gas were further compressed by passing through a second compressor having the same pressure ratio and efficiency and with no
cooling between the two compressors, what would be the overall efficiency of the complete process?
Solution
T02
T02
p02
p01
γ−1
γ
=
T01
=
=
306.6 K
=
T01 +
T02 − T01
ηc
=
273 +
306.6 − 273
0.83
273 ×
=
1.05
0.7
313.5 K
0.286
Ans
⇐=
Centrifugal Compressors
317
03
03’
T
02’ 02
01
s
Fig. 8.40
Wc
=
1.005 × (313.5 − 273) = 40.7 kJ/kg
Ans
⇐=
With additional compressor
1.05
0.7
0.286
T03
=
313.5 ×
=
T03
=
T02 +
=
313.5 +
T03
=
273 × (1.5 × 1.5)
ηoverall
=
344.3 − 273
× 100
359.9 − 273
352 K
T03 − T02
ηc
352.0 − 313.5
0.83
=
0.286
=
=
359.9 K
344.3 K
82%
Ans
⇐=
8.5 Determine the impeller diameters and the width at the impeller exit
and the power required to drive the compressor, from the following
given data:
Speed (N )
Mass flow rate (ṁ)
Pressure ratio (r)
Isentropic efficiency (ηc )
Slip factor (μ)
Flow coefficient at impeller exit(φ)
Hub diameter of the eye
Axial velocity of air at entry to and
exit from the impeller
Stagnation temperature at inlet
Stagnation pressure at inlet
:
:
:
:
:
:
:
12,500 rev/min
15 kg/s
4:1
75%
0.9
0.3
15 cm
:
:
:
150 m/s
295 K
1.0 bar
Assume equal pressure ratio in the impeller and diffuser.
318
Gas Turbines
Solution
u2
=
ca2
φ
150
0.3
=
500 m/s
ṁμu22
=
15 × 0.9 × 5002
1000
=
3375 kW
u2
=
πD2
D2
=
500 × 60
π × 12500
=
0.7639 m
T1
=
T01 −
c2a
2Cp
=
295 −
=
283.8 K
=
p01
T1
T01
=
0.8733 bar
ρ1
=
0.8733 × 105
287 × 283.8
A1
=
ṁ
ρ1 ca1
=
0.0934 m2
=
0.0934
D1
=
0.376 m
p3
p1
=
4
p2
p1
=
p3
p2
p22
=
4p21
p2
=
2p1
T2
=
283.8 × 20.286
T2
=
283.8 +
=
Power input
p1
π 2
D − 0.152
4 1
N
60
γ
γ−1
=
1502
2 × 1005
283.8
295
3.5
3
1.07 kg/m
15
1.07 × 150
=
=
=
Ans
⇐=
Ans
⇐=
2 × 0.8733 = 1.7466 bar
=
346 − 283.8
0.75
346 K
=
366.73 K
Centrifugal Compressors
ρ2
=
1.7466 × 105
284.6 × 366.73
A2
=
πD2 W2
W2
=
1
15
×
1.67 × 150 π × 0.7639
=
0.025 m
=
319
1.67 kg/m3
ṁ
ρ2 ca2
=
=
Ans
⇐=
2.5 cm
8.6 A single sided centrifugal compressor is to deliver 14 kg of air per
second when operating at a pressure ratio of 4:1 and a speed of 12,000
rpm. The total head inlet conditions may be taken as 288 K and 1.033
kgf/cm2 . Assuming a slip factor as 0.9, a power input factor of 1.04
and an isentropic efficiency (based on total head) of 80%, estimate the
overall diameter of the impeller. If the Mach number is not to exceed
unity at the impeller tip and 50% of the losses are assumed to occur
in the impeller, find the minimum possible depth of the diffuser.
Solution
c2
ca2
w2
β2
α2
ct2
cw2
u2
Fig. 8.41
p03
p01
=
ηc (T03 − T01 )
1+
T01
4
=
1+
T03
=
462.95 K
T03 − T01
=
ΨμU 2
Cp
U
=
πDN
60
=
γ
γ−1
0.8 × (T03 − 288)
288
3.5
(462.95 − 288) × 1.005 × 103
1.04 × 0.9
433.4
1
2
= 433.4 m/s
320
Gas Turbines
D
=
60 × U
πN
=
69 cm
=
60 × 433.4
π × 12000
=
0.69 m
Ans
⇐=
Minimum depth of the diffuser will be at impeller tip so if we find for the
impeller, radial velocity at tip and density at tip and knowing W , we can
calculate the area of passage and thus the depth of diffuser which is also
the depth of impeller tip.
For Mach number, M at impeller tip
M
=
c2
√
γRT3
c22
=
γRT3
T03
=
T3 + Temperature equivalent of c2
=
T3 +
c22
2 × 1.005 × 103
=
T3 +
γRT3
2 × 1.005 × 103
=
T3 1 +
γR
2 × 1.005 × 103
=
T3 1 +
1.4 × 287
2 × 1.005 × 103
T3
=
T03
1.2
c2
=
γRT3
=
393.71 m/s
(since M = 1)
=
=
1.2 × T3
462.95
= 385.8 K
1.2
√
=
1.4 × 287 × 385.8
Now, we know that
c22
=
c2t2 + c2a2
c2a2
=
c22 − c2t2
=
393.712 − (0.9 × 433.4)
ca2
=
53.48 m/s
T02 − T01
=
ηc (T03 − T01 )
T02
=
139.96 + T01 = 139.96 + 288 = 427.96 K
=
393.712 − (μU )2
2
=
0.8 × (462.95 − 288)
Centrifugal Compressors
321
The overall loss
=
T03 − T02
=
462.95 − 427.96
=
34.99 K
=
2.37
Since, 50% of the loss occurs in the impeller, so
=
T3 − Loss in impeller
=
385.8 −
p2
p01
=
T2
T01
p2
=
1 × 2.37
ρ2
=
p2
2.37 × 105
3
= 2.24 kg/m
=
RT2
287 × 368.305
A
=
ṁ
ρ2 ca2
T2
34.99
2
=
γ
γ−1
=
3.5
368.305
288
=
=
368.305 K
2.37 bar
14
2.24 × 53.48
=
0.117 m2
Total area of flow in impeller
A
=
2πr × Depth of diffuser
=
A
2πD/2
=
5.4 cm
Depth of diffuser
=
0.117
π × 0.69
=
0.054 m
Ans
⇐=
8.7 A centrifugal compressor runs at 10,000 rpm and delivers 600 m3 /min
of free air at a pressure ratio of 4:1. The isentropic efficiency of
compressor is 82%. The outer radius of impeller (which has radial
blades) is twice the inner one and neglect the slip coefficient. Assume
that the ambient air conditions are 1 bar and 293 K. The axial velocity
of flow is 60 m/s and is constant throughout. Determine
(i) power input to the compressor,
(ii) impeller diameters at inlet and outlet and width at inlet, and
(iii) impeller and diffuser blade angles at inlet.
Solution
γ−1
γ
T03
=
T01 × (4)
T03
=
T01 +
=
293 + 173.85
= 293 × 40.286 = 435.56 K
T03 − T01
ηc
=
=
293 +
466.85 K
435.56 − 293
0.82
322
Gas Turbines
03
03’
T
c1
w1 = ca1
01
α root
1
s
u1
Fig. 8.42
u22
=
Cp ΔT
=
=
174719.25 m2 /s2
u2
=
418 m/s
WC
=
u22
D2
=
418 × 60
π × 10000
D1
=
D2
2
=
T1
=
T01 −
c2a
602
= 291.2 K
= 293 −
2Cp
2 × 1005
=
1005 × 173.85
Ans
174.719kW/kg/s
T1
T01
=
⇐=
0.8 m
Ans
0.4 m
⇐=
γ
γ−1
291.2
293
3.5
p1
=
p01
ρ1
=
p1
RT1
=
1.171 kg/m
Q̇
=
ca Aflow at root
Wroot
=
1
10
×
60 π × 0.4
u1
=
π × 10000
πN
×D =
× 0.4 = 209.4 m/s
60
60
αroot
=
tan−1
60
209.4
αtip
=
tan−1
60
418
=
= 0.9787 bar
0.9787 × 105
287 × 291.2
=
3
=
=
=
=
ca πD1 Wroot
0.133 m
Ans
⇐=
Ans
16◦
⇐=
8.17◦
⇐=
Ans
Centrifugal Compressors
323
8.8 A centrifugal compressor has a pressure ratio of 4:1 with an isentropic
efficiency of 80% when running at 15000 rpm and inducing air at
293 K. Curved vanes at inlet give the air a prewhirl of 25◦ to the
axial direction at all radii and the mean dia of eye is 250 mm. The
absolute air velocity at inlet is 150 m/s. Impeller tip dia is 600 mm.
Calculate the slip factor.
Solution
c1
w1
ο
25
ca1
ο
β1
65
ct1
cw1
u1
Fig. 8.43
T02
T01
=
p02
p01
T02
=
293 ×
γ−1
γ
4
1
0.286
=
435.56 K
Isentropic temperature rise
T02 − T01
=
435.56 − 293
=
142.56 K
=
Isentropic temperature rise
Isentropic efficiency
=
142.56
0.8
Actual temperature rise,
ΔT
=
178.2 K
Power input per unit mass flow rate
=
Cp × ΔT = 1.005 × 178.2 = 179 KJ/kg
ct1
At exit,
=
c1
=
150 m/s
u1
=
π × Mean dia of eye × 15000
60
=
π × 0.250 × 15000
60
=
150 × sin 25
c1 sin 25
=
=
196.35 m/s
63.4 m/s
324
Gas Turbines
u2
=
π × Impeller tip dia × 15000
60
=
π × 0.6 × 15000
60
Power input unit mass flow rate
=
=
471.24 m/s
u2 ct2 − u1 ct1
3
=
471.24 × ct2 − 196.35 × 63.4
ct2
=
406.27 m/s
μ
=
ct2
u2
179 × 10
=
406.27
471.24
=
0.862
Ans
⇐=
8.9 Determine the number of radial impeller vanes using Stanitz formulae
for a centrifugal compressor which requires 180 kJ of power input per
unit mass flow rate and is running at 15000 rpm. Guide vanes at inlet
give the air a prewhirl of 25◦ to the axial direction at all radii and the
mean dia of eye is 250 mm. Impeller tip dia is 600 mm. The absolute
air velocity at inlet is 150 m/s.
Solution
w1
ο
c1
25
ο
ca1
β1
65
ct1
cw1
u1
Fig. 8.44
Power input per unit mass flow rate
u1
u2
ct1
=
u2 ct2 − u1 ct1
=
π × Mean dia of eye × N
60
=
π × 0.25 × 15000
60
=
π × Impeller tip dia × N
60
=
π × 0.6 × 15000
60
=
c1 sin 25 = 150 × sin 25 = 63.39 m/s
=
=
196.35 m/s
471.24 m/s
Centrifugal Compressors
180 × 103
=
471.24ct2 − (196.35 × 63.39)
ct2
=
196.35 × 63.39 + 180 × 103
471.24
=
408.38 m/s
325
Stanitz formulae for radial impellers is given by
μ
=
1−
1.98
z
where z is number of impeller vanes.
μ
=
ct2
u2
0.866
=
1−
z
≈
15
Therefore,
=
408.38
471.24
=
0.866
0.63 × π
z
Ans
⇐=
8.10 A centrifugal compressor compresses 30 kg of air per second at a
rotational speed of 15000 rpm. The air enters the compressor axially,
and the conditions at the exit sections are radius = 0.3 m, relative
velocity of air at the tip = 100 m/s at an angle of 80◦ C with respect
to plane of rotation. Take p01 = 1 bar and T01 = 300 K.
Find the torque and power required to drive the compressor and also
the head developed.
Solution
c2
cr2
w2
β2
ct2
cw2
u2
Fig. 8.45
N
15000
= π × 0.6 ×
= 471.24 m/s
60
60
u2
=
πD2
ct2
=
u2 − w2 cos β1 = 471.24 − 100 × cos 80
=
453.88 m/s
326
Gas Turbines
Torque
Power
W
W
=
F r = mct2 r = 30 × 453.88 × 0.3
=
4084.92 Nm
=
T ω = 4084.92 × 2 × π ×
=
6.417 × 106 W
=
6.417 × 103 kW
=
u2 ct2
=
213886.41
=
Cp T01
=
Ans
⇐=
15000
60
Ans
⇐=
471.24 × 453.88
p02
p01
213886.41
=
1005 × 300 ×
p02
p01
=
6.531
p02
=
6.531 bar
γ−1
γ
−1
p02
p01
γ−1
γ
−1
Ans
⇐=
Review Questions
8.1 What is a centrifugal compressor and what are its advantages?
8.2 Where do the centrifugal compressors find application and why?
8.3 With a neat sketch explain the essential parts of a centrifugal compressor.
8.4 With a suitable diagram explain the working principle of a centrifugal
compressor.
8.5 What are the three types of blade shapes possible and how they are
classified?
8.6 With a neat sketch explain the inlet and exit velocity triangles for
various types of blades.
8.7 Briefly explain the flow through the following components:
(i) the inlet casing,
(ii) the inducer,
Centrifugal Compressors
327
(iii) the impeller, and
(iv) the impeller channel.
8.8 Briefly explain with suitable diagram how does the blade shape affect
the performance of the compressor.
8.9 What is meant by slip?
8.10 Define slip factor and derive an expression for the same.
8.11 Why diffusers are necessary in a centrifugal compressor?
8.12 Explain the details of vaned and vaneless diffusers?
8.13 What is meant by volute? Explain the purpose of volute casing.
8.14 Explain briefly the following performance parameters:
(i) power input factor,
(ii) pressure coefficients, and
(iii) compressor efficiency.
8.15 Briefly explain the phenomena of surge and chocking in centrifugal
compressors.
Exercise
8.1 Find the actual shaft power required to the compressor of Jumbo
turbo-jet engine from the following data:
Pressure ratio : 3:1
γ : 1.4
Adiabatic efficiency based on shaft power : 75%
Mass flow of the air compressed : 21 kg/s
Temperature of inlet air : 45◦ C
Neglect any ram effect due to the forward speed of turbojet.
Ans: 3303.56 kW
8.2 A centrifugal compressor of 40.6 cm diameter revolving at 18000 rpm
delivers air at an isentropic efficiency of 0.78. What would be the
approximate pressure ratio expected if the machine was at 6000 m
altitude where p0 = 35 cm of Hg and T0 = 248 K. Calculate the
actual delivery temperature and the power required to deliver air at
the rate of 0.5 kg/s. Neglect effects of inlet and exit velocities.
Ans: (i) 5 (ii) 434.8 K (iii) 73.2 kW
8.3 A single stage turbo blower compresses 1000 m3 /min of air at 1 bar
and 15◦ C through a pressure ratio of 1.33 with an index of compression as 1.6, assuming the power input to the blower to be 800 kW and
Cp = 1.005, calculate the isentropic efficiency of the blower. Assume
R = 287 kJ/kg K.
Ans: 82.35%
328
Gas Turbines
8.4 A centrifugal air compressor delivers 20 kg/s of air with a total head
pressure ratio of 4:1. The speed of the compressor is 12,000 rpm.
Inlet total temperature is 15◦ C, slip factor 0.9, power inlet factor
1.04, and the total head isentropic efficiency as 80%. Calculate the
overall diameter of the impeller.
Ans: 0.69 m
8.5 For a single-sided impeller of a centrifugal compressor, the following
conditions are given:
Speed
Ambient stagnation temperature
Ambient stagnation pressure
Hub diameter at eye
Outer diameter of the eye
Impeller tip dia
Axial velocity at inlet to and exit
from compressor
:
:
:
:
:
:
:
3600
−7◦ C
1 bar
12.5 cm
20 cm
100 cm
130 m/s
Establish the principal dimensions of the compressor.
Ans: (i) 0.0305 m (ii) 0.00516 m (iii) 73.8◦ (iv) 34.6◦
8.6 A single-sided centrifugal compressor has the internal dia of eye 15 cm.
The compressor delivers air at the rate of 9 kg/s with a pressure ratio
of 4.4 to 1 at 20,000 rpm. The axial velocity is 150 m/s with no
pre-whirl. Initial condition of air are pressure 1 bar and temperature
20◦ C. Assuming adiabatic efficiency as 80%, the ratio of whirl speed
to tip speed as 0.95 and neglecting all other losses, calculate the rise
of total temperature, tip speed, tip dia and external dia of eye.
Ans: (i) 200.64 K (ii) 460.71 m/s (iii) 44 cm (iv) 29.35 cm
8.7 A double-sided centrifugal compressor has impeller eye root and tip
dia of 18 cm and 32 cm respectively and is required to rotate at
16000 rpm. Find values for the impeller vane angles at root and tip
of the eye if the angle is given 70◦ of prewhirl at all radii. The axial
component of the inlet velocity is constant over the eye and is about
150 m/s. Find also the maximum Mach number at the eye. Inlet
conditions may be assumed as 288 K and pressure 1 bar.
Ans: (i) 57.33◦ (ii) 35.11◦ (iii) 0.784
8.8 Air enters axially in a centrifugal compressor fitted with radial blades
at head of 1.033 bar and temperature of 288 K. The air leaves the
diffuser with negligible velocity. The tip diameter of the impeller is
30 cm and the speed 25000 rpm. Neglect all losses. Calculate the
temperature and pressure of the air as it leaves the compressor.
Ans: (i) 441.45 K (ii) 4.61 bar
8.9 A centrifugal compressor has an impeller tip speed of 366 m/s. Determine the absolute Mach number of the flow leaving the radial vanes of
the impeller when the radial component of velocity at impeller exit is
Centrifugal Compressors
329
30 m/s and the slip factor is 0.9. Given that the flow area at impeller
exit is 0.1 m2 and ηi = 90%, calculate the mass flow rate. Assume
p01 = 1.033 bar.
Ans: (i) 0.878 (ii) 5.65 kg/s
8.10 In a centrifugal compressor, air leaving the guide vanes has a velocity
of 90 m/s at 20◦ to the axial direction at the outer radius of the eye.
Assuming frictionless flow through the guide vanes, determine the
inlet relative Mach number and the impeller efficiency. Other details
of the compressor and its operating conditions are
Impeller entry tip dia, Dti
Impeller exit tip dia, Dte
Slip factor μ
Radial component of velocity at
impeller exit, cr2
Rotational speed of impeller, N
Static pressure at impeller exit, p2
Ambient total temperature at inlet
Ambient total pressure at inlet
Absolute angle
:
:
:
:
0.45 m
0.76 m
0.9
53 m/s
:
:
:
:
:
11000 rpm
2.3 bar
15◦ C
1.033 bar
70◦
Ans: (i) 0.721 (ii) 91.6%
8.11 Air enters the impeller of a centrifugal compressor in an axial direction
at a rate of 2 kg/s with a pressure of 1.033 bar and a temperature
of 288 K. Measured angle of air entry is 55◦ from the axial direction
and leaves in a radial direction. The diameters of the eye root and tip
at inlet are one-fifth and one-half respectively of the impeller outside
dia, if the impeller is driven at 20000 rpm, determine
(i) the diameter of the impeller, and
(ii) the temperature rise through the impeller.
State the assumptions made.
Ans: (i) 29.8 cm (ii) 96.9 K
8.12 Determine the slip factor in a single-sided centrifugal compressor fitted in the aircraft flying with a speed of 230 m/s at an altitude where
the pressure is 0.25 bar and the static temperature is 220 K. The
mean dia of eye is 25.5 cm and the impeller tip dia is 54 cm. Rotational speed of the compressor is 16000 rpm and the inlet duct of
the impeller eye contains fixed vanes which give the air prewhirl of
65◦ with respect to prewhirl speed at all radii. Stagnation pressure
at the compressor outlet is 1.75 bar. Take the power input factor as
1.04 and isentropic efficiency as 0.8.
Ans: 0.91
8.13 The following data refers to a design of a single-sided centrifugal compressor:
330
Gas Turbines
Air mass flow (ṁ)
Eye tip dia (Dc )
Eye root dia (Dr )
Overall dia of the impeller (Do )
Slip factor (μ)
Isentropic efficiency (ηc )
Power input factor (Pif )
Rotational speed (N )
Inlet stagnation pressure (p01 )
Inlet stagnation temperature (T01 )
:
:
:
:
:
:
:
:
:
:
9 kg/s
300 mm
150 mm
500 mm
0.9
80%
1.04
18000
1.1 bar
295 K
Compute
(i) pressure ratio of the compressor,
(ii) inlet angle of the impeller vane at the root and tip radii of the
eye, and
(iii) the axial depth of the impeller channel at the periphery of the
impeller. Assume that half the total loss occurs with impeller.
Ans: (i) 4.75 (ii) 45.16◦ (iii) 26.70◦ (iv) 1.65 cm
8.14 A centrifugal compressor operates with prewhirl and is run with a
rotor tip speed of 500 m/s. If slip factor is 0.95, isentropic efficiency
of compressor is 0.85 and power input factor is 1.035, calculate the
following for operation in standard sea level air:
(i) pressure ratio,
(ii) the work required per kg of air and
(iii) the power required for a flow of 30 kg/s.
Take Cp = 1.005 kJ/kg K, ambient temperature 17◦ C.
Ans: (i) 6.63 (ii) 246.09 kJ/kg (iii) 7382.7 kW
8.15 An aircraft engine is fitted with a single-sided centrifugal compressor.
The aircraft flies with a speed of 850 km/h at an altitude where the
pressure is 0.23 bar and the temperature 217 K. The inlet duct of the
impeller eye contains fixed vanes which give the air pre-whirl of 65◦
at all radii. These inner and out diameters of the eye are 180 and 330
mm respectively, the diameter of the impeller tip is 540 mm and the
rotational speed 16000 rpm. Estimate the stagnation pressure at the
compressor outlet when the mass flow is 216 kg per minute.
Neglect losses in inlet duct and fixed vanes, and assume that the
isentropic efficiency of the compressor is 0.80. Take the slip factor as
0.9 and the power input factor as 1.04.
Ans: 1.908 bar
Centrifugal Compressors
331
Multiple Choice Questions (choose the most appropriate answer)
1. A single state modern centrifugal compressor have a pressure ratio
(a) 2:1
(b) 4:1
(c) 6:1
(d) 8:1
2. A centrifugal compressor is best suited for the
(a) simple cycle
(b) intercooled cycle
(c) reheat cycle
(d) heat exchange cycle
3. In the inlet casing of the centrifugal compressor, pressure
(a) increases
(b) remains constant
(c) decreases
(d) initially increases and then remains constant
4. In order to have ideal energy transfer in the centrifugal compressor
(a) heat transfer should be high
(b) the air should enter the compressor with high tangential component of the absolute velocity
(c) the air should leave the impeller with a tangential velocity equal
to the impeller velocity
(d) none of the above
5. The exit blade angle (β2 ) for a forward curved blade should be
(a) > 90◦
(b) = 90◦
(c) < 90◦
(d) can be any angle
6. The pressure rise in a centrifugal compressor is due to
(a) internal effect
(b) external effect
(c) both internal and external effect
(d) none of the above
332
Gas Turbines
7. For better performance of the centrifugal compressor, the slip factor
μ should be
(a) close to 1
(b) close to 0.5
(c) close to 0
(d) can be of any value
8. The compressor isentropic efficiency in terms of pressure coefficient
(ψp ), slip factor (μ) and power input factor (Pif ) is given by
(a) ηc =
Pif ×μ
ψp
(b) ηc =
Pif
μ×ψp
(c) ηc =
μ×ψp
Pif
(d) ηc =
ψp
Pif ×μ
9. With respect to mass flow rate the frictional losses are
(a) more than incidence losses
(b) less than incidence losses
(c) equal to incidence losses
(d) initially lower and after a certain mass flow rate it is higher
10. It is better to operate a centrifugal compressor with respect to mass
flow rate
(a) close to the surge line
(b) left side of the surge line
(c) right side of the surge line
(d) both at left and right side of the surge line
Ans:
1. – (b)
6. – (c)
2. – (d)
7. – (a)
3. – (c)
8. – (d)
4. – (c)
9. – (d)
5. – (a)
10. – (c)
9
AXIAL FLOW
COMPRESSORS
INTRODUCTION
The principal type of compressor being used nowadays, in majority
of the gas turbine power plants and especially in aircraft applications, is
the axial flow compressor. Although in olden days, the turboprop engines
incorporated the centrifugal compressors, the recent trend, particularly for
high-speed and long-range applications, is towards the axial flow type. This
dominance is mainly due to the ability of the axial flow compressor to satisfy
the basic requirements of the aircraft gas turbine.
The basic requirements of compressors for aircraft gas-turbine application are well-known. In general, they include
(i) high air-flow capacity per unit frontal area,
(ii) high pressure ratio per stage,
(iii) high efficiency, and
(iv) discharge direction suitable for multistaging.
Because of the demand for rapid engine acceleration and for operation over
a wide range of flight conditions, a high level of aerodynamic performance
must be maintained over a wide range of mass flow rates and speeds. Physically, the compressor should be designed in such a way to have a minimum
length and also its weight must be as low as possible. The mechanical
design should be simple, so as to reduce manufacturing time and cost. Further, the resulting structure should be mechanically rugged and must have
high reliability.
9.1
HISTORICAL BACKGROUND
The basic concepts of multistage axial flow compressor operation have been
known for approximately 100 years, being presented to the French Academic
334
Gas Turbines
des Sciences in 1853 by Tournaire. One of the earliest experimental axial
flow compressors (1884) was obtained by C. A. Parsons by running a multistage reaction-type turbine in reverse. Efficiencies for this type of unit
were quite low. It was mainly because the blading was not designed for
the condition of a pressure rise in the direction of flow. Beginning at the
turn of twentieth century, a number of axial flow compressors were built, in
some cases with the blade design based on propeller theory. However, the
efficiency of these units was still low (50 to 60 per cent). Further, development of the axial flow compressor was retarded by the lack of sufficient
knowledge of fluid mechanics at that time.
The advances in aviation during the period of World War I and the
rapidly developing background in fluid mechanics and aerodynamics gave
new impetus to research on compressors. The performance of axial flow
compressors was considerably improved by the use of isolated-airfoil theory.
As long as moderate pressure ratios per stage were desired, isolated-airfoil
theory was quite capable of producing compressors with reasonably high
efficiency. Compressors of this class were used in machinery as ventilating
fans, air conditioning units, and steam generator fans.
Beginning in the middle of 1930’s, interest in the axial flow compressor
was greatly increased as the result of the quest for air superiority. Efficient
superchargers were necessary for reciprocating engines in order to increase
engine power output as well as improved high-altitude aircraft performance.
With the development of efficient compressor and turbine components, turbojet engines for aircraft also began receiving attention.
In 1936 the Royal Aircraft Establishment in England began the development of axial flow compressor for jet propulsion. A series of high
performance compressors was developed, culminating in the F.2 engine in
1941. In Germany, concentrated research work ultimately resulted in the
use of axial flow compressors in the Jumo 004 and the B.M.W. 003 turbojet
engines. In the United States, aerodynamic research results were applied
to obtain high-performance axial flow units.
In the development of all these units, increased stage pressure ratios
were sought by utilizing high blade cambers and closer blade spacings. Under these conditions the flow patterns about the blades began to affect
each other, and it became apparent that the isolated-airfoil approach was
inadequate. Aerodynamic theory was therefore, developed specifically for
the case of a lattice or cascade of airfoils. In addition to theoretical studies, systematic experimental investigations of the performance of airfoils in
cascade were conducted to provide the required design information.
By 1945, compressors of high efficiency could be developed by incorporating aerodynamic principles in design and development. Since that time,
considerable research has been directed at extending aerodynamic limits in
an attempt to maximize compressor and gas-turbine performance. One of
the major developments in this direction has been the successful extension
of allowable relative inlet Mach numbers without sacrificing efficiency. The
subject of allowable blade loading, or blade surface diffusion, has also been
attacked with a degree of success. Accompanying improvements led to an
Axial Flow Compressors
335
increase in the understanding of the physics of flow through axial flow compressor blading. This resulted in corresponding improvements in techniques
of aerodynamic design and high efficiencies and axial flow machines became
a reality.
9.2
GEOMETRY AND WORKING PRINCIPLE
As already stated elsewhere, it may be noted that an axial compressor is a
pressure producing machine. The energy level of air or gas flowing through
it, is increased by the action of the rotor blades which exert a torque on
the fluid. This torque is supplied by an external source – an electric motor
or a gas-turbine. Besides its applications in the industrial gas turbine units
the multistage axial compressor is the principal element of all gas-turbine
power plants for land and aeronautical applications.
An axial flow compressor [Fig.9.1(a) and (b)] consists of an alternating
sequence of fixed and moving sets of blades. The sets of fixed blades are
spaced around an outer stationary casing, called the stator. The sets of
moving blades are fixed to a spindle called the rotor. The rotor and stator
banks must be as close as possible for smooth and efficient flow. The radius
of the rotor hub and the length of the blades are designed so that there is
only a very small tip clearance at the end of the stator and rotor blades.
One set of stator blades and one set of rotor blades constitute a stage.
(a) Drum type rotor
(b) Disc type rotor
336
Gas Turbines
kinetic energy imparted to the working fluid. It also redirects the fluid at
an angle suitable for entry into the rotating blades of the following stage.
Usually at entry one more stator is provided to guide the air correctly into
the first rotor. These blades are sometimes referred to as the Inlet Guide
Vanes (IGV). The details are shown in Fig.9.2. In many compressors there
are one to three rows of diffuser or straightener blades installed after the
last stage to straighten and slow down the air before it enters into the
combustion chamber.
Inlet guide vanes (IGV)
Stage
Rotor blades
Stator blades
(diffuser)
Fig. 9.2 An axial flow compressor stage
9.3
STAGE VELOCITY TRIANGLES
The flow geometry at the entry and exit of a compressor stage is described
by the velocity triangles at these stations. A minimum number of data on
velocity vectors and their directions are required to draw a complete set of
velocity triangles.
Compressors have a finite cross-section at the entry and exit. Therefore, the magnitude of velocity vectors and their directions vary over these
sections. On account of this, an infinite number of velocity triangles are
required to fully describe the flow. This is obviously not possible. On
the other hand, a single pair of velocity triangles will only represent onedimensional flow through the stage.
In view of this, mean values of velocity vectors and their directions are
defined for blade rows of given geometries and flow conditions. These values
make it possible to draw the mean velocity triangles for the stage.
The velocity triangles for a compressor stage contain, besides the peripheral velocity (u) of the rotor blades both the absolute (c) and relative
(w) fluid velocity vectors. These velocities are related by the following
well-known vector equation:
c
=
u+w
(9.1)
where c is absolute velocity vector, u is peripheral velocity vector and w is
Axial Flow Compressors
337
relative velocity vector. This simple relation is frequently used and is very
useful in drawing the velocity triangles for turbomachines.
For instance the velocity triangles shown in Fig.9.3 are for a general
stage which receives air or gas with an absolute velocity c1 and angle α1
(from the axial direction) from the previous stage. In the case of the first
stage in a multistage machine the axial direction of the approaching flow is
changed to the desired direction (α1 ) by providing a row of blades upstream
of the rotor which are called inlet guide vanes (IGV) or upstream guide
vanes (UGV). Therefore, the first stage experiences additional losses arising
from flow through the guide vanes.
Upstream guide vanes
Entry velocity triangle w1
1
w2
β2
β1 α1
ca1
wt1
Rotor blades
h1 , p1
h 01,p01
c1
ct1
u
α2
c2
Exit velocity triangle
2
ca2
h 2 , p2
h 02, p02
c t2
wt2
u
Diffuser blades
3
ca3
α
3
c3
h3 , p3
h 03,p03
Fig. 9.3 Velocity triangles for a compressor stage
For a general stage, the entry to the rotor, exit from the rotor and the
diffuser blade row (stator) are designated as stations 1,2 and 3 respectively.
The air angles in the absolute and the relative systems are denoted by
α1 , α2 , α3 and β1 , β2 respectively as seen in Fig.9.3. If the flow is repeated
in another stage
c1
=
c3
and
α1
=
α3
Subscripts a and t (Fig.9.3) denote axial and tangential directions respectively. Thus the absolute swirl or whirl vectors ct1 and ct2 are the tangential
components of absolute velocities c1 and c2 respectively. Similarly, wt1 and
wt2 are the tangential components of the relative velocities w1 and w2 respectively.
338
Gas Turbines
The following trigonometrical relations obtained from velocity triangles
(Fig.9.3) will be used throughout this chapter.
From velocity triangles at the entry:
ca1
= c1 cos α1
=
w1 cos β1
(9.2)
ct1
= c1 sin α1
=
ca1 tan α1
(9.3)
wt1
= w1 sin β1
=
ca1 tan β1
(9.4)
u
= ct1 + wt1
(9.5)
u
= c1 sin α1 + w1 sin β1
(9.6)
u
= ca1 (tan α1 + tan β1 )
(9.7)
From velocity triangles at the exit:
ca2
= c2 cos α2
=
w2 cos β2
(9.8)
ct2
= c2 sin α2
=
ca2 tan α2
(9.9)
wt2
= w2 sin β2
=
ca2 tan β2
(9.10)
u
= ct2 + wt2
(9.11)
u
= c2 sin α2 + w2 sin β2
(9.12)
u
= ca2 (tan α2 + tan β2 )
(9.13)
For constant axial velocity through the stage:
ca1
ca
= ca2
=
ca3
=
ca
= c1 cos α1
=
w1 cos β1
= c2 cos α2
=
w2 cos β2
(9.14)
(9.15)
Equations 9.7 and 9.13 give
u
ca
=
1
φ
=
tan α1 + tan β1
=
tan α2 + tan β2
(9.16)
This relation can also be presented in another form using Eqs. 9.5 and 9.11,
ct1 + wt1 = ct2 + wt2
ct2 − ct1
=
ca (tan α2 − tan α1 ) =
wt1 − wt2
(9.17)
ca (tan β1 − tan β2 )
(9.18)
Equations 9.17 and 9.18 give the change in the swirl components across the
rotor blade row. For steady flow in an axial machine, this is proportional
to the torque exerted on the fluid by the rotor.
Axial Flow Compressors
339
Table 9.1 illustrates what happen to the flow velocity and pressure when
air passes through the stage of the compressor.
Table 9.1 Variations occurring in an axial flow compressor
Absolute
velocity, c
Relative
velocity, w
Flow
width
Static
pressure, p
Total
pressure, p0
Rotor
Increases
Decreases
Increases
Increases
Increases
Stator
Decreases
–
Increases
Increases
About constant
9.3.1
Work Input to the Compressor
Now let us derive an expression for work input in terms of velocity and
blade angles. The derivation is based on the assumption that the axial
velocity, ca , remains constant throughout the machine. This assumption
simplifies the design calculations as well as expression for work input and
pressure rise. To maintain the axial velocity from the first stage to the last
stage (Fig. 9.1) constant throughout the machine, the area of flow is made
converging as pressure is increasing in every stage. Now, from Eq. 9.16
u =
ca (tan α1 + tan β1 )
=
ca (tan α2 + tan β2 )
(9.19)
also
W
=
u(ct2 − ct1 )
=
u(ca tan α2 − ca tan α1 )
=
uca (tan α2 − tan α1 )
(9.20)
(9.21)
In terms of angle, β, it can be written as
W
=
uca (tan β1 − tan β2 )
(9.22)
According the Euler’s⎡ energy equation (Eq. 3.13)
E
=
⎤
1⎢ 2
2
2
2
2
2 ⎥
⎣ c1 − c2 + u1 − u2 + w2 − w1 ⎦
2
I
II
III
For axial flow compressors (u = u1 = u2 ), the above equation will reduce
to
1 2
1
c − c21 + w12 − w22
W =
(9.23)
2 2
2
To obtain higher efficiencies the work input should be as minimum as possible. To achieve this, proper care in the design of blade and flow geometries
are essential.
340
9.4
Gas Turbines
WORK DONE FACTOR
Owing to secondary flows and the growth of boundary layers on the hub
and casing of the compressor annulus, the axial velocity along the blade
height is far from uniform. This effect is not so prominent in the first stage
of a multistage machine but is quite significant in the subsequent stages.
Figure 9.4 depicts the typical axial velocity distributions in the first and
last stages of a multistage axial compressor. The degree of distortion of the
axial velocity distributions in the last stage will depend on the number of
stages. On account of this, the axial velocity in the hub and tip regions is
Casing
Actual
Flow
Annulus
height
Mean
First
Last
Hub
Axial Flow Compressors
341
The air angles β2 and α1 are fixed by the cascade geometry of the rotor
blades and the upstream blade row. Therefore, assuming (tan α1 + tan β2 )
and u as constant, Eq. 9.28 relates work to the axial velocity at various
sections along the blade height.
The velocity triangles of Fig.9.3 are redrawn in Fig.9.5 for the design
value (mean value shown in Fig.9.4), and the reduced (ca − Δca ) and increased (ca + Δca ) values of the axial velocity.
Increased incidence
w1
α1 c1
u
u
u
Reduced incidence
c2
w
c - Δca 2 β
ca a
2
ca+Δca
Reduced ca
Design ca
Increased ca
Reduced c a
Design ca
Increased ca
u
u
u
Fig. 9.5 Effect of axial velocity on the stage velocity triangles and work
It is seen from the velocity triangles that the work absorbing capacity
decreases with an increase in the axial velocity and vice versa. Therefore,
the work absorbing capacity of the stage is reduced in the central region
of the annulus and increased in the hub and tip regions. However, the
expected increase in the work at the hub and tip is not obtained in actual
practice on account of higher losses. Therefore, the net result is that the
stage work absorbing capacity is less than that given by Euler’s equation
based on a constant value of the axial velocity along the blade height. This
reduction in the work absorbing capacity of the stage is taken into account
by a factor known as “workdone factor”. This varies from 0.98 to 0.85
depending on the number of stages. Thus, the work done factor accounts
for the effect of boundary layer and tip clearance. It is an empirical factor
which reduces the capacity of compressor. It is denoted by Ω. It takes
into account the axial velocity distribution also which is otherwise assumed
constant. Therefore, the workdone on air becomes
W
= Ωuca (tan β1 − tan β2 )
(9.29)
In terms of temperature difference, we have
ΔTs
Cp ΔTs
ΔTs
= T02 − T01
(9.30)
= Ωuca (tan β1 − tan β2 )
=
Ωuca
(tan β1 − tan β2 )
Cp
(9.31)
342
Gas Turbines
In fact, the stage temperature rise will be less due to the three dimensional
effects in the compressor annulus. Experiments have shown that in order
to get the actual energy transfer the result obtained should be multiplied
by a factor Ω which is the work done factor (refer Eq. 9.29).
Work done factor is really a measure of the ratio of the actual work
absorbing capacity of the stage to its ideal value as calculated from the
equation. The explanation of this is based on the fact that the axial velocity
contribution is not constant across the annulus. The magnitude increases
as the flow velocity across the annulus has compensating effects in respect
of work capacity. Unfortunately, the influence of the boundary layer and
tip clearance has an adverse effect on this compensation and the net result
is a loss in total work capacity which is accounted for by the work done
factor, Ω.
9.5
ENTHALPY–ENTROPY DIAGRAM
Figure 9.6 shows the enthalpy–entropy diagram for a general axial flow
compressor stage. Static and stagnation values of pressures and enthalpies
at various stations are as shown in Fig.9.6. 1–2 –3 shows isentropic compression whereas 1–2–3 shows actual compression. The stagnation point
03 corresponds to the final state at the end of isentropic compression.
1/2 c22
02
03
p
3
02
03’
p
Stator
03
p
3’
p
2’
3
h
1/2 c32
2
2
p
01
01’
p
01
1/2 c12
Rotor
1
1
s
Fig. 9.6 Enthalpy–entropy diagram of an axial compressor
From Fig.9.3 it can be seen that air enters the rotor blades with lower
absolute velocity (c1 ) but with large relative velocity (w1 ) whereas it leaves
the rotor with large c2 and lower w2 . However, when it comes out of the
Axial Flow Compressors
343
diffuser blades c3 is reduced which will be close to (very slightly higher)
c1 . Hence the stagnation pressure p01 will be slightly higher than the static
pressure p1 by 12 c21 which is shown in Fig.9.6. However, the stagnation
pressure p02 will be much higher than the static pressure p2 as can be seen
from the Fig.9.6. However, the flow occurs at constant enthalpy as can
been seen in the figure. That is
h02
=
h03
1
h2 + c22
2
=
1
h3 + c23
2
(9.32)
Further, it should be noted that the actual energy transformation process
(1 − 2) and (2 − 3) in the rotor and diffuser blade rows occur with stagnation pressure loss and increase in entropy. However, the relative stagnation
enthalpy remains constant.
h01
9.6
rel
=
h02
1
h1 + w12
2
=
1
h2 + w22
2
rel
(9.33)
COMPRESSOR STAGE EFFICIENCY
The efficiency of the compression process can now be defined based on ideal
and actual process as shown in the h–s diagram (Fig.9.6).
The ideal work input to the stage
Wideal
= h03 − h01
(9.34)
= Cp (T03 − T01 )
(9.35)
This is the minimum stage work input required to obtain a static pressure
rise of p3 − p1 . However, the actual process due to various losses and the
associated irreversibilities will require a higher magnitude of work input for
the same pressure rise. This is given by
Wactual
=
h03 − h01
(9.36)
=
Cp (T03 − T01 )
(9.37)
The compressor stage efficiency pertaining to total (stagnation) conditions
at entry and exit can be written as
ηpc
=
Wideal
Wactual
(9.38)
=
T03 − T01
T03 − T01
(9.39)
The magnitude of the stage work can now be written in terms of actual
velocities and air angles from the velocity triangles. Using Eq. 9.21, we can
344
Gas Turbines
write
Wactual
=
h03 − h01
=
uca (tan α2 − tan α1 )
=
uca (tan β1 − tan β2 )
=
1 2
1
c − c21 + w12 − w22
2 2
2
Knowing the stage pressure ratio, the isentropic enthalpy drop, h03 −
h01 , can be calculated and thereby ηpc can be evaluated from the Eq. 9.38.
9.7
PERFORMANCE COEFFICIENTS
In order to evaluate the performance of the compressor same dimensionless performance coefficients are found useful in various analyses. We will
discuss them briefly in the following section.
9.7.1
Flow Coefficient
It is defined as the ratio of axial velocity to peripheral speed of the blades
φ
=
ca
u
Flow coefficient is sometimes called as compressor-velocity ratio. It may
be noted that φ is sensitive to changes in angle of incidence, and as such
it is a useful parameter for representing the stalling characteristics of the
compressor.
9.7.2
Rotor Pressure Loss Coefficient
It is defined as the ratio of the pressure loss in the rotor due to relative
motion of air to the pressure equivalent of relative inlet velocity
Yrel
9.7.3
=
p01
rel − p02 rel
1
2
2 ρw1
(9.40)
Rotor Enthalpy Loss Coefficient
It is defined as the ratio of the difference between the actual and isentropic
enthalpy to the enthalpy equivalent of relative inlet velocity
ξrel
=
h2 − h2
1 2
2 w1
=
Cp (T2 − T2 )
1 2
2 w1
(9.41)
Because of friction and churning, the enthalpy at the outlet will be more
and thereby more work input will become necessary.
Axial Flow Compressors
9.7.4
345
Stator or Diffuser Pressure Loss Coefficient
It is defined as the ratio of the pressure loss in the diffuser due to flow
velocity to the pressure equivalent of actual inlet velocity of the diffuser.
YD
9.7.5
p02 − p03
1
2
2 ρc2
=
(9.42)
Stator or Diffuser Enthalpy Loss Coefficient
It is defined as the ratio of the difference between the actual and isentropic
enthalpy the enthalpy equivalent of absolute velocity of flow at diffuser
inlet.
ξD
9.7.6
=
h3 − h3
1 2
2 c2
=
Cp (T3 − T3 )
1 2
2 c2
(9.43)
Loading Coefficient
It is defined as the actual stagnation enthalpy rise in the stage to enthalpy
equivalent of peripheral speed of the rotor.
Ψ =
h03 − h01
u2
=
W
u2
Substituting for W from Eq. 9.22,
Ψ =
uca (tan β1 − tan β2 )
u2
=
φ(tan β1 − tan β2 )
=
Cp (T03 − T01 )
u2
=
φ(tan α2 − tan α1 )
(9.44)
(9.45)
Loading coefficient in terms of ηpc can be written as
Ψ =
Cp (T03 − T01 )
u2 ηpc
(9.46)
Some designers define the loading coefficient as the ratio of stage work to
the blade kinetic energy
W
ψ =
(9.47)
1 u2
2
But we will use ψ without the factor 2, i.e., Eq.9.46.
9.8
DEGREE OF REACTION
The degree of reaction prescribes the distribution of the stage pressure rise
between the rotor and the diffuser blade rows. This in turn determines the
346
Gas Turbines
cascade losses in each of these blade rows. The degree of reaction for axial
compressors can also be defined in a number of ways: it can be expressed
either in terms of enthalpies, pressures or flow geometry.
For an actual compressor stage the degree of reaction is defined as
R
=
actual change of enthalpy in the rotor
actual change of enthalpy in the stage
=
h2 − h1
h3 − h1
=
T2 − T1
T3 − T1
(9.48)
For c1 = c3 ,
h3 − h1
= h03 − h01
=
u(ct2 − ct1 )
From Eq. 9.33, we have
h2 − h1
=
1 2
w − w22
2 1
Now, degree of reaction, R, can be written as
R
=
h2 − h 1
h03 − h01
(9.49)
=
w12 − w22
2u(ct2 − ct1 )
(9.50)
Equation 9.50 can be further expressed in terms of air angles.
But
R
=
c2a tan2 β1 − tan2 β2
2uca (tan β1 − tan β2 )
R
=
1 ca
(tan β1 + tan β2 )
2 u
(9.51)
ca
= φ, and
u
1
(tan β1 + tan β2 )
2
= tan βm
Therefore,
R
= φ tan βm
(9.52)
Equation 9.51 can be rearranged to give
1 ca
[(tan β1 + tan α1 ) − (tan α1 − tan β2 )]
R =
2 u
From Eq. 9.27
tan β1 + tan α1
Therefore,
=
u
ca
Axial Flow Compressors
R
1 1 ca
−
(tan α1 − tan β2 )
2 2 u
=
347
(9.53)
This is a useful relation in terms of the geometry of flow and can be used to
study the effect of air angles and the required cascade geometry (to provide
these air angles) on the degree of reaction of an axial compressor stage.
9.8.1
Low Reaction Stages
A low reaction stage has a lesser pressure rise in its rotor compared to
that in the diffuser, i.e., (Δp)rel < (Δp)D . In such a stage the quantity (tan α1 − tan β2 ) is positive or in other words α1 > β2 (Fig.9.3 and
Eq. 9.53). The same effect can be explained in another manner. In Eq. 9.53
ca tan α1
= ct1
=
u − wt1
ca tan β2
= wt2
=
u − ct2
Therefore, after substituting these values in Eq. 9.53
R
=
wt2
1 1 ct1
−
−
2 2 u
u
(9.54)
=
1
1
−
(ct2 − wt1 )
2 2u
(9.55)
This equation relates the degree of reaction to the magnitudes of swirl or
the whirl components approaching the rotor and the diffuser. Thus a low
degree of reaction is obtained when the rotor blade rows remove less swirl
compared to the diffuser blade rows, i.e.,
wt1 < ct2
Figure 9.7 shows the enthalpy–entropy diagram for such a stage. The
swirl removing ability of a blade row is reflected in the static pressure rise
across it. In a low-degree reaction stage the diffuser blade rows are burdened
by a comparatively larger static pressure rise which is not desirable for
obtaining higher efficiencies.
9.8.2
Fifty Per cent Reaction Stages
One of the ways to reduce the burden of a large pressure rise in a blade row
is to divide the stage pressure rise equally between the rotor and diffuser.
To approach this condition (Fig.9.8).
h 2 − h1
= h3 − h2 =
1
(h3 − h1 )
2
This when substituted in Eq. 9.48 gives R = 12 .
Equation 9.50 for R =
1
2
gives
(9.56)
348
Gas Turbines
p
3
3
3’
Enthalpy
Diffuser
h3- h 2
p
2
2
2’
h2 - h 1
Rotor
p
1
1
Entropy
Fig. 9.7 Enthalpy–entropy diagram for a low reaction stage R <
3
1
2
p
3
3’
Enthalpy
Diffuser
h3- h2 =1/2(h3 - h1)
p2
2’
2
Rotor
h2- h1 =1/2(h3 - h1)
p1
1
Entropy
Fig. 9.8 Enthalpy–entropy diagram for a 50% reaction stage R =
1
2
Axial Flow Compressors
349
=
w12 − w22
2u(ct2 − ct1 )
w12 − w22
=
1 2
1
c2 − c21 + w12 − w22
2
2
w12 − w22
= c22 − c21
(9.57)
= β2
(9.58)
1
2
Substituting from Eq. 9.23
For R = 12 , Eq. 9.53 gives
α1
This when substituted along with Eq. 9.27 gives
α2
= β1
(9.59)
Equation 9.58 and 9.59 along with Eq. 9.57 yield
w1
= c2
w2
= c1
(9.60)
These relations show that the velocity triangles at the entry and exit
of the rotor of a fifty per cent reaction stage are symmetrical. The whirl
or swirl components at the entries of the rotor and diffuser blade rows are
also same.
9.8.3
ct1
= wt2
wt1
= ct2
(9.61)
High Reaction Stages
The static pressure rise in the rotor of a high reaction stage is larger compared to that in the diffuser, i.e., (Δp)rel > (Δp)D . Figure 9.9 shows
the enthalpy diagram. For such a stage the quantity (tan α1 − tan β2 ) in
Eq. 9.53 is negative, giving R > 12 . Therefore, for such a stage
α1 < β2
and
wt1 > ct2
(9.62)
Figure 9.10 shows the velocity triangles for such a stage. It can be
observed that the rotor blade row generates a higher static pressure on
account of the larger magnitude of the swirl component wt1 at its entry.
The swirl component ct2 passed on to the diffuser blade row is relatively
smaller, resulting in a lower static pressure rise therein.
Since the rotor blade rows have relatively higher efficiencies, it is advantageous to have a slightly greater pressure rise in them compared to the
diffuser.
350
Gas Turbines
p
3
3
3’
Enthalpy
Diffuser
h3 - h 2
p
2
2
2’
h2- h 1
Rotor
p
1
1
Entropy
Fig. 9.9 Enthalpy–entropy diagram for a high reaction stage R >
β1 α1
w1
u
w2
wt2
β2 α2
u
c2
c t2
u
ct1
wt1
Rotor blades
c1
Diffuser blades
α3
c3
Fig. 9.10 High reaction stage R > 12 , α1 < β2
1
2
Axial Flow Compressors
9.9
351
FLOW THROUGH BLADE ROWS
After studying the geometry and thermodynamics of the flow through a
compressor stage, further insight can be obtained by looking at the flow in
the individual blade rows. Therefore, the two parts of the h–s diagram in
Fig.9.6 (for the stage) are redrawn in Figs.9.11 and 9.12. The similarity
between these diagrams must be noted.
p
p
01 rel
02 rel
Enthalpy
01 rel
02 rel
h 01 rel = h02 rel
1/2 w22
2
1/2 w2’
p
2
2
2’
1/2 w12
p
1
1
Entropy
Fig. 9.11 Enthalpy–entropy diagram for flow through rotor blade row
The flow over a small pressure rise can be considered incompressible, i.e.,
density can be assumed to remain constant with little sacrifice in accuracy.
9.9.1
Rotor Blade Row
The flow process as observed by an observer sitting on the rotor is shown
in Fig.9.11. The initial and final pressures are p1 and p2 for both isentropic
and adiabatic processes.
In the isentropic process the flow will diffuse to a velocity w2 giving the
stagnation enthalpy and pressure as h01 rel and p01 rel respectively.
h2 − h1
=
1
(p2 − p1 )
ρ
=
1
(Δp)rel
ρ
(9.63)
The actual process gives the final velocity w2 and stagnation pressure
p02 rel . Here the same static pressure rise (Δp)rel occurs with a greater
change in the kinetic energy 12 w12 − w22 . In the ideal or isentropic process
this is
1 2
1
w − w22 < w12 − w22
2 1
2
352
Gas Turbines
p
02
02
p
h02= h03
1/2 c 32
1/2 c23’
Enthalpy
03
03
p
3
3
3’
1/2 c 22
p
2
2
Entropy
Fig. 9.12 Enthalpy–entropy diagram for flow through diffuser (stator) blade
row
This difference is due to the losses and the increase in entropy.
The efficiency of the rotor blade row can now be defined by
ηrel
=
h2 − h1
h2 − h1
= 1−
=
(h2 − h1 ) − (h2 − h2 )
h2 − h1
h 2 − h2
h2 − h1
(9.64)
(9.65)
Assuming perfect gas and substituting from Eq. 9.41
ηrel
= 1−
= 1−
w12
ξrel
2Cp (T2 − T1 )
ξrel
1−
(9.66)
(9.67)
w22
w12
The assumption of incompressible flow is not required in Eqs. 9.65, 9.66
and 9.67
h2 − h2
= (h2 − h1 ) − (h2 − h1 )
For incompressible flow, substituting from Eq. 9.63
1 2
1
w − w22 − (p2 − p1 )
h2 − h2 =
2 1
ρ
=
1
ρ
1
p1 + ρw12
2
1
− p2 + ρw22
2
Axial Flow Compressors
1
(p01
ρ
=
rel
− p02
rel )
=
(Δp0 )rel
ρ
353
(9.68)
Substituting this in Eq. 9.65
ηrel
= 1−
(Δp0 )rel
ρ(h2 − h1 )
(9.69)
Substitution from Eq. 9.40 gives
ηrel
9.9.2
= 1−
Yrel
1−
(9.70)
w22
w12
Stator Blade Row
The ideal and actual flow processes occurring in the diffuser blade row are
shown in Fig. 9.12. Its efficiency is again defined similar to that of rotor
blade row.
ηD
=
h3 − h2
h3 − h2
=
1−
=
(h3 − h2 ) − (h3 − h3 )
h3 − h2
h 3 − h3
h3 − h2
(9.71)
(9.72)
Substituting in terms of the enthalpy loss coefficient
ηD
=
1−
c22
ξD
2Cp (T3 − T2 )
=
1−
ξD
1−
c23
c22
(9.73)
For incompressible flow,
h 3 − h3
=
1
(p02 − p03 )
ρ
=
(Δp0 )D
ρ
(9.74)
Therefore, the diffuser efficiency can be expressed in terms of the diffuser
stagnation pressure loss coefficient
9.10
ηD
=
1−
ηD
=
1−
(Δp0 )D
ρ(h3 − h2 )
YD
1−
c23
c22
(9.75)
(9.76)
FLOW LOSSES
Principal aerodynamic losses occurring in most of the turbomachines arise
due to the growth of the boundary layer and its separation on the blade and
passage surfaces. Others occur due to wasteful circulatory flows and the
formation of shock waves. Non-uniform velocity profiles at the exit of the
354
Gas Turbines
cascade lead to another type of loss referred to as the mixing or equalization
loss.
Aerodynamic losses occurring in a turbomachine blade cascade can be
grouped in the following categories.
9.10.1
Profile Loss
As the term indicates, this loss is associated with the growth of the boundary layer on the blade profile. Separation of the boundary layer occurs when
the adverse pressure gradient on the surface or surfaces becomes too steep;
this increases the profile loss. The pattern of the boundary layer growth
and its separation depend on the geometries of the blade and the flow.
Positive and negative stall losses occur on account of increased positive or
negative incidences respectively.
Generally, the suction surface of a blade is more prone to boundary
layer separation. The separation point depends besides the blade profile on
factors like the degree of turbulence, Reynolds number and the incidence.
If the flow is initially supersonic or becomes supersonic on the blade
surface additional losses occur due to the formation of shock waves resulting
from the local deceleration of supersonic flow to subsonic.
9.10.2
Annulus Loss
The majority of blade rows in turbomachines are housed in casings. The
axial compressor stage has a pair of fixed and moving blade rows.
In stationary blade rows a loss of energy occurs due to the growth of
the boundary layer on the end walls. This also occurs in the rotating row
of blades but the flow on the end walls in this case is subjected to effects
associated with the rotation of the cascade. The boundary layer on the
floor (hub) of the blade passages is subjected to centrifugal force, whereas
that on the ceiling (outer casing) is scrapped by the moving blades.
9.10.3
Secondary Loss
This loss occurs in the regions of flow near the end walls owing to the presence of unwanted circulatory or cross flows. Such secondary flows develop
on account of turning of the flow through the blade channel in the presence
of annulus wall boundary layers. Figure 9.13 depicts the pressure gradients across a blade channel and the secondary and trailing vortices. Static
pressure gradients from the suction to the pressure side in the regions away
from the hub and tip are represented by the curve AB. In the vicinity of
the hub and tip or the end walls the pressure gradient curve CD is not so
steep on account of much lower velocities due to the wall boundary layers.
Thus the static pressures at the four corners of the section of flow under
considerations are
pB > pD > pC > pA
These pressure differentials across the flow near the end walls give rise
Axial Flow Compressors
B
D
p
C
A
y
C
A
Tip
355
B
D
C
D
Over deflection
A
B
Under deflectio
Secondary
vortices
Blade wakes
Trailing vortices
Hub
A
B
C
D
Fig. 9.13 Secondary flow in a cascade of blades
to circulatory flows which are superimposed on the main flow through the
blade passage. As a result of this, secondary vortices in the streamwise direction are generated in the blade passages. These vortices, besides wasteful expenditure of fluid’s energy, transport (DC) low energy fluid from the
pressure to the suction side of the blade passage, thus increasing the possibility of an early separation of the boundary layer on the suction side. The
flow nearer the hub and tip is over-deflected while that slightly away from
the end walls is under-deflected as shown in Fig.9.13.
The secondary vortices in the adjacent blade channels induce vortices
in the wake regions (as shown in Fig.9.13). These trailing vortices lead to
additional losses. It is worth observing here that the secondary flows in the
cascade also affect the profile and annulus losses.
The magnitude of the loss due to secondary flow depend on the fraction
of the passage height that is affected by this flow. Blade passages of very low
height (aspect ratio) or high hub-tip ratio are likely to be fully occupied by
secondary vortices as shown in Fig.9.14(a), and experience higher secondary
losses. In contrast to this longer blades [Fig.9.14(b)] have a large proportion
of the flow free of secondary flows and therefore experience comparatively
lower secondary losses.
If the total losses in a blade passage are measured along its height, they
appear as peaks near the hub and tip on account of secondary losses. The
flow in the central region which is outside the influence of secondary flows
(particularly in longer blades) can be assumed to suffer only profile loss.
Figure 9.15 illustrates this pattern of losses along the blade height.
9.10.4
Tip Clearance Loss
This loss arises due to the clearance between a moving blade and the casing.
In a turbine rotor blade ring the suction sides lead and the pressure sides
356
Gas Turbines
Tip
Tip
Hub
Hub
(a) High hub-tip ratio
(b) Low hub-tip ratio
Fig. 9.14 Secondary vortices in short and long blades
Y=
Secondary loss
Δ p0
1
2
ρ c22
Hub
Profile loss
Tip
Blade height
Fig. 9.15 Variation of losses along the blade height
trail. On account of the static pressure difference, the flow leaks from
the pressure side towards the suction side as shown in Fig.9.16. However,
due to the scrapping up of the casing boundary layer by the blade tips,
the scrapped up flow opposes the aforementioned tip leakage. The tip
clearance and secondary flows are closely related to each other and it is
often convenient to estimate them together.
9.11
STAGE LOSSES
Figure 9.17 shows the energy flow diagram for an axial flow compressor
stage. Figures in the brackets indicate the order of energy or loss corresponding to 100 units of energy supplied at the shaft.
The stage work (h03 − h01 ) is less than the energy supplied to the shaft
by the prime mover on account of bearing and disc friction losses. All
the stage work does not appear as energy at the stator entry on account
of aerodynamic losses in the rotor blade row. After deducting the stator
(diffuser) blade row losses from the energy at its entry, the value of the ideal
or isentropic work required to obtain the stage pressure rise is obtained.
The cascade losses in the rotor and stator would depend on the degree
of reaction. The values shown in the energy flow diagram are only to give
Axial Flow Compressors
357
Suction side
Pressure side
Tip leakage
Scraped flow
Motion of blades (scraping)
Fig. 9.16 Flow through tip clearance
Energy from the prime mover or shaft work (100)
Stage work (98)
Rotor aerodynamic
losses (9)
Shaft loses (2)
Disc
Bearing
friction loss
loss
Secondary Profile Annulus
loss
loss
loss
Energy at the stator entry (89)
Tip
leakage
Isentropic work (82) Stator aerodynamic losses(7)
Secondary loss Profile loss Annulus loss
Fig. 9.17 Energy flow diagram for an axial flow compressor stage
an example. The ratio of the isentropic work (82) and the actual stage
work (98) gives the stage efficiency, whereas the overall efficiency is directly
obtained as 82%.
9.12
PRESSURE RISE CALCULATION IN A BLADE RING
In case of aerofoil blading we already know that lift, L, is given by
L =
CL ρw2 A
2
(9.77)
D
CD ρw2 A
2
(9.78)
and the drag, D, is given by
=
358
Gas Turbines
Now
R
=
F
=
L sin θ + D cos θ
where F = R is the resultant force in kg and θ is the angle as shown in
Fig.9.18. R is the force on each blade and if n is the number of blades in
each blade ring then,
(L sin θ + D cos θ) nd
Torque on shaft,τ =
(9.79)
2
and so the power required (W τ )
P
=
2πN nd
(L sin θ + D cos θ)
2
(9.80)
where N is in rev/m and d is diameter of wheel
P
=
nπdN (L sin θ + D cos θ)
and if the isentropic efficiency is known then we can obtain the pressure
ratio as follows :
γ−1
T01
P = ṁ
Cp r γ − 1
(9.81)
ηc
L
f=R
θ
D
Fig. 9.18 Aerofoil blade pressure rise
where
T01
ηc
r
From Eqs. 9.80
: inlet temperature of air
: isentropic efficiency of compression
: pressure ratio
and 9.81
r=
ηc nπdN
(L sin θ + D cos θ) + 1
T01 ṁ
γ
γ−1
(9.82)
Thus, we will get the pressure ratio by substituting the values of other
terms. This will be pressure ratio that can be obtained in a stage or in a
blade ring.
Axial Flow Compressors
9.13
359
PERFORMANCE CHARACTERISTICS
The performance characteristics of axial compressors or their stages at various speeds can be presented in terms of the plots of the following parameters:
(i) Pressure rise vs flow rate,
Δp
= f (Q)
=
f (m)
(ii) Pressure ratio vs non-dimensional flow rate (Fig.9.19),
√
p2
m T01
= f
p1
p01
(iii) Loading coefficient vs flow coefficient (Fig.9.20),
ψ
= f (φ)
(9.83)
The actual performance curve based on measured values is always below
the ideal curve obtained theoretically on account of losses. This is shown in
Fig.9.21. The surge point and stable and unstable flow regimes have been
explained in the following sections.
9.13.1
Off-design Operation
A compressor gives its best performance while operating at its design point,
i.e., at the pressure ratio and flow rate for which it has been designed.
However, like any other machine or system, it is also expected to operate
away from the design point. Therefore, a knowledge about its behaviour at
off-design operation is equally important. Off-design characteristic curves
can be obtained theoretically from Eq. 9.44.
ψ
=
φ(tan β1 − tan β2 )
tan α2
=
1
− tan β2
φ
=
1 − φ(tan β2 + tan α1 )
But,
=
φ(tan α2 − tan α1 )
(9.84)
Therefore,
ψ
(9.85)
The quantity (tan β2 + tan α1 ) can be assumed constant in a wide range
of incidence up to the stalling value is. This is justified in view of small
variations in the air angles at the rotor and stator exits. Therefore, writing
α1 = α3 ,
A =
tan β2 + tan α3
(9.86)
If the design values are identified by the superscript∗ , Eq. 9.85 along with
Eq. 9.86 can be written as
Gas Turbines
Su
rge
lin
e
360
p
N
p
T 01
02
01
m
= constan
T 01
p
01
Fig. 9.19 Performance curves of a compressor
50% reaction stage
Δw/u 2= Δ p / ρ u
2
1
α = β = constant
1
2
0.5
0
0
0.5
φ = u/ca
1.0
Fig. 9.20 Variation of pressure coefficient with flow coefficient for an axial
flow compressor stage
Axial Flow Compressors
361
Pressure rise
Surge point
Stage losses
Ideal
Stable
l
tua
Ac
Unstable
Flow rate
Fig. 9.21 Ideal and actual performance curves for an axialcompressor
ψ∗
=
A =
1 − Aφ∗
(9.87)
1 − ψ∗
φ∗
At off-design conditions
ψ
=
1 − Aφ
=
1 − (1 − ψ ∗ )
φ
φ∗
(9.88)
This equation also gives the off-design characteristic of an axial flow compressor. Figure 9.22 shows theoretical values of A, the curves are falling
while for negative values rising characteristics are obtained. The actual
curves will be similarly modified but slightly on account of losses.
9.13.2
Surging
Unstable flow in axial compressors can be due to two reasons:
(i) separation of flow from the blade surfaces called stalling, and
(ii) complete breakdown of the steady through flow called surging.
Both these phenomena occur due to off-design conditions of operation and
are aerodynamically and mechanically undesirable.
Sometimes, it is difficult to differentiate between operating conditions
leading to stalling and surging. It may be noted that the flow in some regions stalls without surging taking place. Surging affects the whole machine
while stalling is a local phenomenon.
Some typical performance characteristic curves at different speeds (N1 , N2 ,
etc.,) are shown in Fig.9.23. The surge phenomenon is explained with the
362
Gas Turbines
A = negative
Loading coefficient
1.0
A=0
A = positive
0.75
0.50
0.5
0.25
0.0
0.0
0.25
0.50
1.0
1.0
0.75
Flow coefficient
Fig. 9.22 Off-design characteristic curves for axial compressor stage
aid of one of the curves in this figure. Let the operation of the compressor at a given instant of time be represented by point A(pA , ṁA ) on the
characteristic N3 curve. If the flow rate through the machine is reduced to
ṁB by closing a valve on the delivery pipe, the static pressure upstream of
valve is increased. This higher pressure, pB is matched with the increased
delivery pressure (at B) developed by the compressor. With further throttling of the flow (to ṁC and ṁS ), the increased pressures in the delivery
pipe are matched by the compressor delivery pressures at C and S on the
characteristic curve.
Surge cycle
D S
Pressure r se
E
C
B
A
N4
S
N3
Surge line
S
N2
N1
mE mD mS mB mA
Flow rate
Fig. 9.23 Surging in compressors
The characteristic curve at flow rates below ṁS provides lower pressure
as at D and E. However, the pipe pressures due to further closure of the
valve (point D) will be higher than these. This mismatching between the
Axial Flow Compressors
363
pipe pressure and compressor delivery pressure can only exist for a very
short time. This is because the higher pressure in the pipe will blow the
air towards the compressor, thus reversing the flow leading to a complete
breakdown of the normal steady flow from the compressor to the pipe. During this very short period the pressure in the pipe falls and the compressor
regains its normal stable operation (say at point B) delivering higher flow
rate (ṁB ). However, the valve position still corresponds to the flow rate
ṁD . Therefore, the compressor operating conditions return through points
C and S to D. Due to the breakdown of the flow through the compressor,
the pressure falls further to pE and the entire phenomenon, i.e., the surge
cycle EBCSDE is repeated again and again. The frequency and magnitude of this to-and-fro motion of the air (surging) depend on the relative
volumes of the compressor and delivery pipe, and the flow rate below ṁS .
Surging of the compressor leads to vibration of the entire machine which
can ultimately lead to mechanical failure. Therefore, the operation of compressors on the left of the peak of the performance curve is injurious to the
machine and must be avoided.
Surge points (S) on each curve corresponding to different speeds can be
located and a surge line drawn as shown in Fig.9.23. The stable range of
operation of the compressor is on the right-hand side of this line. There is
also a limit of operation on the extreme right of the characteristics when
the mass-flow rate cannot be further increased due to choking. This is
obviously a function of the Mach number which itself depends on the fluid
velocity and its state.
9.13.3
Stalling
As stated earlier, stalling is the separation of flow from the blade surface.
At low flow rates (lower axial velocities), the incidence is increased as shown
in Fig.9.5. At large values of the incidence, flow separation occurs on the
suction side of the blades which is referred to as positive stalling. Negative
stall is due to the separation of flow occurring on the pressure side of the
blade due to large values of negative incidence. However, in a great majority
of cases this is not as significant as the positive stall which is the main
subject under consideration in this section.
In a high pressure ratio multistage compressor the axial velocity is already relatively small in the higher pressure stages on account of higher
densities. In such stages a small deviation from the design point causes the
incidence to exceed its stalling value and stall cells first appear near the hub
and tip regions. The size and number of these stall cells or patches increase
with the decreasing flow rates. At very low flow rates they grow larger and
affect the entire blade height. Large-scale stalling of the blades causes a
significant drop in the delivery pressure which can lead to the reversal of
flow or surge. The stage efficiency also drops considerably on account of
higher losses. The axisymmetric nature of the flow is also destroyed in the
compressor annulus.
364
9.13.4
Gas Turbines
Rotating Stall
Figure 9.24 shows four blades (1, 2, 3 and 4) in a compressor rotor. Owing to
some distortion or non-uniformity of flow one of the blades (say the third)
receives the flow at increased incidence. This causes this blade (number
three) to stall. On account of this the passage between the third and fourth
blades is blocked causing deflection of flow in the neighbouring blades. As
a result, the fourth blade again receives flow at increased incidence and
the second blade at decreased incidence. Therefore, stalling occurs on the
fourth blade also. This progressive deflection of the flow towards the left
clears the blade passages on the right on account of the decreasing incidence
and the resulting unstalling. Thus the stall cells or patches move towards
the left-hand side at a fraction of the blade speed. In the relative system
they appear to move in a direction opposite to that of the rotor blades.
However, on account of their (stall cell) lower speed as compared to that
of the rotor, they move at a certain speed in the direction of the rotation
in the absolute frame of coordinates. Rotating stall cells, develop in a
Increased incidence Reduced incidence
Air
4
3
2
Propagating
1
u
Unstalling
Fig. 9.24 Stall propagation in a compressor blade row
variety of patterns at different off-design conditions as shown in Fig.9.25.
The blades are subjected to forced vibrations on account of their passage
through the stall cells at a certain frequency. The frequency and amplitude
of vibrations depend on the extent of loading and unloading of the blades,
and the number of stall cells. Under these conditions the blades can fail due
to resonance. This occurs when the frequency of the passage of stall cell
through a blade coincides with its natural frequency. Both the efficiency
and delivery pressure drop considerably on account of rotating stall.
Stall cells
Axial Flow Compressors
9.14
365
COMPARISON OF AXIAL AND CENTRIFUGAL COMPRESSORS
In this section, we will compare the axial and centrifugal compressors under
twelve different headings, such as
(i) type of flow
(ii) pressure ratio per stage
(iii) isentropic efficiency
(iv) frontal area
(v) flexibility of operation
(vi) part load performance
(vii) effect of deposits
(viii) starting torque
(ix) suitability for multistaging
(x) delivery pressure
(xi) applications
(xii) efficiency with respect to speed
(i) Type of flow For the axial compressor the flow direction is parallel to
the axis of the machine whereas for the centrifugal compressor it is
radial. There is no difference in radius for the axial flow compressor
between inlet and exit whereas there is large radius difference in the
case of centrifugal compressors.
(ii) Pressure ratio In a single stage the centrifugal compressor is capable
of developing a pressure ratio close to 5, whereas in the case of axial
flow compressor it is only about 1.25.
The supersonic centrifugal compressors can go upto a pressure ratio
of 10.
In order to achieve the pressure ratio of a single stage of centrifugal
compressor, multistaging is required in axial flow machines.
(iii) Isentropic efficiency The isentropic efficiency of axial flow compressors
are higher (86 to 88%) compared to centrifugal compressor (80 to
82%). Higher isentropic efficiency of modern axial flow compressor is
due to the aerofoil blades.
(iv) Frontal area The frontal area of a centrifugal compressor is larger
due to flow entering at one radius and leaving at another larger radius. Since there is no radius difference at entry and exit for an axial
compressor the frontal area is smaller.
366
Gas Turbines
(v) Flexibility of operation Comparatively higher due to adjustable prewhirl
and diffuser vanes in the case of centrifugal compressors. Due to the
absence of above two factors, the flexibility of operation is quite limited in the case of axial flow compressors.
(vi) Part load performance Since variation in mass flow rate can be tolerated to some extent in centrifugal compressors, part load performance
is better whereas it is poor in axial flow compressors.
(vii) Effect of deposits Deposits on the surface of the rotor are a major
factor in the operation and performance of compressors. However,
deposits do not adversely affect the performance in the case of centrifugal compressors whereas it adversely affects the operation and
performance of the axial flow machines.
(viii) Starting torque The starting torque required is low in the case of
centrifugal compressors whereas it is high in the case of axial flow
machines.
(ix) Multistaging Comparatively difficult in centrifugal compressors. Maximum of only two staging is possible. The axial flow machines are
more suitable for multistaging.
(x) Delivery pressure Higher delivery pressure (about 40 bar) is possible
in centrifugal machines whereas only 20 bar is possible in axial flow
compressors.
(xi) Applications Centrifugal compressors are used for blowing in steel
mills, low pressure refrigeration systems, big central air-conditioning
plants, fertilizer industry, supercharging of IC engines and gas pumping over a long distance pipe lines. Axial flow machines are mainly
used in jet engines due to higher isentropic efficiency and lower frontal
area. It is also employed in stationary industrial gas turbine power
plants and steel mills.
(xii) Efficiency with respect to speed The characteristics are more flat in
centrifugal compressors whereas it is less flat in axial flow compressors.
To put it in a nutshell, the comparison of axial and centrifugal machines
are given in Table 9.2.
Worked out Examples
9.1 A 10 stage axial flow compressor provides an overall pressure ratio of
5:1 with an overall isentropic efficiency of 87%. When the temperature
of air at inlet is 15◦ C. The work is equally divided between the stages.
A 50% reaction is used with a blade speed of 210 m/s and a constant
axial velocity of 170 m/s. Estimate the blade angles. Assume a work
done factor of 1.
Axial Flow Compressors
367
Table 9.2 Comparison of Axial Flow Compressors and Centrifugal Compressors
Aspect of comparison
Axial flow compressors
Centrifugal compressors
Type of flow
Parallel to the axis
Radial
Pressure ratio/stage
About 1.25
5
Isentropic efficiency
Higher (86 to 88%)
Lower (80 to 82%)
Frontal area
Smaller
Larger
Flexibility of operation
Limited
Higher
Part load performance
Poor
Better
Effect of deposits
Adverse effect
No adverse effect
Starting torque
High
Low
Suitability for
multistaging
More suitable
Difficult
Delivery pressure
Lower (20 bar)
Higher (about 40 bar)
Applications
Jet engines, stationary
industrial gas turbine
power plants and
steel mills.
Steel mills,
low pressure
refrigeration,
big central
air-conditioning plants,
fertilizer industry,
IC engines, gas pumping
Efficiency with
respect to speed
Less flat
More flat
Solution
Temperature increase per stage
ΔTstage
γ−1
T1
r γ −1
nηc
=
ΔToverall
n
=
288
× 50.286 − 1
10 × 0.87
=
Increase in temperature per stage,
ΔTs
=
Ωuca
(tan β1 − tan β2 )
Cp
=
19.35 K
368
Gas Turbines
β α1
w1
1
c1
ca1
wt1
ct1
u
T
02
02’
β
2
w2
c2
α2
ca2
c t2
wt2
u
01
s
Fig. 9.26
=
ΔTs Cp
19.35 × 1005
= 0.545
=
Ωuca
1 × 210 × 170
tan β1 + tan β2
=
u
ca
tan β1
=
0.89
β1
=
41.67◦
β2
=
tan−1 (1.235 − 0.89)
tan β1 − tan β2
For a 50% reaction,
=
210
170
=
1.235
Ans
⇐=
=
19◦
Ans
⇐=
9.2 Air at 1.0132 bar and 288 K enters an axial flow compressor stage
with an axial velocity 150 m/s. There are no inlet guide vanes. The
rotor stage has a tip diameter of 60 cm and a hub diameter of 50 cm
and rotates at 100 rps. The air enters the rotor and leaves the stator
in the axial direction with no change in velocity or radius. The air
is turned through 30.2◦ as it passes through rotor. Assume a stage
pressure ratio of 1.2. Assuming the constant specific heats and that
the air enters and leaves the blade at the blade angles,
(i) construct the velocity diagram at mean dia for this stage,
(ii) mass flow rate,
(iii) power required, and
(iv) degree of reaction
Solution
To construct velocity diagram
Axial Flow Compressors
β1
w1
α1
c1=ca1
u
c2
β2 α 2
w2
ca2
c t2
w t2
u
Fig. 9.27
u
β1
=
π(dhub + dtip )
×N
2
=
π × (0.50 + 0.60)
× 100
2
=
tan−1
u
ca
= tan−1
=
172.78
150
172.76 m/s
= 49◦ 2
As air is deflected by 30◦ ,
β2
=
49◦ 2 − 30◦ = 19◦
In the figure let Wt2 = x
x
=
ca tan β2
=
150 × tan 19◦ = 51.65 m/s
tan α2
=
u−x
ca
α2
=
38.92◦
ṁ
=
π 2
d − d2hub ca ρ2
4 tip
T1
=
T01 −
T2
=
T1
=
172.78 − 51.65
= 80.75
150
C12
1502
= 276.8 K
= 288 −
2Cp
2 × 1005
p2
p1
γ−1
γ
= 276.8 × 1.20.286 = 291.6 K
369
370
Gas Turbines
p2
=
1.216 bar
ρ2
=
1.216 × 105
287 × 291.6
ṁ
=
π
Ans
× 0.602 − 0.502 × 150 × 1.453 = 18.83 kg/s ⇐=
4
P
=
Ωuca ṁ(tan β1 − tan β2 )
=
1 × 172.76 × 150 × 18.83 × (tan 49.2 − tan 19)
1000
=
397.297 kW
3
=
1.453 kg/m
Ans
⇐=
Degree of reaction
R
=
ca
(tan β1 + tan β2 )
2u
=
150
× (tan 49◦ 2 + tan 19◦ ) = 0.65
2 × 172.76
Ans
⇐=
9.3 An axial flow air compressor of 50% reaction design has blades with
inlet and outlet angles of 45◦ and 10◦ respectively. The compressor is
to produce a pressure ratio of 6:1 with an overall isentropic efficiency
of 0.85 when inlet static temperature is 37◦ C. The blade speed and
axial velocity are constant throughout the compressor. Assuming a
value of 200 m/s for blade speed find the number of stages required
if the work done factor is (i) unity and (ii) 0.87 for all stages.
Solution
β α1
w1
1
c1
ca1
wt1
ct1
u
T
02
02’
β
2
w2
c2
α2
ca2
c t2
wt2
u
01
s
Fig. 9.28
Increase in temperature per stage,
Axial Flow Compressors
ΔTs
=
Ωuca
(tan β1 − tan β2 )
Cp
ca
=
u
tan β1 + tan β2
=
170.02 m/s
=
371
200
tan 45 + tan 10
Number of stages when Ω = 1
1 × 200 × 170.02 × (tan 45 − tan 10)
ΔTs
=
1005
ΔToverall
ΔToverall
ΔTstage
=
27.86 K
=
T1 γ−1
rp γ − 1
ηc
=
244.12 K
=
244.12
27.86
=
=
8.76
310
× 60.286 − 1
0.85
≈
9
Ans
⇐=
Number of stages when Ω = 0.87
0.87 × 200 × 170.02 × (tan 45 − tan 10)
=
ΔTs
1005
ΔToverall
ΔTstage
=
24.24 K
=
244.12
24.24
=
10.07
Ans
⇐=
9.4 Find the polytropic efficiency of an axial flow compressor from the
following data:
The total head pressure ratio
:
4
Overall total head isentropic efficiency
:
85%
Total head inlet temperature
:
290 K
The inlet and outlet air angles from the rotor blades of the above
compressor are 10◦ and 45◦ respectively. The rotor and stator blades
are symmetrical. The mean blade speed and axial velocity remain
constant throughout the compressor. Assuming a value of 220 m/s
for blade speed and the work done factor as 0.86, find the number of
stages required. Also find the inlet Mach number relative to rotor at
the mean blade height of the first stage. Assume R = 284.6 kJ/kg K.
Solution
ηc
r
=
r
γ−1
γ
γ−1
γ
−1
1
ηpc
−1
where ηp is the polytropic efficiency.
372
Gas Turbines
β = α2
1
α =β
2
1
β α
1
1
w1
c1
ca1
wt1
ct1
u
wt2
β2
α2
c2
ca2
ct2
w2
u
Fig. 9.29
0.85
=
40.286 − 1
40.286 η1p − 1
ηp
=
0.286 ×
log 4
× 100 = 87.65%
log 1.572
Ans
⇐=
Since, the stages are symmetrical, degree of reaction is 50% and also α1 = β2
and α2 = β1 .
From the figure,
u
=
ca
=
ca (tan 10 + tan 45)
220
0.17633 + 1
=
187 m/s
The stage temperature rise,
ΔTs
=
Ωuca
(tan α2 − tan α1 )
JCp
=
0.86 × 220 × 187
× (tan 45 − tan 10) = 29 K
1005
T02
T01
=
T02
=
p02
p01
γ−1
γηp
1.572 × 290
=
=
0.286
4 0.8764
455.9 K
=
1.572
Axial Flow Compressors
373
Total temperature rise
T02 − T01
=
455.9 − 290 = 165.9 K ≈ 166 K
Total number of stages
=
Mach number at inlet
166
29
=
Total temperature rise
Temperature rise in one stage
=
=
5.72
≈
Ans
6
⇐=
w1
√
γRT1
w1
=
ca
cos 45
=
187
0.7071
=
264.5 m/s
c1
=
ca
cos 10
=
187
cos 10
=
190 m/s
T1
=
T01 −
M
=
264.5
√
1.4 × 284.6 × 272
c21
1902
= 272 K
= 290 −
2Cp
2 × 1005
=
0.8
Ans
⇐=
9.5 An axial flow compressor takes in 1000 m3 /min of free air at 0.9 bar
and 15◦ C. The blades are of aerofoil type having projected area and
blade length as 19.25 cm2 and 6.75 cm respectively. The blade ring
mean diameter is 60 cm and speed is 6000 rpm. On each blade ring
there are 50 blades and the blades occupy 10% of the axial area of
flow. Values of CL and CD are 0.6 and 0.05 respectively at zero angle
of incidence. Assuming isentropic compression, calculate the pressure
rise per blade ring and the power input per stage. Assume axial inlet.
Solution
β
1
w
c = ca
u
Fig. 9.30
ρ
=
p1
RT1
=
0.9 × 105
3
= 1.09 kg/m
287 × 288
374
Gas Turbines
Q
=
Aca
and A
=
kπDm h
ca
=
Q
60A
=
145.55 m/s
u
=
πDm N
60
tan β1
=
u
188.5
= 1.295
=
ca
145.55
β1
=
52.33◦
w
=
L
=
CL ρw2 Ac
2
=
0.6 × 1.09 × 238.152 × 19.25 × 10−4
= 35.7 N
2
=
CD ρw2 Ac
2
=
0.05 × 1.09 × 238.152 × 19.25 × 10−4
= 2.98 N
2
1000 × 104
60 × (1 − 0.1) × π × 60 × 6.75
=
Blade velocity
D
=
π × 0.60 × 6000
= 188.5 m/s
60
145.552 + 188.52
=
238.15 m/s
Power input/stage
P
=
(L cos β1 + D sin β1 )un
=
(35.7 × cos 52.33 + 2.98 × sin 52.33) ×
188.5 × 50 × 10−3
r
Ans
=
227.85 kW
ṁ
=
1000 × 1.09
60
P
ṁ
=
Cp T1 γ−1
r γ −1
ηc
=
P
ηc
+1
×
ṁ Cp T1
=
227.85 × 1
+1
18.33 × 1.005 × 288
γ−1
γ
r0.286
⇐=
=
18.167 kg/s
Axial Flow Compressors
375
p2
p1
r
=
1.160
=
p2
=
0.98 × 1.160
=
Ans
1.14 bar
⇐=
9.6 Air at a temperature of 290 K enters a ten stage axial flow compressor at the rate of 3 kg/s. The pressure ratio is 6.5 and the isentropic
efficiency is 90%, the compression process being adiabatic. The compressor has symmetrical blades. The axial velocity of 110 m/s is
uniform across the stage and the mean blade speed of each stage is
180 m/s.
Determine the direction of the air at entry to and exit from the rotor
and the stator blades and also the power given to the air. Assume
Cp = 1.005 kJ/kg K and γ = 1.4.
Solution
β
w1
1
α
1
c1
ca1
c t1
wt1
u
β
w2
2
c2
α2
ca2
c t2
w t2
u
Fig. 9.31
Assume that the temperature change is constant in each stage, then the
power may be obtained by considering the overall conditions.
γ−1
γ
T2
T1
=
T2
=
290 × 6.50.286
ηc
=
T2 − T1
T2 − T1
p2
p1
=
495.33 K
376
Gas Turbines
0.90
=
495.33 − 290
T2 − 290
T2
=
518.14 K
Therefore, power given to the air,
=
ṁCp ΔT
=
687.84 kW
=
3 × 1.005 × (518.14 − 290)
Temperature change per stage
ΔTs
=
ΔT
10
518.14 − 290
10
=
=
22.81 K
Work done/kg of air second
=
uΔct
=
180 × Δct
Also work done/per kg of air per second
Δct
=
Cp ΔTs
=
1005 × 22.81
180
=
180Δct
=
127.36 m/s
For symmetrical stages,
Δct
=
ca (tan β1 − tan β2 )
127.36
=
110 × (tan β1 − tan β2 )
R
=
ca
(tan β1 + tan β2 )
2u
u
=
ca (tan β1 + tan β2 )
β1
=
54.41◦
⇐=
β2
=
13.5◦
⇐=
and when R = 0.5
Ans
Ans
9.7 An axial flow compressor has an overall pressure ratio of 4.0 and mass
flow of 3 kg/s. If the polytropic efficiency is 88 per cent and the stagnation temperature rise per stage must not exceed 25 K, calculate the
number of stages required and the pressure ratio of the first and last
stages. Assume equal temperature rise in all stages. If the absolute
velocity approaching the last rotor is 165 m/s at an angle of 20◦ from
the axial direction, the work done factor is 0.83, the velocity diagram
is symmetrical, and the mean diameter of the last stage rotor is 18
cm, calculate the rotational speed and the length of the last stage
Axial Flow Compressors
377
rotor blade at inlet to the stage. Ambient conditions are 1.01 bar and
288 K.
Solution
β = α2
1
α =β
2
1
β α
1
1
w1
c1
ca1
wt1
ct1
u
β2
wt2
α2
c2
ca2
ct2
w2
u
Fig. 9.32
ΔTstage
=
ηpc
=
γ−1
γ
n−1
n
=
γ−1 1
γ
ηpc
=
T01
p02
p01
=
451.92 K
=
T02 − T01
=
163.92 K
Number of stages
=
Temperature rise across
Total temperature rise per stage
Ns
=
163.92
25
T02
ΔToverall
25 K
n
n−1
n−1
n
0.286
= 288 × 4 0.88
=
=
451.92 − 288
6.56
≈
7
Ans
⇐=
378
Gas Turbines
Pressure ratio across first stage
r1
ηc
=
r
=
γ−1
γ
−1
γ−1 1
γ ηpc
r
γ
γ−1
ηc ΔTs
T01
1+
=
0.855
r1
=
1+
ΔTs
=
40.286 − 1
40.325 − 1
=
−1
0.855 × 25
288
3.5
= 1.285
Ans
⇐=
163.92
ΔToverall
=
= 23.42 K
7
7
Pressure ratio across last stage:
Temperature at inlet to last stage,
ΔTs
=
T02 − T0
T0
=
T02 − ΔTs
=
451.92 − 23.42
s
s
r
s
=
ηc ΔT0
1+
T01
r
s
=
1+
s
=
428.5 K
γ
γ−1
0.855 × 23.42
428.5
3.5
= 1.173
For symmetrical blade, R = 0.5
=
20◦
α1
=
β2
α2
=
β1
u
=
ca (tan β1 + tan β2 )
ca
=
C1 cos α1 = 165 × cos 20 = 155.05 m/s
u
=
155.05 × (tan β1 + tan β2 )
ΔT0s
=
ωuca
(tan β1 − tan β2 )
Cp
u(tan β1 − tan β2 )
=
1005 × 23.42
0.83 × 155.05
tan2 β1 − tan2 β2
=
182.90
155.05
=
=
1.1796
182.9
Ans
⇐=
Axial Flow Compressors
tan2 β1
=
1.1796 + tan2 β2
tan1 β1
=
1.1796 + tan2 20
tan β1
=
1.1455
β1
=
48.88◦
u
=
ca (tan β1 + tan β2 )
=
155.05 × (tan 48.88 + tan 20)
=
234.04 m/s
u
=
πDN
60
N
=
60 × u
πD
=
413.87 rps
=
1.3121
Ans
⇐=
60 × 234.04
1
×
π × 0.18
60
=
Ans
⇐=
Pressure ratio across the last stage = 1.173
Total pressure at inlet to the last stage,
=
4
1.173
=
3.41 bar
Total temperature at inlet to the last stage,
=
428.5 K
=
T0 −
=
428.5 −
=
414.96 K
Static temperature,
Tst
pst
c21
2Cp
1652
2 × 1.005 × 1000
p0
=
T0
Tst
379
(γ−1)/γ
=
3.048 bar
ρ
=
3.048 × 105
287 × 414.96
ṁ
=
Aca ρ
=
3.41
=
=
428.5 3.5
414.96
3
2.56 kg/m
ca ρπDm h
380
Gas Turbines
Length of last stage,
h
=
ṁ
ca ρπDm
=
3 × 100
2.56 × π × 0.18 × 155.05
=
1.336 cm
Ans
⇐=
9.8 The first stage of an axial compressor is designed on free vortex principles, with no inlet guide vanes. The rotational speed is 6000 rev/min
and the stagnation temperature rise is 20 K. The hub-tip ratio is 0.60,
the work done factor is 0.93 and the isentropic efficiency of the stage
is 0.89. Assuming an inlet velocity of 140 m/s and ambient conditions
of 1.01 bar and 288 K, calculate
(i) the tip radius and corresponding rotor air angles β1 and β2 , if
the Mach number relative to the tip is limited to 0.95,
(ii) the mass flow entering the stage,
(iii) the stage stagnation pressure ratio and power input, and
(iv) the rotor air angles at the root section.
Take R = 0.287 kJ/kg K, Cp = 1.005 kJ/kg K and γ = 1.4.
Solution
w1
02
T
β
1
ca1= c
1
02’
u
01
s
w2
β
2
c2
α2
ca2
c t2
w t2
u
Fig. 9.33
Axial Flow Compressors
T1
w1
C12
2Cp
=
T01 −
=
278.25 K
=
M1
=
=
288 −
381
1402
2 × 1005
γRT1
√
0.95 × 1.4 × 287 × 278.25 = 317.65 m/s
Assuming axial inlet, the velocity triangle at inlet can be drawn as shown
in Fig.
cos β1
=
c1
w1
140
317.65
=
=
0.4407
Tip radius corresponding rotor angles
β1
=
63.85◦
Ans
⇐=
u
=
w1 sin β1
u
=
285.13 m/s
ΔTs
=
Ω
tan β1 − tan β2
=
Cp ΔTs
uΩca
=
1005 × 20
285.13 × 0.93 × 140
=
0.5414
tan β2
=
tan 63.85 − 0.5414
β2
=
56.23◦
T1
=
278.25 K
p1
=
ρ1
=
p1
RT1
=
1.12 kg/m
Rtip
=
60 × 285.13
60u
=
= 0.454 m
2πN
2 × π × 6000
Rroot
=
0.6 × Rtip = 0.6 × 0.454 = 0.2724 m
=
317.65 × sin 63.85
uca
(tan β1 − tan β2 )
Cp
1.4954
Ans
⇐=
p01
T01
T1
=
3.5
=
1.01
3.5
288
278.25
= 0.8953 bar
0.8953 × 105
0.287 × 103 × 278.25
=
3
Ans
⇐=
382
Gas Turbines
Rm
=
1
(0.454 + 0.2724)
2
h
=
Rtip − Rroot
=
0.454 − 0.2724
=
ρ2πrm hca
=
1.12 × 2 × π × 0.3632 × 0.1816 × 140
=
64.98 kg/s
m
=
Input power
3.5
1+
ηs ΔT03
T01
=
1+
0.89 × 20
288
ṁCp ΔT0
=
1306 kW
0.1816 m
Ans
=
=
0.3632 m
⇐=
Stagnation pressure ratio,
r
=
=
3.5
=
1.234
Ans
⇐=
64.98 × 1.005 × 20
Ans
⇐=
Rotor air angle at root section,
=
2πRroot N
60
=
2 × π × 0.2724 × 6000
= 171.15 m/s
60
tan β1
=
171.15
140
β1
=
50◦ 71
tan β2
=
tan β1 −
Cp ΔT03
Ωuroot ca
=
1.2225 −
1005 × 20
= 0.32
0.93 × 171.15 × 140
=
17.74◦
uroot
β2
=
1.2225
Ans
⇐=
Ans
⇐=
9.9 Determine the stage efficiency, ηs and work done factor Ω of an axial
flow compressor, if the actual pressure ratio developed was 1.35 and
actual temperature rise was 30 K. The blade inlet and outlet angles
are 47◦ and 15◦ respectively. The peripheral and axial velocities are
225 m/s and 180 m/s respectively.
Axial Flow Compressors
383
Solution
rs
γ
γ−1
ΔT0s ηs
1+
T01
=
γ−1
ηs
rs γ − 1 T01
=
=
ΔT0s
1.350.286 − 1 × 300
× 100
30
Ans
⇐=
=
89.6%
ΔT0s
=
Ω
uca (tan β1 − tan β2 )
Cp
Ω
=
ΔT0s Cp
uca (tan β1 − tan β2 )
=
30 × 1005
225 × 180 × (tan 47 − tan 15)
=
0.925
Ans
⇐=
9.10 A 50% reaction, axial flow compressor runs at a mean blade speed
of 250 m/s. The pressure ratio developed by the machine is 1.3.
Determine the blade and air angle if the mean flow velocity was 200
m/s. Condition at inlet are 1 bar and 300 K.
Solution
β α1
w1
1
c1
ca1
wt1
ct1
u
Fig. 9.34
ΔT
=
r
γ−1
γ
− 1 T1
=
1.30.286 − 1 × 300
=
23.38K
Degree of reaction
R
=
ca
(tan β1 + tan β2 )
2u
0.5
=
200
(tan β1 + tan β2 )
2 × 250
Ans
⇐=
384
Gas Turbines
tan β1 + tan β2
=
1.25
ΔT
=
uca
(tan β1 − tan β2 )
Cp
tan β1 − tan β2
=
23.38 × 1005
250 × 200
β1
=
40.7◦
⇐=
β2
=
21.3◦
⇐=
α1
=
β2
α1
=
21.3◦
⇐=
α2
=
40.7◦
⇐=
=
0.47
Ans
Ans
For 50% reaction,
and α2
=
β1
Ans
Ans
9.11 A helicopter gas turbine requires an overall compressor pressure ratio
of 10:1. This is to be obtained using a two-spool layout consisting
of a four-stage axial compressor followed by a single-stage centrifugal
compressor. The polytropic efficiency of the axial compressor is 92
per cent and that of the centrifugal is 83 per cent.
The axial compressor has a stage temperature rise of 30 K, using a
50 per cent reaction design with a stator outlet angle of 20◦ . If the
mean diameter of each stage is 25.0 cm and each stage is identical,
calculate the required rotational speed. Assume a work done factor
of 0.86 and a constant axial velocity of 150 m/s.
Assuming an axial velocity at the eye of the impeller, an impeller tip
diameter of 33.0 cm, a slip factor of 0.90 and a power input factor
of 1.04, calculate the rotational speed required for the centrifugal
compressor. Ambient conditions are 1.01 bar and 288 K.
Solution
β α1
w1
1
c1
ca1
wt1
ct1
u
Fig. 9.35
For symmetrical stages (R = 0.5)
Axial Flow Compressors
u
=
ca (tan β1 + tan β2 )
385
(1)
Temperature rise across the stage is
uca
ΔT0s
=
Ω
(tan β1 − tan β2 )
Cp
=
ΔT0s Cp
Ωca (tan β1 − tan β2 )
Cp ΔT0s
=
Ωc2a tan2 β1 − tan2 β2
tan2 β1
=
Cp ΔT0s
+ tan2 β2
Ωc2a
=
1005 × 30
+ tan2 20
0.86 × 1502
β1
=
52.44
u
=
150 × (tan 52.44 + tan 20)
=
249.66 m/s
=
249.66
π × 0.25
u
(2)
Equating (1) and (2)
Nac
=
=
=
1.6906
πDNac
Ans
317.88 rps
⇐=
Total pressure ratio across the axial compressor is
γ
η
nΔT0s ( pc γ−1 )
=
1+
r1
T1
=
0.92× 1.4
0.4 )
4 × 30 (
1+
288
=
3.07
Centrifugal compressor : It is installed after the axial compressor.
Outlet of axial
=
Inlet to centrifugal
Pressure ratio across centrifugal compressor
10
r2
= 3.2573
=
3.07
T02 is outlet temperature of axial compressor
T02
1 γ−1
ηpc γ
=
T01 [r1 ]
=
0.4
288 × [3.07]( 1.4×0.92 )
=
Temperature at the outlet of centrifugal compressor,
408.02 K
386
Gas Turbines
T03
1 γ−1
ηpc γ
=
T02 [r2 ]
=
0.4
× 1
408.02 × [3.2573]( 1.4 0.83 )
=
612.67 K
=
T03 − T02
=
612.67 − 408.02
Cp ΔTsc
=
pif μu2
u
=
ΔTsc
Ncc
=
204.65 K
204.65 × 1005
1.04 × 0.9
=
468.76 m/s
=
πDNcc
=
468.76
= 452.15 rps
π × 0.33
Ans
⇐=
Review Questions
9.1 What are the basic requirements of compressors for aircraft applications? Do axial flow compressors meet them? Explain.
9.2 With a suitable sketch explain the working principle of an axial flow
compressor.
9.3 What is meant by a stage and explain in detail the stage velocity
triangles.
9.4 Derive an expression for work input to the compressor and explain.
What is meant by work done factor?
9.5 Explain the following performance coefficients:
(i) flow coefficient
(ii) rotor pressure flow coefficient
(iii) rotor enthalpy loss coefficient
(iv) diffuser pressure loss coefficient
(v) diffuser enthalpy loss coefficient
(vi) loading coefficient
9.6 Define degree of reaction and derive an expression for the same.
9.7 What is meant by low degree of reaction and high degree of reaction?
How do you differentiate these two?
9.8 Show that when the degree of reaction is 50% the blades are symmetrical.
Axial Flow Compressors
387
9.9 Briefly explain the flow through a compressor and bring out the details
of various losses in an axial flow compressor.
9.10 Derive an expression to calculate the pressure ratio in a stage.
9.11 Explain briefly the design and off-design characteristics of an axial
flow compressor.
9.12 Explain the phenomena of surging and stalling in an axial flow compressor. Explain also the rotating stall.
Exercise
9.1 Air enters an axial flow compressor at 1 bar and 20◦ C at low velocity. It is compresses through a pressure ratio of 11. Find the final
temperature and pressure at outlet from the compressor. Take the
compressor efficiency as 85%.
Ans: (i) 632.65 K (ii) 11 bar
9.2 An axial flow compressor stage has blade root, mean and tip velocities
of 150, 200 and 250 m/s. The stage is to be designed for a stagnation
temperature rise of 20 K and an axial velocity of 150 m/s, both constant from root to tip. The work done factor is 0.93. Assuming 50%
reaction at mean radius calculate the stage air angles. Also calculate
stage air angles and the degree of reaction at root and tip for a free
vortex design.
Ans: Mid : 45.76◦; 17.04◦;
Root : 45◦ ; 2.256◦ ; 0.52
Tip : 59.04◦; 47.47◦; 0.83
9.3 Recalculate the stage air angles for the same data as in the previous question for a stage with 50 per cent reaction at all radii.
Ans: Root : 44.42◦ ; 1.14◦
Mean : 45.76◦; 17.04◦;
Tip : 48.24◦; 28.59◦
9.4 An axial flow compressor has an overall pressure ratio of 4.0 and mass
flow of 160 kg/min. If the polytropic efficiency is 0.88 and the stagnation temperature rise per stage must not exceed 25◦ C, calculate the
number of stages required and the pressure ratio of the first and last
stages. Assume equal temperature rise in all stages. If the absolute
velocity approaching the last rotor is 155 m/s at an angle of 20◦ from
the axial direction, the work done factor is 0.83, the velocity diagram
is symmetrical, and the mean diameter of the last stage rotor is 180
mm, compute the rotational speed and the length of the last stage
rotor blade at inlet to the stage. Ambient conditions are 1.01 bar and
288 K. Also calculate air angles.
Ans: (i) 7 (ii) 1.265 (iii) 1.173 (iv) 24355 rpm (v) 1.255 cm
(vi) 50.5◦ ; 20◦
388
Gas Turbines
9.5 An axial flow compressor was tested and found that it gave a pressure
ratio of 3 atmospheres and a temperature rise of 125◦ C. A 2000 kW
motor was used to drive the compressor. Determine the compressor
efficiency and the mass flow of air delivered, if the mechanical efficiency to be 95% and pressure and temperature at inlet were 1 atm
and 300 K respectively.
Ans: (i) 88.6% (ii) 15.124 kg/s
9.6 A pressure ratio of 1.5 is to be achieved in a single-stage of an axial
flow compressor. The air approaches the rotor at an absolute velocity
of 200 m/s and an angle of 12◦ . The speed of the compressor is 8000
rpm and has a mean radius 30 cm. Determine the blade angles and
the blade height if the mass flow is 10 kg/s. Conditions at inlet are
stagnation pressure = 1 bar and stagnation temperature = 300 K.
Compressor stage efficiency is 95%.
Ans: (i) 47◦ (ii) 15.5◦ (iii) 28 mm
9.7 A multistage axial flow compressor, with symmetrical blading is to
deliver 10 kg/s with a pressure ratio of 4. The working fluid air enters
the stage with a speed of 200 m/s and at an angle of 12◦ . The mean
blade speed of the rotor is 250 m/s and rotational speed is 5000 rpm.
Determine the blade angles, number of stages required for compressor
and blade height, if the stage efficiency is 90% and inlet stagnation
temperature and pressure are 300 K and 1 bar.
Ans: (i) 46.81◦ (ii) 12◦ (iii) 4 stages (iv) 17.57 mm
9.8 A 50% reaction, axial flow compressor has inlet and exit blade angles
of 45◦ and 15◦ respectively. The axial flow velocity is to be maintained
at 200 m/s. Determine the mass flow, blade height, tangential force,
axial force and the resultant force acting on the blade to produce a
power of 300 kW at a speed of 5000 rpm and a blade spacing of 9 mm.
Take p01 = 1 bar and T01 = 300 K.
Ans: (i) 8.08 kg/s (ii) 13.77 mm (iii) 1183 N
(iv) 4.44 N (v) 1183 N
9.9 From the cascade test of a 50% reaction axial flow compressor, the
loading coefficient ψ was found to be 0.43 with the velocity of air
flow in the axial direction being 200 m/s and the rotor speed 250
m/s. Determine the blades angles and pressure ratio if the maximum
compressor blade efficiency were to be 95%. Assume solidity ratio
(the ratio of chord length to blade spacing) = 1, p01 = 1 bar and
T01 = 300 K.
Ans: (i) 41.8◦ (ii) 19.61◦ (iii) 1.32
9.10 For an axial flow cascade, with usual notation derive the drag and lift
coefficients as
CD
=
Δp0 s cos3 αm
1
2 l
cos2 α1
2 ρc1
Axial Flow Compressors
CL
= 2
s
389
(tan α1 − tan α2 ) cos αm − CD tan αm
From the above show that the maximum diffuser (blade) efficiency
can be written as
D
ηDmax = 1 − 2
L
State the assumption made. Also show that for a 50% reaction axial
compressor, the pressure rise in a rotor blade in a cascade can be
derived as
Δprotor =
1 2
ρc
(CL sin αm − CD cos αm )
2 m s
A 50% reaction, aspect ratio (the ratio of blade height to chord length)
3 and blade height 10 cm, axial compressor cascade was tested and
found to have a blade efficiency of 90% and a lift coefficient of 0.8.
The mean axial velocity was 200 m/s and the inlet and exit blade
angle were 45◦ and 15◦ respectively. Determine the lift and drag
forces exerted on the blade, and pressure rise achieved in the stage
with the flow coefficient of 0.5. Density of air at inlet is 0.9 kg/m3 .
Take solidity ratio as 1.
Ans: (i) 67.3 N (ii) 3.4 N (iii) 0.2 bar
9.11 An axial flow compressor stage was found to have a drag coefficient
of 0.04. The blades are symmetric, the inlet and exit blade angles
were 50◦ and 15◦ respectively. The mean axial velocity is 200 m/s.
Determine the blade efficiency and the actual pressure ratio. Take
solidity ratio as 1 and inlet static temperature as 275 K.
Ans: (i) 96.86% (ii) 1.833
9.12 A three-stage axial flow compressor developing a pressure ratio of
3 delivers 10 kg/s of air. The fluid (air) enters the rotor with a
velocity of 220 m/s and at an angle of 15◦ . The velocity diagram
is symmetrical. Determine the speed of the compressor if the blade
height is restricted to 25 mm at the inlet. Take inlet conditions,
p01 = 1 bar, T01 = 300 K, degree of reaction = 0.5 and polytropic
efficiency = 0.9. Take inlet conditions as stagnation state.
Ans: 8091 rpm
9.13 An axial compressor is fitted with 50% reaction blading, the blade
inlet and outlet angles being 50◦ and 15◦ when measured from the
axial direction. The mean diameter of a certain blade pair is 85
cm and the speed is 5500 rpm. Calculate the axial velocity and the
isentropic efficiency of the stage if the pressure ratio of compression
is to be 1.4 when the air inlet temperature is 25◦ C.
Ans: (i) 167.69 m/s (ii) 79.8%
390
Gas Turbines
9.14 An axial flow compressor has a flow coefficient of 0.8 and the loading
coefficient is 0.88. If the blades are symmetrical, calculate the blade
angles and the speed of the compressor. Take axial velocity as 200 m/s
and mean blade diameter as 47.75 cm.
Ans: (i) 49.6◦ (ii) 4.289◦ (iii) 10000 rpm
9.15 An axial flow compressor runs at 9000 rpm and the mean blade diameter for the fourth stage rotor is 55 cm. The rotor blades are 90 mm
high and the mass flow rate is 45 kg/s. At entry to the pressure are
345 K and 1.7 bar. While the air leaves the previous row of stator
blades at an angle of an angle of 28◦ to the axial direction. Calculate
the stage temperature rise, work input and the pressure ratio for the
compressor stage, given that the rotor blades deflect the air through
18◦ , that the work done factor is 0.88 and isentropic efficiency is 85%.
Ans: (i) 18.8 K (ii) 850.23 kW (iii) 1.163
9.16 From the cascade test of a 50% reaction axial flow compressor, the
lift CL was found to be 0.8 with the velocity of air flow in the axial
direction being 200 m/s and the rotor speed 250 m/s. Determine the
temperature rise and blades angles if the maximum compressor blade
efficiency were to be 95%. Assume solidity ratio (the ratio of chord
length to blade spacing) as 1, p01 = 1 bar and T01 = 300 K.
Ans: (i) 23.83 k (ii) 40.84
◦
(iii) 21.08
Multiple Choice Questions (choose the most appropriate answer)
1. For aircraft application a compressor must have
(a) low air flow capacity
(b) high frontal area
(c) high pressure ratio per stage
(d) low efficiency
2. Modern multistage axial flow compressors have a pressure ratio
(a) 2
(b) 6
(c) 12
(d) 18
3. The three velocities of a compressor c, u and w is related by
(a) u = w − c
(b) c = u + w
(c) w = c − u
(d) c/u = w
◦
Axial Flow Compressors
391
4. In the axial flow compressor the absolute velocity in the stator
(a) increases
(b) decreases
(c) initially increases and then decreases
(d) remains constant
5. In the axial flow compressor the absolute velocity in the rotor
(a) increases
(b) decreases
(c) initially increases and then decreases
(d) remains constant
6. The work absorbing capacity of an axial flow compressor
(a) increases with increase in the axial velocity
(b) decreases with increase in the axial velocity
(c) remain the same with increase in the axial velocity
(d) has no relation between them
7. For a axial flow compressor the loading coefficient for the given stage
work is
(a) directly proportional to u
(b) inversely proportional to u
(c) inversely proportional to u2
(d) directly proportional to u2
8. The flow coefficient is defined as
(a) φ = ca − u
(b) φ = ca + u
(c) φ = ca /u
(d) φ = ca × u
9. The degree of reaction is defined as
actual enthalpy change in rotor
actual enthalpy change in stage
actual enthalpy change in stage
(b)
actual enthalpy change in rotor
actual enthalpy change in rotor
(c)
actual enthalpy change in stator
actual enthalpy change in stator
(d) actual enthalpy change in rotor
(a)
392
Gas Turbines
10. In a multistage axial flow compressor, the axial velocity at higher
stage is
(a) small
(b) high
(c) remains same
(d) none of the above
Ans:
1. – (c)
6. – (b)
2. – (d)
7. – (c)
3. – (b)
8. – (c)
4. – (b)
9. – (a)
5. – (a)
10. – (a)
10
COMBUSTION
SYSTEMS
INTRODUCTION
The combustion process is of critical importance in a gas turbine cycle.
It is because in this process the chemical energy of the fuel is converted to
heat energy, which is later converted into work by the turbine. Therefore
losses incurred in the combustion process will have a direct effect on the
thermal efficiency of the cycle.
The combustion system is not yet amenable to a complete theoretical
analysis as the other components of a gas turbine. Although sufficient
experience has been gained to enable a new design of a combustion system
that would give a qualitative idea of its performance, the design can only
be perfected by testing the component independently and modifying it in
the light of the test results.
This is due to the basic problem inherent in itself. In the gas turbine
combustion system initially a mixing of fuel and air under conditions in
which the resulting flame is self-sustaining should be accomplished first.
Further, the chemical reaction should be complete. Thus, the combustor
design involves the formation of turbulent zones, with the complication of
both aerodynamics and thermo-chemical effects. The details of aerodynamics is available for the design of flow around stream lined bodies and also
the flow through channels. The production of exactly controlled turbulent
mixing zones is so complex that one has to resort to empiricism. Thus, in
spite of the tremendous research effort on combustion, there is still no exact
mechanism for flame stabilization in steady flows which would be generally
agreed upon. The formulation of rules for similarity in combustion is still
only in a tentative form.
However, it may be noted that during the last four or five decades, gas
turbine combustor technology has developed gradually and continuously
rather than through any dramatic change. Despite this, the challenge to
ingenuity in design is greater than ever before. New concepts and tech-
394
Gas Turbines
nology are still needed to satisfy current and projected pollutant emission
regulations. Further, there is a growing emphasis on engines that can utilize a much broader range of fuels. With this in mind, in this chapter we
will discuss the general principles and broad correlations which have been
established over the years for the design of gas turbine combustion system.
10.1
COMBUSTION THEORY APPLIED TO GAS TURBINE COMBUSTOR
In any combustion process obtaining complete reaction between fuel and
air has a chemical aspect and a physical aspect. Chemical aspects are
concerned with rate of reaction etc., whereas physical aspects are concerned
with particle size, injection mixing and evaporation. To understand these
aspects completely, one has to refer to the various combustion theories
presented from time to time in the combustion literature.
There are three recognized postulations as to the combustion mechanism:
(i) Carbon preferential burning which states that carbon in the hydrocarbon fuels burns before the hydrogen.
(ii) Hydrogen preferential burning which states that hydrogen in the hydrocarbon fuels burns before the carbon.
(iii) Hydroxylation which states that there is an initial uniting of oxygen
with the hydrocarbon to form a hydroxylated compound. Through
chain reactions of molecules, atoms and radicals, hydroxylated compound burns to CO, CO2 and H2 O. The hydroxylation theory, although not complete, seems to have met with more general acceptance
than the other two theories.
The modern theory is based on the statistics of probability as well as kinetics. It is known from kinetic theory of gases that the individual molecules
are in motion at some average velocity but with a wide difference between
the velocities of the slowest and fastest molecules. For combustion reaction to take place the process requires the collision of molecules of fuel and
oxygen. The collision must have a sufficiently high energy level so that the
molecules are broken down into atoms and radicals. Since, the temperature
is a function of the molecular activity, raising the temperature increases the
probability and intensity of collision of high velocity molecules. Therefore,
there will be an increase in the intensity of combustion.
Once the initial combustion starts it proceeds by means of chain reactions. Acceleration of the reaction is brought about by chain branching and
is retarded by chain breaking. Free atoms of H and O are the free radicals
and they may act as the chain carriers. Chain carriers may be destroyed
through a surface reaction. In normal combustion, the relationship between
chain branching and chain breaking is such that the reaction proceeds more
or less at a constant rate.
Combustion Systems
395
The process of combustion in a gas turbine combustion involves the
following four important steps:
(i) formation of reactive mixture
(ii) ignition
(iii) flame propagation
(iv) cooling of combustion products with air
Under ideal situations, each of these steps would be completed separately before proceeding to the next step. However, space restrictions
under the normal operating conditions do not permit the achievement of
these ideal conditions.
The intimate and uniform mixing of fuel and air is an essential prerequisite to complete the combustion process. The closer this mixing approaches
molecular dimensions, there is a better possibility for complete combustion.
To achieve this, the fuel should be atomized first, thus providing for good
distribution with large surface area for evaporation at the same time.
Apart from providing for good distribution and vaporization of the fuel,
it is necessary to maintain an optimum air-fuel ratio to ensure ignition and
sufficiently fast combustion. Rich mixture, i.e., insufficient air will result
in cracking of the fuel with the formation of amorphous carbon which is
difficult to burn. Although insufficient air is a cause of carbon formation,
the problem is intimately associated with improper mixing. On the other
hand, lean mixture or poor atomization and poor mixing will lead to failure
of combustion.
In actual practice it is necessary to use mixtures in order to satisfy this
latter requirement, viz., failure of combustion, since the atomization and
distribution can never be achieved on a molecular scale. There is a fixed
range for the air to travel through the combustor for stable combustion.
This limits the range of air-fuel ratios between the rich and lean mixtures.
Operation outside this range results in unstable combustion causing vibrations and combustion failure. Figure 10.1 shows a typical stability curve
as a function of air-fuel ratio and mass flow rate. The stable region lies
between the rich and lean limits. It will be noticed that the stability range
and air-fuel ratio range decreases as the air velocity is increased. Normally
combustion chambers are designed with an inlet air velocity not exceeding
80 m/s at design load.
In order to cool the products of combustion to a temperature acceptable
to turbine blades, it is necessary to use a total air-fuel ratio far in excess
of those permitting stable combustion. This difficulty is usually avoided
by admitting a satisfactory amount of primary air so as to maintain stable
combustion. The products of combustion are then cooled by introducing
additional air called secondary air. The air-fuel ratio calculated with respect
to the sum of primary and secondary air is known as the total air-fuel ratio.
From an analytical point of view it is convenient to consider the gas
turbine combustor flame tube as comprising three main zones: primary, intermediate or secondary and dilution. Combustion takes place mainly in the
396
Gas Turbines
Air-fuel ratio
200
We
ak
150
100
lim
it
Stable region
50
Rich
0
0
t
limi
0.25 0.5 0.75 1.0
Air mass flow kg/s
1.25
Fig. 10.1 Typical stability curve
primary zone. In the intermediate zone a small amount of air is injected into
the hot gases coming out of the primary zone, to lower their temperature
and thereby encourage the recombination of dissociated species. Finally, in
the dilution zone, the combustion gases are mixed with the amount of air
needed to achieve the required turbine inlet temperature. Aerodynamics
plays a key role in all three zones.
For different fuels there are correspondingly different ranges of inflammability within which propagation of combustion will take place once the initial reaction starts. For a given fuel this range is affected by liming rich and
lean mixture beyond which a self-propagation combustion reaction cannot
take place. The range of inflammability is affected mainly by three factors:
(i) pressure, (ii) temperature and (iii) the thoroughness of mixing of fuel
and air.
The above three factors determine the ignition characteristics of the fuel.
Thus, the spontaneous combustion will occur in a reactive mixture only if
the mixture is heated to its self ignition temperature. At pressures below
atmospheric at altitudes, the ignition temperature is much higher than at
sea level condition. This will pose problems for aircraft gas turbines.
It may be noted that the reaction becomes self propagating only when
the reaction rate exceeds the energy transfer to the surrounding unburnt
mixture. A certain minimum time is required for the initial reaction to take
place. It should be noted that this time, called ignition delay, is dependent
on the temperature of the mixture. If the fuel is heavy so that evaporation
of the droplets is slow the rate of chemical reaction will be slow due to poor
mixing. Initially a small spark is sufficient to initiate combustion. When
fuels with high ignition temperatures or where high convective heat-transfer
rates are used especially in a high velocity stream, a pilot flame may be
necessary.
The rate of flame propagation depends primarily on the range of inflammability, the pressure and the temperature of the mixture, and the
shape of the combustion chamber. Hence, these factors must be given due
Combustion Systems
397
consideration. To some extent the rate of flame propagation is also affected
by the expansion of hot gases behind the flame front.
In a quiescent reactive mixture, it is expected that the rate of reaction and the rate of flame propagation would increase. However, in the
constant-pressure combustion process used in the gas turbine power plant
it is found that after the initial acceleration the rate of flame propagation
becomes steady. This steady value is a function of physical conditions enumerated above. Flame speeds in the constant-pressure combustion process
are comparatively low (2 to 5 m/s). Therefore, a region of low stream velocity must be provided in which the required rate of flame propagation can
be achieved.
Turbulence is another important requirement in the process of mixing
and has a marked effect on the rate of flame propagation. However, increased turbulence has a corresponding effect on the heat transfer rate and
it is reasonable to assume that there is an optimum turbulence level beyond
which flame extinction would occur. When the flow velocity exceeds that
of the flame, ignition may still take place in a low velocity region provided
behind the turbulent wake of a flame stabilizer aiding the propagation of
the flame downstream.
10.2
FACTORS AFFECTING COMBUSTION CHAMBER
DESIGN
Experience gained from the design of commercial heating plants, such as
oil-fired boilers is of little use to the designer of combustion systems for gas
turbines. A number of requirements peculiar to the gas turbine make the
design problem quite different, and these may be summarized as follows:
(i) The temperature level of the gases after combustion must be comparatively low to suit the highly stressed turbine blade materials.
(ii) At the exit of the combustion chamber the temperature distribution
must be of known form if a high turbine performance is to be realized
and the blades are not to suffer from local overheating. It need not
be uniform, but may increase with radius over the turbine annulus
because the blade stresses decrease from root to tip.
(iii) Combustion must be maintained in a stream of air moving with a high
velocity in the region of 30–60 m/s, and stable operation is required
over a wide range of air-fuel ratio from full load to idling conditions.
The air-fuel ratio might vary from about 60:1 to 120:1 for simple gas
turbines and from 100:1 to 200:1 if a heat-exchanger is employed.
The heat dilution implied by these figures is necessary to satisfy the
turbine material considerations.
(iv) The formation of carbon deposits (coking) must be avoided, particularly the hard brittle variety. Small particles carried into the turbine
along with the high-velocity gas stream can erode the blades. Further, aerodynamically excited vibration in the combustion chamber
398
Gas Turbines
might cause sizable pieces of carbon to break free resulting in the
irreparable damage to the turbine.
(v) In aircraft gas turbines, combustion must be stable over a wide range
of chamber pressure because this parameter changes with altitude and
forward speed.
Probably the only feature of the gas turbine which eases the combustion
engineer’s problem is the peculiar interdependence of compressor delivery
air density and mass flow which leads to the velocity of the air at entry to
the combustion system being reasonably constant over the operating range.
For aircraft applications, there are additional limitations like small space
and low weight, which are, however, slightly offset by somewhat shorter endurance requirements. Aircraft engine combustion chambers are normally
constructed of light gauge heat-resisting alloy sheet (approximately 0.8 mm
thick), but are only expected to have a life of some 10,000 hours. Combustion chambers for industrial gas turbine plant may be constructed on
much sturdier lines but, on the other hand, a life of about 100,000 hours
is required. Refractory linings may eventually be used in heavy chambers.
However, the effects of hard carbon deposits breaking free, apply with even
greater force to refractory material.
We have already seen that the gas turbine cycle is very sensitive to
component inefficiencies, and it is of the utmost important that the aforementioned requirements should be met without sacrificing combustion efficiency. That is, it is essential that over most of the operating range all
the fuel injected should be completely burnt and the full calorific value
realized. Also, because any pressure drop between inlet and outlet of the
combustion chamber leads to both an increase in specific fuel consumption
and reduction in specific power output, it is essential to keep the pressure
loss to a minimum.
It may be noted that the smaller the space available for combustion
the shorter will be the time available for the necessary chemical reactions.
Hence, it is more difficult to meet all the requirements and still obtain a
high combustion efficiency with low pressure loss. Clearly in this respect
the designer of combustion systems for industrial gas turbines has much
easier task than its counterpart in the aircraft field.
10.3
FACTORS AFFECTING COMBUSTION CHAMBER
PERFORMANCE
The main function of a gas turbine combustor is to effect the chemical
combination of oxygen of the air supplied by the compressor with carbon
and hydrogen components of the fuel in such a manner that a steady stream
of gas at a uniform temperature is produced. Even after years of research
work we have gained only limited experience in the design of combustion
system for gas turbines. A number of factors peculiar to the gas turbine
affect the performance of the combustion chamber. We will briefly discuss
them in this section.
Combustion Systems
10.3.1
399
Pressure Loss
It is evident that the turbulence is necessary for rapid combustion. However, this will cause some pressure drop in the combustion chamber. This
loss is usually be regarded as a parasitic loss and hence should be minimized. The pressure losses are caused by two factors, viz., pressure drop
due to friction and that due to the accelerations accompanying heat addition. Thus, the overall pressure loss in a gas turbine combustion system
arises due to
(i) cold losses, and
(ii) hot losses.
The combined pressure loss due to both heating and friction is the sum
of the pressure losses determined separately as cold losses and hot losses.
Thus the overall stagnation pressure loss can be regarded as the sum of
the fundamental loss (a small component which is a function of T02 /T01 )
and the frictional loss. Our knowledge of friction in ordinary turbulent
pipe flow at high Reynolds number would suggest that when the pressure
loss is expressed non-dimensionally in terms of the dynamic head it will not
vary much over the range of Reynolds number in which combustion systems
operate. Experiments have shown, in fact, that the overall pressure loss can
often be expressed adequately by an equation of the form
Pressure loss factor, P LF
=
Δp0
2
ṁ /2ρ1 A2m
The above pressure loss factor can be brought to the form
P LF
=
K2
T02
−1
T01
(10.1)
where K2 is the constant arising out of hot losses. However, it may be
noted that due to friction and turbulence cold losses also do occur which is
constant. Hence, the total pressure loss factor is given by
P LF
=
Cold loss + Hot loss
=
K1 + K2
T02
−1
T01
(10.2)
Note that rather than ρ1 c21 /2, a conventional dynamic head is used based
on a velocity calculated from the inlet density, air mass flow rate, ṁ, and
maximum cross sectional area Am of the chamber. This velocity – sometimes known as the reference velocity – is more representative of conditions
in the chamber, and the convention is useful when comparing results from
chambers of different shape. Equation 10.2 is illustrated in Fig.10.2. If K1
and K2 are determined from a combustion chamber on a test rig from a cold
run and a hot run, then Eq.10.2 enables the pressure loss to be estimated
when the chamber is operating as part of a gas turbine over a wide range
of conditions of mass flow, pressure ratio and fuel input.
400
Gas Turbines
Pressure loss factor
40
Fundamental pressure loss
30
20
10
Cold loss K1
0
1
2
Temperature ratio
T02
3
T 01
Fig. 10.2 Variation of pressure loss factor
To give an idea of relative orders of magnitude, typical values of P LF
at design operating conditions for tubular, turbo-annular and annular combustion chambers are 35, 25 and 18 respectively. There are two points
which must be remembered when considering pressure loss data. Firstly,
the velocity of the air leaving the last stage of an axial compressor is quite
high – say 150 m/s – and some form of diffusing section is introduced between the compressor and combustion chamber to reduce the velocity to
about 60 m/s. It is a matter of convention, depending upon the layout
of the gas turbine, as to how much of the stagnation pressure loss in this
diffuser is included in the P LF of the combustion system. In other words,
it depends on where the compressor is deemed to end and the combustion
chamber begin.
Any pressure loss due to turbulence and skin friction in an unheated gas
stream is usually quoted in dimensionless form as a friction of the dynamic
head of the flow. It may be shown by dimensional analysis that this dimensionless group, or pressure loss factor is a function of Reynolds number
and Mach number if compressibility effects are considered. Over the range
of operating conditions met within combustion systems the effects of these
two variables are small, so that the pressure loss factor is approximately
constant for any given combustion chamber. Once this factor has been determined experimentally for one particular set of flow conditions, i.e., mass
flow, inlet pressure and temperature, the frictional pressure loss may be
estimated for any other set of inlet conditions.
10.3.2
Combustion Intensity
The size of the combustion chamber is determined primarily by the rate
of heat release required. The nominal heat release rate can be found from
ṁf ΔHc , where ṁ is the mass flow rate, f the fuel-air ratio and ΔHc
the net calorific value of the fuel. Enough has been said for the reader to
appreciate that the larger the volume which can be provided the easier it will
Combustion Systems
401
be to achieve a low pressure drop, high efficiency, good outlet temperature
distribution and satisfactory stability characteristics.
The design problem is also eased by an increase in the pressure and
temperature of the air entering the chamber, for two reasons. Firstly, an
increase will reduce the time necessary for the ‘preparation’ of the fuel
and air mixture (evaporation of droplets etc.,) making more time available
for the combustion process itself. Note that since the compressor delivery
temperature is a function of the compressor delivery pressure, the pressure
(usually expressed in atmospheres) is an adequate measure of both.
Secondly, it may be noted that the combustion chamber pressure is
important because of its effect on the rate at which the chemical reactions
proceed. An indication of the nature of this dependence can be obtained
from chemical kinetics,, i.e., kinetic theory applied to reacting gases. By
calculating the number of molecular collisions per unit time and unit volume
which have an energy exceeding a certain activation value E, it is possible to
obtain the following expression for the ratio at which a simple bimolecular
gas reaction proceeds.
Rate of reaction ∝ mj mk ρ2 σ 2 T 1/2 M −3/2 e−E/R0 T
ρ and T have their usual meaning and the other symbols denote:
mj , mk : local concentrations of molecules of species, j and k
σ
: mean molecular diameter
M
: mean molecular weight
E
: activation energy
R0
: universal gas constant
Substituting for ρ in terms of p and T we can for our purpose simplify
the expression to
Rate of energy ∝ p2 f (T )
Now T is maintained at a high value by having an approximately stoichiometric mixture in the primary zone. However, we are concerned here with
the independent variable p. It is not to be expected that the theoretical
exponent 2 will apply to the complex set of reactions occurring when a
hydrocarbon fuel is burnt in air. Experiments with homogeneous mixtures
in stoichiometric proportions suggest that it should be 1.8. At first sight it
appears therefore that the design problem should be eased as the pressure
is increased according to the law p1.8 . In fact there is reason to believe
that under design operating conditions the chemical reaction rate is not a
limiting factor. In an actual combustion chamber where physical mixing
processes play such an important role, an exponent of unity is more realistic. This is not to say that under extreme conditions – say at high altitude
– the performance will not fall off more in accordance with the p1.8 law.
A quantity known as the combustion intensity has been introduced to
take account of the foregoing effects. One definition used is
Combustion intensity =
Heat release rate
Combustion volume × pressure
402
Gas Turbines
Another definition employs p1.8 , with the units kW/m3 atm1.8 . However, it is defined, certainly the lower the value of the combustion intensity
the easier it is to design a combustion system which will meet all the desired requirements. It is quite inappropriate to compare the performance
of different systems on the basis of efficiency, pressure loss etc., if they are
operating with widely differing orders of combustion intensity. In aircraft
systems, the combustion intensity is in the region 2 − 5 × 104 kW/m3 atm,
while in industrial gas turbines the figure may be a tenth of this owing to
the larger volume of combustion space available and to the fact that when
a heat-exchanger is employed a smaller heat release is required.
10.3.3
Combustion Efficiency
The combustion efficiency is computed from the chemical analysis of the
gases. This is not easy, as not only it is difficult to obtain representative
samples from the high velocity stream, but also, owing to the high air-fuel
ratios used, the constituents to be measured are of a very small proportion of the whole sample. Ordinary apparatus, such as the Orsat, is not
adequate for this purpose and special methods have been evolved. If an
overall combustion efficiency is all that is required, however, and not an
investigation state of the combustion process at any stage, it is possible to
use an efficiency defined by the energy balance equation. Since the mean
specific heat may be assumed to be same for both the actual and theoretical
temperature rise, this efficiency may be written as
Combustion efficiency =
Actual total − head temperature rise
Theoretical total − head temperature rise
The actual temperature-rise is found by direct measurement at the inlet
and outlet of the chamber, while the theoretical temperature rise may be
found from the already available curves, using measured values of air-fuel
ratio and inlet total-head temperature. Now this expression for efficiency
is the result of an energy balance, i.e., a balance of such terms as ma Cp T0 .
Since there is always in practice a variation in velocity as well as a variation
in temperature across the section of a combustion chamber, it is necessary
to use, not the ordinary arithmetic mean of number of readings across a
section, but what is known as the weighted mean.
10.4
FORM OF COMBUSTION SYSTEM
In an open-cycle gas turbine, where compressor sucks fresh air continuously
from the atmosphere, the direct burning fuel in the working stream is always
used. Because the combustion is self-sustaining and continuous an electric
spark is required only for initiating the process.
Frequently the air leaving the compressor is split into a number of
streams, each supplying to a tubular (can type) combustion chamber. These
chambers are spaced around the shaft connecting the compressor and turbine. Each of these chambers has its own fuel injection system fed from a
Combustion Systems
403
common supply line. This type of layout is particularly suitable for gas turbines using centrifugal compressors where the air stream is already divided
by the diffuser vanes.
A single annular combustion chamber surrounding the rotor shaft would
seem to be ideal for use with axial compressors. In this system maximum
use is made of the space between the compressor and turbine with lower
pressure losses and an engine of minimum diameter. However, there are
certain disadvantages in the single annular chamber. They are
(i) It is difficult to obtain a uniform fuel-air distribution and a uniform
outlet temperature distribution, in spite of employing a large number
of fuel jets.
(ii) The annular chamber is structurally weaker and it is difficult to avoid
buckling of the hot flame tube walls, more so in the case of large
engines.
(iii) The development of an annular chamber will have to be carried out
on a single chamber requiring a test facility capable of supplying the
full engine air mass flow. This requires a huge layout and involves
enormous cost.
However, for industrial gas turbines, with no importance for the space
occupied, the combustion may be carried out in one or two large cylindrical
chambers feeding the turbine via a scroll or volute. Large combustion
chambers might enable adjustable valves or dampers to be incorporated for
regulating the airflow to different points in the chamber.
10.5
REQUIREMENTS OF THE COMBUSTION CHAMBER
The main function of a gas turbine combustion chamber is to provide for
the complete combustion of fuel and air, the air being supplied by the compressor and the products of combustion being delivered to the turbine. In
carrying out this function, the combustion chamber must fulfill the following requirements.
(i) Complete combustion of the fuel must be achieved.
Any unburnt fuel, in the sense of the full heating value per kg not
having been realized, is directly reflected in the fuel consumption or
thermal efficiency.
(ii) The total pressure loss must be minimum.
Combustion pressure loss is one of the flow losses which is called the
parasitic losses and its effect has already been discussed.
(iii) Carbon deposits must not be formed under any expected condition of
operation.
As all normal fuels contain carbon, there is a tendency for free carbon
to be deposited on the walls of the combustor or carried through
404
Gas Turbines
will appear as smoke. Any accumulation of the carbon will upset
the designed flow pattern and increase pressure loss due to blockage.
This can cause damage to the turbine blades when it comes out of the
combustion chamber at a temperature and at high velocity. Smoke
is always objectionable, although a light haze is usually allowable, if
not desirable.
(iv) Ignition must be reliable and accomplished with ease over wide range
of atmospheric conditions.
As the gas turbine is a steady flow machine, ignition is required only at
the time of starting, the combustion being self-sustaining once the fuel
is ignited by outside means. This allows a considerable simplification
of the ignition system. Nevertheless, ignition of a liquid fuel under
all conditions is not always easy, taken in conjunction with the fact
that the igniting device is likely to be permanently exposed to the
combustion gases.
(v) Temperature and velocity distribution at turbine inlet must be controlled.
It is very essential that the gas temperature at the turbine inlet be
either uniform or have a designed gradient. At the high general level
of temperature the strength of the blade material may be critical
within small limits. Poor mixing may produce some hot spots at some
sections. The ideal turbine inlet temperature distribution should be
such that the stress levels in the blade are minimum. A uniform
combustor outlet temperature is not necessarily the most desirable.
Control of the temperature distribution is difficult to achieve to the
desirable limits and sometimes the best that can be obtained without
compromising other qualities is to avoid excessive hot spots. Velocity
distribution is not so important, but some attention to this may be
necessary to avoid particularly bad conditions, for instance where the
turbine immediately follows a bend.
(vi) The volume and weight of the combustor must be kept within reasonable limits.
These criteria are always desirable, of course, but in some applications, the aircraft gas turbine being the prime example, they may be
limiting. Even for stationary industrial application, the size and the
weight of the combustion chamber must be suitable for the turbine
as a whole and this imposes restrictions, on design, as compared with
other steady flow combustion devices.
(vii) Reliability and endurance should be ascertained.
These are again desired criteria, but importance of the former is
paramount in gas turbines, because a failure may wreck the turbine.
Owing to the high speed of the turbine, even a small solid element can
cause considerable damage. Reliability and endurance are differentiated, because in an aircraft engine, for example, a relatively short life
Combustion Systems
405
may be acceptable, while the reliability for that period is absolutely
essential.
All these performance criteria must be satisfied to some degree. The importance of each depending on the application and the problem is to arrive
at a satisfactory compromise. It must be noted that in most cases it is not
only the main operating or design point conditions that must be satisfied,
but part-load and idling conditions as well.
Although both pressure and mass flow rate may vary considerably, they
are to a large extent independent. Together with the adiabatic relationship
of compressor delivery pressure, the result is that the air velocity at entry to
the combustor is relatively constant. This is fortunate, because the velocity
is a control parameter in combustion performance.
10.6
THE PROCESS OF COMBUSTION IN A GAS TURBINE
Combustion of a liquid fuel involves the following:
(i) the mixing of a fine spray of fuel droplets with air,
(ii) vapourization of the droplets,
(iii) the breaking down of heavy hydrocarbons into lighter fractions,
(iv) the intimate mixing of molecules of these hydrocarbons with oxygen
molecules, and finally
(v) the chemical reactions themselves.
A high temperature is necessary if all these processes are to occur sufficiently rapidly for combustion in a moving air stream to be completed in
a small space by the combustion of an approximately stoichiometric mixture. It should be clear that the combustion of a gaseous fuel presents fewer
problems, but much of what follows is still applicable.
Since the overall air-fuel ratio is in the region of 100:1, while the stoichiometric ratio is approximately 15:1, the first essential is that the air
should be introduced in stages. Three such stages can be distinguished.
(i) About 15–20 per cent of the air is introduced around the jet of fuel in
the primary zone to provide the necessary high temperature for rapid
combustion.
(ii) Some 30 per cent of the total air is then introduced through holes
in the flame-tube in the secondary zone to complete the combustion.
For high combustion efficiency, this air must be injected carefully at
the right points in the process, to avoid chilling the flame locally and
drastically reducing the reaction rate in that neighbourhood.
(iii) In the tertiary or dilution zone the remaining air is mixed with the
products of combustion to cool them down to the temperature required at inlet to the turbine. Sufficient turbulence must be promoted
406
Gas Turbines
so that the hot and cold streams are thoroughly mixed to give the
desired outlet temperature distribution, with no heat streaks which
would damage the turbine blades.
Having described the way in which the combustion process is accomplished,
it is now possible to see how incomplete combustion and pressure losses
arise. Due to poor fuel injector design leads to fuel droplets being carried
along the flame-tube wall. This causes incomplete combustion due to local
chilling of the flame at points of secondary air entry. This can easily reduce
the reaction rate. Because of this some of the products into which the
fuel has decomposed are left in their partially burnt state. Due to this the
temperature at the downstream end of the chamber is normally below that
at which the burning of these products can be expected to take place. Light
hydrocarbons into which the fuel has decomposed have a higher ignition
temperature than the original fuel. It may be seen that it is difficult to
prevent some chilling from taking place, particularly if space is limited and
the secondary air cannot be introduced gradually enough. Usually devices
are used to increase large-scale turbulence and more uniform distribution
of the secondary air throughout the burning gases. Because of which the
combustion efficiency will be improved but at the expense of increased
pressure loss. A satisfactory compromise must somehow be reached.
10.7
COMBUSTION CHAMBER GEOMETRY
Discussion about the combustion process in the previous section serves to
indicate the necessary features of practical combustion chamber. First of
all, it is necessary for the reaction zone to operate near the stoichiometric
mixture strength in order to develop the highest possible temperature for
a rapid reaction or rate of flame propagation.
The stoichiometric mixture strength for most hydrocarbon is 15:1. With
an overall air-fuel ratio of 60:1 in a gas turbine, only about one quarter of
the total air must be admitted to the reaction zone. Rest of the air is to
be admitted to reduce the gas temperature to the required turbine inlet
temperature. This reaction zone is often called the primary zone.
The primary zone must have flame stabilizers such as baffles to establish
a recirculation zone. But in addition to stabilization, a vigorous mixing
action must be provided in order to mix air and fuel and then to mix
unburnt mixture with burnt gases. The stability parameter indicates that
it is better to provide a small number of large baffles than a large number
of small baffles.
The necessity of high temperature for significant reaction rate shows
that diluting air must either be added only when it is certain that the
reaction has gone to completion or that it must be added in slow degree so
that the reaction is not immediately quenched. Because of the problem of
mixing air and fuel, it is common to introduce some air as secondary air,
i..e, it has a function in the combustion process in promoting mixing and
ensuring adequate oxygen for the fuel for complete combustion downstream
of the nominal reaction zone. This zone in which the secondary air is added
Combustion Systems
407
is called secondary or mixing zone. Finally, to bring down the temperature
to the required level acceptable to the turbine blades the remaining air is
introduced to mix with the products of combustion. This zone is called
the tertiary or dilution zone. The necessary outline structure of many
combustion chambers then appears as in Fig.10.3, yielding a typical annular
form, with a central linear or flame tube containing the primary zone and
baffle, surrounded by the outer casing or air casing. The annular space
serves the purpose not only for separating the required primary air from the
total air, but of providing a cooling air stream which limits the temperature
of the liner, which contains a reaction zone where the gas temperature
may reach locally a value of the order of 2000 K, corresponding to the
stoichiometric combustion temperature. In actual practice, the stabilizing
baffle is often considerably more complex than the simple bluff bodies.
Fuel
Primary
Air or
zone
outer casing
Air
Gap for
cooling air
Gas
Baffle
Liner or flame tube
Diluting and mixing zone
Fig. 10.3 Typical combustion chamber
10.7.1
Primary Zone or Flame Stabilizing Zone
The zonal method of introducing air cannot by itself give a self piloting
flame in air stream which is moving in order of magnitude faster than the
flame speed in a burning mixture. The essential requirement is therefore to
provide a recirculation zone which directs some of the burning mixture in
the primary zone back into the incoming fuel and air to give the primary
air a considerable tangential components at entry, by means of swirl vanes,
which mixes air and fuel over the cross-section and gives a recirculation
pattern by means of vortex action. A free vortex has increasing tangential
velocity with decreasing radius. By the Bernoulli equation, the higher velocity at the center entails a lower static pressure and thus a radial pressure
gradient. The resulting circulation pattern is shown in Fig.10.4. If fresh
air is admitted downstream then this is drawn upstream towards the fuel
injection region. The use of swirl is a long established practice with oil and
gas firing, but if used along, is liable to give rather coarse mixing, with
some regions over-stimulated and other lacking.
408
Gas Turbines
Swirl vanes
Fig. 10.4 Flow pattern in the primary zone
The basic Lucas combustor (Fig.10.5) combines a small swirler around
the fuel injector, with large number of small holes in surrounding cone, the
latter admitting air in jets, which provide finer degree of general turbulence.
Figure 10.5 also shows another necessary feature of many combustors, in
the flared passage preceding the baffle, which acts as a metering device and
flow straightener for the primary air. Often, the velocity distribution in the
duct from the compressor is highly irregular. If uncorrected, would lead to
poor mixing, stabilization and temperature distribution in the combustor.
Such poor distribution may occur due to separated flow in the diffuser or
to a curved passage necessary from the geometry of the combustors.
Fig. 10.5 A Lucas combustion chamber with inlet swirler
Another simple stabilization pattern, basis of G.E. combustors is shown
in Fig.10.6. Here, the recirculation zone is formed by eddies from a row
of holes just downstream of the fuel injection point. The location of these
primary stabilizing jets, together with admission of secondary and diluting
air further downstream, is critical. This requires considerable test work.
Fig. 10.6 A typical G.E. combustor
Combustion Systems
409
Many types of primary zone stabilizing arrangements are possible. Those
described above are simply among those which are used in considerable
numbers. These cases are all for downstream fuel injection, but often
equally good or better results can be obtained by upstream injection, as
shown schematically in Fig.10.7. The advantage of this, is the possibility
of improved mixing, as air may be injected directly into the center of the
fuel injection zone. The disadvantage is the possibility of overheating the
fuel pipe and injector unless they are sufficiently cooled, as they are located
directly in the reaction zone.
Air
Fig. 10.7 Flame stabilization with upstream injection
10.8
MIXING AND DILUTION
The problem of adding approximately three-quarters of the total air to onequarter of reactant gases in the dilution zone is not an easy job, because of
the necessity for a good temperature distribution with minimum pressure
loss. In the typical cylindrical construction, the air is admitted progressively
through holes or slots, with usually a small gap being left between the liner
and the casing to provide a film cooling air for the final contraction to the
nozzle entry. An excellent method of maintaining the liner walls at the
desired temperature is by means of film cooling. This can be achieved by
making the liner in overlapping sections so that each has a layer of cool air
flowing over it.
The size and the number of dilution air holes is a compromise between
a large number of small holes to give fine scale mixing and a small number
of large holes to give better penetration. It is important that the mixing
air must reach the axis of the combustion chamber, as otherwise at the
core a high temperature gas will persist. A slot with major axis in the
flow direction gives good penetration, but tends to weaken the liner. The
liner is not subjected to a large pressure difference, but may distort owing
to local high-temperature gradients. Further, the combustion process is
accompanied by vibrations of high frequency and the whole combustion
chambers is subjected to a vibration fatigue effect too.
10.9
COMBUSTION CHAMBER ARRANGEMENTS
Comparison of various gas turbines and jet engines gives that, at present,
three arrangements of combustion chambers are being practiced, viz.,
410
Gas Turbines
(i)
a large single chamber, (ii) multiple chambers, and
(iii) the annular chamber.
A large single chamber, is usually employed in the case of heavy industrial power plants where space and weight are of secondary importance.
Since, space is not so important, relatively low gas velocities can be used
which tends towards higher combustion efficiency and low pressure loss.
Further, the absence of a weight restriction permits the use of heavy sections, which if properly designed gives long life. Moreover, high operating
temperatures without buckling can be achieved.
The multiple chamber arrangement is common practice for aircraft applications as well as few stationary power plants or marine engines. Its main
advantage is the ease with which the desired total combustion chamber volume can be achieved. This is beneficial from the design point of view since
if any increase in overall dimensions for that of the compressor and turbine
is required this can be achieved with least difficulty. Secondly, individual
chambers can be removed and replaced without disturbing the rest of the
assembly. The design also fits nicely in with the general diffuser design of
centrifugal compressor, and little difficulty is encountered in obtaining a
fairly uniform gas temperature at exit from chambers.
The third type of combustion chamber, the annular chamber, is illustrated in Fig.10.8. It consists of an annular passage connecting the compressor delivery with the turbine nozzle, permitting the formation of a continuous sheet of hot gas which flows from the former to the latter. Inside
this annular chamber is supported the inner flame tube, which in this case
is also annular. Fuel is injected into the inner annular chamber by series of
nozzles located at the compressor end of the chamber.
Burner Basket
To jet
Annulus
Combustion Systems
411
With near perfect mixing and combustion of the fuel and air an ideal
flow situation can be envisaged in such designs. Further, with less pressure
loss and turbine nozzles being fed with a continuous sheet of gas round
the periphery of the machine. The main difficulty associated with such a
chamber is to supply fuel in such a way as to produce the desired temperature distribution. In general the fuel is fed by means of limited number
of nozzles, possibly 15 to 20 arranged around the inlet of the inner flame
holder. It is believed that recent advances have overcome this difficulty.
A further difficulty associated with the annular type of chamber is that
of providing adequate cooling air to the internal surfaces, bearings, etc.,
which may be located inside the inner annular wall, since this space is
completely blocked off from the outside atmosphere by the compressor,
combustion chamber and turbine wheel.
10.10
SOME PRACTICAL PROBLEMS
To conclude this chapter we will discuss some of the problems which have
so far not been mentioned but which are none the less important. These
are concerned with
(i)
flame tube cooling, (ii) fuel injection,
(iii) ignition,
(iv) use of cheaper fuels, and
(v)
pollution.
10.10.1
Flame-tube Cooling
One problem which has assumed greater importance as permissible turbine
inlet temperatures have increased is that of cooling the flame-tube. The
tube receives energy by convection from the hot gases and by radiation
from the flame. It loses energy by convection to the cooler air flowing
along the outer surface and by radiation to the outer casing, but this loss is
not sufficient to maintain the tube wall at a safe temperature. A common
practice is to leave narrow annular gaps between overlapping sections of the
flame tube, as shown in Fig.10.9(a), so that a film of cooling air is swept
along the inner surface. Corrugated ‘wigglestip’, spot welded to successive
lengths of flame-tube, provides a tube of adequate stiffness with annular
gaps which do not vary too much with thermal expansion. Another method,
illustrated in Fig.10.9(b), is to use a ring of small holes with an internal
splash ring to deflect the jets along the inner surface. A film of cool air
both insulates the surface from the hot gases, and removes energy received
by radiation.
Although empirical relations are available from which it is possible to
predict convective heat transfer rates when film cooling a plate of known
temperature, the emissivities of the flame and flame-tube can vary so widely
that prediction of the flame-tube temperature from an energy balance is
not possible with any accuracy. Even in this limited aspect of combustion
chamber design, final development is still a matter of trial and error on the
412
Gas Turbines
(a)
(b)
Fig. 10.9 Film cooling of flame-tube
test rig. The emissivity of the flame varies with the type of fuel, tending
to increase with the specific gravity. Carbon dioxide and water vapour
are the principle radiating components in non-luminous flames, and soot
particles in luminous flames. It is worth noting that vaporizer systems ease
the problem, because flames from premixed fuel vapour-air mixtures have
a lower luminosity than those from sprays of droplets.
Higher turbine inlet temperatures imply the use of lower air-fuel ratios,
with consequently less air available for film cooling. Furthermore, the use
of a higher cycle temperature is usually accompanied by the use of a higher
cycle pressure ratio to obtain the full benefit in terms of cycle efficiency.
Thus the temperature of the air leaving the compressor is increased and
its cooling potential is reduced. If permissible turbine inlet temperatures
increase much beyond 1500 K, some form of porous material may have to
be used for the flame-tube: ‘effusion cooling’, as it is called, is very much
more economical in cooling air than film cooling.
10.10.2
Fuel Injection
The use of liquid fuel requires that it be injected in metered quantity in
a state as finely divided as possible, so that evaporation and mixing are
rapid. Therefore, the injection system must fulfill the requirements, viz.,
(i) meter the fuel flow, and
(ii) atomize the fuel.
The metering function of a plain hole is a combination of applied pressure and the cross-sectional area of the metering orifice, the flow being
determined by an equation of the form
m
t
where
=
ṁ
=
kAρ
2p
ρ
(10.3)
m : kg of fuel
k
: constant
ρ
: weight per cum of fuel
A : nozzle area sq. meter
t
: time of flow, s
p
: injection pressure at nozzle, kg/m2
This points out one of the difficulties of fuel injection. It may be noted
that the flow varies as the square root of the pressure. For example, if
Combustion Systems
413
1.5 bar is assumed as the lowest value at which a spray can be formed
at minimum flow then it can be shown that a top pressure of 375 bar is
necessary for full flow.
The throttle valve, provided in the fuel system, in general determines the
pressure applied to the nozzle side of the system which in conjunction with
the hole size will fix the flow in accordance with the Eq.10.3. When more
than one nozzle is connected with manifold, as is common, it is essential
that they are all calibrated to divide the fuel equally between the various
outlets within a rather narrow tolerance.
The atomization produced by nozzle is necessary to divide the fuel flow
into a large number of small droplets to permit rapid oxidation with a short
length and complete combustion without the production of smoke.
A simple atomizer is a nozzle with a plain hole, which will, if the fuel is
supplied to it under pressure, results in production of a turbulent stream of
fuel form into droplets because of surface tension. This spray is characterized by a rather narrow cone angle, and a high penetrative power suitable
for diesel-engine injectors. The injectors for oil burners, jet engines and for
similar applications, where penetration is not an important consideration,
produce a softer spray but one that is widely dispersed over a large area to
provide additional surface for oxidation. The type of injector generally used
is the centrifugal or swirl type atomizer shown in Fig.10.10. The essential
of this injector are means of feeding the oil to the periphery of a vortex or
swirl chamber in a roughly tangential direction producing a vortex in which
the angular velocity varies inversely as the radius. The discharge orifice is
small compared with the vortex chamber and the rotating liquid spills out
in a thin sheet, the quantity varying with the pressure of supply. Thus,
there are two factors acting on the fuel,
(i) the pressure acting axially, and
(ii) the centrifugal force acting radially.
The result is that when the flow rate is high enough, this spinning sheet
of liquid will break up into ligaments forming drops which will travel from
the orifice forward and outward giving the hallow cone of fuel spray. This
atomization may occur with at little as 0.7 to 1.05 bar in a well designed
nozzle when using kerosene as fuel.
The requirements of good atomizer can be listed as follows:
(i) Atomization must be effected over a wide range of fuel flows and
pressures.
(ii) No moving parts at the nozzle and of simple design.
(iii) It should distribute the fuel uniformly throughout the air flow.
(iv) It should be easily reproducible to high degree of accuracy to enable
multi-injector systems to operate with uniform distribution.
(v) The weight and size should be low to reduce heat absorption to a
minimum.
414
Gas Turbines
Two or three types of swirl atomizers have been developed. One such
is the simple single-outlet type of Fig.10.8 where all the fuel flow leaves via
the one spray orifice.
Air core
Cone angle of spray
Axial force
Resultant direction
Discharge orifice
Swirl chamber
Tangential feed ducts
Fig. 10.10 Swirl type atomizer
Spray
A
A
Inlet
Tangential ducts
Air core
Valve
Split outlet
Section AA’
Combustion Systems
415
Another type of the burner, known as the Duplex, has two orifices in
parallel feeding into the swirl chamber. One of the ports does not come
into operation until the fuel pressure exceeds a certain value. This naturally
gives a stepped characteristic which is not entirely satisfactory.
10.10.3
Ignition
The first step in starting a gas turbine is to accelerate the compressor
to a speed which gives an air flow capable of sustaining combustion. In
some cases this is achieved by feeding compressed air from an external
supply directly to the turbine driving the compressor. If is more common,
however, to employ an electric motor or small auxiliary turbine connected
to the main shaft by a reduction gear and clutch. The auxiliary turbine may
operate from a compressed air supply or, as in the case of military aircraft
where independence of ground equipment is desirable, from a cartridge.
During the period of acceleration the ignition system is switched on and
fuel is fed to the burners. An igniter plug is situated near the primary zone
in one or two flame-tubes or cans. Once the flame is established, suitably
placed interconnecting tubes between the cans permit ‘light round’, i.e.,
flame propagation from one flame-tube to another. Light round presents
few problems in annular chambers. With an aircraft gas turbine, there is the
additional requirement that re-ignition must be possible under windmilling
conditions if for any reason the flame is extinguished at altitude.
The ignition performance can be expressed by an ignition loop which
is similar to the stability loop of Fig.10.1 but lying inside it. That is, at
any given air mass flow the range of air-fuel ratio within which the mixture
can be ignited is smaller than that for which stable combustion is possible
once ignition has occurred. The ignition loop is very dependent on combustion chamber pressure, and the lower the pressure the more difficult is
the problem of ignition. Relight of an aircraft engine at altitude is thus the
most stringent requirement. Although high-tension sparking plugs similar
to those used in piston engines are adequate for ground starting, a spark
of much greater energy is necessary to ensure ignition under adverse conditions. A surface discharge igniter, yielding a spark having an energy of
about three joules at the rate of one spark per second, is probably the most
widely used type for aircraft gas turbines in which fuel is injected as a spray
of droplets.
One example of a surface-discharge igniter is shown in Fig.10.12. It
consists of a central and outer electrode separated by a ceramic insulator
except near the tip where the separation is by a layer of semiconductor
material. When a condenser voltage is applied, current flows through the
semiconductor which becomes incandescent and provides an ionized path
of low resistance for the energy stored in the capacitor. Once ionization has
occurred, the main discharge takes place as an intense flashover. To obtain
good performance and long life the location of the igniter is critical; it must
protrude through the layer of cooling air on the inside of the flame-tube
wall to the outer edge of the fuel spray, but not so far as to be seriously
wetted by the fuel.
416
Gas Turbines
Semiconductor
Ceramic
Earthed electrod
Central electrode
Fig. 10.12 Surface-discharge igniter
For vapourizing combustion chambers, some form of torch igniter is
necessary. This comprises a spark plug and auxiliary spray burner in a
common housing. The location is not critical, but it is bulkier and heavier
system than the surface-discharge type. The torch igniter is particularly
suitable for industrial gas turbines, and has the advantage that the auxiliary
burner can be supplied with distillate fuel from a separate tank when a
heavy oil is used as the main fuel.
10.10.4
Use of Cheap Fuels
The gas turbine has found a use in process industries where the power
output is desired wholly or in part as compressed air, or steam from a
waste-heat boiler. It is particularly suitable when there is a gaseous byproduct of the process which can be used as fuel for the gas turbine. One
obvious example is the use of a gas turbine in a steel mill for supplying air
to the blast furnaces while using blast furnace gas a fuel. There is little
difficulty in designing a combustion system to handle gaseous fuels and it
is not this form of cheap fuel with which we are concerned here.
That the gas turbine has taken so long to compete seriously with other
forms of prime mover, except for aircraft applications where small size and
low weight are vital, is largely due to the difficulties encountered when
trying to use residual oil. This cheap fuel is the residue from crude oil
following the extraction of profitable light fractions. Some of its undesirable
characteristics are
(i) high viscosity requiring heating before delivery to the atomizers.
(ii) tendency to polymerize to form tar or sludge when overheated.
(iii) incompatibility with other fuels with which it might come into contact, leading to jelly-like substances which can clog the fuel system.
(iv) high carbon content leading to excessive carbon deposits in the combustion chamber.
Combustion Systems
417
(v) presence of vanadium, the vanadium compounds formed during combustion causing corrosion in the turbine.
(vi) presence of alkali metals, such as sodium, which combine with sulphur
in the fuel to form corrosive sulphates;
(vii) relatively large amount of ash, causing build up of deposits on the
nozzle blades with consequent reduction in air mass flow and power
output.
Characteristics (i), (ii), (iii) and (iv) have a nuisance value but the problems
which they give rise can be overcome without much difficulty. (v), (vi) and
(vii) on the other hand have proved to be serious.
The rate of corrosion from (v) and (vi) increases with turbine inlet temperature, and early industrial gas turbines designed for residual oil operated
with temperatures not much greater than 900 K to avoid the problem. Such
a low cycle temperature inevitably meant a low cycle efficiency. It has now
been found that the alkali metals can be removed by washing the fuel oil
with water and centrifuging the mixture, and that fuel additives such as
magnesium can be used to neutralize the vanadium. Although the latter
treatment costs very little, the former is expensive particularly when carried out by the consumer on a relatively small scale: the fuel is no longer
‘cheap’ and the gas turbine becomes barely competitive.
With regard to (vii), i.e., build up of deposits, it has been found that
applications involving intermittent operation (e.g., locomotive propulsion)
are more suitable than those involving continuous running (e.g., base load
electric power generation). This is because expansions and contractions
associated with thermal cycling cause the deposits to crack off. No satisfactory solution has been found for continuous running plant, and high
maintenance costs would be an additional factor making the gas turbine
uneconomic.
Finally, various experimental coal burning gas turbines have been built
but without success. The problems of ash removal after combustion, or
treatment of the coal itself to remove the impurities leading to ash, have
prevented the gas turbine from competing successfully with coal burning
steam power plant.
10.10.5
Pollution
From the related facts that a gas turbine uses very high air-fuel ratios and
provides a combustion efficiency of almost 100 per cent over most of the
operating range, it should be clear that the gas turbine is never likely to be
a major contributor to atmospheric pollution. The main trace pollutants
as from any air-breathing engine burning fossil fuel, are
(i) unburnt hydrocarbons and carbon monoxides,
(ii) oxides of nitrogen, and
(iii) oxides of sulphur.
418
Gas Turbines
The concentration is negligible except under idling conditions when the
combustion chamber operating temperature is at its lowest. With turbojet
engines the order of magnitude is about 20–30 ppm (parts per million) by
weight under normal operating conditions. These pollutants arise due to
local chilling of the chemical reactions. The concentration is therefore minimal with an annular combustion chamber or single large tubular chamber
because such configurations provide a smaller area of cool flame-tube wall
per unit volume than do cannular or multi-tubular designs.
In any air breathing engine a small fraction of the nitrogen is oxidized,
and the nitric oxide so formed in the exhaust continues to oxidize slowly
in the atmosphere to form nitrogen dioxide. Although non-toxic, it is undesirable because it plays an important part in the formation of ‘smog’.
The oxidation of nitrogen increases with combustion temperature, and in a
turbojet exhaust the concentration of nitric oxide is likely to be negligible
under idling conditions rising to approximately 100 ppm by weight at full
power.
Oxides of sulphur are detrimental not only to plant and animal life, but
also to the engine itself because of corrosive acids to which they give rise.
The kerosene used as fuel for most gas turbine, however, has a very low
sulphur content and the sulphur emission is negligible. Furthermore, the
latest oil refineries are now able to produce fuel economically with a much
lower sulphur content than has been specified by users in the past.
Some aircraft gas turbines have exhibited a smoky exhaust indicating
the presence of small particles of unburnt carbon. An amount quite negligible from the point of view of atmospheric pollution can have a pronounced
visible effect due to the light scattering properties of the particles. A smoky
exhaust can be virtually eliminated during development testing by modifications to the primary zone and fuel burners, and it is unlikely to arise
at all when the fuel and air are premixed as in vaporizer systems. It may
be concluded that the good characteristics of the gas turbine are likely to
increase its competitiveness in the twenty first century.
Review Questions
10.1 Explain in detail the combustion theory applied to a gas turbine combustion system.
10.2 What are the various factors those affect the combustion chamber performance? Explain.
10.3 Describe briefly the factors affecting the combustion chamber design.
10.4 What are the various forms of a combustion system?
10.5 What are the basic requirements of a combustion chamber? Explain.
10.6 Explain the process of combustion in gas turbine combustion chambers.
Combustion Systems
419
10.7 With a neat sketch explain the combustion chamber geometry bringing
out the various zones that play a part in the process of combustion.
10.8 What are the various possibility of combustion chamber arrangements?
10.9 Mention the various practical problems in the operation of a combustion chamber.
10.10 Write
(i)
(iii)
(v)
short notes on
flame tube cooling,
ignition,
pollution.
(ii)
(iv)
fuel injection,
use of cheaper fuels, and
Multiple Choice Questions (choose the most appropriate answer)
1. Combustion in a gas turbine combustion system is at
(a) constant-volume
(b) constant-pressure
(c) constant temperature
(d) constant enthalpy
2. Combustion process in a combustor is
(a) isobaric
(b) isochoric
(c) isentropic
(d) isothermal
3. The recognized postulations as to the combustion mechanism in a
combustion chamber is due to
(a) carbon preferential burning
(b) hydrogen preferential burning
(c) hydroxylation
(d) all of the above
4. Acceleration of the reaction in a gas turbine combustion system is
brought about by
(a) bombardment of nucleus
(b) chain breaking
(c) chain branching
(d) none of the above
420
Gas Turbines
5. Air-fuel ratio in a gas turbine is in the range of
(a) 20 to 30
(b) 30 to 40
(c) 40 to 60
(d) 60 to 100
6. Large amount of combustion takes place in a combustor in the
(a) primary zone
(b) secondary zone
(c) dilution zone
(d) in all places
7. The outlet temperature of the combustor is around
(a) 1600 K
(b) 1400 K
(c) 1200 K
(d) 1000 K
8. In the primary zone the air-fuel ratio is
(a) 13
(b) 15
(c) 17
(d) 20
9. The major pollutant coming out of the combustion chamber is
(a) CO
(b) SO2
(c) NOx
(d) unburnt hydrocarbons
10. The maximum temperature from the combustor is limited because
(a) it is difficult to burn the fuel
(b) the air-fuel ratio is too lean
(c) combustion chamber walls cannot sustain high temperature
(d) turbine blades cannot accept very high temperatures
Ans:
1. – (b)
6. – (a)
2. – (a)
7. – (a)
3. – (d)
8. – (b)
4. – (c)
9. – (c)
5. – (d)
10. – (d)
11
IMPULSE AND
REACTION TURBINES
INTRODUCTION
Work can be extracted from a gas at a higher inlet pressure to the lower
back pressure by allowing it to flow through a turbine. In a turbine as the
gas passes through, it expands. The work done by the gas is equivalent to
the change of its enthalpy.
It is a well known fact that the turbines operate on the momentum
principle. Part of the energy of the gas during expansion is converted into
kinetic energy in the flow nozzles. The gas leaves these stationary nozzles
at a relatively higher velocity. Then it is made to impinge on the blades
over the turbine rotor or wheel. Momentum imparted to the blades turns
the wheel. Thus, the two primary parts of the turbine are
(i) the stator nozzles, and
(ii) the turbine rotor blades.
Most turbines possess more than one stage with their respective wheels
mounted on a common shaft. Figure 11.1 shows a turbine stage. The stage
consists of a ring of fixed nozzle blades followed by the rotor blade ring.
However, a nozzleless stage with only the rotor is also possible as in the
case of an inward flow radial turbine. In certain requirements, the flow in a
turbine stage may be partly radial and partly axial which is called a mixed
stage. It combines the advantage of both axial and radial types.
Normally, a turbine stage is classified as
(i) an impulse stage, and
(ii) a reaction stage.
An impulse stage is characterized by the expansion of the gas which
occurs only in the stator nozzles. The rotor blades act as directional vanes
422
Gas Turbines
Inlet
Nozzle blade row
Stage
Rotor blade row
Axial clearance
Outlet
Fig. 11.1 A turbine stage
to deflect the direction of the flow. Further, they convert the kinetic energy
of the gas into work by changing the momentum of the gas more or less at
constant-pressure.
A reaction stage is one in which expansion of the gas takes place both
in the stator and in the rotor. The function of the stator is the same as
that in the impulse stage, but the function in the rotor is two fold.
(i) the rotor converts the kinetic energy of the gas into work, and
(ii) contributes a reaction force on the rotor blades.
The reaction force is due to the increase in the velocity of the gas relative
to the blades. This results from the expansion of the gas during its passage
through the rotor.
11.1
A SINGLE IMPULSE STAGE
Impulse machines are those in which there is no change of static or pressure
head of the fluid in the rotor. The rotor blades cause only energy transfer
and there is no energy transformation. The energy transformation from
pressure or static head to kinetic energy or vice versa takes place in fixed
blades only. To give an example, the transfer of kinetic energy to the rotor
in an impulse turbine from a high velocity fluid occurs only due to the
impulsive action of the fluid on the rotor. Figure 11.2 shows an impulse
turbine stage. As can be seen from the figure that in the rotor blade
passages of an impulse turbine there is no acceleration of the fluid, i.e., there
is no energy transformation. Hence, the chances are greater for separation
due to boundary layer growth on the blades surface. Due to this, the rotor
blade passages of the impulse machine suffer greater losses giving lower
stage efficiencies.
The paddle wheel, Pelton wheel and Curtis steam turbine are some of
the examples of impulse machines.
11.2
A SINGLE REACTION STAGE
The reaction machines are those, in which, changes in static or pressure
head occur both in the rotor and stator blade passages. Here, the energy
Impulse and Reaction Turbines
423
Nozzle Rotor
Energy transfer
Energy transformation
Velocity
Pressure
Fig. 11.2 An impulse turbine stage
transformation occurs both in fixed as well as moving blades. The rotor experiences both energy transfer as well as energy transformation. Therefore,
reaction turbines are considered to be more efficient. This is mainly due to
continuous acceleration of flow with lower losses. Figure 11.3 shows a single
stage reaction turbine along with pressure and velocity changes when the
fluid passes through a turbine stage.
The degree of reaction of a turbomachine stage may be defined as the
ratio of the static or pressure head change occurring in the rotor to the total
change across the stage.
A 50% or half degree reaction machine has some special characteristics.
Axial-flow turbines with fifty per cent reaction have symmetrical blades in
their rotors and stators. It may be noted that the velocity triangles at the
entry and exit of a 50% reaction stage are also symmetrical.
Hero’s turbine, the lawn sprinkler and Parson’s steam turbine are some
of the examples of reaction machines.
11.3
MULTISTAGE MACHINES
It may be noted that for a given rotor speed, only a limited change in the
energy level of the fluid can occur in single stage machines. This is true
for both turbines and compressors. Hence, it may become necessary that
when a large change in the energy level is required, more than one stage
becomes a necessity.
Multistage machines may employ either impulse or reaction stages or a
combination of both. Impulse machines may utilize any one of the following:
424
Gas Turbines
Nozzle Rotor
Energy transfer
Energy transformation
Velocity
Pressure
Fig. 11.3 A reaction turbine stage
(i) a large pressure drop in a number of pressure stages
(ii) a high kinetic energy in a number of velocity stages
(iii) a combination of the above two
It may be noted that gas turbine plants have a smaller number of stages
due to comparatively lower values of the pressure ratio employed, compared
to other similar power plants.
11.4
VELOCITY TRIANGLES OF A SINGLE STAGE MACHINE
The flow geometry at the entry and exit of a turbomachine stage is described
by the velocity triangles at these stations.
As already mentioned earlier, the velocity triangles for a turbomachine
contain the following three components
(i) the peripheral velocity, (u), of the rotor blades,
(ii) the absolute velocity, (c), of the fluid, and
(iii) the relative velocity, (w), of the fluid.
These velocities are related by the following well-known vector equation:
c
=
u+w
(11.1)
This simple relation is frequently used and is very useful in drawing the
velocity triangles for turbomachines.
The notation used here to draw velocity triangles correspond to the x-y
coordinates; the suffix (a) identifies components in the axial direction and
the suffix (t) refers to the tangential direction. Air angles in the absolute
system are denoted by alpha (α), where as those in the relative system are
represented by beta (β).
Impulse and Reaction Turbines
425
The velocity triangles at the entry and exit of a general turbine stage are
shown in Fig. 11.4. Since the stage is axial, the change in the mean diameter
between its entry and exit can be neglected. Therefore, the peripheral or
tangential velocity (u) remains constant in the velocity triangles. Axial
and tangential components of both absolute and relative velocities are also
shown in Fig. 11.4. Static and stagnation values of pressure and enthalpy
in the absolute and relative systems are indicated.
α1
c1
Station
1
c a1
p1, p01 , h 1 ,h 01
c t1
Nozzle blades
Station
p2, p02 , p 02 rel
2
ca2
Station
h 2, h 02 ,h 02 rel
β2
2
α2
w t2
u
c t2
Rotor blades
Station
c2
w2
p3, p03 , p 03 rel
3
w3
β3
h 3, h 03 ,h 03 rel
α3
c 3 c a3
c t3
u
w t3
Fig. 11.4 Velocity triangles for a turbine stage
The following trigonometrical relations can be deduced from the velocity
triangles shown in Fig. 11.4.
ca2
= c2 cos α2
=
w2 cos β2
ct2
= c2 sin α2
=
wt2 + u
= w2 sin β2 + u
u
sin (α2 − β2 )
=
c2
sin (90 + β2 )
u
c2
=
sin (α2 − β2 )
cos β2
ca3
ct3
=
c3 sin α3
= w3 cos β3
= wt3 − u
=
=
(11.2)
(11.3)
=
c2
cos β2
(11.4)
c3 cos α3
w3 sin β3 − u
(11.5)
(11.6)
426
Gas Turbines
ct2 + ct3
=
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎪
(w2 sin β2 + u) + (w3 sin β3 − u) ⎪
⎬
=
wt2 + wt3
=
c2 sin α2 + c3 sin α3
=
(11.7)
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎭
w2 sin β2 + w3 sin β3
It is often assumed that the axial velocity component remains constant
through the stage. For such a condition
ca
ca
=
=
ca1
= ca2
=
ca3
c1 cos α1
= c2 cos α2
=
c3 cos α3
ca
= w2 cos β2
=
w3 cos β3
(11.8)
Equation 11.7 for constant axial velocity yields a useful relation,
tan α2 + tan α3
11.5
= tan β2 + tan β3
(11.9)
EXPRESSION FOR WORK OUTPUT
Though force and torque are exerted on both stationary and moving blades
alike, work can only be done on the moving rotor blades. Thus the rotor
blades transfer energy from the fluid to the shaft. The stage work in an
axial turbine (u3 = u2 = u) can be written as
W
=
u2 ct2 − u3 ct3
=
u{ct2 − (−ct3 )}
=
u(ct2 + ct3 )
⎫
⎪
⎪
⎪
⎪
⎬
(11.10)
⎪
⎪
⎪
⎪
⎭
This equation can also be expressed in another useful form.‡
W
=
u2
ct3
ct2
+
u
u
(11.11)
The first term cut2 in the bracket depends on the nozzle or fixed blade
angle (α2 ) and the ratio σ = cu2 . The contribution of the second term cut3
to the work is generally small. It is also observed that the kinetic energy
of the fluid leaving the stage is greater for larger values of ct3 . The leaving
loss from the stage is minimum when ct3 = 0, i.e., when the discharge from
the stage is axial (c3 = ca3 ). However, this condition gives lesser stage work
as can be seen from Eqs. 11.10 and 11.11.
‡ Usually, Eq. 11.10 will be written with a minus sign between c
t2 and ct3 . Whenever
this is written with a plus sign it is implied that ct3 is negative.
Impulse and Reaction Turbines
427
If the swirl at the exit of the stage is zero;, i.e., ct3 = 0, Eq. 11.11 yields
:
⎫
ct2
2:
W =
⎪
u u ct3=0
⎪
⎪
⎪
⎬
:
c2 sin α2
2:
=
u
(11.12)
u
ct3=0 ⎪
⎪
⎪
⎪
:
⎭
sin α2
=
u2 :
σ
ct3=0
Thus, when the exit swirl is zero, the work output is a function of speed
ratio, fixed blade angle and peripheral speed.
11.6
BLADE LOADING AND FLOW COEFFICIENTS
Performance of turbomachines are characterized by various dimensionless
parameters. For example, the loading coefficient (ψ) and the flow coefficient
(φ) have been defined as
ψ
W
u2
=
(11.13)
ca
(11.14)
u
Since the work, (W ), in Eq. 11.13 is frequently referred to as the blade or
stage work, the coefficient (ψ) would also be known as the blade or stage
loading coefficient.
From Eqs. 11.10, 11.13 and 11.14, for constant axial velocity (ca ), it can
be shown that
φ
ψ
=
φ(tan α2 + tan α3 )
=
=
φ(tan β2 + tan β3 )
(11.15)
The φ − ψ plots are useful in comparing the performances of various stages
of different sizes and geometries.
11.7
BLADE AND STAGE EFFICIENCIES
Even though the blade and stage work (outputs) are the same, the blade
and stage efficiencies need not be equal. This is because the energy inputs to
the rotor blades and the stage (fixed blade ring plus the rotor) are different.
The blade efficiency is also known as the utilization factor ( ) which is an
index of the energy utilizing capability of the rotor blades. Thus,
=
ηb
=
Rotor blade work
Energy supplied to the rotor blades
=
W
Erb
(11.16)
The blade work can be written as the sum of changes in the various kinetic
energies:
W
=
u2 ct2 + u3 ct3
=
1 2
1
1
c − c23 + u22 − u23 + w32 − w22
2 2
2
2
(11.17)
428
Gas Turbines
The energy supplied to the rotor blades is the absolute kinetic energy in
the jet at the entry plus the kinetic energy change within the rotor blades.
Erb =
1 2 1 2
1
c + w − w22 + u22 − u23
2 2 2 3
2
(11.18)
Substituting Eqs. 11.17 and 11.18 into Eq. 11.16, we get
=
ηb
=
c22 − c23 + u22 − u23 + w32 − w22
c22 + (w32 − w22 ) + (u22 − u23 )
(11.19)
For axial machines, u = u2 = u3 ,
=
ηb
=
c22 − c23 + w32 − w22
c22 + (w32 − w22 )
(11.20)
To avoid any confusion between the stage and blade efficiencies, the term
utilization factor ( ) will be used in place of blade efficiency in this book.
A stage has already been defined. The rows of nozzle and rotor blades
between stations 1 and 2, and 2 and 3 respectively have been represented
by the blades as shown in Fig. 11.4.
As pointed out earlier, the stage output is the same as the rotor blade
work which is given by Eqs. 11.10, 11.11 and 11.17, but the energy input
to the stage is different from that of the rotor blades. The stage efficiency
(ηst ) takes into account the aerodynamic losses in the fixed and moving
blade rows of the stage.
11.8
MAXIMUM UTILIZATION FACTOR FOR A SINGLE IMPULSE
STAGE
Figure 11.5 shows a single stage impulse turbine. In this, there is no static
pressure change in the rotor. The pressure and velocity variations of the
fluid passing through the stage is also shown in Fig. 11.5. Because of
the pressure drop in the nozzle blade row, the absolute velocity of the
fluid increases correspondingly and transformation of energy occurs in the
nozzle. However, the transfer of energy occurs across the rotor blade row.
Therefore the absolute fluid velocity decreases through the rotor blade row,
as shown in the figure.
The velocity triangles for a single impulse stage are as shown in Fig. 11.6.
For frictionless flow in the absence of any pressure drop through the rotor
blades the relative velocities at their entry and exit are very much the same
(w3 ≈ w2 ). To obtain this condition the rotor blade angles must be equal.
Therefore the utilization factor is given by
=
u(ct2 + ct3 )
1 2
2 c2
Substituting from Eq. 11.7 and noting w2 = w3 and β2 = β3 , we get
Impulse and Reaction Turbines
N
429
R
N = Nozzle
R = Rotor
c
p
Velocity
Pressure
Fig. 11.5 Variation of pressure and velocity through a single stage impulse
turbine
=
4uw2 sin β2
c22
From the velocity triangle at the entry (Fig. 11.6),
=
Substituting σ =
u
c2
ηb
4u(c2 sin α2 − u)
c22
and rearranging,
=
=
4 σ sin α2 − σ 2
(11.21)
This shows that the utilization factor is a function of the blade-to-gas
speed ratio and the nozzle angle.
=
f (σ, α2 )
Now, the value of σ corresponding to maximum utilization can be determined. For given value of α2 , differentiating Eq. 11.21,
d
= sin α2 − 2σ = 0
dσ
σopt
=
u
c2
=
1
sin α2
2
(11.22)
c2 sin α2
=
ct2
=
2u
(11.23)
=
w3 sin β3
Equations 11.3 and 11.23 give
w2 sin β2
=
u
(11.24)
Substituting this in Eq. 11.6 yields ct3 = 0. This means that the exit from
the stage should be axial, i.e., c3 = ca3 . This result can also be obtained
by mere physical interpretation of the mechanics of flow.
Combining Eq. 11.22 and 11.12,
430
Gas Turbines
β2
α2
ca2
w t2
c2
w2
u
c t2
α
w3
w2
=k
u
β3
3
c3 ca3
c t3
wt3
Fig. 11.6 Velocity triangles for a single impulse stage with negative swirl
at exit
Wct3=0
=
Wopt
=
2u2
(11.25)
Velocity triangles to satisfy these conditions are shown in Fig. 11.7. The
combination of Eqs. 11.21 and 11.22 gives
max
=
sin2 α2
(11.26)
The rotor blade angles can also be determined for these conditions
u
(11.27)
tan β3 = tan β2 =
c2 cos α2
Substituting from Eq. 11.22,
tan β3
=
tan β2
=
1
tan α2
2
(11.28)
Thus, for maximum utilization factor, the rotor blade angles are fixed by
the nozzle air angle at exit.
11.9
VELOCITY-COMPOUNDING OF MULTISTAGE IMPULSE
TURBINE
When the pressure drop to be achieved in an impulse turbine is large; it
will not be possible to achieve in one stage. If this has to be achieved in a
single stage then this would lead to either a larger diameter or a very high
rotational speed. Thus, a single stage utilizing a large pressure drop will
have very high peripheral speed of its rotor which is not possible to achieve
in practice. Therefore, machines with large pressure drops employ more
than one stage.
Impulse and Reaction Turbines
431
c1
β2
ca
α2
wt2
w2
c2
u
c t2 = 2u
w 3=
w2
β
3
u
ca = c 3
ct3 = 0
u
Fig. 11.7 Velocity triangles for a single impulse stage with maximum utilization factor
One of the methods that is followed in multistage expansion in impulse
turbines is to generate high velocity of the fluid. This can be accomplished
by expanding the gas through a large pressure drop in the nozzle blade
row. The high velocity fluid then transfers its energy in a number of stages
by incorporating many rotor blade rows separated by rows of fixed guide
blades. Figure 11.8 shows a two-stage velocity compounded impulse turbine. The decrease in the absolute velocity of the fluid across the two rotor
blade rows (R1 and R2 ) is mainly due to the energy transfer. The slight decrease in the fluid velocity through the fixed guide blades (F ) is attributed
to losses. Since, the turbine is of the impulse type, the pressure of the fluid
remains constant after its expansion in the nozzle blade row. Such stages
are referred to as velocity or Curtis stages.
Figure 11.9 shows the velocity triangles for the above mentioned turbine.
To achieve maximum utilization factor, the exit from the last stage (in this
case the second-stage rotor) must be in the axial direction, i.e., α3 = 0 and
c 3 = ca .
The following two assumptions are usually made for this type of turbines:
1. Equiangular flow through rotor and guide blades
β2
=
β3 ;
α3
=
α2 ;
β2
=
β3
c2 ;
w2
=
w3
2. Frictionless flow over the blades
w2
=
w3 ;
c3
=
For maximum utilization factor, ct3 = 0 and
u
=
w3 sin β3
=
w3 sin β2 = w2 sin β2
(11.29)
432
Gas Turbines
N
F
R1
R2
F = Fixed blade
N = Nozzle
R = Rotor
c
p
Velocity
Pressure
Fig. 11.8 Variation of pressure and velocity through a two-stage velocity
compounded impulse turbine
β2
ca
c2
w2
3u
ct2 = 4u
α2
w3=w 2
c3
ct3 =2u
u
β2’
ca
α2’
u
α3
β3 ca
c2’= c 3
w2’
u
u
c t2’ = 2u
’
=0
=’ w 2 β3’ α
c3’3’= ca
w3
ct3’= 0
u
Fig. 11.9 Velocity triangles for a two-stage velocity compounded impulse
turbine with maximum utilization factor
Impulse and Reaction Turbines
ct2
ct2
=
2u =
c2 sin α2
ct3
=
3u =
w3 sin β3
=
c2 sin α2
=
=
=
w2 sin β2 + u
c2 sin α2
=
4u
σopt
=
u
c2
433
c3 sin α3
(11.30)
w2 sin β3
(11.31)
Substituting from Eq. 11.31,
=
1
sin α2
4
(11.32)
Similarly, for a three-stage velocity compounded turbine
σopt
=
1
sin α2
6
=
1
sin α2
2n
Thus for n velocity stages,
σopt
(11.33)
Values of work in the first and second stages, from Eq. 11.7 are given
by
WI
=
u(w2 sin β2 + w3 sin β3 )
WII
=
u(w2 sin β2 + w3 sin β3 )
WI
=
2uw2 sin β2
WI
=
6u2
From assumptions 1 and 2,
From Eq. 11.31,
(11.34)
This expression can also be written directly from the velocity triangles
shown in Fig. 11.9.
Similarly Eq. 11.29 gives
WII
=
2u sin β2
=
2u2
(11.35)
The total work of the velocity compounded impulse turbine (with two
stages) is
WI + WII
=
8u2
(11.36)
The expressions in Eqs. 11.34, 11.35 and 11.36 can be represented by
the following general relations.
If the number of velocity stages is n with individual stages identified by
i = 1, i = 2, . . . , i = n, then the work done by the ith stage is given by
434
Gas Turbines
Table 11.1 Values of stage and turbine work, blade-to-gas speed ratio and
maximum utilization factor for velocity compounded impulse
turbine
Work
n
σopt
i=1
i=2
i=3
i=4
1
2u2
2
6u2
2u2
3
10u2
6u2
2u2
4
14u2
10u2
6u2
2u2
5
18u2
14u2
10u2
6u2
Wi
=
i=5
2u2
max
WT
2u2
1
2
sin α2
sin2 α2
8u2
1
4
sin α2
sin2 α2
18u2
1
6
sin α2
sin2 α2
32u2
1
8
sin α2
sin2 α2
50u2
1
10
sin α2
sin2 α2
2{2(n − i) + 1}u2
(11.37)
The total turbine work with n velocity stages is given by
WT
=
n
)
Wi
=
2n2 u2
(11.38)
i=1
Since the turbine is of the impulse type, the input energy is equal to
Therefore, the utilization factor from Eq. 11.38 is
max
=
WT
u
2
1 2 = 4n
c
c
2
2 2
=
sin2 α2
1 2
2 c2 .
2
= 4n2 σ 2
(11.39)
Equations 11.33 and 11.39 yield
max
This shows that the maximum value of the utilization factor for a given
value of the nozzle angle remains unaltered with the number of velocity
stages under the assumed conditions.
Values in Table 11.1 have been computed from Eqs. 11.33, 11.37 and
11.38. It is observed that the turbine work in the velocity compounded
machines increases drastically by employing more velocity stages. This is
a great advantage because the pressure of the gas is reduced quickly. This
gives a much smaller number of stages which will reduce the overall turbine
length and only a small part of the turbine casing is subjected to high
pressure gas. However, the work in the subsequent velocity stages goes on
decreasing, e.g., the third stage in the three-stage turbine only produces
one-ninth of the total work. Therefore, velocity stages beyond three have
no practical importance.
It may be noted that the pressure ratios across gas turbines are much
lower as compared to steam turbines. Therefore, the problem of reducing
Impulse and Reaction Turbines
435
the gas pressure quickly in the casings of gas turbines is not that serious.
It may be noted that the losses in the velocity stages are higher, compared
to the reaction stages. Therefore, velocity compounding has little to offer
in the field of gas turbines except in some special applications.
11.10
PRESSURE COMPOUNDING OF MULTISTAGE IMPULSE
TURBINE
In the previous sections we have seen the details of velocity compounding of
the impulse turbine. There are two major problems in velocity-compounded
stages:
(i) the nozzles have to be of the convergent-divergent type for generating
high velocity. This results in a more expensive, and difficult design
of the nozzle blade rows; and
(ii) high velocity at the nozzle exit leads to higher cascade losses. Shock
waves are generated if the flow is supersonic which further increase
the losses.
In order to avoid these problems another method of utilizing a high
pressure ratio is employed in which the total pressure drop is divided into
a number of impulse stages. These are known as pressure-compounded or
Rateau stages. On account of the comparatively lower pressure drop, the
nozzle blade rows are subsonic (M < 1). Therefore, such a stage does not
suffer from the disabilities of the velocity stages. Figure 11.10 shows the
variation of pressure and velocity of gas through the two pressure stages
of an impulse turbine. This is equivalent to using two stages of the type
shown in Fig. 11.5 in series. Velocity triangles for two Rateau stages, each
having an axial exit are shown in Fig. 11.11. It may be noted that the
nozzle blades in each stage receive flow in the axial direction.
Some designers employ pressure stages upto the last stage. This gives a
turbine of shorter length as compared to the reaction type, with a penalty
on efficiency.
11.11
THE REACTION TURBINE
In a turbine if the degree of reaction is unity, then we call that machine as
the pure reaction machine. But in practice it is impossible to have a 100%
degree of reaction and therefore a combination of impulse and reaction is
employed. In a reaction turbine there will be number of stages and the
pressure changes through each row of blades, both moving and stationary.
Leakage is a serious problem in reaction blading in the high pressure part
where the blades are short and percentage tip clearance area is relatively
large compared to the total area. Leakage depends upon the clearance, the
velocity ratio and the stage pressure ratio. Hence, the sealing problem at
blade tips is quite critical because the pressure gradient exits across the
blades in reaction machines.
436
Gas Turbines
N1
R1
N2
R2
N = Nozzle
c
p
R = Rotor
Velocity
Pressure
Fig. 11.10 Variation of pressure and velocity through a two-stage pressure
compounded impulse turbine
11.12
MULTISTAGE REACTION TURBINES
A brief introduction of reaction stages and the degree of reaction has been
given already in connection with the axial flow compressor. Figure 11.12
shows two reaction stages and the variation of pressure and velocity of the
gas in them. The gas pressure decreases continuously over both fixed and
moving rows of blades. Since the pressure drop in each stage is smaller
as compared to the impulse stages, the gas velocities are relatively low.
Besides this the flow is accelerating throughout. These factors make the
reaction stages aerodynamically more efficient though the tip leakage loss
increases on account of the relatively higher pressure difference across the
rotor blades.
Multistage reaction turbines employ a large pressure drop by dividing
it into smaller values in individual stages. Thus the reaction stages are like
the pressure-compounded stages (see Section 11.10) with a new element of
‘reaction’ introduced in them, i.e., of accelerating the flow through blade
rows also.
11.12.1
Enthalpy-Entropy Diagram
Figure 11.13 shows the enthalpy-entropy diagram for flow through a general
turbine stage with some degree of reaction. This should be studied with
the velocity triangles in Fig. 11.4. Both static and stagnation values of the
pressures and enthalpies are indicated at various stations.
In general, the gas approaches the stage with a finite velocity (c1 ).
Therefore, the values of pressures p1 and p01 and enthalpies h1 and h01
have been differentiated. The relation between the static and stagnation
Impulse and Reaction Turbines
437
c2
w2
u
=w
w3
2
c3=ca
u
Second stage nozzle ring
c 2’
w’2
u
=’ w
2’
w3
c 3’= ca
u
Fig. 11.11 Velocity triangles for a two-stage pressure-compounded impulse
turbine
F1
R1
F2
R2
F = Fixed blade
R = Rotor
c
p
Velocity
Pressure
Fig. 11.12 Variation of pressure and velocity through a two-stage reaction
turbine
438
Gas Turbines
p
01
01
1/2c21
02 p02
p
1
Enthalpy
1
p
Ws
1/2 c 22
02 rel
0 2 rel
Wa
h01= h 02
0 3 rel p03 rel
= h03 rel
h 02 rel
2
2
1/2 w 2
p
2
2’
2
1/2 w 3
03
Δs
p
0 3"
p
03
1/2 c 23
p
03"
3
3
3’
3"
Entropy
Fig. 11.13 Enthalpy-entropy diagram for flow through a turbine stage
values has already been explained. The reversible adiabatic (isentropic)
expansion of the gas through fixed and moving rows of blades is represented
by processes 1–2 and 2 –3 respectively. Stagnation points O1 and O3 are
defined by the following relations:
h01
h03
1
= h1 + c21
2
(11.40)
1
= h3 + c203
2
(11.41)
The ideal work in the stage is given by
Ws
= h01 − h03
=
Cp (T01 − T03 )
(11.42)
The actual expansion process (irreversible adiabatic) is represented by
the curve 1-2-3, and the actual states of the gas at the exits of the fixed
and moving blades are represented by points 2 and 3 respectively. Since,
there is only energy transformation and no energy transfer (work) between
states 1 and 2, the stagnation enthalpy remains constant.
h01
=
h02
=
1
h2 + c22
2
(11.43)
The loss due to irreversibility (Δs = s2 − s2 ) is given by the enthalpy
loss coefficient
h2 − h2
2Cp
ξN =
=
(T2 − T2 )
(11.44)
1 2
c22
2 c2
Impulse and Reaction Turbines
439
The stagnation pressure loss coefficient corresponding to Eq. 11.44 is
defined by
(Δp0 )N
p01 − p02
YN =
=
(11.45)
1
1
2
2
ρc
2
2
2 ρc2
The expansion process (2–3) in the moving blade rows represents both
transformation and transfer of energy. Therefore, the difference in the
absolute stagnation enthalpies at states 2 and 3 gives the actual value of
the stage work.
Wr
h03
=
h02 − h03
=
h01 − h03
=
Cp (T01 − T03 )
(11.46)
=
1
h3 + c23
2
(11.47)
However, to an observer moving with the rotor the relative flow appears
as the absolute flow in the nozzle to a stationary observer. Therefore, for
him (in the relative system) the stagnation enthalpy in the moving frame
of coordinates remains constant.
h02rel
1
h2 + w22
2
= h03rel
1
= h3 + w32
2
(11.48)
The enthalpy and pressure losses for the moving row of blades are given
by the following coefficients:
ξR
=
h3 − h3
1 2
2 w3
YR
=
p02rel − p03rel
1
2
2 ρw3
=
2Cp
(T3 − T3 )
w32
=
(Δp0 )R
1
2
2 ρw3
(11.49)
(11.50)
It can be proved that for most of the applications (M < 1), there is little
difference between the values of the loss coefficients based on enthalpy and
stagnation pressure, i.e., ξ ≈ Y .
11.12.2
Degree of Reaction
Degree of reaction of a turbine stage can be defined in a number of ways;
it can be expressed in terms of pressures or velocities or enthalpies or the
flow geometry in the stage.
(a) A definition on the basis of the isentropic change in the stage is
given by the following relation:
R
=
Isentropic change of enthalpy in the rotor
Isentropic change of enthalpy in the stage
440
Gas Turbines
=
;2
;31
3
Δhr
=
Δhs
h2 − h3
h1 − h3
(11.51)
Assuming ρ = constant, Eq. 11.51 can be transformed in terms of pressures.
Using Δh = Δp
ρ ,
; 2 Δp
;2
Δp
p2 − p3
ρ
3
≈ ;31
(11.52)
R = ; 1 Δp
=
p1 − p3
Δp
3
ρ
3
(b) Definitions in Eqs. 11.51 and 11.52 are not widely used in practice.
Since the actual enthalpy changes and work are more important for turbine
stages, a definition of the degree of reaction based on these quantities is
more logical and useful.
It may be noted that the change of enthalpy through the rotor and the
total change in the stage are
h2 − h3
h01 − h03
=
1 2
1
w3 − w22 + u22 − u23
2
2
= h02 − h03
=
(u2 ct2 − u3 ct3 )
(11.53)
(11.54)
The ratio of these quantities gives another definition of the degree of reaction.
h2 − h 3
R =
(11.55)
h01 − h03
If c1 ≈ c3 ,
R
=
h2 − h3
h1 − h3
(11.56)
Equations 11.53 and 11.54 when substituted in Eq. 11.55 yield
R
=
1
2
w32 − w22 + 12 u22 − u23
u2 ct2 − u3 ct3
(11.57)
For axial machines, u3 = u2 = u. Therefore, from Fig. 11.4,
R
=
w32 − w22
2u(ct2 + ct3 )
(11.58)
(c) Equation 11.58 can also be expressed in terms of the geometry of
flow through the stage.
From velocity triangles (Fig. 11.4),
w22
2
= c2a2 + wt2
w32
2
= c2a3 + wt3
For constant axial velocity,
Impulse and Reaction Turbines
w32 − w22
2
2
= wt3
− wt2
= c2a tan2 β3 − tan2 β2
w32 − w22
441
(11.59)
= (wt3 + wt2 )(wt3 − wt2 )
Using Eq. 11.7 and the velocity triangles,
w32 − w22
= ca (ct2 + ct3 )(tan β3 − tan β2 )
(11.60)
Equation 11.60 when put into Eq. 11.58 yields a widely used definition for
the degree of reaction.
ca
R =
(tan β3 − tan β2 )
(11.61)
2u
But
φ
=
ca
u
Now, the mean axial velocity cam can be written as
1
(ca1 + ca2 )
cam =
2
Similarly,
(11.62)
ctm
=
1
(ct2 − ct1 )
2
(11.63)
tan αm
=
ctm
cam
(11.64)
Therefore,
Hence, for constant velocity
tan αm
=
1
(tan α2 − tan α1 )
2
(11.65)
tan βm
=
1
(tan β3 − tan β2 )
2
(11.66)
R
= φ tan βm
(11.67)
Two more useful relations are obtained assuming ca to be constant. From
the velocity triangle at the entry, Eq. 11.3 can be written as
ca tan α2
tan β2
= ca tan β2 + u
= tan α2 −
u
ca
(11.68)
Eliminating tan β2 , between Eqs. 11.61 and 11.68, and rearranging,
ca
1
+
(tan β3 − tan α2 )
(11.69)
R =
2 2u
Equation 11.6 can similarly be written as
442
Gas Turbines
p
1
Enthalpy
1
p =p
2
3
h2 = h 3
2,3
Entropy
Fig. 11.14 Zero reaction or impulse stage, enthalpy-entropy diagram for
reversible flow
ca tan α3
tan β3
= ca tan β3 − u
= tan α3 +
u
ca
(11.70)
Equation 11.70 when substituted in Eq. 11.69 gives
ca
(tan α3 − tan α2 )
R = 1+
2u
11.12.3
Zero Degree Reaction Stage
Though a single-stage impulse turbine has already been discussed in Sec.11.8,
it is further discussed here in the light of degree of reaction.
(a) Figure 11.14 shows the enthalpy-entropy diagram for reversible adiabatic expansion in an impulse stage. Expansion of the gas occurs only
in the nozzle blade row and its thermodynamic state remains unchanged
between stations 2 and 3. Therefore the air angles, static pressure and
enthalpy at the entry and exit of the rotor in this stage are the same.
β 2 = β3
p2 = p3
h2 = h3
These conditions when put into Eqs. 11.51, 11.52, 11.55, 11.56 and 11.61,
give the degree of reaction as zero. On account of the reversible nature of
flow, such a stage is of no practical importance.
(b) Figure 11.15 depicts the enthalpy-entropy diagram for irreversible
adiabatic expansion in an impulse stage with equal pressures at the rotor
Impulse and Reaction Turbines
443
p1
Enthalpy
1
h3
h2
2
1/2 w 3
1/2 w 22
= h 03 rel
h 02 rel
p =p
3
2
3
2
2’
Entropy
Fig. 11.15 Impulse stage (negative degree of reaction), enthalpy-entropy
diagram for irreversible flow
entry and exit. On account of losses, both in the nozzle and rotor blade
rows, the gas experiences an increase in entropy. Therefore, for this stage
p2
=
p3
w2
>
w3
h2
<
h3
β2
>
β3
These conditions give the degree of reaction as negative (Eqs. 11.55,
11.56, 11.58 and 11.61). However, Eq. 11.52 defines such a stage as impulse.
(c) Figure 11.16 shows the enthalpy-entropy diagram of a real (adiabatic) and a truly zero reaction stage. Here a small pressure drop in
the rotor compensates for the deceleration of the flow in the rotor blades
(Fig. 11.15),. Thus the enthalpies and relative velocities remain constant
between the rotor entry and exit.
Equation 11.52 for such a stage gives the value of reaction as positive,
whereas Eqs. 11.55, 11.56, 11.58 and 11.61 yield R = 0.
11.12.4
Fifty Per Cent Reaction Stage
For fifty per cent degree of reaction, Eq. 11.56, gives
R
=
h2 − h3
h1 − h3
=
1
2
444
Gas Turbines
p
1
Enthalpy
1
h02 rel= h03 re
p
1/2w32
2
1/2 w 2
2
p
3
2
h2 = h3
3
2’
Entropy
Fig. 11.16 Zero reaction stage, enthalpy-entropy diagram for irreversible
flow
This gives equal values to the actual enthalpy drops in the fixed and rotor
blade rows (Fig. 11.12).
h1 − h2
= h2 − h3
=
1
(h1 − h3 )
2
(11.71)
Figure 11.17 shows the enthalpy-entropy diagram for such a stage. The
flow and cascade geometries in the fixed and moving blade rows are such
that they provide equal enthalpy drops. This does not imply that the
pressure drops in the two rows are also the same.
From Eq. 11.69,
1
2
tan β3
β3
=
1
ca
+
(tan β3 − tan α2 )
2 2u
= tan α2
= α2
(11.72)
This condition when put into Eq. 11.9 yields
tan β2
β2
Equation 11.57 for R =
1
2
= α3
gives
1 2
1
w − w22 + u22 − u23
2 3
2
For axial turbines,
= tan α3
=
1
(u2 ct2 − u3 ct3 )
2
(11.73)
Impulse and Reaction Turbines
445
p
1
1
h 1 - h2 = 1/2 (h1 - h3 )
Enthalpy
p2
2
h 2 - h3 = 1/2 (h1 - h3 )
p
3
3
Entropy
Fig. 11.17 Enthalpy-entropy diagram for a 50% reaction stage
w32 − w22
1
1 2
c − c23 + w32 − w22
2 2
2
=
= c22 − c23
(11.74)
This with Eqs. 11.72 and 11.73 gives
w3
= c2
(11.75)
w2
= c3
(11.76)
Thus, it is observed that the velocity triangles at the entry and exit of a
rotor in a fifty per cent reaction turbineare similar. These are shown in
Fig. 11.18.
Utilization Factor
Equations 11.75 and 11.76 when put into Eq. 11.20 gives the value of the
utilization factor for the fifty per cent reaction stage.
=
=
2 c22 − c23
c22 + (c22 − c23 )
1−
1−
c23
c22
c23
2c22
From the entry velocity triangle in Fig. 11.18,
=
c22 − c23
c22 − 12 c23
(11.77)
446
Gas Turbines
c1 = c3
Fixed blade row
β2
α2
ca
c2 = w 3
w2 = c3
u
Rotor blade row
α3
β3
w3
ca
u
Fig. 11.18 Velocity triangles for a 50% reaction stage
w22
=
c22 + u2 − 2uc2 cos (90 − α2 )
c23
=
c22 + u2 − 2uc2 sin α2
c23
c22
=
1+
σ
=
u
c2
c23
c22
=
1 + σ 2 − 2σ sin α2
c23
c22
=
2σ sin α2 − σ 2
(11.79)
c23
2c22
=
1
1 + 2σ sin α2 − σ 2
2
(11.80)
1−
1−
u
c2
2
−2
u
c2
for R =
2
1
2
sin α2
Equations 11.79 and 11.80 when put into Eq. 11.77 give
=
(11.78)
2σ sin α2 − σ 2
1 + (2σ sin α2 − σ 2 )
Impulse and Reaction Turbines
=
2
1+
1
2σ sin α2 − σ 2
447
(11.81)
The utilization factor is maximum when the factor 2σ sin α2 − σ 2 is
maximum.
d
2σ sin α2 − σ 2
= 0
dσ
2 sin α2
=
2σ
σopt
=
u
c2
=
sin α2
(11.82)
This condition in Eq. 11.81 gives
max
=
2 sin2 α2
1 + sin2 α2
(11.83)
Equation 11.82 gives
u =
c2 sin α2
=
ct2
Since the entry and exit velocity triangles are symmetrical,
u =
w3 sin β3
=
wt3
These relations give right-angled velocity triangles (Fig. 11.19) at the entry
and exit.
β2
=
0
α3
=
0
The exit swirl is zero (ct3 = 0), i.e., the exit from the stage is axial.
The stage work for maximum utilization from Eq. 11.11 or 11.12, or
direct from the velocity triangles (Fig. 11.19) is
Wopt
=
u2
(11.84)
This is half of the value for a single-stage impulse and one-eighth of the
two-stage velocity compounded impulse type.
Equation 11.58 can be written as
1 2
w − w22
= u(ct2 − ct3 )R
2 3
+
,
1 2
1 2
2
2
c − c3 + w3 − w2 R
=
2 2
2
w32 − w22
=
R
c2 − c23
1−R 2
This when put into Eq. 11.20 gives the utilization factor for an axial turbine
as
448
Gas Turbines
ca
α2
c2
ct2=u
w3
β3
ca
ct3 = 0
u
Fig. 11.19 Velocity triangles for a 50% reaction stage at maximum utilization factor
c22 − c23 + 1 +
=
c22 +
R
2
1−R (c2
R
1−R
− c23 )
This on simplification gives a general relation between the utilization factor
and the degree of reaction
=
c22 − c23
c22 − Rc23
For R = 1, this equation gives indeterminate value. Therefore Eq. 11.71 is
used.
11.12.5
Hundred Per Cent Reaction Stage
Hundred per cent degree of reaction implies that the entire change in the
static properties occurs in the rotor. Equation 11.56 for this condition gives
h2 − h3
h1 − h3
R
=
h1
= h2
=
1
when R = 1
(11.85)
Equation 11.58 gives
1 2
w − w22
2 3
= u(ct2 + ct3 )
which shows that the stage work is wholly due to the change in the kinetic
energy occurring in the rotor. Rewriting the quantity on the right-hand
side (refer Eq. 11.17 and note that u2 = u3 ),
Impulse and Reaction Turbines
w2 β α 2
2
u
w3
c2 β
3
c 3=
u
449
c2
ct2
α3=α2
ct2 = ct3
Fig. 11.20 Velocity triangles for a 100% reaction turbine stage
1 2
w − w22
2 3
1 2
1
c − c23 + w32 − w22
2 2
2
=
(11.86)
c2
= c3
when R = 1
(11.87)
α2
= α3
when R = 1
(11.88)
Further Eq. 11.71 gives
These conditions give the velocity triangles as shown in Fig. 11.20. It
is observed that the rotor blades are highly staggered. On account of the
large difference of static pressure across the rotor and high relative velocity
at its exit, the losses are relatively higher.
A stage with a degree of reaction higher than hundred per cent requires
h2
>
h1
c2
<
c1
α3
>
α2
These conditions show that for such a stage the flow is decelerating in the
nozzle or upstream fixed blades. This is obviously undesirable.
11.12.6
Negative Reaction
A negative degree of reaction in a turbine stage, such as R > 100% is also
undesirable. For such a stage
h3
w3
β2
>
<
>
h2
w2
β3
(Eq. 11.55)
(Eq. 11.58)
(Eq. 11.67)
These conditions will require the diffusion of flow in the rotor blade row.
Such a condition may be, accepted only as a necessary evil in long blades
where it is necessary to limit the high degree of reaction at the blade tips.
450
11.13
Gas Turbines
BLADE-TO-GAS SPEED RATIO
It has been seen in the earlier sections that the stage work and utilization
factor are governed by the blade-to-gas speed ratio parameter (velocity
ratio) σ = cu2 . Its value for maximum utilization was determined for some
stages. Therefore, efficiencies of the turbine stages can also be plotted
against this ratio. Such plots for some impulse and reaction stages are
shown in Fig. 11.21. The performance of steam turbines is often presented
in this form. The curves in Fig. 11.21 also show the optimum values of the
velocity ratio and the range of off-design for various types of stages. The
fifty per cent reaction stage shows a wider range. Another important aspect
that is depicted here is that in applications where high gas velocities (due
to high-pressure ratio) are unavoidable, it is advisable to employ impulse
stages to achieve practical and convenient values of the size and speed of the
machine. Sometimes it is more convenient to use an isentropic velocity ratio.
R = 50%
1.0
ε
η
0.8
N
Single-stage
impulse
ηN 0.6
Two-stage
impulse
0.4
0.2
σopt
σopt
0
0.2
0.4
σ = c2
0.6
σopt
0.8
1.0
Fig. 11.21 Variation of utilization factor and stage efficiency with blade-togas speed ratio
This is the ratio of the blade velocity and the isentropic gas velocity that
would be obtained in its isentropic expansion through the stage pressure
ratio.
u
σs =
(11.89)
cs
1 2
c
2 s
=
cs
=
2(h01 − h03 )
cs
=
2Cp T01 1 − r
h01 − h03
(Fig. 11.13)
(11.90)
For a perfect gas,
1−γ
γ
(11.91)
A relation between velocity ratios and the degree of reaction can be derived
for isentropic flow. From Eq. 11.51,
Impulse and Reaction Turbines
=
h2 − h3
h1 − h3
=
1−
h1 − h2
≈
1 2
c
2 2
h1 − h3
≈
1 2
c
2 s
R
h1 − h2
h1 − h3
Therefore,
R
=
451
c2 2
u
cs 2
u
c2
1 − 22
cs
=
1−
σ
=
σs
(11.93)
σ
=
1
σs
√
ηnoz
(11.94)
σ
=
√
2σs
(11.95)
=
1−
σs2
σ2
(11.92)
For R = 0,
With a nozzle efficiency, ηnoz ,
For R = 12 ,
11.14
LOSSES AND EFFICIENCIES
Losses occurring in a turbomachine blade rows (stationary or moving) have
been described in Sec.9.10. Besides these losses, additional losses occur in
an actual turbine due to disc and bearing friction.
Figure 11.22 shows the energy flow diagram for the impulse stage of
an axial turbine. Numbers in brackets indicate the order of energy or
loss corresponding to 100 units of isentropic work (h01 − h03 ). It is seen
that the energy reaching the shaft after accounting for stage cascade losses
(nozzle and rotor blade aerodynamic losses) and leaving loss is about 85% of
the ideal value; shaft losses are a negligible proportion of this value. If the
turbine is of the partial admission type it incurs additional losses as shown
in the figure. The flow losses discussed in section 9.10 are also appropriate
to turbine cascades.
11.15
PERFORMANCE GRAPHS
Performance of various types of gas turbine stages is often presented in
terms of plots of the loading coefficient (ψ) against the values of the flow
coefficient (φ).
452
Gas Turbines
Ideal or isentropic work (100)
Energy nozzle exit (95)
Leaving loss (3)
Ideal stage work (92)
Actual stage work
Shaft work (84)
Disc
friction loss
Nozzle aero losses (5)
Secondary loss
Partial admission losses
Profile loss
Annulus loss
Rotor area losses
Shaft losses (1)
Pumping
Bearing
(for partial admission) loss
Mixing
Loss
Expansion
Loss
Secondary
loss
Shear flow
Loss
Profile
loss
Annulus
loss
Jet dispersion
Loss
Leakage
Loss
Tip leakage
loss
Fig. 11.22 Energy flow diagram for an impulse turbine
11.15.1
Zero Degree Reaction
Equation 11.16 for zero degree reaction (β2 = β3 ) gives
ψ = 2φ tan β2 = 2φ tan β3
(11.96)
At maximum utilization factor (Fig. 11.7),
u
1
tan β3 =
=
ca
φ
ψ
11.15.2
= 2
(11.97)
Fifty Per Cent Reaction
Equation 11.68 for R = 12 (α2 = β3 ) gives
u
tan β2 = tan α2 −
ca
tan β2
= tan β3 −
1
φ
This relation when put in Eq. 11.16 gives
ψ
= 2φ tan β3 − 1
=
2φ tan α2 − 1
(11.98)
Impulse and Reaction Turbines
453
At maximum utilization factor (Fig. 11.19),
ψ
= 1
(11.99)
Equation 11.61 gives
R
ca
ca
tan β3 −
tan β2
2u
2u
=
For axial exit from the stage,
ca
tan β2 = 1
u
Therefore,
R
ca
1
−
tan β2
2 2u
=
ca
tan β2
u
= 1 − 2R
(11.100)
Equation 11.16 gives
ψ
=
ca
ca
tan β2 +
tan β3
u
u
=
ca
tan β2 + 1
u
= φ tan β2 + 1
(11.101)
This is a general equation for a stage with an axial exit. Equation 11.100
when put in Eq. 11.101 yields
ψ
= 2(1 − R)
(11.102)
This for R = 0 and R = 12 gives the same results as in Eqs. 11.97 and 11.99
respectively. This is not valid for R = 1 or R < 0 because then the exit
from the stage cannot remain axial. Equation 11.102 shows that the stage
loading decreases with an increase in the degree of reaction. Figure 11.23
shows the φ-ψ plots for various axial turbine stages.
5.0
β2 = β3
o
o
60
10.0
o
30
2.5
0
0.5
o
70
o
o
60 5.0
o
45
ψ
α = 80
1.0
φ
(a) Zero reaction
β = 60
2
o
45
ψ
ψ
5.0
0
0.5
1.0
φ
(b) 50% reaction
o
30
2.5
0
0.5
1.0
φ
(c) Axial exit
Fig. 11.23 Performance charts for various axial turbine stages
454
Gas Turbines
Worked out Examples
11.1 A multistage gas turbine is to be designed with impulse stages, and
is to operate with an inlet pressure and temperature of 6 bar and
900 K and an outlet pressure of 1 bar. The isentropic efficiency of
the turbine is 85%. All the stages are to have a nozzle outlet angle
of 75◦ and equal outlet and inlet blade angles. Mean blade speed
of 250 m/s and equal inlet and outlet gas velocities. Estimate the
maximum number of stages required. Assume Cp = 1.15 kJ/kg K,
γ = 1.333 and optimum blade speed ratio.
Solution
Let the suffix f s, 1, 2 and 3 represent final stage, exit from the nozzle, inlet
to the moving blade and exit from the moving blade respectively.
Let the isentropic temperature at the exit of the final stage Tf s . Now
T02
T0f s
=
T0f s
=
p02
p0f s
T02
1.56
γ−1
γ
900
1.56
=
0.333
1.333
6
1
=
=
=
1.56
576.9 K
Actual overall temperature drop
ΔToverall
=
ηT T02 − T0f s
=
0.85 × (900 − 576.9)
=
274.6 K
Applying energy equation to the nozzle of any stage,
c21
2
=
h2 +
h 1 − h2
=
1 2
c − c21
2 2
Cp (T1 − T2 )
=
1 2
c − c21
2 2
h1 +
c22
2
(1)
For optimum blade speed ratio,
σopt
=
u
c2
c2
=
2u
sin α1
=
sin α2
2
=
2 × 250
sin 75
=
517.6 m/s
For the given conditions, the inlet and exit velocity triangles are as shown
in the Fig. and also note that c3 = C1 .
Impulse and Reaction Turbines
c1
=
c3
=
c2 cos α2
=
133.96 m/s
=
c22 − c21
2Cp
=
108.68 K
=
455
517.6 × cos 75
Now from Eq.1
T1 − T2
517.62 − 133.962
2 × 1.15 × 103
=
For the impulse stage there is no expansion in rotor. Therefore, T1 − T2
will be the stage temperature drop. Hence, number of stages n is given by
n
=
ΔToverall
ΔTstage
Stages required
=
3
=
274.6
108.68
=
2.57
Ans
⇐=
11.2 A gas turbine having single stage rotates at 10000 rpm. At entry
to the nozzles the total head temperature and pressure of the gas
is 700◦ C and 4.5 bar respectively and at outlet from the nozzle the
static pressure is 2.6 bar. At the turbine outlet annulus the static
pressure is 1.5 bar. Mach number at outlet is limited to 0.5 and gas
leaves in an axial direction. The outlet nozzle angle is 70◦ to the
axial direction and the nozzle friction loss may be assumed to be 3%
of the isentropic temperature drop from total head at entry to static
conditions at outlet nozzle pressure. Calculate
(i) the gas angles at entry and outlet from the wheel showing them
on velocity diagrams for mean blade section,
(ii) output power developed by the turbine shaft.
Assume the mean blades diameter as 64 cm, gas mass flow rate as
22.5 kg/s, turbine mechanical efficiency = 99%, Cp = 1.147 kJ/kg K
and γ = 1.33.
Solution
Let the suffix 1, 2 and 3 represent the entry to the nozzle blade, exit from
the nozzle blade (entry to rotor blade) and exit from the rotor blade respectively. Let T02 be the isentropic temperature after expansion in the
nozzle.
γ−1
γ
T02
T01
=
T02p rime
=
0.8727 × 973
ηnozzle
=
0.97
p02
p01
2.6
4.5
=
=
=
0.333
1.333
=
0.8727
849.14 K
T01 − T02
T01 − T02
=
973 − T02
973 − 849.14
456
Gas Turbines
α1
c1
ca1
w1
β1
u
c t1
w2
c2
β2
u
Fig. 11.24
T02
=
c2
=
2Cp ΔT
=
2 × 1147 × (973 − 852.9)
u
=
From Fig.11.24
wt2
=
852.9 K
πDN
60
=
524.9 m/s
3.14 × 0.64 × 10000
= 334.93 m/s
60
=
c2 sin α2 − u = 524.9 × sin 70 − 334.93 = 158.3 m/s
ca
=
c2 cos α2
tan β2
=
c2 sin α2 − u
ca
β2
=
41.4◦
=
524.9 × cos 70
=
158.3
179.53
=
=
179.53 m/s
0.8815
Ans
⇐=
Assuming rotor losses are negligible
T3
T2
=
p2
p3
c3
γ−1
γ
=
M
γRT3
=
265.33 m/s
=
=
852.9
2.6
1.5
0.330
1.333
0.5 ×
=
744 K
√
1.33 × 284.6 × 744
Impulse and Reaction Turbines
tan β3
=
u
ca
334.93
265.33
β3
=
51.6◦
ct2
=
c2 sin α2 = 524.9 × sin 70 = 493.2 m/s
=
=
457
1.2623
Ans
⇐=
Power developed
=
ηm ṁu(ct2 + ct3 )
=
ṁuct2
=
0.99 × 22.5 × 334.93 × 493.2
1000
=
3679.53 kW
Ans
⇐=
11.3 In a single-stage impulse turbine the nozzle discharges the fluid on
to the blades at an angle of 65◦ to the axial direction and the fluid
leaves the blades with an absolute velocity of 300 m/s at an angle of
30◦ to the axial direction. If the blades have equal inlet and outlet
angles and there is no axial thrust, estimate the blade angle, power
produced per kg/s of the fluid and the blade efficiency.
Solution
Since the axial thrust is zero,
ca3
=
ca2
ca
c
w1
ca1
=
β1
α1
u
ct1
w2 c
2
c a2
β2
α2
u
c t2
Fig. 11.25
w3
=
w2
ca3
=
ca2 = ca = ca3 × cos α3
458
Gas Turbines
=
300 × cos 30
=
c2
=
ca2
cos α2
259.8
= 614.7 m/s
cos 65
u
=
ct2 − wt2
c2 sin α2 − ca2 tan β2
=
ca3 tan β3 − c3 sin α3
=
=
259.8 m/s
wt3 − ct3
Since ca2 = ca3 = ca and β2 = β3 = β,
2ca tan β
=
c2 sin α2 + c3 sin α3
tan β
=
614.7 × sin 65 + 300 × sin 30
2 × 259.8
=
1.3609
β
=
53.7◦
β2
=
β3
u
=
C2 sin α2 − ca2 tan β2
=
614.7 × sin 65 − 259.8 × tan 53.7
=
203.43
=
c2 sin α2
=
557.1 m/s
=
c3 sin α3
=
150 m/s
=
203.43 × (557.1 + 150) × 10−3
=
144 kJ/kg
σ
=
u
c2
Blade efficiency
=
4 σ sin α2 − σ 2
=
4 × 0.33 × sin 65 − 0.332 = 0.761
=
76.1%
ct2
ct3
WT
=
=
Ans
⇐=
Ans
53.7◦
⇐=
=
614.7 × sin 65
=
300 × sin 30
203.43
614.7
=
0.33
Ans
⇐=
11.4 Gas at 7 bar and 300◦ C expands to 3 bar in an impulse turbine stage.
The nozzle angle is 70◦ with reference to the exit direction. The
rotor blades have equal inlet and outlet angles, and the stage operates
Impulse and Reaction Turbines
459
with the optimum blade speed ratio. Assuming that the isentropic
efficiency of the nozzles is 0.9, and that the velocity at entry to the
stage is negligible, deduce the blade angle used and the mass flow
required for this stage to produce 75 kW. Take Cp = 1.15 kJ/kg K.
Solution
c
w1
ca1
β1
α1
u
ct1
w2 c
2
c a2
β2
α2
u
c t2
Fig. 11.26
Let the suffix 1, 2 and 3 represent the entry to the nozzle blade, exit from
the nozzle blade (entry to rotor blade) and exit from the rotor blade respectively. Let T02 be the isentropic temperature after expansion in the
nozzle.
γ−1
γ
0.333
1.333
T02
T01
=
T02p rime
=
573 × 0.8092
T02
=
T01 − ηn × (T01 − T02 )
=
573 − 0.9 × (573 − 463.67)
c2
p02
p01
3
7
=
=
=
463.67 K
=
2Cp ΔT
=
2 × 1150 × (573 − 474.6)
=
474.6 K
=
475.73 m/s
For optimum blade speed ratio
u
1
sin α2
=
c2
2
u
=
475.75 × sin 70
2
0.8092
=
223.52 m/s
460
Gas Turbines
ct2 − u
ca2
c2 sin α2 − u
ca
=
wt2
ca2
=
475.73 × sin 70 − 223.52
475.73 × cos 70
β2
=
54◦
ct2
=
c2 sin α2
=
447 m/s
ca2
=
w2 cos β2
=
c2 cos α2
w2
=
c2 cos α2
cos β2
=
475.73 × cos 70
cos 54
=
276.8 m/s
w
=
w2
ct3
=
w3 sin β3 − u = 276.8 × sin 54 − 223.52
=
0.416 m/s
WT
=
ṁu
(ct2 + ct3 )
1000
ṁ
=
75 × 1000
= 0.75 kg/s
223.52 × (447 + 0.416)
tan β2
=
=
=
1.3737
Ans
⇐=
=
=
w3
475.73 × sin 70
=
276.8 m/s
Ans
⇐=
11.5 The following particulars of a single stage turbine of free vortex type
is given below:
Total head inlet pressure
:
4.6 bar
Total head inlet temperature
:
700◦ C
Static head pressure at mean radius
:
1.6 bar
Mean blade diameter/blade height
:
10
Nozzle loss coefficient
:
0.10
Nozzle outlet angle
:
60◦
Determine the gas temperatures, velocities and discharge angle at the
blade root and tip radii. Take Cp = 1.147 and γ = 1.33 and mass
flow rate = 20 kg/s.
Solution
Let the suffix 1, 2 and 3 represent the entry to the nozzle, exit of the nozzle
(entry to the moving blade) and exit from the moving blade respectively.
Let T01 be the isentropic temperature after expansion in the nozzle.
Impulse and Reaction Turbines
c1
w1
ca1
461
α1
β1
u
ct1
Fig. 11.27
T2
T01
=
T2
=
γ−1
γ
pe
p01
0.77 × 973
0.33
1.33
=
1.6
4.6
=
749.2 K
=
0.77
Nozzle loss coefficient
=
T2 − T2
T02
=
T2 +
c22
2Cp
=
973 − T2
0.1
=
T2 − 749.2
973 − T2
T2
=
769.5 K
c2
=
2Cp ΔT
=
2 × 1147 × (973 − 769.5)
(1)
c22
2Cp
c22
2Cp
=
683.25 m/s
=
c2 cos α2
=
341.63 m/s
=
p2
RT2
=
0.73 kg/m3
Aca ρ
=
ṁ
A
=
20
= 0.08 m2
341.63 × 0.73
ca
ρ
=
=
683.25 × cos 60
1.6 × 105
284.6 × 769.5
462
Gas Turbines
A
=
Dm
=
πDm h
=
πDm
0.8 × 10
10
=
Dm
10
2
πDm
10
=
0.5 m
Mean blade diameter = 50 cm
Blade height =
D
10
= 5 cm
Mean blade radius = 25 cm
At the root
Droot
=
Dm − h
rroot
=
22.5 cm
Dtip
=
Dm + h
rtip
=
27.5 cm
=
50 − 5
=
45 cm
=
50 + 5
=
55 cm
At the tip
For a free vortex ct × r must be constant
ct
root
× rroot
=
ct
ct
=
c2 sin α2
=
591.71 m/s
mean
mean
× rmean
=
683.25 × sin 60
ct2
root
=
591.71 × 0.25
0.225
α2
root
=
tan−1
ct2 root
ca
=
tan−1
657.45
341.63
=
1.924
=
=
ct2 root
sin α2 root
=
741.2 m/s
=
T02 +
=
733.5 K
c2
T2
root
root
=
657.45 m/s
62.5◦
=
Ans
⇐=
657.45
sin 62.5
Ans
⇐=
c22 root
741.22
= 973 +
2Cp
2 × 1147
Ans
⇐=
Impulse and Reaction Turbines
463
At the tip
ct2
α2
c2
T2
tip
tip
tip
tip
=
ct2
× rmean
591.71 × 0.25
=
rtip
0.275
mean
=
537.9 m/s
=
tan−1
=
57.6◦
=
ct2 tip
sin α2 tip
=
537.9
= 637.1 m/s
sin 57.6
=
973 −
ct2 tip
537.9
= tan−1
ca
341.63
Ans
637.12
2 × 1147
⇐=
=
Ans
⇐=
796 K
Ans
⇐=
Review Questions
11.1 Explain the principle of operation of a turbine and what are the primary parts of a turbine?
11.2 How do you differentiate between an impulse and a reaction turbine?
With neat sketches explain the working of an impulse and a reaction
stage.
11.3 With a neat sketch explain a single stage velocity triangle and derive
an expression for the work output.
11.4 What do you understand by blade and stage efficiency? Derive an
expression for blade efficiency.
11.5 Explain what is meant by velocity compounding of a multistage impulse turbine.
11.6 Explain what do you understand by pressure compounding of a multistage impulse turbine.
11.7 Explain with a sketch and h-s diagram, the working of a reaction
turbine.
11.8 Define degree of reaction and derive an expression for the same.
11.9 Explain the following:
(i) zero per cent reaction stage,
(ii) fifty per cent reaction stage, and
464
Gas Turbines
(iii) hundred per cent reaction stage.
11.10 Explain briefly the performance graphs of an reaction turbine.
Exercise
11.1 In a single-stage impulse turbine the nozzle discharges the hot gas on
to the blades at a velocity of 750 m/s. The mass flow rate of gas is
100 kg/s. The turbine rotates at 20000 rpm. The mean diameter of
wheel is 31.5 cm. The nozzles are inclined at an angle of 20◦ to the
plane of wheel rotation. Calculate
(i) power developed by the blades in MW,
(ii) energy lost in the blades in MW, and
(iii) determine the blade efficiency of the turbine.
Assume the blade friction coefficient as 0.92 and outlet blade angle
as 25◦ to the plane of rotation.
Ans: (i) 24.86 MW (ii) 1.58 MW (iii) 93.6%
11.2 The mean diameter of the blades of an impulse turbine with a single
row wheel is 105 cm and the speed is 3000 rpm. The nozzle angle
is 72◦ with respect to axial direction, the ratio of blade speed to gas
speed is 0.42 and the ratio of the relative velocity at outlet from the
blades to that at inlet is 0.84. The outlet angle of the blade is to
be made 3◦ less than the inlet angle. The mass flow rate is 8 kg/s.
Calculate the following:
(i) tangential thrust on blades,
(ii) axial thrust on blades,
(iii) power produced, and
(iv) blade efficiency.
Ans: (i) 3024.8 N (ii) 83.12 N (iii) 499 kW (iv) 91%
11.3 Define utilization factor.
Show that the utilization factor of a stage of an impulse turbine with
single row wheel can be derived as
= 2 σ sin α2 − σ 2 (1 + kc)
where σ =
u
c2 ; k
=
w3
w2
and c =
sin β3
sin β2 .
State the assumptions made.
Calculate the maximum utilization factor for an equiangular blade for
a nozzle angle of 70◦ . Take the velocity coefficient as 0.83.
If the utilization factor is to be 90% of the maximum, what are the
possible ratios of the blade speed to gas speed?
Ans: (i) 80.8% (ii) 0.615 or 0.325
Impulse and Reaction Turbines
465
11.4 Gas with a velocity of 240 m/s relative to blades enters an impulse
moving row at an angle of 60◦ with respect to axial direction. The
tangential velocity of blades is 183 m/s. The work developed in blades
is estimated as 75 kJ/kg of gas. Find the blade efficiency and the blade
friction coefficient for relative velocities. Assume symmetrical blades.
Ans: (i) 89.7% (ii) 0.97
11.5 The nozzles of a two-row velocity compounded stage have outlet angles of 68◦ with reference to the axial direction and utilize an isentropic enthalpy drop of 200 kJ/kg of gas. All moving and guide
blades are symmetrical, and mean blade speed is 150 m/s. Assuming
an isentropic efficiency of the nozzle of 90%, find all the blade angles
and calculate the specific power output produced by the stage. The
velocity at inlet to the stage can be neglected.
Ans: (i) β1 = β2 = 61◦ ; β3 = 25.3◦ ;
α2 = α3 = 48.7◦ (ii) 153.75 kW
Multiple Choice Questions (choose the most appropriate answer)
1. An impulse turbine stage is characterized by the expansion of the
gases in
(a) stator nozzles
(b) rotor nozzles
(c) both stator nozzles and rotor nozzles
(d) none of the above
2. A reaction turbine stage is characterized by the expansion of the gases
in
(a) stator nozzles
(b) rotor nozzles
(c) both stator nozzles and rotor nozzles
(d) none of the above
3. In an impulse turbine stage, pressure
(a) increases in stator and decreases in rotor
(b) decreases in stator and increases in rotor
(c) decreases in stator and remains constant in rotor
(d) remains constant in both stator and rotor
4. In an impulse turbine stage, velocity
(a) increases in stator and decreases in rotor
(b) decreases in stator and increases in rotor
466
Gas Turbines
(c) increases in stator and remains constant
(d) remains constant in both stator and rotor
5. In a reaction stage turbine, the velocity
(a)
(b)
(c)
(d)
increases in stator and remains constant in rotor
increases in stator and decreases in rotor
decreases in stator and remains constant in rotor
remains constant throughout
6. In a reaction stage turbine, the pressure
(a)
(b)
(c)
(d)
increases in stator and remains constant
increases in stator and decreases in rotor
decreases in stator and remains constant in rotor
decreases in stator and also in rotor
7. The loading coefficient for a reaction turbine for constant axial velocity is given by
(a) ψ = φ(tan α2 + tan α3 )
(b) ψ = φ(tan α2 − tan α3 )
(c) ψ = φ(tan α2 × tan α3 )
(d) ψ = φ
tan α2
tan α3
8. Degree of reaction of a reaction turbine is given by
(a) R =
(b) R =
(c) R =
(d) R =
ca
2u (tan β3
ca
2u (tan β3
ca
2u (tan β3
ca tan β3
2u tan β2
+ tan β2 )
− tan β2 )
× tan β2 )
9. Multistage reaction turbine are employed to achieve
(a)
(b)
(c)
(d)
a large pressure drop
a large mass flow rate
a large volume flow rate
none of the above
10. Multistage reaction turbine are employed to achieve
(a)
(b)
(c)
(d)
a high velocity of fluid
a high pressure of fluid
a high temperature of fluid
all of the above
Ans:
1. – (a)
6. – (d)
2. – (c)
7. – (a)
3. – (c)
8. – (b)
4. – (a)
9. – (a)
5. – (b)
10. – (a)
12
TRANSONIC AND
SUPERSONIC
COMPRESSORS AND
TURBINES
INTRODUCTION
The pressure rise in a compressor is due to the conversion of kinetic
energy to potential energy with higher static pressure at the exit. Therefore, the compressor outlet pressure depends upon the inlet velocity of the
compressor. Nowadays, higher pressure ratios are achieved with reasonably
good efficiency.
Based on the Mach number, which is the ratio of fluid velocity to the
velocity of sound, the compressor can be broadly classified into
(i) subsonic compressors
(ii) supersonic compressors
However, the Mach number between 0.8 and 1.2 is considered as transonic. Mach number between 1.2 and 5 is considered as supersonic and
above 5, it is hyposonic.
The recent trend in the design of compressors is to achieve higher and
higher pressure ratio per stage. The pressure rise or the specific work is
proportional to the square of the inlet velocity. The question is why not
increase the velocity at inlet in a subsonic compressor? It should be noted
(i)
(ii)
(iii)
subsonic
sonic
supersonic
(M < 1)
(M = 1)
(M > 1)
468
Gas Turbines
that the pressure ratio attainable in a single stage in a subsonic compressor
is limited by the rapid decrease in efficiency. Further, with increase in inlet
velocity shock waves will form which will increase the losses. There is also a
possibility of flow separation and the boundary layer development, both of
which affect the flow. Hence, a transonic or supersonic compressor should
be designed appropriately from the beginning taking into account all the
above problems.
12.1
THE SUPERSONIC COMPRESSOR
As stated in the Section 12, the increased losses are largely due to the presence of shock waves and the flow separation. This is due to the interaction
of the shock waves with the blade surface boundary layer. On the other
hand, with supersonic velocities compression occurs through shock waves.
This enables a high pressure ratio to be obtained in a machine with short
axial length. Thus the supersonic compressors appear to have an advantage
and may be expected to be compact machines. It may be noted that higher
pressure ratios are obtained at the expense of efficiency. Figure 12.1 shows
the details of subsonic, transonic and supersonic compressors.
Subsonic �
(M = 0 to 0.8)
Transonic
(M = 0.8 to 1.2)
Supersonic
(M = 1.2 to 5.0)
Fig. 12.1 Compressor size for a given pressure ratio and mass flow rate
When supersonic flow approaches a blade passage, depending on the
blade geometry at inlet, a normal shock will form (Fig. 12.2). The normal
shock is considered as a discontinuity. The flow properties will change over
a negligibly small distance. The static pressure, temperature and entropy
will increase while the velocity and stagnation pressure are reduced. Mach
number is reduced from supersonic to subsonic. Higher the inlet Mach
number, higher will be these changes and the shock is considered strong.
The flow does not undergo any turning through a normal shock.
The supersonic stage concept started around nineteen forties. The actual development of a supersonic stage was much delayed because of the
non-availability of proper materials to withstand the high stresses due to
higher rotational speeds. Since, the supersonic compressor operates in the
high subsonic or transonic range during off-design speeds, the distinction
between the transonic and the supersonic compressor is rather vague.
Transonic and Supersonic Compressors and Turbines
469
A compressor with supersonic relative velocity everywhere at inlet to
rotor blade row is by convention called a supersonic compressor. If the
relative velocity is subsonic in some part of the inlet then it is known as a
transonic compressor. In a transonic axial flow compressor the Mach number based on the tip peripheral speed may be as high as 1.4. It is possible
to set a stage pressure ratio about 1.6 to 1.8. In supersonic compressors,
still higher values are possible.
M>1
N
M<1
N - Normal shock
Fig. 12.2 Normal shock in the blade passage
As already explained, increasing the inlet Mach number of a compressor stage from subsonic to transonic to supersonic is normally accompanied
by an increase in the pressure ratio per stage. This means that a reduction in the number of stages for a given pressure ratio can be achieved by
designing a supersonic compressor (Fig. 12.1). Further, by increasing the
axial component of Mach number Max through the stage, the frontal area
is decreased for a given mass flow. Thus a reduction in the size of the compressor for a given power can be achieved. It is to be noted that usually
the Max is subsonic in supersonic compressors. The maximum flow rate
through an annulus of given size is attained when Max is unity. Therefore,
there is no great advantage in having Max greater than unity.
12.2
SUPERSONIC AXIAL FLOW COMPRESSORS
In the case of subsonic compressors, the deceleration and the pressure rise
that could be achieved is partly due to the flow deflection, which is limited
to 20 - 23 degrees. If higher deflections are attempted then the increased
boundary layer separation will result. This will cause a net reduction in
deflection with increased losses. The ’de Haller criterion’ fixes the deceleration ratio of the exit relative velocity from the blade row to the inlet
velocity, which should not be less than 0.7. On the other hand in the case
of supersonic flows, a higher compression in the stage can be achieved.
This is due to the characteristic high pressure rise that takes place across
a stabilized shock.
470
12.2.1
Gas Turbines
Types of Supersonic compressors
The design of blades for a supersonic compressor should take care of the
following tasks for better performance.
(i) Deceleration of the supersonic flow to subsonic velocities
(ii) Deflection of the incoming flow towards the axial direction
(iii) Progressive transformation of kinetic energy of the subsonic flow into
pressure energy
(iv) Increase of total enthalpy of the fluid in the absolute frame of reference
Supersonic compressors are divided according to the type of flow in the
rotor blade row and the stationary blade row or diffuser. Assume that the
relative velocity leaving the rotor blade row is subsonic. Then the change
from supersonic to subsonic flow in the rotor blade row is assumed to take
place with shock. Such a case is known as shock-in-rotor compressor. When
the absolute velocity into a stationary blade row is supersonic which is
changed to subsonic at the exit then it is called a shock in stationary blade
compressor.
(a) Shock-in-rotor and the subsonic stationary blade row
The shock is stabilized in the long blade channels at low supersonic inlet
Mach numbers. The inlet axial Mach number M1ax is subsonic. A normal
shock is stabilized in the rotor at low supersonic Mach number with the
help of high solidity blading1
The rotor exit absolute Mach number is subsonic. A conventional stationary blade row is used to diffuse the flow further. Fig. 12.3 shows such
an arrangement.
The part of the leading edge shock, held inside the passage is normal
to flow. It is reflected on the suction surface of the adjacent blade. Blade
surface boundary layer is thickened after the shock and it could even separate.
(b) Shock-in-rotor and shock-in-stationary blades
This is conventionally described as the distributed shock compressor.
The inlet relative Mach number to the rotor is supersonic and the shock
gets stabilized near the exit of the rotor. The supersonic absolute Mach
number at the rotor stationary blade row interface, places a heavy duty on
the downstream stationary blade row which must deflect and diffuse the
flow. Figure 12.4 describes the distributed shock configuration.
(c) Impulse rotor and shock in stationary blades
The impulse rotor achieves a supersonic deflection. The pressure increase is obtained only in the stationary blades. The flow through the
rotor is entirely in a supersonic condition with the exit axial Mach number
1 Blade solidity is the ratio of the blade chord, represented by its length, over the blade
pitch, which is the circumferential spacing of the blades at a given radius or diameter
from the axial centerline axis. In other words, blade pitch is the circumferential length
at a given diameter divided by the number of blades in the full fan blade row.
Transonic and Supersonic Compressors and Turbines
471
π= w/a
M2r
π2
1
M2
M3 <1
2
π1 >1
Mtr > 1
3
M1 <1
2
4
M1 = M1x
1. Normal shock in rotor�
2. Subsonic
3. Supersonic
4. Direction of rotation
Fig. 12.3 Shock-in-rotor and subsonic stationary blades
M2r<1
M2r>1
M3 <1
Mtr > 1
π >1
1
1
2
3
2
M1 <1
4
5
1. Shock in rotor�
2. Supersonic
3. Subsonic �
4. Shock in stationary blades �
5. Direction of rotation
Fig. 12.4 Shock-in-rotor and Shock-in-stationary blades
472
Gas Turbines
also becoming supersonic. Figure 12.5 shows such a supersonic stage. This
arrangement is not advantageous as the pressure rise takes place only in
the stationary blade row and has little practical use.
π = u/a
M 2r= M1r
π 2= π 1
M 2>1
M 3<1
π 1> 1
M1r>1
M 1<1
Direction �
of rotation
Fig. 12.5 Impulse rotor and shock in stationary blades
12.2.2
Blade Profiles
As already stated, in the case of subsonic axial flow compressors the pressure rise is mainly due to flow turning, whereas for supersonic stages flow
turning is not critical. Therefore, the blade profiles will be less cambered.
The maximum thickness value will also be less in supersonic compressors.
The requirement of the blade profile is to accelerate the inlet velocity to
reasonable supersonic velocity, say a Mach number of 1.3 or 1.4. A shock of
reasonable strength will be produced without appreciably increasing losses.
If the inlet Mach number is say 0.9 then the blade profile should accelerate
the flow to about 1.3 Mach number before the shock. If the inlet Mach
number is slightly supersonic, say M = 1.1, then the blade profile should
provide less acceleration to achieve 1.3 Mach number. The blade suction
surface curvature near the inlet should be changed suitably depending on
the inlet Mach number.
For high subsonic inlet Mach numbers, say M = 0.9, double circular arc
blades (DCA) are normally employed. The flow gets accelerated at the en-
Transonic and Supersonic Compressors and Turbines
473
trance region until a normal shock is formed at the entrance, Fig. 12.6. The
Mach number on the sonic line is 1. The flow gets decelerated to subsonic
velocities across the shock. Use of DCA profiles for slightly supersonic inlet
Mach numbers would lead to higher Mach numbers at inlet to shock. This
will result in stronger shocks with increased losses.
M<1
S o n i c li n
Normal �
shock
M>1
e
M= 0.9
Fig. 12.6 DCA profile
In order to avoid this, the concept of multiple circular arc blades (MCA)
were evolved. In this case, the suction surface is made out of two or more
circular arcs of different curvature. The upstream suction surface arc at the
entrance region is of lower camber. This reduces the acceleration on the
suction surface at the entrance. Further reduction in the suction surface
curvature at the entrance leads to a circular wedge profile (CW) with the
wedge at inlet as shown in Fig. 12.7. This will allow higher Mach number
at the entrance.
Almost in all application, a high static pressure rise is what is required
and not high velocity. The axial velocity leaving a supersonic stage will
be quite high even though it is subsonic. Therefore in a compressor with
supersonic stages the final stage must be subsonic. The matching the two
is not an easy task.
474
Gas Turbines
Bo w s h
oc
k
M > 1.0
Fig. 12.7 CW profile
12.3
SUPERSONIC RADIAL COMPRESSORS
The axial velocity of the air at the entrance of the compressor is mostly
subsonic. However, depending on the rotational speed of the impeller, the
relative velocity into the inducer becomes supersonic. In most of the supersonic radial compressors used for gas turbines, the supersonic conditions
have been mostly limited to the inlet of the inducer and the exit of the
impeller. This could cause a surge condition. Most of the results presented
in literature on supersonic radial compressors, report that the peak compressor efficiency is obtained with inlet Mach numbers in the range of 0.6
to 0.85. Further increase in Mach number reduces the efficiency drastically.
There has been hardly any mention of shocks being stabilized efficiently
in rotors of centrifugal compressors. However, the Mach number leaving
the impeller of a 12:1 pressure ratio impeller has been found to be 1.30.
At this Mach number high fluctuations and non-uniform inlet conditions
are found to develop. It has been reported that a Mach number of unity
is almost exceeded in the case of even 4:1 pressure ratio impellers. The
traditional method to overcome such high supersonic inlet velocities in the
diffuser is to provide a short vaneless diffuser at the entry. This allows
sufficient diffusion to a subsonic velocity before entering any vane system.
It has been observed that leaving such a vaneless diffusing space is a disadvantage. It is due to the fact that the boundary layer gets skewed relative
to the main stream flow. Further, the flow which is subjected to a strong
pressure gradient separates and invariably the diffuser stalls.
Transonic and Supersonic Compressors and Turbines
12.4
475
SUPERSONIC AXIAL FLOW TURBINE STAGES
Similar to the supersonic compressors, supersonic axial flow turbines have
also been developed to obtain higher specific work. In the following sections
the details are discussed.
12.4.1
Turbine stages with Supersonic velocities
Multistage steam turbines are usually provided with a Curtis velocity compounded turbine with large specific work. The maximum pressure drop
takes place across the first nozzle blade row. As a result of this high pressure drop, the absolute velocity leaving the nozzle and often the relative
velocity entering the rotor blade row will be supersonic. Gas turbines used
in rocket engines to drive the liquid fuel pumps also operate under very high
pressure ratios resulting in supersonic velocities. In large output steam turbines, the rotor blades are very long in the last stage and they operate with
high tip velocities with relative flow Mach number around the range of 1.5
to 1.8 while leaving the rotor blades. Typical blade passages for such velocities are shown in Fig. 12.8. Such turbine stages with supersonic velocities
can be classified as follows:
c2
w2
Th
roa
t
u
Convergent divergent passage
w1
c1
M1r > 1.0
u
Fig. 12.8 Typical tip section of a steam turbine last stage
(i) Supersonic stages with supersonic absolute velocity at nozzle exit and
supersonic relative velocity in the rotor blade row. For such conditions, the rotor is invariably, of the impulse type and the axial velocity
would be subsonic:
(ii) Transonic stages with supersonic absolute velocities at nozzle exit and
supersonic relative velocities at rotor exit. For such conditions, the
rotor is often designed with a high degree of reaction.
The essential difference between a subsonic and supersonic turbine can
be stated as follows: In subsonic turbines the flow incidence at the rotor
476
Gas Turbines
blade is mainly controlled by the upstream nozzle outlet angle and the rotor
blade speed u. In supersonic turbines, the relative velocity is supersonic.
Therefore, the incidence is mainly controlled by the rotor blade leading
edge. Thus the rotor blade of a supersonic turbine fixes its own incidence.
Thereby, the static pressure and the flow angle downstream of the preceding
nozzle is controlled.
12.4.2
Effect of Turbine Blade Speed
Let us examine the supersonic flow in such a turbine (Fig. 12.9) under
design conditions. Neglecting the nozzle trailing edge thickness and rotor
blade leading edge thickness, the incidence on to the rotor would be zero.
The flow through the nozzle and rotor would be shock free. The Mach
number of the flow leaving the nozzle will be equal to the design value.
The flow direction will be the same as that of the nozzle angle with almost
zero deviation, Fig. 12.9(a).
c2
w2
u
(a) u - design
Rotor blade
c2
u
c2
w2
u
(b) u - increased
w2
(c) u - decreased
Fig. 12.9 Effect of turbine blade speed on the velocity triangle
If the blade speed u is now increased the velocity triangle will now
change as in Fig. 12.9(b). However, the relative flow direction will be the
same as the rotor blade direction at inlet resulting in zero incidence. This
is obtained by deflecting the absolute flow leaving the nozzle through an
increased angle or decreased α2 . This is caused by the creation of shock
and expansion waves at the trailing end of the nozzle blades. The flow from
the nozzle first passes through an oblique trailing edge shock wave. Then
there is a reflection of the shock wave from the adjacent blade and again
through an expansion fan which comes off the trailing edge of the adjacent
blade. The mass flow will remain the same as per the design, because, the
inlet conditions remain unchanged. The pressure and density downstream
Transonic and Supersonic Compressors and Turbines
477
of the nozzle will be higher than that of design conditions. Therefore, the
axial velocity will reduce. The exit Mach number M2 will be less than the
design M2 .
Similarly, if the blade speed is reduced it would result in flow condition
shown in Fig. 12.9(c). The pressure at the exit of the nozzle will be less than
its design value and M2 would be greater than the design M2 . The nozzle
will operate at a higher pressure ratio (P3 /P2 ) than the design condition.
Such conditions of operation are termed ’supercritical’ conditions. The
above explanation is based on the assumptions that
(i) the trailing edge of the nozzle blades are very thin, and
(ii) the rotor blade leading edges are also very thin.
Definite wedge angle of the nozzle trailing edge even with sharp edge
will give rise to two oblique shock waves at the trailing edge. Definite
thickness of the trailing edge will produce more disturbance at the trailing
edge. If the rotor blades are not very thin then at the leading edge, the
operating incidence is not zero, but, is slightly positive measured relative
to the flat region of the suction leading edge. This incidence is a function of
the inlet Mach number and the blade leading edge geometry and is called
the unique incidence. Shocks originate from the leading edge and expansion
waves from the chamfer end.
The entering relative flow passes through a wave pattern made up of
a shock and expansion waves as shown in Fig. 12.10. As the relative flow
M2oo
β 2 oo
M 2r oo
S
M2
β =β
2b
β2b
β2
tu
Chamfer wedge angle
2
(S-tu) sin (180 - β2b)
2 inlet to blade row
2 oo head of waves or eixt of nozzle
Shock
Expansion wave
Fig. 12.10 Supersonic impulse rotor incidence
478
Gas Turbines
reaches the wedge end of the blade chamfer a shock waves is produced.
This turns the flow parallel to the tangential direction, i.e., parallel to the
blade chamfered surface. As the flow reaches the chamfered end, it has, to
be turned parallel to the blade suction surface inlet region. This is done
by the expansion waves emanating from the chamfer end. If the deflection
through the waves is small then the flow can be considered to be isentropic
and then it is a problem of simple area change.
Review Questions
12.1 Based on Mach number how are compressors classified?
12.2 Explain subsonic and supersonic compressors with figures.
12.3 With sketches explain the different types of supersonic compressors.
12.4 Explain the blade profile requirement of supersonic compressors.
12.5 With a diagram explain the turbine stages with supersonic velocities.
12.6 With a neat sketch explain the effect of turbine blade speed on velocity
triangle.
Multiple Choice Questions (choose the most appropriate answer)
1. The Mach number between 0.8 and 1.2 is called
(a) subsonic
(b) supersonic
(c) hypersonic
(d) transonic
2. In supersonic compressors, high pressure ratio is obtained due to
shock which makes the machine size to be
(a) long
(b) medium
(c) short
(d) none of the above
3. In supersonic compressor if the relative velocity is subsonic at some
part then it is known as
(a) subsonic compressor
(b) transonic compressor
(c) supersonic compressor
(d) hypersonic compressor
Transonic and Supersonic Compressors and Turbines
479
4. In a supersonic compressor the pressure ratio that can be obtained
per stage is
(a) 1.7
(b) 1.5
(c) 1.4
(d) 1.2
5. A supersonic compressor with shock in rotor and shock in stationary
blade is called
(a) normal shock compressor
(b) oblique shock compressor
(c) distributed shock compressor
(d) none of the above
6. For high subsonic inlet Mach numbers (M = 0.9)
(a) circular arc blades are preferred
(b) double circular arc blades are preferred
(c) straight blades are preferred
(d) slightly cambered blades are preferred
7. The peak efficiency of a supersonic compressor is obtained with inlet
Mach number in the range of
(a) M = 0.6 to 0.8
(b) M = 1.0 to 1.2
(c) M = 1.4 to 1.6
(d) M = 1.8 to 2.0
8. Supersonic stages in turbines with supersonic absolute velocity at
nozzle exit and supersonic relative velocity at the rotor blade row
must be of
(a) reaction type
(b) impulse type
(c) mixed type
(d) none of the above
9. Transonic stages in turbines with supersonic absolute velocities at
nozzle exit and supersonic relative velocities at rotor exit must be of
(a) reaction type
(b) impulse type
480
Gas Turbines
(c) mixed type
(d) none of the above
10. If the turbine nozzle operates at a higher pressure ratio than the
design value then it is called
(a) super critical condition
(b) critical condition
(c) subcritical condition
(d) abnormal condition
11. By increasing the axial component of Mach number, Max , through
the stage, the frontal area
(a) decreases
(b) increases
(c) remains the same
(d) does not depend on Mach number
12. Peak efficiency of a supersonic compressor is obtained with inlet Mach
number in the range of
(a) 0.4 to 0.6
(b) 0.6 to 0.8
(c) 0.8 to 1.0
(d) 1.0 to 1.2
Ans:
1. – (a)
6. – (b)
11. – (a)
2. – (c)
7. – (a)
12. – (b)
3. – (b)
8. – (b)
4. – (a)
9. – (a)
5. – (c)
10. – (a)
13
INLETS AND NOZZLES
INTRODUCTION
Compressors, combustion systems and turbines were discussed in detail
in the last four chapters. There are two more additional components, viz.,
inlets and nozzles to be considered especially in the case of power plants for
jet propulsion. The inlet and exhaust nozzle are the two engine components
that directly interfere with the internal air flow and the flow about the
aircraft. In fact, integration of the engine and the airframe is one of the most
complex problems and has a major impact on the performance especially
in aircraft system.
The purpose of any aircraft gas turbine engine inlet is to provide a
sufficient air supply to the compressor with as minimum a pressure loss as
possible. This is to be achieved with as small a drag force on the airplane
as possible. It also has the purpose of being a diffuser that reduces the
velocity of the entering air as efficiently as possible.
The purpose of the exhaust nozzle is to increase the velocity of the
exhaust gas before discharge from the nozzle. Further, it should collect and
straighten the gas flow. For large values of specific thrust, the kinetic energy
of the exhaust gas must be high. This requires a high exhaust velocity.
The pressure ratio across the nozzle controls the expansion process. The
maximum thrust for a given engine is obtained when the exit pressure pe
equals the ambient pressure p0 .
13.1
INLETS
The inlet interchanges the organized kinetic and random thermal energies
of the gas in an essentially adiabatic process. However, the perfect (no-loss)
inlet would correspond to an isentropic process. The primary purpose of the
inlet is to bring the air required by the engine from free-stream conditions
to the conditions required at the entrance of the compressor with minimum
total pressure loss. The best work output from a compressor is obtained
when the flow of air is at a Mach number of about 0.5. It is a known fact
that the engine performance heavily depends on the installation losses in
482
Gas Turbines
the inlet such as additive drag, forebody or cowl drag, bypass air, boundarylayer bleed air, etc., Hence, the design of the inlet should be done with great
care so as to minimize these losses. The performance of an inlet is related
to the following characteristics:
(i) high total pressure ratio,
(ii) controllable flow matching requirements,
(iii) good uniformity of flow,
(iv) low installation drag,
(v) good starting and stability,
(vi) low signatures (acoustic, radar, etc.,), and minimum weight, and
(vii) low cost while meeting life and reliability goals.
An inlet’s overall performance must be determined by simultaneously evaluating all these characteristics. It may be noted that improvement in one,
is often achieved at the expense of another. The design and operation of
subsonic and supersonic inlets differ considerably due to the characteristics
of the flow. For the subsonic inlets, near-isentropic internal diffusion can
be easily achieved, and the inlet flow rate adjusts to the demand. For a
supersonic inlet, the internal aerodynamic performance is a major design
problem. It should be appreciated that achieving efficient and stable supersonic diffusion over a wide range of Mach numbers is very difficult. In
addition, the supersonic inlet must be able to capture its required mass flow
rate. This requires variable geometry to minimize inlet loss and drag and
provide stable operation. Because of these differences, in the following sections we consider the subsonic and supersonic inlets separately, beginning
with the subsonic inlet and diffuser.
13.2
SUBSONIC INLETS
Most subsonic aircraft have their engines placed in nacelles. Hence, in
this section we may not deal with the inlet alone but include the nacelle at
subsonic Mach numbers. The cross section of a typical subsonic inlet and its
geometric parameters are shown in Fig.13.1. The inlet area A1 , is based on
the flow cross section at the inlet height. It may be noted that the subsonic
inlet can draw in airflow whose free-stream area A0 is larger than the inlet
area A1 . Variable inlet geometry is not required (except sometimes blow-in
doors are used to reduce installation drag during takeoff). The details in
this section on subsonic inlets are based on a fixed-geometry inlet.
The operating conditions of an inlet depend on the flight velocity and
mass flow requirements. Figure 13.2 shows the streamline pattern for three
typical subsonic conditions. Figure 13.2(a) shows acceleration of the fluid
external to the inlet which will occur when the inlet operates at a velocity
lower than the design value or at a mass flow higher than the design value.
Inlets and Nozzles
483
Highlight (hl)
d fan d max
A l d hl d t
2
Al=
π d hl
4
L
Fig. 13.1 Subsonic inlet nomenclature
Figure 13.2(b) show the configuration for correct mass flow for the design
conditions. Figure 13.2(c) shows deceleration of the fluid external to the
inlet which will occur at a velocity higher than design or a mass flow lower
than design.
(a) Static operation (b) Low-speed operation (c) High-speed operation
(or insufficient air
(correct engine air)
(more air than the
for engine)
engine needs)
Fig. 13.2 Typical streamline patterns for subsonic inlet
A list of the major design variables for the inlet and nacelle includes the
following:
(i) Inlet total pressure ratio and drag during cruise
(ii) Engine location on wing or fuselage
(iii) Aircraft attitude envelope
(iv) Inlet total pressure ratio and distortion levels required for engine operation
(v) Engine-out windmilling airflow and drag
(vi) Integration of diffuser and fan flow path contour
484
Gas Turbines
(vii) Integration of external nacelle contour with thrust reverser and accessories
(viii) Flow field interaction between nacelle, pylon, and wing
(ix) Noise suppression requirements
13.3
DIFFUSER
The flow within the inlet is required to undergo diffusion in a divergent
duct. This reduction in flow velocity creates an increase in static pressure
that interacts with the boundary layer. If the pressure rise due to diffusion
occurs more rapidly than turbulent mixing can re-energize the boundary
layer, the boundary layer will assume the configurations shown in Fig.13.3.
If the flow in an inlet separates, the total and static pressures are reduced
from their corresponding nonseparated flow values.
(a) Turbulent(b) Intermediate (c) Separation (d) Reversed flow
Fig. 13.3 Boundary layer development under adverse pressure gradient
As shown in Fig.13.4, the rate of area increase in a diffuser has a direct
effect on the behavior of flow in the diffuser. If the rate of area increase is
greater than that needed to keep the boundary layer energized and attached,
the flow may be characterized by unsteady zones of stall. The turbulent
mixing is no longer able to overcome the pressure forces at all points in
the flow, and local separation occurs at some points. The total pressure
decreases markedly due to the irreversible mixing of a fairly large portion of
low-velocity fluid with the main flow. If the diffuser walls diverge rapidly,
the flow will separate completely and behave much as a jet, as shown in
Fig.13.4(d).
The rate of area increase without stall for a diffuser depends on the
characteristics of the flow at the entrance and on the length of the divergent
section. Figure 13.5 shows the results for two-dimensional straight-walled
diffusers. The results are for incompressible flow, and they do not give
a qualitatively valid indication of the sensitivity of any diffuser to rapid
divergence of flow area.
For the design of an optimum diffuser, research has shown that the
boundary layer profile should maintain a constant shape. The boundary
layer thickness will, of course, increase as the flow moves down the diffuser.
The stipulation of a constant shape for the boundary layer profile implies
Inlets and Nozzles
(a) Well-behaved flow
485
(b) Large transitory stall
(c) Steady stall
(d) Jet flow
Fig. 13.4 Types of flow in straight-walled diffusers
40
2θ
Flow separation
H
20
2θ
L
10
0
Without flow separation
1
2
3
4
6
8
20
L/H
Fig. 13.5 Flow separation limits in two-dimensional straight-walled diffusers
the assumption that mixing re-energizes the profile at the same rate as the
static pressure depletes it.
In the presence of an adverse pressure gradient (static pressure increasing in the direction of flow), boundary layers tend to separate when the
boundary layer is not re-energized rapidly enough by turbulent mixing. If
vortices are generated by vortex generators in pairs, regions of inflow and
outflow exist. These carry high-energy air into the boundary layer and
low-energy air out. Figure 13.6 shows how vortex generators re-energize a
boundary layer.
By using vortex generators together with a short, wide-angle diffuser, it
may be possible to have a lower total pressure loss than with a long diffuser
without vortex generators. Here, the reduced skin friction losses associated
with flow separation are traded against vortex losses. The use of shorter
diffusers may reduce weight and facilitate engine installation.
486
Gas Turbines
After
reenergizing
Trailing vortex
Before
Vortex generator
Fig. 13.6 The role of vortex generator
13.4
SUPERSONIC INLETS
It is much more difficult to design supersonic inlets than the subsonic inlets.
The inlet used on a supersonic aircraft is the design that optimizes performance for the mission for which the aircraft is designed. The supersonic
inlet must have the following characteristics:
(i) provide adequate subsonic performance,
(ii) good pressure distribution at the compressor inlet,
(iii) high pressure recovery ratios, and
(iv) must be able to operate efficiently at all ambient pressures and temperatures during take-off, subsonic flight, as well as its supersonic
design condition.
Both axisymmetric and two-dimensional inlets have been designed and
used. Variable geometry center bodies, which change the inlet geometry,
for better off-design operation, and boundary-layer bleed have been incorporated into supersonic inlets.
The inlet performance characteristics that are used to assess the performance of supersonic inlets and those with the largest influence on aircraft
performance and range are
(i) total pressure recovery,
(ii) cowl drag,
(iii) boundary layer bleed flow,
(iv) capture-area ratio (mass flow ratio), and
(v) weight.
Inlets and Nozzles
487
Supersonic inlets are usually classified by their percent of internal compression. Internal compression refers to the amount of supersonic area
change that takes place between the cowl lip and the throat. This is illustrated in Figure 13.7 which identifies the center body, cowl lip, capture area
and throat for a supersonic inlet.
External
compression
Internal
compression
Cowl
Cowl lip
Throat area
ck
Capture area
liq
ho
es
u
Ob
Centre body
488
Gas Turbines
Inlets and Nozzles
Nozzle entrance
489
Nozzle throat
Fig. 13.10 Convergent exhaust nozzle
13.5.2
Convergent-Divergent (C-D) Nozzle
The convergent-divergent nozzle has two sections (i) convergent duct and
(ii) divergent duct. The section of the nozzle where the cross-sectional area
is minimum is called the throat of the nozzle. Most convergent-divergent
nozzles used in aircraft are not simple ducts. They incorporate variable
geometry and other aerodynamic features. The convergent-divergent nozzle
is used if the nozzle pressure ratio pte /p0 is high (greater than about 6).
High-performance engines in supersonic aircraft generally have some form
of a convergent-divergent nozzle. If the engine incorporates an afterburner,
the nozzle throat is made in such a way that the operating conditions of
the engine upstream of the afterburner remain unchanged (in other words,
vary the exit nozzle area so that the engine performance does not change
when the afterburner is in operation. Further, the exit area must be varied
to match the flow conditions in order to produce the maximum available
thrust.
Earlier supersonic aircraft used ejector nozzles in their high performance
turbojets. Use of such a nozzle permitted the following:
(i) bypassing varying amounts of inlet air around the engine,
(ii) providing engine cooling,
(iii) good inlet recovery, and
(iv) reduced boat-tail drag.
Ejector nozzles can also receive air from outside the nacelle directly into the
nozzle for better overall nozzle matching – these are called two-stage ejector
nozzles. For the modern high performance afterburning turbofan engines,
simple convergent-divergent nozzles are used without secondary air.
13.5.3
Nozzle Functions
An exhaust nozzle may be thought of as a device dividing the power available from the main burner exit gas between the requirements of the turbine
and the jet power. Thus the nozzle serves as a back-pressure control for the
engine and an acceleration device converting gas thermal energy to kinetic
490
Gas Turbines
energy. A secondary function of the nozzle is to provide required thrust
reversing and/or thrust vectoring.
13.5.4
Engine Backpressure Control
The throat area of the nozzle is one of the main means available to control
the thrust and fuel consumption characteristics of an existing engine. In
preliminary engine cycle analysis, the throat area of the nozzle is fixed by
selection of specific values for the engine design parameters and the design
mass flow rate. Assumption of constant areas can establish the off-design
operating characteristics of the engine and the resulting operating lines for
each major component. Changing the nozzle throat area from its original
design value will change the engine design and the operating characteristics
of the engine at both on- and off-design conditions.
At times, it is necessary to change the off-design operation of an engine in only a few operating regions, and variation of the throat area of
the exhaust nozzle may provide the needed change. At reduced engine corrected mass flow rates (normally corresponding to reduced engine throttle
settings), the operating line of a multistage compressor moves closer to the
stall or surge line. This has already been discussed in Sections 8.11 and
9.13. Steady-state operation close to the stall or surge line is not desirable
since transient operation may cause the compressor to stall or surge. The
operating line can be moved away from the stall or surge line by increasing
the exhaust nozzle throat area. This increase in nozzle throat area reduces
the engine backpressure and increases the corrected mass flow rate through
the compressor. Large changes in the exhaust nozzle throat area are required for afterburning engines to compensate for the large changes in total
temperature leaving the afterburner. The variable-area nozzle required for
an afterburning engine can also be used for back pressure control at its
nonafterburning settings.
One advantage of the variable-area exhaust nozzle is that it improves the
starting of the engine. Opening the nozzle throat area to its maximum value
reduces the backpressure on the turbine and increases its expansion ratio.
Thus the necessary turbine power for starting operation may be produced
at a lower turbine inlet temperature. Also, since the backpressure on the
gas generator is reduced, the compressor may be started at a lower engine
speed, which reduces the required size of the engine starter.
13.5.5
Exhaust Nozzle Area Ratio
Maximum engine thrust is realized for ideal flow when the exhaust nozzle
flow is expanded to ambient pressure (pe = p0 ). When the nozzle pressure ratio is above choking, supersonic expansion occurs between aft-facing
surfaces. A small amount of under-expansion is less harmful to aircraft
and engine performance than overexpansion. Overexpansion can produce
regions of separated flow in the nozzle and on the aft end of the aircraft,
reducing aircraft performance.
Inlets and Nozzles
491
The exhaust nozzle pressure ratio (pte = p0 ) is a strong function of flight
Mach number. Whereas convergent nozzles are usually used on subsonic
aircraft, convergent-divergent nozzles are usually used for supersonic aircraft. When afterburning engine operation is required, complex variablegeometry nozzles must be used. Most of the nozzles shown in Fig.13.11
are convergent-divergent nozzles with variable throat and exit areas. The
throat area of the nozzle is controlled to satisfy engine backpressure requirements, and the exit area is scheduled with the throat area.
13.5.6
Thrust Reversing and Thrust Vectoring
The need for thrust reversing and thrust vectoring is normally determined
by the required aircraft and engine system performance. Thrust reversers
are used on commercial transports to supplement the brakes. In-flight
thrust reversal has been shown to enhance combat effectiveness of fighter
aircraft.
Two basic types of thrust reversers are used: the cascade-blocker type
and the clamshell type (Fig.13.12). In the cascade-blocker type, the primary
nozzle exit is blocked off, and cascades are opened in the upstream portion
of the nozzle duct to reverse the flow. In the clamshell type, the exhaust jet
is split and reversed by the clamshell. Since both types usually provide a
change in effective throat area during deployment or when deployed, most
reversers are designed such that the effective nozzle throat area increases
during the brief transitory period, thus preventing compressor stall. The
exhaust system of the Concorde, the supersonic passenger aircraft, has two
nozzles, a primary nozzle and a secondary nozzle. The secondary nozzle is
positioned as a convergent nozzle for takeoff and as a divergent nozzle for
supersonic cruise.
Attempts to develop thrust vectoring nozzles for combat aircraft has
increased in the last decade. Vectoring nozzles have been used on vertical
takeoff and landing (VTOL) aircraft, and are proposed for future fighters
to improve maneuvering and augment lift in combat. Thrust vectoring
at augmented power settings is being developed for use in future fighters. However, cooling of the nozzle walls in contact with the hot turning
or stagnating flows is very difficult and will require increased amounts of
nozzle-cooling airflow.
13.5.7
Nozzle Coefficients
Nozzle performance is ordinarily evaluated by two dimensionless coefficients; the gross thrust coefficient and the discharge or flow coefficient.
Figure 13.13 shows a convergent-divergent exhaust nozzle with the geometric parameters used in the following definitions of nozzle coefficients. Only
total losses downstream of station 8 (as marked in the figure) are included
in the gross thrust coefficients.
492
Gas Turbines
Short convergent
Con-di iris
Iris
Simple ejector
Fully variable ejector
Blown-in-door ejector
Plug
Isentropic ramp
Fig. 13.11 Typical nozzle concepts for afterburning engines
Inlets and Nozzles
Primary
propulsion nozzle
493
Deployed clamshell
*
*
Blocker
Cascade reverse
Clamshell reverser
Fig. 13.12 Thrust reversers
13.5.8
Gross Thrust Coefficient
The gross thrust coefficient Cgtc is the ratio of actual gross thrust Factual
to the ideal gross thrust Fideal or
Cgtc
≈
Factual
Fideal
(13.1)
Empirically derived coefficients are applied to Eq.13.1 to account for the
losses and directionality of the actual nozzle flow. Each engine organization
uses somewhat different coefficients, but each of the following basic losses
are accounted for:
(i) Thrust loss due to exhaust velocity vector angularity
(ii) Thrust loss due to the reduction in velocity magnitude caused by
friction in the boundary layers
(iii) Thrust loss due to loss of mass flow between stations 7 and 9 from
leakage through the nozzle walls
(iv) Thrust loss due to flow nonuniformities
13.5.9
Discharge or Flow Coefficient
The ratio of the actual mass flow ṁ8 to the ideal mass flow ṁ8i is called
the discharge coefficient CD (refer Fig.13.13):
CD
≈
ṁ8
ṁ8i
(13.2)
This coefficient can be shown to be equal to the ratio of the effective
one-dimensional flow area required to pass the total actual nozzle flow A8e ,
to nozzle physical throat area A8 as follows:
CD
=
ṁ8
ṁ8i
=
ρ8 c8 A8e
ρ8 c8 A8
=
A8e
A8
(13.3)
494
Gas Turbines
θ
α
A9
Ls
A8
Station
7
8
9
A8
Primary nozzle throat area
A9
Secondary nozzle exit area
α
Secondary nozzle half angle
θ
Primary nozzle half angle
Ls
Secondary nozzle length
Fig. 13.13 Nozzle geometric parameters
The variation of the discharge coefficient with nozzle pressure ratio is
shown in Fig.13.14(a) for a conical convergent nozzle. When the nozzle
is choked, the discharge coefficient reaches a maximum value CDmax . The
value of CDmax is a function of the primary nozzle half-angle θ, as shown in
Fig.13.14(b). Figure 13.14(c) shows the variation in discharge coefficient for
a convergent-divergent nozzle with nozzle pressure ratio. Note the change
in behavior of CD between that of the convergent-divergent nozzle and that
of the convergent nozzle as the nozzle pressure ratio drops below choking.
This is due to the venturi behavior of the convergent-divergent nozzle.
13.5.10
Velocity Coefficient
The velocity coefficient Cv is the ratio of the actual exit velocity c9 to the
ideal exit velocity c9i corresponding to pt9 = pt8 , or
Cv
≈
c9
c9i
(13.4)
and represents the effect of frictional loss in the boundary layer of the
nozzle. It is mainly a function of the nozzle ratio A8 /A9 and the half-angle
α, as shown in Fig.13.15.
13.5.11
Angularity Coefficient
The angularity coefficient CA represents the axial friction of the nozzle
momentum; thus it is proportional to the thrust loss due to the nonaxial
Inlets and Nozzles
1.00
C D max
0.98
CD
0.96
0.94
pt8 /p
0
(a) Convergent nozzle
0.92
0
10 20 30 40
(b) CD max vs θ
C D max
CD
1
2
pt8 /p
3
4
0
(c) C-D nozzle
Fig. 13.14 Nozzle discharge coefficient
α
2
4
6
8
1
10
CA
12
14
0.97
1
A 9 /A 8
Fig. 13.15 C-D nozzle velocity coefficient
2
495
496
Gas Turbines
V9
α
αj
Fig. 13.16 Local angularity coefficient
exit of the exhaust gas (see Fig.13.16). For a differential element of flow,
this coefficient is the cosine of the local exit flow angle αj .
The local flow angle αj varies from zero at the centerline to α at the
outer wall; thus, the nozzle angularity coefficient is the integral of αj . across
the nozzle.
1
CA ≈
cos αj dṁ
(13.5)
ṁ
Figure 13.17 presents the correlation of the angularity coefficient with
the area ratio A8 /A9 and half-angle α. The details are based on analytical
evaluations of the inviscid flow field in convergent-divergent nozzles for
practical nozzle geometries.
o
α =o 15
12
o
9
CV
6
o
o
4
o
1
o
2
1
o
3
A 9 /A 8
4
Fig. 13.17 C-D nozzle angularity coefficient
13.5.12
Nozzle Performance
Many nozzle coefficients simplify to algebraic expressions or become unity
for the special case of one-dimensional adiabatic flow. This is a useful limit
Inlets and Nozzles
497
for understanding each coefficient and for preliminary analysis of nozzle
performance using engine cycle performance data.
The gross thrust coefficient for one-dimensional flow of a calorically
perfect gas can be shown to be
Cgtc
.
/
/
/1 −
= CD Cv /
0
1−
p9i
pt8
p0
pt8
(γ−1)/γ
⎡
γ−1
⎢
1+
2γ
(γ−1)/γ ⎣
pt9
p9
1 − pp09
(γ−1)/γ
−1
⎤
⎥
⎦
(13.6)
The extent of overexpansion in nozzles is limited by flow separation
resulting from the interaction of the nozzle boundary layer and the strong
oblique shock waves at the exit of the nozzle. In extreme overexpansion,
it is observed that the oblique shock waves moved from the exit lip into
the nozzle, the flow downstream of the shock waves was separated in the
vicinity of the wall, and as a result, the wall static pressure downstream
of the shock waves was nearly equal to the ambient pressure, p0 . A simple
estimate for the ratio of the pressure just preceding the shock waves, ps , to
the ambient pressure, p0 , is given by
ps
p0
≈
0.37
(13.7)
This flow separation limit can be included in the one-dimensional gross
thrust coefficient of Eq.13.6 for isentropic flow by considering the effective
exit pressure (p9 = p9i ) to be the pressure just preceding the shock wave
(ps ). The design area ratio A9 /A8 of convergent-divergent nozzles is selected such that the nozzle flow does not separate due to overexpansion
for most throttle settings. This is because the increase in gross thrust
coefficient associated with flow separation does not normally offset the accompanying increase in installation loss.
Nozzle pressure ratios are 3 to 5 in the subsonic cruise speed range of
turbofan and turbojet engines. Typically, a subsonic engine uses a convergent exhaust nozzle. This is because, in the nozzle pressure range of 3 to 5,
the convergent nozzle gross thrust (interception of lines with vertical axis,
A9 /A8 = 1) is 1 to 3 percent below the peak gross thrust (p9 = p0 ). Consequently, there may be insufficient gross thrust increase available in going
to a convergent-divergent nozzle on a subsonic cruise turbofan or turbojet
engine to pay for the added drag and weight of such a nozzle. In some
applications, this loss in gross thrust coefficient of a convergent nozzle is
too much, and a C-D nozzle is used.
The design pressure ratio across the nozzle increases rapidly with supersonic flight Mach number. At Mach 2, a pressure ratio of about 12 is
typical. At this pressure ratio, the convergent nozzle gross thrust penalty
will be of the order of 10 percent. Substitution of convergent-divergent
nozzles for convergent nozzles provides large thrust gains for supersonic
aircraft.
498
Gas Turbines
Review Questions
13.1 What is the purpose of an aircraft gas turbine inlet and nozzle?
13.2 Briefly explain inlets.
13.3 What is the purpose of a diffuser in the inlet? Explain.
13.4 Briefly explain the supersonic inlets.
13.5 What are the various functions of an exhaust nozzle?
13.6 What are the two types of nozzles used in an aircraft engine. Briefly
explain them.
13.7 Summarize the functions of an exhaust nozzle.
13.8 Write short notes on
(i) engine back pressure control,
(ii) exhaust nozzle area ratio, and
(iii) thrust reversing.
13.9 Explain the following performance coefficients:
(i) nozzle coefficient,
(ii) cross thrust coefficient,
(iii) flow coefficient,
(iv) velocity coefficient, and
(v) angularity coefficient.
13.10 Briefly describe the nozzle performance.
Multiple Choice Questions (choose the most appropriate answer)
1. The purpose of any aircraft gas turbine inlet is to provide
(a) sufficient air supply to the compressor
(b) to have minimum drag force
(c) reduction in velocity
(d) all of the above
2. The best work output from a compressor is obtained when air flows
at a Mach number of about
(a) 0.1
(b) 0.3
Inlets and Nozzles
499
(c) 0.5
(d) 0.6
3. Because of the diffusion in the inlet casing
(a) air velocity decreases and pressure increases
(b) air velocity and pressure decreases
(c) air velocity and pressure increases
(d) air velocity increases and pressure decreases
4. As the flow takes place through an inlet diffuser, the boundary layer
thickness
(a) remains constant
(b) increases
(c) decreases
(d) initially increases and then remains constant
5. A supersonic inlet must have
(a) adequate subsonic performance
(b) good pressure distribution at the compressor inlet
(c) must be able to operate at all ambient conditions
(d) all of the above
6. The function of an exhaust nozzle
(a) decelerate the flow
(b) allow for cooling of walls
(c) increase jet noise
(d) none of the above
7. Supersonic aircrafts are provided with
(a) convergent nozzle
(b) divergent nozzle
(c) convergent divergent nozzle
(d) none of the above
8. The exhaust nozzle pressure ratio is a strong function of
(a) Reynolds number
(b) Mach number
(c) Prandtl number
(d) Schmidt number
500
Gas Turbines
9. Subsonic aircrafts are provided with
(a) convergent nozzle
(b) divergent nozzle
(c) convergent divergent nozzle
(d) none of the above
10. Performance of a exhaust nozzle is a function of
(a) gross thrust coefficient
(b) flow coefficient
(c) velocity and angularity coefficient
(d) all of the above
Ans:
1. – (d)
6. – (b)
2. – (c)
7. – (c)
3. – (a)
8. – (b)
4. – (b)
9. – (a)
5. – (d)
10. – (d)
14
BLADES
INTRODUCTION
Turbine blades are one of the most important components in a gas turbine power plant. These are components across which flow of high pressure
gases takes place to produce work. A blade can be defined as the medium
of transfer of energy from the gases to the turbine rotor.
The blade as an entity, is subjected to a large number of forces, some
are inevitable and some are caused by the rotation of blade. The blade is
subjected to forces in the three directions, viz.,
(i) the rotor driving force along the radial direction,
(ii) the axial forces caused by the gas flow, and
(iii) the forces acting normal to the turbine shaft due to the centrifugal
forces.
Further, the blade is subjected to differential thermal stresses, erosioncorrosion and a host of other hostile parameters hampering its smooth
functioning. In this chapter we will briefly discuss the details of blade
materials, manufacturing, fixing and cooling.
14.1
BLADE MATERIALS
Researchers around the world are trying to develop new materials which
have high strength and stability at high temperatures to meet the demands
of the turbine designers. Different alloy compositions have been developed
which have a good stability to withstand the thermal stresses. These alloys
have been modified to give good erosion- corrosion characteristics to the
blades. The method of manufacture of the blades plays an important role
in determining the strength, uniformity of micro-structure and properties
of the blades.
Among the material that have been found to be suitable for use in
blades are steels, titanium alloys and nickel base alloys. All the three types
502
Gas Turbines
of alloys which are mainly used, have varying proportions of chromium and
aluminium to improve the strength and corrosion at high temperatures.
Titanium alloys are preferred to steel because of its lower density (nearly
50%). Titanium has superior oxidation resistance and like other materials
it also has a decreasing strength with temperature. However, by alloying
with other elements its strength can be stabilized. Latest types of titanium
alloys have good properties upto about 550◦C (IMI 829). Titanium also
has a lower coefficient of thermal expansion which helps to reduce thermal
stresses. The main disadvantage with titanium alloys has been the high
reactivity at very high temperatures. Titanium dissolves oxygen at high
temperatures. It even reduces the refractory lining. Hence, usually titanium is worked upon with vacuum or with an inert gas atmosphere (argon).
The above mentioned alloy IMI 829 has composition Ti–5.5, Al–3.5, Sn–3,
Zr–1, Nb–0.3 and Mo–0.3.
Nickel alloys have also been developed extensively and are currently being used for turbines. These alloys have superior strength and oxidation
resistance even though nickel by itself has poor oxidation resistance. This
weakness is overcome by alloying with chromium. In alloys, chromium is
generally 15–30% and forms chromium oxide (Cr2 O3 ) a protective layer
and chromium carbide. Other elements generally added are aluminium, titanium and niobium to improve the strength at high temperatures. Current
alloys also use cobalt, hafnium, boron, zirconia, molybdenum etc., Nimonic
118 the most advanced alloy developed has working capacity approximately
upto 925◦ C and has composition 9.0% (Ti + Al + Nb), 3.5% Mo, 14.8% Cr,
14.9% cobalt. For working with nickel alloys it is necessary to use vacuum
melting techniques to encourage the use of hardened metals and reduce
trace elements which affect the ductility.
Materials working under high stress and high temperature have a particular rate of creep. It means elongation under stress is not a fixed quantity
as it is in case of normal temperature. Elongation continues to increase
with time and blades gradually reduce the original gap provided at their
tips. Therefore contacts with casing results in failure. The repeated heating
and cooling of the material affect the physical properties of the materials.
14.1.1
Factors to be Considered in the Selection of Materials
Since, the turbine blades are the most conditioned members of a gas turbine,
they must withstand
(i) high operating temperatures; Material should possess good strength
at high temperatures
(ii) centrifugal tensile stresses due to rotational speeds as high as 30000
rpm
(iii) bending stresses due to equivalent impulse load of the gases acting
on the cantilever blade at a certain distance from fixing, also, it must
have high creep strength.
Blades
503
(iv) hot erosive and corrosive effects due to the high temperature combustion products, for example, CO2 and CO with water vapour
(v) varying temperature and should have structural stability when exposed to varying temperatures
Further, it should posses the following characteristics:
(i) castability or forgeability,
(ii) weldability,
(iii) machinability, and
(iv) no embrittlement during operation.
Turbine blade material must therefore be selected from the consideration
of the working conditions, taking into account the above factors.
14.2
MANUFACTURING TECHNIQUES
The above discussion brings out the fact that there are many restrictions
and requirements which will affect the manufacturing process.
The various techniques of manufacturing blades are
(i) forging,
(ii) casting,
(iii) fabrications, and
(iv) powder metallurgy.
In the following sections we will briefly discuss the various manufacturing techniques employed.
14.2.1
Forging
Forging is essentially a process of impacting a piece of metal to give it a
certain shape. In a gas turbine forging was used in the beginning for the
manufacture of both compressor and turbine blades. But, presently it is
mostly used for compressor and low pressure stages of the turbine blades.
This is because, as the need arises for alloys of greater strength becomes
difficult to forge blades. Also, the required section of holes in hollow blades
could not be achieved by this method. Blades being manufactured by this
technique can be categorized as
(i) precision forging, and
(ii) oversized forging.
504
Gas Turbines
In the first method, direct blade profile is obtained during the process of
forging. In the second method, oversized blades are manufactured and then
machined to obtain the final shape. The amount of machining is generally
1 to 1.5 mm
Either type of forging can be made by any one of the four processes
(i) extrusion,
(ii) coining,
(iii) upsetting, and
(iv) final forging.
The bar stock which is initially circular, is suitably modified to ensure
that it fits into the forging die. The bar stock is usually called use.
In the upsetting operation the part which will later play the role of
shroud is developed and its shape is also achieved. There are essentially
two types of upsetters:
(i) hydraulic, and
(ii) electric.
The section obtained is usually not tapered and hence this process is usually
split up into two processes to achieve the taper.
Before the upsetting process the use is coined. The head of the blade is
produced to the required shape.
After the use undergoes the above operations it is checked and corrected
for various defects such as removal of flash around the heads, removal of
scale and forging lubricants by blasting or chemical means. Controlled
etching is also done to remove the surface layers of metals contaminated
during the heating operations and application of forging lubricant etc.,
In the final forging operation the blade shape is developed with a controlled application of force and with minimum flash. To remove the flash
a certain amount of dimensional accuracy is attained through preforming.
The removal of the flash becomes necessary to avoid wear due to dies. The
various materials used for forging compressor blades are the following:
(i) Aluminium alloy such as RR59 is used in the Dart and Avon engines.
Aluminium is forged at 450◦C although no coating is as such necessary
to prevent oxidation but graphite is often used. The blades are then
solution treated at about 510◦ C and quenched to avoid precipitation
during cooling. Because the blades are distorted, they are corrected
for profile and then aged at 150 – 200◦ C. The blades are then anodized
to produce a corrosion resistant surface.
(ii) Steel blades are produced by coating with a thin layer of nickel and
forging at 1100◦C. The nickel is necessary to prevent decarburization
by the reducing atmosphere and to act as a lubricant. The nickel
Blades
505
layer is stripped off after forging. The blade is then heated to about
1050◦C to dissolve the hardening metals. The distortions are removed
by cold coining operations. In recent times the final forge heating is
also done as a heat treatment of the blade. Then the blade is rapidly
cooled and tempered to give the required hardness. The blades are
then checked and polished by blasting or barrelling.
Titanium blades are manufactured by first coating with a thin layer of
glass or by electroplating static coating of glass. This is done to prevent
sticking of the blades to the dies. During forging operations it is found that
oxygen is often dissolved. This is removed by etching with hydrofluoric
acid mixed with either nitric acid or ferric sulphate. Then distortion of
the blades is corrected between hot dies. Hydrogen absorption is another
problem and very careful monitoring is required in the entire manufacturing process and excess hydrogen is removed by heating in vacuum. Some
commonly used alloys are IMI 679, IMI 550, Ti 811 and Ti 6242.
Blades with nickel based alloys are extensively used in the last stages of
compressor and turbine blades. As the temperature requirement increases,
the strength requirement of the alloys also goes on increasing. New manufacturing techniques are to be developed. The introduction of vacuum
melting helps to reduce the trace elements in the material and improved
forgeability. Increased homogenization also helps to improve forgeability.
Besides the problem encountered in forging, nickel base alloys have limited
ductility in a narrow forgeability range. This makes it necessary to heat
the entire forging uniformly to produce a uniform grain structure. Further,
there is a need for avoiding differential flow of metal in the forging to avoid
shear and cracking.
The forging is accomplished in small steps so that there is uniform
distribution of energy to the blade. Both nickel plating or glass powder
coating is given to ensure a lubricant action and non-sticking quality before
oxidation. After the final forging, the lubricant is removed and the blade is
age hardened. The oversize blades are subjected to blasting, to remove any
oxide layers and then ready for machining. Close finish blades are subjected
to finishing operations very carefully, such as etching.
With the use of the higher and higher temperatures which are close to
the thermal yield strength of the blades – blade cooling becomes necessary. To achieve this, holes are drilled in the blades through which either
compressed air or water would flow to cool the blade.
The simplest technique of blade manufacture to meet this requirement
is to forge the blade and then drill the holes in it. But better techniques
have been developed which produce better results. They are (i) tadpole
method, and (ii) extrusion method.
In the tadpole method a complex form is first formed. The blade
thus becomes a solid piece. This is achieved by the regular extrusion and
preforming. The blade is checked very closely for dimensional accuracy
and holes are drilled (3 in number) along the cross-section of the blade.
The blades are then subjected to coining operations after being heated to
1100◦ C. The coining operation is performed in a screw press. The operation
506
Gas Turbines
converts the section of the blade to that of the required aerofoil shape. The
nickel coating is then removed and finishing operation is done on the blade.
This technique is extensively used for the high pressure turbine blades of
Rolls-Royce, Tyne and some Avon engines. The alloys generally used are
Nimonic 105.
In the extrusion method a certain length of a circular bar is taken and
subjected to extrusion process to create a certain shape and cross-section.
Holes are drilled and filled with nickel rods and then firmly welded. The
bar is then once again subjected to extrusion to create not only the aerofoil
section but also the head. The blade is once again upset to create the shroud
if found necessary. This technique requires a certain amount of machining.
14.2.2
Casting
The blades manufactured by this technique are intricate and any amount
of complexity can be adopted. This technique uses either precision casting
or the lost wax method and is mainly used for nickel and cobalt alloys. The
technique also employs vacuum casting methods. The main advantage of
this process is the high dimensional accuracy that can be achieved (± 0.06
mm). The investment casting technique essentially involves 7 main stages.
(i) Producing an injection wax pattern which may be removed by heating
(ii) Assembling the wax patterns into a runner system
(iii) Formation of refractory castings around the wax pattern
(iv) Removing the wax from the refractory castings
(v) Hardening the refractory castings by firing in a kiln
(vi) Melting and casting the molten metal
(vii) Removing the part from the casting material and inspecting the part
The process is started with the formation of the wax pattern. This is
achieved by solidifying wax in a die. Since the overall process efficiency is
dependent on the dimensional accuracy of the die, the dies are machined
very very accurately (of the order of 0.005 mm on CNC machines). The
essential properties that are to be satisfied by a wax material are
(i) low ash content,
(ii) dimensional and chemical stability,
(iii) good joining and fabrication conditions,
(iv) predictable expansion and contracting characteristics,
(v) minimum amount of lead and bismuth, and
(vi) no residue formation.
Blades
507
Usually the waxes used are a blend of natural and mineral waxes available. This is to achieve suitable characteristics. The natural waxes (carnauba and candellia) are obtained from South American plants while mineral or montan waxes are obtained from European lignite. A recent trend
is the use of polystyrene because of its low contraction and excellent die
reproduction.
The wax is injected in the die under pressure and allowed to partially
dry in the die. This is achieved by using an injection machine.
The cores used are of three types.
(i) Soluble cores These are made from a mixture of mica (silica), sodium
carbonate and polyethylene glycol. In this method the core is inserted
in a wax pattern die and wax is poured in the die. The core is removed
by immersion in agitated water or dilute acid. The cavity left is filled
with investment material. This is a very simple technique.
(ii) Ceramic cores These are prepared from silicon or zircon and small
quantities of other materials such as cristobalite or urea may be added
to improve the strength at room temperature operations. The cores
are pressed together.
(iii) Silica rods These are used where it is difficult to get ceramic shapes
such as radial cooling holes. These are produced by using extruded
or drawn vitreous silica rods as cores. Silica has advantages due to
easiness of its removal by leaching. The wax patterns are formed by
injecting wax around the core but with about 3 mm space above to
help the core to be supported by the investment material when it is
formed around the wax pattern. Varnish or polystyrene are coated
over one point to allow the use to float and prevent distortion due to
differential core and shell thermal expansion.
The next step is to join the runner and riser assembly which is also
made of wax into the pattern. This is done carefully to prevent distortion
of the wax pattern. The wax pattern is then dipped in a refractory slurry
and the slurry covering is allowed to dry. A layer of refractory grit is then
stuck on the refractory layer which is still drying. The process is repeated
until a certain thickness of the investment is achieved. The number of
coatings applied depends on the required shell strength and on the weight
of the casting besides depending on the properties. Current investment
materials use silica or hydrolyzed ethyl silicate and refractory materials such
as alumina, zircon, alumino silicates, silica etc., To improve consistency and
good coating properties of the investment material it is necessary to add
wetting, defoaming, deaerating and suspension agents to the slurry mix.
Where grain refinement of the casting is important, materials such as cobalt
oxide or cobalt aluminate are added to the primary face coat of the shell.
The removal of the wax from the investment is achieved by subjecting
the mould to high pressure steam or by direct heating at temperatures in
the range of 1000◦ C (flash dewaxing). The outer layer of the wax is melted
and absorbed by the investment material to allow for expansion of the later
508
Gas Turbines
melting wax. Steam dewaxing is preferred as it allows the removal of the
wax material. The moulds are then fired at temperatures above 900◦ C to
form a ceramic bond and this is done in a slightly oxidizing atmosphere to
allow the last traces of wax to be removed.
The alloys are prepared by either air melting or vacuum melting. The
alloys may be prepared either from virgin materials or from scrap material.
In the virgin material method the various elements such as nickel, cobalt,
chromium, molybdenum and tungsten and elements as boron in nickelboron and carbon in chromium-carbon are added. Melting techniques are
followed with great care to prevent excessive melt temperatures. For this
protective stage fluxes are often used.
The alloys prepared from the scrap material available from the previous
casting are usually adjusted for chemical composition.
In the vacuum casting technique virgin alloy is prepared by careful layering of individual materials like nickel chromium, cobalt and refractory
metals as carbon. The elements are melted slowly to let out the gas from
the furnace by carbon boil (carbon getting oxidized). Reaction elements
like aluminium and titanium are then added with good furnace temperature control. The elements like zirconium, hafnium, boron are added near
the end in the melting process.
Integrally cast turbine wheels are also being produced off late. The
casting presents difficult problems for the foundry-man as high casting temperature are required to produce good dimension and decrease the porosity
of the hub. For good blades the grain structure should be uniform for creep
resistance and the hub should have fine grains, free from porosity. This is
achieved using grain refining agents while the microporosity is controlled
only by hot isostatic pressing.
14.3
BLADE FIXING
The blades of a gas turbine are to be fitted carefully to the discs. This
is served by the blade root which is either welded or fixed to the disc
circumference. While considering the root the following aspects are to be
kept in mind.
(i) The root should be able to transmit the centrifugal force and vibrations to the turbine disc.
(ii) It should be easy to mount the blades on the turbine discs.
(iii) The manufacturing process should be affordable.
(iv) The temperature effects on the blade and the disc should not cause
expansion due to high temperature and should not lead to any looseness or deformation of the attachment. Consideration should also be
given to the need for cooling the blade. Some degree of freedom is
usually given to the blades to counteract vibration.
Blades
509
For turbines the configuration of the blade root disc assembly plays an
important role in fixing. Early turbine blades were fitted to the discs by
the Delaval root fixing which was later superseded by the fir tree fixing
which are used in majority of the gas turbines nowadays. The machining of
serrations is very important to ensure the blade load to be shared equally
by all the serrations. The root is manufactured by broaching the blade and
the disc separately. The blade is free in the serrations when the turbine is
stationary and is stiffened by the centrifugal forces when the disc rotates.
Sufficient clearance is to be obviously provided at the tip of the blade.
Various methods are used to develop a good fit for the blade and prevent
it from sliding out of the serration. Some of the blades are having screws
between the bulb root and the groove in the turbine disc. Hollow blades
are occasionally held by pins introduced from one side and a common plate
on the other side some times holes are drilled radially through the shoulder
of the disc rim upto the blade seating groove when it is aligned with the
rim hole and a steel pin driven through the rim hole into the blade.
The following are some of the methods for fixing blade to the wheel:
(i) Bulb Root Method A bulb is made in the root of the blade which is
fixed to the wheel with the help of steel pins welded and inserted
in the holes. This method is employed in Metro Vick gas turbine.
Endwise movement of the blade is usually prevented by swaging over
the metal of the rotor around the root portions of the blades. Figure
14.1 shows the details.
Fig. 14.1 Bulb root method of blade fixing
(ii) Tee and Double Tee Figure 14.2 illustrates the method of fixing the
blade using Tee and double Tee technique. A T–slot is made in the
wheel and after keeping the blade in position a wedge is placed to
keep it in tact.
(iii) Anchor Pin As shown in the Fig. 14.3 the blades are fixed by inserting
the pin through the hole in the wheel and in the blade root and
screwing the nut over the threaded portion.
510
Gas Turbines
(a) Tee
(b) Double Tee
Fig. 14.2 Blade fixing by T
Fig. 14.3 Anchor pin method of blade fixing
(iv) Fir Tree Method The details are shown in Fig. 14.4. This method has
certain advantages, viz.,
(a) rotor is not weakened,
(b) contact is distributed,
(c) raper is provided, and
(d) allowance is given for thermal expansion.
This type of fixing is used on the W2 B turbine and subsequently on
the Rolls-Royce engines. The blade is driven into the rotor teeth and
locked by turning over the lip at the apex of the fir cone on the blade.
(v) Grub Screw Method A screw is fitted half in the rotor drum and half
in the blade root. This method is used on German gas turbines for
securing the blades in the rotor (Fig. 14.5). In this case the blade is
a dovetail section and is secured by Grub-screws against end movements.
(vi) Welding The blade can also be fixed by welding as shown in Fig. 14.6.
14.4
PROBLEMS OF HIGH TEMPERATURE OPERATION
It is always the aim of a gas turbine designer to increase power output,
reduce weight and also to reduce the fuel consumption. Improvement in
the Carnot efficiency η = 1 − TT21 of a heat engine with an increase in the
Blades
Pitch at tip
Pitch at root
A gap of 100 microns
Fig. 14.4 ’Fir-tree root’ method of blade fixing
Rotor
Grub screw
Fig. 14.5 Grub screw method of blade fixing
Blade
Welding material
Wheel
Fig. 14.6 Blade fixing by welding
511
512
Gas Turbines
entry temperature, T1 is well known. Hence, attempts to increase the power
to weight ratio have lead to the development of gas turbines operating at
higher and higher entry temperatures.
The maximum temperature of a gas turbine plant cycle occurs at the
entry of the first stage of the turbine. We already know the effect of the
maximum cycle temperature at various values of the pressure ratio on the
thermal efficiency (refer Chapter 6). Employment of high temperature for
a given pressure ratio leads to a higher thermal efficiency, higher power
to weight ratio and lower specific fuel consumption. Referring to Chapter
6, Figs.6.7 and 6.8 show the effect of pressure ratio and maximum cycle
temperature on the specific power output and efficiency of gas turbine power
plants. From these two figures, the advantages of operating the power plant
at higher gas temperature is evident.
However, employment of high temperature gas, in gas turbines require
materials which can withstand the effects of high temperature operation.
Further, high temperature operation affects all the components in the engine with varying degree. However, the life-limiting component is usually
the high pressure turbine blade, where failure is associated with thermal fatigue, oxidation or corrosion and creep. If a turbine blade is heated rapidly
to a high temperature it causes uneven temperature distribution and as a
result of this, severe thermal stresses are developed within the material.
Beyond certain temperature (650 - 800◦ C) the blade material does not remain elastic and continues to stretch under the applied forces. This is called
creep. If this phenomenon exists for a long time fracture can occur.
There are two options open to the designer to overcame the problems
of high temperature.
(i) A new material may be sought which is capable of operating at high
temperatures.
(ii) By blade cooling which maintains the temperature of the blade at a
value low enough to preserve the desired material properties.
From the point of view of new materials, many alloys have been specially
developed for gas turbine blades operating at high temperatures having
manganese, molybdenum, copper, columbium, silicon, tungsten, vanadium
and zirconium in them. They are used in various proportions to obtain
certain desired properties in the alloys.
Ceramic materials and mixtures of ceramics with metals, whilst having
some very desirable properties have been shunned because they have equally
disagreeable shortcomings such as brittleness and proneness to failure in
thermal shock.
The use of high gas temperatures at entry is intimately linked with the
characteristics of materials that can be used in such applications. We have
already discussed the properties required in the high temperature materials
employed in gas turbines. However, we will recall them once again. The
requirements are:
(i) high strength at the maximum possible temperature,
Blades
513
(ii) low creep rate,
(iii) resistance to corrosion and oxidation,
(iv) resistance to fatigue, and
(v) ease in manufacture, i.e., machinability, castability, weldability, etc.,
14.5
BLADE COOLING
Blade cooling is the most effective way of maintaining high operating temperatures making use of the available material. Blade cooling may be classified based on the cooling site as external cooling and internal cooling.
Another classification based on the cooling medium is liquid cooling and
air cooling.
14.5.1
External Cooling
As the name implies, the external surface of the gas turbine blade is cooled
by making use of compressed air form the compressor. The quantity of
air required for this purpose is from 1 to 3% of the main flow entering the
turbine stage by which blade metal temperatures can be reduced by about
200–300◦C. By employing suitable material (nickel-based alloys) an average
blade temperature of 800◦ C can be used. This can permit maximum gas
temperature of about 1400 K. Still higher temperatures can be employed
with nickel, chromium and cobalt based alloys.
Other methods of external cooling are film cooling and transpiration or
effusion cooling. Film cooling to local areas can be applied by drilling holes
on precision cast blades by the process of Electrical Discharge Machining,
EDM, which removes material by multiple spark discharge action at the end
of a special electrode. By this process very accurate, smooth, low residual
stress holes can be produced upto a dia. of 0.15 mm. The cooling air
flowing out of these small holes forms a thin film over the blade surfaces.
Besides cooling the blade surface it decreases the heat transfer from the hot
gases to the blade metal. This is shown in Fig. 14.7.
Cooling air passages
Fig. 14.7 Film cooling
In the transpiration cooling (Fig. 14.8) the air is allowed to effuse or
sweat form the pores of the porous blade metal. This provides a blanket of
cool air, insulating the metal of the turbine blade from the hot gas. Effusion
514
Gas Turbines
of the coolant over the entire blade surface causes uniform cooling of the
blade. Sintered wire composites are the most commonly used transpiring
materials. Some of the disadvantages of this method are, low strength at
high temperature, low modulus of elasticity, very small pore size and poor
oxidation resistance.
Envelope of film
Porous wall
Fig. 14.8 Transpiration or effusion cooling
14.5.2
Internal Cooling
Internal cooling of blades is achieved by passing air or liquid through internal cooling passages from hub towards the blade tip. The internal passages
may be circular or elliptical as shown in Fig. 14.9 and are distributed over
the entire surface of the blade. The cooling of the blades is achieved by
conduction and convection. Relatively hotter air escapes to the main flow
from the blade tips after traversing the entire blade length in the cooling
passages.
Internal cooling passages
Fig. 14.9 Convection cooling
Hollow blades can also be manufactured with a core and internal cooling
passage as in Fig. 14.10. The cooling air is first admitted into the inner shell
(1) from which it comes out through narrow and long slots (2) and (3) and
impinges the inner surface of the leading and trailing edges respectively.
The air form (2) after impinging, flows over the inner shell and comes out
by the holes (5) on the concave surface near the trailing edge. This type
of cooling is called impingement cooling. Based on the cooling medium
employed, blade cooling may be classified into liquid cooling and air cooling.
Blades
Convection
Cooling air
515
3
4
2
Jet impingement
1 Core
5
Fig. 14.10 Impingement cooling
14.6
LIQUID COOLING
Because of the higher rates of heat transfer the use of water or other liquids
as coolants have been tried. Liquid cooling has been tried on long life
turbines as the heat transfer rates are higher for these turbines. Even
though this is one of the most effective methods of removing heat, it must
be noted that it may overcool the blade which is not desirable.
When liquid is used as coolant the equipment becomes additionally complex. When water is used, the disadvantage is that either the vapourization
must be allowed for or water must be circulated at high pressure above its
vapour pressure so as to prevent vapourization. There is another disadvantage in using water as coolant since, it has a high heat transfer coefficient
and because of this the heat transfer rate becomes quite high as already
mentioned. For this reason organic liquids with low vapour pressure than
water for the same temperature have been tried. If organic coolants are used
there is a possibility that it could be mixed with the fuel before burning.
A liquid cooling system is more advantageous when the liquid is recirculated, so that it can be used over and over again. Because of the
complexity this method is not suitable for aircraft applications. The use of
water as coolant has been tried on gas turbines for power generation. Figure
14.11 shows a water cooled blade in which water flows through internal
passages in the blade. It is almost impossible to eliminate corrosion or the
formation of deposits in this system.
14.7
AIR COOLING
Most of the mechanical complexities that are present with liquid cooling
virtually disappear when air is used as the cooling medium. So much so,
all present day aircraft gas turbine featuring turbine cooling fall into this
category. In all these systems air is bled from the high pressure end of the
compressor and delivered to the blades and vanes to be cooled. Quantity of
the coolant required is about 1 to 3% of engine air flow per turbine blade
row. Advantage is taken of the fact that the cooling air is some 600◦C
cooler than the gas surrounding the blades. After passing through the
516
Gas Turbines
Top cover
Plug
Plug
Inlet
Outlet
Blades
14.7.2
517
Requirements for Efficient Blade Cooling
In a conventional cooled blade, cooling is obtained due to convection by
passing cooling air through internal passages within the blade. The success
in obtaining the large reduction in metal temperature at the expense of
a small quantity of cooling flow is governed by the skill in devising and
machining the cooling passages. Because the internal cooling relies on the
cooling air scrubbing against the cooling surface, the internal surface area
must be large and the velocity of the cooling air must be high. This implies
that the cross-sectional flow area of the passages must be small.
The design of the blade internal geometry for cooling is more complex
because of the various aerodynamic, heat transfer, stress and mechanical
design criteria that must be satisfied. The most successful designs have
incorporated radial passages through which cooling air passes, escaping at
the tip.
14.7.3
Heat Transfer in Cooling Passages
As already mentioned, the cooling passage along the blade span must be
devised so as to give a large internal surface area. Also for efficient heat
transfer the cooling air passing through the passages must have high velocities which implies that the cross-sectional area of the passages must be
small. Taking these two factors into consideration, a factor termed Z factor
is introduced which could be used to compare the relative merits of various
cooling passage configurations.
The coolant geometric parameter
Z
=
Sc1.2
C2
×
C 1.2
Ac
(14.1)
where Sc is internal surface area per unit length of the blade, Ac is total
cross sectional area of the cooling passage in one blade, C is true chord of
the blade.
The Z factor determines the relative blade temperature. A high value
of Z means the ratio of heat transferred per unit temperature difference to
the coolant to the total heat transferred per unit temperature difference to
the blade from gas is high. Increasing Z would mean a reduction in the
mean blade temperature. The blade temperature varies rapidly when Z is
small, but is progressively less sensitive to changes in Z for large values.
Values of Z above 150 are required for good cooling but there is little point
in going beyond 250 for values of Z.
One way of achieving a high value of Z is to have a large number of
small spanwise holes. By using elliptically shaped holes, rather than round
ones, the number required can be kept to a minimum. The influence of Z
on the reduction in midspan average temperature for various percentages
of cooling air flow is shown in Fig. 14.13.
Gas Turbines
Reduction in midspan
average metal temperature (o C)
518
300
200
150
100
Z=50
200
100
0
0
1
2
3
4
5
Percentage cooling air flow
6
Fig. 14.13 The effect of Z factor on midspan cooling
14.8
PRACTICAL AIR COOLED BLADES
It is the problem of manufacturing more than anything else which has
practically restricted the development of the air cooled blades. Even the
simplest cooled blades are so expensive to manufacture that to justify their
use they must be doubly reliable. In fact a cooled blade which will not last
a major proportion of the aircraft life, say 10000 hours, almost prices out
the engine out of the market on the basis of high overhaul costs. Different
cooled blade designs are conveniently classified according to their method
of manufacture as follows: (i) fabricated, (ii) forged, and (iii) cast.
14.8.1
Fabricated Designs
Fabricated blades were the fore runners in the field of blade cooling. The
blade was nothing more than a plain hollow blade. The amount of cooling
obtained was trivial because of the very poor Z factor.
14.8.2
Forged Blades
Instead of one hole as in previous one, this employs three spanwise holes.
This cooling configuration was conceived in the early 1950. Tyne, Conway
and Spey blades are the most important ones belonging to this category
made by Rolls-Royce.
The Tyne Blade Cooling air enters at the blade root, passes up the
trailing edge hole, down the centre hole and up the leading edge hole, to
exhaust at the blade tip (Fig. 14.14).
The cooling air flow, expressed as a percentage of the main turbine flow
is just under 1% and this cools the mid-span section of the blade by an
average of 40◦ C. Blade life is increased from 2,000 hours uncooled to an
achieved life of 6,500 hours Z factor is 33.
Conway Blade In this design cooling air flow from one side of the blade
root to both leading and trailing edges, turns over at the blade tip and
flows down the centre passage and then exhausts at the blade root on the
Blades
3 cm
519
Z = 33.0
20
Variation from average temp.
10
0
-10
Span
Take-off temperature
Cooling air flow
Mid span cooling
Material
Predicted life
Achieved life
Life without cooling
1242 K
0.8%
Data from
40 oC Canadair CL44 operation
N 105
7000 h
6500+ h
2000 h
Fig. 14.14 Tyne triple pass turbine blade
opposite side of the blade (Fig. 14.15). Cooling air used is 1.4% of main
turbine flow and this reduces the midspan average temperature by 120◦C.
From a life of 75 hours at uncooled stage life is increased to 13,000 hours
after cooling. Z factor is 31. The low Z value of both Tyne and Conway
blade implies that if the air is passed straight through the blade the cooling
would be very inefficient. Therefore more than one pass of the cooling
air is used. Spey Blade This type of blade has elliptical shaped cooling
7 cm
Z = 31.0
100
Variation from average temp.
50
0
-50
Span
Take-off temperature
Cooling air flow
Mid span cooling
Material
Predicted life
Achieved life
Life without cooling
1310 K
1.4%
120 oC Boeing 707-420 and
N 105
DCB-40 operation
13000+ h
15000 h
75 h
Fig. 14.15 Conway double pass turbine blade
holes formed by a unique manufacturing method. Cooling air enters from
both sides of the root and exhausts from the tip of the blade, thus giving
the single pass cooling system (Fig. 14.16). The cooling air flow used is
2% of the main turbine flow and this reduces the midspan average blade
temperature by 220◦ C. If uncooled this blade would last for 12 minutes,
with air cooling the predicted life is 10000 hours.
520
Gas Turbines
300
200
Variation from average temp.
100
0
-100
Cooling air flow
Mid span cooling
Material
Predicted life
Life without cooling
2.0%
200 o C
N 108
10000 h
0.2 h or 12 min
Span
Data from
BAC I-II operation
Blades
521
Review Questions
14.1 What are the forces the blades are subjected to? Explain.
14.2 Explain the various materials that can be used for blades.
14.3 Explain the factors to be considered in the selection of blade materials.
14.4 Briefly explain the various manufacturing techniques for blades.
14.5 Give an account of various blade fixing techniques.
14.6 Why the blades are to be cooled?
14.7 Explain the problems encountered in high temperature operation.
14.8 What is meant by internal and external cooling? Explain.
14.9 What are the requirements for blade cooling?
14.10 Give an account of practical air-cooled blades.
Multiple Choice Questions (choose the most appropriate answer)
1. A turbine rotor blade is one, which transfers energy
(a) from gases to the turbine rotor
(b) from turbine to the rotor gases
(c) there is no energy transfer
(d) to the compressor
2. The most preferred material for blades is
(a) steel alloys
(b) titanium alloys
(c) brass alloys
(d) aluminium alloys
3. Turbine blades must withstand
(a) high operating pressure
(b) high operating temperature
(c) high operating volume
(d) all of the above
4. Modern blades are manufactured by
(a) forging
(b) turning
522
Gas Turbines
(c) machining
(d) wire cutting
5. Forging can be done by
(a)
(b)
(c)
(d)
upsetting
extrusion
coining
any of the above
6. For the casting the cores are usually
(a)
(b)
(c)
(d)
soluble cores
ceramic cores
silica cores
any of the above
7. The materials used for gas turbine blades must have
(a)
(b)
(c)
(d)
high strength
high resistance to corrosion
high resistance to fatigue
all of the above
8. External cooling of gas turbine blade is achieved using
(a)
(b)
(c)
(d)
liquid
air
exhaust gas
none of the above
9. When liquid cooling is employed in internal cooling of blades, the
water should be pressurized
(a)
(b)
(c)
(d)
below the vapour pressure
above the vapour pressure
upto the vapour pressure
to any pressure
10. The most stressed part in the blade is
(a)
(b)
(c)
(d)
tip
middle section
root
side
Ans:
1. – (a)
6. – (d)
2. – (b)
7. – (d)
3. – (b)
8. – (a)
4. – (a)
9. – (b)
5. – (d)
10. – (c)
15
COMPONENT
MATCHING AND
PERFORMANCE
EVALUATION
INTRODUCTION
In the last six chapters, we dealt with in detail the various components
of a gas turbine power plant. For example, Chapter 8 and 9 are concerned
with compressors whereas Chapter 10 is devoted for combustion systems.
In Chapter 11 there is a detailed discussion on turbines and in Chapter 13
we have analyzed the inlets and nozzles. In Chapter 14 we have discussed
briefly the blade materials, manufacturing, fixing and cooling.
It may be recalled that we have already discussed the various possible
cycle arrangements and also their analysis both at ideal and practical conditions. How the efficiency of each component plays a role in determining
the overall performance of the system was brought to light.
From practical cycle analysis (Chapter 6) it is possible to calculate, for
instance, the pressure-ratio, which, for any given maximum cycle temperature will give the highest overall efficiency, and the mass-flow required to
give the desired output. From such preliminary calculations it is possible to
choose the most suitable design data for any particular application. Then,
it becomes easier to design the individual components of a gas turbine so
that the complete unit will give the required performance at the design
point. However, the problem remains to examine more critically the characteristics of the entire plant as a unit, the methods of control, governing,
and the operation at a condition other than design load.
To a large extent the desired performance characteristics of the gas
turbine power plant can be accomplished by the control of various design
features of the components. Therefore, design decisions for the components
524
Gas Turbines
must be governed by what is best for the whole plant rather than what is
necessarily the best for the individual component. What is best for the
whole plant is dictated by economic considerations, the purpose the plant
is to serve, and such matters as weight and space restrictions.
This chapter includes how the performance or the characteristic curves
for different components such as turbine and compressor are represented
graphically and then by means of graphs, how to link these components so
as to make the combination run in perfect equilibrium.
A matching study is an investigation of the interplay of the engine geometry and engine parameters such as pressure ratio, airflow, rotor speed,
component efficiencies, pressure drops, areas and so on. Such a study must
be conducted to answer questions about the steady-state and transient operation of a gas turbine engine.
Most of the discussion in this chapter deals with the steady-state operation of a single-spool aircraft engines. If one has to effect perfect matching of
components, then, equilibrium running diagram should be constructed taking into account all the parameters mentioned in the previous paragraphs.
In this connection the following questions are to be answered:
(i) When a gas turbine is operated at an off-design condition with a
fixed engine geometry, what happens to the component parameters
and resulting component match?
(ii) How does the magnitude of turbine inlet temperature influence the
resulting component match?
(iii) When the jet nozzle area of a turbojet engine is changed, does the
component match point change? If so, how to arrive at the new match
point?
(iv) What effect does afterburner operation have on the match point for
the gas generator of an afterburning turbojet engine? Is it possible for
an afterburning turbojet engine gas generator to operate at the same
match point with the afterburner and without afterburner turning
on?
(v) How does diffuser water injection or operation with a low calorific
value fuel influence the engine match point?
(vi) How does variation of the turbine nozzle area influence the engine
match?
(vii) How does engine bleed, power extraction and/or turbine blade cooling, influence the engine match?
Answers to these questions involve the interplay of a large number of
variables and would require a detailed computer code. Since, it is out of preview of this chapter, our analysis will be based on simplifying assumptions.
The discussion regarding engine matching in this chapter is intended to give
answers, in a general way, to many of the above questions. The equations
Component Matching and Performance Evaluation
525
developed to answer these questions contain a number of assumptions that
may be questionable in an actual engine. They are made here to simplify the
problem so that one can see the trends without becoming deeply involved
in a number of finer details. Engine matching, because of its complexity is
done almost exclusively on high speed digital computers but, since all new
computer programs must be checked for accuracy and are constantly being
modified, one must understand the principles of matching techniques.
The geometry and flow areas of a given power plant are established
at the “design point”. At all other operating conditions, the components
must be “matched” to determine the pressure ratio, airflow, rotor speed,
efficiency and so on. The “match point” is defined as the steady-state operating point for a gas turbine when the compressor and turbine are balanced
in rotor speed, power, and flow, the operating points at the various power
settings defining the operating line for the given engine configuration.
No matter what type of engine is being considered, the conservation of
mass, energy, and momentum must be satisfied. The first two lead to the
“match point”, the third to the thrust developed by the engines for aircraft
application.
For satisfying the principle of conservation of mass requires:
(i) The flow through the turbine must equal to the flow through the compressor plus the fuel added, minus any air extracted. Care must be
taken into account for any extracted air from the system and exactly
where it re-enters the system.
(ii) The exhaust system (nozzle) flow characteristics must also be satisfied.
For satisfying the principle of conservation of energy requires that the
power developed by each turbine equal the power required to drive each
compressor plus losses and power extracted.
When dealing with the general matching trends of a single-spool turbojet engine, several simplifying assumptions are made. These assumptions
hold true unless stated otherwise and include the following:
(i) No burner or exhaust system pressure losses will be included; that is,
it is assumed that pressure remains constant throughout the combustion chamber and also from the exit of the turbine to the inlet of the
exhaust nozzle.
(ii) The mass rate of flow is assumed to be the same through the compressor, turbine, and exhaust nozzle. This means that the mass of fuel
added is being neglected, no air is extracted from or after the compressor, and turbine cooling is not being used on the engine being
studied.
(iii) No power is extracted and no losses occur between the turbine and
the compressor, i.e., bearing losses are neglected.
526
Gas Turbines
(iv) The turbine nozzle and jet nozzle areas are constant values as determined for the design point conditions.
(v) It is assumed that air is the working fluid throughout and that the
specific heat ratio, γ, has a constant value of 1.4.
Using the above assumptions, we can construct the equilibrium running
diagram. In order to draw the same, first we should know the compressor
and turbine performance characteristics.
15.1
PERFORMANCE CHARACTERISTICS
Performance curves of compressor and turbine may be plotted for delivery pressure and temperature against the mass flow for different speeds.
These curves will reveal how the compressor and turbine will respond to
the changes in different variables. But these are not the only variables
which are responsible for performance variations. There are other parameters such as inlet pressure and temperature and also the physical properties
of the working fluid. If full variation of all these variables is allowed over
the entire working range of the components, it would involve large number
of tedious experiments and thus precise presentation of the results will be
next to impossible.
To overcome this complication, the technique of dimensional analysis
is helpful. This consists of arranging the variable in picture in particular
groups of variables which are dimensionless. This way the complete performance characteristics of any compressor or turbine may be represented only
by two sets of curves. These details have already been dealt in Chapter 3.
15.2
EQUILIBRIUM RUNNING DIAGRAM
It may be recalled that using dimensional analysis a set of performance
characteristics can be obtained and plotted. Typical performance characteristic curves are as shown in Fig. 15.1. When the components are to
be linked together in an engine use of these characteristic curves are very
helpful. The problem is to find corresponding operating points on the characteristic curves of each component when the unit is running at a uniform
speed.
The equilibrium running points for a series of speeds may be plotted on
the compressor characteristics and joined up to form an equilibrium running
diagram. Once the operating conditions have been determined it is simpler
to find performance data such as power output or thrust and specific fuel
consumption.
There are two schemes for which the equilibrium running, points may
be obtained.
(i) When a separate power-turbine is provided to take the load the compressor turbine runs the compressor only (Fig. 15.2).
Component Matching and Performance Evaluation
(a)
t
A
N
T 01
Sur
ge
Y
line
line
(b)
Sur
ge
p03
p01
527
T 02 Y
T01
X
m T01
A
N
T 01
X
p01
m T01
p01
Compressor characteristics
p03
p04
N
T03
m T03
p03
(d)
Choking mass flow
Choking mass flow
(c)
T 03
T04
N
T03
m T03
p03
Turbine characteristics
Fig. 15.1 Compressor and turbine characteristics
(ii) Only one turbine is there which runs both the compressors as well as
the load. (Fig. 15.3).
Torque-speed characteristics for these two schemes are as shown in
Fig. 15.4. The characteristic is poor when the turbine is coupled to compressor as well as load. Curve 1 is for the 1st scheme in which turbine
is coupled to the compressor only and there is a separate power turbine.
Curve 2 is for 2nd scheme in which a single turbine drives the compressor
and the load.
15.3
TO FIND THE EQUILIBRIUM POINTS
Now, let us see how to find the equilibrium points from the characteristic
curves. First of all for equilibrium running, the following conditions must
be satisfied.
(i) Speed of the turbine and compressor is same. Therefore,
√
N
T
N
√
√ 03
= √
T01
T03 T01
(15.1)
Gas Turbines
Heat exchanger
Combustion
chamber
Air
Fuel
HPT
HPC
Compressor
Turbine
Fuel
Combustion chamber
Power
LPT
Separate power turbine
output
Fig. 15.2 open-cycle gas turbine with separate power turbine
Combustion chamber
1
Air
Product of
combustion
3
2
4
Fuel
Power
output
Compressor
Turbine
Fig. 15.3 Turbine running the compressor and load
Sepa
rate p
ower
1
turbi
ne
Torque
528
sor
s
pre
2
m
co d
to loa
d
e
l
nd
up a
Co
Speed
Fig. 15.4 Torque–speed characteristics
Component Matching and Performance Evaluation
=
N √
√
t
T03
529
(15.2)
where t = T03 /T01 .
(ii) Mass flow through the compressor and turbine are assumed to be
same. Hence,
√
√
√
ṁ T01
ṁ T03
p03 T01
√
=
×
(15.3)
p01
p03
p01 T03
or
√
ṁ T01
p01
=
√
ṁ T03 1
√ R
p03
t
(15.4)
where R is the pressure ratio p03 /p01 .
It may be noted that the fuel supply increases the mass of gas in
turbine but some amount of the compressed air is also taken out for
cooling which is approximately equal to the amount of fuel supply.
(iii) Power output of the turbine must be equal to the work input required
to run the compressor. Hence
Cpa (T02 − T01 )
= Cpg (T03 − T04 )ηmech
or
or
(T02 − T01 )
=
Cpg
(T03 − T04 )ηmech
Cpa
T02 − T01
T01
=
Cpg T03 − T04 T03
ηmech
Cpa
T03
T01
T02 − T01
T01
=
Cpg T03 − T04
t ηmech
Cpa
T03
(15.5)
where ηmech is the mechanical efficiency of the transmission.
To satisfy the above conditions, the compressor and turbine characteristics diagrams are to be slightly changed. The changes to be effected are
the following:
(i) Instead of TT02
and TT03
in y-axis in Fig. 15.1, graphs should be plotted
01
04
T02 −T01
04
for T01 and T03T−T
in the y axis.
03
(ii) Instead of pp02
the graphs should be plotted for pp03
where p03 ac01
01
counts for pressure losses in the piping combustion chamber and heat
exchanger.
530
Gas Turbines
Because of these changes in the variables in the y-axis, the shape of the
curves will not change, but, the values will change. The revised graphs are
as shown in Fig. 15.5.
(a)
A
Z
T02 - T01
T01
N
T 01
Sur
ge
Y
t
Sur
ge
p03
p01
line
line
(b)
X
m T01
A
N
T 01
X
p01
m T01
p01
Compressor characteristics
B
N
T03
T 03 - T04
T03
B
P
N
T03
p
m T03
Choking mass flow
p03
p04
q
(d)
Choking mass flow
(c)
p03
p
m T03
p03
Turbine characteristics
Fig. 15.5 Revised characteristic diagram of compressor and turbine
15.4
PROCEDURE TO FIND EQUILIBRIUM POINT
To find out the equilibrium points and lines, the following procedure is to
be followed:
(i) Assume some value of t and N .
(ii) On the speed curve of particular value of √N
choose a point A, on
T01
the compressor√characteristics [Fig. 15.5(a)] and find
out the values,
√
p03
ṁ T01
ṁ T01
viz., pp03
and
.
Let
us
denote
=
Y
and
= X.
p01
p01
p01
01
√
01
(iii) On the same speed curve of √N
and with X = ṁp01T01 find T02T−T
T01
01
[Fig. 15.5(b)] of the compressor characteristics and let it be Z.
(iv) Using Eq. 15.2, find the value of
√N
T01
and t. Similarly find
√
ṁ T03
p03
√N
T03
corresponding to the value of
from Eq. 15.4.
Component Matching and Performance Evaluation
(v) On the speed curve
and let it be P .
√N
T03
and with
√
ṁ T03
p03
find
T03 −T04
T03
531
in [Fig. 15.5(d)]
(vi) After getting Z and P substitute in the Eq. 15.5 and see whether the
equation is satisfied or not.
(vii) If this equation is satisfied, point A is the equilibrium running point
for that N and t.
(viii) If Eq. 15.5 is not satisfied, choose another point on the same speed
curve and again find the values of Z and P . Carry out this trial till
Eq. 15.5 is satisfied. That point will be the point A which satisfies
Eq. 15.5.
(ix) Similar points can be tried on other speed curves.
(x) The line joining all such points on all speed curves is known as equilibrium running line for that particular value of t.
(xi) For some other value, the equilibrium running line will be different
and thus many lines can be drawn for many values, of t. This will be
known as equilibrium running diagram which is as show in Fig. 15.6.
03
p 01
Su
p
rge
lin
e
t
N
T01
m T01
p01
Fig. 15.6 Equilibrium running diagram for compressor-turbine combination
15.5
PERFORMANCE EVALUATION OF SINGLE-SPOOL TURBOJET ENGINE
To understand the details of engine matching a single-spool turbojet engine
will be considered for performance evaluation. This study can be used as
a building block, to help the understanding of other types.
532
Gas Turbines
Figure 15.7 is a schematic diagram of a single-spool turbojet engine
showing various components and station numbers.
CC
Air
Exhaust gases
T
C
CC
(0)
(1)
(2)
(3)
(4)
(5)
0-1 Inlet
1-2 Compressor 2-3 Combustion chambe
3-4 Turbine 4-5 Nozzle
Fig. 15.7 Schematic diagram of a single-spool turbojet engine
The turbine flow characteristics that will be used in this analysis are
shown in Fig. 15.8. Note that in Fig. 15.8, the variation of turbine flow
parameter as a function of expansion ratio which is shown as a single
curve. This has been done without effecting corrections for turbine rotational speed. This type of turbine flow characteristics is considered mainly:
(i) to simplify the analysis, and
Turbine flow parameter, m T03
p03
(ii) because of the negligible effect that turbine rotational speed has on
the turbine flow parameter and efficiency.
A
B
C
D
Choking expansion ratio
Turbine expansion ratio, p03 / p04
Fig. 15.8 Turbine flow characteristics
From Fig. 15.8 it can be seen that the turbine of the single-spool turbojet
engine operates at choked condition over a wide region of its operating
spectrum. Thus, most of the time, the turbine operating point will be to
the right of point A (choking, expansion ratio) as shown in Fig. 15.8.
Component Matching and Performance Evaluation
533
When the turbine flow parameter becomes constant (the turbine is
choked), the turbine expansion ratio can still continue to increase. For
a fixed turbine inlet temperature and turbine efficiency, the turbine work
is dependent on the turbine expansion ratio. Whether the “match point”
for the turbine is at A, B, C, D, or another point depends on the flow characteristics of what follows the turbine For the single-spool turbojet engine
we should consider the flow characteristics of the exhaust nozzle also.
The relationship between the turbine flow characteristics and compressor flow characteristics can be expressed algebraically by the following
equality:
⎤
⎡
√
√
ṁg3 T03
1
ṁa1 T01 T03 ṁg3 ⎣
⎦
=
(15.6)
p03
p02
p03 A3
δ01
θ01 ṁa1
p p A
p02
p01
01 std
3
01
01
and θ01 = TTstd
where δ01 = ppstd
For a constant-pressure in the combustion chamber and assuming mass
flow rate through the compressor to be same as through the turbine ṁg3 =
ṁa1 , √fixed turbine nozzle area (A3 = constant), and a choked turbine
ṁg3 T03
p03 A3
is also constant, Eq. 15.6 becomes
p02
p01
=
C1
√
ṁa1 θ01
δ01
T03
θ01
(15.7)
where C1 is a constant.
For a fixed T03 /θ01 , Eq. 15.7 reduces to
p02
p01
=
C2
√
ṁa1 θ01
δ01
(15.8)
which is the equation of a straight line passing through the point (0,0). The
slope of this straight line, C2 , increases for increasing values of (T03 /T01 ),
that is, turbine inlet temperature over compressor inlet temperature.
Plotting Eq. 15.8 on a typical compressor map yields the results shown
in Fig. 15.9.
Equation 15.8 shows that the pressure ratio is zero when the airflow
is zero. This, of course, is impossible. What is wrong? Equation 15.8
assumes a choked turbine. At low air flows, the turbine unchokes so that
the constant T03 /θ01 lines curve into a pressure ratio of 1.0 at zero airflow.
This is shown in Fig. 15.9 by the dashed lines.
For a given T03 /θ01 , where is the operating point? Is it at A, B, C, or
D as shown in Fig. 15.9. This is fixed by, the compressor work required,
which, for an energy balance, is fixed by the turbine work, the turbine
work being fixed by the turbine inlet temperature and expansion ratio, the
turbine expansion ratio being fixed by the nozzle flow characteristics. These
details are discussed in the following paragraphs.
For the single-spool turbojet engine illustrated in Fig. 15.7 the ideal
work of compression for constant specific heat and unit mass flow is given
e
in
l
ge
ur
T 03 =
θ 01
s
con
tan
t
Gas Turbines
Compressor pressure ratio, p02 / p01
534
D
B
S
C
A
nt
nsta
T 03 = co
θ 01
1
0
Corrected air flow, m T01
δ 01
0
0
Fig. 15.9 Equation 15.8 plotted on a typical compressor map
by
wC,i
=
h02 − h01
=
Δh0C,i
(15.9)
=
Cpa (T02,i − T01 )
(15.10)
Equation 15.10, when divided by compressor inlet temperature over the
standard temperature, becomes, for constant specific heats
Δh0C,i
θ01
=
p02
p01
Cpa Tstd
(k−1)/k
−1
(15.11)
Equation 15.11 shows that the ideal work of compression divided by θ01 is a
function of pressure ratio only. When variable specific heats are considered,
the ideal work of compression divided by θ01 is approximately a function
of pressure ratio.
The actual compressor work depends on the compressor efficiency. Therefore,
Δh0Ca
θ01
=
Δh0C,i
θ01
Cpa Tstd
1
ηC
(k−1)/k
p02
p04
=
−1
ηC
(15.12)
The ideal turbine work for turbine of the single-spool turbojet engine
illustrated in Fig. 15.7 is, for constant specific heats,
wT,i
= Δh0T,i
=
(T03 − T04,i )
(15.13)
Component Matching and Performance Evaluation
= Cpg T03 1 −
(1−k)/k
p03
p04
(15.14)
or
Δh0T,i
θ03
p03
p04
= Cpg Tstd 1 −
535
(1−k)/k
(15.15)
Equation 15.14 illustrates that the ideal work developed by a turbine is a
function of the expansion ratio and the turbine inlet temperature. Equation
15.15 illustrates that the ideal work developed by a turbine divided by θ03
is a function of the expansion ratio only.
The actual turbine work depends on the turbine efficiency, or
Δh0T,a
θ03
=
Δh0T,i
ηT
θ03
(15.16)
p03
p04
= Cpg Tstd ηT 1 −
(1−k)/k
(15.17)
We have already seen that for a choked turbine, the compressor pressure
ratio is a function of turbine inlet temperature over theta and the compressor corrected mass rate of flow is given by Eq. 15.7 which is reproduced
here.
√
p02
T03
ṁa1 θ01
= C1
(15.18)
p01
δ01
θ01
This results in several straight lines as illustrated in Fig. 15.9, each straight
line being for a different turbine inlet temperature over theta. Equation 15.7
is based on conservation of mass and, as illustrated in Fig. 15.9, does not
give any due consideration regarding, whether the engine will be operating
at point A, B, C, or D.
The actual steady-state operating point occurs where the turbine power
is equal to the power required to drive the compressor; that is, it is determined by, an energy balance. Based on the assumptions stated earlier (no power extraction, no losses between compressor and turbine, and
ṁa1 = ṁg3 ),
Δh0Ca
θ01
=
Δh0T,a
θ03
T03
T01
(15.19)
Combining Eqs. 15.12, 15.17 and 15.19 yields, for the case where Cpa = cpg ,
and a = (k − 1)/k.
p02
p01
a
−1
= ηC ηT 1 −
p03
p04
−a
T03
T01
(15.20)
This equation illustrates that the compressor pressure ratio is a function
of compressor efficiency, turbine efficiency, turbine expansion ratio, and the
ratio of turbine inlet temperature to compressor inlet temperature, or
Gas Turbines
Choked
Unchoked
Choking expansion ratio
m 4 T04
Exhaust nozzle flow parameter
536
Exhaust nozzle expansion ratio, p04 / pamb
Fig. 15.10 Typical exhaust nozzle flow characteristics
p02
p01
= f ηC , ηT ,
p03
,
p04
T03
T01
(15.21)
Equation 15.21 illustrates that for a fixed T03 /T01 and constant values of ηC
and ηT , the compressor pressure ratio is a function of the turbine expansion
ratio. The turbine expansion ratio for a specified T03 /T01 is fixed by the
exhaust nozzle flow characteristics.
A typical nozzle flow characteristic is given in Fig. 15.10, in which the
exhaust expansion ratio, is specified as p04 /pamb instead of p0 /pstd .
The turbine and exhaust nozzle flow parameters, when combined, will
yield the following expression:
√
ṁg4 T04
p04 A4
=
√
ṁg3 T03
p03 A3
p03
p04
T04
T03
A3
A4
(15.22)
For a given turbine polytropic efficiency, the turbine actual temperature
ratio is related to the turbine expansion ratio according to the following
equation:
T04
=
T03
p04
p03
ηp (k−1)/k
(15.23)
Combining Eqs. 15.22 and 15.23 and assuming that ṁg3 = ṁg4 = ṁg yields
√
ṁg T04
p04 A4
=
√
ṁg T03
p03 A3
p03
p04
[1−0.5ηp (k−1)/k]
A3
A4
(15.24)
Equation 15.24 shows that for exhaust and turbine nozzles of fixed area,
the expansion ratio has a constant value once the turbine and exhaust nozzles become choked. This holds true only if the turbine polytropic efficiency
(and therefore the overall turbine efficiency) is a constant.
Component Matching and Performance Evaluation
Compressor pressure ratio, p02 / p01
6.5
537
Steady-state operating line
6.0
e
5.5
e
urg
5.0
lin
4.16
t=
3.82
3.47
S
4.5
η
4.0
%
95
=
c
0%
3.5
3.0
ηc
%
= 92
ηc
η
=9
12
14
16
18
%
80
η
2.5
10
=
c
20
%
c
=
70
22
24
Corrected mass rate of flow, m 1 θ 01 / δ01 (kg/s)
Fig. 15.11 Hypothetical compressor map with steady-state operating line
15.6
OPERATING LINE
It is now important to evaluate what happens to the steady-state operating conditions of an engine as the turbine inlet temperature and/or flight
conditions are changed. This will be done by assuming a hypothetical
compressor map for a single-spool turbojet engine, fixing the areas at the
design point, then determining what happens to the engine operation as
the turbine inlet temperature is varied. This can be better illustrated by
a numerical example, with the results shown on a hypothetical compressor
map in Fig. 15.11.
A single-spool turbojet engine, having the compressor performance illustrated in Fig. 15.11, has the following design
point conditions at sea level
√
ṁa1 θ01
static on a standard day. pp02
=
5.0;
=
22.7 kg/s; T03 = 1200 K;
δ01
01
ηT = 92% (constant).
Calculate, neglecting the mass of fuel added, the turbine expansion ratio.
Assume no pressure drop in the combustion chamber and constant specific
heats with Cp = 1.004 kJ/kg K and k = 1.40.
From the compressor map (Fig. 15.11) at the design point conditions,
it is determined that the compressor efficiency ηc is 90%. Therefore, from
Eq. 15.12, since θ01 = 1.0.
Wca
ΔhT,i
= Δhca
=
1 × 1.004 × 288.15 × 5(0.4/1.4) − 1
= 187.9 kJ/kg
0.90
=
ΔhT,a
ηT
=
ΔhC,a
ηT
538
Gas Turbines
θ03
=
187.9
0.92
=
1200
288.15
From Eq. 15.15
p03
=
p04
1−
=
204.24
=
≈
204 kJ/kg
4.16
204.24
4.16 × 1.004 × 288.15
−(1.4/0.4)
= 1.91
Note that the expansion ratio across the exhaust nozzle, for the conditions
assumed, is
p04
5.0
= 2.62
=
pamb
1.91
which is well above the pressure ratio required for the exhaust nozzle to be
choked.
It is next important to determine the operating line for this engine. This
is illustrated by means of another example below.
Determine, for the engine defined in the above example, the steady-state
operating points when the turbine inlet temperature is (i) 1100 K (ii) 1000
K.
For T03 = 1100 K; T01 = 288.15 K; Tstd = 288.15 K; θ01 = 1.0; θ03 =
3.82.
From Eq. 15.17
ΔhT,a
θ01
=
3.82 × 1.004 × 288.15 × 0.92
1 − 1.91(−0.4/1.4)
1.0
= 171.8 kJ/kg
At the design point conditions, using Eq. 15.7 one may determine C1 ,
C1
=
5.0 × 1
√
22.7 × 1200
=
0.00636
Therefore. the relationship between compressor pressure ratio and corrected mass rate of flow is,
√
p02
T03
ṁa1 θ01
×
= 0.00636 ×
p01
δ01
θ01
or, for θ03 = 3.82, i.e., T03 = 1100 K, the relationship becomes
√
p02
ṁa1 θ01
= 0.211 ×
p01
δ01
(15.25)
Equation 15.25, which is an expression of the conservation of mass, is plotted on Fig. 15.11 and is labelled T03 /T01 = 3.82. Since both the turbine
and exhaust nozzle are choked and the turbine efficiency is a constant, the
turbine expansion ratio is a constant and is equal to 1.91.
Component Matching and Performance Evaluation
539
The actual turbine work, as determined by Eq. 15.17, is
ΔhT,a
θ03
=
1.004 × 288.15 × 0.92 × 1 − 1.91(−0.4/1.4)
=
44.97 kJ/kg
Note that this is a constant value for this engine as long as the turbine and
exhaust nozzle are choked. For T03 = 1100 K, i.e., θ03 = 3.82
ΔhT,a
θ01
=
ΔhC,a
θ01
=
44.97 × 3.82
1
=
ΔhT,a × θ03
θ01
=
171.79 kJ/kg
It is now necessary to determine the compressor pressure ratio that satisfies
ΔhC,i
θ01
=
171.79
The ideal compressor work over θ01 is determined from Eq. 15.11. For a
pressure ratio of 4.4. The result is
ΔhC,a
θ01
=
1.004 × 288.15 × 4.4(0.4/1.4) − 1
=
152.65 kJ/kg
The compressor efficiency, for an ideal work of 152.65 and an actual work
of 171.79 is
152.65
= 0.888
ηC =
171.79
The results for several pressure ratios are, for an actual work of 171.79
ΔhC,a
are given in Table 15.1. Therefore, a θ01
= 171.79 line may be constructed on the compressor map. This is shown in Fig. 15.12, which is a
portion of the compressor map.
ΔhC,a
The point of intersection between the TT03
= 3.82 line and θ01
=
01
171.79 line is the steady-state operating point at sea level condition on a
standard day for a turbine inlet temperature of 1100 K. From the plot, this
point is
p02
p01
ηC
√
ṁa1 θ01
δ01
For T03 = 1000 K
= 4.49
= 90.4%
= 21.4 kg/s
540
Gas Turbines
Table 15.1
p02 /p01
ΔhC,a /θ01 (kJ/kg)
ηC
155.3
152.5
149.6
146.6
143.6
0.905
0.888
0.872
0.854
0.837
Steady-state operating line
5.0
p /p
02 01
Compressor pressure ratio
4.5
4.4
4.3
4.2
4.1
t=
3.82
2%
4.5
ηc
=9
9
1.7
17
=
Δh
%
0%
90
θ1
=
=8
ηc
ηc
4.0
c,a
3.5
%
η
c
=
70
23
20
21.5
(kg/s)
/
m
Corrected mass rate of flow, 1 θ 01 δ01
Fig. 15.12 Compressor map for t = 3.82
θ03
=
1000
288.15
=
3.47
ΔhT,a
θ01
=
ΔhC,a
θ01
=
44.9 × 3.47
=
155.8 kJ/kg
Once again, plotting on a portion of the compressor map (Fig. 15.13),
one determines that the steady-state operating point is
p02
= 4.05
p01
ηC
√
ṁa1 θ01
δ01
= 90.3%
= 20.1 kg/s
Three points of the steady-state operating line have now been determined. The steady-state operating line is shown on the hypothetical compressor map (Fig. 15.11).
The exhaust nozzle is choked above a compressor pressure ratio of approximately 3.6. Below this compressor pressure ratio the exhaust nozzle
Component Matching and Performance Evaluation
541
Table 15.2
p02 /p01
ΔhC,a /θ01 (kJ/kg)
ηC
143.6
140.6
137.5
134.3
0.921
0.902
0.882
0.862
5.0
Steady-state operating line
t=
3.47
%
p /p
02 01
Compressor pressure ratio
4.1
4.0
3.9
3.8
4.5
4.0
3.5
η
=
c
92
5.8
15
c,a =
h
Δ
θ 01
%
η
=
c
90
ηc
%
%
0
=8
η
=
c
70
20
22.5
25
/
(kg/s)
m
θ
δ
Corrected mass rate of flow, 1 01 01
Fig. 15.13 Compressor map for t = 3.47
will not be choked. This mean that even though the turbine may be operating choked, the turbine expansion ratio will change as the turbine inlet
temperature varies if the steady-state operating point is below a compressor
pressure ratio of 3.6 for the engine illustrated in Fig. 15.11.
It is important to keep in mind the many simplifying assumptions in
the preceding analysis. These included the following:
(i) The mass rate of flow through each component is the same.
(ii) There is no pressure drop in the combustion chamber or between the
turbine exit and exhaust nozzle inlet.
(iii) No power is extracted and bearing losses are neglected.
(iv) The turbine and exhaust nozzle areas have fixed values.
(v) There are constant specific heats with k = 1.4.
(vi) There is constant turbine efficiency.
(vii) The turbine and exhaust nozzle are choked.
542
Gas Turbines
One should next consider what happens when one or more of these simplifying assumptions is changed.
How should the match point for a specified T03 /T01 change if the mass
of fuel added is taken into account?
In Eq. 15.6, ṁg3 /ṁa1 > 1.0. Therefore, the constant C1 of Eq. 15.7 will
increase, since all other quantities in Eq. 15.6 remain constant. This means
that the slope of each of the T03 /T01 lines will increase. Since, the power
developed by the turbine must be equal to the power required to drive the
compressor,
ṁa1 ΔhC,a
=
ṁg3 ΔhT,a
(15.26)
or Eq. 15.19 becomes
ΔhC,a
θ01
=
ΔhT,a
θ03
ṁg3
ṁa1
T03
T01
(15.27)
This means that the match point for specified T03 /T01 will shift as illustrated in Fig. 15.14, where the original conservation of mass and energy
lines are shown as solid lines, the new ones as dashed lines.
t
tan
t=
ns
co
Δ h c,a
θ1
= constant
Fig. 15.14 Effect of fuel addition on T03 /T01 line
How would the match point for a specified T03 /T01 change if there is a
5% pressure drop in the combustion chamber?
In Eq. 15.6, p03 /p02 < 1.0. Therefore, since all other quantities in
Eq. 15.6 remain constant, the constant C1 , in Eq. 15.7 will increase. Equation 15.19 will not change. Therefore, the match point will shift as illustrated in Fig. 15.15.
The preceding two examples illustrate how the conservation of mass
and conservation of energy “lines” change when one of the assumptions
is changed. The reader now should be able to determine how the match
points, and therefore the operating line, will shift if the exhaust nozzle
area is increased, what must be changed and how for the gas generator to
operate at the same “match point” when an afterburner is in operation.
and/or how diffuser water injection or operation with a low calorific value
fuel will influence the engine match point.
Component Matching and Performance Evaluation
543
tant
t=
s
con
Fig. 15.15 Effect of combustion chamber pressure drop on T03 /T01 line
15.7
GENERAL MATCHING PROCEDURE
The preceding section discussed component matching of a single-spool turbojet with many simplifying assumptions. The general matching procedure
must take into account variable specific heats, the actual products of combustion, the fact that air may be extracted at an intermediate stage or at
the exit from the compressor, turbine cooling may be used, the mass of fuel
added, the pressure drop in the combustion chamber, turbine flow characteristics as a function of rotor speed, the fact that turbine efficiency is not
a constant, and so on. It is obvious that the general matching problem is
very complicated, yet it must be done to select the best engine design for
a given application.
A general matching procedure is outlined below for a two-spool turbojet
engine. It is assumed that the two-spool turbojet engine has the station
numbers shown in Fig. 15.16 and that the compressor, turbine, and nozzle
performance characteristics as shown in Fig. 15.17 are known and stored in
a high end computer where the matching study is being performed.
Air
Exhaust
gases
(0)
(1) (1.5) (2)
0-1 Inlet
2-3 Combustion chamber
4-5 Nozzle
(3) (3.5) (4)
(5)
1-2 Compressor (L.P.C. + H.P.C.)
3-4 Turbine (L.P.T. + H.P.T.)
Fig. 15.16 Schematic diagram of a two-spool turbojet engine
The following are the step-by-step procedure to obtain matching points.
It is assumed that the operating conditions (altitude and speed of the aircraft) and turbine inlet temperature are known.
Gas Turbines
p
p
02
01.5
544
m 1 θ 01 / δ1
m 1.5
(b) High-pressure compressor
η HPT
m 3 T03 / p03 A 3
(a) Low-pressure compressor
θ 01.5 / δ 01.5
p03.5 / p
p03 / p
03.5
04
(d) High-pressure turbine
η LPT
m 3.5 T03.5 / p03.5A 3.5
(c) High-pressure turbine
p03.5 / p
p03.5 / p
04
04
(f) Low-pressure turbine
m 4 T04 / p A 4
04
(e) Low-pressure turbine
p04 / p
amb
(g) Exhaust nozzle
Fig. 15.17 Compressor, turbine and nozzle characteristics for a two-spool
turbojet engine
Component Matching and Performance Evaluation
545
(i) Knowing the flight condition fix T01 and p01 .
√
and ṁa1δ01θ01 . Since
(ii) Assume an LPC operating point (assume pp01.5
01
C
ηLP C and N√LP
are known from the LPC map [Fig. 15.17(a)], NLP C ,
θ01
ṁa1 , ΔhLP C , ṁa1,5 , T01.5a and p01.5 can be calculated.
ṁ
√
θ
01.5
02
(iii) Assume an LPC pressure ratio, pp01.5
. Calculate a1,5
then read
δ01.5
N
HP
C
from the HPC map [Fig. 15.17(b)] ηHP C , √θ . Calculate NHP C ,
01.5
ΔhLP C,a , ṁg2 , T02,a and p02 .
(iv) Since T03 is known, determine the fuel-air ratio and combustion chamber pressure drop. Calculate ṁg3 and p03 .
(v) Assume a HPT expansion ratio. Calculate N√HP T . Determine from the
θg3
turbine performance characteristics ηHP T [Fig. 15.17(d)] and
[Fig. 15.17(c)]. Calculate ΔhHP T,a and ṁg3 .
√
ṁg3 T03
p03 A3
(vi) Check to determine if ΔhHP T,a and ṁg3 are within a preset tolerance.
If not repeat steps 3 through 5 until a match is obtained.
(vii) Once the high-pressure spool has been matched, ṁg3.5 p03.5 T03.5a and
NLP C are known. Assume an LPT expansion ratio, pp03.5
. Deter04
mine ηLP C [Fig. 15.17(f)] and
ΔhLP T,a and ṁg3.5 .
√
ṁg3.5 T03.5
p03.5 A3.5
[Fig. 15.17(e)]. Calculate
(viii) Check to determine if ΔhLP T,a and ṁg3 are within a preset tolerance.
If not repeat steps 2 through 7 until a match is obtained.
(ix) Once the low-pressure spool has been matched ṁg4 p04 and T04,a
are known. Determine
from the exhaust nozzle flow characteristics
√
ṁ
T
[Fig. 15.17(g)] pg404 A404 . Calculate A4 and compare with the known
exhaust nozzle A4 . An engine match exists and the thrust, thrustspecific fuel consumption, and other desired values may be calculated
with a preset tolerance. If a match does not exist, repeat 2 through
9 until a match has been achieved.
It should be obvious, from the matching procedure described above,
that the component matching study for a gas turbine is quite a complex
problem. The procedure described is for a two-spool turbojet engine. It
should be obvious that the problem of complexity increases as one studies
a turbofan engine. Keep in mind that three-spool static pressure-balanced
engines are currently being built. A matching procedure for this type of
engine is left to the reader.
15.8
TRANSIENT OPERATION
The discussion so far in this chapter has dealt with the stead- state operation of a gas turbine. It is next important to determine what happens when
the engine undergoes a power change.
546
Gas Turbines
T 03 = constant
Steady state operating line
N / θ01 = const.
p
02
e
rg
B
e
lin
Su
A
m 1 θ 01
δ1
Fig. 15.18 Transient operation
Consider the case where a single-spool gas turbine engine, with the
steady-state operating line shown in Fig. 15.18, is to be accelerated from
point A (low power setting) to point B (high power setting).
To change the steady-state operating point from A to B, the mass rate
of flow of fuel must be increased. Increasing the mass rate of flow of fuel
will initially increase the engine to a higher T03 /T01 line than the initial
equilibrium
√ value without appreciably changing the√ corrected mass flow
rate (ṁa1 θ01 /δ01 ) or the corrected rotor speed N/ θ01 ).
Next, the engine will start to increase in rotor speed (N ), which will
increase the corrected mass rate of flow. The exact path that the engine will
follow in accelerating from A to B depends on the design characteristics
of the engine components and the manner in which the fuel flow to the
combustion chamber is changed. One possible acceleration path between
steady-state operating points A and B is shown in Fig. 15.18. Acceleration
and deceleration paths will take the compressor to opposite sides of the
steady-state operating line.
One must be careful during transient operations to schedule the fuel
flow change so that the compressor does not surge during the change.
Review Questions
15.1 What do you understand by the term ‘matching of components’?
15.2 Does inlet temperature influence the component match? Explain.
15.3 Define match point and explain the effect of exit nozzle area on the
match point.
Component Matching and Performance Evaluation
547
15.4 Explain the details of component match with and without afterburner
operation.
15.5 Mention the various assumptions in the general matching trends.
15.6 What is an equilibrium diagram?
15.7 Explain with sketches the method of finding the equilibrium points.
15.8 Give a brief account of performance evaluation of a single spool turbojet engine.
15.9 What is meant by an operating line and what are the assumptions
made? Explain how to determine the operating line.
15.10 Explain the general matching procedure for a two-spool turbojet engine.
Multiple Choice Questions (choose the most appropriate answer)
1. At match point the compressor and turbine are balanced for
(a) rotor speed
(b) power
(c) flow rate
(d) all of the above
2. In simplified matching technique
(a) bearing losses are neglected
(b) compressor efficiency is neglected
(c) turbine efficiency is neglected
(d) varying mass flow consideration is considered
3. The performance map of a compressor is a funtion of
(a) mass flow rate and pressure ratio
(b) pressure ratio and temperature ratio
(c) pressure ratio and specific heat ratio
(d) pressure ratio and work ratio
4. The performance map of a turbine is a funtion of
(a) pressure ratio and temperature ratio
(b) pressure ratio and specific heat ratio
(c) pressure ratio and work ratio
(d) mass flow rate and pressure ratio
548
Gas Turbines
5. Simplifying assumption used in matching procedure is that
(a) constant specific heat
(b) constant turbine efficiency
(c) chocked exhaust nozzle
(d) all of the above
Ans:
1. – (d)
2. – (a)
3. – (a)
4. – (d)
5. – (d)
16
ENVIRONMENTAL
CONSIDERATIONS
AND APPLICATIONS
INTRODUCTION
The fourteen preceding chapters have covered the various details of the
gas turbine, examined possible gas turbine cycles, considered the various
components and factors that influence their performance, and then examined the manner in which these components influence one another to determine the engine operating point.
One last area will be considered, this being a brief examination of environmental aspects of gas turbines, viz., current gas turbine air and noise
regulations and engine modifications that can be effected to reduce the
quantity of air pollutants and noise emitted by a gas turbine engine. Further, typical applications of gas turbines will also be discussed.
16.1
AIR POLLUTION
Most of the early air pollution episodes, such as those that occurred in
London in 1873, 1952 and 1956, in the Meuse Valley of Belgium in 1930,
or in Donora, Pennsylvania in 1948, were associated with sulphur dioxide
and particulate emissions. This was due to atmospheric inversions in highly
industrialized areas.
The first legislation enacted was the Air Pollution Control Act of 1955
by USA. It was quite narrow in scope. It mainly considered prevention and
control of air pollution. Further, it placed the primary responsibility on
state and local governments.
Further, American Congress, because of worsening conditions in urban
areas due to air pollution by mobile sources, directed the surgeon general
to conduct a thorough study of motor vehicle exhaust effects on human
health. As a result, 1955 Act was amended.
550
Gas Turbines
The two acts were the Air Pollution Control Act Amendments of 1960
and Amendments of 1962. The Clean Air Act of 1963 provided for the
development of air quality criteria. The Act encouraged state, regional, and
local programmes, for the control and abatement of air pollution. Further,
for the first time, provided for federal financial aid for research and technical
assistance.
The Air Quality Act of 1967 initiated a two-year study on the concept
of national emission standards from stationary sources which served as the
basis for the 1970 legislative act. The 1967 act also required:
”. . . development and issue to the States such criteria of air quality that
may be required for the protection of the public health and welfare.”
This provision led to a series of documents for common air pollutants
such as hydrocarbons, carbon monoxide, sulphur oxides, nitrogen oxides,
and so on. Further, a separate document on each pollutant was prepared
regarding the air quality criteria which summarized what science at that
time was able to measure of the effects of air pollution on humans and the
environment. Also, there was the second document summarizing information on techniques for controlling certain emissions.
The 1967 act also required that
”The Secretary shall conduct a full and complete investigation and study
on the feasibility and practicability of controlling emissions from jet and
piston aircraft engines and of establishing national emission standards with
and report to Congress the results of such study and investigation within
one year from the date of enactment of the Air Quality Act of 1967, together
with his recommendations.”
The major provisions of the Clean Air Amendments of 1970 included:
(i) Primary responsibility for assuring air quality within the entire geographic area comprising of each state lies with such state.
(ii) A requirement that national ambient air quality standards (both primary and secondary) are to be established by the Environmental Protection Agency (EPA).
(iii) A requirement that standards of performance for new station be established and implemented.
(iv) A requirement that industry was to monitor and maintain emission
records and make them available to EPA officials.
(v) A requirement establishing, aircraft emission standards.
The Clean Air Act Amendments of 1977 included many amendments.
The one affecting gas turbines was Sec. 225, which stated:
”Any regulations in effect under this section on date of enactment of
the Air Act Amendments of 1977 or proposed or promulgated thereafter,
or amendments thereto, with respect to aircraft shall not apply if disapproved by the President, after notice and opportunity for public hearing,
on the basis of a finding by the Secretary of Transportation that any such
Environmental Considerations and Applications
551
regulation would create a hazard to aircraft safety. Any such finding shall
include a reasonably specific statement of the basis upon which the finding,
was made.”
16.2
AIRCRAFT EMISSION STANDARDS
The Clean Air Amendments of 1970 required that national and primary
ambient air quality standards were to be established by EPA. These standards were published in the Federal Register in 1971 (1) and set maximum
concentration limits for carbon monoxide, hydrocarbons, nitrogen dioxide,
sulphur dioxide, particulate and oxidant.
The Clean Air Amendments of 1970 also required the establishment of
aircraft emission standards. Proposed emission standards were published
on December 12, 1972 with the final emission standards and test procedures
being published on 1973. The standards published in 1973 have done minor
modifications over the last twenty years. It is recommended that any reader
needing, the latest standards consult the Code of Federal Regulations. The
original and current standards are discussed below.
Of the pollutants generated in any combustion process, only carbon
monoxide (CO), hydrocarbons (HC), nitrogen oxides (NOx ), and smoke
have created the most concern in aircraft gas turbine engines and for which
emission standards have been issued.
The first step in developing aircraft and aircraft engine standards was
deciding to classify the engines. They had to be classified in a manner
consistent with their design, performance, and construction, giving consideration to their potential for reducing their emissions and the need to do
so.
Eight classes were originally defined. The details are listed in Table 16.1.
A separate class was selected for turboprop engines because the proposed
equivalency between shaft and jet thrust was not considered acceptable over
the landing-take-off cycle.
The Pratt and Whitney JT3D and JT8D engines were given separate categories in order to be able to require separate schedules of smoke
retrofits. Supersonic gas turbine engines were included in a separate category because:
(i) Some employ afterburners during take-off and acceleration, because of
which the combustion pressure and mixing methods, result in higher
hydrocarbon and carbon monoxide emissions.
(ii) The cycle pressure ratios for engines used in supersonic aircraft usually operate at lower pressure ratios compared to engines designed
for subsonic flight. At low flight speed and at low altitude, these engines do not benefit from the ram effect they experience at high flight
speed.
(iii) They usually do not a employ high-bypass-ratio turbofan engines.
552
Gas Turbines
Table 16.1 Original Engine Classification System for EPA Standard
EPA Class
Description
TI
All aircraft turbofan and turbojet engines of rated
power less than 3636 kg thrust, except engines in Class
T5.
T2
All aircraft turbofan and turbojet engines of rated
power of 3636 kg thrust or greater, except engine in
Class T3, T4, or T5.
T3
All aircraft gas turbine engines of the JT3D model
family.
T4
All aircraft gas turbine engines of the JT8D model
family.
T5
All aircraft gas turbine engines for propulsion of aircraft
designed to operate at supersonic flight speeds.
P1
All aircraft piston engines except radial engines.
P2
All aircraft turboprop engines.
APU
Any engine installed in or on an aircraft exclusive of
the propulsion engines.
The current engine classes are slightly different from those listed in
Table 16.1. The current classifications are listed in Table 16.21. The next
major step before aircraft engine emission standards could be set was to
determine the engine operating conditions which will enable determining
the air pollutants. The current exhaust emission test selected is designed
to measure hydrocarbons, carbon monoxide and carbon dioxide emissions
for a simulated aircraft landing-takeoff (LTO). This is based on the time,
in each mode during a high-activity period at major airports.
The details of the standards, regarding the gas turbine engine test in
each of the operating modes are listed in Table 16.3 The values given in Table 16.3 are the percentages of rated power settings on a standard day. The
standard day is defined as a day with a temperature of 15◦ C, a pressure of
101325 Pa, and a specific humidity of 0.0 kg H2 O/kg dry air. Emission tests
are to be conducted on a warmed-up engine, which has achieved a steady
operating temperature. The time in each mode is listed in Table 16.4.
Gaseous emission standards for hydrocarbons, carbon monoxide, nitric
oxide, and smoke have been set in the 1979 standards. The 1994 standards
have details regarding gaseous emission standards for hydrocarbons, smoke
exhaust emissions, and fuel venting. The 1994 engine fuel venting emission
1 Code of Federal Regulations, Title 14, Part 34 U.S. Government Printing Office,
Washington D.C. Jan.1,1994.
Environmental Considerations and Applications
553
Table 16.2 Current Engine Classification System for EPA Standards
EPA Class
Description
TP
All aircraft with turboprop engines
TF
All turbofan or turbojet aircraft engines except engines
of Class T3, T8, and TSS
T3
All aircraft gas turbine engines of JT3D model family
T8
All aircraft gas turbine engines of the JT8D model
family
TSS
All aircraft gas turbine engines employed for propulsion of aircraft designed to operate at supersonic flight
speeds
Table 16.3 Gas Turbine Engine Power Settings for Emission Measurements
Aircraft
Operating mode
Taxi/idle
Take off
Climb out
Descent
Approach
TP
7% rated
thrust
100%
90%
NA
30%
Engine Class
TF, T3, T8
TSS
7% rated
7% rated
thrust
thrust
100%
100%
85%
65%
NA
15%
30%
34%
standards apply to all new aircraft gas turbine engines of classes T3, T8,
TF and TSS with rated output equal to or greater than 36 kilonewtons
manufactured after January 1, 1974. For all in-use aircraft gas turbine engines of classes T3, T8, TF and TSS with rated output equal to or greater
than 36 kilonewtons manufactured after February 1, 1974 also these standards apply. The engine fuel venting emission standards also apply to all
new aircraft gas turbine engines of class TF with rated outputs less than
36 kilonewtons, all turboprop engines of class TP manufactured on or after January 1, 1975 and all aircraft gas turbine engines of class TF with
rated outputs less than 36 kilonewtons and class TP turboprop engines
manufactured after January 1, 1975.
The engine fuel venting emission standards require that no fuel be discharged into the atmosphere from any new or in-use gas turbine engines
after dates listed above. The purpose of this standard is to eliminate the
direct discharge into the atmosphere of fuel drained from the engine fuel
nozzle manifolds after the engine has been shut down. The details of current
smoke exhaust emission standards are listed in Table 16.5. The effective
date, maximum smoke number, engine class, and rated output where the
standards apply are also listed in this table.
554
Gas Turbines
Table 16.4 Landing-Take-off Cycle for Aircraft Engine Emission Standards
Aircraft
Operating
Mode
Taxi/idle
Take off
Climb out
Descent
Approach
TP
(min)
26.0
0.5
2.5
NA
4.5
Engine Class
TF, T3, T8
TSS
(min)
(min)
26.0
26.0
0.7
1.2
2.2
2.0
NA
1.2
4.0
2.3
Table 16.5 Smoke Exhaust Emission Standard–New Aircraft Gas Turbine
Engines
Engine
class
T8
TF
T3
T3, T8,
TSS, TF
TF
Rated
output
all
≥ 129 kN
all
≥ 26.7 kN
≥ 26.7 kN
≥ 26.7 kN
Manufactured
on or after
Feb.1, 1974
Jan. 1, 1976
Jan. 1, 1978
Jan. 1, 1984
Jan.19, 1984
Aug. 9, 1985
TP
≥ 1000 kW
Jan. 1, 1984
Maximum
smoke number
20
83.6 (RO∗ )−0.274
25
83.6 (RO∗ )−0.274
with max of 50
83.6 (RO∗ )−0.274
with max of 50
187 (RO∗∗ )−0.168
* RO is rated output in kilonewtons
** RO is rated output in kilowatts
The 1979 standards are listed in Tables 16.6 and 16.7. In the tables
“T” standards are pounds pollutants/1000 lb-thrust hours/cycle and “P”
standards are pounds pollutant/1000 hp-hours/cycle. Note at that time
there were different standards for newly manufactured and newly certified
engines and that there were HC, CO and NO standards. The 1994 gaseous
exhaust emission hydrocarbon limits are listed in Table 16.8. rPR in Table
16.8 is the engine rated pressure ratio. It should be noted that the hydrocarbon emission standards are based on mass of pollutant emitted to
thrust-hours over the landing take-off cycle that is specified in Table 16.4
and at the power settings that are listed in Table 16.3. Other possible ways
of expressing gaseous emissions are
(i) Pollutant concentration : This has the advantage of being easy to use
but does not provide any guide regarding to the mass of pollutant
being emitted by the gas turbine engines.
(ii) Ratio of mass of pollutants emitted to mass of fuel consumed : This
provides a guide as to how “clean” the combustion system is in the
gas turbine engine. However, it does not reveal the air pollution
Environmental Considerations and Applications
555
Table 16.6 Gaseous Emission Standards Applicable to Newly Manufactured Aircraft Turbine Engines
Engine class
T1
T2, T3, T4
P2
T5
HC
1.6
0.8
4.9
3.9
CO
9.4
4.3
26.8
30.1
NOx
3.7
3.0
12.9
9.0
Table 16.7 Gaseous Emission Standards Applicable to Newly Certified Aircraft Gas Turbine Engines
Engine class
T2, T3, T4
T5
HC
1.0
1.0
CO
7.8
7.8
NOx
5.0
5.0
Effective date
Jan.1, 1983
Jan.1, 1984
effects of the complete engine since different engines have different
fuel consumption characteristics.
(iii) Total mass of pollutants emitted over the LTO cycle : This would
be the most useful one if one is interested in estimating total airport
emissions.
The one selected by EPA, normalizes emissions over the LTO cycle.
16.3
STATIONARY ENGINE EMISSION STANDARDS
The primary federal environmental laws applicable to power generating gas
turbine engines are the Clean Air Act Amendments of 1970, 1977, and
1990. The national and primary ambient air quality standards established
by the Clean Air Amendments of 1970 which have been discussed in Section
16.2 are very important when considering standards that apply to power
generating gas turbine engines.
The current standards are published in the Code of Federal Regulations
and will be discussed in this section. Subpart CG-Standards of Performance
for stationary Gas Turbines of Part 60-Standards of Performance of New
Stationary Sources apply to all gas turbine engines whose heat input at
peak-load is equal to or greater than 10.7 gigajoules per hour (GJ/h) for gas
turbine engine installations, which commenced construction, modification,
or reconstruction after October 3, 1977. Maximum emission limits for the
oxides of nitrogen, NOx , and sulphur dioxide SO2 , are specified in the
standard based on input energy to the gas turbine unit and intended use;
that is, utility or industrial application.
A unit is considered by EPA as an industrial gas turbine engine if less
than one-third of the potential electric output capacity is sold. A unit is
556
Gas Turbines
Table 16.8 Hydrocarbon Gaseous Exhaust Emission Standards – New
Commercial Aircraft Gas Turbine Engines Manufactured on or
after January 1, 1984
Engine class
T3, T8, TF
TSS
Rated output
≥ 26.7 kilonewtons
all
Maximum emissions
grams/kilonewton
rated output
19.6
140 (0.92)rP R
considered as a utility gas turbine engine if more than one-third of the
potential electric output capacity is sold through a utility system.
Current gas turbine engine New Source Performance Standards (NSPS)
are listed, in Table 16.9. The emission limits as listed in Table 16.9 are the
NSPS limits when on a dry basis and converted to 15% oxygen. The limits
are calculated using the following formulae:
(i) For electric utility applications when the heat input is greater than
107.2 GJ/h
14.4
NOx,max = 0.0075
+X
(16.1)
qr
(ii) For all uses where the heat input is between 10.7 and 107.2 GJ or for
base loaded non-utility applications where the output is less than 30
MW.
14.4
NOx,max = 0.0150
+X
(16.2)
qr
In Eqs. 16.1 and 16.2 qr is the manufacturer’s rated heat release rate
at rated load, kilojoules per watt hour, and X is an allowance factor for
fuel-bound nitrogen.
The value of qr cannot exceed 14.4 kilojoules per watt hour (kJ/wh).
The value of 14.4 assumes that the gas turbine engine has a thermal efficiency of 25%. This means that gas turbine engines with thermal efficiencies
higher than 25% will be allowed to emit more NOx than is listed in Table
16.9. No fuel bound allowance is made if the percent by weight nitrogen in
the fuel is 0.015% or less. It is recommended that the readers consult the
latest standards if they are interested in the fuel-bound nitrogen allowance
for fuels with higher nitrogen contents.
The sulphur dioxide standards are the same for all gas turbine engines.
All are limited to 150 ppmv at 15% oxygen and on a dry basis. All operators
shall never burn fuel that contains sulphur in excess of 0.8% by weight.
There are several exceptions to the emission limits listed in Table 16.9.
Two of these are
(i) Stationary gas turbine engines which use water or steam injection for
control of NOx emissions are exempt when ice fog is deemed a traffic
hazard.
Environmental Considerations and Applications
557
Table 16.9 Gas Turbine Engine New Source Performance Standards
Input energy (based
on lower calorific
value of fuel)
Greater than 107.2 gigajoules/h
Use
Electric utility
Emission limit∗
NOx
SO2
(ppmv)
(ppmv)
75∗∗
150
Between 10.7 and 107.2 gigajoules/h
All uses
150∗∗∗
150
Base load equal to or less
than 30 megawatts
Nonutility
150∗∗∗
150
Less than 10.7 gigajoules/h
All uses
None
150
Greater than 30 megawatts
Nonutility
None
150
Regenerative cycle, input
less than 107.2 gigajoules/h
All uses
None
150
All input levels
Emergency,
Fire fighting, Military
None
150
* All values corrected to 15% oxygen, dry basis.
** Use Eq. 16.1
*** Use Eq. 16.2
(ii) Stationary gas turbine engines with a peak load heat input equal to
or greater than 10.7 GJ/h and less than 107.2 GJ/h that commenced
construction prior to October 3, 1982.
The emission limits as listed in Table 16.9 are the maximum levels for new
engines. Levels can be set at lower levels to prevent significant deterioration
in areas that meet National Ambient Air Quality Standards (NAAQS) or
in the areas that do not meet the NAAQS air quality levels.
16.4
NOx FORMATION
It might have been noted in the previous section that the oxides of nitrogen,
are the predominant emissions from stationary gas turbine engines and
the one that is controlled by the standards. The most prevalent NOx
emissions are nitric oxide, NO, and nitrogen dioxide, NO2 . Nitric oxide is
the one mainly formed in the combustion chamber. Factors that influence
the amount of NO formed are
(i) peak temperature,
558
Gas Turbines
(ii) percentage of excess air,
(iii) pressure,
(iv) residence time at peak temperature, and
(v) fuel bound nitrogen.
The peak temperature is attained when the fuel is burned with the
stoichiometric (chemically correct) amount of air. Higher the temperature
of the air at the inlet to the combustion chamber, higher the resulting
equilibrium adiabatic flame temperature.
Burning the fuel with excess air lowers the maximum temperature but
increases the availability of oxygen and nitrogen in the products of combustion. It is known that for a fixed air supply temperature and combustion
chamber pressure, the amount of NO formed for equilibrium conditions increases form 0% excess air to 30% excess air, then starts to decrease even
though the adiabatic equilibrium flame temperature decreases continuously.
It is a known fact that increasing the combustion temperature, pressure,
increases the equilibrium adiabatic flame temperature but decreases the
amount of NO formed.
The preceding discussion assumes that equilibrium has been reached.
The next important thing is to determine the rate at which the products
will reach equilibrium. The basic mechanism presently used to predict
the formation of NO had its origin in the work of Zeldovich and coworkers
around 1946. The reader is referred to Wark and Warner2 for a development
of one mechanism that can be used to determine the rate at which NO is
formed. The equation developed by Wark and Warner is
c+1
(1 − Y )
c−1
(1 + Y )
e−Mt
=
(16.3)
where
Y
=
c =
M
In the above
[NO]
=
[NO]e =
xN2
=
xO2
=
T
=
p
=
=
[N O]
[N O]e
(16.4)
0.5 −7750/T
2.1 × 104 [xN2 ]
e
0.5
T [xO2 ]
5.4 × 1015 (p)
0.5
0.5 −58330/T
[xN2 ]
T
e
(16.5)
(16.6)
equations
concentration of NO
concentration of NO at equilibrium
mole fraction of N2 in the products
mole fraction of O2 in the products
temperature, K
pressure, atm
2 Wark, W. and Warner, C., Air Pollution, Its Origin and Control, EIP A DunDonnelley Publisher, New York, NY, 1976.
Environmental Considerations and Applications
16.5
559
NOx REDUCTION IN STATIONARY ENGINES
It has been shown in Section 16.4 that the higher the temperature and
longer the gases are at that temperature, more nitric oxide is formed. As
shown by the emission limits in Table 16.9, NOx is the main pollutant from
stationary gas turbine engines. The amount of SO2 emitted is limited by
the amount of sulphur in the fuel since this is the only source of sulphur.
Prior to NOx emission controls, gas turbine engine combustion chambers were designed so that the fuel-air ratio in the primary zone was approximately the stoichiometric value; that is, the percent excess air in the
primary zone was 0%. This resulted in maximum temperature.
The maximum temperature can be reduced by designing the combustion
chamber so that the primary zone either operates fuel rich (insufficient air
for complete combustion) or fuel lean (excess air). Both of these conditions
can result in increased smoke (fuel rich) or increased carbon monoxide and
total hydrocarbon emissions (fuel lean). Several methods can be used to reduce NOx emissions such as water or steam injection or staged combustion
or selective catalytic reduction.
The most commonly used method of controlling NOx emissions is with
water or steam injection into the primary zone of the combustion chamber.
The water (or steam) injected acts as a heat sink, resulting in a lower
maximum temperature, thereby reducing the amount of NOx formed. The
rate at which water is injected is approximately 50% of the fuel flow. Steam
rates are usually 100–200% of the fuel flow. An advantage of this method
is that it increases the output from the power turbine due to the increased
flow through the gas generator and power turbines and the higher specific
heat of water. Disadvantages of using water or steam injection are
(i) need of a constant supply of pure water
(ii) possibility of increase in CO emissions
(iii) pressure oscillations in combustion chamber especially with water injection
(iv) penalty on heat release rate
The following results have been reported in literature when water and
steam injection have been tried.
(i) An NOx level of 75 ppmvd for an oil-fired simple cycle with water
flow which is 50% of the fuel flow. Output was increased 3% when
compared with no water injection, and the unit had a heat release
rate penalty of 1.8%.
(ii) An NOx level of 42 ppmvd for a natural gas fired simple cycle with
water flow equal to the fuel flow. Output increased by 5% and the
heat release rate penalty was 3%.
560
Gas Turbines
(iii) An NOx level of 42 ppmvd for a natural gas fired combined cycle with
steam flow of 1.4 times the fuel flow. Output increased 5%, and the
heat release rate penalty was 2%.
(iv) An NOx level of 25 ppmvd for a natural gas fired simple cycle with
GE’s Quiet combustion chamber. Water injection was used with water flow 1.2 times fuel flow. The output increased by 6%, and the
heat release rate penalty was 4%.
(v) An NOx level of 25 ppmvd for a natural gas fired combined cycle with
steam flow 1.3 times fuel flow. Output increased by 5.5% with a heat
release rate penalty of 3%.
Staged combustion is currently being tested by a number of manufacturers. It provides a way of achieving NO emission levels of 25 ppmvd or less
at 15% oxygen without using water or steam injection. Most of the systems
being tested use a two-stage premixed combustor for use with natural gas.
The resulting mixture is lean so the amount of NOx is low.
Selective catalytic reduction involves injecting ammonia into the gas turbine engine exhaust stream. The exhaust gases then pass over a catalyst
where the NOx reacts with the ammonia, (NH3 ), oxygen, (O2 ) and nitrogen, (N2 ) to form water, (H2 O), and nitrogen, (N2 ). When combined with
water or steam injection, it is reported that NOx levels of 10 ppm or less
can be achieved.
One major disadvantage is that the reaction is very much temperature dependent. For a vanadium pentoxide type catalyst, the exhaust gas
temperature range for best operation is 600–750 ◦ F. For this reason, the
selective catalytic reduction method for reducing NOx emissions is limited
to combined cycles only.
16.6
NOISE
Since the introduction of jet-powered commercial airplanes, aircraft noise
has been of concern. A continuous effort has been made to develop the
technology to design a quiet gas turbine engine. The main concern of the
public about aircraft noise is the large number of major airports around
the world that have noise restrictions in the form of curfews, night-time
limitations, flight restrictions, and/or preferred run ways or routings that
take aircraft over water or sparsely populated areas. The major problem
with aircraft noise, in terms of number of people exposed and the frequency
with which they are exposed, occurs in the vicinity of airports.
Unlike emissions, whose sole source is the combustion chamber of the
gas turbine engine, noise has many sources. The two major sources of noise
are
(i) Propulsion system noise
(ii) Aircraft noise other than propulsion system noise, including sonic
boom, flap noise, etc.,
Environmental Considerations and Applications
561
We will consider only propulsion system noise. Propulsion system noise
may be classified into two categories, viz.,
(i) externally generated noise associated with the exhaust gases from the
propulsion system, the propeller of a turboprop powered aircraft, and
(ii) internally generated noise associated with the rotating machinery and
the combustion process.
An important distinction between internally generated and externally
generated noise is that internally generated noise can be suppressed, whereas
externally generated noise cannot be suppressed.
The sources of noise will be divided into three main groups:
(i) Jet or exhaust noise
(ii) Fan noise
(iii) Core noise
Each of these groups is discussed below.
Jet noise results from the mixing of the high-velocity exhaust stream with
the ambient air. A considerable amount of turbulence is generated when
these two streams at different velocities mix. With the increase in intensity
of the turbulence, the noise starts increasing. Researchers have found that
the magnitude of the jet noise increases as the eighth power of the velocity.
The dominating noise of the early turbojet engines, the jet roar, was
generated behind the jet engine exhaust nozzles where the high exhaust
stream mixes with the ambient air.
With the introduction of the nonmixed turbofan engine, there were two
exhaust streams, therefore two sources of external noise. One source was
the turbulent mixing of the fan exhaust steam with the ambient air. The
other source was the turbulent mixing of the core exhaust stream with the
fan exhaust stream and the ambient air.
When a turbojet engine is converted to a turbofan engine with the
same core pressure ratio and turbine inlet temperature, the core velocity
decreases. The amount of decrease and difference in velocity between the
core and fan exhaust streams depends on the fan pressure ratio and bypass
ratio.
Fan noise is one of the major, if not the predominant, sources of noise in
a high-bypass-ratio turbofan engine. Fan noise has different characteristics
depending on whether the tip speed of the fan rotor blades is subsonic or
supersonic. Fan noise usually is separated into three categories:
(i) Broad-band noise, which is essentially the noise generated from the
turbulence in the air and by the air load fluctuations as it passes
across the blade (rotor and stator) surfaces.
(ii) Discrete tone noise, which is noise generated by the fluctuating pressures generated by the interaction between the rotor blades and the
562
Gas Turbines
stationary blades. The frequency of this noise may be predicted by
the rotor rotational speed.
(iii) Multiple-tone noise, which is associated with the shock waves on the
rotor blades caused by supersonic relative flow over the blades.
As higher-bypass-ratio engines are built, the exhaust velocity and therefore the jet noise are reduced. Under these conditions, the turbofan core
noise becomes more important. Core noise consists of compressor noise,
combustion noise, and turbine noise.
Compressor and turbine noise are similar to fan noise, resulting mainly
from the interaction between the rotating and stationary blades. Combustion noise results from the turbulence generated by the burning of the
fuel.
16.7
NOISE STANDARDS
Increased use of jet-powered commercial aircraft, along with a growing concern for the quality of the environment, has resulted in considerable emphasis on the reduction of noise from gas turbine engines since the mid
1960s.
The initial national noise regulations were those in Public Law of USA
in 1969, which are commonly known as FAR 36. These were modified in
1973. It is suggested that the reader consult the Code of Federal Regulations
under Title 14, Part 36, for the latest regulations. Below is an abridged
version of these regulations. The noise regulations depend on the number
of engines, when application is made, and other engine design conditions.
For subsonic aircraft with turbofan engines with a bypass ratio of 2 or
more, the current noise limits are as follows:
(i) If application was made before January 1, 1967, it had to meet stage
2 noise limits or be reduced to the lowest levels economically and
technologically possible.
(ii) If application was made on or after January 1, 1967, and before
November 5, 1975, it had to meet stage 2 noise limits.
(iii) If application was made on or after November 5, 1975, it must meet
stage 3 noise levels.
For aircraft with turbofan engines with a bypass ratio less than 2, the
current noise limits are as follows:
(i) If application was made before December 1, 1969, the noise level had
to be the lowest level reasonably obtainable through use of procedures
and information developed for the flight crew.
(ii) If application was made on or after December 1, 1969, and before
November 5, 1975, the noise levels has to be no greater than stage 2
noise limits.
Environmental Considerations and Applications
563
(iii) If application is made after November 5, 1975, it must meet stage 3
noise limits.
For the Concorde airplane, the noise levels must be reduced to the
lowest levels that are economically reasonable, technologically practicable,
and appropriate for the Concorde design.
Section A36 of the code prescribes in detail the conditions under which
the aircraft noise certification tests must be conducted, the measurement
procedures that must be used to measure the noise levels, and the corrections that must be applied for variations in atmospheric conditions or flight
path.
The measurement points are as follows:
(i) For take-off, at a point 7000 m from the start of the take-off roll on
the extended centerline of the runaway. A typical profile is shown
in Figure 16.1. The take-off profile is defined by five parameters,
including length of take-off roll, climb angle, and thrust change points.
Microphone
Start of take-off roll
7000 m
Fig. 16.1 Typical take-off profile for noise measurements
(ii) For approach, measurements are taken at a point 2000 m from the
touchdown point on the extended centerline of the runaway. A typical
profile is shown in Figure 16.2.
(iii) For sideline, measurements are made at a point parallel to and 450
meters from the extended centerline of the runaway where the noise
level after liftoff is the greatest. The exception is that for Stage 1 and 2
compliance for aircraft powered by more than three turbojet engines,
measurements are made at 0.35 nautical miles from the centerline.
Stage 2 take-off, sideline, and approach noise limits are shown in Figure
16.3. The Stage 2 noise limits are independent of the number of engines.
Stage 3 take-off noise limits are shown in Figure 16.4. Note that for
maximum weights of about 390 ton, the limits are constant but depend on
the number of engines. In all cases the allowable noise limits are reduced by
4 EPNdB for each halving of the 390 ton maximum weight until a maximum
noise limit of 89 EPNdB is reached. This occurs for maximum weights of
Gas Turbines
Microphone
Threshold
2200 m
Effective perceived noise level, EPNdB
Fig. 16.2 Typical approach profile for noise measurements
110
105
100
95
90
10
50 100
500 1000 2000
Maximum aircraft weight, 500 kg
Fig. 16.3 Stage 2 take-off, sideline and approach noise limits
Effective perceived noise level
EPNdB
564
110
105
s
100
re
mo
95
90
ine
ng
e
n3
tha
s
ine
es
gin eng
3
n
tha
n
3e
er
few
10
50
500
2000
Maximum aircraft weight, 500 kg
Fig. 16.4 Stage 3 take-off noise limits
Environmental Considerations and Applications
565
Effective perceived noise level, EPNdB
20 ton for aircraft with more than three engines, 28 ton for aircraft with
three engines, and 48 ton for aircraft with fewer than three engines.
Stage 3 sideline noise limits are shown in Figure 16.5. The allowable
sideline noise limits are independent of the number of engines.
110
105
100
95
90
10
50
500
2000
Maximum aircraft weight, 500 kg
Fig. 16.5 Stage 3 sideline noise limits
Effective perceived noise level, EPNdB
Stage 3 approach noise limits are shown in Figure 16.6. These noise
limits, like the sideline limits, are independent of the number of engines.
110
105
100
95
90
10
50
500
2000
Maximum aircraft weight, 500 kg
Fig. 16.6 Stage 3 approach noise limits
16.8
NOISE REDUCTION
Noise sources and noise standards were discussed in the last two preceding
sections. It is now important to examine what has or can be done to reduce
the noise emitted from a gas turbine engine.
One must keep in mind that design changes in the engine that reduce
noise emissions usually add weight, length and cost. Quite often, features
desirable in reducing noise are in conflict with the best aerodynamic design.
566
Gas Turbines
External noise is caused by the mixing of the exhaust stream with the
ambient air, with the noise level increasing by approximately the eighth
power of velocity. Noise suppressors were used in early commercial airplanes
powered by turbojet engines. Shortly thereafter, low-bypass-ratio turbofan
(BPR ≡ 1) were installed on commercial aircraft, which slightly lowered the
exhaust stream velocity. Most of these aircraft were certified and entered
service long before the current noise regulations. For these aircraft, exhaust
(external) noise dominates. The only effective way to reduce the noise level
of these aircraft is to decrease the exhaust stream velocity. This, of course,
can be done by replacing the engine on the aircraft with a turbofan engine
with a higher bypass ratio.
An effective way to reduce the maximum exhaust velocity, and therefore
the exhaust jet noise, on mixed (pressure-balanced) turbofan engines is
to use an exhaust gas mixer behind the turbine. An exhaust-gas mixer
mixes the high-velocity core exhaust gas stream with the lower-velocity
cold stream that has passed through the fan but bypassed the combustion
chamber and turbine. It reduces the maximum exhaust stream velocity and
eliminates the external turbulent mixing of these two streams. However, it
does add weight to the engine.
Turbofan engines with bypass ratios of 5 or higher have been installed
on many of the commercial aircraft in the last twenty years. With highbypass-ratio turbofan engines, the exhaust velocity, and therefore the exhaust noise, is reduced considerably. With the reduction in the exhaust
noise, additional internal noise is heard and becomes a problem. Internal
noise, unlike external noise, may either be eliminated or confined.
One way to confine the noise is to install an acoustical liner at the
inlet to the engine and/or in the exhaust duct. Conventional acoustical
treatment is quite effective and is currently used but does add weight to
the airplane.
Other ways to reduce internal noise and the disadvantages are to:
(i) Decrease the fan tip speed. The lowest-noise fans are generally subsonic. Low-speed fans usually require more compressor stages, which
leads to a heavier and more costly engine.
(ii) Increase the spacing between the rotor and stator. Noise is reduced
as the spacing increases, but large spacing tend to increase the length
and weight of the engine.
(iii) Eliminate the inlet guide vanes, which eliminates one of the rotorstator interactions.
(iv) Change the number of rotor and/or stator blades. Changing the
ratio changes the frequency of the noise. If the number of blades is
decreased, it may affect the turbomachinery performance.
Environmental Considerations and Applications
16.9
567
ASSESSMENT OF THE GAS TURBINE
The most outstanding characteristic of the gas turbine is that of flexibility
of arrangement, with respect to choice of cycle, choice of plant arrangement,
type of component and type of fuel. This is borne out by actual developments over the past many years, which have shown gas turbines in power
output from a few kilowatts to outputs of 30,000 kW in straightforward
shaft units and of even greater equivalent thrust horsepower in jet engines.
The most outstanding physical characteristics are its small size and low
weight per unit power output. The values are much lower than those of
other prime movers. Within limits the gas turbines can burn a wide range
of fuels with little modification and it is possible to develop a gas turbine
which can burn any fuel from a gas to residual oils.
It is these factors above all which have led to the use of the gas turbines as it stands at the present time, even though the theoretical analysis
has shown that the efficiency, particularly at part load, and the range of
satisfactory operation are not very attractive except for more complex arrangements. There are, however, other features which lead to the choice
of the gas turbine when an overall balance sheet is made of operation and
which in some particular applications are overriding.
Below, an attempt is made to summarize the potentialities of the gas
turbine, but it should be noted that the relative values of a particular
characteristic vary with application.
(i) High Specific Output The power output on weight and volume basis
is quite high except the most complex cycle.
(ii) Relatively Lower Efficiency than Comparable Primemovers Diesel
and petrol engines are usually more efficient than gas turbine in the
lower output range. This is true both at the design point and, more
particularly at part-load. In the higher output range where the diesel
engine becomes unacceptable due to size, weight and number of separate unit required, the gas turbine has to compete with the steam
turbine. The efficiency of the steam turbine is highly variable depending upon the size and upon the complexity of its cycle (condensing
or non-condensing, reheat, etc.,), but it is comparatively high. A
closed-cycle gas turbine can compete with all but the largest and
most complex steam plants.
(iii) High Air Rate This may pose problems in the air intake and disposal
of exhaust gas particularly for mobile units (marine and automotive).
If advantage can be taken of large volume of exhaust gas at relatively
high temperature for use as an auxiliary heating medium, then the
high air-rate is useful and the overall efficiency of the complete thermal plant is high. No such problem with respect to air and gas ducts
occurs in the closed-cycle plant.
(iv) Low Maintenance Cost Here the gas turbines are comparable to the
steam turbines in lubricating oil consumption, and is very superior to
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Gas Turbines
the reciprocating IC engines. With respect to inspection and repair
or replacement of components, the gas turbine has advantage, as it
is a much simpler plant. It has many fewer accessories than either a
steam power plant, a diesel or a petrol engine. It should also be noted
that because gas turbine cycle processes are carried out by discrete
components, modification is usually easier than in other plants. Thus
improved components, such as compressors, combustors or turbines,
can be introduced with minimum disturbance to the rest of the plant.
(v) Freedom from Auxiliary Services For most of the applications, water
supply is not required, which is considered as an advantage with respect to location. Ignition system is simple as it is necessary only for
starting. However, for aircraft units, the control and electrical systems are complex, owing to the large variation of ambient conditions.
However, this is true for any power plant for an aircraft. The fuel
control system for the gas turbine is no more complicated than the
carburettor in its developed form for aero-engines.
(vi) Low Installation Cost Because of its high output per unit weight and
volume, both the building for housing the plant and actual floor foundation required are much smaller and simpler than for other power
plants of similar output. This may be an important point from economic considerations.
(vii) Starting of a Gas Turbine Starting has both favourable and unfavourable aspects. On the favourable side is that except for very
large units, it can be started and put on load in very short time. Usually, the time is much shorter than for the alternative prime mover for
the particular application. Turbojets of enormous power are ready for
aircraft take-off in a matter of seconds. Gas turbines for large output
for electrical generation may require minutes of warm-up time. It is
usually much less than the corresponding steam turbine, even assuming steam available at throttle. This characteristic is due to the light
construction, which requires the minimum of large metal masses of
great thermal capacity.
On the unfavourable side is the power required from rest to selfsustaining or idling speed. Not only component efficiencies are very
poor at low speed, but the turbine temperature must be limited for
the sake of metal properties and to avoid compressor stalling. It is
necessary to accelerate the unit after combustion is initiated up to a
speed at which it is self-driving at a safe turbine temperature. This
speed is usually about one-third of the design speed.
The electric motors of some fair size are required which will sustain high current for comparatively larger time. Cartridge or ‘rocket’
starters are used successfully for aircraft turbines when self contained
starting means are required otherwise auxiliary electric generators are
needed. Again on credit side, starting at low ambient temperatures
is usually easier than with reciprocating engines, because of the lack
of contact resistance.
Environmental Considerations and Applications
569
(viii) Effect of Ambient Conditions The effect of variation of pressure is not
serious and comparable to that on reciprocating engines. The effect
of temperature variation is very marked. The plant must be designed
so that a certain minimum output is maintained at the highest intake
temperature, thus probably having excess capacity for major part of
its operation. However, there are many applications when increased
output is welcome at low inlet temperatures.
(ix) Multifuel Capability A gas turbine can be designed to utilize gaseous
fuels or distillate of liquid hydrocarbons with ease. The problem of
combustion of heavier and cheaper fuels is not completely solved. If
the problem of deposit and corrosion due to ash content can be solved,
then there is no inherent reason why any liquid fuel cannot be used.
(x) Capital Cost In this case no generalization can be made. Because it
is largely number of units which determine the first cost and the industrial production is not yet sufficiently great to have warranted the
standardized and mass production methods like other prime movers.
For small units, the cost of gas turbines is so much greater than that
of reciprocating engines. Therefore, it may seem to be improbable for
it ever to compete.
16.10
TYPICAL APPLICATIONS OF GAS TURBINES
Research efforts and investigations have gone into various aspects of gas
turbines in the constant endeavour to improve the performance, increase
the overall efficiencies, extend the applications and, in general, to overcome
the practical problems and difficulties associated with these power units.
Majority of these aspects have already been dealt in the last 14 chapters.
Finally in the following sections, an account is given about typical applications of gas turbine to locomotive, aircraft, marine and stationary plant.
The possible application of gas turbine to automobiles is also discussed.
16.11
THE SMALL GAS TURBINE APPLICATIONS
About sixty years ago it was clear that in the aircraft field the gas turbine
has established itself for military purposes, and even on the commercial
side the reciprocating engine was fast becoming obsolete. It was, therefore,
natural to examine how far the turbine invasion could be expected to penetrate into the automobile and small industrial outputs, say 200 kW and
lower. Even at first sight, this area does not appear promising because of
the advanced state of development of petrol and diesel engines in this range,
together with their relatively very low capital cost due to mass production
for automotive use. A gas turbine of small size is likely to be less efficient
than that of large ones. It is mainly due to Reynolds number effects and
difficulty in maintaining the same relative accuracy of part dimensions and
of small clearances. Nevertheless, numerically there are probably many
more gas turbines of 200 kW and less in use than those of larger output.
570
Gas Turbines
It is true that their use is predominantly in the military field. However,
their application in shaft-power units, a field in which existing reciprocating
engines dominate is not yet predominant.
16.12
ELECTRIC POWER GENERATION APPLICATIONS
The efficiency of the gas turbine is not high enough to allow it to compete,
with the steam turbine for continuous power generation. However, its other
characteristics have led to its use in number of instances.
Many gas turbine installations have been built since 1947, for standby
and peak-load purposes. These vary from about 4,000 kW to 24,000 kW
output, the 10,000 kW to 15,000 kW installations being the most favoured.
In January 1948, fifteen Brown Boveri plants, with a total capacity of 92,000
kW, were in operation for power generation purposes in India.
Most of the existing gas turbines are operated only on oil or gas. The
10,000 kW gas turbine, working with a normal air inlet at 20◦ C and a cheap
fuel oil, is considered superior to the steam turbine, but above this output
the latter scores on the ground of efficiency.
If the exhaust gases from the gas turbine power station can be used
for a waste heat boiler or a water heater, it should prove a little more
advantageous in its field, provided that a cheap fuel oil or coal can be
burned.
Many open-cycle plants are being used for electrical generation, a number of them being only simple cycle plants with efficiencies of 18–22%. The
reasons are many. One of the most important reasons being that the location where water is either non-existent or very sparse. Thus several are in
use, in desert areas, notably in the Middle East. Many of the installations
are in the oil-fields or at oil refineries. In such installations natural or process gas is either ‘free’ or relatively cheap. This provides the best possible
fuel for trouble-free combustion and compensating for low efficiency. Some
generating units have heat exchangers, thus yielding higher efficiencies.
Gas turbines are particularly useful for peak-load or stand-by units.
They are self-contained, require no additional steam capacity. They can
be put on load with the minimum of delay. They occupy less floor space
than other plants for the same capacity. They can also increase the station
capacity beyond their own nominal output, because the exhaust gas can be
used for air or water preheating, thus raising the existing boiler capacity.
16.13
MARINE APPLICATION
The application of the gas turbine to marine use has been slower than originally anticipated. The slow development in this field is natural on account
of number of ships being built is less. The high capital cost of single unit is
quite high. The great loss involved in breakdown of equipment above and
beyond the direct repair cost involved also inhibit marine applications. The
relatively high fuel consumption of simple gas turbines, coupled with the
Environmental Considerations and Applications
571
difficulties associated with combustion of cheaper fuels, has counteracted
expectations. Nevertheless, there are apparent advantages and these are
gradually being demonstrated in test vessels.
The major advantage once again is low weight and compactness of the
gas turbine. Engine rooms of all vessels are always crowded, owing to the
importance of each tonne and cubic metre of capacity for pay load. Gas
turbine can ‘offer’ either more room or more power in the same space. Other
attractive features are the low maintenance costs and the ability to replace
individual components with minimum effort and loss of time.
There is considerable difference in the requirements of mercantile and
naval vessels. The former require a power plant which operates at full
speed at most of the time, reduced speed being necessary only during severe
weather and for maneuvering in close quarters. A naval vessel, on the other
hand, spends a large portion of its life at relatively low speed, requiring top
speed only in emergency. Thus part-load efficiency, while highly desirable,
is not so important for mercantile marine vessels. It would seem that for
the former use, a very high degree of reliability while in latter case, it is
high part-load fuel consumption which is more important.
The gas turbine also has the advantages of availability and absence of
transmitted vibration. The high air-rate poses some problems of intake and
exhaust ducts, which must be of considerable size for the open-cycle. On
the other hand, closed-cycle seems very well suited to marine propulsion.
Another problem lies in the provision of power for astern movement.
It is not possible to utilize a condenser vacuum as does the steam turbine
installation. A separate reversing gas turbine leads to windage and disc
friction losses. Reversing gear mechanism with torque converter is possible, though difficult for high power. Electric propulsion, though readily
reversible, adds more cost and maintenance. The reversible-pitch propeller
has not yet been extensively used.
For naval gas turbines, one solution for part-load efficiency is the use of
the closed-cycle, and another is the use of multiple units. For the later, the
long-term low output can be supplied by a relatively small unit with good
efficiency. Several other simple cycle units can be made to standby which
can be coupled immediately to the main shaft when required for high speed
operation. For small naval craft, such as motor torpedo-boats or air rescue,
boats, the light weight, high-output gas turbine is an excellent proposition.
It enables much higher speeds to be attained than would be possible with
other types of prime movers. The first marine installation of a gas turbine
was successful and has been followed by several others in many countries.
16.14
GAS PUMPING APPLICATIONS
Gas turbines coupled to centrifugal compressors for pumping natural gas
have proved to be of one of the most successful uses. The major pumping
application is in cross-country pipe-lines, in which booster stations are required at frequent intervals. Such locations are often in isolated regions,
in which water is difficult to supply. Furthermore, the gas turbine speed is
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Gas Turbines
perfectly suited to the centrifugal compressor for the natural gas. The low
maintenance and low installation cost make it considerably more attractive
than a reciprocating gas or oil engine. Again, the available fuel, that in the
pipe line itself, is very suitable. The majority of such units are simple cycle,
single-shaft units, although for some uses, a free turbine is advantageous.
The addition of a heat exchanger improves efficiency at the expense of first
cost.
16.15
LOCOMOTIVES APPLICATIONS
With one outstanding exception, the gas-turbine has not to-date proved
popular for locomotive use, although certain apparent advantages led to
earlier prediction that this would indeed be so. Almost any power plant
would be an advantage over the reciprocating steam locomotive. In almost
all countries the universal replacement has been the diesel engine. In India,
steam locomotives are being replaced by diesel locomotives and by electric
locomotives on all trunk routes. But due to the oil shortage, felt severely
after 1973, Government of India is revising its policy of switching over
completely to diesel traction. It is quite likely that all the trunk routes
may be completely electrified and central power stations based on coal may
be built all over the country for meeting the severe oil shortage.
The gas turbine has advantages of great power in a small volume and of
immediate availability. The major disadvantage is low efficiency, even with
a heat exchanger, which cannot be of high effectiveness without having a
large space. The gas turbine is also not suited to short journeys, because of
its poor part-load efficiency. Thus in India, the gas turbine locomotive does
not seem likely to prosper. In USA, on the other hand, loads are higher
and distances are greater.
Diesel power requires electrical traction and it is possible for the gas
turbine to use a direct mechanical drive, thus eliminating some initial cost
and complication. Diesel engines require considerably more costly maintenance than gas turbines and the reduction of this item would allow the gas
turbine to be considered very favourably if, at the same time, the fuel cost
could be made comparable. Here the solution of the problem of heavy fuel
combustion would be important or, alternatively, the introduction of coal
burning. In the USA, development of a direct coal-burning unit has been
developed using pulverized fuel. The major problem is of the erosive effect
on the turbine of the inevitable large quantity of ash. In Canada, effort has
been directed to the exhaust-heated cycle, eliminating the turbine erosion,
but requiring a rather large heat exchanger.
16.16
AUTOMOTIVE APPLICATIONS
Ever since the gas turbine came to public notice, one of the most often
asked questions has been its adaptability for automotive use. Its apparent
advantages are again weight and volume, simplicity (simple ignition system
and absence of water cooling), good starting torque characteristics (with
Environmental Considerations and Applications
573
a power turbine), and absence of vibrations. Against these assets there
are disadvantages like poor part-load fuel consumption and necessity for a
reduction gear of large ratio.
In spite of the ultimate advantages of lightness, simplicity and compactness, absence of necessity for a cooling system and negligible oil consumption, the use of gas turbines for road transport is much less favourable.
Even then, most of the major automobile manufacturers in Europe and the
USA have engaged in development on suitable gas turbines, from 100 to
250 kW.
Considerations of efficiency and load control requirements rule out the
simple compressor-turbine coupled arrangement with power take-off from
the turbine shaft. Apart from this, the problem of mechanical transmission
will be difficult. For automobile application it becomes necessary to use
either electrical transmission which is expensive and heavy, or a separate
power turbine independent of turbocompressor shaft. This power turbine
is geared to the road wheels, while the compressor unit can run upto the
speed with the car stationary. The exhaust from the compressor turbine
would thus develop a fluid drive effect on the power turbine and exert on
it a starting torque.
Low thermal efficiency remains the greatest hurdle in the development
of a small gas turbine for an automobile. Unless a heat exchanger is used,
the specific fuel consumption will probably be twice that of the average
petrol engine and three times that of the diesel. While the commercial user
will assess lower capital and maintenance costs a set-off against higher fuel
consumption and cost, this aspect will appeal less to the private motorist,
whose accounts rarely include depreciation and interest.
16.17
AIRCRAFT APPLICATIONS
Nearly all military aircraft are powered by turbojets, from trainers to latest
supersonic fighters. The age of the turbojet and turboprop commercial airliner has long been started. The turbojet is satisfactory up to a flight Mach
number of 2.5, above which the ram-pressure developed leads to diminishing returns. For military use, it would appear that the area of turbojet is
about two decades old, with rockets already taking over.
16.18
PROCESS APPLICATIONS
By process applications, we mean, those in which the major purpose of
the gas turbine is not supplying shaft power, although some of the power
extraction is possible, but in supplying compressed air or gas for physicochemical processes. The air may be bled off from the compressor for some
process use, this representing the excess power of the turbine over the compressor. Alternatively, all the compressed air may be passed, through a
‘reactor’ of some nature, in which it is required to perform a chemical or
physical operation, the air then passing to, the turbine, with possibly some
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Gas Turbines
fuel being required to raise its temperature sufficiently for an energy balance. In other cases, a normal combustor with outside fuel supply is used
in the ordinary way, however, the gases are not completely expanded in
the turbine, passing to the process at elevated temperature and pressure.
This last case is analogous to the turbojet, but with the gases used for
chemical process rather than for propulsion. There are many such different
processes in the petrochemical field, catalytic cracking, oxygen production
and so forth.
Included in this category are many different ways of utilizing the gas
turbine as an auxiliary component in conjunction with steam cycle. The
possible arrangements are many and will undoubtedly be tried out in future. Perhaps one factor militates against their use in new steam power
installations is that the latter are now usually of very large output and of
high efficiency, utilizing elevated temperatures and pressures. The latter
are now quite, often in the supercritical region and represent a considerable
advance in technology, so that the inclusion of another new element, the
gas turbine is not required at the present time.
16.19
ADDITIONAL FEATURES OF GAS TURBINE ENGINES
Having known a power unit which functionally competes with the petrol
and diesel engines, let us look on other features which characterize the gas
turbine.
16.19.1
Exhaust
This aspect is of the greatest importance and will undoubtedly play a crucial part in the general acceptance and introduction of gas turbine for road
Vehicles. In USA, particularly at present, there is growing concern over
the quality of vehicle exhaust gases and newer and newer legalizations are
introduced which makes it obligatory for any exhaust system to control
such emissions as carbon monoxide, oxides of nitrogen and unburnt hydrocarbons. The turbine exhaust, due to high efficiency possible with its
combustion system and lack of flame-chilling, can meet these stringent demands in the near future.
16.19.2
Easy Starting
Due to very low compression ratio during the starting cycle the main resistance to spinning over is the bearing drag. These can be very low and
hence the gas turbine is noted for its ability to start reliably at very low
temperatures without any additional starting aids. Whilst this is essential
for military duties it is also one of the most useful attribute for general
operation. Starting at ambient temperatures of −30 ◦ C are quite common
and −60 ◦ C has also been achieved in some conventional small gas turbines.
Environmental Considerations and Applications
16.19.3
575
Smoothness
This applies in two ways :
(i) in the lack of reciprocating masses which cannot be fully balanced,
and
(ii) in the two shaft characteristic preventing any torque ‘snatch’ on acceleration or deceleration.
The lack of any physical vibration makes the operation of any vehicle
less fatiguing and when the comparison is made with diesel trucks this
is an important factor in reducing driver fatigue. The excellent torque
characteristic permits fewer gear changes, all of which will be very smooth as
in the best standard of automobile automatic transmissions. This all adds
upto a general driving smoothness which probably has to be experienced
to be believed.
16.19.4
Cooling and Heating Aspects
In piston engines it is imperative that cooling is arranged around the combustion areas to avoid excessive metal temperatures, and this plus the general heat to oil by churning and from friction, requires a considerable cooling
system. With the gas turbine no cooling system around the combustion
chamber or indeed anywhere in the gas paths is required, and it is only
necessary to cool the oil which is keeping the bearings and associated gear
trains cool and lubricated. Hence, the amount of parasitic weight of the
cooling system is greatly reduced.
On the other hand, the delay in bringing that engine temperatures upto
normal is very short and heaters are able to operate as soon as the engine
is upto the speed. This is of considerable advantage in cold climate where
devising of screens or equivalent is frequently required before a vehicle can
commence operation.
16.19.5
Multifuel Ability
Whilst this matters little for normal automotive use where a suitable fuel
will always be available, it is of great benefit in military application. To be
able to run on any fuel, ranging from heavy oils to aviation fuels and including paraffin and diesel, is highly desirable as virtually any fuel available in
any part of the world will permit the engine to operate satisfactorily. Some
fuel systems may require simple adjustments to cover the wide range but
others need none and can change fuel while running.
16.19.6
Filtration and Noise
The gas turbine uses large quantity of air than its piston counterpart of
similar power and this results in large air inlet and exhaust ducts. Atmospheric pollution has, in time, detrimental effect on the efficiency of the
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Gas Turbines
compressor by depositing a thin layer of oil, dust rubber, etc., within the
diffuser passages. To reduce, this it has been found expedient to filter the
air before it enters the engines These filters are, therefore, bulky compared
with those on a piston engine but are usually very light as paper elements
are used, and also have a considerable intake silencing effect.
The exhaust noise level of an engine without heat exchangers is high,
but the inclusion of heat exchanger greatly decreases this. If the engine
compartment is well designed to avoid gaps and holes and is lined with
suitable absorbing material, the general level of noise can be as low as
comparable piston engine installations.
16.19.7
Engine Life
There has been no mention yet of an important feature of the gas turbine
engine, namely its excellent life between overhauls. The aero-gas turbine
has proved conclusively that in that application at least the turbine easily
surpasses piston engines. This is principally because, the turbine components work under steady state conditions, i.e., a turbine stress remains
constant and in one direction, whilst the piston engine equivalent, say a
crankshafts is subjected to alternating stresses continually. There is also
the aspect of the sealing by contact, in the piston engine against sealing
by close clearances, avoiding friction, in turbine. In addition aero engines
work at a high power for most of their lives which is very demanding on
lightweight piston engines.
For road use the picture is very different. The amount of time the
automotive engine is near full power is very small. Over half of the engine’s
life is spent below quarter power. This has permitted piston engines to be
developed with acceptable lives.
In fact, engines can now be said to have sufficient life under normal conditions to outlast the rest of the vehicle. This means that the advantages of
long life for gas turbine as applied to automobiles are diminished. However,
there are benefits :
(i) the complete lack of vibration will reduce failures in other components
of the cars, transmissions, etc., and
(ii) the likelihood of premature failure will be reduced, which should permit improved warranties to be offered to the public, buying gas turbine cars.
The case for the small gas turbine in vehicles has been made by many
knowledgeable people-extolling its virtues and showing how its limitations
can be overcome. Why then are we not all driving around smoothly in gas
turbine cars? The answer lies in two reasons :
(i) the very high investment of money, plant and engineers, in existing
reciprocating piston engines, and
(ii) the economic aspects of manufacturing the gas turbine engine.
Environmental Considerations and Applications
577
The first will be overcome when the second is acceptable and by some public
demand such as non-polluting exhaust, or even simply a sales demand.
The real problem facing the introduction of the gas turbine engine is its
manufacturing cost. True cost must include considerations of the life of the
engine in service, and the running costs including any repairs, in addition
to the basic manufacturing cost. We already know that the operating costs
will be comparable to, or little lower than the piston engine so the real issue
lies with the manufacturing cost of the engine.
Considered in overall terms, there are less components and a lower overall weight of the engineering hardware, and this should automatically result
in a lower overall cost of the unit. The difficulties lie in two fields:
(i) the greater proportion of more exotic and hence expensive materials,
and
(ii) the development of different manufacturing techniques.
The gas turbine ‘art’ is at the most interesting stage. The learning
curve has overcome the majority of technical problems and is now fairly
and squarely on price reduction aspects and tremendous strides are being
taken by all the companies involved with small gas turbines. When such
very large amount of money is being invested in the development of gas
turbines by big American automotive companies, there is bound to be a
rapid progress towards mass production of their engine and vehicles using
it. As in all competitive manufacturing, components can only be made for
their minimum price by mass production, and hence the break-through of
gas turbines is likely to be sudden when it occurs. There is no justification
for expecting that piston engines will be replaced overnight, there being
many fields where the gas turbine will show insufficient benefit to warrant
the change.
In higher powers, over 250 kW the turbine is already becoming competitive in cost with piston engines but it is unlikely that, below 50–100
kW, will it ever be commercially competitive to replace the established
reciprocating engines?
The final aspect which must be remembered at all times is the potential
of development available in the small gas turbines. Power will increase
for the same weight and bulk, fuel consumption will drop further, and
technical developments will further improve the torque characteristics and
the control aspects. Ceramics will, and are beginning to replace the metals
used in the hottest parts of the engine. They have tremendous potential
not only because of their ability to withstand very high temperatures but
also because of their intrinsic low cost and general availability.
With the maximum permitted loading of vehicles likely to become something of the order of 50 tonnes, engines of about 300 kW will be required
for commercial vehicles. For this output, the gas turbine is worthy of serious consideration. Another favourable factor is widespread development
throughout the world of the motorway Net works on which relatively constant speeds of operation for long periods are practicable. The stage is set
578
Gas Turbines
for the gas turbine to make a big impression on the power units of the world,
and when it does it should make the air breathe cleaner and transportation
more comfortable.
16.20
TRENDS IN THE FUTURE DEVELOPMENT
(i) There will be higher compressor airflows, pressure ratios and efficiencies with fewer compressor stages, fewer parts and lower cost.
Variable-pitch fan or compressor blades will provide reverse thrust
for braking.
(ii) Engines will be developed to have a lower specific fuel consumption
resulting from component design improvements and other changes.
(iii) Increased turbine efficiencies with fewer stages to do the necessary
work, less weight, lower cost and decreased cooling air requirements
will be developed.
(iv) Increased turbine temperatures using better materials and improved
cooling along with new manufacturing techniques will be used.
(v) Less use of magnesium, aluminium and iron alloys but more of nickel
and cobalt-base alloys plus increased use of composite materials.
(vi) Burning fuel more cleanly with less pollution which will run more
quietly. It will make this type of power plant less hostile to the
environment.
(vii) More airborne and ground engine-condition-monitoring equipment
will be used, such as vibration and oil analyzers, and radiometer
sensors to measure turbine blade temperature while the engine is operating.
Review Questions
16.1 Give a brief account of aircraft emission standards.
16.2 What are the standards for the emission of pollutants from a stationary engine?
16.3 Explain the mechanism of NOx formation and also the methods for
its reduction.
16.4 What is noise? What are the sources of noise in a gas turbine engine?
16.5 Explain the noise standards in connection with gas turbine power
plants.
16.6 What are the various methods adopted for noise reduction?
16.7 Give a brief account of the assessment of gas turbine power plant.
Environmental Considerations and Applications
579
16.8 Bring out clearly the role of gas turbine for the following applications:
(i) locomotive, (ii) aircraft, (iii) marine, and (iv) stationary power
plant.
16.9 What are the additional features of gas turbine engine?
16.10 Explain the trend in the future of gas turbine.
Multiple Choice Questions (choose the most appropriate answer)
1. The first legislation enacted to control air pollution was by
(a) UK
(b) Germany
(c) USA
(d) Japan
2. The ever first act for pollution control was enacted in the year
(a) 1940
(b) 1955
(c) 1960
(d) 1962
3. The APU classification of EP1 class of engine is for
(a) all aircraft with piston engine
(b) all aircraft with turboprop engine
(c) all aircraft for supersonic flights
(d) all engines installed in or on an aircraft exclusive of the propulsion engine
4. The TP classification for engine of EPA class for
(a) all aircraft with turboprop engine
(b) all aircraft with turbojet engine
(c) all aircraft with turbofan engine
(d) any engine used in aircraft
5. Gaseous emission standards were set in the year
(a) 1969
(b) 1975
(c) 1979
(d) 1985
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Gas Turbines
6. The emission standard applicable to newly manufactured aircraft turbine engines of class T1 is
(a) HC < 2%
(b) CO < 10%
(c) NOx < 5%
(d) all of the above
7. The dominant factor that influence the amount of NOx in a gas turbine engine is
(a) pressure
(b) peak temperature and excess air
(c) fuel nitrogen
(d) residence time
8. Disadvantage of using water injection to reduce NOx in aircraft gas
turbine engines is
(a) carrying water is very heavy
(b) increased heat release which increases the temperature
(c) steady increase in pressure in the combustion chamber
(d) possibility of increase in CO emission
9. Jet is the result of
(a) mixing of fuel and air
(b) due to altitude
(c) mixing of high velocity exhaust stream with ambient air
(d) due to vortex generation behind wings
10. The major application of gas turbine is for
(a) aircraft application
(b) automotive application
(c) locomotive application
(d) gas pumping
Ans:
1. – (c)
6. – (d)
2. – (b)
7. – (b)
3. – (d)
8. – (d)
4. – (a)
9. – (c)
5. – (c)
10. – (a)
17
ROCKET PROPULSION
INTRODUCTION
The history of rocket propulsion can be traced back to eighth century
when the Greeks learnt the art of flying fire. Though the actual period
regarding the use of the principle of rocket is quite uncertain, the Chinese
were, perhaps, the first to use this principle in fire arrows as early as 1234.
In spite of the fact that a large number of investigators were working on
rockets, it was only during the two World Wars that rocket was given its
due importance for warfare and other uses.
It was in 1919 that Dr. R.H. Goddard from USA published what is
called the first report on rockets. After that, with the advent of German V-1
and V-2 rockets, the rockets started establishing. First came the V-1 rocket
working on constant-volume explosion type combustion; it was actually a
pulsejet. Then came, the V-2 rocket a liquid propulsion device, using liquid
oxygen and ethyl alcohol in water as main propellants. During and after
the Second World War, Americans studied the captured V-2 rockets and
afterwards rockets started establishing themselves for war purposes. The
role of the rocket as an aircraft power plant is becoming important day by
day. Whereas the primary application of the use of rockets is at present
military, they do offer promise for long-range high-speed transport aircraft
and also as a power plant for space travel.
All the post war efforts culminated into the magnificent moon landing
by Neil Armstrong in 1969; and the success of the Russian Lunar vehicle,
“Lunokhod” has opened new horizons for the exploration of interstellar
systems and a great boost to the insatiable explorative instinct of the man.
In many ways the rocket is similar to the other jet power plants that
have been discussed in the chapter on Jet Propulsion. Many of the principles that have been developed, apply directly to the rockets. One major
difference between the rocket engine power plant and other jet propulsion
systems is that it carries its entire propellant (oxidizer and fuel) with it.
Other jet propulsion systems depend upon atmospheric air. Thus the basic difference is that the air breathing jet propulsion engine utilizes oxygen
from its surrounding medium, whereas the rocket engine utilizes oxidizer,
582
Gas Turbines
which it carries in its own tanks. The fact that it does not depend upon
atmospheric oxygen allows it to operate at very high altitudes and even
in vacuum. This fundamental distinction between the two types of the
power plants produces radically different performance capabilities, which
are summarized in Table 17.1.
Table 17.1 Performance Differences of the air-breathing engine and the
rocket engines
Air-breathing engine
Rocket engine
Altitude limitation.
Thrust decreases with
No altitude limitation,
travel possible.
space
altitude.
Thrust increases slightly with
altitude.
Rate of climb decreases with altitude.
Rate of climb increases with
altitude.
Engine ram drag increases with
flight speed.
Engine has no ram drag; constant
thrust with speed.
Flight speed always less than jet
velocity.
Flight speed not limited, can be
greater than jet velocity.
Reasonable efficiency and reasonable flight duration.
Low efficiency except at extremely high flight speed for
small duration.
It is noted that all the advantages of rocket engine may be offset by its
low efficiency. Overall propellant consumption is high because the rocket
must carry its oxidizer as well as fuel. For these reasons, the rocket power
plant should not be considered as a direct competitor to air-breathing engines. The rocket is really a power plant that can attain performance
capabilities beyond the scope of the air-breathing engines.
17.1
CLASSIFICATION OF ROCKETS
The necessary energy and the momentum that can be imparted to a rocket
propulsion device by a propellant can be achieved in many ways. For example, chemical, nuclear, or solar energy can be used or an electrostatic or
electro magnetic force can also impart the momentum. Alternatively, the
propellant can be heated with the help of electric power and expelled in
a nozzle. Figure 17.1 gives the classification of rockets on the basis of the
methods used for accelerating the propulsion device.
Chemical rockets are so far the most widely used rockets. They depend
upon the burning of the fuel inside the combustion chamber and expanding
it through a nozzle to obtain thrust. The important character of chemical
rockets is that the propulsive energy is obtained from within the fuel and
the oxidant. The propellant may be solid or liquid. In the free radical
propellant system, the propellant has certain free radicals in them, so that
Rocket Propulsion
583
Rockets
Chemical
Solid
propellant
Liquid
propellant
Nuclear
Free Fusion
radical
Fission
Electrodynamic
Ion
rocket
Plasma
rocket
Photon
rocket
Fig. 17.1 Classification of rockets
when they recombine enormous amount of energy is released which can be
used for obtaining propulsion.
The vast store of the atomic energy can be utilized in case of nuclear
propulsion. Fission or fusion both can be used to increase the energy of the
propellant.
In case of electrodynamics propulsion devices a separate energy source
is used and electrostatic or electromagnetic forces accelerate the propellant.
The propellant is either discrete charged particles that are accelerated by
electrostatic forces or a stream of electrically conducting material which
can be accelerated by electromagnetic forces.
The first type is called electrostatic or ion rocket, and the second type is
called plasma jet. Photon propulsion seems to be the ultimate. In this the
propellant being expelled in the form of photons traveling with the velocity
of light. As on date photon propulsion does not seem to be technically
feasible.
17.2
PRINCIPLE OF ROCKET PROPULSION
The rocket engine in its simplest form consists of a combustion chamber
and an expanding nozzle. The fuel and the oxidant, when ignited cause the
combustion to proceed at a very fast rate. The exhaust gases produce the
required propulsive forces. In other words propellant gases, that are generated in the combustion chamber by burning propellants, are expanded in
a nozzle to a supersonic velocity, thereby, requiring a convergent-divergent
nozzle. These high velocity gases going out of the nozzle produce the thrust
and propel the rocket. Now, the power required to produce an exhaust jet
velocity, cj , can be written as
P
=
1 2
ṁc
2 j
(17.1)
Further, the thrust can be written as
F
=
ṁcj = mα
(17.2)
584
Gas Turbines
where
m = mass of the rocket
ṁ = mass rate of consumption of propellant
cj = exhaust velocity from the nozzle
F = thrust
P = power required to give an exhaust jet velocity cj
α = acceleration of the rocket
Combining Eq. 17.1 and 17.2, we get,
αcj
P
=
m
2
(17.3)
From the above equation it is clear that the velocity and the thrust that can
be obtained from a given type of rocket propulsion system greatly depend
upon the power which is available for imparting kinetic energy and the mass
of the rocket vehicle.
In the case of chemical rockets this energy is extracted from the propellant itself and depends greatly upon the characteristics of the propellant.
Such rockets are energy-limited. The exhaust velocity obtained by the use
of present day propellants is low which limits their performance. It is same
in the case for nuclear rockets also.
In the case of ion, plasma, and photon propulsion, this energy is supplied
from a separate power source which is usually an electric power obtained
from nuclear or solar source. Such rockets are power limited as it is very
difficult to produce such a huge amount of power within the small mass of
the rocket.
In order to obtain high exhaust jet velocity, i.e., high specific impulse
for a given thrust, very large amount of power is required. In the case
of nuclear rockets the acceptable wall temperature limits the power. In
electrodynamic propulsion huge size equipments are needed to convert such
large amount of power into kinetic energy. This in turn increases the mass
of the vehicle beyond economic and practical limits. If exhaust velocity is
low then the mass flow required will be high. Since this mass of propellants
has to be carried within the rocket, a compromise has to be made in order
to obtain best economy. For a given power, very high velocities may not be
of great use as the thrust will be low which will not allow to take any big
payload. It must be noted that the useful output of all propulsive devices is
thrust, the pull which motivates the craft. Figure 17.2 shows the range of
the accelerations and the exhaust velocities obtainable from different types
of rocket propulsion systems.
For going out of the earth’s gravitational field high acceleration is needed.
Either chemical or nuclear systems may be used. Once the gravity limit is
crossed the low acceleration available by ion or plasma devices is sufficient
for inter-planetary travels.
17.3
ANALYSIS OF AN IDEAL CHEMICAL ROCKET
Figure 17.3 shows the schematic diagram of an ideal chemical rocket and
its corresponding T -s diagram.
Rocket Propulsion
585
Chemical
Earth surface
gravitational acceleration
10
Acceleration m/s
2
Solid core
reactor
Fusion
reactor
Solar arc
Ion and plasma
Photon
10
0
10
1
10
2
10
3
4
10
10
5
10
6
10
7
Exhaust velocity km/s
Fig. 17.2 Range of accelerations and exhaust velocities obtainable by different rocket propulsion systems
p
2
02
p
3
p
T
4
Exhaust
Nozzle
gases (c j )
3
4
1
4
1
2
3
s
Fig. 17.3 Schematic diagram of a chemical rocket and its corresponding T -s
diagram
586
Gas Turbines
The propellant is stored in the combustion chamber at pressure and
it is assumed that the burning occurs at constant-pressure. The products
of combustion enter the convergence divergent nozzle at point 2 (point 3
corresponds to nozzle throat) and get expanded isentropically in the nozzle from stagnation pressure, p02 to static pressure, p4 , leaving it with an
exhaust velocity cj . We also assume that the gases behave as perfect gas.
Since the expansion is isentropic, we get
.
/
γ−1
/
γ
p4
0
cj =
(17.4)
2Cp T02 1 −
p02
Substituting Cp in terms of universal gas constant R, molecular weight
M , and ratio of specific heats γ, we get
.
/
γ−1
/ 2γR
γ
p4
0
cj =
T02 1 −
(17.5)
(γ − 1)M
p02
If q is the heat supplied in the form of chemical energy per unit mass of
propellant, we get
q
= Cp (T02 − T01 )
Then, from Eq.17.5, it follows that
.
/
/ 2γR
q
T01 +
cj = 0
(γ − 1)M
Cp
1−
p4
p02
γ−1
γ
(17.6)
If At is the area of the throat of the nozzle and pt the pressure at throat it
can be shown that the mass flow rate is given by
.
/
γ+1
γ−1
At p02 /
2
0
ṁ =
γ
(17.7)
γ+1
R
T
2
M
The thrust produced is given by
F
= ṁcj + (p4 − pa )A4
(17.8)
where pa is the atmospheric pressure. It should be noted that the term
(p4 − pa )A4 has not been neglected because the exhaust velocity cj is supersonic.
By substituting Eqs.17.6 and 17.7 into Eq.17.8 the thrust produced by
a chemical rocket can be written as
.
⎡
⎤
/
γ+1
/
γ−1
2
⎢ At p02 0
⎥
F = ⎣
γ
⎦
γ+1
R
T
02
M
Rocket Propulsion
!
2γR
q
T01 +
×
(γ − 1)M
Cp
1−
p4
p02
γ−1
γ
+(p4 − pa )A4
587
"
(17.9)
As is clear from the above equation, the thrust depends upon the pressure in the combustion chamber, the properties of the propellant and the
geometrical shape of the rocket.
To obtain high thrust the molecular weight of the propellants must be
as low as possible (see Eq.17.9). Most of the propellants, presently in use,
produce CO, H2 O, CO2 , N2 and H2 and the average molecular weight of
such species is about 25 and seldom less than 20 or so. It means that a limit
on the performance of a chemical rocket is imposed by the fact that the
molecular weight of the products of combustion cannot be reduced below
certain limit. Alternative fuels such as hydrogen and fluorine (molecular
weight 8.9) or hydrogen and oxygen (molecular weight 9) or even some light
metals like lithium with a molecular weight of 7 can be used. Even with the
use of such light materials this limit, cannot be lowered indefinitely because
the use of lower molecular weight reactants do not necessarily give lighter
products.
Another factor, which limits the thrust obtainable, is the maximum allowable temperature as well as the maximum temperature, which can be
produced by chemical reactions. At very high temperatures, dissociation
does not allow the whole of the heat energy to be converted into the kinetic energy and the maximum obtainable temperatures are limited. The
pressurization of combustion chamber, to a certain extent, curbs the dissociation. However, above a pressure of about 20 bar this curbing tendency
almost vanishes.
Thus, the chemical rocket is energy limited in the sense that in the first
instance a large amount of heat release from the propellant is difficult to
obtain, and even if obtained, it is still more difficult to utilize it due to
dissociation. The effect of variation in the ratio of specific heats is not very
much due to the presence of dissociation. The value of γ varies between 1.2
and 1.3 for almost all rocket propulsion fuels. Table 17.2 gives theoretical
flame temperatures that can be achieved in chemical rockets and the specific
impulse obtainable with different fluids.
Apart from the characteristics of the propellant the area ratio of the
nozzle and its shape also play an important role in dictating the performance of the chemical rocket. They affect the velocities obtainable as well
as the drag on the rocket.
17.4
OPTIMUM EXPANSION RATIO FOR ROCKET
The thrust developed by a rocket is given by
F
=
ṁcj + (pj − pa )Aj
(17.10)
By increasing the divergent portion of the convergent-divergent nozzle
588
Gas Turbines
Table 17.2 Theoretical Heat Release that can be obtained in Chemical
Rockets
System
Cyanogen and
ozone
Hydrogen
and fluorine
Hydrogen
peroxide
and gasoline
Specific
impulse, (s)
≈ 310
≈ 410
≈ 290
Highest
temperature
≈ 5000
≈ 4000 K
≈ 3000 K
Lowest
molecular
weight
between 8
and 9
between 8
and 9
between 8
and 9
Lowest
ratio of
specific heats
between 1.2
and 1.3
between 1.2
and 1.3
between 1.2
and 1.3
of the rocket the area Aj , velocity cj , and static pressure pj , at nozzle exit
can be varied. By differentiating Eq.17.10 with respect to pj and equating
to zero; the condition for maximum thrust can be obtained as
dF
dcj
dAj
= ṁ
+ Aj + (pj − pa )
=0
dpj
dpj
dpj
(17.11)
Euler’s equation can be written as
dpj
dcj
=
−ρj cj
(17.12)
By continuity equation ṁ = ρj cj Aj , then we have
dpj
dcj
=
−
ṁ
Aj
(17.13)
From Eqs.17.11and 17.13, we have
−ṁ
Aj
ṁ
+ Aj + (pj − pa )
dAj
=0
dpj
(17.14)
or
(pj − pa )
Since
dAj
dpj
dAj
dpj
= 0
(17.15)
= 0, the condition for maximum thrust is given by
pj = pa
(17.16)
Rocket Propulsion
589
i.e., the optimum expansion condition is that when the static pressure at
the exit of the rocket nozzle is the same as the ambient pressure. Simple physical reasoning can also arrive at this conclusion. Suppose that the
length of the diverging nozzle passage is increased over that corresponding
to the optimum expansion ratio (pa = pj ); this will result in further expansion in the nozzle and pressure at exit of the nozzle will be less than
ambient pressure pa and the second term in the Eq.17.10 will become negative and thrust will decrease. On the other hand, if the length of the nozzle
is less than that corresponding to pj = pa will result in lesser expansion
and hence in reduced exhaust velocity and thrust will again be reduced. So
the condition pj = pa gives the best expansion condition for a rocket nozzle
with a given throat diameter. Hence, it is imperative that the design of
rocket nozzle is very crucial for producing the required thrust.
17.5
THE CHEMICAL ROCKET
Chemical rockets are classified on the basis of the type of propellant used:
(i) Solid propellant rockets
(ii) Liquid propellant rockets
(iii) Free radical rockets
The nature of the propellant used greatly affects the overall shape and
size of the rocket. However, major differences in details occur. One typical
example is that in the case of solid propellant rockets no fuel-pump and
injector is required as in liquid propellant rockets.
17.5.1
Solid Propellant Rocket Engines
Figure 17.4 shows a schematic diagram of a solid propellant rocket. It consists of a combustion chamber, an expansion nozzle, and an igniter. The
peculiar characteristic of this type of rocket is that entire propellant is contained in the combustion chamber which has its oxidant within the fuel
itself. Therefore, the normal duration of combustion in such a motor is
relatively very short (usually less than a minute). Both fuel and oxidizer
are homogeneously mixed within the solid propellant grain. The figure indicates that the major components are the propellant grain the case, the
nozzle and the igniter (not shown in figure). The igniter, which is usually
a pyrotechnic mixture triggered by an electrical signal, initiates combustion over the entire open face of the grain. After ignition the propellant
burns more or less evenly and at a steady rate, thus, the name ‘Cigarette
burning’ has been used to describe the type of burning for the rocket motor
shown in Fig.17.4. This burning produces gases at high pressure and high
temperature.
The burning causes rapid decomposition and heat release. This heat is
conducted back to the propellant and thus the process is self-sustaining.
Such a process is called burning by deflagration. The burning rate depends
590
Gas Turbines
Solid
propellant
Combustion
chamber
Motor wall
Exhaust gases
}
Flexible
liner
Burning
surface
Nozzle
Fig. 17.4 Typical solid propellant motor
upon the pressure of the chamber and increases with an increase in pressure.
Burning may be controlled or uncontrolled. To obtain maximum thrust
whole of the combustion space is filled with the propellant and it is allowed
to burn like a cigarette.
These gases then flow through the throat area within the grain and are
expanded in the exhaust nozzle to a high velocity jet of 1500 to 2400 m/s,
depending upon the propellant combination used. The area of the propellant grain over which the combustion takes place determines the magnitude of the exhaust gas generation and hence the thrust of the engine. The
thickness of the grain determines the burning time or, duration of thrust
production. An active cooling system is not provided since the propellant
grain itself frequently insulates the structural case throughout the combustion process. However, insulation is often plated on the inner surface of the
case. Therefore, the burning rate is usually controlled by pressure as well
as by giving different shapes and sizes to the propellant. This propellant
shape is called grain. The various types of grains are shown in Fig.17.5.
The pressures developed in a solid propellant rocket are about 80 to 200
bar. The thrust and duration of operation depends upon the shape, size
and the material of the propellant. Usually the propellant used is cast or
extruded.
A solid propellant must release its thermochemical energy without, any
additional oxidizer, and must have large energy contents which is to be released rapidly. Hence, a rapid burning substance which contains hydrogen,
carbon and oxygen is required. Thus, solid propellants which contains explosives such a nitroglycerin C3 H5 (ONO2 )3 , nitromethane CH3 NO2 , trinitrotoluene C6 H2 (CH3 ) (NO2 )3 are typical.
It should be recognized that the solid-propellant rocket motor cannot be
shut off once started and that its duration is short. Hence, it is used where
the propulsion requirements meet these criteria. Typical areas of solidpropellant rocket motors are rocket-assisted take-off of aircraft (RATO)
(units producing 5 to 10 kN of thrust for 10 to 30 seconds), large missile
boosters (units with thrust values up to several hundred thousand kN for
Rocket Propulsion
Tubular
Rod and tube
Double anchor
Star
Multifin
Double composition
591
Fig. 17.5 Internal-burning charge designs
several seconds), and the main power plant for a multiplicity of missiles
including the relatively small projectile like air-to-air missiles (AAM), the
large artillery type surface-to-surface missiles (SSM), the surface-to-air missiles (SAM) and some air-to-surface missiles (ASM). Still other uses might
include application to one or several stages of long range ballistic missiles
and even satellite vehicles. From the foregoing discussion it is apparent
that even though the solid-propellant rocket motor has limitations, it is
used quite extensively and for varied purposes. The rather widespread use
of solid propellant rocket motor is relatively recent, the trend in greater
usage is continuing. This trend is primarily due to three factors, viz.,
(i) Increased use of missiles at the expense of aircraft in the modern air
weapons
(ii) Improved technology in the solid propellant rocket motor field. Specific impulse during the last fifteen years has increased significantly;
new case-bonding techniques have developed to the extent that with
proper design, the latter portions of the propellant that burn act as
a heat insulator
(iii) Inherent advantages of solid-propellant motors over the liquid propellant motor from the military point of view is the storage stability of
the solid-propellant and its relatively minor logistics problems
Although the use of solid-propellant rocket motors is extensive, the application of liquid-propellant-rocket motor is currently even more extensive
in the high thrust units. Also for long duration applications, the liquidpropellant rocket motor has greater potential.
A modification of solid-propellant engine is the hybrid engine, which
incorporates one solid propellant and one liquid-propellant. Normally, the
592
Gas Turbines
solid propellant is the fuel because of its higher energy potential, while
the liquid propellant is the oxidizer. Figure 17.6 schematically illustrates a
typical hybrid rocket engine.
Liquid oxidizer
Injector
Solid propellant gas generator
Solid fuel
Fig. 17.6 Typical hybrid rocket engine
The hybrid engine tends to combine the advantages of both solid and
liquid propellant engines. It can be operated intermittently and throttled
to a degree by varying the oxidizer flow rate. It tends to have simplicity
and hence the reliability of the solid-propellant engine.
Hybrid engines are still under development. Their advantages would
seem to offer good potential for application in the future.
17.5.2
Solid Propellants
In general, the solid propellants can be classified as composite propellants.
and double-base propellants.
Composite propellants consist of an oxidizer in a ground crystalline
shape mixed in the fuel of plastic nature acting as a binder to hold the mixture and to keep a uniform composition. Typical oxidizers are finely ground
crystals of perchlorate or nitrate of potassium, lithium or ammonium. The
binder may be rubber, asphalt or elastomers. Composite propellants are
comparatively difficult to cast. Further, it will have a high oxidizer content,
which results in high density of the exhaust. The mechanical properties of
the composite propellants depend upon the nature of the binder used. For
example, use of polystyrene as a binder results in hard structure of the
grain. The burning rate of the propellant can be controlled by controlling
the ratio of the mass of the fuel to the oxidizer, by addition of catalyst or
by the shape and size of the oxidizer crystals. Double-base propellants or
the homogenous propellants are characterized by nitrocellulose containing
[C6 H72 (ONO3 )3 ] compounds, which act as a plastic and gives the fuel a colloidal characteristic. These are true monopropellants in that each molecule
of the propellant has, in it, the necessary amount of fuel and oxidizer.
Double-base propellants, combinations of nitroglycerin [C3 H5 (NO3 )3 ] and
nitrocellulose with small quantities of additives are very commonly used.
The additives are provided to impart stability of combustion and freedom
from ageing which are necessary for achieving high performance.
Such propellants are plastic in nature, have a very high viscosity and
their appearance is smooth and waxy. These are made in required shape
by casting or extrusion. Nowadays this classification of solid propellant has
become arbitrary in nature because of the development of such nitrocel-
Rocket Propulsion
593
lulose and nitroglycerin based propellants, which also have oxidizers like
potassium perchlorate.
In Table 17.3 the typical solid propellants and their chemical formulae
along with the important properties are given.
Table 17.3 Solid Propellants and their Properties
Solid propellant
system
Flame
temperature
Molecular
weight
(K)
KClO4 and C2 H2 O
NH4 ClO4 and C2 H4 O
Nitrocellulose
and Nitroglycerine
Asphalt and
Perchlorate
17.5.3
Specific
impulse
(s)
1800–3000
1800–2150
2350–3150
25–35
22–25
22–28
165–210
175–240
105–195
2350–2650
–
180–195
Propellants and their Desirable Characteristics
A great variety of propellants have been developed for rocket systems. A
few of the more widely used propellants will be discussed in this section
along with their various desirable characteristics.
(i) Monopropellant A monopropellant, as the name indicates, is one,
which is capable of releasing its chemical energy without the addition
of an oxidizer. Monopropellants have been used in liquid-propellant
rockets. Among the materials that have been used for monopropellants are hydrazine, N2 H4 ; nitromethane, CH3 NO3 ; and hydrogen
peroxide, H2 O2 .
(ii) Bipropellant fuels Bipropellant systems have been most widely used
in liquid propellant rockets. Among the fuels that have been used
are ethyl alcohol (ethanol), C2 H5 OH; methyl alcohol (methanol),
CH3 OH; gasoline, C8 H18 ; hydrazine, N2 H4 ; furfural alcohol,
C4 H3 OCH2 OH; aniline C6 H5 NH2 ; ammonia, NH3 ; and hydrogen, H2 .
(iii) Bipropellant-oxidizers Among the liquid oxidizers that have been used
for bipropellant systems are oxygen, O2 ; nitric acid, HNO3 ; and hydrogen peroxide, H2 O2 . Table 17.4 gives the several combinations of
oxidizers and fuels which have been used for bipropellant systems.
The desirable characteristics of propellants are the following:
(i) Ability to produce a high chamber temperature. In general, this
requirement means a high calorific value per unit mass of propellant.
594
Gas Turbines
Table 17.4 Liquid Bipropellant Combinations
Oxidizer
Fuel
Oxygen
gasoline,
methane,
ethanol,
75% ethanol, 25% water
methanol
ammonia
hydrazine
hydrogen
Red-fuming nitric acid
aniline
ethanol
White-fuming nitric acid
furfural alcohol
Hydrogen peroxide
hydrazine
ethanol
methanol
(ii) Low molecular weight of products of combustion. This is one of the
most important characteristics in order to obtain a high jet velocity
and high specific thrust.
(iii) High density to reduce the physical dimensions of the overall missile
by allowing a large propellant weight in a given space. In general,
requirements (ii) and (iii) conflict and a balance between the two
must be reached for any particular application.
(iv) Ease of storage and handling.
(v) No chemical reaction with the system including tanks, piping, valves,
and injection nozzles.
(vi) Readily ignitable.
(vii) Availability.
An ideal propellant which satisfies all of the above requirements is yet to
be developed. For example, rocket motors using liquid hydrogen as a fuel
are capable of the highest jet velocities but the low density necessitates
large fuel tanks; hence greater overall size and increased aerodynamic drag.
Further, the extreme low temperatures necessary to maintain hydrogen in
the liquid state impose difficult handling and mechanical problems.
Rocket Propulsion
17.5.4
595
Liquid Propellant Rocket Engines
Liquid propellant rocket engines may be divided into several categories.
These include cryogenic and storable bipropellant systems, monopropellant
systems, and the hybrid systems, which are really a cross between a liquid
and a solid propellant systems. Although significantly simpler than bipropellant engines, the performance of monopropellant engines is sufficiently
low. It may be noted that these systems are currently used only for relatively small applications. Since the liquid bipropellant system has greater
potentialities it will be discussed in details.
Figure 17.7 illustrates the principal elements of a liquid bipropellant
rocket engine. The rocket engine consists of an oxidizer tank, a fuel tank,
various control valve lines and the rocket motor. The rocket motor is the
heart of the rocket engine and, therefore, it deserves additional discussion.
The basic rocket motor comprises of an injection plate I, a combustion
chamber B, and a discharge nozzle N. The function of the injection plate
is to receive the liquid oxidizer and fuel and direct them in liquid streams
so that they mix up with one another and produce a chemical reaction in
the combustion chamber. When the reaction takes place in the combustion
chamber, very high pressure and temperature gases are produced which
in turn are expanded in nozzle to produce a high supersonic exit velocity
(1500 to 4500 m/s). The net thrust is equal to the product of the difference
between the exit gas velocities, and zero velocity (because the initial velocity
of the fuel and oxidizer with respect to the engine is zero) and mass flow
rate. An ignition system is not shown in the engine because the suitable
liquids can be chosen which react spontaneously upon mutual impingement.
These are called hypergolic combinations1 . A cooling system (not shown in
diagram) is also necessary to prevent the walls of the motor from melting
because the temperature of reaction often exceeds 2700◦C.
Oxidizer tank
Fuel tank
Control valve
B
cj
N
I
Control valve
Fig. 17.7 Principal elements of a liquid bipropellant rocket engine
In order that the rocket motor might function properly the combustion
chamber pressure must be considerably higher than the surrounding pressure (typical values of internal pressures are 20 to 40 bar). The source of
1 Meaning propellant combinations that ignite spontaneously upon contact such as
nitric acid and aniline.
596
Gas Turbines
this high combustion chamber pressure must originate upstream of the injector, i.e., a source of pressure is required to force the propellants through
the injector nozzles (which requires a pressure drop of about 7 bar or more)
and into the combustion chamber. At present there are two general systems of transporting the propellants from their storage tanks to the rocket
motors. They are
(i) the gas pressurization system, and
(ii) pump pressurization system .
17.5.5
Gas Pressurization System
This system is the simplest type of liquid-propellant rocket motor. The
details are shown schematically in Fig.17.8.
System energizing valve
Inert gas supply usually
nitrogen of 154 bar
31.5 bar
21 bar
Fuel tank
Pressure regulators
28 bar
Oxidizer tank
Bi-propellant control valve
35 bar
Fig. 17.8 Liquid bipropellant gas pressurization system
As illustrated in Fig.17.8 an inert gas (usually nitrogen), a gas which
will not react nor be excessively soluble with the fuel or oxidizer, is stored
in tanks at, as high a pressure as possible and is supplied through pressureregulator valves to force the liquid propellants through the lines, bipropellant control valves, injector plate and into the combustion chamber.
Typical values of pressures are shown at various flow points to show the
relative magnitude of pressures required. Since typical mixture ratios (oxidizer flow to fuel flow) are 3 to 5, more pressure and/or larger flow lines
are used in the oxidizer flow system. The system is first energized by opening the system-energizing valve, and then the rocket motor is turned on by
opening the bipropellant valve. The motor can be stopped and restarted
at will by closing or opening the latter valve.
The gas pressurization system is a relatively simpler power plant. Its
big advantage, however, is the fact that both oxidizer and fuel tanks are
pressurized. But the point to note is that when the tanks are of considerable
Rocket Propulsion
597
volume (as they must be for long range applications), the tank walls, which
must withstand such high pressures, are excessively thick and heavy.
In addition to the increase of propellant tank weight with motor duration, the inert gas volume, and consequently the inert gas tank weight will
also increase. Thus, the major limitation of the gas pressurization system
is its heavy weight when used for long duration operation. This limitation makes the gas pressurization system applicable only to short duration
operations.
17.5.6
Pump-pressurization System
The limitation of gas pressurization system to the application of long-range
rocket flight stimulated the development of the pump-pressurization system. A typical liquid bipropellant pump pressurization rocket engine is
shown in Fig.17.9. In this system the liquid oxidizer and fuel are stored in
tanks at low pressure (thus the tanks can be made light) and forced into the
rocket at high pressure by the fuel and oxidizer pumps. The power required
for driving the pumps is supplied by a gas turbine which is supplied with
steam and oxygen (as shown in this illustration) obtained by decomposing
liquid hydrogen peroxide by a catalyst such as calcium or sodium permanganate. The gases used to drive the gas turbine need not be generated from
a separate propellant as shown in Fig.17.9, but the use of separate hydrogen
peroxide system is popular in rocket motor design, because the gases generated are at relatively low temperature (about 478◦ C), therefore, the problem of high turbine blade temperature is not there. Sometimes the turbine
is driven by the gases generated from the main propellant system, which
at first examination, may look simpler and more logical. However, when
it is remembered that the main propellants are chosen for their high thermochemical activity and consequently react at temperature up to 3100◦C,
that portion which is used to operate the turbine must be combined to “run
fuel-rich” to keep the turbine temperature at acceptable value, or a third
liquid, usually water, is required to dilute the turbine gases down to the
temperature at about 875◦C.
Because of the third liquid, the gas turbine and the pumps additional
lines are necessary. This makes the pump-pressurization system considerably more complex than the gas-pressurization system. The extra complexity is not only due to extra pieces of equipment; but also additional
problems arise because of the pumping plant. Perhaps the greatest problem is the design of pumps that will handle the liquids safely and without
leaks. Leaks cannot be tolerated because of the possibility of fires or explosions. Since, propellants are selected on the basis of their vigorous reacting
ability in the combustion chamber, the risk of fire or explosion due to leaks
is very high. To eliminate leaks, special seal design and special seal materials are required. Some of the newer plastics such as teflon have great
promise in this application. Liquid oxidizers are generally acids, liquid oxygen, or concentrated hydrogen peroxide; therefore special pump materials
are required for the oxidizer pump impeller and casing. The pumping problems have already been satisfactorily solved, and successful applications of
598
Gas Turbines
H2 O2
Oxidizer tank
Fuel tank
High pressure gas
Steam & oxygen generated �
in the presence of a catalyst
Pump
Turbine
Pump
Waste
Bi-propellant control valves
Fig. 17.9 Typical pump-pressurization rocket motor
pump pressurization system have been used even in piloted aircrafts.
17.5.7
Liquid Propellants
As already discussed there are two basic types of liquid propellants, namely:
(i) monopropellant, and
(ii) bipropellant.
Monopropellants are those, which do not require an oxidizer and can produce thermal energy by decomposition either by a catalyst or by ignition. Typical examples are hydrazine, nitromethane and hydrogen peroxide. Monopropellants are usually hazardous and are not preferred. They
are used for limited purposes such as for a source of power to the turbine
in a pump feed system, etc.,
Bipropellant, as the name implies, is a combination of a fuel and an
oxidizer, which when combines produces thermal energy. Typical examples
of such combinations are hydrogen peroxide and alcohol, liquid oxygen and
liquid hydrogen, etc., Similar combinations are given in Table17.5
The above combinations may or may not be hypergolic. Nowadays, a
combination of metallic fuels like lithium or beryllium with either oxygen
or fluorine and liquid hydrogen, what is called as a tripropellant, is also
being used to improve the performance of chemical rockets.
Table 17.5 shows the common liquid propellants used along with the
approximate obtainable specific impulse from them. Fluorine is the most
powerful oxidizer, next is the chlorine trifluoride, which is so active that
even the glass-fibre cotton, a fireproof material burns in it. Hydrogen peroxide and ozone are also good oxidizers. Liquid hydrogen has maximum
Rocket Propulsion
599
Table 17.5 Common Bipropellants used in Liquid Propellant Rockets
Oxidizer
Fuel
Temperature (K)
Molecular
weight
Specific
impulse (s)
Hydrogen
peroxide(H2 O2 )
Gasoline
2950
21
250
Red fuming
nitric acid
(HNO3 )
Gasoline
3125
25
240
Alcohol
3350
22
260
Liquid oxygen
∗
Liquid oxygen
Liquid
ammonia
—
—
—
Liquid oxygen
Hydrazine
(N2 H4 )
3400
18
310
Liquid oxygen
Liquid
hydrogen
3000
9
390
Liquid fluorine
Liquid
hydrogen
4000
8.9
410
Liquid fluorine
Ammonia
4600
19
360
Nitric acid
Aniline
(COH5 NH2 )
—
—
Nitrogen
tetroxide
Hydrazine
3260
—
290
Ozone
Cyanogen
520
—
270
Hydrogen
peroxide
Ethyl
diamene
(C2 H4 N2 H5
—
—
—
* Not commonly adopted
600
Gas Turbines
heat value but is highly explosive. Germans used alcohol and liquid oxygen
as propellants for V-2 rocket. Kerosene is also used with some powdered
metals as rocket fuel. Hydrazine requires comparatively smaller quantities
of oxidizer for burning.
17.5.8
Requirements of a Liquid Propellant
Any liquid or a combination of liquids, if capable of producing mainly
gaseous products by an exothermic chemical reaction, can be regarded as
a liquid propellant. However, to be ideally suitable for liquid propellant
rocket it must have certain specific characteristics such as
(i) High calorific value.
(ii) High density.
(iii) Ease of storage and handling.
(iv) Low corrosion characteristics.
(v) Less toxicity.
(vi) Performance independent of ambient temperature.
(vii) Smooth ignition and high reliability.
(viii) Chemical stability and long shelf life.
(ix) Negligible viscosity change with temperature.
As far as known none of the presently available propellants meets all
these requirements and still lot of development work is needed in this field
even after so much successful launchings of rockets and satellites.
17.6
ADVANTAGES OF LIQUID PROPELLANT ROCKETS
OVER SOLID PROPELLANT ROCKETS
The following are some of the advantages of liquid-propellant rockets over
the solid-propellant rockets:
(i) Duration of operation can be increased.
(ii) Use of cooling allows the metal walls to keep their strength.
(iii) Use of lower pressures in conjunction with cooling makes it possible
to use less expensive or non-critical materials.
(iv) Size of the combustion chamber is smaller as whole of the fuel need
not be stored in it.
(v) Comparatively easier to control by controlling the propellant flow.
Rocket Propulsion
601
(vi) Heat losses to the wall are reduced due to preheating of fuel during
cooling process.
Liquid propellant rockets are extensively used with aircraft, for assisted
take-off, ballistic missiles, thrust augmentation, and in space flight.
17.7
FREE RADICAL PROPULSION
If a stable chemical material is supplied with sufficient energy to break the
energy bonds some unstable free radicals will be produced. If these free
radicals are used in a rocket so that they recombine and release a vast
amount of energy, the limitations of a chemical rocket can be partially
overcome. Molecular hydrogen with certain concentration of unstable free
radicals in it, seems to be the best propellant due to its lower molecular
weight. However, it is long time before such a system can be used in practice
due to difficulties in generating free radicals, preserving them for sufficient
time and controlling the recombination rates.
17.8
NUCLEAR PROPULSION
Work on nuclear rocket engines was begun in 1956 in USA. With the significant increase in emphasis on space flight activities beginning in 1958,
National Aeronautics and Space Administration, USA (NASA) took over
the development of nuclear technology for nuclear rocket-propulsion systems for manned or unmanned space exploration. Nuclear rocket engine
produces very small thrust and thus have quite low thrust-to-weight ratio
which are only useful for space operations in gravity-free fields.
The maximum exhaust velocity, that can be obtained in a chemical
rocket is limited to less than 4.5 km/s because of the lower energy release
from the propellant and other limitations such as attainment of a high
temperature in the combustion chamber. It is well known that in a nuclear
reaction vast amounts of energy, 106 to 108 times that of a chemical reaction
can be obtained. This high rate of energy release will result in very low
mass of the propelling device and a wider latitude in selection of propellant
material. Detailed study of these reactions and related equipment is beyond
the scope of this book and only a simple description is given below.
There are two types of nuclear reactions which can be used to produce
energy, viz., fission and fusion.
In fission a heavy mo1ecule is broken into fragments by the bombardment of neutron on its nucleus. When the neutron is absorbed into the
nucleus, the heavy molecule splits up into a number of fragments and energy is released in the process. This energy is in the form of heat energy.
In addition to this, the reaction also produces about 2.5 neutrons which
makes the reaction self-sustaining.
Figure 17.10 shows a schematic diagram of a nuclear propulsion unit. It
consists of a nuclear core having 92 U235 as the nuclear fuel. The control rods
control the reaction. A reflector is provided to avoid leakage of neutrons.
602
Gas Turbines
Hydrogen is used as a propellant, which takes up heat from the reactor and
expands in the nozzle.
Control rods
Reflector
H 2 from pump
Exhaust
Reactor core
Fig. 17.10 Schematic diagram of solid core nuclear-heated hydrogen rocket
In the second type of reaction, i.e., thermonuclear reaction or fusion,
charted nuclei of light elements are fused into a heavy nucleus and in the
process vast amount of energy, is released. To start fission reaction, coulomb
repulsion forces between the nuclei must be overcome, which presents a
formidable problem as it requires temperatures of the order of 1060 K.
Nevertheless, if nuclear fusion is achieved in near future then the future
propulsion devices are very likely to use nuclear energy. Nuclear energy has
been successfully used in submarines, ships and some experimental rockets.
17.9
ELECTRO DYNAMIC PROPULSION
The fundamental requirement for space propulsion is the generation of very
high exhaust velocities in order to minimize propellant consumption. Electric propulsion engines generate exhaust velocities more than 4 to 10 times
those of chemical rocket engines, and thus they have considerable potential for space propulsion. There are three basic types of electric rocket
propulsion engines:
(i) ion rocket propulsion,
(ii) the plasma rocket propulsion, and
(iii) the photon propulsion .
For each of these engines, the energy is supplied from an energy source
separate from propulsion device.
17.9.1
Ion Rocket Propulsion Engines
Figure 17.11 represents a schematic diagram of Electrostatic or ion rocket
engine. The major components are
(i) propellant supply tank and propellant feed mechanism,
Rocket Propulsion
603
(ii) thrust chamber, and
(iii) electric power supply.
The thrust chamber incorporates a vapourizing chamber, an ionization
chamber, an accelerating grid and an electron emitter. The propellant is
first heated and vaporized and then passed on to the ionization chamber,
where electrons are stripped to ionize it. The resultant ions are then accelerated electrostatically with a d.c. voltage drop (in excess of 1000 volts)
and discharged to produce thrust. The electrons stripped in the ionization
chamber are collected and injected into the ion stream aft of the accelerating grid to neutralize the ion beam and prevent the ions from being
attracted back to the grid. Removal of electrons is also necessary to maintain a neutral charge on the space vehicle.
Electric power supply
Electron
emitter
Ion flow
Propellant�
tank
Pump
Vaporizing �
chamber
Ionization �
grid
Accelerating �
grid
Neutral
plasma
flow
Fig. 17.11 Schematic diagram of an ion rocket engine
17.9.2
Plasma Rocket Propulsion
Plasma propulsion differs from ion propulsion in that the electrical forces
are not directly used to accelerate the propellant; instead the propellant
is first heated to a very high temperature and then expanded through a
nozzle. It is a hybrid thermal-electrical propulsion system. It relies on
expansion of hot plasma for bulk of its thrust but its energy is supplied
electrically rather than by chemical combination or by nuclear fusion or
fission.
A high voltage source, which again may be a nuclear reactor, is used to
produce an electric arc to heat up the propellant to a high temperature. The
propellant is either hydrogen or helium as both have low molecular weight
and high maximum stagnation temperature before they dissociate. Helium
is preferable as it has a higher stable temperature. Once the propellant is
heated to a high temperature it is expanded in a nozzle to get propulsive
force.
For starting, electrodes are brought closer than the normal position to
produce a spark. The chamber core becomes a good conductor in the presence of positive ions. After the arc is established, the electrodes are taken
604
Gas Turbines
apart to their normal position. Energy leaves the arc in electromagnetic
radiation and is consumed because of the increase in the potential energy of
ionization and dissociation. The temperature of the material in the region
of arc is increased. The propellant is fed in the form of vortex to concentrate the arc into a narrow region. This allows a very high temperature
(50,000 K) in the central core-much more than the wall material can withstand since the propellant vortex also cools the wall. This high temperature
plasma escapes through the nozzle and a propulsive force is created.
Since ionization of hydrogen takes place only at about 10,000 K a large
amount of energy supplied to the arc is not available for producing thrust.
Unless this energy is used back by proper recombination, the efficiency will
be very low.
Production of an efficient plasma arc, the difficulty of cooling the rocket
wall and the need of very high power for arcing are the major limitations
of plasma jet propulsion system. Hot plasma erodes the nozzle, and the
life of electrodes is limited to a few minutes. Not much information is
available about it but may have a great future potential. The plasma
rocket propulsion can be of two types:
(i) Arc plasma rocket engine, and
(ii) Magneto plasma rocket engine.
Arc Plasma Rocket Engine
The arc plasma rocket engine is one of the simplest types of electrical
propulsion systems being developed. Its schematic diagram is shown in
Fig.17.12. The major components are
(i) propellant supply tank and propellant feed mechanism,
(ii) thrust chamber,
(iii) cooling system, and
(iv) electric power supply.
The thrust chamber contains two electrodes. The nozzle walls serve as
the cathode, and an electrode within the chamber is the anode. An arc is
formed between these electrodes, and the propellant is heated to an electrically neutral plasma in passing through the arc. Thrust results from the
expansion of the heated plasma through the nozzle. The velocity produced
is 4000 to 19,500 m/s. The propellant can be used to cool the chamber
regeneratively before passing into the chamber. If the rate of propellant
consumption is too low to provide enough cooling, a supplementary cooling
system with a radiator to reject the excess heat must be provided.
The arc plasma or electrothermal rocket is quite similar to the chemical
racket, the primary difference being that electrical rather than chemical
energy is used for heating the propellant.
Rocket Propulsion
Propellant tank
605
Pump
Electric power supply
Cathode
Vj
Anode
Pump
Radiator for cooling system
Fig. 17.12 Schematic diagram of an arc plasma rocket engine
Magneto-Plasma Rocket Engine
Another type of electrodynamic propulsion system is the magnetohydrodynamic (MHD) or magneto-plasma rocket engine. Figure 17.13 shows
principal elements of such a system. MHD engines produce thrust from the
interaction between magnetic and electric fields. The force which results is
customarily called the J × B force, where J is the current density and B is
the applied magnetic field. The nature of this resulting force, which accelerates the propellant, is analogous to the force which rotates electric motors.
Although a great amount of research and development work remains to be
done, the ultimate development of an efficient and reliable MHD engine
appears assured. The magneto-plasma rocket engine uses an arc chamber
Magnetic field
Propeller
Tank
Ve
Pump
Arc chamber
Accelerator
Plasma
(+,-)
Electric supply
Fig. 17.13 Principal elements of a magneto-plasma rocket engine
to produce a conductive plasma, similar to arc plasma engine. The plasma
then passes through an accelerator. Two different methods can be used to
accelerate the plasma :
(i) changing or collapsing of magnetic fields, or
(ii) interaction of an electric current and a steady magnetic field.
606
Gas Turbines
The first method is cyclic and can impart extremely high velocities to the
plasma (up to 2,40,000 m/s). The second method is continuous but imparts
somewhat lower to the plasma (up to 15,000 m/s).
17.10
PHOTON PROPULSION
Photons are electromagnetic quanta of energy which do not carry an intrinsic mass but carry a momentum equal to hf /c where h is Plank’s constant,
f the frequency of radiation, and c is the velocity of light. When photons are produced some mass according to Einstein’s equation m = E/c2 ,
disappears, i.e., in other words if photons are emitted some mass is transferred. Figure 17.14 shows a schematic diagram of a photon propulsion unit
in which a hot nuclear power source generates a flux of photons, which is
shaped by a photon absorber and then escapes through the nozzle producing a thrust equal to
d
F
=
hf
c
=
dt
1000Pe
c
where Pe is the power in jet in kilowatts. As on date such rockets are not
feasible, but probably have a great future potential!
Nuclear photon source
Pay load
Photons
Nozzle
Biological shield
Shaped absorber
Fig. 17.14 Schematic diagram of a photon propulsion unit
17.11
COMPARISON OF VARIOUS TYPES OF ROCKETS
So far we have discussed various types of rockets. In summary, there are
two general measures of the performance of a rocket engine.
(i) specific impulse, which will determine the amount of propellant that
must be used to accomplish a given task.
(ii) fixed weight of the engine, including the necessary tankage, power
supply, and structure.
The chemical rocket engine is a fairly lightweight device. However, the
specific impulse is not high. Solid and liquid propellants in present use
Rocket Propulsion
607
deliver an impulse of around 250 seconds. The best liquid propellants so
far conceived and evaluated yield an impulse of about 350 seconds. Certain
solid propellants, proposed on the basis of theory alone might yield 300
seconds. The fundamental theory of chemical binding energies precludes
the possibility of any substantial gains over these numbers, Even some asyet-undiscovered superfuel is unlikely to raise the specific impulse beyond
400 seconds.
Nuclear rocket is not limited by propellant binding energies. However,
it is constrained by the temperature limitations of wall materials. Using
hydrogen as a propellant, values of specific impulse of perhaps 1,000 seconds
or more are feasible. If gaseous containment of the fissioning fuel becomes
possible, specific impulses of several thousand seconds might be achieved.
This type of rocket engine appears very promising. Research on nuclear
rockets and controlled thermonuclear power reactors may yield information
useful to the construction of such a device.
The primary consideration in obtaining useful thrust from ion or plasma
rockets is the construction of lightweight electric power supplies. A gross
reduction in electrical generation equipment is required to make the electric
rocket useful for flight in the solar system. In any event, the electric rocket
is likely to remain a low-thrust device. Therefore, large chemical or nuclear
rockets would still be required to boost a space ship from the surface of the
earth.
17.12
STAGING
Often, the required velocity (delta-v) for a mission is unattainable by any
single rocket. This is because the propellant, tankage, structure, guidance,
valves and engines and so on, take a particular minimum percentage of takeoff mass. This becomes too high for the propellant it carries to achieve that
velocity.
For example the first stage of the Saturn V, carrying the weight of the
upper stages, was able to achieve a mass ratio of about 10, and achieved
a specific impulse of 263 seconds. This gives a delta-v of around 5.9 km/s
whereas around 9.4 km/s delta-v is needed to achieve orbit with all losses
allowed for.
This problem is frequently solved by staging - the rocket sheds excess
weight (usually empty tankage and associated engines) during launch. Staging is either serial where the rockets light after the previous stage has fallen
away, or parallel, where rockets are burning together and then detach when
they burn out.
The maximum speeds that can be achieved with staging is theoretically
limited only by the speed of light. However the payload that can be carried
goes down geometrically with each extra stage needed, while the additional
delta-v for each stage is simply additive.
608
Gas Turbines
17.13
MULTISTAGE ROCKET
A multistage (or multi-stage) rocket is a rocket that uses two or more stages,
each of which contains its own engines and propellant. A tandem or serial
stage is mounted on top of another stage; a parallel stage is attached alongside another stage. The result is effectively two or more rockets stacked on
top of or attached next to each other. Taken together these are sometimes
called a launch vehicle. Two stage rockets are quite common, but rockets
with as many as five separate stages have been successfully launched.
By jettisoning stages when they run out of propellant, the mass of the
remaining rocket decreases. This staging allows the thrust of the remaining
stages to more easily accelerate the rocket to its final speed and height.
In serial or tandem staging schemes, the first stage is at the bottom
and is usually the largest, the second stage and subsequent upper stages
are above it, usually decreasing in size. In parallel staging schemes solid or
liquid rocket boosters are used to assist with lift-off. These are sometimes
referred to as ’stage 0’. In the typical case, the first stage and booster
engines fire to propel the entire rocket upwards. When the boosters run
out of fuel, they are detached from the rest of the rocket (usually with some
kind of small explosive charge) and fall away. The first stage then burns to
completion and falls off. This leaves a smaller rocket, with the second stage
on the bottom, which then fires. Known in rocketry circles as staging, this
process is repeated until the final stage’s motor burns to completion.
In some cases with serial staging, the upper stage ignites before the
separation - the interstage ring is designed with this in mind, and the
thrust is used to help positively separate the two vehicles. This is known
as ”fire in the hole”.
17.13.1
Upper Stages
The upper stages of space launch vehicles are designed to operate at high
altitude, and thus under little or no atmospheric pressure. This allows
them to use lower pressure combustion chambers and still obtain nearoptimum nozzle expansion ratios with nozzles of reasonable size. In many
low pressure liquid rocket upper stage engines, such as the Aerojet AJ10, propellants are pressure fed without need for complex turbomachinery.
Low chamber pressures also generate lower heat transfer rates, which allow
ablative cooling of the combustion chambers rather than more elaborate
regenerative cooling.
17.13.2
Advantages
The main reason for multi-stage rockets and boosters is that once the fuel
is burned, the space and structure which contained it and the motors themselves are useless and only add weight to the vehicle which slows down its
future acceleration. By dropping the stages which are no longer useful, the
rocket lightens itself. The thrust of future stages is able to provide more
acceleration than if the earlier stage were still attached, or a single, large
Rocket Propulsion
609
rocket would be capable of. When a stage drops off, the rest of the rocket
is still traveling near the speed that the whole assembly reached at burnout time. This means that it needs less total fuel to reach a given velocity
and/or altitude.
A further advantage is that each stage can use a different type of rocket
motor each tuned for its particular operating conditions. Thus the lower
stage motors are designed for use at atmospheric pressure, while the upper
stages can use motors suited to near vacuum conditions. Lower stages tend
to require more structure than upper as they need to bear their own weight
plus that of the stages above them, optimizing the structure of each stage
decreases the weight of the total vehicle and provides further advantage.
17.13.3
Disadvantages
Staging requires the vehicle to lift motors which are used only at a later
stage. Further, it makes entire system more complex and harder to build.
Nevertheless the savings are so great that every rocket ever used to deliver
a payload into orbit had staging of some sort.
In more recent times the usefulness of the technique has come into question due to developments in technology. In the case of the Space Shuttle the
costs of space launches appear to be mostly composed of the operational
costs of the people involved, as opposed to fuel or equipment. Reducing
these costs appears to be the best way to lower the overall launch costs.
New technology that is mainly in the theoretical and developmental stages
is being looked at to lower the costs of launch vehicles.
17.14
COMPARISON OF VARIOUS PROPULSION SYSTEMS
We have seen various propulsive devices, which include air breathing engines and rockets. In this section, we will make the comparison of various
propulsion systems which respect to propulsive efficiency, overall efficiency
and performance details (Figs.17.15, 17.16 and 17.17).
The considerations of thrust and propulsive efficiency are by no means
the whole story in comparing various propulsion systems. We must consider, for example, the thermodynamic efficiency of the engine cycle, or
how efficiently the chemical energy of the fuel is converted to useful work
such as driving the propeller and compressor, or accelerating the exhaust
jet. The weight and size of the engine, its complexity, and its service life
are other factors to be considered.
17.15
PROPULSIVE EFFICIENCY
To go deeper into the subject of propulsive effectiveness, we can consider
the propulsive efficiency. Marine engineers call it the Froude efficiency
after William Froude (1810-1879) who first used it. Propulsive efficiency is
defined as the ratio of useful power output to the rate of energy input. For
610
Ro
ll
pe
ro
Propulsive efficiency
P
cke
t
Tu
rbo
jet
Gas Turbines
er
Flight velocity, V
0
Fig. 17.15 Propulsive efficiency at different speeds of various
system
propulsion
50
Overall efficiency, %
40
Rocket
Ramjet
Reciprocating
aircraft
30
Turbojet
20
Automobile
10
0
0
800
1600
2400
3200
8000
9000
Speed, kph
Fig. 17.16 Efficiency of various propulsion systems at different speeds
Rocket Propulsion
Power plant
Thrust per unit �
engine weight
611
Frontal area� Specific fuel �
per unit thrust consumption
Rocket
Ramjet
Turbojet
Piston
Fig. 17.17 Performance details of different propulsive systems
flight, propulsive efficiency is expressed as:
thrust × flight speed
It is to be noted that the various propulsion systems operate best in different speed ranges. This is shown qualitatively in Fig.17.15. The propeller
is the most efficient propulsive method at low speeds, while the jet engine
achieves best efficiency only at relatively high flight speeds. The very high
exhaust velocities of the rocket make its propulsive efficiency high only at
very high flight speeds. Figure 17.16 shows the overall efficiency of various
propulsive devices. Figure 17.17 shows the performance details of different
propulsive systems. It is clear that the maximum thrust is obtained from
the rockets whereas it is least for piston engines. At the same time it may
be noted that the specific fuel consumption is the highest for rockets and
lowest for the piston engines. However, from the point of view of frontal
area the rocket has the advantage of being the least front area whereas it
is maximum for the piston engines.
Review Questions
17.1 What is the basic difference between rocket propulsion and jet propulsion? Can rockets work in vacuum?
17.2 How rockets are classified? What is the stage of development of each
type?
17.3 Derive a general expression for the thrust produced by a chemical
612
Gas Turbines
rocket and hence discuss the importance of the molecular weight of
the propellants.
17.4 What are the factors which limit the thrust obtainable from chemical
rockets ?
17.5 What is the condition for the maximum thrust in a chemical rocket?
Derive an expression for the same.
17.6 Draw a schematic diagram of a solid propellant rocket and explain its
working. What are the applications of this type of rocket?
17.7 Briefly describe the two types of solid propellant rockets.
17.8 Draw a schematic diagram of a liquid propellant rocket. What are the
different systems of injecting liquid propellants into the combustion
chamber?
17.9 Discuss the mechanism of burning of various types of liquid propellants.
17.10 What is meant by monopropellant and bipropellant fuels for rockets?
Give important examples of each type. What is the difference in the
application of a monopropellant and a bipropellant rocket engine?
17.11 What are the desirable requirements of a liquid propellant for rockets?
17.12 Compare the advantages and disadvantages of solid and liquid propellants?
17.13 Explain the concept of free radical propulsion.
17.14 Describe how nuclear energy can be used for propulsion of rockets.
Has it been used so far?
17.15 Discuss, with suitable sketches, the three systems of electrodynamic
rocket propulsion.
17.16 What factors are important in the comparison of propulsive devices?
State the optimum operational range, specific fuel consumption and
relative weights for various propulsion devices.
17.17 What are the advantages and disadvantages of rocket engines?
17.18 What are the various types of oxidizers in use in rockets?
17.19 Describe the use of hydrogen peroxide as a monopropellant. What is
the function of hydrogen peroxide in a bipropellant rocket engine?
17.20 Discuss the relative merits of gas-pressurization rocket power plant
and the pump-pressurization system.
17.21 Briefly compare the various rocket propulsion devices.
17.22 What is meant by staging? Explain why multi-staging is required.
Rocket Propulsion
613
17.23 What are the advantages and disadvantages of multistage rockets?
17.24 With a neat graph compare the various propulsive and overall efficiency of various propulsion devices (air breathing and rocket) with
respect to flight speed.
17.25 With a neat diagram compare the performance parameters of rocket,
Ramjet, turbojet and piton engines.
Multiple Choice Questions (choose the most appropriate answer)
1. The first report on rockets was published in the year
(a) 1909
(b) 1919
(c) 1929
(d) 1939
2. The magnificent moon landing by Neil Armstrong was in the year
(a) 1949
(b) 1959
(c) 1969
(d) 1979
3. For rocket propulsion the nozzle used is
(a) convergent type
(b) divergent type
(c) convergent divergent type
(d) any type can be used
4. Burning of propellant occurs at
(a) constant-volume
(b) constant-pressure
(c) constant temperature
(d) none of the above
5. Thrust in a rocket depends on
(a) pressure in the combustion chamber
(b) properties of the propellant
(c) geometrical shape of the rocket
(d) all of the above
614
Gas Turbines
6. Chemical rockets use propellant which is
(a) solid
(b) liquid
(c) free radical
(d) all of the above
7. Burning of a propellant in a rocket is due to
(a) subsonic combustion
(b) supersonic combustion
(c) deflagration
(d) chemical decomposition
8. Mono propellant is one which requires
(a) fuel
(b) fuel and oxidizer
(c) oxidizer
(d) none of the above
9. Plasma rocket propulsion and ion rocket propulsion comes under the
category of
(a) electrical propulsion
(b) mechanical propulsion
(c) chemical propulsion
(d) nuclear propulsion
10. If a stable chemical material is supplied with sufficient energy to break
the energy bonds and if it is used for rocket propulsion, it is called
(a) nuclear propulsion
(b) free radical propulsion
(c) electrodynamic propulsion
(d) plasma rocket propulsion
Ans:
1. – (b)
6. – (d)
2. – (c)
7. – (c)
3. – (c)
8. – (c)
4. – (b)
9. – (a)
5. – (d)
10. – (b)
Appendix
1. THE SIMPLE GAS TURBINE CYCLE
2. THE HEAT EXCHANGE CYCLE
616 Gas Turbines
3. THE REHEAT CYCLE
4. THE REHEAT AND HEAT EXCHANGE CYCLE
Appendix
5. THE INTERCOOLED CYCLE
6. THE INTERCOOLED CYCLE WITH HEAT EXCHANGER
617
618
Gas Turbines
Index
actual cycle
performance, 156
adiabatic flow
one-dimensional, 496
adiabatic process, 13
aero-thermodynamic theory, 25
aerodynamic loss
annulus, 354
profile, 354
secondary, 354
tip clearance, 355
aerofoil
cambered, 56
uncambered, 56
afterburner, 229, 247
air angles, 337, 341
air cooled blades, 518
air pollution, 549
air-fuel ratio, 151
aircraft engine
single-spool, 524
airfoil, 54
elementary theory, 54
angle of incidence, 56
angularity coefficient, 494
annulus loss, 354
application
aircraft, 573
automotive, 572
gas pumping, 571
gas turbine, 72, 569
locomotive, 572
marine, 570
power generation, 570
process, 573
pulse jet, 224
ramjet, 220
small gas turbine, 569
arc plasma rocket engine, 604
area-velocity relation, 26
arrangement
closed-cycle, 63
open-cycle, 63
single-shaft, 63
augmentation
thrust, 246, 247
axial flow compressor, 333, 335
geometry, 335
historical background, 333
working principle, 335
backward swept blades, 289
Balje’s formula, 302
basic definitions, 19
binder, 592
bipropellant, 598
bipropellant fuels, 593
bipropellant-oxidizers, 593
blade, 501
air cooled, 518
backward swept, 289
backward-curved, 294
cast, 520
conway, 518
cooling, 513, 514
effect of shape, 294
efficiency, 427
fabricated, 518
fixing, 508
forged, 518
forward-curved, 294
loading, 427
manufacture, 505
materials, 501
radial, 295
radial-curved, 294
radial-tipped, 289
selection of materials, 502
619
620
INDEX
spey, 519
turbine, 501
turbine rotor, 421
tyne, 518
blade cooling
requirements, 517
blade fixing, 508
anchor pin , 509
bulb root method, 509
fir tree method, 510
grub screw method, 510
tee and double tee, 509
welding, 510
blade manufacture, 505
extrusion method, 505
tadpole method, 505
blade profiles, 472
blade row
rotor, 351
stator, 353
blade shapes, 287
bleed burn cycle, 247, 252
boundary layer, 35, 340, 485
laminar, 36
separation, 36
turbulent, 36
Brayton cycle, 229
camber line, 54
carbon deposits, 403
carbon preferential burning, 394
centrifugal compressor, 279
essential parts of, 280
losses, 310
principle of operation, 282
centrifugal machines, 44
centripetal machines, 44
chemical kinetics, 401
chemical rocket, 589
free radical, 589
liquid propellant, 589
solid propellant, 589
choking, 311
classification
rocket, 582
Clausius statement, 12
closed-cycle, 68
advantages, 71
basic requirement of working medium,
71
disadvantages, 71
coefficient
angularity, 494
discharge, 493, 494
flow, 493
gross thrust, 493, 497
nozzle, 491
velocity, 494
combustion, 4
efficiency, 402
form of system, 402
intensity, 400, 401
process in gas turbine, 405
spontaneous, 396
combustion chamber, 3
annular, 403
arrangements, 409
geometry, 406
performance, 398
requirement, 403
combustion efficiency, 150
combustion process, 393
combustion systems, 3, 393
combustion theory, 394
complex cycle, 99, 165
component matching, 523
composite propellants, 592
compression, 4
compressor, 3, 47
axial flow, 333
centrifugal, 279, 333
efficiency, 140
flow analysis, 290
low-pressure, 279
medium-pressure, 279
requirement, 333
rotor, 42
stage efficiency, 343
supersonic, 468
turbo-, 279
work input, 339
continuity equation, 26
control surface, 8
cooling
air, 515
aspects, 575
INDEX
external, 513
internal, 514
liquid, 515
passage, 517
cycle, 8
actual, 156
aircraft propulsion, 79
arrangement, 63
bleed burn, 247, 252
Brayton, 229
closed, 63, 68
comparison, 101
complex, 99
efficiency, 151
Ericsson, 103
heat exchange, 83, 159
ideal, 79
intercooled, 92, 163
intercooled and reheat, 96
intercooled with heat exchange
and reheat, 99, 165
intercooled with heat exchanger,
94
intercooled with reheat, 165
jet propulsion, 213
Joule, 229
open, 63
performance, 79
practical, 3
reheat, 86, 159
reheat and heat exchange, 89,
159
simple, 157
simple gas turbine, 80
blade ring, 303, 305
vaned, 303
vaneless, 302
discharge coefficient, 493, 494
double-base propellants, 592
drag, 56
drag force, 59
dynamic temperature, 138
effect
altitude, 246
centrifugal, 44
external, 43
impulse, 43
internal, 44
reaction, 44
suction, 56
efficiency, 451
blade, 427
combustion, 150
compressor, 140, 309
compressor stage, 343
cycle, 151
isentropic, 47
overall, 3, 245
polytropic, 152, 155
propeller, 241
propulsive, 242, 244
ram, 238
small-stage, 152
stage, 427
transmission, 242
turbine, 140, 141
electro dynamic propulsion, 602
plasma rocket, 602, 603
ion rocket, 602
definitions and laws, 7
photon, 602
deflagration, 589
degree of reaction, 45, 345, 423, 439– electrodynamics, 583
emission standards
441
aircraft, 551
fifty per cent, 347
hydrocarbon, 554
high reaction stage, 349
smoke exhaust, 553
hundred per cent, 448
stationary engine, 555
low reaction stage, 347
energy, 9
negative, 449
equation, 15, 42
zero, 452
gravitational potential, 15
degree of turbulence, 35
internal, 10, 15
density, 8
kinetic, 15
diffuser, 280, 294, 302, 484
621
622
INDEX
transfer, 16, 39, 43, 293
transformation, 16, 39, 423
energy equation, 15
engine
afterburning, 490
air breathing, 214
backpressure control, 490
gas turbine, 214, 215
hybrid, 591
liquid propellant rocket, 595
NOx reduction, 559
performance, 516
propeller, 214
pulse jet, 215, 220
ramjet, 215, 216
rocket, 214, 581
turbofan, 215
turbojet, 213, 215, 227
turboprop, 213, 215, 224
enthalpy, 10
enthalpy loss coefficient, 438
enthalpy stagnation, 17
enthalpy-entropy diagram, 284, 342,
436, 442
entropy, 12
equilibrium running diagram, 526
Ericsson cycle, 103
Euler’s energy equation, 286
exhaust, 574
exhaust nozzle area ratio, 490
expansion waves, 30
hypersonic, 25
inviscid, 21
isentropic, 34
laminar, 34
losses, 143, 353
mixed, 44
nozzles, 421
quasi-one-dimensional, 26
radial, 44
Rayleigh-type, 34
rotor blade row, 351
secondary, 340, 356
stator blade row, 353
steady, 20
transonic, 25
turbulent, 34
unsteady, 20
flow coefficient, 301, 344, 451
flow process, 14
flow through blade rows, 351
fluid, 19
velocity, 19
fluid dynamic equation, 7
fluid dynamics analysis, 39
forward speed, 246
free radical propulsion, 601
free radical rocket, 589
friction coefficient, 33
friction factor, 36
future development, 578
gas
Fanno line flow, 32
Fanno-type flow, 34
fifty per cent reaction, 452
fifty per cent reaction stage, 347
finite disturbance, 25
flame stabilization, 393
flame stabilizing zone, 407
flow
axial, 44
coefficients, 427
compressible, 20
compressor, 290
constant-area duct, 32
Fanno-type, 34
friction, 32
heat transfer, 32
ideal, 11
perfect, 11
real, 11
semi-perfect, 11
gas pressurization system, 596
gas turbine, 2, 4
applications, 72, 569
assessment, 567
combustor, 398
comparison, 73
components, 3, 4
engine, 214, 215
features, 574
parallel flow, 64
practical cycle, 137
process of combustion, 405
INDEX
series flow, 64
simple, 2
technology, 215, 279
gross thrust, 237
gross thrust coefficient, 497
heat exchange cycle, 83
heat exchanger, 63
effectiveness, 147, 148
high reaction stage, 349
high temperature operation, 510
hybrid engine, 591
hydraulic diameter, 33
hydro-turbomachines, 16
hydrogen preferential burning, 394
hydroxylation, 394
hypersonic flow, 25
ideal cycle, 79
assumptions, 80
ideal energy transfer, 286
ideal gas, 11
ignition, 404
ignition delay, 396
impeller, 280, 293
blade shape on performance, 294
channel, 296
components, 280
hub, 281
inducer, 281
radial-vaned, 310
shroud, 281
vanes, 281
impulse
machines, 422
turbine, 428
impulse stage, 421
single, 422
impulse turbine
pressure compounding of multistage, 435
inducer, 281, 293
inlet casing, 280, 291
inlet diffuser, 238
inlet drag, 237
inlet guide vanes, 287, 336, 337
inlet momentum, 237
inlets, 481
623
design variables, 483
performance of, 482
subsonic, 482
supersonic, 486
intercooled and reheat cycle, 96
intercooled cycle, 92
intercooled cycle with heat exchange
and reheat, 99
intercooled cycle with heat exchanger,
94
ion rocket, 583
ion rocket propulsion, 602
irreversible process, 13
isentropic flow, 34
isentropic process, 13
isentropic velocity ratio, 450
jet propulsion, 213
analysis, 213
cycle, 213
principle, 213
Joule cycle, 229
Kelvin-Planck’s statement, 12
laminar flow, 34
leaving losses, 243
lift, 56
lift force, 59
liquid propellant, 598
advantages, 600
bipropellant, 598
monopropellant, 598
requirement, 600
liquid propellant rocket, 589
liquid propellant rocket engine, 595
gas pressurization system, 596
pump pressurization system, 596,
597
loading coefficient, 345, 451
loss, 451
cascade, 356, 451
coefficient, 439
cold, 399
frictional, 310
hot, 399
leaving, 243, 451
mechanical, 150
parasitic, 144
624
INDEX
pressure, 143, 399
profile, 354
stage, 356
tip clearance, 355
total pressure, 403
low reaction stage, 347
Mach angle, 22
Mach cone, 22
Mach lines, 22
Mach number, 17, 22
Mach waves, 22
machine
half degree reaction, 423
multistage, 423
machines
axial, 44
classification of, 44
impulse, 45
reaction, 45
magneto plasma rocket engine, 604,
605
mass velocity, 32
match point, 525
matching procedure, 543
maximum utilization factor, 430
momentum equation, 7
momentum thrust, 236
monopropellant, 593
multistage rocket, 608
negative energy transfer, 42
negative reaction, 449
net thrust, 236
Newton’s laws
second law of, 213
third law of, 213
Newtonian fluid, 21
NOx formation, 557
NOx reduction, 559
noise, 560, 575
broad-band, 561
core, 561, 562
fan, 561
jet, 561
multiple-tone, 562
reduction, 565
sources of, 560
standards, 562
non-flow process, 13
normal shock wave, 28
normal shock waves, 28
nozzle, 435, 481
coefficients, 491
convergent, 488
convergent-divergent, 489
efficiency, 451
ejector, 489
exhaust, 488
flow, 421
flow characteristic, 536
functions, 489
performance, 491, 496
stator, 421
variable-area, 490
nuclear propulsion, 583, 601
nuclear propulsion unit, 601
oblique shock, 30, 31
off-design characteristic, 361
off-design operation, 359
one-dimensional flow area, 493
optimum expansion ratio, 587
overall efficiency, 245
performance characteristics, 310, 523,
526
performance coefficients, 344
performance evaluation, 523
performance graphs, 451
performance parameter, 307
photon propulsion, 602, 606
pi theorem, 50
piloted operation, 215
pilotless operation, 215
plasma jet, 583
plasma rocket propulsion, 602, 603
polytropic efficiency, 152, 155
positive energy transfer, 42
power
propulsive, 240, 243
thrust, 244
power input factor, 307
practical cycle, 137
pressure, 8
pressure coefficients, 308
INDEX
pressure drop coefficient, 239
pressure loss, 143
pressure loss factor, 399
pressure thrust, 236
primary zone, 407, 409
process, 8
adiabatic, 13
flow, 14
irreversible, 12, 13
non-flow, 13
reversible, 12, 14
steady flow, 45
profile loss, 354
propellant, 581, 593
characteristics, 593
free radical, 589
grain, 589, 590
liquid, 589
solid, 589, 592, 593
propellant grain, 589, 590
propeller
engine, 214
propulsion
arc plasma rocket engine, 604
comparison systems, 609
electro dynamic, 602
electrodynamics, 583
free radical, 601
ion rocket, 602
magneto plasma rocket engine,
604, 605
nuclear, 583, 601
photon, 583, 602, 606
plasma rocket, 602, 603
principle, 583
propulsive
efficiency, 242, 244, 609
power, 240
propulsive power, 243
pulse jet
advantages, 223
applications, 224
disadvantages, 223
engine, 220
pump pressurization system, 596, 597
quasi-one-dimensional flow, 25, 26
radial-tipped blade, 289
625
ram effect, 216, 246
ram efficiency, 238
ram pressure, 216
ramjet
advantages, 219
applications, 220
characteristics, 220
disadvantages, 220
engine, 216
performance, 218
thermodynamic cycle, 217
Rayleigh-type flow, 34
reaction machine, 422
pure, 435
reaction stage, 421, 422
fifty per cent, 443
hundred per cent, 448
single, 422
zero degree, 442
reaction turbine, 421, 435
fifty per cent, 445
multistage, 436
reciprocating engine, 214
relative eddy, 296, 297
reversible process, 12
Reynolds number, 21
rocket, 581
advantages, 582
chemical, 582, 584, 589
classification, 582
comparison, 606
engine, 581
free radical, 589
hybrid engine, 591
ion, 583
liquid propellant, 589
optimum expansion ratio, 587
performance capabilities, 582
solid propellant, 589
rocket engine, 214, 581
liquid propellant, 595
rocket propulsion, 581
principle, 583
rotating machines, 39
dimensional analysis, 48
non-dimensional parameters, 51
rotating stall, 364
rotational effects, 41
626
INDEX
rotor blade row, 351
rotor enthalpy loss coefficient, 344
rotor pressure loss coefficient, 344
secondary loss, 354
shock, 470
shock wave, 28, 30
shroud, 281
single-spool turbojet engine, 533
skin friction, 36
slip factor, 297, 299
slip velocity, 297, 299
small-stage efficiency, 152
solid propellant, 592, 593
advantages, 591
composite, 592
double-base, 592
solid propellant rocket, 589
specific heat
constant-pressure, 10
constant-volume, 10
specific impulse, 237
specific static thrust, 237
specific thrust, 237
stage
efficiency, 427
pressure ratio, 450
stage losses, 356
stage velocity triangles, 336
stage work, 356
staging, 607
stagnation
density, 18
enthalpy, 17
pressure, 18, 139
pressure loss coefficient, 439
properties, 138
state, 19
temperature, 17, 138
values, 18
velocity sound, 18
stalling, 361, 363
Stanitz’s formula, 301
state, 8
stator blade row, 353
steady-state operating point, 535
Stodola’s formula, 300
stream tube, 20
streamline, 19
streamtube area-velocity relation, 25
subsonic compressor, 467
suction side, 56
supersonic axial flow compressors, 469
supersonic axial flow turbines, 475
supersonic compressor, 467–469
supersonic radial compressor, 474
surge line, 490
surging, 311, 361
swirl atomizer, 414
system, 7
closed, 8
open, 8
temperature, 8
thermal efficiency, 240
thermal jet, 214
thermodynamic
first law of, 7, 9
second law of, 7, 12
throat, 27
thrust, 41, 235, 236, 586
augmentation, 246, 247
basic laws of, 235
equation, 235
gross, 237
net, 236
power, 242, 244
pressure, 236
reversers, 491
specific, 237
tip clearance, 356
tip clearance loss, 355
torque-speed characteristics, 527
transmission efficiency, 242
transonic flow, 25
trigonometrical relations, 425
turbine, 3, 47, 421
blades, 501
efficiency, 140, 141
impulse, 421
reaction, 421
turbine blades, 501
turbine rotor, 42
turbo-compressor, 279
turbojet, 213
advantages, 234
INDEX
627
components, 227
disadvantages, 235
efficiency, 238
engine, 227
performance, 232
thermal efficiency, 239, 240
thermodynamics cycle, 229
turboprop, 213
advantages, 226
applications, 227
components, 224
disadvantages, 227
engine, 224
performance, 226
thermodynamic cycle, 225
turboprop engines, 280
turbulence, 35
turbulent flow, 34
inlet guide, 287, 336, 337
upstream guide, 337
variable specific heat, 149
velocity coefficient, 494
velocity triangle, 287, 297, 336, 341,
349, 424, 428, 431
entry, 287
exit, 289, 295
velocity-compounding, 430
virtual pressure rise, 43
viscosity, 21
dynamic, 21
kinematic, 21
volute casing, 306
volute passage, 307
cross-section, 307
vortex generators, 485
vanes, 283
curved, 283
zone of action, 22
zone of silence, 22
water-alcohol mixture, 250
waves propagation, 23
unique incidence, 477
utilization factor, 427–429, 434, 445, work done factor, 340
work ratio, 143
447