Fundamentals of Turbomachinery 9788120337756

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Fundamentals of Turbomachinery

Fundamentals of

TURBOMACHINERY B.K. VENKANNA Professor Department of Mechanical Engineering Basaveshwar Engineering College Bagalkot

New Delhi-110001 2009

FUNDAMENTALS OF TURBOMACHINERY B.K. Venkanna © 2009 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-3775-6 The export rights of this book are vested solely with the publisher. Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Rajkamal Electric Press, Plot No. 2, Phase IV, HSIDC, Kundli-131028, Sonepat, Haryana.

To my wife

Swati for her unflinching support in every walk of my life and standing by me during my hard times

Contents

Preface Acknowledgements

1.

xv xvii

Introduction to Turbomachines 1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.8 1.9 1.10 1.11 1.12

Introduction 1 1.1.1 Solids 1 1.1.2 Liquids and Gases 1 Fluid Machines 2 Functional Classification of Fluid Machines 2 Turbomachines 3 Parts of a Turbomachine 3 Comparison between Positive Displacement Machines and Turbomachines 5 Basic Laws and Equations 6 1.7.1 Continuity 6 1.7.2 Steady Flow Energy Equation (First Law of Thermodynamics) 1.7.3 Entropy (Second Law of Thermodynamics) 8 Types of Turbomachines 9 Turbines 10 Pumps and Compressors 10 Fans and Blowers 10 Dimensionless Parameters and Their Physical Significance 11 vii

1–57

6

viii

Contents

1.13

Dimensional Analysis 11 1.13.1 Fundamental Quantities 11 1.13.2 Secondary Quantities or Derived Quantities 11 1.13.3 Dimensional Homogenity 11 1.14 Buckingham’s p-Theorem 12 1.15 Procedure for Applying Buckingham’s p-Theorem 12 1.16 Application of Dimensional Analysis to a General Fluid Flow Problem 14 1.16.1 Physical Significance of p Terms 17 1.17 Application of Dimensional Analysis to Turbomachines 19 1.18 Significance of p Terms 21 1.18.1 Capacity Coefficient or Flow Coefficient or Specific Capacity or Discharge Coefficient 21 1.18.2 Head Coefficient or Specific Head 22 1.18.3 Power Coefficient or Specific Power 22 1.18.4 Reynold’s Number 23 1.18.5 Effect of Reynold’s Number 23 1.18.6 Specific Speed 24 1.18.7 Definition of Specific Speed 25 1.19 Examples 26 Important Equations 54 Review Questions 56 Exercises 57

2.

Energy Transfer in Turbomachines 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

58–157

Euler Turbine Equation 58 Alternate Forms of Euler Turbine Equation 61 Components of Energy Transfer 62 The Steady Flow Equation of the First Law of Thermodynamics 63 Degree of Reaction 64 General Analysis of a Turbomachine 65 2.6.1 Effect of Blade Discharge Angle b2 on Energy Transfer and Degree of Reaction 65 General Analysis of Centrifugal Pumps and Compressors 69 2.7.1 Effect of Blade Discharge Angle on Performance 69 2.7.2 Theoretical Head Capacity Relationship 69 General Analysis of Axial Flow Compressors and Pumps 76 2.8.1 General Expression for Degree of Reaction 76 2.8.2 Velocity Triangles for Different Values of Degree of Reaction General Analysis of Turbines 83 2.9.1 Utilization Factor (e) 83 2.9.2 Axial Flow Turbines 85 2.9.3 Radial Flow Turbines 92

81

Contents

ix

2.10

Condition for Maximum Utilization: Axial Turbine 93 2.10.1 Reaction Turbine 93 2.10.2 Impulse Turbine 95 2.11 Optimum Blade Speed Ratio (f OPT) for Different Types of Turbines 97 for Maximum Energy Transfer (W.D.)max 2.11.1 Reaction Turbine 97 2.11.2 Impulse Turbine 97 2.12 Examples 97 Important Equations 152 Review Questions 155 Exercises 157

3.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes 3.1 3.2 3.3 3.4

3.5

3.6

158–246

Velocity of Sound or Sonic Velocity 158 Mach Number 161 Classification of Fluid Flow 161 Stagnation and Static Properties 162 3.4.1 Static State 162 3.4.2 Stagnation State 162 162 3.4.3 Stagnation Enthalpy (h0) 3.4.4 Stagnation Temperature (T0) 163 3.4.5 Stagnation Pressure or Total Pressure (p0) 163 164 3.4.6 Stagnation Density (r0) 3.4.7 Stagnation Velocity of Sound (a0) 164 Compression Process 165 3.5.1 Isentropic Efficiency or Adiabatic Efficiency or Isothermal Efficiency or Compression Efficiency 166 3.5.2 Overall Isentropic Efficiency, Stage Efficiency, Comparison and Relation between Overall Efficiency and Stage Efficiency 168 3.5.3 Polytropic Efficiency or Infinitesimal Stage Efficiency (hp) of a Compression Process 171 3.5.4 Constant Stage Pressure Ratio 174 3.5.5 Preheat Factor (PF) 176 Expansion Process 176 3.6.1 Isentropic Efficiency or Adiabatic Efficiency or Expansion Efficiency 176 3.6.2 Overall Isentropic Efficiency, Stage Efficiency and Comparison and Relation between Stage Efficiency and Overall Efficiency for Expansion Process 179 3.6.3 Polytropic Efficiency or Infinitesimal Stage Efficiency (hp) of an Expansion Process 182

x

4.

Contents

3.6.4 Multistage Machine with Constant Stage Pressure Ratio 3.6.5 Reheat Factor for Expansion Process (RF) 188 3.7 Examples 188 Important Equations 239 Review Questions 243 Exercises 245

185

Centrifugal Compressors and Pumps

247–347

CENTRIFUGAL COMPRESSORS 4.1 Working Principle, Components and Description 247 4.2 Work Done and Pressure Rise 248 4.2.1 Enthalpy–Entropy Diagram 252 4.2.2 Overall Pressure Ratio 253 4.2.3 Limiting Inlet Velocity 255 255 4.3 Pressure Coefficient (fp) 4.4 Blade Angles at Eye Root and Eye Tip 256 4.5 Eye Conditions for an Impeller 257 4.6 Influence of Impeller Blade Shape 258 4.7 Slip Factor (s ) 259 4.8 Power Factor (j) 260 4.9 Prewhirl and Inlet Guide Vanes 261 4.10 Diffuser 262 4.10.1 Vanless Diffuser 262 4.10.2 Determination of Diffuser Inlet Angle, Width and Length of the Diffuser Passages 263 4.10.3 Width of the Impeller Channel 264 4.11 Surging of Centrifugal Compressors 265 CENTRIFUGAL PUMPS 4.12 Introduction 266 4.13 Centrifugal Pumps 266 4.14 Working Principle 266 4.15 Main Parts of a Centrifugal Pump 267 4.15.1 Impeller 267 4.15.2 Casing 268 4.15.3 Suction Pipe, Foot Valve and a Strainer 268 4.15.4 Delivery Pipe 268 4.15.5 Delivery Valve or Check Valve or Regulating Valve 268 4.16 Classification of Centrifugal Pumps 268 4.16.1 According to the Working Head 268 4.16.2 According to the Type of Casing 269 4.16.3 According to Fluid Entrance to the Impeller 270 4.16.4 According to the Direction of Flow of Water through the Impeller 270 4.16.5 According to Number of Impellers 271

Contents

4.16.6 According to Liquid Handled 271 4.16.7 According to Specific Speed 272 4.17 Heads of a Centrifugal Pump 273 4.17.1 Static Head (HS) 273 273 4.17.2 Manometric Head (Hm) 4.18 Efficiencies of Centrifugal Pump 274 4.18.1 Manometric Efficiency (hmano) 274 4.18.2 Mechanical Efficiency (hm) 275 4.18.3 Hydraulic Efficiency (hH) 275 275 4.18.4 Volumetric Efficiency (hv) 4.18.5 Overall Efficiency (ho) 275 4.19 Work Done by the Pump 275 4.20 Pressure Rise in Pump, Impeller and Manometric Head 4.21 Minimum Starting Speed 278 4.22 Multistage Pumps 279 4.23 Cavitation 279 4.24 Examples (Centrifugal Compressors) 280 4.25 Examples (Centrifugal Pumps) 315 Important Equations 342 Review Questions 344 Exercises 344

5.

278

Axial Flow Compressors 5.1 5.2 5.3 5.4 5.5

xi

Introduction 348 Description and Principle of Operation 349 Stage Velocity Triangle 350 Work Done 352 Temperature and Entropy Diagram for a Stage of an Axial Flow Compressor 352 5.6 Overall Pressure Ratio per Stage (pR0) 354 5.7 Work Done Factor (y) 355 5.8 Flow Coefficient (f) 356 5.9 Pressure Coefficient (fp) 356 5.10 Degree of Reaction (R) 356 5.11 Combined Velocity Triangles for Different Values of R 358 5.12 Radial Equilibrium Conditions 360 5.13 Air Angle Distribution 362 5.13.1 Free Vertex Flow 362 5.13.2 Constant Reaction Design 366 5.14 Examples 367 Important Equations 413 Review Questions 414 Exercises 415

348–417

xii

6.

Contents

Steam and Gas Turbines Introduction 418 Classification of Steam Turbines 418 Principle of Operation of Steam Turbines 419 6.3.1 Impulse Turbine 419 6.3.2 Reaction Turbine 420 6.3.3 Impulse Reaction Turbine 420 6.4 Impulse Staging and Need for Compounding 421 6.5 Methods of Compounding of Steam Turbine 422 6.5.1 Velocity Compounding 422 6.5.2 Pressure Compounding 423 6.5.3 Pressure and Velocity Compounding 424 6.6 Differences between Impulse and Reaction Turbines 425 6.7 Advantages of Steam Turbine over Other Prime Movers 425 6.8 Velocity Triangles for Impulse Turbine 425 6.9 Performance Parameters of Impulse Turbine 427 6.10 Effects of Friction and Blade Angles and Blade Efficiency 429 6.11 Condition for Maximum Efficiency 430 6.12 Multistage Impulse Turbine 431 6.13 Condition for Maximum Efficiency for Two-Stage Impulse Turbine 6.13.1 Utilization Factor 434 6.13.2 Degree of Reaction 434 6.14 Advantages and Disadvantages of Velocity Compounding 435 6.15 Impulse Reaction Turbines or Reaction Turbines 435 6.16 Condition for Maximum Efficiency for Reaction Turbine 438 6.17 Reheat Factor (RF) 442 6.18 Blade Design Parameters 443 6.19 Examples 445 Important Equations 479 Review Questions 484 Exercises 485

418–486

6.1 6.2 6.3

7.

Hydraulic Turbines 7.1 7.2 7.3

7.4

Introduction 487 Classification of Hydraulic Turbines 487 Unit Quantities 489 7.3.1 Significance of Unit Quantities 489 7.3.2 Unit Speed (Nu) 489 7.3.3 Unit Discharge (Qu) 490 7.3.4 Unit Power (Pu) 490 Pelton Wheel 490 7.4.1 Terminology 491

434

487–581

Contents

xiii

7.5

Components of the Pelton Wheel 492 7.5.1 Nozzle and Flow Regulating Mechanism 493 7.5.2 Jet Deflector 494 7.5.3 Braking Jet 494 7.5.4 Runner and Bucket 494 7.5.5 Casing 495 7.6 Velocity Triangles and Power for Pelton Wheel 495 7.6.1 Condition for Maximum Hydaulic Efficiency 498 7.6.2 Maximum Efficiency of Pelton Wheel 498 7.6.3 Turbine Efficiency 499 7.7 Francis Turbine 501 7.7.1 Types of Reaction Turbine 501 7.8 Components of Francis Turbine 502 7.9 Velocity Triangles, Power and Efficiency 503 7.10 Important Design Parameters of Francis Turbine 505 7.11 Draft Tube 506 7.11.1 Types of Draft Tube 506 7.11.2 Design of Draft Tube 507 7.11.3 Functions of Draft Tube 508 7.12 Kaplan Turbine 509 7.12.1 Design Parameters of Kaplan Turbine 509 7.13 Comparison between Impulse and Reaction Turbines 511 7.14 Examples 512 Important Equations 576 Review Questions 579 Exercises 579

Bibliography

583

Model Question Papers (with Answers)

585–642

Index

643–645

Preface

Power is one of the main needs for the development of any country. The major share of power generated is through the use of steam and gas turbines. Similarly, fans, compressors and blowers are some of the power absorbing machines used in almost all the industries. Hence, understanding the concepts of turbomachines is very important. Educationalists and scientists considered this subject to be very much essential for mechanical engineering students. Study of the principles of tuerbomachinery is therefore offered as a core subject for the mechanical engineering stream in all the universities. Many books are available exclusively on water turbines, on steam turbines, gas turbines and power absorbing turbomachines. There are also books on the market which deal with general aspects of turbomachines. But such books are meant for higher level studies. Therefore a need is felt by all for a book which offers the fundamentals of all the aspects of turbomachines. The present book is aimed at fulfilling this need and the author sincerely hopes that the book would serve this purpose. Hence, this book is intended to serve as a textbook for the undergraduate students in mechanical engineering as well as for the students in civil engineering who undergo a similar course. Generally, the books are of three types. First, written exclusively for practising engineers. Second, written exclusively for research purposes where rigorous mathematical treatment is provided with little stress on the art of problem solving and learning basic principles. The third category presents a huge number of problems without much emphasis on fundamental principles. This book has been written keeping in mind the requirement to clearly understand the fundamentals, the need to grasp the physics of different phenomena and the necessity to acquire the art of solving the problems. Hence it tries to provide a solid foundation in the theory through the application of the basic principles to solve practical problems. The book is the outcome of my long experience of teaching this subject and thermodynamics to undergraduate xv

xvi

Preface

engineering students. The discussions/interactions with the students during the class helped a lot while writing this book. The subject is presented in an elegant and simple manner, so that the students can grasp the essential concepts easily and quickly. A large number of problems, graded in the order of increasing complexity, have been solved to help the students in developing confidence in this subject. Despite best efforts, the book may not be free from errors, technical or otherwise. The author, will be grateful to all those who bring errors to his notice. Suggestions for the improvement of the book will be thankfully acknowledged. B.K.Venkanna

Acknowledgements

At the outset I am deeply indebted to my parents Late Sri B.V. Krishnamurthy and Smt. B.K. Vasuvamba who always encouraged me to take up challenging tasks in life. I recall the advice of my father, who used to say, “Money comes and goes, but morality stays and grows”. I thank members of Sri B.V.V. Sangha and Principal of Basaveshwar Engineering College, Bagalkot, Karnataka, for excellent and congenial academic environment of the college that inspired me to accept this challenge. I also wish to thank my friends, colleagues and especially Dr. V.R. Kabadi, H.O.D. of mechanical engineering department of the college, for moral support and encouragement received to take up this work. I acknowledge the students whose active interactions, comments and thought-provoking questions prompted me to present the topics in a more elegant manner. I owe a debt of gratitude to my wife, Swati, who is also my colleague in the department of mechanical engineering, Sri. Basaveshwar Engineering College, Baglkot, as without her continued encouragement, patience and understanding, I would not have reached the milestone of completion of this text. Interactions with her contributed to the overall improvement of the book. Finally, I and my wife thank profusely our beloved children, Arpita and Vinayak, who both allowed us to devote time to this book foregoing their due moments of love and care from us. B.K.Venkanna

xvii

1

Introduction to Turbomachines

1.1 INTRODUCTION There are many devices which produce mechanical power, out of which the turbine in many respects is the most satisfactory machine. The turbomachine in a crude form is said to have been used many centuries ago, but its development as a really practical form of prime mover has only taken place during the last 100 years. Today, it stands as the most important prime mover in existence. Similarly, there are many other devices such as compressors, fans, blowers, etc. which take power in, to boost either the pressure of the working fluid or the velocity of the working fluid. In general, turbomachines, turbines and compressors are now being used in electric power generation, aircraft propulsion and a wide variety of medium and heavy industries.

1.1.1 Solids A solid is a substance that has a definite shape, and which retains its shape until some external force is applied to change it. Under a pure compressive load, it undergoes an infinitesimal change in volume. It offers resistance to change in shape without a change in volume under the application of tangential forces.

1.1.2 Liquids and Gases A liquid takes the shape of a vessel into which it is poured, i.e. liquids do not have their own shape and size. On the other hand, a gas completely fills up the vessel which contains it. Liquids and gases do not offer any resistance to change in shape when a tangential force is applied. 1

2

Fundamentals of Turbomachinery

Even with a small force they undergo deformation. The continuous deformation under the action of tangential force causes liquids and gases to flow. It is because of this similarity in the dynamic behaviour of liquids and gases that they are called fluids.

1.2 FLUID MACHINES ‘Fluid machine’ is a general term used for all devices/machines that handle liquids and gases, i.e. fluids. A general classification of fluid machines is as follows: (i) Turbomachines: Generally known as rotadynamic machines. For example, rotary pumps, compressors, fans, blowers and turbines. (ii) Reciprocating machines: For example, reciprocating pumps and compressors, i.e. positive displacement machines. (iii) Various fluid lifting devices: For example, jet pumps, air lift pumps, pulsometer pumps and hydraulic ram. (iv) Pumps transmitting fluids (oil): Use fluids under pressure to operate and control systems. For example, gear pumps, constant delivery and variable delivery pumps, various applications and accessories relating to fluid systems.

1.3 FUNCTIONAL CLASSIFICATION OF FLUID MACHINES (i) According to energy consideration (a) Machines that supply energy to the fluids. Examples: Pumps and compressors. (b) Machines that extract energy from the fluids. Examples: Turbines (steam, gas, water). (c) Machines that are combination of both (a) and (b) Examples: Energy transmitters and torque converters. (ii) According to increase or decrease in pressure (a) Increase in pressure. Examples: Pumps (rotary or reciprocating), fans, compressors (rotary or reciprocating), propellers. (b) Decrease in pressure. Examples: Turbines (water, steam, gas) and wind mills. (iii) According to the direction of flow (a) (b) (c) (d)

Linear flow: Reciprocating pumps Radial flow: Centrifugal pumps Axial flow: Axial flow pumps and turbines Diagonal flow: Diagonal flow pumps

Introduction to Turbomachines

3

(iv) According to the direction of movement of machines (a) Rotadynamic machines where the main parts rotate. (b) Reciprocating machines where a piston or plunger moves forward and backward. (vi) According to the principle on which they are designed (a) Positive displacement machines. Examples: Plunger, rotary, vane and pneumatic pumps. (b) Centrifugal and propeller action. Examples: Radial, axial, mixed flow pumps and turbines. (c) Density difference. Examples: Air lifts, thermosiphon pumps and coolers. (d) Momentum transfer. Examples: Water jet pumps and ejectors. (e) Wave transmission of energy. Examples: Hydraulic ram, air or hydraulic brakes. (f) Viscosity difference pumps. (g) Gravity type pumps. Example: Persain pumps. (h) Mechanical advantage machines. Example: Screw pump. (i) Elasticity principle. Example: Diaphragam pumps. (j) Miscellaneous machines. Example: Heat pumps.

1.4 TURBOMACHINES Generally in turbomachines the transfer of mechanical energy occurs into or out of machine in a steady flow process. Turbo means spinning or whirling around. Turbomachines include those types of machines which produce head or pressure such as centrifugal pumps and compressors and those which produce power such as turbines. In turbomachines, the fluid is not positively contained but steadily flows through the machine undergoing changes in pressure by means of dynamic effects. Hence, the other name for turbomachines is ‘dynamic’ machines.

General definition A turbomachine is a device in which energy transfer occurs between a flowing fluid and a rotating element due to dynamic action and results in a change in pressure and momentum of the fluid.

1.5 PARTS OF A TURBOMACHINE A turbomachine is comprised of the following parts: (i) Rotor or impeller or runner (ii) Guide blade or stationary or fixed element or nozzle (iii) Shaft

4

Fundamentals of Turbomachinery

(iv) Housing or casing (v) Diffuser (vi) Draft tube

Rotor or impeller or runner Rotor is the rotating element of a turbomachine. It is fixed with blades or vanes and also called the impeller or runner depending upon the particular machine. For example, the rotating member of centrifugal pumps and centrifugal compressors is called the impeller. The rotating member of radial flow hydraulic turbines and pumps is called the runner. In contrast, the rotating member of axial flow gas and steam turbines is called the rotor. Energy transfer occurs between the fluid and the rotating member due to exchange of momentum between the two.

Guide blade or stationary or fixed element or nozzle The stationary element or guide blade is arranged depending upon the kind of flow required. The stationary element is not a compulsory part of every turbomachine. The ceiling fan is a turbomachine and there is no stationary element.

Shaft Either input shaft or output shaft or both may be necessary depending upon the type of turbomachines. For example: (a) Power absorbing turbomachine: only input shaft (b) Power generating turbomachine: only output shaft (c) Power transmitting turbomachine: both input and output shaft

Housing or casing The housing is not a compulsory part of every turbomachine. When the housing is present, it restricts the fluid so that it flows in a given space and does not escape in directions other than those required for energy transfer. A turbomachine having housing is called enclosed machine and the one having no housing is called extended machine. Volute: It is a type of casing where a spiral passage is used for the collection of the diffused fluid of a compressor or pump. The volute casing is used in hydraulic turbines to increase the velocity of the fluid before it enters the runner.

Diffuser A passage with increase in cross-sectional area in the direction of fluid flow, and which converts kinetic energy into static pressure head. It is usually situated at the outlet of a compressor, for example, axial flow compressor.

Draft tube It is a diffuser placed at the outlet of a hydraulic turbine. For example, Francis and Kaplan turbines.

Introduction to Turbomachines

5

1.6 COMPARISON BETWEEN POSITIVE DISPLACEMENT MACHINES AND TURBOMACHINES In a positive displacement machine, the interaction between the moving part and the fluid involves a change in volume or translation of the fluid or both. Fluid expansion or compression occurs without an appreciable change in the mass centre of gravity of the contained fluid. Hence, in these machines the macroscopic kinetic energy change and momentum change can be neglected. Because of the positive containment, the moving surface changes the fluid volume, i.e. the fluid cannot escape from the boundary except by leakage. Turbomachine

Positive displacement machine

1. Action (a) Dynamic (b) Pressure and momentum of the fluid changes.

(a) Nearly static (b) Volume of the fluid changes.

2. Operation (a) Pure rotary motion of the mechanical element.

(b) Steady flow of fluid. (c) The fluid state will be the same as that of the surroundings when the machine is stopped.

(a) Usually it is the reciprocating motion of the mechanical element but some rotary positive displacement machines are also built. Examples: Gear pump, vane pump. (b) Unsteady flow of fluid. (c) Entrapped fluid state is different from the surroundings when the machine is stopped and if heat transfers and leakage are avoided.

3. Mechanical features (a) Rotating masses can be completely balanced and vibrations eliminated. Hence high speeds can be adopted. (b) Light foundations suffice. (c) Design is simple. (d) Weight per unit output is less.

(a) Because of the reciprocating masses, vibrations are more. Hence low speeds are adopted. (b) Heavy foundations are required. (c) Mechanical design is complex because of valves. (d) Weight per unit output is more.

4. Efficiency of conversion process (a) Efficiency is low because of dynamic energy transfer. (b) The efficiency of the compression process is low.

(a) High efficiency because of static energy transfer. (b) The efficiencies of the compression and expansion processes are almost the same. (Contd.)

6

Fundamentals of Turbomachinery

Turbomachine

Positive displacement machine

5. Volumetric efficiency (a) It is almost 100%. (b) High fluid handling capacity per unit weight of machine.

(a) Much below that of a turbomachine because of valves. (b) Low fluid handling capacity per unit weight of machine.

6. Fluid phase change and surging (a) (b) (c) (d)

Causes cavitation in pumps and turbines. Erodes steam turbine blades. Deteriorates performance. Surging or pulsation leads to unstable flow. (e) Causes vibrations and may destroy the machine.

7. Operates between a moving fluid and a rotating element, resulting in thermodynamic and dynamic action.

No serious problems are encountered.

Operates between a near static fluid and a slow moving surface resulting in thermodynamic and mechanical action.

1.7 BASIC LAWS AND EQUATIONS The basic laws of thermodynamics and fluid mechanics are used in turbomachines. All or some may be used depending upon the situation.

1.7.1 Continuity  , remains constant, i.e. For steady flow through the control volume, the mass flow rate, m m

U1V1 A1

U2V2 A2

where the velocity vector ( V ) is perpendicular to the cross-sectional area A. For compressible flow machines, the mass flow rate (kg/s) is used and for incompressible flow machines (hydraulic machines), the volume flow rate (m3/s) is preferred.

1.7.2 Steady Flow Energy Equation (First Law of Thermodynamics) Considering unit mass of the working fluid and writing the general energy equation, we get from Figure 1.1 (u1„  p1X1 ) 

V12 z1 g  q 2 gc gc

w  (u2 „  p2 X2 ) 

V22 z2 g  2 gc gc

(1.1)

Introduction to Turbomachines

7

where u1¢ and u2¢ p1v1 and p2v2 v1 and v2 V1 and V2 z1 and z2 q and w h01 and h02

are the inlet and exit values of internal energy (kJ/kg) are the flow work done on or by fluid (kN-m) are the specific volumes (m3/kg) are the inlet and exit values of velocity of the fluid (m/s) are the inlet and exit values of potential energy (m) are the heat and work interactions between the surroundings and the system (control flow) (kJ/kg) are the inlet and exit values of stagnation enthalpy (total enthalpy). q

w Control surface

2

1 V1 A1 p1 u1

V2 A2 p2 u2

Turbomachine

Control volume

1 z1

Figure 1.1

2 z2

Steady flow energy process for a control volume.

In case of turbomachines, the rate of flow of working fluid is very high, the surface area available for transfer of heat is quite small and therefore, the process may be assumed to be adiabatic, i.e. q = 0. Equation (1.1) then becomes (cvT1  RT1 )  T1 (cv  R) 

\

w

V12 z1 g  2 gc gc

V12 z1 g  2 gc gc

w  (cvT2  RT2 )  w  T2 (cv  R) 

(cv  R)(T1  T2 ) 

V22 z2 g  2 gc gc

V12  V22 ( z1  z2 ) g  2 gc gc

If z1 = z2, w

V22 z2 g  2 gc gc

(cv  R)(T1  T2 ) 

V12  V22 2 gc

8

Fundamentals of Turbomachinery

c p (T1  T2 )  (h1  h2 ) 

V12  V22 2 gc

V12  V22 2 gc

È V12 Ø È V22 Ø É h1  2 g Ù  É h2  2 g Ù Ê Ê cÚ cÚ

w = h01 – h02 = –Dh0 or

dw = dh0

(1.2)

If we neglect the kinetic energy, then w = –Dh

(1.3)

The suffix ‘0’ represents the total head or stagnation conditions. In power generating turbomachines, w is positive so that Dh0 is negative, i.e. the total enthalpy of flowing fluid decreases from inlet to exit. In power absorbing turbomachines, mechanical energy input occurs so that the stagnation enthalpy of the fluid increases from inlet to exit. We know that in turbomachines, the energy transfer between the fluid and the blades can occur only by dynamic action, i.e. all the work is done when the fluid flows over the rotor blades (not over the stator). Therefore, the stator is a flow directing element. In the absence of frictional losses, there is no stagnation enthalpy change when the fluid flows over the stator, only static enthalpy and kinetic energy or potential energy changes can occur.

1.7.3 Entropy (Second Law of Thermodynamics) The second law of thermodynamics states that for a fluid undergoing a reversible adiabatic process, the entropy change is zero. Entropy increases from inlet to exit, if the fluid undergoes an adiabatic or any other process. Due to the increase in entropy, the power developed by a turbine is less than the ideal isentropic power developed. Similarly, the work input to a pump is greater than the isentropic or ideal work input. The second law of thermodynamics can be written as: Gq T

ds

(1.4)

From the first law of thermodynamics, q – w = m(u2¢ – u1¢) dq – dw = du¢

i.e.

ds × T = dw + du¢ \

Tds = pdv + du¢ We have h = u¢ + pv

(1.5)

9

Introduction to Turbomachines

dh = du¢ + pdv + vdp

or

(1.6)

By using Eqs. (1.5) and (1.6), we have Tds = dh – vdp \

T0ds0 = dh0 – v0dp0

\

dh0 = v0dp + T0ds = –dw

(1.7)

1.8 TYPES OF TURBOMACHINES The turbomachines shown in Figure 1.2 are classified as follows: 1. According to energy consideration (a) Turbomachines transferring rotor energy to fluid energy, i.e. machines supply energy to the fluid as in pumps and compressors.

Figure 1.2

Types of turbomachines: (a) Centrifugal pump. (b) Axial flow pump. (c) Steam turbine. (d) Pelton wheel. (e) Francis turbine. (f) Kaplan turbine.

10

Fundamentals of Turbomachinery

(b) Machines transferring fluid energy to a rotor energy, i.e. machines that extract energy from the fluid as in turbines (steam, gas, water). 2. According to the direction of flow (a) (b) (c) (d)

Radial flow as in centrifugal pumps, fans, turbines and compressors Axial flow as in axial flow pumps, compressors, fans, and turbines Mixed flow as in Francis turbine. Tangential flow as in Pelton wheel

3. According to the action of fluid on the moving blade (a) Impulse machines Fluid energy is converted into impulsive force by changing the direction of fluid as in a steam turbine (De-Laval) and Pelton wheel. (b) Reaction machines The pressure energy of fluid continuously drops as it flows over the blades and the velocity increases. The fluid leaving the blades will exert a reactive force in the backward direction of its flow. This reactive force sets the blades in motion as in a steam turbine (Parson’s reaction turbine), Francis turbine, Kaplan turbine and Propeller turbine. 4. (a) Open or extended machines such as propeller, wind mill, fan. (b) Enclosed machines such as turbines, pumps and compressors. 5. According to the type of fluid handled (a) Water : Examples: Pumps, hydraulic turbines (b) Steam : Example: Steam turbines (c) Air or gas : Examples: Fans, compressors, blowers, turbines.

1.9 TURBINES The power generating turbomachines decrease the head or energy level of the working fluid passing through them. These machines are called turbines.

1.10 PUMPS AND COMPRESSORS The power absorbing turbomachines increase the head or energy level of the working fluid passing through them. These machines are called pumps and compressors.

1.11 FANS AND BLOWERS The power absorbing turbomachines which continuously move mass of air, gas or vapour at the desired velocity by the action of the rotor. Also, there may be a slight rise in pressure across the rotor.

Introduction to Turbomachines

11

1.12 DIMENSIONLESS PARAMETERS AND THEIR PHYSICAL SIGNIFICANCE A large number of variables are involved in describing the performance characteristics of turbomachines. Because of these large number of variables, more number of experiments have to be conducted in a performance test. It involves more time and the cost of design and experimentation will be more too. In order to reduce time and cost, these variables are grouped into non-dimensional quantities, which are less in number (manageable number) than the variable themselves. Then, in design or experimentation these non-dimensional quantities are varied instead of the large number of parameters forming these groups. Dimensionless quantities also have two other advantages. (i) The prediction of a prototype performance from the tests conducted on a scale model. (ii) To find the most suitable type of machine on the basis of maximum efficiency for a specified range of head, speed and flow rate.

1.13 DIMENSIONAL ANALYSIS From the previous description, we know that it is advantageous to reduce the number of variables into manageable numbers by grouping these variables into non-dimensional quantities. The technique or procedure used to get non-dimensional quantities is known as dimensional analysis. The dimensional analysis is a mathematical technique that deals with the dimensions of the quantities (see Table 1.1) involved in the process. It is based on the assumption that the process can be expressed by a dimensionally homogeneous equation with certain variables.

1.13.1 Fundamental Quantities Mass, length, time and temperature are called fundamental quantities since there is no direct relation between these quantities.

1.13.2 Secondary Quantities or Derived Quantities The quantities derived from fundamental quantities are called derived quantities or secondary quantities. Some examples of derived quantities are area, volume, velocity, force, acceleration, etc.

1.13.3 Dimensional Homogenity An equation is said to be dimensionally homogeneous if the fundamental dimensions have identical powers of M, L, T on both sides. Let us take an example: Q = AV i.e.

L3 L L3 = L2 ¹ = T T T

12

Fundamentals of Turbomachinery

1.14 BUCKINGHAM’S p-THEOREM The Buckingham’s p-theorem states that if there are n variables (independent and dependent variables) in a dimensionally homogeneous expression (i.e. in a physical phenomenon) and if these variables contain m fundamental dimensions (M-L-T), then the variables are grouped into (n-m) non-dimensional independent terms. Each term is called p term. Suppose a variable X1 depends upon independent variables X2, X3, ..., Xn, then the functional equation may be written as X1 = f(X2, X3, ..., Xn)

(1.8)

Equation (1.8) can also be written as f1(X1, X2, X3, ..., Xn) = C

(1.9)

where C = constant f1 = some function. In the above Eq. (1.9) there are n variables. If these variables contain m fundamental dimensions, then according to Buckingham’s p-theorem, f1(p1, p2, p3, ..., pn-m) = C

(1.10)

1.15 PROCEDURE FOR APPLYING BUCKINGHAM’S p-THEOREM 1. With the given data, write the functional relationship. 2. Write the equation in its general form. 3. Choose m repeating variables and write separate expressions for each p term, every p term must contain the repeating variables and one of the remaining variables. The number of repeating variables is equal to the number of fundamental dimensions of the problem. Following points must be considered while selecting repeating variables. (a) As far as possible independent variables should be selected. (b) The selection should be in such a way that one variable contains the geometric property, second variable contains the flow property and the third variable contains the fluid property. Variables with geometric property in turbomachines are: (i) (ii) (iii) (iv)

Length (l) Diameter (d) Thickness (t) Height (h)

i.e. i.e. i.e. i.e.

blade chord or blade length rotor diameter blade thickness blade height

Variables with flow property, i.e. kinematic variables are: (i) Velocity (u)

i.e. blade velocity

Ë Ìu Í

S dN Û 60 ÜÝ

Introduction to Turbomachines

(ii) (iii) (iv) (v) (vi) (vii)

13

Velocity (V) i.e. flow velocity Speed (N) i.e. rotation speed Volume flow rate (Q) ) Mass flow rate (m Acceleration (a) Angular velocity (w)

Variables with fluid property, i.e. dynamic variables affecting the performance of turbomachines are: (i) (iii) (v) (vii) (ix) (xi)

Gas density (r) Bulk modulus (K) Force (F) Elasticity (e) Specific weight (w) Resistance (W)

(ii) (iv) (vi) (viii) (x)

Dynamic viscosity (m) Pressure difference (Dp) Power (P) Surface tension (s) Stress

For geometric similarity, the ratios of the linear dimensions and the shape of the bodies in the model and prototype are the same, the values of the individual dimensions are immaterial. Kinematic similarity states that the ratios of velocities are the same in the model and prototype regardless of the individual values. This gives similar velocity triangles in both the model and prototype. In dynamic similarity the ratios of the various forces should be the same regardless of the individual values. 4. The repeating variables are written in exponential form. 5. With the help of dimensional homogeneity, find the values of a, b and c, etc. by obtaining simultaneous equations as explained in the subsequent Section 1.16. 6. Now, substitute the value of these exponents in the p terms. 7. Write the functional relation in the required form. Table 1.1

Dimensions of some physical quantities

Symbol

Quantity (Variable)

Dimensions

A v r V a M F p W.D. P I

Surface Volume Density Velocity Acceleration Momentum Force Pressure and Stress Energy and Work Power Moment of inertia

L2 L3 M/L3 L/T L/T2 ML2/T ML/T2 M/LT2 ML2/T2 ML2/T3 ML2 (Contd.)

14

Fundamentals of Turbomachinery

Table 1.1

Dimensions of some physical quantities (contd.)

Symbol

Quantity (Variable)

w a T e s m n t m Q h d a,b,c... w

Angular velocity Angular acceleration Torque Modulus of elasticity Surface tension Viscosity (absolute) Viscosity (kinematic) Angular momentum Mass flow rate Volume flow rate Enthalpy per unit mass Rotor diameter Machine dimensions Specific weight

Dimensions 1/T 1/T2 ML2/T2 M/LT2 M/T2 M/LT L2/T ML2T M/T L3/T L2/T2 L L M/L2T2

1.16 APPLICATION OF DIMENSIONAL ANALYSIS TO A GENERAL FLUID FLOW PROBLEM The detailed analysis of turbomachines involves the flow of fluid through a duct. It is useful to examine the general fluid flow problem by dimensional analysis. Consider a steady flow in a simple stationary system. Two geometrical variables t (thickness) and d (diameter) are chosen although more can be included. From the kinematic variables, either velocity or volume flow rate may be included. All the fluid properties r, m, w, e, s are included. Now we can the write functional relationship of these variables with pressure (pressure increment). Ddp = f(t, d, Q, r, m, w, e, s)

(1.11)

There are 9 variables and with 3 primary dimensions or repeating variables, 6 ps are necessary to describe the physical relationship completely. d, Q, r may be selected as repeating variables since they do not form a dimensionless product and they are then combined with t, Dp, m, s, e and w in turn. p1 term:

S1 M 0 L0 T 0

d a1 Q b1 U c1 t [L]a1 [L3 T 1 ]b1 [ML3 ]c1 L1

Equating the powers of M-L-T on both sides, For M :

c1 = 0

For L :

a1 + 3b1 – 3c1 + 1 = 0

For T :

–b1 = 0

Introduction to Turbomachines

15

Solving the above equations, a1 = –1, b1 = 0, c1 = 0 Substituting for a1, b1 and c1 in p1 term,

p2 term :

S1

d 1 Q 0 U 0 t

S2

d a2 Q b2 U c2 'p

M 0 L0 T 0

t d

(1.12)

[L]a2 [L3 T 1 ]b2 [ML3 ]c2 [ML1T 2 ]

For M :

c2 + 1 = 0

For L :

a2 + 3b2 – 3c2 – 1 = 0

For T :

–b2 – 2 = 0

Solving the above equations, c2 = –1, b2 = –2, a2 = 4 Substituting for a2, b2 and c2 in p2 term, S2

d 4 Q 2 U 1 'p

d 4 'p Q2 p

Q µ Vd2 \

Q2 µ V2d4

\

S2

p3 term:

S3 M 0 L0 T0

d 4 'p 2

4

V d U

'p

(1.13)

V 2U

d a3 Q b3 U c3 P [L]a3 [L3 T 1 ]b3 [ML3 ]c3 [ML1T 1 ]

For M :

c3 + 1 = 0

For L :

a3 + 3b3 – 3c3 – 1 = 0

For T :

–b3 – 1 = 0

Solving the above equations, b3 = –1, c3 = –1, a3 = 1 Substituting for a3, b3, c3 in p3 term, S3

We know that Q µ V × d2 Pd S3 \ V d 2U

d 1 Q 1 U 1 P

Pd U –Q

P U –d –V or U – d –V P

(1.14)

16

Fundamentals of Turbomachinery

p4 term :

S4 M 0 L0 T0

d a4 Q b4 U c4 V [L]a4 [L3 T 1 ]b4 [ML3 ]c4 [MT 2 ]

For M :

c4 + 1 = 0

For L :

a4 + 3b4 – 3c4 = 0

For T :

–b4 – 2 = 0

Solving the above equations, b4 = –2, c4 = –1, a4 = 3 Substituting for a4, b4, c4 in p4 term, S4

V d3

d 3 Q 2 U 1 V V d3 2

4

V d U

Q2 U V

UV 2 d

The p4 term can also be written as

p5 term :

S4

UV 2 d V

S5

d a5 Q b5 U c5 e

M 0 L0 T 0

(1.15)

[L]a5 [L3 T 1 ]b5 [ML3 ]c5 [ML1T 2 ]

For M :

c5 + 1 = 0

For L :

a5 + 3b5 – 3c5 – 1 = 0

For T :

–b5 – 2 = 0

Solving the above equations, b5 = –2, c5 = –1, a5 = 4 Substituting for a5, b5, c5 in p5 term, S5

ed 4

d 4 Q 2 U 1 e ed4 2

4

V d U

Q2 U e

UV 2

The p5 term can also be written as S5

UV 2 e

(1.16)

Introduction to Turbomachines

p6 term :

17

d a6 Q b6 U c6 w

S6

[L]a6 [L3 T 1 ]b6 [ML3 ]c6 [ML2 T 2 ]

M 0 L0 T 0

For M : c6 + 1 = 0 For L : a6 + 3b6 – 3c6 – 2 = 0 For T : –b6 – 2 = 0 Solving the above equations, b6 = –2, c6 = –1, a6 = 5 Substituting for a6, b6, c6 in p6 term, d 5 Q 2 U 1 w

S6

wd 5

wd

2 4

UV 2

V d U

wd 5 Q2 U

The p6 term can also be written as UV 2 (1.17) wd Note: Equations (1.14) to (1.17) represent dimensionless quantities. Even if these equations are inverted, the individual dimensions are not affected. The equations are therefore inverted merely for convenience. S6

1.16.1 Physical Significance of

p

Terms

From Eq. (1.12), we note that the relation S1

t d

simply states the geometrical relationship. The remaining p terms contain either rV or rV2. The term rV 2 = rV(V) is proportional to inertia force or force necessary to accelerate a mass. We know that,  inertia force, FI µ ma µ mV m = mass flow rate = AVr

FI µ AV2r

\

UV 2 —

\

FI F — I2 A d

where d is a characteristic dimension. From Eq. (1.13), S2

'p UV 2

—

'pd 2 Pressure force — FI Inertia force

(1.18)

18

Fundamentals of Turbomachinery

p2 is the ratio of the force due to change in pressure to the inertia force due to the motion of the fluid. From Eq. (1.14), UVd S3 (1.19) P We know that

P

\

S3

Viscous shear per unit area FV /d 2 FV — — V/d Vd Unit rate of shear UVd FI Vd Inertia force — – — P V FV Viscous force

(1.20)

p3 is the Reynold’s number and expresses the ratio of inertia force to viscous force. From Eq. (1.15),

where

S4

F UV 2 d FI d d Inertia force — 2 — I — V F Surface tension force d FS S

V—

FS Surface tension force — d Length

(1.21)

p4 is the Weber number and expresses the ratio of inertia force to surface tension force. From Eq. (1.16), S5

F UV 2 d 2 FI Inertia force — I2 – — — e Fe Fe Elastic force d

e µ elastic force per unit area —

where

S5

\

Fe d

2

—

'p 'V V

UV 2 V 2 V 2 — — 2 — M2 e e a U

where a = sound velocity M = Mach number p5 is also known as the Mach number (M2)

M2 \

M

V2 a2 V Fluid velocity a Sound velocity

(1.22)

Introduction to Turbomachines

19

From Eq. (1.17),

S6

wd

F UV 2 d2 Inertia force — I2 – — wd Fg Gravitational force d

wd 3

d p6 is known as Froude number.

2

—

Weight d

2

—

Gravitational force Fg — 2 d d

1.17 APPLICATION OF DIMENSIONAL ANALYSIS TO TURBOMACHINES Incompressible flow machines Figure 1.3 shows a control volume through which an incompressible fluid is flowing. Following are the variables considered. Flow rate Speed Power Head or Pressure Fluid density Fluid viscosity Diameter

Figure 1.3

Q N P H r m d

m3/s rps W m kg/m3 N-s/m2 m

A generalized turbomachine for the purpose of dimensional analysis.

P = f [r N m d Q H] Select the repeating variables as d N r \ m = 3 \ Total variables n = 7

(1.23)

20

Fundamentals of Turbomachinery

\

(n – m) = (7 – 3) = Four p terms.

p1 term: For M : For L : For T :

S1

d a1 N b1 U c1 Q

M 0 L0 T 0 [L]a1 [T 1 ]b1 [ML3 ]c1 L3 T 1 –3c1 = 0 a1 – 3c1 + 3 = 0 –b1 – 1 = 0

Solving the above equations, c1 = 0, b1 = –1, a1 = –3 Substituting for a1, b1 and c1 in p1 term,

p2 term : For M : For L : For T :

S1

d 3 N 1 U 0 Q

S2

d a2 N b2 U c2 H

Q Nd 3

(1.24)

M 0 L0 T 0 [L]a2 [T 1 ]b2 [ML3 ]c2 L2 T 2 c2 = 0 a2 – 3c2 + 2 = 0 –b2 – 2 = 0

Solving the above equations, c2 = 0, b2 = –2, a2 = –2 Substituting for a2, b2 and c2 in p2 term,

p3 term : For M : For L : For T :

S2

d 2 N 2 U 0 H

S3

d a3 N b3 U c3 P

H N 2d2

(1.25)

M 0 L0 T0 [L]a3 [T 1 ]b3 [ML3 ]c3 ML2 T 3 c3 + 1 = 0 a3 – 3c3 + 2 = 0 –b3 – 3 = 0

Solving the above equations, c3 = –1, b3 = –3, a3 = –5 Substituting for a3, b3 and c3 in p3 term,

p4 term :

S3

d 5 N 3 U 1 P

S4

d a4 N b4 U c4 P

P UN 3d 5

(1.26)

Introduction to Turbomachines

21

[L]a4 [T 1 ]b4 [ML3 ]c4 ML1T 1

M 0 L0 T 0

For M :

c4 + 1 = 0

For L :

a4 – 3c4 – 1 = 0

For T :

–b4 – 1 = 0

Solving the above equations, c4 = –1, b4 = –1, a4 = –2 Substituting for a4, b4 and c4 in p4 term, S4

d 2 N 1 U 1 P

1.18 SIGNIFICANCE OF

p

P U Nd 2

or

U Nd 2 P

(1.27)

TERMS

1.18.1 Capacity Coefficient or Flow Coefficient or Specific Capacity or Discharge Coefficient S1

From Eq. (1.24),

Q Nd 3

= capacity coefficient

(1.28)

We know that, Q µ AV µ d2V; Nd µ u; \ where

S1 I

Q Nd

3

—

d 2V Nd

speed ratio

3

—

Nd3 = Nd(d2) µ ud2

V V 1 — — Nd u I

(1.28a)

u V

Runner tangential speed Theortical jet spouting speed

For a given value of S1

Q

, it signifies that the ratio of blade velocity to jet velocity Nd 3 is fixed. Therefore, the shape of the velocity triangle can be determined for any given machine. The flow coefficient signifies that the volume flow rate of fluid through turbomachines of unit runner diameter and running at unit speed is same for both the model and prototype. The flow coefficient is constant for similar rotors. (a) Consider a turbomachine, fan or pump, having a certain runner diameter, the discharge is proportional to the speed. (b) Consider two fans or pumps, running at same speed, with different runner diameters, say, d1 and d2, where d1 > d2. Then, discharge is proportional to runner diameter, i.e. fan with d1 diameter discharges more air than the fan with d2 diameter.

22

Fundamentals of Turbomachinery

We have Nd ( d 2 ) — ud 2 — Vd 2 — d 2 H

\

Q

S1

Nd

3

—

Q d

2

= specific discharge

H

(1.29)

Specific discharge signifies that when the turbine is operated under unit head and the chosen characteristic dimension is unity, then the specific discharge remains the same both for the model and prototype.

1.18.2 Head Coefficient or Specific Head From Eq. (1.25),

H

S2

—

2 2

N d

H u

2

—

H V

2

—

H uV

(1.30)

We know that, Nd µ u (Nd)2 µ u2 ;

\

Vµu

Head coefficient is the ratio of the kinetic energy of the fluid (due to H) to the kinetic energy of the fluid running at the rotor tangential speed or it is the ratio of fluid head to kinetic energy of the rotor. H

\

S2

'p 'p — Ug U

(r = constant in incompressible fluids) 'p

H 2 2

UV 2

N d

(1.31)

, which is essentially a flow relationship.

Comparing Eqs. (1.18) and (1.31), both are same. Therefore, p2 is a parameter that expresses the dynamic condition.

1.18.3 Power Coefficient or Specific Power From Eq. (1.26), the parameter S3

P

is known as power coefficient or specific power.

UN 3d 5

This parameter can also be obtained by the product of p1 and p2. \ Now, \

S3

S1 – S 2

P

H – m

S3

Q Nd

3

–

Q Nd

3

–

H

(1.32)

( Nd )2

UQ g H P m ( Nd )

2

—

d 2V Nd

3

–

P 2 2

U AVN d

—

P UN 3d 5

Introduction to Turbomachines

P

S3

\

23 (1.33)

UN 3d 5

Now, N3d5 µ u3d2 µ V3d2 µ V2Vd2 µ Vd2H µ d 2 H H or \

N3d5 µ d2H3/2 S3

P 3 5

UN d

—

P 2

Ud H

3/2

—

P 2

d H3/2

(1.33a)

Particularly in water turbines, r can be omitted. The specific power of both the model and prototype will remain the same if their efficiencies are the same.

1.18.4 Reynold’s Number From Eq. (1.27), the parameter S4

U Nd 2 P

contains the viscosity term, that is something like a Reynold’s number. We know that, Nd µ u µ V \

S4

UVd P

(1.34)

Comparing Eqs. (1.20) and (1.34), both are same, hence the parameter p4 is Reynold’s number.

1.18.5 Effect of Reynold’s Number In pipe flow the Reynold’s number is an important parameter, which represents the nature of flow. If the Reynold’s number is greater than 3000, the flow is termed turbulent, if it is less than zero the flow is called laminar. The values of Reynold’s number in turbomachines are much higher than the critical values. Most of the turbomachines use relatively low viscosity fluids like air, water and light oil. Therefore, the viscous action of the fluid has very little effect on the power output of the machine. But, Reynold’s number is an important parameter for small pumps, compressors, fans and blowers. Their performance improves with an increase in Reynold’s number. In centrifugal pumps handling very heavy fluids the Reynold’s number may become as low as 10, which is of the order of one million machines handling light fluids at very high velocities. Even though viscosity effects are negligible, machines handling light fluids undergo efficiency changes under varying load conditions and with varying sizes. Therefore, the efficiency of a turbomachine can be considered as

24

Fundamentals of Turbomachinery 1/ 5 ÎÑ Ë d „ Û ÞÑ   „ K 1 (1 ) Ï ÌdÜ ß Í Ý Ñà ÑÐ

K

(1.35)

where h = efficiency of the prototype of diameter d h¢ = efficiency of the model of diameter d¢.

1.18.6 Specific Speed The parameter p5 can be obtained by the combination of p1 and p2. \

S5

or

S5

1/ 2

S1

Ë Q Û Ì 3Ü Í Nd Ý

(S 2 )3 / 4

Q1 / 2 N 3 / 2 d 3 / 2

Ë N 2d2 Û –Ì Ü ÍÌ H ÝÜ

3/4

Q1 / 2 N

= specific speed = NS (1.36) N1 / 2 d 3 / 2 H 3 / 4 H3/ 4 Among many p s treated so far, p5 is the only non-dimensional parameter that does not contain the linear dimension d of the runner wheel. The parameter p5 is known as specific speed (NS). It is the parameter of greatest importance in incompressible machines. The significance of specific speed is that, it is same for geometrically similar machines of all sizes, while operating under the same conditions of flow and head. The specific speed of Eq. (1.36) is used for pumps, fans and compressors. Hence,

NQ1 / 2

N SP

H3/ 4 [Here the suffix P denotes pumps, fans, and compressors.] Another specific speed can be obtained by combining p1 and p3 as follows: \

S6

or

N ST

1/ 2

S 31 / 2

N ST

Ë P Û Ì 3 5Ü Í UN d Ý

S 25 / 4

P1 / 2 N 5 / 2 d 5 / 2 U

1/ 2

N

3/2 5/2

d

H

5/4

Ë N 2d 2 Û Ì Ü ÌÍ H ÜÝ

5/ 4

PN 1/2

U

(1.37)

H5/ 4

(1.38)

Exclusively, Eq. (1.38) is used for incompressible fluids, particularly the water turbine where r is omitted. \

N ST

PN

H5/ 4 Following information can be obtained for specific speed of turbines. (i) High speed propeller turbines (Kaplan, propeller) have high specific speeds. (ii) High head machines (Pelton wheel) have low specific speeds.

(1.39)

Introduction to Turbomachines

25

1.18.7 Definition of Specific Speed

Specific speed of turbines (NST) It is the speed of a geometrically similar turbine working under unit head and developing unit power, i.e. S dN u — dN — V — H 60 \

d—

H N

P = Power = \ \

\

N2 —

UQgH H HH H 5 / 2 — QH — d 2VH — d 2 H H — — 2 K0 N2 N

H5/ 2 H5/ 4 or N — P P kH 5 / 4

, where k is constant of proportionality P If H = unit head, P = unit power, then N = NST N

N ST

k 15 / 4

(1.41)

k 1 Substituting NST = k in Eq. (1.40), N

N ST H 5 / 4 P

(1.40)

; ? N ST

N P H5/ 4

(1.42)

Equations (1.39) and (1.42) are same. Different types of turbines have different ranges of specific speeds. Impulse turbines have low specific speed, Francis turbine medium and Kaplan turbine high. High specific speed makes the size of turbine and powerhouse small. For low head and high output, a high specific speed turbine should be used. Thus, based on the existing conditions, the type of turbine can be decided. Generally NST is used as a guide to select a type of turbine under the given conditions of head and flow (i.e. site conditions). However, such a rule of thumb is used to ensure maximum efficiency. Thus, when NST is very high, Kaplan is the best selection to get a very high efficiency. When NST is very low, higher efficiencies are possible only if the Pelton wheel is selected.

Specific speed of pumps (NSP) The specific speed of a centrifugal pump is defined as the speed of a geometrically similar pump delivering unit quantity (1 m3/s) of liquid against a head of one metre.

26

Fundamentals of Turbomachinery

Q = area × velocity flow µ pdbV µ d2V µ d 2 H m

Hm

Q

Hm —

N2

H m3 / 2

(1.43)

N2

H m3 / 2

where k is the constant of proportionality N2 Hm = 1 m, Q = 1 m3/s, then N = NSP

Q

If \

1

\

2 N SP

k

k – 13 / 2 2 N SP

k

2 Substituting N SP

k in Eq. (1.43), Q

2 N SP H m3 / 2

N

2

; ? N SP

QN 2

N Q

H m3 / 2

H m3 / 4

(1.44)

Equations (1.37) and (1.44) are same.

1.19 EXAMPLES EXAMPLE 1.1 The resisting force F of a supersonic plane during flight can be considered as dependent upon the length of the aircraft L, velocity V, air viscosity m, air density r and bulk modulus of air K. Express the functional relationship between these variables and the resisting force. Solution:

Let the functional relationship be F = f [L, V, m, r, K]

The above equation in its general form may be written as f1 [F, L, V, m, r, K] = 0 Here n = 6, m = 3, (n – m) = 6 – 3 = three p terms; f1(p1, p2, p3) = 0 Let L, V, and r be the repeating variables. Then, we get S1

Hence:

M 0 L0 T 0

[ L ]a1 [V ]b1 [ U ]c1 F (L) a1 (LT 1 )b1 (ML3 )c1 (MLT) 2

M: c1 + 1 = 0 L : a1 + b1 – 3c1 + 1 = 0 T: –b1 – 2 = 0

\ c1 = –1 \ a1 = –2 \ b1 = –2

(i)

Introduction to Turbomachines

27

Substituting a1, b1 and c1 in the p1 term,

Now, Hence: M: L: T:

F

S1

L2 V 2 U 1 F

S2

( L ) a2 (V )b2 ( U)c2 P

2

L V 2U

(ii)

(L) a2 (LT 1 )b2 (ML3 )c2 (ML1T) 1

M 0 L0 T 0

c2 + 1 = 0 a2 + b2 – 3c2 – 1 = 0 –b2 – 1 = 0

\ c2 = –1 \ a2 = –1 \ b2 = –1

Substituting a2, b2 and c2 in the p2 term, P LV U

S2

L1 V 1 U 1 P

Again,

S3

( L )a3 (V )b3 ( U)c3 K

Hence:

M 0 L0 T 0

M:

(iii)

(L) a3 (LT 1 )b3 (ML3 )c3 (ML1T) 2

\ c3 = –1

c3 + 1 = 0

L : a3 + b3 – 3c3 – 1 = 0

\ a3 = 0

T:

\ b3 = –2

–b3 – 2 = 0

Substituting a3, b3 and c3 in the p3 term, L0 V 2 U 1 K

S3

K

(iv)

V 2U

Substituting p1, p2, p3 in Eq. (i),

Ë F P K Û f1 Ì 2 2 , , 2 Ü Í L V U LV U V U Ý \

F

0

Ë P K Û [ L2V 2 U ] I Ì , 2 Ü Í LV U V U Ý

Ans.

EXAMPLE 1.2 The efficiency h of a fan depends upon density r, dynamic viscosity m of the fluid, angular velocity w, diameter D of the rotor and discharge Q. Express h in terms of dimensionless parameters. Solution:

Let the functional relationship be h = f [r, m, w, D, Q]

(i)

The above equation in its general form may be written as f1 [h, r, m, w, D, Q] = 0

(ii)

28

Fundamentals of Turbomachinery

Here n = 6, m = 3, and let r, w and D be the repeating variables, \

(n – m) = (6 – 3) = three p terms.

\ Equation (ii) can be written as f1(p1, p2, p3) = 0 S1

(iii)

U a1 Z b1 D c1 K

p1 = h, since h itself is dimensionless S2 M 0 L0 T0

U a2 Z b2 D c2 P (ML3 )2 (T 1 )b2 (L)c2 (ML1T) 1

Hence: a2 + 1 = 0

\ a2 = –1

L : –3a2 + c2 – 1 = 0

\ c2 = –2

T:

\ b2 = –1

M:

–b2 – 1 = 0

Substituting a2, b2 and c2 in the p2 term,

Now, Hence:

S2

U 1 Z 1 D 2 P

S3

U a3 Z b3 D c3 Q

M 0 L0 T 0

P UZ D 2

(ML3 ) a3 (T 1 )b3 (L)c3 (L3 T) 1

\ a3 = 0

M:

a3 = 0

L:

–3a3 + c3 + 3 = 0

\ c3 = –3

T:

–b3 – 1 = 0

\ b3 = –1

Substituting a3, b3 and c3 in the p3 term,

S3

U 0 Z 1 D 3 Q

Q Z D3

Substituting the values of all ps in Eq. (iii),

Ë P Q Û f1 ÌK, , 2 3Ü Í UZ D Z D Ý \

Ë P Q Û , K IÌ 2 3Ü Í UZ D Z D Ý

0

Ans.

EXAMPLE 1.3 The pressure drop Dp in a pipe depends upon the mean velocity of flow V, length of pipe L, diameter of pipe d, viscosity of fluid m, average height of roughness projections on inside pipe surface h and mass density of fluid r. By using the Buckingham p-theorem, obtain a dimensionless expression for Dp.

Introduction to Turbomachines

Solution:

29

Let the functional relationship be Dp = f [V, L, d, m, h, r]

(i)

The general form of the above equation is f1 [Dp, V, L, d, m, h, r] = 0

(ii)

n = 7, m = 3, and let V, d, r be the repeating variables. \

(n – m) = (7 – 3) = 4p terms

\ Equation (ii) can be written as f1(p1, p2, p3, p4) = 0 S1 M 0 L0 T0

V a1 d b1 U c1 'p (LT 1 )a1 (L)b1 (ML3 )c1 (ML1T) 2

Hence: For M :

c1 + 1 = 0

L:

a1 + b1 – 3c1 – 1 = 0

T:

–a1 – 2 = 0

\

S1

V 2 d 0 U 1 'p

Now,

S2

V a2 d b2 U c2 L

M 0 L0 T 0

\ c1 = –1 \ b1 = 0 \ a1 = –2 'p UV 2

(LT 1 ) a2 (L)b2 (ML3 )c2 (L)

Hence: M:

c2 = 0

L:

a2 + b2 – 3c2 + 1 = 0

T:

–a2 = 0

\

S2

V 0 d 1 U 0 L

Also,

S3

V a3 d b3 U c3 P

M 0 L0 T 0

\ c2 = 0 \ b2 = –1 \ a2 = 0 L d

(LT 1 )a3 (L)b3 (ML3 )c3 (ML1T) 1

Hence,

\

M:

c2 + 1 = 0

\ c2 = –1

L:

a3 + b3 – 3c2 – 1 = 0

\ b3 = –1

T:

–a3 – 1 = 0

\ a3 = –1

S3

V 1 d 1 U 1 P

P UVd

(iii)

30

Fundamentals of Turbomachinery

Again,

S4 M 0 L0 T 0

V a4 d b4 U c4 h (LT 1 )a4 (L)b4 (ML3 )c4 (L)

Hence: M:

c4 + 0 = 0

L:

a4 + b4 – 3c4 + 1 = 0

T:

–a4 = 0

\ c4 = 0 \ b4 = –1 \ a4 = 0

h d Substituting the values of all the p terms in Eq. (iii), S4

\

V 0 d 1 U 0 h

Ë 'p L P h Û f1 Ì 2 , , , Ü Í UV d UVd d Ý 'p

\

0

ËL P hÛ UV 2I Ì , , Ü Í d UVd d Ý

Ans.

Show that the discharge of a centrifugal pump is given by Q = ND3

EXAMPLE 1.4

Ë gH P Û where N is the speed of the pump in rpm, D the diameter of the impeller, IÌ 2 2 , 2 Ü Í N D ND U Ý g the acceleration due to gravity, H the manometric head, m viscosity of fluid and r the density of the fluid. Solution:

Let the functional relationship be Q = f [N, D, g, H, m, r]

(i)

The general form of the above relation can be written as f1 [Q, N, D, g, H, m, r] = 0

(ii)

n = 7, m = 3, and let N, D, r be the repeating variables. \

(n – m) = (7 – 3) = 4p terms

Equation (ii) can be written as f1(p1, p2, p3, p4) = 0 Here,

S1 M 0 L0 T 0

N a1 D b1 U c1 Q (T 1 )a1 (L)b1 (ML3 )c1 (L3 T) 1

Hence: \ c1 = 0

M:

c1 = 0

L:

b1 – 3c1 + 3 = 0

\ b1 = –3

T:

–a1 – 1 = 0

\ a1 = –1

(iii)

Introduction to Turbomachines

\

S1

N 1 D 3 U 0Q

Now,

S2

N a2 D b2 U c2 g

M 0 L0 T0

31

Q ND3

(T 1 ) a2 (L)b2 (ML3 )c2 (LT) 2

Hence: \ c2 = 0

M:

c2 = 0

L:

b2 – 3c2 + 1 = 0

\ b2 = –1

T:

–a2 – 2 = 0

\ a2 = –2

Also,

S2

N 2 D 1 e0 g

S3

N a3 D b3 U c3 H

M 0 L0 T 0

g N2D

(T 1 ) a1 (L)b1 (ML3 )c1 (L)

Hence: \ c3 = 0

M:

c3 = 0

L:

b3 – 3c3 + 1 = 0

T:

–a3 = 0

\ b3 = –1 \ a3 = 0

\

S3

N 0 D 1 U 0 H

Again,

S4

N a4 D b4 U c4 P

M 0 L0 T0

H D

(L)a4 (ML3 )b4 (ML3 )c4 (ML1T) 1

Hence:

\

M:

c4 + 1 = 0

\ c4 = –1

L:

b4 – 3c4 – 1 = 0

\ b4 = –2

T:

–a4 – 1 = 0

\ a4 = –1

S4

N 1 D 2 U 1 P

P ND 2 U

Substituting the values of all ps in Eq. (iii), we get

Q ND3 \

Q

Ë g H P Û IÌ 2 , , 2 Ü Í N D D ND U Ý Ë gH P Û ND3 I Ì 2 2 , 2 Ü Í N D ND U Ý

Ans.

32

Fundamentals of Turbomachinery

EXAMPLE 1.5 Derive on the basis of dimensional analysis suitable parameters to present the thrust developed by a property. Assume that the thrust P depends upon the angular velocity w, speed of advance V, diameter D, dynamic viscosity m, mass density r, elasticity of the fluid medium which can be denoted by the speed of sound C in the medium. Solution:

Let the functional relationship be P = f [w, V, D, m, r, C]

(i)

f1[P, w, V, D, m, r, C] = 0 n = 7, m = 3, \

(ii)

\ number of p terms = 4

f [p1, p2, p3, p4] = 0

(iii)

Let D, V, r be the repeating variables. S1 M 0 L0 T 0

D a1 V b1 U c1 P La1 (LT 1 )b1 (ML3 )c1 (MLT) 2

Hence: M:

c1 + 1 = 0

\ c1 = –1

L:

a1 + b1 – 3c1 + 1 = 0

\ a1 = –2

T:

–b1 – 2c1 = 0

\ b1 = –2

S1

\ Now,

S2 M 0 L0 T 0

D 2 V 2 U 1 P

P UV 2 D 2

D a2 V b2 U c2 Z La2 (LT 1 )b2 (ML3 )c2 (T 1 )

Hence: \ c2 = 0

M:

c2 = 0

L:

a2 + b2 – 3c2 = 0

\ a2 = +1

T:

–b2 – 1 = 0

\ b2 = –1

\

S2

D 1 V 1 V 1 U 0 Z

Also,

S3

D a3 V b3 U c3 P

M 0 L0 T 0

ZD V

(L) a3 (LT 1 )b3 (ML3 )c3 (ML1T 1 )

Hence: M:

c3 + 1 = 0

\ c3 = –1

L:

a3 + b3 – 3c3 – 1 = 0

\ a3 = –1

T:

–b3 – 1 = 0

\ b3 = –1

Introduction to Turbomachines

\

S3

D 1 V 1 U 1 P

Again,

S4

D a4 V b4 U c4 C

M 0 L0 T 0

33

P UVD

(L) a4 (LT 1 )b4 (ML3 )c4 (LT 1 )

Hence: M:

c4 = 0

\ c4 = 0

L:

a4 + b4 – 3c4 + 1 = 0

\ a4 = 0

T:

–b4 – 1 = 0

\

S4

\ b4 = –1

D 0V 1 U 0 C

C V

Substituting the values of all the ps in Eq. (iii),

Ë P DZ P CÛ f1 Ì 2 2 , , , Ü Í D V U V DV U V Ý \

0

Ë DZ P CÛ D 2V 2 U I Ì , , Ü Í V DV U V Ý

P

Ans.

EXAMPLE 1.6 Prove that the frictional torque T of a disc of diameter D rotating at speed N in a fluid of viscosity m and density r in a turbulent flow is given by

T Solution:

Ë P Û D5 N 2 UI Ì 2 Ü Í D NU Ý

The functional relationship can be written as T = f¢[D, N, m, r]

\

(i)

f [T, D, N, m, r] = 0 n = 5, m = 3

\

\ there are two p terms

f [p1, p2] = 0

Let D, N, r be the repeating variables. S1 M 0 L0 T 0

\

D a1 N b1 U c1 T La1 (T 1 )b1 (ML3 )c1 (ML2 T 2 )

Hence: M:

c1 + 1 = 0

\ c1 = –1

L:

a1 – 3c1 + 2 = 0

\ a1 = –5

T:

–b1 – 2 = 0

\ b1 = –2

(ii)

34

Fundamentals of Turbomachinery

\

S1

D 5 N 2 U 1 T

Now,

S2

D a2 N b2 U c2 P

M 0 L0 T0

7 U N 2 D5

La2 (T 1 )b2 (ML3 )c2 (ML1T 1 )

Hence: M:

c2 + 1 = 0

\ c2 = –1

L:

–b2 – 1 = 0

\ b2 = –1

T:

a2 – 3c2 – 1 = 0

\ a2 = –2

\

S2

D 2 N 1 U 1 P

P U ND 2

Substituting the values of all the ps in Eq. (iii),

Ë T P Û IÌ 5 2 , 2 Ü Í D N U D NP Ý \

T

0

Ë P Û D 5 N 2 UI1 Ì 2 Ü Í D NP Ý

Ans.

EXAMPLE 1.7 The performance of a gas turbine may be assumed to depend on the mass flow rate, initial temperature, initial and final pressures, the turbine temperature drop, outer diameter of the rotor and speed of rotation. Using the fundamental dimensions of mass, length and time, obtain the dimensionless parameters which describe the performance of the gas turbine. Solution: only. We have

Dimensionless parameters should be developed, using mass, length and time cpT1 = h1; cpT2 = h2 cpDT1 = Dh1; cpDT2 = Dh2

\ Functional relationship can be written as P

\

f [ m , 'p, 'h, D, N ]

f „ [ P, m , 'p, 'h, D, N ] 0

n = 6, m = 3 \

\ There are three p terms

f ¢ [p1, p2, p3] = 0

Let m , D, N be the repeating variables. \

S1 M 0 L0 T 0

m a1 D b1 N c1 P (MT 1 )a1 (L)b1 (T 1 )c1 (ML2 T 3 )

(i) (ii)

Introduction to Turbomachines

35

Hence: M:

a1 + 1 = 0

\ a1 = –1

L:

b1 + 2 = 0

\ b1 = –2

T:

–a1 – c1 – 3 = 0

\ c1 = –2

\

S1

m 1 D 2 N 2 P

Now,

S2

m a2 D b2 N c2 'p

M 0 L0 T0

P  2N2 mD

(MT 1 ) a2 (L)b2 (T 1 )c2 (ML1T 2 )

Hence: M:

a2 + 1 = 0

\ a2 = –1

L:

b2 – 1 = 0

\ b2 = –1

T:

–a2 – c2 – 2 = 0

\ c2 = –1

\

S2

m 1 D1 N 1 'p

Also,

S3

m a3 D b3 N c3 'h

M 0 L0 T 0

'p D N m

(MT 1 ) a3 (L)b3 (T 1 )c2 (L2 T 1 )

Hence: \ a3 = 0

M:

a3 = 0

L:

b3 + 2 = 0

\ b3 = –2

T:

–a3 – c3 – 2 = 0

\ c3 = –2

S3

\

m 0 D 2 N 2 'h

'h N 2 D2

Substituting p1, p2, p3 in Eq. (iii), P 2

 N mD

\ where

2

P

Ë D 'P ' h Û IÌ , 2 2Ü  D N Ý Í mN Ë D 'P ' h Û  2 N 2I Ì , mD Ü Í mN D 2 N 2 Ý

Ans.

Dh = cpDT

EXAMPLE 1.8 The performance of a lubricating oil ring depends on the inside diameter of the ring D, shaft speed N, oil discharge Q, density r, viscosity m, surface tension s and specific weight w of the fluid. Find a functional relationship in terms of dimensionless parameters.

36

Fundamentals of Turbomachinery

Solution:

The functional relationship can be written as P = f [D, N, Q, r, m, s, w]

(i)

The above equation can be written in general form as f ¢[P, D, N, Q, r, m, s, w] = 0 \

n = 8, m = 3,

\

\ There are five p terms

f ¢ [p1, p2, p3, p4, p5] = 0

Let D, N, r be the repeating variables. Now,

S1 M 0 L0 T 0

D a1 N b1 U c1 P La1 (T 1 )b1 (ML3 )c1 (MLT 2 )

Hence: M:

c1 + 1 = 0

\ c1 = –1

L:

a1 – 3c1 + 1 = 0

\ a1 = –4

T:

–b1 – 2 = 0

\ b1 = –2

\

S1

D 4 N 2 U 1 P

Now,

S2

D a2 N b2 U c2 Q

M 0 L0 T0

P U N 2 D4

(L)a2 (T 1 )b2 (ML3 )c2 (L3 T 1 )

Hence: \ c2 = 0

M:

c2 = 0

L:

a2 – 3c2 + 3 = 0

\ a2 = –3

T:

–b2 – 1 = 0

\ b2 = –1

\

S2

D 3 N 1 U 0 Q

Also,

S3

D a3 N b3 U c3 P

M 0 L0 T 0

Q ND3

La3 (T 1 )b3 (ML3 )c3 (ML1T 1 )

Hence:

\

M:

c3 + 1 = 0

\ c3 = –1

L:

a3 – 3c3 – 1 = 0

\ a3 = –2

T:

–b3 – 1 = 0

\ b3 = –1

S3

D 2 N 1 U 1 P

P U ND 2

(ii)

Introduction to Turbomachines

Again,

37

D a4 N b4 U c4 V

S4

La4 (T 1 )b4 (ML3 )c3 (MT 2 )

M 0 L0 T 0

Hence: M: L: T:

\ c4 = –1 \ a4 = –3 \ b4 = –2

c4 + 1 = 0 a4 – 3c4 = 0 –b4 – 2 = 0

\

S4

D 3 N 2 U 1 V

Again,

S5

D a5 N b5 U c5 w

V U N 2 D3

La5 (T 1 )b5 (ML3 )c5 (ML2 T 2 )

M 0 L0 T0

Hence: M: L: T:

\ c5 = –1 \ a5 = –1 \ b5 = –2

c5 + 1 = 0 a5 – 3c5 – 2 = 0 –b5 – 2 = 0

\

D 1 N 2 U 1 w

S5

w UN 2D

Substituting the values of all ps in Eq. (iii),

Ë P Q P V w Û f„Ì 4 2 , , , , 3 2 2 3 2 Ü Í D N U ND U ND U N D U N D Ý \

P

0

Ë Q w Û P V D 4 N 2 UI Ì , , , 3 2 2 3 2 Ü Í ND U ND U N D U N D Ý

Ans.

EXAMPLE 1.9 A quarter-scale turbine model is tested under a head of 10 m. The full-scale turbine is required to work under a head of 28.5 m and 415 rpm. (a) At what speed must the model be run if it develops 94 kW and uses 0.96 m3/s at this speed? (b) What power will be obtained from the full-scale turbine? (c) Name the type of turbine. Solution: Given that

(a) Speed of the model (Nm): Lm Lp

m  model p  prototype (full-scale turbine)

1 4

We have the speed relations as follows: Um Up

1/ 2

Ë Hm Û Ì Ü ÌÍ H p ÜÝ

Dm N m Dp N p

1 Nm – 4 Np

38

Fundamentals of Turbomachinery 1/ 2

1/ 2 ËH Û Ë 10 Û Nm 4N p Ì m Ü 4–Ì Ü – 415 983.3 rpm Í 28.5 Ý ÌÍ H p ÜÝ (b) Power developed by the full scale turbine, i.e. prototype (Pp):

\

[ N ST ]m

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ m

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ p

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ m

415 P

983.3 – 94

536.11 rpm

10 5 / 4

536.11; ? Pp

28.55 / 4 (c) Type of turbine:

Ans.

7236.29 kW

Ans.

[NST]P = 536.11 rpm Hence, it is Kaplan turbine.

Ans.

EXAMPLE 1.10 A quarter-scale turbine model is tested under a head of 30 m. The full-scale turbine is required to work under a head of 130 m and to run at 450 rpm. Calculate (a) the speed of the model if it produces 130 kW when the discharge is 0.6 m3/s and (b) the power produced by the prototype if its efficiency is 6% more than that of the model. Solution: We know that the speed ratio of the model and prototype is the same. (a) The speed of the model (Nm): Dm Dp

1 4

We have the relation as follows: Ë u Û Ì Ü Í 2 gH Ý m

Ë u Û Ë DN Û Ì Ü or Ì Ü Í H Ým Í 2 gH Ý p 1/ 2

\

Nm

D p Ë Hm Û Ì Ü Dm ÌÍ H p ÜÝ

Ë DN Û Ì Ü Í H Ýp 1/ 2

– Np

Ë 30 Û 4 – 450 – Ì Ü Í 130 Ý

= 864.69 rpm

Ans.

(b) Power produced by the prototype (Pp): hm = overall efficiency of the model =

Power UQgH

130 – 1000 W 1000

kg m3 m – 0.6 – 9.81 2 – 30 m 3 s m s

Introduction to Turbomachines

130 × 1000 W 1000 × 0.6 × 9.81 × 30 × kg m 2 s

=

39

N m s

130 × 1000 W

=

176580

= 0.7362

N

m s

W

or 73.62%

The specific discharge of model and prototype, Qp D 2p

Qm Dm2

Hp

(Eq. (1.29))

Hm 2

\

Qp

1/ 2

Ë 130 Û 0.6 – 16 – Ì Ü Í 30 Ý

Ë P Û Ì Ü Í UQgH Ý P

Kp

\

1/ 2

Ë Dp Û Ë H p Û Qm Ì Ü Ì Ü Í Dm Ý Í H m Ý = 19.984 m3/s

Pp = hp × (rQgH)p hp = hm + 0.06 = 0.7362 + 0.06 = 0.7962

\

Pp = 0.7962 × 1000 × 19.984 × 9.81 × 130

N 3 kg × ms × m2 × m s m3

=

20291630.9

N-m or W s

Pp = 20.29 MW

Ans.

EXAMPLE 1.11 A Francis turbine model is built to scale of 1 : 5. Model data

Prototype data

P = 4 kW N = 350 rpm H=2m

P=? N=? H=6m

Assume that the overall efficiency of the model is 70%. Calculate (a) the speed of the prototype and (b) the power.

40

Fundamentals of Turbomachinery

Solution: (a) Speed of the prototype (Np): We have the speed ratio relation as follows: Ë u Û Ì Ü Í 2 gH Ý m

Ë u Û Ë DN Û Ì Ü or Ì Ü Í H Ým Í 2 gH Ý p

\

Np

Ë DN Û Ì Ü Í H Ým

or

Np

1 Ë6Û – 5 ÌÍ 2 ÜÝ

Ë HÛ Ì Ü ÌÍ D ÜÝ p

Ë DN Û Ì Ü Í H Ýp 1/ 2

Ë Dm Û Ë H p Û Ì ÜÌ Ü ÌÍ D p ÜÝ Í H m Ý

Nm

1/ 2

– 350 121.24 rpm

Ans.

(b) Power of the prototype (Pp): hm = overall efficiency of the model

\

0.7

Ë P Û Ì Ü Í UQgH Ý m

Qm

0.2912 m 3 /s

4000 1000 – Qm – 9.81 – 2

We have the capacity coefficient from Eq. (1.28) for model and prototype, Ë Q Û Ì Ü Í ND3 Ý m

\

Qp

Ë Q Û Ì Ü Í ND3 Ý p Ë N p Û Ë Dp Û Qm Ì ÜÌ Ü Í N m Ý Í Dm Ý 0.2912 –

or

Qp

3

121.24 – (5)3 350

12.61 m 3 /s

From Eq. (1.35), we have

\ \

Ë 1  Km Û Ì Ü ÍÌ 1  K p ÝÜ

Ë Dp Û Ì Ü Í Dm Ý

(1  0.7) (1  K p )

(5)0.2

0.2

hp = 0.7826

Ë P Û Ì Ü Í UQgH Ý p

Pp – 1000

Now,

Kp

\

Pp = 580.9 kW Ans.

1000 – 12.61 – 9.81 – 6

0.7826 (i)

Introduction to Turbomachines

41

Alternatively: We have specific speed relation as follows:

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ m

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ p

( N ST ) m

ËN P Û Ì 5/ 4 Ü ÍÌ H ÝÜ

( N ST )m

( N ST ) p

294.32 rpm

25 / 4

294.32

ËN P Û Ì 5/4 Ü ÍÌ H ÝÜ p

121.24 Pp

294.32

\

350 – 4

65 / 4 Pp = 519.64 kW

Ans.

This is not the correct power of the prototype. A correction factor is used to get the correct power of the prototype. Ppc

corrected power of the prototype.

We have the following relation.

\

Pp

Km Kp

Ppc

Ppc

Pp

Ppc

580.95 kW

Kp Km

519.64 –

0.7826 0.7 (ii)

\ Equations (i) and (ii) are same.

Ans.

EXAMPLE 1.12 Calculate (a) the diameter and (b) the output of second turbine for the following data. Assume the efficiency and gate opening to be the same in both the cases. Turbine 1 (Model)

Turbine 2 Prototype

P 25 kW D 0.5 m N 1000 rpm H 20 m Solution: (a) Diameter of the 2nd turbine, i.e. From Eq. (1.30), we have head coefficient as Ë H Û Ì 2 2Ü Í N D Ým

Ë H Û Ì 2 2Ü Í N D Ýp

P=? D=? N = 200 rpm H = 160 m prototype (Dp):

42

Fundamentals of Turbomachinery 2

\

D 2p

Ë H p Û Ë Nm Û 2 Ü – Dm Ì Ü–Ì H N Ì Ü m p Í Ý Í Ý

or

D2p

50 ? D p

2

Ë 160 Û Ë 1000 Û 2 Ì 20 Ü – Ì 200 Ü – (0.5) Í Ý Í Ý

7.071 m

Ans.

(b) Output of the 2nd turbine, i.e. prototype (Pp): From Eq. (1.33), we have the power coefficient, Ë P Û Ì 3 5Ü Í UN D Ým

Ë P Û Ì 3 5Ü Í UN D Ý p 3

\

Pp

Ë N p Û Ë Dp Û Pm – Ì Ü Ì Ü Í N m Ý Í Dm Ý

5

3

Ë 200 Û Ë 7.071 Û 25 – Ì Ü – Ì 0.5 Ü 1000 Í Ý Í Ý

5

113131.7 kW

Ans.

Alternatively: We have specific relation as follows: (NST)m = (NST)p \

( N ST ) m

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ m

\

( N ST ) p

118.22

\

1000 – 25 20 5 / 4

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ p

118.22

200 – P 160 5 / 4

Pp = 113137.1 kW

Ans.

EXAMPLE 1.13 A turbine model working under a head of 2 m runs at 170 rpm and has a diameter of 1 m. A prototype turbine develops 22 MW under a head of 250 m with a specific speed of 100. Calculate (a) the scale ratio and (b) the power developed by the model. Solution: (a) Scale ratio (DR): We have the specific speed relation as

( N ST ) p \

100

ËN P Û Ì 5/4 Ü ÍÌ H ÝÜ p

N p 22 – 1000 250 5 / 4

Np = 670.21 rpm

We have the speed ratio relation between the model and prototype as follows:

È DN Ø ÉÊ Ù H Úm

È DN Ø ÉÊ Ù H Úp

Introduction to Turbomachines

\

Dm Dp

DR

43

1/ 2

Ë N p Û Ë Hm Û Ü Ì ÜÌ Í N m Ý ÌÍ H p ÜÝ

1/ 2

Ë 670.21 Û Ë 2 Û Ì 170 Ü – Ì 250 Ü Í Ý Í Ý

0.3526

1 2.84

Ans.

(b) Power developed by the model (Pm): ( N ST ) m

\

ËN P Û Ì 5/4 Ü ÌÍ H ÜÝ m

100

Pm = 1.9574 kW

170 – P (2)5 / 4

Ans.

(i)

Alternatively: By using Eq. (1.33a), specific speed relation, Ë P Û Ì 2 3/2 Ü Í D H Ým

\

Ë P Û Ì 2 3/ 2 Ü Í D H Ýp ËD Û Pp Ì m Ü ÌÍ D p ÜÝ

Pm

2

Ë Hm Û Ì Ü ÌÍ H p ÜÝ

3/ 2

Ë 2 Û 22 – 1000 – (0.3526)2 – Ì Ü Í 250 Ý

3/ 2

= 1.9571 kW Ans.

(ii) Or

or

Ë P Û Ì 3 5Ü Í UN D Ým

Ë P Û Ì 3 5Ü Í UN D Ý p

Pm

ËN Û Pp – Ì m Ü ÌÍ N p ÜÝ

Pm = 1.957 kW

3

(Eq. (1.33)) Ë Dm Û Ì Ü ÌÍ D p ÜÝ

5

3

Ë 170 Û 5 22 – 1000 – Ì Ü – (0.3526) Í 670.21 Ý

Ans.

(iii)

EXAMPLE 1.14 A multistage (full scale) centrifugal pump running at 500 rpm delivers 5 m3/s, against a head of 100 m. A model of this pump delivers 0.3 m3/s and the power input is 100 kW at an efficiency of 90%. Calculate (a) the speed of the model and (b) the scale ratio. Solution: We have the relation for the centrifugal pump Ë H U gQ Û hm = efficiency of the pump = Ì Ü Í p Ým

\

Hm

Kp U gQ

0.9 – 100 – 1000 1000 – 9.81 – 0.3

30.58 m

44

Fundamentals of Turbomachinery

(a) Speed of the model (Nm): Specific speed relations, (NSP)m = (NSP)p ËN Q Û Ì 3/4 Ü ÍÌ H ÝÜ m

ËN Q Û Ì 3/4 Ü ÍÌ H ÝÜ p 1/ 2

Ë Qp Û Np – Ì Ü Í Qm Ý

or

Nm

\

Nm = 839.41 rpm

ËH Û –Ì mÜ ÍÌ H p ÝÜ

3/4

1/ 2

Ë 5 Û 500 – Ì Ü Í 0.3 Ý

Ë 30.58 Û –Ì Ü Í 100 Ý

3/4

Ans.

(b) Scale ratio (DR): We have the capacity coefficient from Eq. (1.28), Ë Q Û Ì Ü Í ND3 Ý m

Ë Q Û Ì Ü Í ND3 Ý p 1/ 3

ÎÑË N p Û Ë Q Û ÞÑ m Üß ÏÌ ÜÌ N Q ÐÑÍ m Ý ÍÌ p ÝÜ àÑ

or

DR

Ë Dm Û Ì Ü ÍÌ D p ÝÜ

\

DR

0.3 Û Ë 500 Ì 839.41 – 5 Ü Í Ý

1/3

0.33

Ans.

EXAMPLE 1.15 A centrifugal pump works against a manometric head of 30 m with an impeller diameter of 300 mm. When the pump was running at 1500 rpm, it produced a head of 22 m. What changes in design do you suggest to get the designed head, i.e. 30 m? Solution:

We have the head coefficient from Eq. (1.30), Ë H Û Ì 2Ü Í ( ND) Ý

C

Suggestions: Case I:

Speed should change keeping the diameter same.

Ë H Û Ì 2Ü ÌÍ ( N1 D) ÜÝ N1 = Old design; N2 = New design; D1

\

22 1500

Ë H Û Ì 2Ü ÌÍ ( N 2 D) ÜÝ = D2 = No change in diameter 30

2

N 22

Introduction to Turbomachines

\ Case II:

30 – 1500 1751.62 rpm 22 Impeller diameter should change, keeping the speed constant. N2

\

N

or

H

H

2

2

D12

N D22

25

30

300

\

2

D2

45 Ans.

(N is kept constant)

D22 30 – 300 25 = 328.63 mm

Ans.

EXAMPLE 1.16 A single-stage centrifugal pump works against a height of 30 m, running at 2000 rpm, supplies 3 m3/s and has an impeller diameter of 300 mm. Calculate (a) the number of stages and (b) the diameter of each impeller required to pump 6 m3/s of water to a height of 220 m when running at 1500 rpm. Solution: Single stage pump (1)

Multistage pump (2)

H m1 30 m N1 = 2000 rpm Q1 = 3 m3/s D1 = 300 mm

H m2 ? Ht = 220 m N2 = 1500 rpm Q2 = 6 m3/s D2 = ?

(a) Number of stages (Ns): We have the specific speed of the pump as ËN QÛ Ì 3/4 Ü ÌÍ H m ÜÝ1

ËN QÛ Ì 3/4 Ü ÌÍ H m ÜÝ 2

H m3 /2 4

N 2 Ë Q2 Û Ì Ü N1 Í Q1 Ý

1/ 2

\

– H m3 1/ 4 1/ 2

1500 Ë 6 Û – 2000 ÌÍ 3 ÜÝ

\

H m2

\

Ns

– 30 3 / 4

13.596 m

32.45 m Ht H m2

220 32.45

6.779  7

(b) Diameter of each impeller (D2):

Ë Hm Û Ì 2Ü Í ( ND) Ý1

Ë Hm Û Ì 2Ü Í ( ND) Ý2

(Head coefficient, Eq. (1.30))

Ans.

46

Fundamentals of Turbomachinery 2

\

D22

H m2 Ë N1 Û 2 Ì Ü D1 H 1m Í N 2 Ý

or

D2

N1 D1 – – N2 N 2

H m2 H m1

2000 32.45 – 300 – 1500 30 \

D2 = 416 mm

Ans.

EXAMPLE 1.17 A centrifugal pump running at 1500 rpm, with impeller diameter 200 mm, discharges 0.12 m3/s of water working against a head of 40 m with an efficiency of 90%. Calculate (a) the specific speed (b) the performance of a similar pump twice its size keeping the speed constant and (c) the performance of a similar pump at twice the speed keeping the diameter constant, and (d) the performance of a similar pump if speed is doubled as well as the size is doubled. Solution: (a) Specific speed (NSP):

N Q

1500 0.12

Ans. 32.67 rpm H 403 / 4 (b) The performance of a similar pump twice its size (H2, Q2, P2) keeping the speed constant, i.e. N1 = N2:

N SP

3/4

Ë U gQH Û h1 = efficiency of pump1 = Ì Ü Í P Ý1 \

P

U gQH K

1000

kg

– 9.81

m

– 0.12

m3 1 – 40 m – s 0.9

m3 s2 N W m m m = 52320 kg 2 s = 52320 N s s

or P1 = 52.320 kW \ We have the capacity coefficient relation from Eq. (1.28) as

Ë Q Û Ì Ü Í ND3 Ý1 \

Q2

Ë Q Û Ì Ü Í ND3 Ý2

(Q N1 = N2 as per given data)

ËN Û ËD Û Q1 Ì 2 Ü Ì 2 Ü Í N1 Ý Í D1 Ý

3

ËD Û Q1 Ì 2 Ü Í D1 Ý

3

Ans.

47

Introduction to Turbomachines

Ë2 D Û Q1 Ì 1 Ü Í D1 Ý

or

3

0.12 – (2)3

(Q D2 = 2D1 as per given data)

0.96 m 3 /s

Q2

Ans.

We have head coefficient from Eq. (1.30) as

Ë H Û Ì 2Ü Í ( ND) Ý1

Ë H Û Ì 2Ü Í ( ND) Ý2 ËD Û H1 – Ì 2 Ü Í D1 Ý

\

H2

or

H2 = 160 m

2

40 – (2)2

(Q N1 = N2; D2 = 2D1)

We have power coefficient from Eq. (1.33) as \

Ë P Û Ì 5 3Ü Í U D N Ý1

\

P2

Ë P Û Ì 5 3Ü Í U D N Ý2 ËD Û P1 Ì 2 Ü Í D1 Ý

5

Ë2 D Û 52.320 – Ì 1 Ü Í D1 Ý

2

(Q N1 = N2; D2 = 2D1; r1 = r2) P2 = 1674.2 kW

or

Ans.

(c) The performance (Q2, H2, P) of a similar pump twice the speed keeping the diameter constant, i.e. D1 = D2: \ From Eq. (1.28), Ë Q Û Ì 3Ü Í ND Ý1

\

Q2

Ë Q Û Ì 3Ü Í ND Ý 2

ËN Û Q1 Ì 2 Ü Í N1 Ý

(Q D1 = D2; N2 = 2N1) Ë2 N Û 0.12 – Ì 1 Ü Í N1 Ý

0.24 m 3 /s

Ans.

From Eq. (1.30),

Ë H Û Ì 2Ü Í ( ND ) Ý1 \

H2

Ë H Û Ì 2Ü Í ( ND ) Ý2 ËN Û H1 Ì 2 Ü Í N1 Ý

2

(Q D1 = D2; N2 = 2N1) Ë2 N Û 40 Ì 1 Ü Í N1 Ý

2

160 m

Ans.

48

Fundamentals of Turbomachinery

From Eq. (1.33), Ë P Û Ì 5 3Ü Í U D N Ý1

Ë P Û Ì 5 3Ü Í U D N Ý2

(Q N2 = 2N1; D1 = D2; r1 = r2)

3

ËN Û Ë 2N Û P1 Ì 2 Ü 52.32 – Ì 1 Ü N Í 1Ý Í N1 Ý P2 = 418.56 kW

3

P2

\ or

Ans.

(d) The performance Q2, H2, r2 of a similar pump if the speed is doubled (N2 = 2N1) as well as the size is doubled (D2 = 2D1): From Eq. (1.28),

Ë Q Û Ì Ü Í ND3 Ý1

Ë Q Û Ì Ü Í ND3 Ý2

(Q N2 = 2N1; D2 = 2D1)

3

3

ËD Û ËN Û Q1 Ì 2 Ü Ì 2 Ü Í D1 Ý Í N1 Ý Q2 = 1.92 m3/s

\

Ë2 D Û Ë2 N Û 0.12 – Ì 1 Ü Ì 1 Ü Í D1 Ý Í N1 Ý

Q2

or

Ans.

From Eq. (1.30),

Ë H Û Ì 2Ü Í ( ND) Ý1

Ë H Û Ì 2Ü Í ( ND) Ý2

(Q N2 = 2N1; D2 = 2D1)

2

ËD Û ËN Û H1 Ì 2 Ü Ì 2 Ü Í D1 Ý Í N1 Ý H2 = 320 m

\

H2

or

2

2

Ë2 D Û Ë2 N Û 40 – Ì 1 Ü – Ì 1 Ü Í D1 Ý Í N1 Ý

Ans.

From Eq. (1.33), Ë P Û Ì Ü 5 3 ÌÍ U D N ÜÝ1

Ë P Û Ì Ü 5 3 ÌÍ U D N ÜÝ2 5

\ or

ËD Û ËN Û P1 Ì 2 Ü Ì 2 Ü Í D1 Ý Í N1 Ý P2 = 13393.92 kW P2

(Q N2 = 2N1; D2 = 2D1) 3

5

Ë2 D Û Ë2 N Û 52.32 – Ì 1 Ü Ì 1 Ü Í D1 Ý Í N1 Ý

3

Ans.

EXAMPLE 1.18 A model of a centrifugal pump absorbs 5 kW at a speed of 1500 rpm, pumping water against a head of 6 m. The large prototype pump is required to pump water to a head of 30 m. The scale ratio of diameter is 4. Assuming the same efficiency and similarities, find (a) the speed, (b) the power of prototype and (c) the ratio of discharge of prototype and model.

Introduction to Turbomachines

Solution:

49

(a) The speed (Np): Ë H Û Ì 2Ü Í ( ND) Ý m

\

Np

Ë H Û Ì 2Ü Í ( ND) Ý p Ë Hp Û Ì Ü Í Hm Ý

(Eq. (1.30))

Ë Dm Û Ì Ü – Nm ÌÍ D p ÜÝ

1500 –

1 30 – 4 6

838.5 rpm

Ans.

(b) Power (Pp): Ë P Û Ì 5 3Ü Í UD N Ým

Ë P Û Ì 5 3Ü Í UD N Ý p

[' U p

5

Ë Dp Û Ë N p Û Pm Ì Ü Ì Ü Í Dm Ý Í N m Ý

\

Pp

or

Ë 838.5 Û Pp 5 – [4]5 Ì Ü Í 1500 Ý

Um ]

(Eq. (1.33))

3

3

Ans.

894.34 kW

(c) The ratio of discharge of prototype and model (Qp/Qm): Ë Q Û Ì Ü Í ND3 Ý m

\

Qp Qm

Ë Q Û Ì Ü Í ND3 Ý p

(Eq. (1.28))

Ë N p Û Ë Dp Û Ì ÜÌ Ü Í N m Ý Í Dm Ý

2

Ë 838.5 Û 3 Ì 1500 Ü – [4] Í Ý

35.776

Ans.

EXAMPLE 1.19 A single-stage centrifugal pump with impeller diameter of 30 cm rotates at 2000 rpm and lifts 3 m3 of water per second to a height of 30 m with an efficiency of 75%. Find (a) the number of stages and (b) the diameter of each impeller of a similar multistage pump to lift 5 m3 of water per second to a height of 200 m, when rotating at 1500 rpm. Solution: (a) Number of stages (Ns): We have the specific speed relation as ËN QÛ Ì 3/4 Ü ÍÌ H ÝÜ1

i.e.

Ë 2000 3 Û Ì 3/4 Ü ÍÌ 30 ÝÜ1

ËN QÛ Ì 3/4 Ü ÍÌ H ÝÜ 2 Ë 1500 5 Û Ì 3/ 4 Ü ÍÌ H ÝÜ 2

\

H2 = 28.73 m

Hence

Ns

Ht H2

200 28.73

6.96  7

Ans.

50

Fundamentals of Turbomachinery

(b) Diameter of each impeller (D2): From Eq. (1.30),

Ë H Û Ì 2Ü Í ( ND) Ý1 Ë Û 30 Ì 2Ü Í (2000 – 300) Ý1

or \

Ë H Û Ì 2Ü Í ( ND) Ý2 Ë 28.73 Û Ì 2Ü Í (1500 – D) Ý 2

D2 = 39 m

Ans.

EXAMPLE 1.20 An axial flow pump with a rotor diameter of 30 cm handles liquid water at the rate of 2.7 m3/min while operating at 1500 rpm. The corresponding energy input is 125 J/kg. The total-to-total efficiency is 75%. If a second geometrically similar pump with a diameter of 20 cm operates at 3000 rpm, find (a) its flow rate, (b) power input and (c) change in total pressure. Solution: (a) Flow rate (Q2): From Eq. (1.28),

Ë Q Û Ì Ü Í ND3 Ý1 \

2.7 Ë Q Û ; Ì 3Ü 3 Í ND Ý2 30 – 1500

Q2 = 1.6

Q2 3

20 – 3000

m3/min

Ans.

(b) Power input (P2): From Eq. (1.33), Ë P Û Ì 5 3Ü Í U D N Ý1

\

Ë P Û 125 Ì 5 3Ü ; 5 3 Í U D N Ý2 30 – 1500

P2 5

20 – 3000 3

P2 = 131.69 J/kg

Ans.

(c) Change in total pressure (Hm2/Hm1): From Eq. (1.30), Ë Hm Û Ì 2 2Ü Í N D Ý1

Hm2 Ë Hm Û Ì 2 2Ü ; H Í N D Ý2 m1

\ \ Total pressure increases by 77%.

Hm2 H m1

2

Ë D2 Û Ë N 2 Û Ì Ü Ì Ü Í D1 Ý Í N1 Ý

2

Ë 20 3000 Û Ì 30 – 1500 Ü Í Ý

2

1.77

Ans.

EXAMPLE 1.21 The following data were obtained from the main characteristics of a Kaplan turbine of runner diameter 1 m. Pu = 30.695, Qu = 108.6, Nu = 63.6. Estimate (a) the runner diameter, (b) the discharge and (c) the speed of a similar runner working under a head of 30 m and developing 2000 kW. Also (d) determine the specific speed of the runner.

Introduction to Turbomachines

51

Solution: (a) The runner diameter (D2): From Eq. (1.33), Power coefficient —

P 2

2

N D – ND – D

—

2

P 2

V VD 2

(ND)2 µ u2 µ V2

ND µ u, V— H

P

\

\

5

D N

3

—

Ë P Û Ì 3/2 2 Ü Í H D Ý1

\

P 3

V D

2

P

—

H

3/2

Pu

30.695

D12

2

D2 Ë P2 Û Ì 2 3/2 Ü ÍÌ D2 H ÝÜ 2

1

Ë 2000 Û Ì 2 3/2 Ü Í D 30 Ý 2

D2 = 0.63 m

Ans.

(b) The discharge (Q2): From Eq. (1.28), Q ND

\

Q 3

Ë Q Û Ì 2Ü Í H D Ý1

\

Q

ND D

2

H D2

Qu

108.65

D12

2

1

Ë Q Û Ì 2Ü Í H D Ý2

Ë Û Q2 Ì Ü 2 Í (0.63) – 30 Ý2

Q2 = 236.25 m3/s

Ans.

(c) The speed (N2): From Eq. (1.30), Ë ND Û Ì Ü Í H Ý1

\

D1 Nu

1 – 63.6

D2 N 2 H2

0.63 – N 2 30

N2 = 553 rpm

Ans.

(d) Specific speed (NST):

N ST

N P H

5/ 4

553 – 2000 30 5 / 4

352.24 rpm

Ans.

EXAMPLE 1.22 An axial flow compressor is designed to run at 5000 rpm when atmospheric pressure and temperature are 101.325 kPa and 20°C respectively. During the test day, the atmospheric temperature is 30°C. Calculate the speed at which the compressor must run.

52

Fundamentals of Turbomachinery

Solution: Speed of the compressor (N2): We have the relation as follows: Ë N Û Ì Ü ÌÍ T0 ÜÝ1

Ë N Û Ì Ü ÌÍ T0 ÜÝ 2 1/ 2

\

1/ 2

ËT Û N1 Ì 02 Ü Í T01 Ý

N2

Ë (273  30) Û 5000 – Ì Ü Í (273  20) Ý

5084.6 rpm

Ans.

EXAMPLE 1.23 Following data refers to a centrifugal pump. The prototype pump has to handle liquids whose kinematic viscosity (n) varies from 2 to 6 times that of water. Calculate (a) the range of speeds of the model, (b) flow rates of the model and (c) the range of heads of model if the model uses water. Take 1/5 scale model. Prototype

Model

n = 2 to 6 times that of water NSP = 0.2 rps

N=? H=? Dm 1 Dp 5

Q = 2.2 m3/s H = 16 m Solution: We have the Reynold’s number relation, UVD P

\

Ë ND 2 Û Ì Ü ÌÍ Q ÜÝ m

VD P/U

VD uD ( DN ) D — — Q Q Q

ND 2 Q

Ë ND 2 Û Ì Ü ÌÍ Q ÜÝ p

First consider np = 2nm (given data): \

Np Nm

2

Ë Dm Û Ë Q p Û Ì Ü Ì Ü ÍÌ D p ÝÜ Í Q m Ý

2 Ë 2Q m Û Ë1 Û Ì5Ü – Ì Q Ü Í Ý Í m Ý

0.08

(i)

We have the flow coefficient relation from Eq. (1.28) as

Ë Q Û Ì Ü Í ND3 Ý m \

Qp Qm

Ë Q Û Ì Ü Í ND3 Ý p Ë N p Û Ë Dp Û Ì ÜÌ Ü Í N m Ý Í Dm Ý

3

0.08 – [5]3

10

(ii)

Introduction to Turbomachines

53

From Eq. (1.30), Ë H Û Ì 2Ü Í ( ND) Ý m

\

Hp Hm

Ë H Û Ì 2Ü Í ( ND) Ý p 2

ÑÎË N p Û Ë D p Û ÑÞ ÏÌ Ü–Ì Üß ÑÐÍ N m Ý Í Dm Ý Ñà

(0.08 – 5)2

0.16

(iii)

From specific speed relations, N SP

ËN Q Û Ì 3/4 Ü ÍÌ H ÝÜ p

0.2

N – 2.2 163 / 4

\ Np = 1.0787 rps Case I: i.e. viscosity of liquid is 2 times that of water (a) Speed of model if viscosity of liquid is 2 times that of water (Nm): \ From Eqs. (iv) and (i), Np

Nm

1.0787 0.08

0.08 (b) Flow rate of model (Qm): From Eq. (ii) and given data, Qm

Qp

2.2 10

10

(iv)

13.48375 rps

(v)

0.22 m 3 /s

(vi)

(c) Head of model (Hm): From Eq. (iii) and given data, Hp

16 100 m 0.16 0.16 Case II: i.e. viscosity of liquid is 6 times that of water Hm

Np Nm Qp Qm Hp

2

2

È Dm Ø Ë Q p Û É D Ù ÌQ Ü Ê pÚ Í mÝ

Ë1Û Ì 5 Ü – (6) Í Ý Ë N p Û Ë Dp Û Ì ÜÌ Ü Í N m Ý Í Dm Ý

(vii)

0.24

(viii)

30

(ix)

3

Ë N p Dp Û – Ì Ü H m Í N m Dm Ý Np = 1.0787 rps

0.24 – (5)3 2

(0.24 – 5)2 1.44

(x) (xi)

54

Fundamentals of Turbomachinery

(a) Speed of model (Nm): From Eqs. (viii) and (xi), Np

1.0787 0.24

4.5 rps

(xii)

0.0733 m 3 /s

(xiii)

11.11 m

(xiv)

4.5 < model speed < 13.48

Ans.

Nm

0.24

(b) Flow rate of model (Qm): From Eq. (ix) and data, Qm

Qp 30

2.2 30

(c) Head of the model (Hm): From Eq. (x) and data, Hm

Hp

1.44 The speed range, Eqs. (v) and (xii), is

16 1.44

The flow rate range, Eqs. (vi) and (xiii), is 0.0733 < model < 0.22 The head range, Eqs. (vii) and (xiv), is 11.11 < model head < 100

Ans.

EXAMPLE 1.24 A Francis turbine is to be designed to run at 200 rpm under a head of 150 m with an efficiency of 95%. Calculate the power developed from the turbine. Solution: Power developed from the turbine (P): Assume a specific speed of 20 (NST) N P

\

N ST

or

20

\

P = 2755.67 kW

( H )5 / 4 200 P (150)5 / 4

IMPORTANT EQUATIONS • Continuity equation m

U1 V1 A1

where V1 and V2 are velocities of fluid.

U2 V2 A

Ans.

Introduction to Turbomachines

55

• Dimensionless parameters (General fluid flow)

â

S1

â

S2

â

S3

UVd P

â

S4

UV 2 d V

â

S5

V a

â

S6

UV 2 wd

t d

Geometrical relationship

'p UV

Pressure force Inertia force

2

Inertia force Viscous force

Reynolds number

Inertia force Surface tension force

Fluid velocity Sound velocity

Weber number

Mach number

Inertia force Gravitational force

Froude number

• Dimensionless parameters (applied to turbomachines)

â

S1

â

I

â

S2

â

S2

â

S3

â

S5

â

S6

â

u

S dN — dN — V — H 60

â

P

Power =

Q

Q

Nd

3

u V

d

2

H

= Capacity coefficient or Flow coefficient or Specific capacity or Discharge coefficient

Speed ratio =

H N 2d2

Runner tangential speed Theoretical jet spouting speed

= Head coefficient or Specific head

'p UV 2 P 3 5

UN d

N Q H3/ 4 N P H5/ 4

P 2

d H3/2

= Power coefficient or Specific power

N SP

Specific speed of pump

N ST

Specific speed of turbine

H H H H5/ 2 UQgH — QH — d 2VH — d 2 H H — — 2 K0 N2 N

56

Fundamentals of Turbomachinery

â â â

P = Power = H – m N1

N2

T01

T02

m1 T01

m2 T02

P01

P02

U Q gH

REVIEW QUESTIONS 1. Define specific speed of a turbine and a pump. Obtain an expression for the specific speed of turbine. Explain its significance. 2. Briefly discuss how the turbomachines are classified. 3. What is a positive displacement machine? How does it compare with a turbomachine in respect of the following aspects? (i) Mode of action (iii) Mechanical features

(ii) Operation (iv) Efficiency of conversion of energy.

4. A turbomachine is a device in which energy transfer takes place between a flowing fluid and a rotating element due to dynamic action resulting in a change of pressure and momentum of the fluid. Discuss. 5. Define the following for a turbomachine (i) Specific speed (iii) Power coefficient

(ii) Flow coefficient (iv) Capacity coefficient

6. Explain how the principle of dimensional analysis is applied to turbomachines. 7. Define turbomachines and explain the different types of turbomachines. 8. Compare turbomachines and positive displacement machines. 9. Classify the fluid machines. 10. Briefly explain the major parts of turbomachines. 11. Mention dimensionless parameters which are based in turbomachines and explain their significance. 12. Briefly discuss the effect of Reynold’s number. 13. Define and derive the specific speed of turbines and pumps. 14. Derive the specific speed of pumps and turbines by using the dimensional analysis technique [p theorem].

Introduction to Turbomachines

57

EXERCISES 1.1 The discharge Q through a rotating machine such as pump, turbine or compressor depends on the gH (g is acceleration due to gravity and H head over the machine), power supplied P, speed of rotation N, characteristic length D, mass density r and viscosity m of the fluid. Obtain a functional expression for the discharge Q in terms of non-dimensional numbers. 1.2 Four water turbines of specific speed 890 each, are installed in a hydel station. Each of the turbines runs at 50 rpm and shares equally a discharge of 260 m3/s available under a head of 1.73. Assume that each turbine has an efficiency of 82.5%. Find the power of each turbine in kW. 1.3 The resistance R experienced by a partially submerged body depends upon the velocity V, the length of the body L, the viscosity of the fluid m, the density of the fluid r and the gravitational acceleration g. Obtain a dimensionless expression for R. 1.4 A large body of a fluid stream is flowing with a velocity V over a thin rectangular plate of length L, and width b. The plane of the plate is at right angles to the fluid stream velocity V. Find the factor influencing its drag resistance. 1.5 The efficiency of a ceiling fan depends upon the following variables; density r, dynamic viscosity m, the discharge through the fan Q, the diameter D, and the speed N. Derive an expression for the efficiency using dimensional analysis. 1.6 The pressure change across the convergent-divergent nozzle depends upon the discharge Q, length of the nozzle L, the outlet diameter D, the density r, viscosity m, and local velocity of sound C. Determine the dimensionless number. 1.7 A turbine has a scale ratio 1/10. Following data refers to model and prototype. Calculate the discharge, speed and overall efficiency of the prototype. Model P = 25 kW N = 500 rpm H = 10 m h0 = 0.8

Prototype P = 50 kW H = 130 m

1.8 Calculate the number of pumps required to take water from a deep well under a total head of 90 m. All the pumps are identical and are running at 800 rpm. The specific speed of each pump is given as 30 while the rated capacity of each pump is 0.2 m3/s. 1.9 A storage unit has a head of 30 m and a discharge 30 m3/s through the pipe which is connected to the storage unit. The speed of the rotor is 200 rpm. Suggest a turbine suitable for this data.

2

Energy Transfer in Turbomachines

2.1 EULER TURBINE EQUATION According to Newton’s second law of motion, the sum of all the forces acting on a control volume in a particular direction is equal to the rate of change of linear momentum of the fluid across the control volume. Suppose m = mass of the body (kg) V1 = initial velocity of fluid (m/s) V2 = final velocity of fluid (m/s) \ or

m(V2  V1 ) m (V2  V1 ) dt SF × dt = m(V2 – V1) 6F

This equation is a modified form of Newton’s second law of motion. The left hand side of this equation (SF × dt) represents the impulse acting on the body. The right hand side [m(V2 – V1)] represents the change in momentum of the body in the time period dt (short time interval). Hence, this equation is known as impulse momentum equation. It is used to study the impact of fluid jet striking a stationary or moving plate and also to study the flow characteristics, namely the head loss in a pipe due to change in area, hydraulic jump, etc. When the flowing fluid with initial velocity (V1) is obstructed by a surface (vanes, blades, etc.), the fluid undergoes a change in momentum. The impulsive force acting on the fluid by the surface is given by F m (V2  V1 ) 58

Energy Transfer in Turbomachines

59

According to Newton’s third law of motion—for every action there is equal and opposite reaction. Therefore, the fluid reacts to this force thereby exerting equal but opposite force on the surface, i.e. the force exerted by the fluid on the surface is given by F

m (V1  V2 )

Similarly, we can state that the summation of all the torques acting on the system is equal to the rate of change of angular momentum. Figure 2.1 shows a rotor of a generalized turbomachine, where a–a represents the shaft of the machine, i.e. axis of rotation with an angular velocity w. Consider flow of fluid over the rotor, i.e. entering the rotor. The distance of a water particle from the centre of rotation changes every instant. Fluid enters the rotor at 1, passes through the rotor by any path and exits the rotor at 2. The angle of entry and exit may be considered arbitrary.

Figure 2.1

Rotor of a generalized turbomachine: 1. is the inlet point, 2. is the exit point, V1 is the Inlet absolute velocity of fluid, V2 is the exit absolute velocity of fluid, and w is the angular velocity (rad/s).

Let V be the absolute velocity of fluid entering the rotor at 1 and at any angle. This velocity vector may be resolved into three mutually perpendicular components. • • •

Axial component Radial component Tangential component

Va VR Vw

Axial component: This is parallel to the axis of rotation. Axial force is produced due to the change in magnitude of this component. Axial force is taken care by thrust bearing and finally transferred to the casing. Radial component: This is parallel to the radius of the rotor. Radial force is produced due to the change in magnitude of this component. This force is taken care by journal bearing. No torque is exerted on the rotor due to these two forces, i.e. axial and radial.

60

Fundamentals of Turbomachinery

Tangential component: The torque is exerted on the rotor only due to the change in the angular momentum of tangential component.

Assumptions 1. Fluid enters and leaves the vane in a direction tangential to the vane tip at inlet and outlet. 2. There is no frictional resistance as the fluid flows over the vane. Let V = absolute velocity of fluid (m/s) N = speed of the wheel (rpm) r = radius of the wheel (m) w = angular velocity of wheel (rotational speed) (rad/s) = 2pN/60 u = linear velocity of vane tip (peripheral velocity) (m/s) = p dN/60 m = mass flow rate of the fluid (kg/s) d = rotor or drum diameter (m) Suffixes ‘1’ and ‘2’ refer to the values at the inlet and the outlet respectively. Tangential momentum of fluid at entry

Vw1 m / gc

Moment of momentum or angular momentum at entry Similarly,

Vw1 m r1 / gc N-m

N

Vw 2 m r2 / gc N-m

Angular momentum at outlet

T = torque on the wheel = change of angular momentum (Vw1r1  Vw 2 r2 ) m N-m gc

W.D. = work done = rate of energy transfered = torque × angular velocity (Vw1r1  Vw 2 r2 ) Z m N-m/s or W gc

But, we have, wr1 = u1, w r2 = u2 \ or

W.D. W.D. Unit mass flow rate

(Vw1u1  Vw 2 u2 ) m W gc (Vw1u1  Vw 2 u2 ) J/kg gc

(2.1) (2.2)

Equations (2.1) and (2.2) are the forms of Euler turbine equation or Euler equation. This is applied to all turbomachines like pumps, fans, blowers, turbines (gas, steam, water) and compressors. If Vw1u1 > Vw2u2, the right hand side of Eq. (2.2) is positive, then the machine is called turbine. If Vw2u2 > Vw1u1, the right hand side of Eq. (2.2) is negative, then the machine is called pump, fan, compressor or blower.

Energy Transfer in Turbomachines

61

If Vw1u1 > Vw2u2 and if Vw2 is negative, i.e. Vw2 is opposite to that of Vw1, then Eq. (2.2) can be written as W.D. =

Vw1u1  ( Vw 2 u2 ) gc

Vw1u1  Vw 2 u2 J/kg gc

(2.2a)

2.2 ALTERNATE FORMS OF EULER TURBINE EQUATION Let V = absolute velocity of fluid Vr = relative velocity (relative to the rotor) Vf = flow velocity. This is one component of absolute velocity V. It is called radial velocity in case of radial flow machines and axial velocity in case of axial flow machines. Vw = tangential velocity, i.e. tangential component of absolute velocity V.

Figure 2.2

Parts of rotor of generalized turbomachine with inlet and outlet velocity triangles.

From inlet velocity triangle ABD (Figure 2.2),

V f21

V12  Vw21

Now consider the triangle BCD,

V f21

Vr21  (u1  Vw1 )2

(2.3)

62

Fundamentals of Turbomachinery

V f21

or

Vr21  u12  Vw21  2u1Vw1

(2.4)

Equating Eqs. (2.3) and (2.4), V12  Vw21

or

Vr21  u12  Vw21  2u1Vw1

u1Vw1

(V12  u12  Vr21 ) 2

(2.5)

u2Vw2

(V22  u22  Vr22 ) 2

(2.6)

Similarly,

Substituting Eqs. (2.5) and (2.6) in (2.2), W.D. Unit mass flow rate

(V12  u12  Vr21 ) (V22  u22  Vr22 )  2 gc 2 gc (V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

(2.7)

2.3 COMPONENTS OF ENERGY TRANSFER Equation (2.7) is made up of three components as follows:

First component (V12  V22 ) / 2 gc is the change in absolute kinetic energy. Due to this, a change in the dynamic head or dynamic pressure of the fluid takes place through the machine. The exit velocity V2, i.e. exit K.E. is negligible in some turbomachines and considerable in other turbomachines, particularly in power absorbing turbomachines like pumps and compressors. In power absorbing turbomachines, energy is transferred from rotor to fluid, therefore there is an increase in K.E. at the rotor exit. A diffuser converts this K.E. into static pressure rise.

Second component (u12  u22 ) / 2 gc is the change in centrifugal energy of the fluid in the motion. This is due to the change in the radius of rotation of the fluid. This causes a change in static head of the fluid through the rotor.

Third component (Vr22  Vr21 ) / 2 gc is the change in relative kinetic energy due to the change in relative velocity. This causes a change in static head of the fluid across the rotor.

63

Energy Transfer in Turbomachines

2.4 THE STEADY FLOW EQUATION OF THE FIRST LAW OF THERMODYNAMICS The steady flow equation of the first law of thermodynamics is Ë V 2 gz Û Q  m Ì h1  1  1 Ü 2 gc gc ÜÝ ÌÍ

Ë V 2 gz Û W.D.  m Ì h2  2  2 Ü 2 gc gc ÜÝ ÌÍ

(2.8)

where Q = rate of heat transfer (J/s) W.D. = work output (J/s) V2 = kinetic energy (J/kg) 2 gc gz = potential energy (J/kg) gc

Suffixes ‘1’ and ‘2’ refer to the inlet and outlet values, respectively. If h0 = stagnation or total enthalpy (J/kg) Dh0 = total enthalpy change (J/kg) then

h0 Q W.D.  m m

\

h

V 2 gz  2 gc gc

h02  h01

(2.9)

'h0

q – w = Dh0

or For isentropic process, q = 0 \

Dh0 = –w

(2.9a)

In differential form, –dh0 = w (V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

(2.10)

Equations (2.2) and (2.7) represent the ideal work with perfect flow of fluid by the blade and reversible process. This is known as Euler work. This is denoted by WiE (ideal Euler work). Figure 2.3 shows Euler, stage and actual work on T(or h)–s diagram. If the flow is not perfect and reversible, the work done is known as stage work done and denoted by Wst. This is also known as isentropic work done. If the flow is not perfect and the process is irreversible, then the work done is called the acual work done, Wa (adiabatic work done). However, the pressure drop takes place between

64

Fundamentals of Turbomachinery

the same pressure limits as that of stage work done. This is also known as adiabatic work stage done. \

WiE > Wst > Wa

(2.11)

p1 = Static condition

T or h 1

Isentropic Wa

Adiabatic

p02 = Stagnation condition

Wst

p2 = Static condition

WiE 2¢ 02

2 s

Figure 2.3

Euler, stage (isentropic), actual (adiabatic) work on T(or h)–s diagram.

2.5 DEGREE OF REACTION Work done (Eq. (2.7)) depends upon the quantity of change of static pressure and change of dynamic pressure factors. Changes in relative proportions of dynamic and static pressures are important. Depending upon the relative proportions, turbomachines are classified. A parameter is used to classify the turbomachines based on the relative proportions of dynamic and static pressure changes. This parameter is known as degree of reaction and denoted by R. R=

=

Energy transfer due to the change of static pressure in the rotor Total energy transfer in the rotor

(2.12)

(u12  u22 )  (Vr22  Vr21 ) 2 gc [(V12  V22 )  (u12  u22 )  (Vr22  Vr21 )] 2 gc (u12  u22 )  (Vr22  Vr21 ) (V12  V22 )  (u12  u22 )  (Vr22  Vr21 )

(2.13)

65

Energy Transfer in Turbomachines

(u12  u22 )  (Vr22  Vr21 ) 2 . gc . W.D. h1  h2 h01  h02

(2.14)

For an axial flow machine, centrifugal effects can be neglected (i.e. u1 = u2). Therefore, Eq. (2.13) reduces to R

(Vr22  Vr21 ) (V12  V22 )  (Vr22  Vr21 )

(2.15)

If there is no change in static pressure in the rotor and u1 = u2, then such a machine is called an impulse type. Therefore, Eq. (2.13) becomes R=0

(2.15a)

Note: In an impulse type machine, if the fluid enters and leaves the rotor at different radii a change of static pressure occurs due to centrifugal effects in one direction. An equal amount of change in static pressure is produced in the opposite direction. A machine with any degree of reaction must have the rotor enclosed in order to avoid expansion of fluid in all directions. This is called a reaction machine. In a machine with zero degree of reaction, the rotor can be of open type, i.e. an open jet of fluid with no connection with the rotor.

2.6 GENERAL ANALYSIS OF A TURBOMACHINE 2.6.1 Effect of Blade Discharge Angle Degree of Reaction

b2

on Energy Transfer and

Figure 2.4 shows velocity triangles for various values of discharge angle b2. The analysis is based on the following assumptions. •

Centrifugal effect at outlet = 2 × centrifugal effect at inlet, i.e. u2 = 2u1

•

Radial velocity is constant (flow velocity) i.e.

•

No tangential component at inlet i.e.

•

Vf 1 = Vf 2 = Vf Vw1 = 0;

a1 = 90°; Vf1 = V1

Inlet fluid angle, i.e. inlet blade angle is 45°. \

Vf 1 = Vf 2 = u1 = V1 = Vf

•

The outlet blade angle b2 (outlet fluid angle) is variable.

•

Applying the 3rd condition to Eq. (2.2), we get

66

Fundamentals of Turbomachinery

Figure 2.4

Velocity triangles for various values of blade discharge angle b2.

Energy Transfer in Turbomachines

Vw 2 u2 J/kg gc

W.D.

67 (2.16)

From Figure 2.4(a),

W.D. = W.D. =

u2 (u2  V f cot E2 ) gc 2V f (2V f  V f cot E2 ) gc

(u2 = 2u1 = 2Vf)

2V f2 (2  cot E 2 )

(2.16a)

gc 2V f2 (cot E2  2)

(2.17)

gc

For the purpose of comparison, Vf is kept constant for all angles b2 and it is unity. \

W.D.

2 (cot E 2  2) gc

(2.18)

From Figures 2.4(a) and (b) (Figure 2.4(b) inlet triangle and Figure 2.4(a) or (b) exit triangle),

Vr22

V f22  (V f 2 cot E2 )2

Vr21

V f21  u12

2V f21

V f 2 (1  cot 2 E2 )

2V f2

(Q Vf1 = u1)

Substituting the above data and Eq. (2.16a) in Eq. (2.13), we have R

(V f2  4V f2 )  V f2 (1  cot 2 E 2 )  2V f2 2 gc –

2V f2 (2  cot E2 )

5V f2  V f2 (1  cot 2 E2 )

V f2 ( 5  1  cot 2 E2 )

2 – [ 2V f2 (2  cot E2 )]

2 – 2V f2 (cot E2  2)

cot 2 E  4 4 (cot E2  2)

\

R

gc

2  cot E2 4

(cot E 2  2) (cot E2  2) 4(cot E 2  2)

(2.19)

The following results (Table 2.1) are obtained after substituting different b2 values in Eq. (2.18) and Eq. (2.19).

68

Fundamentals of Turbomachinery

Table 2.1

Effect of blade discharge angle b2 on energy transfer and degree of reaction

Discharge angle, b2

W.D. (energy transfer)

Reaction

Remarks

Less than 26.5° Figure 2.4(b)

+ve

+ve reaction

Machine is a turbine, reaction type; the whirl component (tangential component) is in a direction opposite to the direction of rotation.

Equal to 26.5° Figure 2.4(c)

0

+ve reaction

No work done (no energy transfer), reaction type, fluid leaves radially, Vw2 = 0, (whirl component is zero), V1 = V2 = Vf 1 = Vf 2 = u1, increase in static head due to centrifugal effect is counterbalanced by the decrease in static head due to increase in relative velocity. This is a case where the indeterminate quantity, 0/0, has a value of unity.

Greater than 26.5° Figure 2.4(a)

–ve

+ve reaction

Machine is a work-absorbing device, like the pump, compressor etc.; reaction type, both Vw2 and rotation of wheel are in the same direction.

Equal to 90° Figure 2.4(d)

–ve

+ve reaction

Machine is a work-absorbing device, like pump and compressor etc.; reaction type, both Vw2 and rotation of wheel are in the same direction.

Equal to 153.5° Figure 2.4(e)

–ve

zero (no reaction)

Machine is a work-absorbing type, no reaction, impulse type, both Vw2 and rotation of wheel, are in the same direction.

Greater than 153.5° Figure 2.4(f)

–ve

–ve reaction

Machine is a work-absorbing type, reaction type, discharge velocity V2 is very high, static head is less at the outlet than that at the inlet.

Figure 2.4(A) shows the variation of W.D. and reaction R with reference to the blade discharge angle b2. Figure 2.4(A) can be summarized as follows: • • • •

If b2 = 26.5°, W.D. = 0 If b2 > 26.5° W.D. is –ve, therefore, the machine is a work-absorbing device, like the pump, compressor, fan, etc. If b2 < 26.5°, W.D. is +ve, for example, a turbine If b2 = 26.5°, R = 1 R

• •

Change in static pressure head in rotor Total energy transfer in the rotor

\ If b2 = 26.5°, it means there is no change in static pressure head. If b2 < 26.5, R > 1, machine acts like a turbine. If 26.5° £ b2 £ 153.5°, it means 1 ³ R £ 0

Energy Transfer in Turbomachines

Figure 2.4(A)

69

Variation of W.D. and R with reference to the blade discharge angle.

2.7 GENERAL ANALYSIS OF CENTRIFUGAL PUMPS AND COMPRESSORS 2.7.1 Effect of Blade Discharge Angle on Performance This effect is the same as discussed in Section 2.6.1 for the turbomachine.

2.7.2 Theoretical Head Capacity Relationship For pumps and compressors Vw2u2 > Vw1u1, therefore, Eqs. (2.2) and (2.7) can be written as W.D. = W.D. =

Vw 2 u2  Vw1u1 gc

(work done on the fluid)

(V22  V12 )  (u22  u12 )  (Vr21  Vr22 ) 2 gc

(2.20) (2.21)

Figure 2.5 shows the velocity triangles for different values of discharge angle b2 for centrifugal pumps and compressors.

70

Fundamentals of Turbomachinery

Figure 2.5

(contd.)

Energy Transfer in Turbomachines

Figure 2.5

71

Velocity triangles for different discharge angles b2 for centrifugal pumps and compressors.

Inlet conditions are: Radial inlet, a1 = 90°, Vw1 = 0 Equation (2.20) reduces to W.D. = H = head =

u2Vw 2 g

(2.21a)

u2 (u2  V f 2 cot E2 ) g

(Refer to Figure 2.5(c))

(2.22)

We know that volume flow rate is given by Q = A2Vf 2 \ Vf 2 = Q/A2 \

H

u2 g

È Q cot E2 Ø ÉÊ u2  ÙÚ A

(2.23)

2

For a given pump or compressor, u, A, b2 are fixed, and the only variables are H and Q. Centrifugal pumps and compressors can be classified as follows (Figure 2.6): • • •

Radial curved vanes Forward curved vanes Backward curved vanes

b2 = 90° b2 > 90° b2 < 90°

72

Fundamentals of Turbomachinery

Head (H ) Forward

°

90 b2 >

b2 = 90°

Shut off head

Radial

b2 < 90°

Backward

Q

Flow

Figure 2.6

H–Q diagram for various b2 angles.

Equation (2.23) can be written as H

u22 u2Q cot E2  g A2 g

H = K1 – K2Q where

K2

u2 cot E2 ; A2 g

Radial curved vanes (Figure 2.5(b)): a1 = 90°, Vw1 = 0, b2 = 90°, \ K2 = 0 \

K1

u22 g

Vr2 = Vf 2, Vw2 = u2, Vf1 = V1

H = K1 = u22 / g = constant

Hence: • Head is constant for all rates of flow. • Flow direction and wheel rotation direction are same. • Outlet tip of the blade is in the radial direction. Backward curved vanes (Figure 2.5(c)): b2 < 90°, Hence: • • • •

a1 = 90°,

Vw1 = 0, Vf 1 = V1

K2 is positive. H–Q line has negative slope. Outlet tip of the blade is in the direction opposite to that of rotation of wheel. Flow and wheel rotation are in the same direction.

Energy Transfer in Turbomachines

73

Forward curved vanes (Figure 2.5(a)): Hence: • • • •

b2 > 90°,

a1 = 90°,

Vw1 = 0, V1 = Vf 1

K2 is negative. H–Q line has positive slope. Outlet tip of the blade is in the direction of motion. Flow and rotation of wheel are in the same direction.

Majority of centrifugal pumps have backward curved vanes having b2 = 20° to 25°. Radial curved vanes are used for very high speed compressors to get the highest possible pressure head.

Centrifugal machines stage parameters Here centrifugal machines mean, centrifugal fans, blowers, and pumps. Forward curved vanes (Refer to Figure 2.5(a)): It will be noted that for all vane angles, H reduces to u22 / g for zero flow (Q = 0), this condition is known as the shut-off head (obtained by substituting Q = 0 in Eq. (2.23)) We know that m = r1Q1 = r2Q2 Q1 = A1Vf 1; Q2 = A2Vf 2 A1 = pd1b1; A2 = pd2b2 where m = mass flow rate (kg/s) d1 = inlet diameter (m) d2 = outlet diameter (m) r = density of the fluid (kg/m3) b1 = inlet width of the impeller (m) Q = discharge (m3/s) 2 b2 = exit width of the impeller (m) A = area (m ) Vf = flow velocity (m/s) \

m = r1A1Vf 1 = r1pd1b1Vf 1 = r2pd2b2Vf 2

Any density change in the flow is negligible for a small pressure rise through the stage, hence the fluid is assumed to be incompressible. If r = constant and Vf 1 = Vf 2 = Vf , then the above equation reduces to m = d1b1 = d2b2 \

b1 b2

d2 d1

Work done or stage work (W.D.) In the absence of inlet guide vanes, it is assumed that the whirl velocity is ZERO at entry, i.e. a1 = 90°, Vw1 = 0, V1 = Vf1. Work done in an adiabatic process, W.D. = u2Vw2

ÈV Ø u22 É w 2 Ù Ê u2 Ú

'h0

(2.24a)

74

Fundamentals of Turbomachinery

For constant radial velocity (flow velocity), V1 = Vf 1 = Vf 2 = u1 tan b1

(2.24b)

From the exit velocity triangle of Figure 2.5(c), u2 = Vw2 + Vf 2 cot b2 u2 – Vf 2 cot b2 = Vw2

or or

1

Vf 2

cot E2

u2

Vw 2 u2

(2.24c)

From Eqs. (2.24a) and (2.24c), Vf 2 È Ø W.D. u22 É1  cot E2 Ù u2 Ê Ú V2 sin E2

(2.24d)

u2 sin (D 2  E2 ) u2 sin E2 cos D 2 sin (D 2  E2 )

Vw 2

V2 cos D 2

Vw 2 u2

sin E2 cos D 2 (sin D 2 cos E2  cos D 2 sin E2 )

tan E2 tan D 2  tan E2

(2.24e)

\ From Eqs. (2.24a), and (2.24d), ÈV Ø u22 É w 2 Ù Ê u2 Ú Vw 2 u2

Vf 2 È Ø 2 ÉÊ 1  u cot E2 ÙÚ u2 2 V f 2 cot E2 Ø È ÉÊ 1  ÙÚ u 2

tan E2 tan D 2  tan E 2

(From Eq. (2.24e))

(2.24f)

P = m × W.D. = mDh0 = mcpDT0 = mu2Vw2/gc

(2.24g)

Stage pressure rise (stagnation pressure rise through the stage)

(D 0)s

Power (P):

p

If the compression process is isentropic, then 'h0

\

( 'p0 )s

( 'p0 )s U

U'h0

U u2Vw 2 gc

U u22 gc

Vf 2 È Ø 1 cot E2 Ù  ÉÊ u Ú

(2.24h)

2

We know that the static pressure rise (Dpr) through the impeller is due to the change in centrifugal energy and the diffusion of the relative flow. \

( p2  p1 )

( 'p)r

U U (u22  u12 )  (Vr21  Vr22 ) 2 gc 2 gc

(2.24i)

Energy Transfer in Turbomachines

U(u22  u12 ) U(Vr21  Vr22 ) U(V22  V12 )   2 gc 2 gc 2 gc

( 'p0 )s

75 (2.24j)

From Eqs. (2.24i) and (2.24j),

or

( 'p0 ) s

( p2  p1 ) 

( 'p0 ) s

( 'p)r 

U(V22  V12 ) 2 gc

U (V22  V12 ) 2 gc

(2.24k)

Degree of reaction By definition, R

Pressure rise in the rotor (impeller) Pressure rise in the stage (total pressure rise in the stage)

( 'p)r ( 'p0 ) s

From inlet velocity triangle of Figure 2.5(a), Vr21  u12

V12

Substituting the above relation in Eq. (2.24i), ( 'p)r

U (u22  Vr22  V f21 )

U (u22  Vr22  V12 ) 2 gc

(a1 = 90°

2 gc

(2.24l)

\ V1 = Vf 1 = Vf 2)

From exit velocity triangle of Figure 2.5(a),

Vr22  V f22

u22  Vr22  V f22

(Vw 2  u2 )2

2u2Vw2  Vw22

Substituting the above relation in Eq. (2.24l), ( 'p)r

U (2u2Vw 2  Vw22 ) 2 gc

(2.24m)

Substituting Eqs. (2.24h) and (2.24m) in degree of reaction, R

\

U (2u2Vw 2  Vw22 ) 2 gc

–

R

gc

Vw 2 (2u2  Vw 2 )

U u2 Vw 2

Vw 2 2u2

È Vw 2 Ø ÉÊ 1  2u ÙÚ 2

(2.24n)

76

Fundamentals of Turbomachinery

Degree of reaction for different vanes: (a) Backward curved vanes (b2 < 90°) V For backward curved vanes, w 2  1 u2 \ R is always less than unity (b) Radial blades (b2 = 90°) For radial blades, Vw2 = u2 \ R = 0.5 (c) Forward curved vanes (b2 > 90°) For forward curved vanes, Vw2 > u2

\ R < 0.5

Stage efficiency (hs) Ks

Isentropic stagnation work Actual work

( 'p0 )s U u2 Vw 2

( 'p0 )s U u2 Vw 2

[Vw2 is the actual value obtained in a real machine. This is less than the Eulerian value. Therefore ideal work input = (Dp0)s/r and actual work input = u2Vw2.]

2.8 GENERAL ANALYSIS OF AXIAL FLOW COMPRESSORS AND PUMPS Power absorbing turbomachines raise the stagnation pressure of a fluid through mechanical energy intake. In the power absorbing turbomachines, the quantity of interest is the change in stagnation enthalpy of the fluid. Due to the possibilities of separation, it is more difficult to design the blades. Normally the blade turning angles are limited to very small values in compressors, but in turbines they may be 160° or even more. In turbines, due to large turning anlges, a large value of change in tangential component of velocity occurs, keeping the number of stages small for a particular amount of energy transfer. But in compressors the turning angle does not exceed 20°, a small value of change in tangential component of velocity occurs, making the number of stages more for a particular pressure rise.

2.8.1 General Expression for Degree of Reaction Figure 2.7 shows the velocity triangle and rotor and stator blade arrangements. Air enters axially into the inlet guide vanes, leaves with certain angles and impinges on the first rotor with proper angle of attack. Here, kinetic energy is added to the fluid and a small pressure rise takes place. The air leaving from the rotor enters into the stator, where the pressure is further increased. This complets the first stage. Now the fluid is directed into the rotor of the second stage and the same process is continued in the remaining stages. For axial flow pumps and compressors, u1 = u2 = u and flow velocity Vf is termed axial velocity Va.

Energy Transfer in Turbomachines

77

Equations (2.2) and (2.7) can be written as u (Vw 2  Vw1 ) (2.25) gc For power absorbing machine, therefore, Vw2 > Vw1 and the work done on the machine is given by

W.D.

W.D. =

Figure 2.7

(V22  V12 )  (Vr21  Vr22 ) 2 gc

(2.26)

Velocity triangles and rotor and stator blade arrangements for axial flow compressors and pumps: (a) Rotor and stator blade arrangements. (b) Exit of the guide vanes. (c) Inlet to the rotor. (d) Exit from the rotor. (e) Inlet to the stator. (f) Exit from the stator.

78

Fundamentals of Turbomachinery

The absolute velocity V2 increases and the relative velocity decreases Vr2 in the first row of rotors. Due to decrease in relative velocity, there is an increase in static pressure in the rotor. The first row of rotating blades imparts kinetic energy of an amount equal to (2.25) or (2.26), which is also increase in the total pressure. The stator of the first stage decreases the absolute velocity (V3) and in turn causes a static pressure rise. From Figure 2.7, Vf 1 = Vf 2

or

Va1 = Va2

u1 = u2 = u = BC + Vw1 = FG + Vw2 \

Vw2 – Vw1 = BC – FG

(2.27)

Substituting Eq. (2.27) in (2.25), W.D.

u (BC  FG) gc u (Va1 cot E1  Va 2 cot E2 ) gc uVa (cot E1  cot E 2 ) gc

Vw2 = Va cot a2;

or

Vw1 = Va cot a1

(2.28) (From Figure 2.7)

Substituting the above data in Eq. (2.25), W.D.

(Va cot D 2  Va cot D1 ) u gc uVa (cot D 2  cot D1 ) gc

(2.29)

An axial flow stage may have any degree of reaction and there is no centrifugal effect (u1 = u2). Therefore, the change in static head or pressure in the rotor is given by 'h1

Vr21  Vr22 = increase in enthalpy in the rotor 2 gc

Vr21

Va2  BC 2

Vr22

Va2  FG 2

Va2  Va2 cot 2 E1

Va2  Va2 cot 2 E2

Va2 (1  cot 2 E1 )

Va2 (1  cot 2 E 2 )

(2.30)

Energy Transfer in Turbomachines

\

'h1

Now, \

Va2 (1  cot 2 E1  1  cot 2 E2 ) 2 gc

Va2 (cot 2 E1  cot 2 E2 ) 2 gc

R

Relative kinetic energy change in the rotor Total enthalpy change

R

'h1 'h1  'h2

79 (2.31)

Increase in enthalpy in the rotor Total increase in enthalpy in the stage

Equation (2.13) can be written for axial flow pumps and compressors as R

(Vr21  Vr22 ) (V22  V12 )  (Vr21  Vr22 )

(Vr21  Vr22 ) u(Vw 2  Vw1 ) 2 gc gc

'h1 'h1  'h2

'h1 W.D.

'h1 'h0

(2.32)

Substituting Eqs. (2.28) and (2.31) in (2.32),

R

Va2 (cot 2 E1  cot 2 E 2 ) uV (cot E1  cot E2 ) 2 gc – a gc

or

R

Va (cot 2 E1  cot 2 E 2 ) 2u (cot E1  cot E 2 )

or

R

Va (cot E1  cot E2 ) 2u Va 2u

or

R

Ë 1 1 Û  Ì Ü Í tan E1 tan E 2 Ý

'h1 'h1  'h2

Ë Ì' cot E m Í

Va cot E m u

'T1 'T1  'T2

Va 2u

(cot E1  cot E 2 ) Û Ü 2 Ý

Ë tan E 2  tan E1 Û Ì Ü Í tan E1 – tan E 2 Ý 'TR 'TR  'TS

(2.33a)

'hR 'hR  'hS

where DTR and DTS are the temperature rises in the rotor and stator respectively. From the first law of thermodynamics applied to flow processes, h1

V12 W 2 gc

h2 

V22 2 gc

(2.33)

(W = W.D.)

80

Fundamentals of Turbomachinery

'hR

'h1

h2  h1

W

(V22  V12 ) 2 gc

(2.33b)

Applying SFEE to stator, h2 

V22 2 gc

h3 

V32 2 gc

h3 

V12 2 gc

Assuming exit velocity from the stator as V3 = V1, 'hS

'h2

h3  h2

(V22  V12 ) 2 gc

DhS = cpDTS = gain in temperature or pressure in stator is equivalent to the loss of K.E. W

\

R W

1

(V22  V12 ) 2 gc

W

(V22  V12 ) (V22  V12 )  2 gc 2 gc

(V22  V12 ) 2 gc W

V22  V12 2Wgc

(2.34)

From Figure 2.7, V22

Va2  Vw22

V12

Va2  Vw21

Va2  Va2 cot 2D 2 Û Ü Va2  Va2 cot 2D1 ÜÝ

(2.35)

Substituting Eqs. (2.35) and (2.29) in (2.34), R 1

1

(Va2  Va2 cot 2D 2  Va2  Va2 cot 2D1 ) uVa (cot D 2  cot D1 ) 2 gc gc

Va2 (cot 2D 2  cot 2D1 ) 2uVa (cot D 2  cot D1 )

1

Va (cot 2D 2  cot 2D1 ) 2u (cot D 2  cot D1 )

or

R 1

Va (cot D 2  cot D1 ) (cot D 2  cot D1 ) 2u (cot D 2  cot D1 )

or

R 1

Va (cot D 2  cot D1 ) 2u

'hR

c p 'TR

W

(V22  V12 ) 2 gc

(2.36) (Eq. (2.33b))

Energy Transfer in Turbomachines

Now, V1 = Va cosec a1, V2 = Va cosec a2 'hR

uVa

81

(Refer to Figure 2.7)

(cot D 2  cot D1 ) 1 (Va2 cosec 2D 2  Va2 cosec 2D1 )  2 gc gc

uVa (cot D 2  cot D1 ) Va2  (cot 2D 2  cot 2D1 ) gc 2 gc 2uVa (cot D 2  cot D1 )  Va2 (cot 2D 2  cot 2D1 ) 2 gc Va (cot D 2  cot D1 ) [2u  Va (cot D 2  cot D1 )] 2 gc

R

'hR 'hR  'hS

'hR W

Va (cot D 2  cot D1 ) [2u  Va (cot D 2  cot D1 )] uV (cot D 2  cot D1 ) 2 gc a gc

Ë V Û 2u Ì1  a (cot D 2  cot D1 ) Ü 2 u Í Ý 2u Equations (2.36) and (2.36a) are same.

1

Va (cot D 2  cot D1 ) 2u

(2.36a)

2.8.2 Velocity Triangles for Different Values of Degree of Reaction It is assumed that the axial velocity Va(Vf) is constant as the cross-sectional area and density are constant. Separate inlet and exit velocity triangles are shown in Figure 2.7. These two velocity triangles are superimposed on the common base of the blade speed as shown in Figures 2.8, 2.9 and 2.10. Instead of common base, the diagrams may be superimposed with a common apex as shown in Figure 2.10(a). For R = 50% (refer to Figure 2.8):

Figure 2.8

Velocity triangles for R = 50%.

82

Fundamentals of Turbomachinery

Substituting R = 50% in Eqs. (2.33) and (2.36), u Va

\

(cot E1  cot E 2 )

(cot D 2  cot D 2 )

(2.37)

We can say that, if R = 50%, symmetrical blades, b1 = a2,

b2 = a1,

V1 = Vr2,

V2 = Vr1

Static enthalpy and temperature increase in the rotor and stator are equal and the efficiency is maximum. For R > 0.5 (Figure 2.9):

Figure 2.9

• •

Velocity triangles for R > 50%.

Diagram is moving towards left, since b2 < a1. The static enthalpy rise in the rotor is greater than that in the stator (static pressure rise is also greater in the rotor than that in the stator). \

FC = u, DC = Vw1, EC = Vw2, ED = EC – DC = Vw2 – Vw1 = Vw

For R < 0.5 (Figure 2.10): A

B Vr 1

Vr 2 b2 F

Va

V1

Va

a2

b1 E

Figure 2.10

• •

V2

D

a1 C

Velocity triangles for R < 0.5.

Diagram is moving towards right. The static enthalpy and pressure rise are greater in the stator than those in the rotor. \

b2 > a1

Energy Transfer in Turbomachines

Figure 2.10(a)

Combined velocity diagrams with a common apex.

2.9 GENERAL ANALYSIS OF TURBINES 2.9.1 Utilization Factor (e) W.D. = Energy utilized =

(V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

83

84

Fundamentals of Turbomachinery

Energy available to the rotor =

V12  (u12  u22 )  (Vr22  Vr21 ) 2 gc

(V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) V22  2 gc 2 gc

H

W.D.  V22 / 2 gc

W.D. + losses

Ideal work Energy supplied

Energy utilized Energy available to the rotor

(2.38)

Ideal work output Ideal energy available for conversion into work Work developed by the rotor Ideal energy available for conversion into work

Since exit absolute K.E. is lost, i.e. is not available for work development, (V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

H

V12  (u12  u22 )  (Vr22  Vr21 ) 2 gc (V12  V22 )  (u12  u22 )  (Vr22  Vr21 )

(2.39a)

V12  (u12  u22 )  (Vr22  Vr21 )

W.D. W.D.  V22

/ 2 gc

W.D. W.D. + losses

E ideal E ideal + losses

(2.39b)

The only loss in the absence of fluid friction is that due to the K.E. at the exit V22 / 2 gc , this energy represents the energy not utilized by the rotor. \

H

(Vw1u1  Vw 2 u2 ) / gc [V12



(u12



u22 )



(Vr22

2 – (Vw1u1  Vw 2 u2 )

 Vr21 )] / 2 gc

V12

 (u12  u22 )  (Vr22  Vr21 )

(2.39c)

Equation (2.13) can be re-arranged as (u12  u22 )  (Vr22  Vr21 )   

‘

R (V12  V22 )  [(u12  u22 )  (Vr22  Vr21 )] R  

‘

x

or or

x

x  Rx

R (V12

 V22 ) ;

x (1  R)

(u12  u22 )  (Vr22  Vr21 )

R (V12

 V22 )

R(V12  V22 ) /(1  R)

(2.39d)

Energy Transfer in Turbomachines

85

Substituting Eq. (2.39d) in (2.39a), R(V12  V22 ) (1  R) 2 R(V1  V22 ) V12  (1  R)

(V12  V22 )  H

(1  R)(V12  V22 )  R(V12  V22 )

V12  V22

V12 (1  R)  R(V12  V22 )

V12  RV22

(2.39e)

Equation (2.39e) holds good for a single rotor of any turbine, under the conditions where the Euler turbine equations are expected to hold.

2.9.2 Axial Flow Turbines Axial flow turbines include the famous steam and gas turbines using compressible fluids. The analysis is simplified since the centrifugal effect is absent, u1 = u2 = u.

Degree of reaction Substituting u1 = u2 = u in Eqs. (2.13) and (2.39a), (Vr22  Vr21 )

R

Vr22  Vr21 2 – W.D. – gc

(V12  V22 )  (Vr22  Vr21 )

or

(Vr22  Vr21 )

R(V12  V22 )  R(Vr22  Vr21 )

or

(Vr22  Vr21 )

R

(V12  V22 ) (1  R)

(2.40)

(2.41)

Utilization factor H

(V12  V22 )  (Vr22  Vr21 )

(2.42)

V12  (Vr22  Vr21 )

Relationship between utilization factor and degree of reaction Substituting Eq. (2.41) in (2.42), R(V12  V22 ) (1  R) 2 R(V1  V22 ) V12  (1  R)

(V12  V22 )  H

(1  R)(V12  V22 )  RV12  RV22 V12 (1  R)  RV12  RV22

V12  V22  RV12  RV22  RV12  RV22

V12  V22

V12  RV12  RV12  RV22

V12  RV22

(2.43)

86

Fundamentals of Turbomachinery

Case I

Impulse Turbine Vx 2 Vw 2 F Vf 2 (Va 2)

u

A a1

a2

B b2

Vw 1

Vx 1 E

b1

Vf 1 (Va 1)

Vr 1

V2 Vr 2

V1

D

C

Figure 2.11 Combined velocity diagram for steam turbine.

Let

Vr1 = Vr2, i.e. no frictional losses, b1 = b2

\

Va1 = Va2, u1 = u2 = u (axial flow), R = 0 (Impulse turbine) V12

Va21  Vw21 ; V22

Va22  Vw22

Substituting the above data in Eq. (2.43), H

H

or

Va21  Vw21  Va22  Vw22

Vw21  Vw22

V12

V12

(Vw1  Vw 2 ) (Vw1  Vw 2 )

(Vw1  Vw 2 ) 2u

V12

V12

(2.43a)

From Figure 2.11, cot b1 = Vx1/Va1; cot b2 = (Vw2 + u)/Va2; cot b1 = cot b2 \

Vx1/Va1 = (Vw2 + u)/Va2

Also, (Vw1 – Vw2) = Vw1 – Vx1 + u = Vw1 – (Vw1 – u) + u = Vw1 – Vw1 + u + u = 2u Case II Reaction Turbine Let R = 50%, then a1 = b2, a2 = b1, Vr1 = V2, Vr2 = V1 Equation (2.43) can be written as 1 H 1

V22 V12

V22 1 – V12 2

(2.43b)

Energy Transfer in Turbomachines

Vr21

Va21  Vx21

Va21  (Vw1  u)2

Va21  Vw21  u2  2uVw1

\

(Figure 2.11, triangle BEC)

V12  u2  2uV1 cos D1

V22

V12  u2  2uV1 cos D1

V22

ÈuØ u 2u 1É Ù  cos D1 ; Let V1 V1 Ê V1 Ú

V12 V22 V12

87

(2.43c)

2

I

1  I 2  2I cos D1

Substituting the above data in Eq. (2.43b), H

H

or

1  (1  I 2  2I cos D1 )

2(2I cos D1  I 2 )

1  (1  I 2  2I cos D1 ) – 1/ 2

1  2I cos D1  I 2

(2  4I cos D1  2I 2  2)

2(1  2I cos D1  I 2 )  2

1  2I cos D1  I 2

1  2I cos D1  I 2

2

2

W.D.

u(Vw1  Vw 2 ) gc

W.D.

u (2V1 cos D1  u) gc

W.D.

V12 È 2uV1 cos D1 u2 Ø  2Ù gc ÉÊ V12 V1 Ú

V12 / 2 gc (Vr22  Vr21 ) / 2 gc V12 (Vr22  Vr21 )  2 gc 2 gc V12 V12  Vr21  2 gc 2 gc

(2.43d)

1  2I cos D1  I 2 u [V1 cos D1  (Vr 2 cos E2  u)] gc

2uV1 cos D1  u2 gc V12 (2I cos D1  I  ) gc

K.E. supplied to the fixed blade = Dh1 K.E. supplied to the moving blade = Dh2 'h 'h0

total energy supplied to the stage (rotor and stator) V12 Vr21  gc 2 gc

Substituting Eq. (2.43c) in the above equation, 'h0

V12 (V12  u2  2uV1 cos D1 )  gc 2 gc

(2.43e)

88

Fundamentals of Turbomachinery

V12  2V1u cos D1  u2 2 gc

or

'h0 Kb

W.D. 'h0

blading efficiency =

gc V12

(2I cos D1  I 2 )

(2.43f) Equation (2.43e) Equation (2.43f)

2(2I cos D1  I 2 ) 2

2 gc

Kb

È u u2 Ø 1  2 cos D  1 É Ù V1 V12 Ú Ê

V12 (1  2I cos D1  I 2 ) 2 gc

V12

Kb

V12 2 gc

(1  2I cos D1  I )

(1  2I cos D1  I 2 )

(2  4I cos D1  2I 2  2)

2(1  2I cos D1  I 2 )  2

(1  2I cos D1  I 2 )

(1  2I cos D1  I 2 )

2

2

(2.43g)

1  2I cos D1  I 2

Equations (2.43d) and (2.43g) are same, hence the utilization factor and blading efficiency are same.

Condition for maximum efficiency Differentiating Eq. (2.43g) with respect to f and equating to zero to get the condition for maximum efficiency, d (1  2I cos D1  I 2 ) 1 dI

0; ? 2 cos D1  2I

0; Iopt

cos D1

u/V1

Substituting fopt in Eq, (2.43g) to get the maximum utilization factor (emax) or maximum blading efficiency (hbmax),

\

2

Kb max

2

Kbmax

Ë 1  cos2D1  1 Û 2Ì Ü 2 ÌÍ 1  cos D1 ÜÝ

2

2

1  2 cos D1  cos D1

È Ø 1 2 É1  Ù 2 Ê 1  cos D1 Ú

2 cos2D1 1  cos2D1

(2.43h)

The above equation (2.43h) holds good for any degree of reaction except R = 1; If R = 1, V1 = V2. Substituting R = 1 in Eqs. (2.41) and (2.43), e reduces to ¥, this requires special analysis.

R

Vr22  Vr21 2 gc – W.D.

Vr22  Vr21 2 gc – u(Vw1  Vw 2 ) gc

(2.43i)

Energy Transfer in Turbomachines

89

It is assumed that the axial velocity Va is constant through the stage Vr22

Va2  Vx22 ; Vr21

Va2  Vx21

(Refer to Figure 2.11)

Substituting the above data in Eq. (2.43i), (Va2  Vx22  Va2  Vx21 ) 2u(Vx 2  Vx1 )

(Vx 2  Vx1 ) (Vx 2  Vx1 )

(Va cot E2  Va cot E1 ) 2u

Va (cot E2  cot E1 ) 2u

R

2u (Vx 2  Vx1 )

Va cot E2 Va cot E1  2u 2u

\

cot E1

Vx1 ; cot D1 Va

cot E1

cot D1 

u  Vx1 Va

(From Figure 2.11) u Vx1  Va Va

(2.43j) (2.43k)

u  cot E1 Va

u Va

Substituting the above data in Eq. (2.43k), Va cot E 2 È uØ V  É cot D1  Ù – a Va Ú 2u 2u Ê

R

Va cot E2  Va cot D1  u 2u R

\

0.5 

Va cot E 2 (Va cot D1  u)  2u 2u

u Va (cot E2  cot D1 )  2u 2u

Va (cot E2  cot D1 ) 2u

(2.43l)

From Figure 2.11,

\

cot a 1 =

Vw1 u + Vw 2 u Vw 2 u ; cot b 2 = = + = + cot a 2 Va Va Va Va Va

cot E2

cot D 2 

u Va

Substituting the above data in Eq. (2.43l), R

È Ø u ÉÊ cot D 2  V  cot D1 ÙÚ a

0.5 

Va 2u

0.5 

Va V u Va cot D 2   cot D1 – a Va 2 u 2u 2u

1

Va (cot D 2  cot D1 ) 2u

(2.43m)

Vr2 = Vr1

0

V1 = Vr2, V2 = Vr1

a1 = b2; a2 = b1

Symmetrical triangles,

\

0.5

(Vr22  Vr21 ) (V12  V22 )  (Vr22  Vr21 ) 2 V1  V22 Vr22  Vr21

Eq. (2.40) and (2.43l),

\ b1 = b2 3. R = 50%; substituting R = 50% in

\

Vr22  Vr21

Eqs. (2.40) and (2.43j),

Substituting R = 0 in

Vr22  Vr21 R – W.D. R is negative W.D. is2positive; \ Vr1 > Vr2 2. R = 0; i.e. impulse stage

From Eq. (2.40),

1. R < 0, i.e. reaction is negative

Velocity diagrams for various values of ‘R’ (axial flow type turbines) (Figure 2.12) 90 Fundamentals of Turbomachinery

(u12  u22 )  (Vr22  Vr21 )  V22 )  (u12  u22 )  (Vr22  Vr21 )

V 1 = V2 a1 = a2

Figure 2.12 Combined velocity diagrams for different R values: (1) R < 0, (2) R = 0, (3) R = 0.5, (4) R = 1, (5) R > 1.

5. R > 1 \ V2 > V1

\ \

(u12  u22 )  (Vr22  Vr21 )

(V12  V22 )  (u12  u22 ) 

(V12

(Vr22  Vr21 )

1

4. R = 1 Substituting R = 1 in Eqs. (2.40) and (2.43m),

Energy Transfer in Turbomachines

91

92

Fundamentals of Turbomachinery

2.9.3 Radial Flow Turbines A simple example of a radial flow reaction turbine is the lawn sprinkler as shown in Figure 2.13. Water enters this rotor at very high pressure and the pressure energy is transformed into a velocity energy in a nozzle, which is part of the rotor itself.

Figure 2.13

Radial flow reaction turbine and velocity diagram.

Utilization factor Radial entry

\ a1 = 90°, Vw1 = 0, fluid enters with zero radius. \ Vr1 and u1 = 0 \ V2 = Vw2

Water leaves tangentially,

u2 Vw 2 gc

W.D. V22 2 gc

u2V2 gc

(Vr 2  u2 )2 2 gc

u2 (Vr 2  u2 ) Û Ü gc Ü Ü Ü ÜÝ

(2.44)

Substituting the above relations (2.44) in utilization Eq. (2.39b),

H

u2 (Vr 2  u2 )

W.D. W.D. 

V22 2 gc

Ë u (V  u2 ) (Vr 2  u2 )2 Û gc Ì 2 r 2  Ü gc 2 gc Í Ý

u2 (Vr 2  u2 ) u2 (Vr 2  u2 ) 

(Vr 2  u2 ) 2

2u2 2u2  (Vr 2  u2 )

u2 2

2u2 u2  Vr 2

u2 (Vr 2  u2 ) (Vr 2  u2 )2  (Vr 2  u2 ) (Vr 2  u2 ) – 2 2 V 1  r2 u2

(2.45)

Energy Transfer in Turbomachines

93

Degree of reaction for radial flow turbine Substituting Eq. (2.44) in degree of reaction Eq. (2.13), R

[(u12  u22 )  (Vr22  Vr21 )] 2 gc – W.D.

( u22  Vr22 ) Ë u2 (Vr 2  u2 ) Û Ì 2 gc – Ü gc Í Ý \

R

(Vr 2  u2 )(Vr 2  u2 ) 2u2 (Vr 2  u2 )

(Vr22  u22 ) 2u2 (Vr 2  u2 ) Vr 2  u2 2u2

(2.45a)

Relation between utilization factor and degree of reaction Comparing Eqs. (2.45) and (2.45a), we have R

1 H

(2.45b)

2.10 CONDITION FOR MAXIMUM UTILIZATION: AXIAL TURBINE 2.10.1 Reaction Turbine In the expression of utilization factor, the discharge kinetic energy = V22 / 2 gc is not utilized by the rotor. For the maximum utilization, V2 must be minimum, i.e. the maximum possible portion of the fluid energy at the rotor inlet must be utilized for conversion into work. From the velocity triangle it is apparent that V2 is minimum when it is axial. In this case V2 = V1 sin a1. The combined velocity diagram for 50% reaction is shown in Figure 2.14 for maximum utilization or V2 minimum.

Figure 2.14

Combined velocity diagram for 50% reaction and for maximum utilization condition.

94

Fundamentals of Turbomachinery

Substituting V2 = V1 sin a1 in Eq. (2.39e), H max

V12  V12 sin 2D1

V12 (1  sin 2D1 )

V12  RV12 sin 2D1

V12 (1  R sin 2D1 )

H max

\

cos2D1

(2.46)

1  R sin 2D1

Substituting Eq. (2.39d) in Eq. (2.7),

Ë R(V12  V22 ) Û 1 W.D. (V12  V22 )  Ì Ü– Í (1  R) Ý 2 gc (V12  V22 ) (1  R)  R(V12  V22 ) 2(1  R) gc

or

W.D.

\

u1

V12  V22 2(1  R) gc

(V12 cos2D1  V f21 )  (V22 cos2D 2  V f22 ) 2(1  R) gc V12 cos2D1  (V f21  V f22 )

u1Vw1  u2Vw 2 gc

[Q Vw2 = 0, a2 = 90°, Vw1 = V1 cos a1]

2(1  R) V1 cos D1

Let f, known as blade speed ratio (which influences turbine theory, design and efficiency), be defined as

I

u1 V1

V12 cos2D1  (V f21  V f22 ) V12 2(1  R) cos D1

V f22

cos2D1  sin 2D1  or

I

cos2D1 

V12

2(1  R) cos D1 V12 sin 2D1 V12 2(1  R) cos D1 1

1

V f21

V f22

V12 V12 2(1  R) cos D1

V f22

V12 2(1  R) cos D1

cos2D1 2(1  R) cos D1



cos D1 2(1  R)

(2.46a)

(2.46b)

From Eq. (2.46a), we get

V12  V f22 or

V f22

V12 2 I (1  R) cos D1 V12 [1  2I (1  R) cos D1 ]

(2.46c)

Substituting the condition of maximum utilization, i.e. V2 = Vf 2 in Eq. (2.39e), we get H max

V12  V f22 V12  RV f22

Energy Transfer in Turbomachines

or

H max V12  H max RV f22

or

H max V12  V12

or

V12 (H max  1) V f22 (H max R  1); ? V f22

95

V12  V f22 H max RV f22  V f22

V f22 (H max R  1) V12 (H max  1) (H max R  1)

(2.46d)

Equating Eqs. (2.46c) and (2.46d), we get (H max  1) V12 (H max R  1)

or

(H max  1)

or

H max (1  R)

H max R  2I (1  R) cos D1 H max R  1  2I (1  R) cos D1  2I (1  R) cos D1 (H max R  1)

emax = –2f cos a1emax R + 2f cos a1

or or

V12 [1  2I (1  R) cos D1 ]

H max (1  2I cos D1 R)

2I cos D1 ; ? H max

2I cos D1 1  2 RI cos D1

(2.46e)

Substituting Eq. (2.46b) in Eq. (2.46e) Hmax

2 cos2D1 1 – 2(1  R) [1  2 R cos D1 cos D1 / 2(1  R)] cos2D1 1 – (1  R) [(1  R)  R cos2D1 ] /(1  R) cos2D1

cos2D1

cos2D1

1  R  R cos2D1

1  R(1  cos2D1 )

1  R sin 2D1

(2.46f)

A 50% reaction turbine is most widely used. For R = 50%, the diagram is symmetrical with V2 axial for maximum utilization and it is shown in Figure 2.14. \ Substituting R = 50% in Eq. (2.46), we get H max

cos2D1

(2.46g)

1  0.5 sin 2D1 2 cos2D1

2 cos2D1

2 cos2D1

2  sin 2D1

1  1  sin 2D1

1  cos2D1

(2.46h)

Equations (2.43h) and (2.46h) are same.

2.10.2 Impulse Turbine Utilization factor is absolute maximum and equal to unity when a1 = 0°. Then V2 = V1 sin a1 = 0 results in zero angle turbine. This represents the ideal turbine, impossible to attain, because although a1 can be zero, but a finite velocity, V2, with an axial component is necessary to

96

Fundamentals of Turbomachinery

provide steady flow. However, this shows that the nozzle angle should be as small as possible. The velocity triangle for zero reaction is shown in Figure 2.15. The combined velocity diagram for maximum utilization condition is shown in Figure 2.16. Substituting u1 = u2 = u (axial turbine) and R = 0 in Eq. (2.13), we have Vr1 = Vr2 Substituting the above conditions in Eqs. (2.39a), (2.39c) and (2.43), H

(V12  V22 ) / V12

2u(Vw1  Vw 2 ) / V12

(2.47)

Substituting R = 0, V2 = axial (condition for maximum utilization) = V1 sin a1, a2 = 90°, Vw2 = 0, V2 = Vf 2 in Eq. (2.43),

Figure 2.15

Figure 2.16

Combined velocity diagram for maximum utilization condition for impulse turbine.

H max

\

Velocity triangle for zero reaction.

V12  V12 sin 2D1

V12 (1  sin 2D1 )

V12  R sin 2D1

V12 (1  R sin 2D1 )

cos2D1

(2.48)

Substituting Eq. (2.51) in Eq. (2.46e), we get Hmax

Here,

2 cos D1 cos D1 / 2 10

cos2D1

(Q impulse stage, R = 0)

V2 = Vf2 = Vf1, Vf1 = Vr2; \ AB = BE = u

(2.49)

Energy Transfer in Turbomachines

97

2.11 OPTIMUM BLADE SPEED (fOPT) RATIO FOR DIFFERENT TYPES OF TURBINES FOR MAXIMUM ENERGY TRANSFER (W.D.)max 2.11.1 Reaction Turbine For maximum utilization, Vw1 V1

cos D1

u1 V1

I

Iopt

(Refer to Figure 2.14)

(2.50)

u1Vw1  u1Vw 2 gc

W.D.

u2 gc

u1V1I gc

u1Vw1 gc

(W.D.) max

(2.51)

2.11.2 Impulse Turbine For maximum utilization, triangles ABD and BEC of Figure 2.16 are same. Therefore, AB = BE = u. cos D1

Vw1 V1

AB  BE V1

I

Iopt

u V1

(W.D.) max tan E1

u1Vw1 gc Vf 1 Vw1  u

uu V1

2u V1

cos D1 2

u1V1 cos D1 gc

(2.52) 2u12 gc

u1 2u1 gc

V1 sin D1 (V1 cos D1  u)

(2.52a)

1 V1 cos D1 u  V1 sin D1 V1 sin D1

Substituting Eq. (2.52) in the above relation for optimum condition,

tan E1

or

1

1

cos D1 cot D1  2 sin D1

cot D1 cot D1  2

tan b1 = 2 tan a1

1 cot D1 2

2 cot D1 (2.53)

2.12 EXAMPLES EXAMPLE 2.1 A turbine has the following data. Water is directed at an angle of 30° to the tangent. Degree of reaction is 0.45, utilization factor is 0.895. The absolute velocity at exit is axial, water enters the rotor with a static pressure 500 kPa and stagnation pressure of

98

Fundamentals of Turbomachinery

750 kPa. Calculate (a) the inlet blade angle and (b) the work output for a mass flow rate of 10 kg/s. Solution: Device: Turbine Fluid: Water Nozzle angle: a1 = 30° Utilization factor: e = 0.895 Degree of reaction: R = 0.45 Absolute velocity at exit is axial = V2 = Vf 2, Vw2 = 0, a2 = 90° Water enters the blade with static pressure: p1 = 500 kPa Inlet stagnation pressure: p01 = 750 kPa To determine b1 and power output: r = constant

We have

p01

\

V12 U 2 gc – 1000

p1 

V12 U 2 gc

p01  p1

750  500 kg -m

1000 m 3 1000 kg

\

V12

250 kN

\

V12

500 m 2 /s2 ; V1 = 22.36 m/s

m2

–2–

kN -s2

–

250 kPa

From Eq. (2.43), H

\

V12  V22 V12  RV22

0.895

500  V22 500  0.45V22

V2 = 9.4 m/s Vr22

V22  u2

We can write (u12  u22 )  (Vr21  Vr22 )

(u12  Vr21 )  V22

Energy Transfer in Turbomachines

99

Equation (2.13) can be written as (u12  Vr21 )  V22

R[(V12  V22 )  (u12  u22 )  (Vr22  Vr21 )] R [(V12  V22 )  u12  Vr21  V22 ]

(u12  Vr21 )

\

( RV12  V22 )  R (u12  Vr21 )

(u12  Vr21 )  R(u12  Vr21 )

( RV12  V22 )

(u12  Vr21 ) (1  R)

( RV12  V22 ) RV12  V22 1 R

u12  Vr21

0.45 – 500  9.42 1  0.45

248.44

From inlet velocity triangle,

Vw21

V12  V f21

Vw21

V12  Vr21  Vw21  u12  2u1Vw1

V12  u12  Vr21 u1

\

V12  [Vr12  (Vw1  u1 )2 ]

2u1 V1 cos D1

2u1 Vw1

V12  (u12  Vr21 ) 2V1 cos D1

500  248.44 2 – 22.36 – cos 30’

19.33 m/s

(a) Inlet blade angle (b1):

sin D1 \

Vf 1 V1

Vf 1 = V1 sin a1 = 22.36 × sin 30° = 11.18 m/s Vw1 = V1 cos a1 = 22.36 × cos 30° = 19.36 m/s

tan E1 \

Vf 1

11.18 19.36  19.33

Vw1  u1

372.66

b1 = 89.85°

Ans.

(b) Power output (P): P

mu1Vw1 gc

10

kg s

– 19.33

= 3742.288 W or

m m N s2 – 19.36 – s s kg m

3.742 kW

Ans.

100

Fundamentals of Turbomachinery

EXAMPLE 2.2 The following data refers to an axial flow device. Flow velocity from exit of the nozzle: Degree of reaction: Blade speed: Entry stagnation temperature: Entry static temperature:

Vf1 = 190 m/s R = 50% u1 = u2 = u = 180 m/s T01 = 380 K T1 = 300 K

Calculate (a) the maximum utilization factor, (b) the rotor blade angle at inlet and exit, and (c) the power, if the mass flow rate is 10 kg/s. Solution:

a2 = 90°,

Vw2 = 0,

V2 = Vf 2 (Q Exit is axial)

We have

T01 V12 2 gc c p \

V1

T1 

V12 2 gc c p

T01  T1

(380  300) 80 K

80 – 2 – 1000 – 1.005

400.99 m/s

(a) Maximum utilization factor (emax): We have from Eq. (2.46f), H max

\

cos2D1

cos2 28.28’

1  R sin 2D1

1  0.5 sin 2 28.28’

emax = 0.8736

From inlet triangle, Vw1 = V1 cos a1 = 400.99 × cos 28.28° = 353.13 m/s

Ans.

Energy Transfer in Turbomachines

101

(b) Inlet and exit blade angle (b1, b2):

Vf 1

tan E1

190 353.13  180

Vw1  u

\ b1 = 47.66° From exit velocity triangle, u 180 cos E2 Vr 2 V1 \

Ans. 80 400.99

0.449

b2 = 63.33°

Ans.

(c) Power (P):

P

muVw1 gc

10

kg s

– 180

= 635634 W or

m m N- s2 – 353.13 s s kg-m

635.634 kW

Ans.

EXAMPLE 2.3 The discharge blade angles are 20° each for both the stator and rotor. The steam exit from the fixed blade is 150 m/s. The mass flow rate of the steam is 3.5 kg/s. The ratio of Vf /u at the exit is 0.75. Calculate: (a) Inlet rotor blade angle (b) The power developed (c) The degree of reaction (d) Utilization factor for optimum speed ratio. Solution: Type of turbine: Discharge blade angle of stator: Discharge blade angle of rotor: Steam exit speed from the fixed blade: Mass flow rate of steam: The ratio of Vf /u at exit: Assume optimum speed ratio

Axial a1 = b2 = V1 = m= =

reaction 20° 20° 150 m/s 3.5 kg/s 0.75

Iopt

u / V1

cos D1

To determine, b1, power (P), R and e : We have from Eq. (2.51), Iopt

\

u / V1

cos D1

cos 20’

0.9396

u = V1 × 0.9396 = 150 × 0.9396 = 140.95 m/s Vf1 = V1 sin a1 = 150 × sin 20° = 51.3 m/s

Vr21

V f21  (Vw1  u)2

V f21  (V1 cos D1  u)2

= 51.3 + (150 × cos 20° – 140.95)2

102

Fundamentals of Turbomachinery

Vr1 = 51.3 m/s Vw1 = V1 cos a1 = 150 × cos 20° = 140.95 m/s \

Vw1 = u1 = u2 = u

(axial steam turbine, u = u1 = u2)

(a) Inlet rotor blade angle (b1):

sin E1

Vf 1 Vr1

51.3 1 51.3

\ b1 = 90°

\ The combined velocity diagram changes to

(b) Power (P): Given or

Vf 2

\ Vf 2 = 0.75u 0.75 u Vf2 = 0.75 × 140.95 = 105.71 m/s

From exit velocity triangle,

sin E2 \

Vr 2

Vf 2 Vr 2 Vf 2 sin E 2

105.71 sin 20’

309.1 m/s

Ans.

Energy Transfer in Turbomachines

103

From exit velocity triangle, Vw2 = Vr2 cos b2 – u = 309.1 cos 20° – 140.95 = 149.95 V f22  Vw22

V2

105.712  149.512

183.11 m/s

mu (Vw1  Vw 2 ) gc

P

3.5

kg s

– 140.95

P = 143291.2 W

m m N- s2 – (140.95  149.51) – s s kg -m

or 143.29 kW

Ans.

(c) The degree of reaction (R): Vr22  Vr21 2 gc – W.D.

R

Vr22  Vr21 2 – gc – u(Vw1  Vw 2 )

309.12  51.32 2 – 1 – 140.95 – (140.95  149.51)

(Eq. (2.40)) 1.1347

Ans.

(d) Utilization factor (e): From Eq. (2.43), H

(V12  V22 )

150 2  183.112

(V12  RV22 )

150 2  1.1347 – 183.112

0.709

Ans.

EXAMPLE 2.4 An axial flow turbine stage has a flow coefficient of 0.7, a constant axial velocity and the gas leaves the stator blades at an angle of 65° to the axial direction. Calculate (a) the rotor inlet and exit angles and (b) the degree of reaction. Solution: Flow coefficient

Vf /u = 0.7

Constant axial velocity

Vf 1 = Vf 2 = Vf = constant

Gas inlet angle

a1 = 90° – 65° = 25°

[The given data (65°) is with respect to axial direction, here it is considered with respect to tangential direction.] To determine b1, b2, R: Assume that fluid leaves axially, i.e. a2 = 90°, Vw2 = 0; V2 = Vf 2

104

Fundamentals of Turbomachinery

(a) Rotor blade angles (b1, b2): tan E2

Vf 2

0.7 u b2 = 34.99° » 35°

Ans.

From inlet and exit velocity triangles, W.D. =

W.D. = V f – We have

Now, \

(Vw1 (Vw1

Vw1 Vw2 + Vw2) + Vw2)

gc

u (cot D1  cot D 2 ) gc

Vw1  u ; cot E2 Vf

cot E1

\ \

u(V f 1 cot D1  V f 2 cot D 2 )

u(Vw1  Vw 2 ) gc

= = = =

(Vw1  Vw 2 ) W.D. =

Vf Vf Vf Vf

(i)

Vw 2  u Vf

cot b1 + u cot b2 – u cot b1 + u + Vf cot b2 – u (cot b1 + cot b2)

W.D. – gc u uV f (cot E1  cot E2 ) gc

(ii)

Energy Transfer in Turbomachines

Equating Eqs. (i) and (ii), uVf (cot a1 + cot a2) = uVf (cot b1 + cot b2) \ cot 25° = cot b1 + cot 35° cot b1 = cot 25° – cot 35° = 0.7163 b1 = 54.38° (b) The degree of reaction (R): Vf (Eq. (2.36)) R 1 (cot D 2  cot D1 ) 2u 0.7 1 (cot 90’  cot 25’) 0.249 2 Alternatively: Vf (Eq. (2.43l)) R 0.5  (cot E 2  cot D1 ) 2u 0.7 0.5  (cot 35’  cot 25’) 0.2493 2

105

Ans.

Ans.

EXAMPLE 2.5 The following data refers to a zero reaction turbine. Blade speed = 300 m/s, gas leaves the nozzle ring at an angle of 65° to the axial direction. Static temperature at inlet is 950 K. Assume, cp = 1.2 kJ/kg-K, ideal exit stagnation temperature = 960 K. Calculate the total-to-total efficiency based on actual work developed. Assume that the stage inlet stagnation temperature is 1200 K. Solution: Zero reaction turbine:

R=0

Blade speed:

u = 300 m/s a1 = 90° – 65° = 25°

Fluid leaves the nozzle:

[Given: 65° to the axial direction. Refer to the figure of Example 2.5] Inlet stagnation temperature:

T01 = 1200 K

Ideal exit stagnation temperature:

T02

„

960 K

cp = 1.2 kJ/kg-K Static temperature at inlet:

T1 = 950 K

To determine: ht(t–t): For zero reaction, there is no temperature drop (enthalpy drop) through the rotor. \

h1 = h2;

T2 = T1 = T2¢;

Va = Vf ;

From Eq. (2.33),

R

Va (cot E1  cot E2 ) 2u

f = Vf /u

106

Fundamentals of Turbomachinery

I (cot E1  cot E 2 ) 2

R

I (cot E1  cot E2 ) 2 b1 = b2 0

\ We have

T1 

T01

V12 2 gc c p

\

V12 2 gc c p

\

V12

\

V1 = 774.596 m/s

T01  T1

1200  950

250 K

2 – 250 – 1 – 1000 – 1.2

From inlet velocity triangle, Vw1 = V1 cos a1 = 774.596 × cos 25° = 702.02 m/s Vf 1

V12  Vw21

774.62  702.022

327.4 m/s

(Vw1 – u) = 702.02 – 300 = 402.02 m/s \

tan E1

Vf 1 Vw1  u

327.4 402.02

0.814

\

b1 = 39.16° = b2 (zero reaction)

Assume

Vf 1 = Vf 2 = Vf = constant throughout the rotor.

From the exit velocity triangle,

tan E2

Vf 2

Vf

u  Vw 2

u  Vw 2

327.4 300  Vw 2

tan 39.16’

Energy Transfer in Turbomachines

\

107

Vw2 = 102.00 m/s V f 2  Vw22

V2

327.4 2  102.0 2

342.9 m/s

Total-to-total efficiency based on actual work developed [ht(t–t)]: Wa

(h01  h02 ) „

u(Vw1  Vw 2 ) 300 – (702.02  102.00) gc gc = 241.026 J/kg

c p (T01  T02 ) „

1.2 Kt (t t )

c p (1200  960)

kJ – 240 K kg-K

Wa h01  h02„

288 kJ/kg

241.206 – 100 288.00

83.75%

Ans.

EXAMPLE 2.6 An axial flow gas turbine has axial flow velocity of 250 m/s which is constant. The mass flow rate of gas is 15 kg/s. Blade speed is 350 m/s. The nozzle inlet angle is 30° and the gas exit angle is 75°. Calculate: (a) The blade angles (b) The degree of reaction (c) The blade loading coefficient (d) The power output. Solution: Data: Axial flow gas turbine: Axial velocity (constant): Mass flow rate of gas: Blade speed: Nozzle inlet angle: Gas exit angle:

u1 = u2 = u Vf 1 = Vf 2 = Vf = 250 m/s m = 15 kg/s u = 350 m/s a1 = 30° a2 = 75°

(a) The rotor blade angles (b1, b2): From obtained velocity triangles, Vw1 = Vf tan a1 = 250 × tan 60° = 433.0 m/s Vw2 = Vf tan a2 = 250 × tan 15° = 66.99 m/s

tan E1 \

Vw1  u

250 433.0  350

3.01

b1 = 71.6°

tan E 2 \

Vf 1

Vf u  Vw 2

b2 = 30.94°

Ans.

250 350  66.99

0.5995 Ans.

108

Fundamentals of Turbomachinery

(b) The degree of reaction (R): From Eq. (2.43j) or (2.43l) or (2.43m), Vf R (cot E 2  cot E1 ) 2u 250 (cot 30.94’  cot 71.6’) 2 – 350

0.478

(c) The blade loading coefficient (y): \

W.D.

Vf

u2

u

(cot E1  cot E2 )

(For derivation refer to Example 2.4)

250 (cot 71.6’  cot 30.94’) 1.43 350 (d) Power output (P):

P

mu(Vw1  Vw 2 ) gc

15

kg s

– 350

Ans.

m (433.0  66.99) m – s s kg -m N- s2

= 2624947 N-m/s = 2624.947 kW

Ans.

EXAMPLE 2.7 A single-stage steam turbine has a diameter of 100 cm, and speed of 2500 rpm. The flow rate of steam is 75 kg/s. The steam nozzle angle is 30°. Calculate (a) the rotor blade angles, (b) blade heights, (c) power developed, (d) and the isentropic enthalpy drop in the stage. Assume maximum utilization factor and stage efficiency equal to 0.8. Solution: Steam turbine: Blade diameter: Speed: Mass flow rate of steam: Nozzle angle: Stage efficiency: cp of steam:

Single stage d = 100 cm N = 2500 rpm m = 75 kg/s a1 = 30° ht = 0.8 = 0.6 kJ/kg-K

Energy Transfer in Turbomachines

109

To determine: b1, b2, b, power, Dhs Assume, equiangular blades, i.e. b1 = b2 emax = cos2 a1

(Eq. (2.49)),

Vf1 = Vf 2 = Vf = Va

a2 = 90°, Vw2 = 0

For maximum utilization: From Eq. (2.50),

\

cos D1 2

u V1

Iopt

u

S dN 60

V1

u 0.433

cos 30’ 2

S – 1 – 2500 60 130.9 0.433

0.433

130.9 m/s

302.31 m/s

Vf 1 = V1 sin a1 = 302.31 × sin 30° = 151.2 m/s (a) The rotor blade angles (b1, b2): From Eq. (2.53), we have tan b1 = 2 tan a1 = 2 tan 30° = 1.1547 \

b1 = b2 = 49.11°

Ans.

Alternatively: Vw1

tan E1 \

V12  V f21

302.312  151.2 2

Vf 1

Vf 1

Vw1  u

u

261.82 m/s

151.2 130.9

b1 = 49.11°

Ans. Vw1

T or h u

1 a2 Wa

V2 = Vf 2

Dhs

2 2¢ s

a1

u b2

b1 Vr 1

Vr 2

V1

Vf 1

110

Fundamentals of Turbomachinery

(b) Blade height (b): Q = p d bVf m = p d bVf r \

75 kg

m S d Vf U

b

s – S – 1 m – 151.2

kg m – 1000 s m3

= 0.015789 cm

Ans.

(c) Power developed (P): P

For optimum speed ratio, P

m u (Vw1  Vw 2 ) gc

Vw1 = 2u mu2u gc

mu2 – 2 gc

kg m 2 N-s2 – 2 – 130.92 2 – s kg-m s = 2570221.5 W 75

Ans.

(d) Isentropic enthalpy drop in the stage (Dhs): Kt

Ks

(h1  h2 ) ( h1  h2 )

Wa 'hs

„

'hs

u(Vw1  Vw 2 ) gc – 0.8

u 2u gc – 0.8

0.8

2 – u2 0.8gc

2(130.9)2 m 2 N-s2 – 0.8 – 1 s2 kg-m 42837.03 J/kg or 42.837 kJ/kg

EXAMPLE 2.8

Repeat Example 2.7 considering a two-stage Curtis steam turbine.

Solution: T or h 1 2a

2¢

2b

2 s

Ans.

Energy Transfer in Turbomachines

fopt = optimum speed ratio for two stage = u V1

\ Assume \

cos D1 4

cos 30’ 4

111

cos D1 4

0.2165

u 130.9 604.6 m/s 0.2165 0.2165 Vf 1 = Vf 2 = Vf 3 = constant Vf = V1 sin a1 = 604.6 × sin 30° = 302.3 m/s V1

(a) The rotor blade angles (b1, b2): Vw1 = V1 cos a1 = 604.6 × cos 30° = 523.6 m/s x = Vw1 – u = 523.6 – 130.9 = 392.7 m/s = 3 × u = 3 × 130.9 = 392.7 m/s For optimum speed ratio and for two stage, x = 3u and Vw2 = 2u tan E1

tan E2

Vf

3u b1 = b2 = 36.6°

\

302.3 392.7

0.7698

Ans.

(b) Stator angles (a2, a3): Vr1

V f21  (3u)2

302.32  392.72

495.6 m/s

112

Fundamentals of Turbomachinery

Assume

Vr1 = Vr2 cos E2

\

u  Vw 2 Vr 2

u  Vw 2 Vr1

130.9  Vw 2 495.6

cos 37.6’

Vw2 = 261.76 m/s = 2u

tan D 2

Vf 2

Vf 1

Vw 2

Vw 2

302.3 261.76

1.155

a2 = 49.11° = a3

Ans.

a4 = 90° (axial discharge) (c) 2nd stage rotor blade angles (b3, b4): b3 = b4

Assume \

AB = BC = u tan E3

Vf 3

Vf

u

u

302.3 130.9

2.309

b3 = b4 = 66.59°

Ans.

(d) Power output (P): W1 = u(Vw1 + Vw2) = 6u × u = 6u2 = 6 × (130.9)2 = 102808.86 J/kg W2 = u(V3 + V4) = 2u × u = 2u2 = 2 × (130.9)2 = 34269.62 J/kg P

m(W1  W2 ) gc

75 – (102808.86  34269.62) 1000

10280.886 kW

Ans.

(e) Isentropic enthalpy drop (Dhs): 'hs

W1  W2 0.8

102808.86  34269.62 1000 – 0.8

171.35 kJ/kg

Ans.

EXAMPLE 2.9 Modify Example 2.7 in such a way that the turbine is a two-stage Rateau wheel. Solution: Each stage has the same geometry. Therefore, the two-stage Rateau wheel is nothing but TWO single-stage turbines. The results of Example 2.7 can be used here directly. (a) Blade angles (b1 = b2): b1 = b2 = 49.110°

Ans.

b = 0.015789 cm

Ans.

(b) Blade height (b):

Energy Transfer in Turbomachines

113

(c) Power developed (P): m u (W1  W2 ) gc

P

m u (2u  2u) gc

m 4u 2 gc

75 – 4 – (130.9)2 1 – 1000

5140.443 kW

Ans.

EXAMPLE 2.10 The internal and external diameters of the impeller of a centrifugal pump are 20 cm and 40 cm respectively. The pump is running at 1200 rpm. The vane angle of impeller at inlet is 20°. The water enters the impeller radially and the velocity of flow is constant. Calculate the work done by the impeller per kg of water for the following three cases. (a) b2 = 30°, (b) b2 = 90°, (c) b2 = 100° Solution:

Data: Internal diameter:

d1 = 0.2 m

External diameter:

d2 = 0.4 m

Speed:

N = 1200 rpm

Vane angle at inlet:

b1 = 20°

Water enters radially:

a1 = 90°, Vw1 = 0, V1 = Vf1

Flow is constant:

Vf 1 = Vf 2 = Vf

To determine: W.D. for the three cases

S d1 N 60

S – 0.2 – 1200 60

12.56 m/s

S d2 N 60 From inlet velocity triangle,

S – 0.4 – 1200 60

25.13 m/s

u1 u2

tan E1 \

Vf 1

Vf 1

u1

12.56

Vf 1 = Vf 2 = Vf = 4.57 m/s

114

Fundamentals of Turbomachinery

tan E2 tan 30’

Vf 2

u2

(u2  Vw 2 )

Vw 2

25.13 m/s

tan (180’  100’)

4.57 (25.13  Vw 2 )

\

Vw2 = 17.215 m/s

\

W.D.

Vw 2u2 gc

tan 80’

\ W.D.

= 17.215 × 25.13 = 432.6 J/kg

25.13 – 25.13 gc

(Vw 2  u2 ) 4.57 (Vw 2  25.13)

Vw2 = 25.94 m/s W.D.

= 631.52 J/kg

Vf 2

25.94 – 25.13 gc

= 651.87 J/kg

EXAMPLE 2.11 The mean rotor blade speed of an axial flow turbine with 50% reaction is 210 m/s. Steam emerges from the nozzle inclined at 28° to the plane of the wheel with axial component equal to blade speed. Assuming symmetric inlet and outlet velocity triangles, find (a) the rotor blade angles and (b) utilization factor. Find also the (c) degree of reaction to make the utilization factor maximum, if the axial velocity, blade speed as well as nozzle angle remain the same. Solution: Axial steam turbine: Mean rotor speed: u = 210 m/s Degree of reaction: R = 50% Nozzle angle (inlet): a1 = 28° Axial component = Blade speed: Vf1 = Vf2 = u Symmetric inlet and outlet triangles (if utilization is maximum) a1 = b2, To determine: Case I Case II

b1, b2, e R if emax

a2 = b1,

V1 = Vr2,

V2 = Vr1

Energy Transfer in Turbomachines

115

Case I: 50% reaction

From inlet velocity triangle,

Vf 1

sin D1 \

V1 Vf 1

V1

u sin D1

sin D1

210 sin 28’

447.31 m/s

Vw1 = V1 cos a1 = 447.31 × cos 28° = 394.95 m/s (a) Rotor blade angle (b1):

Vf 1

tan E1 \

Vw1  u

210 394.95  210

1.13544

b1 = 48.63°

Ans.

(b) Utilization factor (e):

Vr1

Vf 1 sin E1

u sin E1

210 sin 48.63’

279.8 m/s V2

From Eq. (2.43), H

Case II:

V12  V22

447.312  279.82

V12  RV22

447.312  0.5 – 279.82

0.7568

Degree of reaction if utilization is maximum

For maximum utilization a2 = 90°, i.e. fluid exits axially, Vw2 = 0, V2 = Vf 2 = Vf1 = u = 210 m/s From exit velocity triangle, Vr 2

V22  u2

Vr1 = 279.8 m/s

210 2  210 2

296.98 m/s

(From Case I)

Ans.

116

Fundamentals of Turbomachinery

V2 = 210 m/s \

R

(Vr22  Vr21 ) (V12  V22 )  (Vr22  Vr21 ) [296.982  279.82 ] (447.32  210 2 )  (296.982  279.82 )

0.0597

Ans.

R can be calculated from Eqs. (2.43j) or (2.43l) or (2.43m). EXAMPLE 2.12 Solution: a1 = 90°

Derive the degree of reaction equation for radial flow machines.

Radial flow machines are always of centrifugal type and generally Vw1 = 0, i.e. W.D.

Refer to Figure 2.5(c). From Eq. (2.34),

R

\

u2Vw 2 gc

u2 (u2  V f 2 cot E 2 ) J/kg gc

u2Vw 2 V22  V12  2 gc gc u2Vw 2 gc

R 1

1 Vw22 2 u2Vw 2

u2Vw 2 

(V f22  Vw22  V12 ) 2 u2Vw 2

1

1 Vw 2 2 u2

Energy Transfer in Turbomachines

117

We know that, Vf1 = Vf2 = Vf R 1

1 1 2

1 2

V f 2 cot E2 Ø È ÉÊ 1  ÙÚ u 2

1 1 V f 2 cot E2  2 2 u2

V f 2 cot E 2 Ø È ÉÊ 1  ÙÚ u

Ans.

2

From the above equation, the following conclusions can be made: 1. If b2 = 90°, R = 0.5, means that the static pressure rise inside the rotor is equal to the change in absolute kinetic energy. 2. If b2 is between 0 and 90°; then cot b2 is positive and hence the degree of reaction is greater than 0.5. 3. If b2 is between 90° and 180°; then cot b2 is positive and hence the degree of reaction is less than 0.5. 4. When R = 0, there is no change of static pressure possible inside the machine, i.e. the machine is having impulse blades. EXAMPLE 2.13 A fluid flows through one stage of a turbomachine. The velocity diagram is as shown in the following figure. (a) Is this a power generating or power absorbing turbomachine? (b) What is the change in total enthalpy across the stage? (c) What is the degree of reaction?

Solution: V1

200 sin 45’

282.8 m/s

V2

200 sin 60’

230.94 m/s

118

Fundamentals of Turbomachinery

u = V1 cos 45° + Vr1 cos 60° = V1 cos 45° + V2 cos 60° = 315.45 m/s Vw1 = AC = V1 cos 45° = 282.8 cos 45° = 199.96 m/s Vw2 = BC = V2 cos 60° = 230.94 cos 60° = 115.45 m/s u(Vw1  Vw 2 ) gc

W.D.

315.43 (199.96  115.45) 1

26656 J/kg

(a) Since W.D. is +ve, the turbomachine is a power generating machine. (b) Enthalpy drop across the stage = W.D. = 26.6 kJ/kg (c) Since V1 = Vr2, V2 = Vr1, a1 = b2, a2 = b1, the velocity triangles are symmetrical. \ The machine is a 50% reaction machine.

Ans.

EXAMPLE 2.14 In a certain machine the fluid enters the rotor with an absolute velocity having an axial component of 10 m/s and tangential component in the direction of rotor motion is equal to 15 m/s. The tangential speed at the exit is 7.5 m/s. The absolute velocity of fluid at exit is 16 m/s in the axial direction. The rotor speed at inlet is 35 m/s. Calculate: (a) The energy transfer between the fluid and the rotor (b) The type of machine. Solution:

Data:

Axial component of flow velocity: Vf1 = 10 m/s Tangential component of flow velocity: Vw1 = 15 m/s Speed of the rotor at inlet: u1 = 35 m/s Rotor speed at exit: u2 = 7.5 m/s Absolute velocity at exit: V2 = 16 m/s Absolute velocity at exit is in the axial direction: a2 = 90° \ Vw2 = 0 To determine: W.D. type of machine (a) Energy transfer (W.D.): W.D. =

(u1Vw1  u2Vw 2 ) gc

35 – 15  7.5 – 0 1

525 J/kg

Ans.

Energy Transfer in Turbomachines

119

(b) Type of machine: Since W.D. is positive, the machine is a power generating machine.

Ans.

EXAMPLE 2.15 Fluid enters a rotor in axial direction at an absolute velocity of 10 m/s. The rotational speed of the rotor is 20 m/s. Calculate (a) the relative velocity and (b) its direction. Solution: Data: Axial direction of fluid entry: Absolute velocity at inlet: Rotational speed of the rotor:

a1 = 90°, Vw1 = 0, V1 = 10 m/s V1 = 20 m/s

V1 = Vf1

To determine: Vr1, b1 (a) Relative velocity (Vr1): V12  u12

Vr1

10 2  20 2

22.36 m/s

Ans.

(b) Direction (b1): tan E1

V1 u1

10 20

0.5; ? E1

26.57’

Ans.

EXAMPLE 2.16 In an inward flow radial turbine, water enters at an angle of 68° to the axial direction and leaves the turbine radially. The rotor speed is 300 rpm. The flow velocity is 3 m/s and remains constant. The inner and exit diameters of the turbine are 0.3 and 0.6 m respectively. The width of the wheel at the inlet is 15 cm. Calculate (a) the rotor blade angles and (b) the power developed. Neglect the thickness of the blades. Solution:

Data:

Machine:

Radial flow turbine

Outer diameter: Inner diameter: Width at inlet: Speed of the rotor: Water enters at an angle:

d2 = 0.3 m d1 = 0.6 m b1 = 15 cm N = 300 rpm a1 = 90° – 68° = 22°

[Given data: 68° to the axial direction. Here, a1 is considered with respect to tangential direction.] Flow velocity or radial velocity

Vf1 = Vf 2 = Vf = 3 m/s

Water leaves radially

a2 = 90°, Vw2 = 0, V2 = Vf 2

To determine: b1, b2, and the power developed

u1

S d1 N 60

S – 0.6 – 300 60

9.42 m/s

120

Fundamentals of Turbomachinery

u2 V1

S d2 N 60

S – 0.3 – 300 60

Vf 1

3 sin 22’

sin 22’

4.71 m/s

8 m/s

Vw1 = V1 cos 22° = 8 × cos 22° = 7.42 m/s

(a) The rotor blade angles (b2, b2):

Vf 1

tan E1 \

u1  Vw1

3 9.42  7.42

1.5

b1 = 56.31°

Vf 2

tan E2 \

3 4.71

u2

Ans.

0.6369

b2 = 32.495°

Ans.

(b) Power developed (P): m = mass flow rate = ra1Vf 1 = rpd1b1Vf 1

S – 1000 P

kg m3

– 0.6 m – 0.15 m – 3

m s

848.2 kg/s

m (Vw1u1  Vw 2 u2 ) gc 848.2

kg s

P = 59280 W

– (9.42 – 7.42)

m2

s2 or 59.28 kW

–

N- s2 1 kg - m

EXAMPLE 2.17 The following data refers to an inward flow turbine: Inlet absolute velocity: Speed of the runner:

V1 = 35 m/s N = 300 rpm

(Q Vw2 = 0) Ans.

Energy Transfer in Turbomachines

Inlet diameter: Outlet diameter:

121

d1 = 1.8 m d2 = 0.9 m

The angle which the guide vanes make with the periphery of the runner is a1 = 30° V2 = 3 m/s a2 = 120° Q = 0.3 m3/s

Discharge velocity at exit: Discharge angle at exit: Volume flow rate: Calculate the power developed. Solution:

S d1 N S – 1.8 – 300 29.27 m/s 60 60 S d2 N S – 0.9 – 300 u2 14.13 m/s 60 60 Vw1 = V1 cos 30° = 30.3 m/s u1

Vw2 = V2 cos (180° – 120°) = 1.5 m/s

Power developed (P): P

m (Vw1u1  Vw 2 u2 ) gc UQ (Vw1u1  Vw 2 u2 ) gc 1000

kg m3

– 0.3

= 263330 W

(Q m = rQ)

m3 m2 N- s2 (30.3 – 29.27  1.5 – 14.13) – s kg - m s2

or 263.33 kW

Ans.

[Vw2 is in opposite direction, hence +ve] EXAMPLE 2.18 An inward flow radial reaction turbine has axial discharge at the outlet with outlet blade angle of 45°. The radial velocity of flow is constant. The blade speed at the inlet is twice that the exit. Express the energy transfer per unit mass and degree of reaction in terms

122

Fundamentals of Turbomachinery

2gc . At what values of a1 will the degree of reaction of inlet nozzle angle a1. Assume V f be zero and unity? What are the corresponding values of energy transfer for unit mass? Solution:

Data:

Machine:

Inward flow radial reaction turbine a2 = 90°, Vw2 = 0, V2 = Vf2 b2 = 45° = Vf1 = Vf2 = Vf = constant u1 = 2u2

Axial discharge: Exit blade angle: Radial velocity of flow: Blade speed at inlet: Assume,

Vf

2gc

Energy transfer (W.D.) in terms of a1: W.D.

u1Vw1  u2Vw 2 gc

u1Vw1 gc

u1V f

2u2V f

gc tan D1

gc tan D1

2V f2 cot D1

(Q u2 = V2 = Vf2)

gc

2 – 2 gc cot D1 gc

4 cot D1 J/kg

Degree of reaction (R) in terms of a1: (V12  V22 ) 2 gc W.D.

W.D.  R

Now,

(V12  V22 ) 2 gc

V f2 cosec 2D1  V f2 2 gc

(Eq. (2.34))

Energy Transfer in Turbomachines

or

V f2 (cosec 2D1  1) 2 gc

R

\

2 gc (cosec2D1  1) 2 gc

123

cot 2D1

(4 cot D1  cot 2D1 ) 4 cot D1 cot D1 Ø È 4 cot D1 É 1  Ê 4 ÙÚ 4 cot D1

or

R

4  cot D1 4

At what values of a1 will the degree of reaction be equal to zero and unity: Case I:

If

a1 = 90°, then R = 1

W.D. = 4 cot a1 = 4 × 0 = 0 Case II:

R will be zero if, a1 = If

cot–1

Ans.

4 = 14.036°

a1 = 14.036°, then R = 0

W.D. = 4 cot a1 = 4 × cot 14.036° = 16.00 J/kg

Ans.

EXAMPLE 2.19 For the following conditions of an axial turbine, show that (a) the degree of reaction is 0.5. (b) Find an expression for the utilization factor in terms of a1. b1 = 90°, Vw1 = u1 Vf1 = Vf2 = Vf = constant a2 = 90°, Vw2 = 0, Vf 2 = V2

Radial blade at inlet: Radial velocity: Radial discharge at exit: Solution:

(a) To show that the degree of reaction R is 0.5: (V12  V22 ) 2 – gc W.D.

W.D.  R

124

Fundamentals of Turbomachinery

Now,

Vw1u1  u2Vw 2 gc

W.D.

u12 gc

u1u1 gc

From inlet triangle, V12

Vr21  u12

or

V12  V22

Vr21  u12  V22

or

(V12  V22 )

Vr21  u12  Vr21

u12

[Q Given data: Vf 1 = Vf 2; Vf 2 = V2 = Vf 1 = Vr1; refer to the figure]

\

R

u12 u2  1 gc 2 gc

or

R

u12

(2u12  u12 ) gc – 2 2 gc u1

u12 gc 1 2

2u12

0.5

Ans.

(b) Expression for utilization factor (e): V2 = Vf2 = Vf1 = Vr1 = u1 tan a1 Now,

W.D.

H

W.D. 

V22 2 gc

u12 gc u12 u12 tan 2D1  gc 2 gc

u12 gc

1 (2  tan 2D1 ) 2

È tan D1 Ø 1 É 2 ÙÚ gc Ê

u12

2

2

2

2  tan 2D1

H

2

sin 2D1

2 cos2D1 2 cos2D1  sin 2D1

cos2D1

2 cos2D1

2 cos2D1

cos2D1  cos2D1  sin 2D1

cos2D1  1

Ans.

Energy Transfer in Turbomachines

125

EXAMPLE 2.20 Show that with the help of velocity triangles for maximum utilization and same amount of energy transfer in impulse and reaction axial flow turbines, uR 2ui2 where, uR is the blade speed of 50% reaction turbine and ui is the blade speed of impulse turbine. Solution:

Data: (W.D.)impulse = (W.D.)50% reaction Maximum utilization

To determine:

uR

2ui2

Velocity triangle for maximum utilization:

Impulse

Reaction

Vw1 = 2ui

a1 = b2, a2 = b1, Vw1 = uR V1 = Vr2, V2 = Vr1 = Vf2 = Vf1

a2 = 90°; Vw2 = 0; V2 = Vf2

a2 = 90°, Vw2 = 0; V2 = Vf2

(W.D.) impulse

Vw1u1  Vw 2 u2 gc

Vw1ui gc

2ui2 gc

(i)

(W.D.)50% reaction

Vw1u1  Vw 2 u2 gc

uR uR gc

uR2 gc

(ii)

As per the given condition, equating Eqs. (i) and (ii), we have 2ui2 gc

\

uR

uR2 gc 2ui2

Ans.

EXAMPLE 2.21 Prove that for maximum utilization for an axial flow turbine with degree of reaction R =

1 , the speed ratio = 0.75 cos a1. 3

126

Fundamentals of Turbomachinery

Solution:

Data:

Maximum utilization:

V2 = Axial,

Vw2 = 0, V2 = Vf2, Vf1 = Vf2 = Vf, V2 = V1 sin a1 Degree of reaction

R

1 3

From the exit velocity triangle ABD Vr22

u2  V22

u2  V12 sin 2D1

Vr21

V f21  (Vw1  u)2 V22  (V1 cos D1  u)2

Vr21

or

(Q V2 = V1 sin a1)

(Q Vf 1 = Vf 2 = V2)

V12 sin 2D1  V12 cos2D1  u2  2uV1 cos D1

(Q V2 = V1 sin a1)

Given data: R

1 3

(Vr22  Vr21 ) (V12  V22 )  (Vr22  Vr21 )

or

(V12  V22 )

or

V12  V12 sin 2D1

( u 2  V12 sin 2D1 )  ( V12 sin 2D1  V12 cos2D1  u 2  2uV1 cos D1 )

or

V12 (1  sin 2 D1 )

2( V12 cos2D1  2uV1 cos D1 )

or

V12 cos2D1

or

4uV1 cos D1

or \

u V1 V12 u V1

3(Vr22  Vr21 )  (Vr22  Vr21 )

2(Vr22  Vr21 )

4uV1 cos D1  2V12 cos2D1 3V12 cos2D1 2 3 cos D1 4 cos D1

I

3 cos D1 4

speed ratio =

3 cos D1 4

0.75 cos D1

Ans.

Energy Transfer in Turbomachines

127

EXAMPLE 2.22 At a stage of 50% reaction axial flow turbine running at 3000 rpm, the mean blade diameter is 68.5 cm. If the maximum utilization factor for the stage is 0.915, calculate (a) the inlet and outlet absolute velocities and (b) the power output. Also, find the power developed for a steam flow rate of 15 kg/s. Solution:

Data: Degree of reaction: Axial flow turbine: Speed: Mean blade diameter: Maximum utilization factor: Mass flow rate:

R u1 N d

= 50% = u2 = u = 3000 rpm = 68.5 cm emax = 0.915 m = 15 kg/s

To determine: V1, V2, Power (a) Inlet and exit absolute velocities (V1, V2): u

S dN 60

S – 0.685 – 3000 60

For 50% reaction a1 = b2, a2 = b1, H

(V12  V22 ) (V12



RV22 )

V1 = Vr2,

(V12  V22 ) (V12



0.5V22 )

107.6 m/s

Vr1 = V2

0.915 ;

\

V12

6.38V22

For maximum utilization (refer to the figure above), u2

Vr22  V22

V12  V22

or

u2 (107.6)2 5.38 5.38 V2 = 46.69 m/s

\

V1

or

6.38V22  V22

5.38V22

V22

6.38 V2

6.38 – 46.69 = 118.03 m/s

Ans. Ans.

128

Fundamentals of Turbomachinery

(b) Power output (P): P

mVw1u1 gc

15

kg s

– 107.6

= 173666.4 W or

m m N-s2 – 107.6 – s s kg - m

173.67 kW

Ans.

EXAMPLE 2.23 Air flows through the rotor of a power absorbing machine at a mean radius of 20 cm. If the tangential component of velocity increases by +20 m/s, calculate (a) the torque exerted on air for a flow rate of 100 m3/s at a point where the temperature and pressure are respectively 20°C and 1 bar. (b) Find the power input in kW if the machine rotates at 2400 rpm. Solution: Data: Axial flow machine: u1 = u2 = u Speed: N = 2400 rpm Diameter of the rotor: d = 2r = 40 cm Tangential component increases by (Vw2 – Vw1): = 20 m/s Volume flow rate: V = Q = 100 m3/s Pressure: p = 1 bar Temperature: T = 20 + 273 = 293 K To determine: Torque, Power input u

S dN 60

1 – 10 5 – 100 287 – 293

pV RT (a) Torque exerted (T): m

T

(b) Power input (P): P

S – 0.4 – 2400 60

50.27 m/s

118.92 kg/s

m r (Vw 2  Vw1 ) gc 118.92 – 0.2 – 20 1

m u (Vw 2  Vw1 ) gc

475.68 N-m

118.92 50.27 – 20 – 1 1000

119.56 kW

Ans.

Ans.

EXAMPLE 2.24 In an inward radial flow hydraulic turbine, water enters with an absolute velocity of 15 m/s with a nozzle angle of 15°. The speed of the rotor is 400 rpm. The diameter of the rotor at inlet is 75 cm and the diameter at the outlet is 50 cm. The fluid leaves the rotor radially with an absolute velocity of 5 m/s. Determine (a) the blade angles, (b) the energy transfer per unit mass, and (c) the degree of reaction and utilization factor.

Energy Transfer in Turbomachines

Solution:

Data: Entry absolute velocity: Nozzle angle: Speed: Inlet diameter of rotor: Exit diameter of rotor: Fluid leaves radially: Exit absolute velocity:

129

V1 = 15 m/s a1 = 15° N = 400 rpm d1 = 75 cm d2 = 50 cm a2 = 90° V2 = 5 m/s

To determine: b1, b2, W.D., R, e

(a) Rotor blade angles (b1, b2):

u1

S d1 N 60

S – 0.75 – 400 60

u2

S d2 N 60

S – 0.5 – 400 60

V1 cos D1

Vw1

tan 1

E1

tan 1

E2

10.47 m/s

15 cos 15’ 14.49 m/s

V1 sin D1 u1  Vw1

V2 u2

15.71 m/s

tan 1

15 – sin 15’ 15.71  14.49 5 10.47

72.54’

25.52’

Ans. Ans.

(b) Work done (W.D.): W.D.

u1Vw1  u2Vw 2 gc

15.71 – 14.49  0 1

227.63 J/kg

(c) Degree of reaction (R): Vr1

u1  Vw1 cos E1

15.71  14.49 cos 72.54’

4.066 m/s

Ans.

130

Fundamentals of Turbomachinery

5 sin E2

Vr 2

5 sin 25.52’

11.60 m/s

(u12  u22 )  (Vr22  Vr21 ) 2 gc W.D.

R

(15.712  10.472 )  (11.62  4.0662 ) 2 – 1 – 227.63

= 0.56

Ans.

(d) Utilization factor (e): W.D.

H

W.D. 

227.63

V22

0.9479

52 227.63  2 –1

2 gc

Ans.

EXAMPLE 2.25 At a stage of 50% reaction axial flow turbine running at 3000 rpm the power output is 265 kJ/kg, utilization factor being 0.615. Find the absolute velocities. Assume symmetric velocity triangles at inlet and outlet. Solution:

Data: Speed: Work done: Utilization factor: Degree of reaction:

N = 3000 rpm W.D. = 265 kJ/kg e = 0.615 R = 0.5

To determine: V1 and V2 The absolute velocities (V1, V2): H

utilization factor =

W.D. W.D. +

or

0.615

265000 265000 

\ Now, \

V22 2 gc

V22 2 gc

V2 = 576 m/s H

Ans.

(V12  V22 )

(V12  5762 )

(V12  RV22 )

(V12  0.5 – 5762 )

V1 = 772.5 m/s

0.615

Ans.

Energy Transfer in Turbomachines

EXAMPLE 2.26

Derive the reaction

RD 1 R

131

S

where S = static component of energy transfer D = dynamic component of energy transfer R = degree of reaction Solution: Static component of energy transfer (Dynamic + Static) components of energy transfer

R

S DS RD + RS = S RD = S – RS = S(1 – R) R

\

RD 1 R

S

Ans.

EXAMPLE 2.27 In a DeLaval steam turbine, the nozzle angle at the inlet is 18°. Relative velocity is reduced to the extent of 6% when steam flows over the moving blades. The output of the turbine is 120 kJ/kg flow of steam. If the blades are equiangular, find (a) the speed ratio, (b) the absolute velocity of steam and (c) the blade speed for maximum utilization factor. Solution:

Data: Machine:

DeLaval turbine

Nozzle angle: Relative velocity reduced: Power output: Blades are equiangular:

a1 = 18° K = 0.94 W.D. = 120 kJ/kg b1 = b2

To determine: r, u, V1 for maximum utilization For maximum utilization, a2 = 90°, V2 is minimum or axial. \ Vw2 = 0 (a) Speed ratio (f): Iopt

speed ratio =

u V1

cos D1 2

cos 18’ 2

0.4755

(Eq. (2.52))

Ans.

(b) Velocity (V1): W.D.

or \

u(Vw1  Vw 2 ) gc

0.4755V1 (V1 cos D1 ) 1 V1 = 515.12 m/s

120000 J/kg =

0.4755 – V1 – V1 cos 18’ 1 Ans.

132

Fundamentals of Turbomachinery

(c) Blade speed (u): u = 0.4755 × V1 = 0.4755 × 515.12 = 244.94 m/s

Ans.

EXAMPLE 2.28 A radial outward flow turbomachine has no inlet whirl. The blade speed at the exit is twice that at the inlet. Radial velocity is constant throughout. Taking the inlet blade angle as 45°, show that the degree of reaction

2  cot E2 4 where b2 is the blade angle at the exit with respect to the tangential direction. R

Solution: No inlet whirl: Blade speed at exit: Radial velocity: Inlet blade angle: To determine: To show that

R

Vw1 = 0, a1 = 90° u2 = 2u1 Vf 1 = Vf 2 = Vf = constant b1 = 45° 2  cot E2 4

Vf1 = Vf2 = V1 = u1 W.D. =

Vw1u1  Vw 2 u2 gc 

 or

W.D. = 

u2Vw 2 gc

(Q Vw1 = 0)

u2 (u2  V f 2 cot E2 ) gc 2V f2 gc

(2  cot E2 )



2V f gc

(2V f  V f cot E2 )

Energy Transfer in Turbomachines

R

u12

(u12  u22 )  (Vr22  Vr21 ) 2 gc – W.D.

V f2 ; u22

4V f2

Vr22

V f2  V f2 cot 2 E2

Vr21

V f2

 V f2

2V f2

133 (i)

Û Ü V f2 (1  cot 2 E 2 ) Ü Ü Ü Ý

(ii)

Substituting (ii) in (i), R

(V f2  4V f2 )  V f2 (1  cot 2 E2 )  2V f2 2 gc – W.D. ( 3V f2 )  V f2 (1  cot 2 E2  2) 2 gc – W.D.

V f2

(cot 2 E2  4) 2 gc



2V f2 gc

(2  cot E2 )

V f2 (cot E2  2) (cot E 2  2) 2 gc

R 2

V f2 gc

cot E 2  2 4

Ans.

(cot E2  2)

EXAMPLE 2.29 The following data refers to a turbomachine. Inlet velocity of whirl = 16 m/s, velocity of flow = 10 m/s, blade speed = 33 m/s, outlet blade speed = 8 m/s. Discharge is radial with an absolute velocity of 16 m/s. If water is the working fluid flowing at the rate of 1 m3/s, calculate (a) the power in kW, (b) the change in total pressure in kN/m2, (c) the degree of reaction, and (d) the utilization factor. Solution: Data: Velocity of whirl at inlet: Velocity of flow at inlet: Blade speed at inlet: Blade speed at exit: Discharge is radial: Absolute velocity at exit: Volume flow rate:

Vw1 = 16 m/s Vf1 = 10 m/s u1 = 33 m/s u2 = 8 m/s a2 = 90°, Vw2 = 0, V2 = Vf2 V2 = 16 m/s Q = 1 m3/s

134

Fundamentals of Turbomachinery

To determine: P, Dp = (p1 – p2), R, e

(a) Power (P): U QVw1 u1 gc

mVw1 u1 gc

P

1000

kg m3

–1

m3 m m N- s2 – 16 – 33 – s s s 1 kg - m

P = 528000 W or

528 kW

Ans.

(b) Change in total pressure (p1 – p2 = Dp) W.D. = H–g gc m–

m s2

–

N- s2 1 kg- m

Vw1 u1 gc

Vw1 u1 gc m m N- s2 – – kg-m s s

N-m N-m = kg kg

\

H

Vw1 u1 gc – gc g

or

H

16 – 33 9.81

H

p1  p2 w

\

p1 – p2 = Hw

53.82 m of water

(w = specific weight)

Energy Transfer in Turbomachines

or

135

(p1 – p2) = 53.82 m × (1000 × 9.81) N/m3 = 52.8 × 104 N/m2 or

5.28 bar

Ans.

(c) Degree of reaction (R): V1

Vw21  V f21

162  10 2 W.D.  R

(V12  V22 ) – U 2 gc – 1000 W.D.

528 kW  or

R

\

R = 0.9055

18.86 m/s

(18.862  162 ) – 1000 kW 2 – 1 – 1000 528 kW Ans.

(d) Utilization factor (e): W.D

H W.D +

\

H

V22 – U 2 gc – 1000

528 162 – 1000 528  2 – 1 – 1000

= 0.8048

Ans.

EXAMPLE 2.30 An inward flow turbine has 0.6 reaction. The blade speed at entry is 12 m/s and the radial velocity of flow is constant at 2.4 m/s. The rotor diameter at entry is twice that at exit. Find the utilization factor angles of the blades at entry and exit assuming there is no exit whirl velocity and no friction loss. Is the utilization factor maximum? Solution:

Data: Machine: Degree of reaction: Entry blade speed: Radial velocity of flow: Inlet rotor diameter: No friction loss:

Inward turbine R = 0.6 u1 = 12 m/s Vf 1 = Vf 2 = 2.4 m/s d1 = 2d2 Vr1 = Vr2

Assume no whirl at exit, i.e. a2 = 90°, Vw2 = 0, V2 = Vf 2 To determine e, b1, b2. Is e maximum?

136

Fundamentals of Turbomachinery

u1

12 m/s ; ? u2

(given data: d1 = 2d2)

6 m/s

From exit velocity triangle, Vr 2 X

\

Vw1

\

V1

u22  V22

36  2.42

Vr21  V f21 u1  X

6.462 m/s Vr1

6.462 2  2.4 2

12  6

Vw21  V f21

6 m/s

(inlet triangle)

6 m/s

62  2.42

6.462 m/s

(a) Utilization factor (e): H

V12  V22

6.462 2  2.42

V12  RV22

6.462 2  0.6 – 2.4 2

0.9399

(Eq. (2.43))

Ans.

(b) Inlet and exit blade angles (b1, b2): From the inlet velocity triangle, Vf 1

tan E1

X

2.4 ; ? E1 6

21.8’

Ans.

V2 u2

2.4 ; ? E2 6

21.8’

Ans.

From the exit velocity triangle, tan E2

(c) Is the utilization factor maximum?

tan D1 H max

Vf 1 Vw1

2.4 ; ? D1 6

cos2D1 1  R sin 2D1

21.80’

(Eq. (2.46f))

Energy Transfer in Turbomachines

cos2 21.8’ 1  0.6 – sin 21.8’

137

0.9399

\ Utilization factor is maximum.

Ans.

EXAMPLE 2.31 A radial outward flow turbomachine has no inlet whirl. The blade speed at the exit is twice that at the inlet. Radial velocity is constant throughout. Taking the inlet blade angle as 45°, show that the degree of reaction is given by

2  cot E2 4

R

where b2 is the blade angle at the exit with respect to the tangential direction. Solution: Vw1 = 0, a1 = 90° u2 = 2u1 Vf 1 = Vf 2 = Vf = constant b1 = 45°

No inlet whirl: Blade speed at exit: Radial velocity: Inlet blade angle: To determine:

To show that R

Vf 1

W.D. =

Vf 2

u1

(Vw1 u1  Vw 2 u2 ) gc



W.D. =  R

V1

2  cot E2 4

u2 Vw 2 gc

u2 (u2  V f2 cot E 2 ) gc 2V f2 gc



2V f gc

(2V f  V f cot E2 )

(2  cot E2 )

(u12  u22 )  (Vr22  Vr21 ) 2 gc – W.D

(i)

138

Fundamentals of Turbomachinery

u12

V f2 ; u22

4V f2

Vr22

V f2  V f2 cot 2 E2

Vr21

V f2

 V f2

2V f2

Û Ü V f2 (1  cot 2 E2 ) Ü Ü Ü Ý

(ii)

Substituting (ii) in (i), R

(V f2  4V f2 )  V f2 (1  cot 2 E 2 )  2V f2 2 gc – W.D. ( 3V f2 )  V f2 (1  cot 2 E2  2) 2 gc – W.D.

V f2 (cot 2 E2  4) 2 gc 2

V f2 gc

(2  cot E2 )

V f2 (cot E 2  2) (cot E2  2) 2 gc 2

V f2 gc

(cot E2  2)

2  cot E2 4

Ans.

EXAMPLE 2.32 The following data refers to a turbomachine. Inlet velocity of whirl = 16 m/s, velocity of flow = 10 m/s, blade speed = 33 m/s, outlet blade speed = 8 m/s. Discharge is radial with an absolute velocity of 16 m/s. If water is the working fluid flowing at the rate of 1 m3/s, calculate the following: (a) Power in kW (b) Change in total pressure in kN/m2 (c) Degree of reaction (d) Utilization factor. Solution: Data: Velocity of whirl at inlet: Velocity of flow at inlet: Blade speed at inlet: Blade speed at exit: Discharge is radial: Absolute velocity at exit: Volume flow rate: To determine: P, (p1 – p2), R, e

Vw1 = 16 m/s Vf1 = 10 m/s u1 = 33 m/s u2 = 8 m/s a2 = 90°, Vw2 = 0; V2 = Vf 2 V2 = 16 m/s Q = 1 m3/s

Energy Transfer in Turbomachines

139

(a) Power (P): P

U QVw1 u1 gc

mVw1u1 gc

1000

kg m3

–1

m3 m m N-s2 – 16 – 33 – s s s kg-m

528,000 W = 528 kW

Ans.

(b) Change in total pressure (p1 – p2): W.D. =

Vw1u1 gc

Hg gc

or

H

Vw1 u1 gc – gc g

16 – 33 9.81

Also,

H

p1  p2 w

(w = specific weight)

\

( p1  p2 )

53.82 m of water

53.82 m – (1000 – 9.81) N/m 3

52.8 – 10 4 N/m 2

Ans.

(c) Degree of reaction (R): V1

Vw21  V f21

162  10 2 W.D.  R

18.86 m/s

(V12  V22 ) – U 2 gc – 1000 W.D.

528 kW 

(18.862  162 ) – 1000 kW 2 – 1 – 1000 528 kW

0.9055

Ans.

140

Fundamentals of Turbomachinery

(d) Utilization factor (e): W.D.

H

528

V22

–U W.D.  2 gc – 1000

162 – 1000 528  2 – 1 – 1000

0.8048

Ans.

EXAMPLE 2.33 At a stage of an impulse turbine the mean blade diameter is 80 cm; its rotational speed 3000 rpm. The absolute velocity of fluid is 300 m/s. A nozzle is inclined at 70° to the axial direction. Find the inlet and exit blade angles of the rotor. Also find the power output from the stage for a mass flow rate of 2 kg/s and the axial thrust on the shaft. Utilization factor is 0.85 and the relative velocity at exit is equal to that at the inlet. Solution:

Data: Machine: Mean blade diameter: Speed: Absolute velocity: Nozzle inclination: Mass flow rate of steam: Utilization factor:

Impulse steam turbine d = 80 cm N = 3000 rpm V1 = 300 m/s a1 = 90° – 70° = 20° m = 2 kg/s e = 0.85

Given data is with respect to axial direction. Here considered for tangential direction. To determine b1, b2, Power, Fa Solution: \

ÐBED = 70° ÐEBD = 90° – 70° = 20° u

S dN 60

S – 0.8 – 3000 60

(given data) (with respect to axial direction) (with respect to tangential direction) 125.66 m/s

Vw1 = V1 cos 20° = 300 cos 20° = 281.9 m/s Vf 1 = 300 sin 20° = 102.6 m/s Vr1

V f21  (Vw1  u)2

102.62  (281.9  125.66)2

186.9 m/s

Energy Transfer in Turbomachines

141

(a) Blade angles (b1, b2):

tan E1 H

\

V2

Vf 1

102.6 281.9  125.66

Vw1  u V12  V22

V12  V22

V12  RV22

V12

V12  HV12

0.6567; ? E1

33.29’

Ans.

(R = 0, impulse turbine)

300 2  0.85 – 300 2

116.19 m/s

Vr1 = Vr2 = 186.9 m/s

V22

Vw22  V f22

(Vr 2 cos E2  u)2  V f22

Vr22 cos2 E2  u2  2uVr 2 cos E2  V f22 or

V22

\

cos E2

Vr22  u2  2uVr 2 cos E2 u2  Vr22  V22

(Q Vf 2 = Vr2 sin b2)

125.662  186.92  116.192 2 – 125.66 – 186.9

2uVr22

b2 = 37.58°

or

Ans.

(b) Power output (P): Vw2 = Vr2 cos b2 – u = 186.9 cos 37.58° – 125.66 = 22.46 m/s P

m u (Vw1  Vw 2 ) gc

2 – 125.66(281.9  22.46) 1

76, 490 W

Ans.

(c) Axial thrust (Fa): Vf 2 = Vr2 sin b2 = 186.9 sin 37.580° = 113.9 m/s

Fa

m

(V f 2  V f 1 ) gc

2–

(113.98  102.6) 1

22.76 N

Ans.

EXAMPLE 2.34 The impeller of a centrifugal pump of outer diameter of 1.2 m is used to lift water at a rate of 1800 kg/s. The blade makes an angle of 150° with the direction of motion at outlet and the speed is 2000 rpm. If the radial velocity of flow is 2.5 m/s, find the impeller power. Solution:

Data: Machine: Outer diameter: Mass flow rate of water: Outlet blade angle:

Centrifugal pump d2 = 1.2 m m = 1800 kg/s b2 = 150°

142

Fundamentals of Turbomachinery

Speed: Radial velocity = Flow velocity

N = 2000 rpm Vf 2 = 2.5 m/s

To determine: P

Exit triangle

Assuming radial inlet, we have a1 = 90°, Vw1 = 0; u2

S – 1.2 – 2000 60

12.56 m/s

Vw2 = u2 – Vf 2 cot 30° = 12.56 – 2.5 cot 30° = 8.23 m/s P

m u2 Vw 2 gc – 1000

1800 – 12.56 – 8.23 1 – 1000

186 kW

Ans.

EXAMPLE 2.35 Combustion products approach an axial flow turbine rotor with an absolute velocity of 600 m/s and 70° to the axial direction. The tangential component of this absolute velocity is in the same direction as the wheel velocity. The mass flow rate is 30 kg/s. The blade speed is 250 m/s and absolute velocity is exited axially. Calculate (a) the power output, (b) utilization factor of reaction, and (c) the degree of reaction. Solution:

Data: Machine: Inlet absolute velocity: Inlet nozzle angle: Mass flow rate of gases: Blade speed: Absolute velocity exited axially, i.e.

Axial turbine V1 = 600 m/s a1 = 90° – 70° = 20° m = 30 kg/s u = 250 m/s a2 = 90°, V2 = Vf 2, Vw2 = 0

Given data is with respect to axial direction, here converted with respect to tangential direction. Vw1 = V1 cos a1 = 600 cos 20° = 563.8 m/s Vf1 = V1 sin a1 = 600 sin 20° = 205.2 m/s

tan E1 Vr1

Vf 1 Vw1  u Vf 1 sin E1

205.2 563.8  250 205.2 sin 33.18’

0.6539; ? E1

374.94 m/s

33.18’

Energy Transfer in Turbomachines

143

Vf 2

205.2 0.8208 250 u b2 = 39.38° (assuming Vf1 = Vf 2)

tan E2

\

V2 sin E2

Vr 2

205.2 sin 39.38’

323.43 m/s

(a) Power output (P): 30 – 250 – 563.8 1 – 1000

m uVw1 gc – 1000

P

4228.5 kW

Ans.

(b) Utilization factor (e): W.D.

uVw1 gc

250 – 563.8 1

W.D.

H

W.D. 

140950 kJ/kg

140950

V22 2 gc

140950 

205.22 2 –1

0.87

Ans.

(c) Degree of reaction (R): H

or

0.87

V12  V22 V12  RV22 600 2  205.22 600 2  R 205.22

; ? R

Ans.

 0.1281

Alternatively: R

(Vr22  Vr21 ) (V12

 V22 )  (Vr22  Vr21 ) (323.432  374.942 )

(600 2  205.22 )  (323.432  374.942 )

 0.1276

Ans.

144

Fundamentals of Turbomachinery

EXAMPLE 2.36 In an axial flow turbine, the discharge blade angles are 22° each for both stator and rotor. The steam speed at the exit of the fixed blade is 150 m/s. The ratio Vf /u is 0.7 at the entry and 0.75 at the exit of the rotor blade. The mass flow rate of steam is 2.5 kg/s. Calculate (a) the inlet rotor blade angle, (b) the degree of reaction, and (c) the power developed by the turbine. Solution:

Data: Machine: Discharge (a) Rotor: Blade angles (b) Stator: Steam speed at exit of fixed blade: Mass flow rate of steam: b1, R,

To determine:

b2 a1 V1 (Vf /u)1 m

Axial = 22° = 22° = 150 m/s = 0.7, (Vf /u)2 = 0.75 = 2.5 kg/s

Power.

Vf 1 = V1 sin a1 = 150 sin 22° = 56.19 m/s

È Vf Ø ÉÊ u ÙÚ 1

0.7 ? u

Vf 1 0.7

56.19 0.7

80.27 m/s

Vr1

(V1 cos D1  u)2  V f21

Vr1

(150 cos 22’  80.27)2  56.192

81.34 m/s

(a) Rotor inlet blade angle (b1): sin E1

Vf 1 Vr 1

56.19 81.34

0.690.8; ? E1

43.69’

(b) Power developed (P):

È Vf Ø ÉÊ u ÙÚ 2

sin E2

0.75,

Vf 2 Vr 2

\ Vf 2 = 0.75u = 0.75 × 80.27 = 60.203 m/s

, ? Vr 2

Vf 2 sin E 2

60.203 sin 22’

160.71 m/s

Ans.

Energy Transfer in Turbomachines

or

Vr1

(V1 cos D1  u)2  V f21

V2

(160.71 cos 22’  80.27)2  60.2032

145

91.37 m/s

Vw1 = V1 cos a1 = 150 cos 22° = 139.08 m/s Vw2 = Vr2 cos b2 – u = 160.71 cos 22° – 80.27 = 68.74 m/s P

m u (Vw1  Vw 2 ) gc – 1000

2.5 – 80.27 – (139.08  68.74) 1 – 1000

41.7 kW

Ans.

(c) Degree of reaction (R): R

Vr22  Vr21 2u (Vw1  Vw 2 )

160.712  81.342 2 – 80.27 – (139.08  68.74)

0.5758

Ans.

EXAMPLE 2.37 The following data refers to a mixed flow pump, where the fluid absolute velocity at the inlet is axial while at the outlet, the relative velocity is radial. The inlet hub diameter is 9 cm, impeller tip diameter = 30 cm, speed 3000 rpm. Axial velocity at inlet is equal to the radial velocity at exit. Calculate (a) the degree of reaction and (b) the energy input to the fluid, if the relative velocity at the exit equals the inlet tangential blade speed. Solution: Machine: Mixed flow pump Absolute velocity at inlet is axial, i.e. V1 is axial: a1 = 90°, V1 = Vf1, Vw1 = 0 Relative velocity at exit is radial, Vr2 is radial: b2 = 90°, Vr2 = Vf 2, u2 = Vw2 Inlet hub diameter: d1 = 0.09 m Impeller tip diameter: d2 = 0.30 m Blade speed: N = 3000 rpm Vf 1 (Va1) = Vr2 = u1

To determine: R, Energy input (a) Degree of reaction (R):

u1

S d1 N 60

S – 0.09 – 3000 60

14.137 m/s

146

Fundamentals of Turbomachinery

u2 R

S d2 N 60

S – 0.3 – 6000 60

Ë Vf 2 Û 0.5 Ì1  cot E2 Ü u2 Í Ý

47.123 m/s

Vf 2 Ë Û 0.5 Ì1  cot 90’Ü 0.5 Ans. 47.123 Í Ý (For derivation, see Example 2.12)

(b) Energy input (W.D.): W.D. = EXAMPLE 2.38

u2 Vw 2 gc

47.123 – 47.123 1 – 1000

2.22 kJ/kg

The following data refers to an axial flow compressor. Machine: Axial flow compressor Degree of reaction: R = 0.5 Inlet blade angle: b1 = 45° Axial flow is constant: Vf 1 = Vf 2 = 100 m/s Speed of blade: N = 6000 rpm Diameter of the blade: d = 0.5 m Blade speed: u1 = u2 Mass of air: m = 2 kg/s

Calculate (a) the fluid angles at inlet and outlet and (b) the power required. Solution:

(a) Angles at inlet and outlet (a1, a2, b2): u

tan E1

\ \

u tan D1  V f

S dN 60

S – 0.5 – 6000 60

Vf u  Vw1 Vf

tan D1 tan E1

u Vf

Vf Vf

157.08 m/s

V f tan D1 u tan D1  V f

tan D1 tan D1 1

Vf = u tan a1 – Vf tan a1 = tan a1 (u – Vf)

Ans.

Energy Transfer in Turbomachines

\

tan D1

Vf u  Vf

100 157.08  100

147

1.7519

a1 = 60.28° = b2

(Q R = 50%)

Ans.

b1 = 45° = a2

(given data and R = 50%)

Ans.

(b) Power required (P):

Vw1 Vw1 P

Vf tan D1 Vf tan D 2

100 tan 60.28’

57.086 m/s

100 100 m/s tan 45’

m u (Vw 2  Vw1 ) gc

ËVw 2 and Vw1 are in the same direction;Û Ì Ü Í hence –ve Ý

2 – 157.08 – (100  57.086) 1

13481.86 W

Ans.

EXAMPLE 2.39 A centrifugal pump delivers water against a head of 25 m. The radial velocity of flow is 3.5 m/s and it is constant. The flow rate of water is 0.05 m3/s. The blades are radial at the tip and pump runs at 1500 rpm. Calculate (a) the diameter at the tip, (b) the width of the blade at the tip, and (c) inlet diffuser angle at the impeller exit. Solution:

Data: Machine: Manometric head: Radial velocity of flow: Water flow rate: Blades are radial at tip: Speed:

Centrifugal pump Hm = 25 m Vf 1 = Vf 2 = 3.5 m/s Q = 0.05 m3/s b2 = 90°, u2 = Vw2, Vf 2 = Vr2 N = 1500 rpm

To determine : d2, b2, a2 Assume, radial or axial entry, i.e. a1 = 90°, Vw1 = 0, V = Vf 1

148

Fundamentals of Turbomachinery

(a) The diameter at the tip (d2): g Hm gc 9.8

m s2

u22 gc

u2Vw 2 gc – 25 m –

N-s2 1 kg-m

u22

m2 s2

–

N-s2 1 kg-m

u22

or

9.8 × 25 =

\

u2

\

d2 = 0.1994 m

15.66 m/s

S d2 N 60

S – d2 – 1500 60 Ans.

(b) Width of the blade at tip (b2): Q = p d2 b2 Vf 2 0.05 m 3 /s

S – 0.1994 m – b2 m – 3.5

m ; ? b2 s

0.02281 m

Ans.

(c) Inlet diffuser angle at the impeller exit (a2):

tan D 2

Vf 2 u2

3.5 15.66

0.2235; ? D 2

12.598’

Ans.

EXAMPLE 2.40 A jet of water in an impulse turbine impinges the blades without shock at a velocity of 50 m/s. The blade speed is 20 m/s. The direction of nozzle is 20° to that of the jet. The exit absolute velocity of the water is normal to the motion of the blades. Calculate (a) the blade angles at inlet and exit, (b) the work done, and (c) the utilization factor. Solution: Data: Machine: Water impinges without shock: Impinging velocity of water: Blade speed: Nozzle angle: Exit absolute velocity: To determine: b1, b2, W.D., e

Impulse turbine Vr1 = Vr2 V1 = 50 m/s u = 20 m/s a1 = 20° V2 = axial, a2 = 90°, V2 = Vf 2, Vw2 = 0

Energy Transfer in Turbomachines

149

Vw1 = V1 cos a1 = 50 cos 20° = 46.985 m/s Vf 1 = V1 sin a1 = 50 sin 20° = 17.1 m/s (a) Blade angles (b1, b2):

tan E1 sin E1 cos E2

Vf 1

17.1 46.985  20

Vw1  u Vf Vr1

; Vr1

u Vr 2

Vf

17.1 sin 32.363’

sin E1

20 31.95

0.6337; ? E1

0.6261; ? E2

32.363’

Ans.

31.95 m/s Vr 2 51.24’

Ans.

(b) Work done (W.D.): W.D.

uVw1 gc

20 – 46.985 1

939.7 W

(c) Utilization factor (e): V2 = u tan b2 = 20 × tan 51.24° = 24.91 m/s H

\

H

V12  V22

50 2  24.912

V12  RV22

50 2  0

0.7517

W.D.

939.7

V2 W.D.  2 2 gc

24.912 939.7  2 –1

(Q R = 0, impulse)

0.7518

Ans.

EXAMPLE 2.41 The mean diameter of an axial flow steam turbine is 50 cm. The maximum utilization is 0.90, and the degree of reaction is 50%. The mass flow rate of steam is 10 kg/s. The speed of the blade is 2000 rpm. Calculate (a) the inlet and exit absolute velocities and (b) the power developed. Solution: Data: Machine: Mean diameter: Maximum utilization factor, em = 0.90: Degree of reaction: Mass flow rate of steam: Speed:

Axial flow steam turbine d = 50 cm \ V2 = minimum, a2 = 90°, V2 = Vf 2 R = 50% m = 10 kg/s N = 2000 rpm

R = 50%, \ a1 = b2, a2 = b1, Vr1 = V2, Vr2 = V1

150

Fundamentals of Turbomachinery

u = Vw1 a1

b1 Vr 1 = Vf 1

b2

V2 = Vf 2

a2

V1

V1

Vr 2

V2=Vr 1 Vr 2 Vf 1=Vf 2 b1

a1 Vw1 u

S dN 60

u

S – 0.5 – 2000 60

u

52.36 m/s

Vr1 = V2 = V1 sin a1 Hm

\

0.90

V12  V22

V12  V12 sin 2D1

1  sin 2D1

V12  RV22

V12  0.5V12 sin 2D1

1  0.5 sin 2D1

a1 = 25.4°

(a) Inlet absolute velocity (V1): H max

\

cos D1

u V1

cos 25.24’

52.36 V1

V1 = 57.89 m/s

Ans.

(b) Exit absolute velocity (V2): V2 = V1 sin a1 = 57.89 × sin 25.24° = 24.68 m/s (c) Power developed (P): [Vw2 = 0, Vw1 = u] \

P

m u (Vw1  Vw 2 ) gc – 1000

10 – 52.36 – 52.36 1 – 1000

27.42 kW

Ans.

EXAMPLE 2.42 In an inward flow radial turbine, water leaves radially. The inlet angle is 20°. The speed of the wheel is 400 rpm. The flow velocity remains constant and is 4 m/s. The inner and outer diameters of the turbine are 60 cm and 30 cm respectively. The width at inlet is 10 cm. Calculate (a) the blade angles and (b) the power developed. Solution:

Data: Machine: Water leaves radially: Speed of the wheel: Flow velocity constant:

Inward flow radial turbine a2 = 90°, V2 = Vf 2, Vw2 = 0 N = 400 rpm Vf1 = Vf 2 = 4 m/s

Energy Transfer in Turbomachines

Inner diameter: Outer diameter: Width at inlet: Inlet angle:

151

d1 = 60 cm d2 = 30 cm b1 = 10 cm a1 = 20°

To determine: b1, b2, power developed

S d1 N 60

u2

S – 0.3 – 400 60

S d2 N 60

u1

S – 0.6 – 400 60

6.283 m/s 12.57 m/s

(a) Blade angles (b1, b2):

sin D1

Vf 1 V1

? V1

Vf 1 sin D1

4 sin 20’

11.695 m/s

Vw1 = V1 cos a1 = 11.695 cos 20° = 10.99 m/s

tan E1

Vf

4 10.99  12.57

Vw1  u

 2.53; ? E1

111.55’

Ans.

Therefore the inlet velocity triangle changes to as shown in Figure (b) above.

tan E 2

Vf 2 u2

4 6.283

0.6366; ? E2

32.48’

Ans.

(b) Power developed (P): Q = pd1b1Vf1 = p × 0.6 × 0.1 × 4.0 = 0.754 m3/s Q U (u1Vw1  u2Vw 2 ) gc

m3 kg m m N-s2 – 1000 3 – 12.57 – 10.99 – s s s kg-m m

\

P

\

P = 104158.348 W » 104.16 kW

0.754

Ans.

152

Fundamentals of Turbomachinery

IMPORTANT EQUATIONS • Euler turbine equation: For power generating machines

â

W.D.

â

P

(Vw1 u1  Vw 2 u2 ) J/kg gc

(For unit mass flow rate)

(2.2)

m(Vw1 u1  Vw 2 u2 ) W gc

If Vw2 is opposite to Vw1, then

â

W.D.

Vw1 u1  Vw 2 u2 J/kg gc

(2.2a)

For axial flow machines u1 = u2 = u, then

â

W.D.

(Vw1  Vw 2 ) u J/kg gc

• Alternate forms of Euler energy equation

â

W.D. Unit mass flow rate

â

V12

â

u12  u22 2 gc

â

Vr22  Vr21 2 gc

(V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

(2.7)  V22

2 gc

change in dynamic head of the fluid through the machine change in static head of the fluid through the rotor due to the change in the radius of rotation. This is also called the change in centrifugal energy.

change in static head of the fluid across the rotor

• From SFEE Dh0 = –w

â

(2.9a)

(V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc • Degree of reaction (R)  dh0

R

(2.10)

Energy transfer due to the change of static pressure in the rotor Total energy transfer in the rotor h1  h2 h01  h02

(u12  u22 )  (Vr22  Vr21 ) (V12  V22 )  (u12  u22 )  (Vr22  Vr21 )

(2.14), (2.13)

Energy Transfer in Turbomachines

153

For axial flow machines, u = u1 = u2, then

â

R

(Vr22  Vr21 )

(2.15)

(V12  V22 )  (Vr22  Vr21 )

For impulse machines, there is no static pressure change in the rotor, then R=0 • Centrifugal pumps and compressors

â â â â

W.D.

(Vw 2 u2  Vw1 u1 ) gc

(2.20)

W.D.

(V22  V12 )  (u22  u12 )  (Vr21  Vr22 ) 2 gc

(2.21)

Usually the inlet is radial for centrifugal pumps and compressors. \ a1 = 90°, Vw1 = 0, V1 = Vf1 \

W.D.

Vw 2 u2 g

H

â

(2.15a)

head

(2.21a)

u2 g

Q cot E 2 Ø È ÉÊ u2  A2 ÙÚ

(2.23)

b2 = discharge blade angle with respect to tangential direction b2 = 90° b2 > 90° b2 < 90°

Radial curved vanes: Forward curved vanes: Backward curved vanes:

• Axial flow compressors and pumps u1 = u2 = u, Va = Vf

â

W.D.

(Vw 2  Vw1 ) u gc

(For unit mass flow rate)

(V22  V12 )  (Vr21  Vr22 ) 2 gc uVa (cot D 2  cot D1 ) gc a1 and a2 are with respect to tangential direction

â

R

R

(2.26) (2.29)

Increase in enthalpy in the rotor Total increase in enthalpy in the stage 'h1 'h1  'h2

â

(2.25)

'h R 'hR  'hS

Va (cot E1  cot E2 ) 2u

'h1 W.D.

(2.32) (2.33)

154

Fundamentals of Turbomachinery

b1 and b2 are with respect to tangential direction.

â

R 1

Va (cot D1  cot D 2 ) 2u

(2.36)

• Turbines e = utilization factor

â

Ideal work Energy supplied

(2.38)

W.D.

(2.39b)

V2 W.D.  2 2 gc

2 – (Vw1 u1  Vw 2 u2 ) V12

 (u12  u22 )  (Vr22  Vr21 )

(2.39c)

For axial flow turbines:

â

H

â

R

â

H

V12  V22

(2.43)

V12  RV22 (Vr22  Vr21 ) (V12  V22 )  (Vr22  Vr21 )

(V12  V22 )  (Vr22  Vr21 ) V12  (Vr22  Vr21 )

(2.40) (2.42)

For radial flow turbines:

â â

Radial entry a1 = 90°, Vw1 = 0, V1 = Vf 1 W.D. =

Vw 2 u2 gc

W.D. =

u2 V2 gc

â

(Water leaves tangentially V2 = Vw2)

W.D.

H

W.D. 

V22 2 gc

2 V 1  r2 u2

(2.45)

â

R

(u12  u22 )  (Vr 2  Vr21 ) 2 gc – W.D.

â

R

Vr 2  u2 2u2

(2.45a)

â

R

1 H

(2.45b)

Energy Transfer in Turbomachines

155

Condition for maximum utilization: (a) Reaction type turbine:

â

H max

cos2D1

(2.46)

1  R sin 2D1

(b) Impulse turbine:

â

emax = cos2 a1

(2.49)

Optimum blade speed ratio for different types of turbines: (a) Reaction turbine:

â

fopt = u = cos a1

(2.50)

(b) Impulse turbine:

â â

Iopt

u V1

cos D1 2

(2.52)

tan b1 = 2 tan a1

(2.53)

REVIEW QUESTIONS 1. With the help of inlet and outlet velocity triangles, show that the degree of reaction for an axial flow compressor is given by R

Va (cot E1  cot E2 ) u–2

where Va = axial flow velocity u = blade velocity. 2. Sketch the velocity triangles at inlet and outlet for a centrifugal pump with radial inlet for (i) forward curved, (ii) radial (iii) backward curved vanes. 3. Define the utilization factor of turbomachines. Obtain an expression for the utilization factor of a turbomachine in terms of inlet and outlet absolute velocity of the fluid and the degree of reaction. 4. “The energy transfer as work per unit mass flow is numerically equal to change in stagnation enthalpy of the fluid between the turbomachine inlet and outlet”. Discuss the above in the light of first and second laws of thermodynamics for a turbomachine. 5. Show that the maximum value of utilization factor for an axial flow impulse turbine is (single stage) emax = cos2 a1 where a1 is the nozzle angle at inlet.

156

Fundamentals of Turbomachinery

6. Derive the relation S

D 1 R

where S = static component of energy transfer D = dynamic component of energy transfer in any turbomachine R = degree of reaction. 7. Derive an expression for the maximum utilization factor and maximum energy transfer for an impulse turbine. 8. Derive an equation for the degree of reaction in a radial flow machine. 9. Derive the Euler-turbine equation. 10. Derive the alternate form of Euler equation and explain each component in that equation. 11. Define the degree of reaction and derive a general expression for the degree of reaction. 12. Explain the effect of blade discharge angle on energy transfer and the degree of reaction in a turbomachine. 13. Derive the theoretical head capacity relation in case of centrifugal pumps, i.e. u22 u22 Q cot E2  gc A2 gc

H

where b2 is the discharge blade angle with respect to the tangential direction. 14. Write the combined velocity triangle for different values of the degree of reaction i.e. R > 0.5, R < 0.5, R = 0.5. 15. (a) Define the utilization factor and derive a general expression for the same. (b) Derive an expression where the utilization factor and the degree of reaction are involved for (a) axial turbine, (b) radial turbine. 16. What is the condition for maximum utilization factor? (a) Impulse turbine (b) Reaction turbine 17. Derive the optimum blade speed ratio for (a) Impulse turbine, i.e. Iopt (b) Reaction turbine, i.e. Iopt

u V1

u V1

cos D1 2

cos D1

where a1 is the nozzle angle with respect to the tangential direction. 18. Write the combined velocity diagram for: (a) 50% reaction and maximum utilization

(b) Pure reaction.

Energy Transfer in Turbomachines

157

EXERCISES 2.1 At a stage of an impulse turbine the mean blade diameter is 75 cm, rotational speed 3500 rpm. The absolute velocity of fluid discharging from a nozzle inclined at 20° to the plane of the wheel is 275 m/s. If the utilization factor is 0.9 and the relative velocity at the rotor exit is 0.9 times that at the inlet, find the inlet and exit rotor angles. Also find the power output from the stage for a mass flow rate of 2 kg/s and the axial thrust on the shaft. 2.2 Draw the velocity triangles at the inlet and outlet for the axial flow compressor with the data given below and compute the absolute velocity at the rotor inlet, the mean blade tip speed and temperature change as air flows through the compressor. (a) Inlet pressure and temperature 1.0 bar and 306 K (b) Axial flow speed of air = 110 m/s (c) Degree of reaction = 50% (d) Inlet blade angle = 20° 2.3 The following data refers to a turbomachine: Inlet: Velocity of whirl = 16 m/s Velocity of flow = 10 m/s Blade speed = 33 m/s Exit: Blade speed = 8 m/s Discharge is radial with an absolute velocity of 16 m/s. If water is the working fluid flowing at the rate of 1 m3/s, compute the following: (a) power in kW, (b) change in total pressure in bar, (c) degree of reaction, and (d) utilization factor. Sketch the velocity triangles to scale. 2.4 The following data refers to a hydraulic reaction turbine of radial type. (a) Head of water = 160 m (b) Rotor blade angle at entry = 119° (c) Diameter at entry = 3.65 m (d) Diameter at exit = 2.45 m (e) Discharge angle at exit = 30° radial with a velocity of 15.5 m/s. (f) Radial component at inlet = 10.3 m/s Find the power developed in kW, degree of reaction and utilization factor for a flow rate of 110 m3/s. 2.5 At a stage in a 50% reaction axial flow turbine running at 3000 rpm, the power output is 265 kW, utilization factor being 0.615. Find the absolute velocities V1 and V2. Assume symmetric velocity triangles at inlet and outlet. 2.6 In a DeLaval steam turbine, the nozzle angle at the inlet 18°, and the relative velocity is reduced to the extent of 6% when steam flows over the moving blades. The output of the turbine is 120 kW/kg flow of steam. If the blades are equiangular, find the speed ratio, the absolute velocity of steam and the blade speed for the maximum utilization factor.

3

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

3.1 VELOCITY OF SOUND OR SONIC VELOCITY

Travelling or moving wavefront, velocity, a p dV

r + dr

r Stationary observer

r + dr

p, r x dV

(b)

(a)

Figure 3.1

p + dp

0

Velocity

p + dp

Density Pressure

Sound is propagated in fluids in the form of waves. Acoustic (sonic) velocity is the velocity of sound in a fluid medium. It is the speed with which a small disturbance is propagated through the fluid. Acoustic velocity is usually considered a property of the fluid. Consider a frictionless piston-and-cylinder arrangement as shown in Figure 3.1 filled with a fluid and initially at rest. Sound is propagated by an infinitesimal movement of the piston moving with a speed of dV to the right. Therefore, the velocity of the fluid before the wavefront is also dV. The wave moves with a velocity ‘a’ towards the right. The propagation of sound through the fluid causes a small change in pressure and density.

x

Propagation of sound with observer at rest: (a) Piston-and-cylinder arrangement. (b) Pressure, density and velocity distribution. 158

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

159

Density Pressure

Therefore, the fluid (gas) immediately adjacent to the piston will experience a slight rise in pressure or in other words it will be compressed. The pressure and density before the wave front are given by (p + dp) and (r + dr) respectively. Pressure rise is so small and since the waves move fast, there is hardly any time for heat transfer. Hence, isentropic conditions may be assumed, i.e. Q = 0. The velocity of the fluid ahead of the pressure wave is zero and pressure and density are p and r. Pressure, density, velocity of the gas just before and after the wavefront for an observer at rest are shown in Figure 3.1(b).

Stationary wave front

p, r

p + dp r + dr

x

Velocity

p

p + dp

a

a – dV r + dr

r

a a – dV x (b)

(a)

Figure 3.2

Propagation of sound with an observer travelling with a velocity: (a) Piston-and-cylinder arrangement. (b) Pressure, density and velocity distribution.

Figure 3.2 shows the flow phenomena for an observer travelling with the pressure wave with the speed of sound ‘a’. The relative velocity between them is zero. Then, for an observer the wavefront becomes stationary. As soon as the fluid passes through the wavefront, the velocity of the fluid reduces from a to a – dV, density increases from r to r + dr and pressure increases from p to p + dp. Applying continuity equation to both sides of wavefront m = r A a = (r + dr)(a – dV)A

(3.1)

where A is the cross-sectional area of the cylinder, we get Ua

U a  U dV  a – d U  d U – dV

The term dr × dV is neglected. Hence dV

dU – a U

(3.2)

Applying the momentum equation and neglecting the shear forces and elevation changes, we have pA – (p + dp)A = raA[(a – dV) – a] \

dp = r a dV

(3.3)

160

Fundamentals of Turbomachinery

Substituting Eq. (3.2) in Eq. (3.3), U – a – dU – a

dp

\

a2

or

a

a2 d U

U dp dU dp dU

dp – gc dU

or

(3.4)

For an isentropic process, p

Differentiating the above equation, dp dU

U

J

c or p

J c U J 1

J U J 1

cUJ

p U

J

Jp U

(3.5)

Substituting Eq. (3.5) in (3.4), a

gc

gcJ

Jp U

(Q p = rRT)

gc J RT

È ÉÊ' R

RT M

RØ M ÙÚ

(3.6)

(3.7)

For an isothermal process, p = constant = pv = RT U

On differentiating the above equation,  pU 2 d U  U 1 dp

or

dp

or

dp dU

0 p U 2 d U U

p U

1

p dU U

RT

Substituting the above equation in Eq. (3.4), a

dp / d U

RT

Bulk modulus of elasticity (K) K

Increase in pressure Relative change in volume

(3.7a)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

161

Let the pressure change from p to p + Dp in a system consisting of a fluid; the change in volume is –Dv/v. \

K

v

\ \

K

'p dp v 'p  0 'v dv 1 dU ; dv  2 U U lim  v

v dp d U / U

2

v

dp 2 U dU

v

dp U dU v

dp dU Substituting the above equation in Eq. (3.4), we get K

U

U a2 ; a

K

(3.7b)

K /U

Equation (3.4) shows that the velocity of sound is lowered in compressible fluids in which there is a large density change for a given pressure change compared to incomressible fluids (liquids). For example, the velocity of sound in air at normal ambient temperature is about 340 m/s compared to 1700 m/s in water. Equation (3.7) shows that the velocity of sound depends only on the nature of the gas, i.e. its molecular weight and the temperature. The velocity of sound is higher if the gas has a smaller molecular weight.

3.2 MACH NUMBER The Mach number (M) is a dimensionless number and it is the ratio of the local velocity of fluid to the velocity of sound in the fluid, i.e. Mach number of a moving object (example, aircraft) is the ratio of its velocity and the velocity of sound in the medium into which it is moving. The Mach number of a flowing fluid is the ratio of its velocity V (fluid velocity) and the velocity of sound a at the prevailing temperature in the fluid. \

V a Mach number can also be defined as the ratio of inertia force to elastic force. M

M2

Inertia force ; K Elastic force

M2

U AV 2 KA

U AV 2 U Aa2

Û modulus of elasticity Ü Ü Ü V2 V ; ? M Ü 2 a a ÜÝ

(3.8)

3.3 CLASSIFICATION OF FLUID FLOW Compressible flow can be classified as (i) Subsonic (iii) Transonic

M1 M>5

(3.9)

162

Fundamentals of Turbomachinery

3.4 STAGNATION AND STATIC PROPERTIES 3.4.1 Static State The properties of fluid measured by using instruments which are at rest relative to the fluid are known as static properties. The static temperature of any fluid particle moving with a uniform speed can be measured by using the thermometer (measuring instrument) such that the thermometer also moves at the same speed as that of the fluid particle is known as static temperature. Figure 3.3(a) shows a flow system with a moving thermometer. The temperature T recorded by the thermometer when both the fluid particle and thermometer move with the same speed is known as static temperature. T

Thermometer Velocity = V

Reservoir p0, T0,

A (stagnant point) p0, T0, h0, V = 0

h0 V=0 Fluid particle (a)

Figure 3.3

(b)

Fluid system: (a) A flow system with a moving thermometer. (b) A flow system with reservoir and stagnant point.

3.4.2 Stagnation State When a flowing fluid is insentropically decelerated to zero velocity (brought to rest) at zero elevation, the resulting state is called the total or stagnation state and the corresponding values of the properties are called total or stagnation properties. Consider a reservoir shown in Figure 3.3(b). The properties of the fluid in the reservoir are represented by p0, T0, h0. As soon as the fluid starts flowing from the reservoir, it converts some of its energy into velocity energy (K.E.). The velocity of fluid in the pipe line is V. Let the point A be the stagnant point. At A the velocity of fluid is zero (as per the definition of stagnant state). Fluid experiences some compression and, thereby, an increase in pressure and temperature of the fluid. In the absence of heat transfer, friction and any other loss restores this energy (kinetic energy). Therefore, the state properties of fluid at A and at reservoir are same and constant. Usually, the stagnation properties are represented by the subscript ‘0’ (zero).

3.4.3 Stagnation Enthalpy (h0) At state point A, the stagnation enthalpy h0 is given by h0

h

V 2 gz  2 gc gc

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

163

h = static enthalpy; z can be neglected \

h0

h

V2 2 gc

c pT 

V2 2 gc

(3.10)

3.4.4 Stagnation Temperature (T0) If a thermometer is inserted in the flow system and the velocity of gas brought to rest in the vicinity of the thermometer, thus converting kinetic energy into thermal energy, then, the temperature in the vicinity of thermometer is called stagnation temperature. For a perfect gas,

\ Equation (3.10) can be written as

h = cpT, c pT0

\

T0

c pT 

T

V2 2 gc

V2 2c p gc

(3.11)

V2/2cp is called the dynamic temperature (velocity temperature) and T the static temperature. From the practical point of view, it is easier to measure stagnation temperature than the static temperature. If we know the stagnation temperature there is no need to refer to the kinetic energy term. Dividing Eq. (3.11) by T,

T0 T

1 1

1

or

T0 T

1

V2 2c p gc T V2 2T J Rgc (J  1)

È ÉÊ' c p

V 2 (J  1)

' a

2 a 2 gc (J  1) 2 M 2 gc

È ÉÊ' M

JR Ø J  1 ÙÚ

J RT



VØ Ù aÚ

3.4.5 Stagnation Pressure or Total Pressure (p0)

Û Ü Ü Ü Ü Ü Ü Ü Ü Ü Ü Ü Ü Ü Ü ÜÝ

(3.12)

Refer to Figure 3.4. Kinetic energy is converted into pressure energy when a flowing fluid is brought to rest isentropically. p0 p

J

È T0 Ø J 1 ÉÊ T ÙÚ

(3.13)

164

Fundamentals of Turbomachinery

Figure 3.4

Stagnation pressure (p0) and static pressure (p).

Substituting Eq. (3.12) in (3.13), J

p0 p

Ë (J  1) M 2 Û J 1  1 Ì Ü 2 ÍÌ ÝÜ

(3.14)

When the pressure changes are small (r » constant), 1 UV 2 2 gc

p

p0

(3.14a)

3.4.6 Stagnation Density (r0) 1

1

Ë T0 Û J 1 Ì Ü Í T1 Ý

U0 U

Ë p0 Û J Ì pÜ Í Ý

U0 U

Ë (J  1) M 2 Û J 1 Ì1  Ü 2 ÌÍ ÜÝ

È ÉÊ' U0

p0 Ø RT0 ÙÚ

1

(3.15)

3.4.7 Stagnation Velocity of Sound (a0) a0

R

a0

J RT0

c p (J  1) J

(J  1)c pT0 (J  1)h0

(3.16)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

165

3.5 Compression Process We have SFEE as u1 

V12 gz1  q 2 gc gc

u2 

V22 gz1  W 2 gc gc

W = Wf + WS Wf = flow work = (p2v2 – p1v1) WS = shaft work = useful work u1  p1v1 

V12 gz1  q gc 2 gc

u2  p2v2 

V12 gz1  q 2 gc gc

h2 

Û Ü Ü Ý

(3.17)

(3.18)

V22 gz2   WS gc 2 gc

For perfect gases, u + pv = h h1 

V22 gz2   WS 2 gc gc

Usually P.E. due to position is neglected. \ z1 = z2 \

h1 

V12 q 2 gc

\

WS

h2 

V22  WS 2 gc

( h1  h2 ) 

(V12  V22 ) q 2 gc

(3.19)

where h1 and h2 are the values of static enthalpy. Equation (3.19) gives the shaft work, i.e. the energy transfer between the shaft (rotor) and fluid. This is equal to Euler turbine equation. \

h1  h2 

WS

(V12  V22 ) q 2 gc

(Vw1u1  Vw 2 u2 ) gc

Ë V12 Û Ë V22 Û Ì h1  Ü  Ì h2  Üq 2 gc ÜÝ ÌÍ 2 gc ÜÝ ÌÍ

For perfect gases, WS

Ë V12 Û Ë V22 Û c T c T    Ì p 1 Ü Ì p 2 Üq 2 gc ÝÜ ÍÌ 2 gc ÝÜ ÍÌ

From the concepts of total temperature (Eq. (3.11), WS

(c pT01  c pT02 )  q

Vw1u1  Vw 2 u2 gc

(3.20)

166

Fundamentals of Turbomachinery

3.5.1 Isentropic Efficiency or Adiabatic Efficiency or Isothermal Efficiency or Compression Efficiency Figure 3.5 shows the compression process in compressors and pumps. Figure 3.6 shows the T–s diagram for a compression process.

Figure 3.5

Compression process in compressors and pumps. p02

T or h 02

(T02¢, h02¢) 02¢

T2

p2

T2¢ Wa

iab

ati

c

2

p01

Ad

2¢

Isentropic

Wisen

) (T 2¢, h 2¢

01 T1

p1

1 s

Figure 3.6

Point 1 01 Point 2 02

= = = =

Ideal and actual compression processes in a stage.

initial state of the fluid (static) stagnation point corresponding to initial state exit state of the fluid (static) stagnation point corresponding to exit state Superscript ( ¢ ) dash = ideal conditions Wisen = isentropic work transfer Wa = actual work transfer

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

V12 ; T2 2c p gc

T1

V22 ; T2 2c p gc

„

Wa = h02 – h01 = cp(T02 – T01)

167

(V2 )2 2c p gc

(3.21)

(for a perfect gas)

(3.22)

„

Ideal work can be calculated by any one of the following: Wisen t-t = h02¢ – h01 ;

Wisen t-s = h2¢ – h01

Wisen s-t = h02¢ – h1 ;

Wisen s-s = h2¢ – h1

(3.23)

Efficiencies can be calculated as follows: hct-t = isentropic compression efficiency based on total to total (h02  h01 ) (h02  h01 )

(T02  T01 ) (T02  T01 )

„

„

(3.24a)

hct-s = isentropic compression efficiency based on total to static (h2  h01 ) (h02  h01 ) „

(3.24b)

hcs-t = isentropic compression efficiency based on static to total ( h02  h1 ) ( h02  h01 ) „

(3.24c)

hcs-s = isentropic compression efficiency based on static to static ( h2  h1 ) (h02  h01 ) „

(3.24d)

We know that p02 p  02 p01 p01

pR 0 = stagnation pressure ratio

„

p2

„

p1



p2 p1

(3.25)

pR = static pressure ratio J

pR 0 T02

T01 > pR 0 @

„

J 1

„

T02  T01 „

Ë T02 Û J 1 Ì Ü Í T01 Ý

p02 p01

T01 > pR 0 @

J

J

1 J

 T01

J 1 Ë Û T01 Ì> pR 0 @ J  1Ü Í Ý

(3.26)

168

Fundamentals of Turbomachinery

Substituting Eqs. (3.22) and (3.26) in (3.24a),

Kct t

J 1 Ë Û T01 Ì> pR 0 @ J  1Ü Í Ý T02  T01

J 1 Ë Û T01 Ì> pR 0 @ J  1Ü c p Í Ý (T02  T01 ) c p

J 1 Ë Û T01 Ì> pR 0 @ J  1Ü Í Ý –c p Wa

J 1 Ë Û T01 Ì> pR 0 @ J  1Ü c p Í Ý Kct t

(3.27)

(3.27a)

Equation (3.27) is used in compressor stages where the gas velocities at entry and exit are significant and where T1 and T2 cannot be ignored. Gas velocities at entry and exit of a stage are almost same and can be neglected. Then, Wa = (h02 – h01) = (h2 – h1) = cp (T2 – T1) Wisen s-s = h2  h1

(3.28)

c p (T2  T1 )

„

„

Substituting Eq. (3.28) in (3.24d), c p (T2  T1 )

Kcs  s

„

c p (T2  T1 )

(T2  T1 ) (T2  T1 ) „

(3.29)

Similar to Eq. (3.26), we can write (T2  T1 ) „

J 1 Ë Û T1 Ì> pR @ J  1Ü Í Ý

(3.29a)

Substituting the above equation in (3.29),

Kcs  s

J 1 Ë Û T1 Ì> pR @ J  1Ü Í Ý T2  T1

(3.30)

From Eqs. (3.28) and (3.30), we have

Wa

c p (T2  T1 )

J 1 Ë Û T1c p Ì> pR @ J  1Ü Í Ý Kcs  s

(3.31)

3.5.2 Overall Isentropic Efficiency, Stage Efficiency, Comparison and Relation between Overall Efficiency and Stage Efficiency In Figure 3.7, the line 1–2¢ represents isentropic compression from pressure p1 to p2. The line 1–2 represents actual compression from pressure p1 to p2.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes T(K+1) 2

T or h

169

p2

F

T(K+1)¢

DWisen3

2¢

E T3

DWisen2

T3¢

pB

Wisen

T2¢ A

pA C T2

DWisen1 1

Wa

DWa2

DWa1

K = number of stages

D

B

DWa3

p1

T1 s

Figure 3.7

Effect of preheating on compression.

Each stage is known as finite stage (a compressor stage with a finite pressure rise is known as a finite stage). Consider a multistage compressor as shown in Figure 3.7 operating between pressures p1 and p2. This is split into three stages of equal pressure ratios. Let

hcs-s = hco = overall isentropic compression efficiency between pressure limits of p1 to p2. hs = stage efficiency (totally 3 stages) (isentropic efficiency) = hs1 = hs2 = hs3 ps = stage pressure ratio = ps1(p1 to pA) = ps2(pA to pB) = ps3(pB to p2) = pA/p1 = pB/pA = p2/pB pR = static-to-static overall pressure ratio between p1 to p2 (considered as a single stage). = p2/p1 pR0 = total-to-total (stagnation) overall pressure ratio between p01 to p02 (considered as a single stage) = p02/p01 DWisen = isentropic work in each stage = DWisen1 = DWisen2 = DWisen3

170

Fundamentals of Turbomachinery

Wa = actual work developed considered as a single stage between p1 to p2 Wisen = isentropic work developed considered as a single stage between p1 to p2 Wa1 = Wa2 = Wa3 = actual work developed in each stage (considered as a multistage compression) Wa1

'Wisen Ks

p1 to p2 = p1 to pA (1st stage), pA to pB (2nd stage), pB to p2 (3rd stage) The working medium leaves the 1st stage and enters the 2nd stage at C. Isentropic compression in the 2nd stage is represented by CD and actual compression by CE. The temperature difference between the ideal process (TD – TC) of 2nd stage is greater than, if the compression would have taken place between A and B, i.e. process between A and B. This is due to the losses which appear as enthalpy, in turn to temperature. \

(TD – TC) > (TB – TA)

Therefore, the isentropic work of the 2nd stage is more. This is due to the inefficiency of the previous stage. This effect is called preheat. Hence, isentropic work as a single stage (between p1 to p2) is smaller than the sum of isentropic work of individual stages (p1 to pA, pA to pB, pB to p2). Hence, we can say that, “in a multistage compressor, each succeeding stage is penalized by the inefficiency of the previous stage.” Let us consider only one stage (i.e. 1st stage), i.e. between the pressure limit p1 to pA. \

Ks1

Kc1

'Wisen1 'Wa1

Process 1–A Process 1–C

DWa1 = actual work required for one stage Similarly,

Ks

Ks 2

Kc 2

'Wisen2 'Wa 2

Process C –D Process C –E

Ks 3

Kc 3

'Wisen3 'Wa3

Process E –F Process E –2

3

Ç Ksi

i 1

'Wisen1 'Wisen2 'Wisen3   'Wa1 'Wa2 'Wa3 6'Wisen 6'Wa

6'Wisen Wa

6'Wisen Process 1–2

(3.32)

Now consider a single stage instead of three stages, Kcs  s

Wisen Wa

Process 1–2 „ Process 1–2

(3.33)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

171

From Eqs. (3.32) and (3.33), we have Wa

6'Wisen Ks

(3.34)

Wa

Wisen Kcs  s

(3.35)

Equating Eqs. (3.34) and (3.35), we have

We know that

Wisen Kcs  s

6'Wisen Ks

Kcs  s Ks

Wisen 6'Wisen

(3.35a)

therefore, hs > hcs-s

(3.35b)

SDWisen > Wisen ;

Thus, for compression process the overall isentropic efficiency is less than the stage efficiency.

3.5.3 Polytropic Efficiency or Infinitesimal Stage Efficiency (hp) of Compression Process A finite stage can be considered as made up of an infinite number of small stages (infinitesimal stages). Each small stage has an efficiency, hp, called the small stage efficiency or infinitesimal stage efficiency or polytropic efficiency. One stage (p1 to pA) of Figure 3.7 is considered to correlate polytropic efficiency and stage efficiency. This is shown in Figure 3.8. Process 1–C = actual finite stage, process 1–A = ideal finite stage Process 3–4 = actual infinitesimal stage Process 3–4¢ = ideal infinitesimal stage. If a gas is compressed from p to (p + dp), and T to (T + dT), where dT is the increment of temperature for an infinitesimal stage, then, we have similar to Eq. (3.29a), (T  dT „)  T

\

Kp

J 1 Ë Û È p  dp Ø J Ì Ü  1Ü T ÌÉ Ê p ÙÚ ÌÍ ÜÝ

(3.36)

(T  dT „  T ) (T  dT  T )

(3.37)

(Q static to static)

The superscript dash ( ¢ ) represents the ideal condition. Substituting Eq. (3.36) in (3.37),

dT T

È dp Ø ÉÊ1  p ÙÚ Kp

J

1 J

1

172

Fundamentals of Turbomachinery T or h pA = pC

C A

p + dp 4

4¢

3¢

p

dT

dT¢ 3, T

p1

1 s

Figure 3.8

Infinitesimal and finite compression process.

This equation can be expanded and dropping higher order terms, we have

1 (J  1) dp – Kp J p

dT T

Integrating between 1 and C, and taking g as constant, ln

or

TC T1

1 (J  1) pC Kp J p1

(3.38)

Kp

p (J  1) ln C J p1 TC ln T1

(3.39)

Equation (3.38) can be written as

TC T1 H

where

Now,

Ë pA Û Ì Ü Í p1 Ý

J

1 J

–

1

Kp

1 (J 1)

Ë pC Û Kp Ì Ü Í p1 Ý

J

H

Ë pA Û K p Ì Ü Í p1 Ý

J 1 J

Ë pC Û Ì Ü Í p1 Ý

n 1 n

(Q pA = pC)

H

> ps @K

p

(3.39a)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

\

Kp

or

n

173

(J  1) n – J (n  1) J Kp

(3.39b)

1  J (1  K p )

The actual temperature rise in the stage is H

È p Ø Kp T1 É C Ù  1 Êp Ú

TC  T1

(3.40)

1

From Eq. (3.31), we can write ËÈ p Ø H Û T1 Ì É C Ù  1Ü ÌÍ Ê p1 Ú ÜÝ Ks

TC  T1

(3.41)

The efficiency of a finite compression stage can be related to the small stage efficiency (infinitesimal stage efficiency). Equating Eqs. (3.40) and (3.41), we get

T1 Ks

ËÈ p Ø H Û Ì É A Ù  1Ü Ì Ê p1 Ú Ü Í Ý

H Ë Û Ì È pA Ø K p Ü T1 Ì É Ù  1Ü Êp Ú ÌÍ 1 ÜÝ H

or

Ks

È pA Ø ÉÊ p ÙÚ  1 1

(3.42)

H

È pA Ø Kp ÉÊ p ÙÚ  1 1

For multistage compression, hs is replaced by the overall efficiency hcs-s of the compressor and pR by the overall pressure ratio pRO. J

\

Kcs  s

1

 pR J

H

 pR K

p

1 1

J

; Kct t

1

 pR 0 J

H

 pR 0 K

p

1 1

(3.42b)

174

Fundamentals of Turbomachinery

3.5.4 Constant Stage Pressure Ratio Let there be K stages. \

ps

pA p1

pB pA

pK 1 pK

"

C = stage pressure ratio per stage pK 1 p1

pR = overall pressure ratio =

psK

(3.42c)

Substituting pR = psK in Eq. (3.42b), J

Kcs  s

ps

1 J

H K

K

1

(3.42d)

K

ps p

1

For the first stage (Figure 3.7), applying Eq. (3.39a), 'T1

T2  T1

ËT Û T1 Ì 2  1Ü Í T1 Ý

Ë H Û T1 Ì Kp Ü  1 p Í s Ý

CT1

Ë H Û Ì p K p  1Ü Í s Ý DT2 = T3 – T2 = T2C = C(T1 + CT1) = CT1(1 + C) C

where Similarly,

DT3 = C(1 + C)2T1 \

DTK = CT1(1 + C)K–1 (DT)T = total temperature rise (overall actual temperature rise) = DT1 + DT2 + DT3 + ... + DTK K

Ç CT1 (1  C)i 1

[(1  C)K  1] T1

(3.42e)

i 1

Substituting

C

( 'T ) T

or

( 'T ) T

Ë H Û Ì p K p  1Ü in Eq. (3.42e), Í s Ý K ÎË Þ H Û ÑÌ Ñ Kp Ü Ï 1  ps  1  1ß T1 Ü Ñ ÌÌ Ñ ÝÜ ÐÍ à

Ë HK Û Ì p K p  1Ü T1 Í s Ý

(3.42f)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

175

Equation (3.42f) gives the overall temperature rise (DT)T in the machine. Similar to Section 3.5.3, we can prove that hs < hp. Applying Eq. (3.41) to the 1st stage of Figure 3.7, Stage 1 \

'T1

where

a

(T2  T1 )

T1 ( psH  1) Ks

psH  1 ; H Ks

T1a

J 1 J

Stage 2 'T2

(T3  T2 )

T2 ( psH  1) Ks

aT2

DT2 = a(T1 + T2 – T1) = a(T1 + DT1) = a(T1 + aT1) = aT1(1 + a) Stage 3 'T3

(T4  T3 )

T3 ( psH  1) Ks

aT3

= a(T2 + T3 – T2) = a(T2 + DT2) = a(T1 + T2 – T1 + T3 – T2) = a(T1 + DT1 + DT2) = a[T1 + aT1 + aT1(1 + a)] = a[T1(1 + a) + aT1(1 + a)] or

DT3 = aT1(1 + a + a + a2) = aT1(1 + 2a + a2) = aT1(1 + a)2

Stage K DTK = aT1(1 + a)K–1 (DT)T = total temperature rise (overall actual temperature rise) = DT1 + DT2 + DT3 + ... + DTK K

Ç 'T

aT1 [1  (1  a)  (1  a)2  ...  (1  a) K 1 ]

i 1

= T1 [(1 + a)K – 1] Substituting

a

( 'T0 )T

psH  1 , Ks K ËÈ Û H Ø  p 1 s Ü  T1 Ì É 1  1 ÌÊ Ü Ks ÙÚ Í Ý

Overall isentropic efficiency (hcs–s) between the state point 1 to K + 1, i.e. for K stages,

176

Fundamentals of Turbomachinery

Kcs  s

(T( K 1) „  T1 ) isen

(T( K 1) „  T1 )isen

(T( K 1) „  T1 )isen

K

(TK 1  T1 ) a

( 'T )T

Ç 'T

(3.42g)

i 1

Applying isentropic relation between the state point 1 and K + 1, Ë T( K 1) „ Û Ì Ü Í T1 Ý

Ë pK 1 Û Ì Ü Í p1 Ý

H

From Eq. (3.42c), Ë pK 1 Û Ì Ü Í p1 Ý

H

psK H

Substituting the above relation in Eq. (3.42g),

Kcs  s

T1 ( psK H  1) ËÈ Û pH  1 Ø T1 Ì É1  s 1  Ü Ks ÙÚ ÌÍ Ê ÜÝ

psK H  1 È Ø 1 H ÉÊ1  K ( ps  1)ÙÚ s

(3.42h)

K

1

3.5.5 Preheat Factor (PF) From Eq. (3.35b), we have overall isentropic efficiency less than the stage efficiency. This is due to the pre-reheating. The pre-reheating is only an internal phenomenon and the compression process still remains adiabatic. Equation (3.35a) can be written as PF

Kcs  s Ks

Wisen 6'Wisen

(3.42i)

We know that Wisen < SDWisen, hence PF is always less than 1.

3.6 EXPANSION PROCESS 3.6.1 Isentropic Efficiency or Adiabatic Efficiency or Expansion Efficiency Figure 3.9 shows a steady state steady flow process through a turbine and Figure 3.10 shows the T-s diagram for an expansion process. Point 1 01 Point 2 02

= = = =

initial state of the fluid (static) stagnation point corresponding to initial state exit state of the fluid (static) stagnation point corresponding to exit state

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

177

Figure 3.9 Expansion process in turbine. p01 T or h 01

p1

T1

Wisen

Isentropic

1 Ad

ia

ba

Wa tic

02 T2

02¢

p02 = p02¢

p2 = p2¢ 2

T2¢ 2¢

s

Figure 3.10

Ideal and actual expansion processes in a stage.

Superscript dash ( ¢ ) = ideal conditions Wisen = isentropic work transfer Wa = actual work transfer or that developed by the shaft

T1

V12 ; T2 2c p gc

V22 ; T2 2c p gc

(V2 )2 2c p gc „

„

Wa = (h02 – h01) = cp(T01 – T02)

(for perfect gases)

(3.43)

178

Fundamentals of Turbomachinery

Ideal work can be calculated by any one of the following: Wisen t t

h01  h02 ; Wisen t

Wisen s t

h1  h02 ; Wisen s

„

„





s

s

h01  h2 h1  h2

„

(3.44)

„

Efficiency can be calculated as follows: ht t–t = isentropic expansion efficiency based on total to total (overall) c p (T01  T02 )

( h01  h02 ) (h01  h02 )

(3.45a)

c p (T01  T02 )

„

„

ht t–s = isentropic expansion efficiency based on total to static c p (T01  T02 )

(h01  h02 ) ( h01  h2 )

Wa Wisen

c p (T01  T2 )

„

„

(3.45b)

ht s–t = isentropic expansion efficiency based on static to total ( h01  h02 ) ( h1  h02 ) = isentropic expansion efficiency based on static to static

(3.45c)

„

ht s–s

( h01  h02 ) (h1  h2 )

(3.45d)

„

We know that J

pR 0

Ë T01 Û J  1 = stagnation pressure ratio Ì Ü Í T02 Ý

p01 p01  p02 p02 „

pR

„

p1 p  1 = static pressure ratio p2 p2 „



J

1

T02 T01

pR 0 J

or

T02

„



J

1

T01 pR 0 J

„

T01  T02

„



J 1

T01  T01 pR 0

J

J 1 Ë Û  T01 Ì1  pR 0J Ü Ì Ü Í Ý

(3.46)

Substituting Eq. (3.46) in (3.45a),

Kt t t

(T01  T02 ) c p È J 1 Ø Û Ë É Ù Ê J ÚÜ Ì T01 1  pR 0 c Ì Ü p ÌÍ ÝÜ

Wa Wisen

(3.47)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

179

From Eqs. (3.43) and (3.47), we have c p (T01  T02 )

Wa

J 1 Ë Û  c p Kt t t T01 Ì1  pR 0 J Ü Ì Ü Í Ý

(3.48)

Equation (3.47) is used in aircraft turbine, which exhausts into the main propulsion, i.e. nozzle. Here, kinetic energy is part of the energy supplied at the entry of the propulsion nozzle. Fluid velocities at entry and exit of a stage are almost same and can be neglected. Then, J

pR

p01 p2

„

„

or or

 pR „

È J 1Ø É Ù Ê J Ú

T2„ T01  T2„

Ë T01 Û J 1 = pressure ratio between total to static Ì Ü Í T2 Ý

(3.49)

„

T2„ T01

T01  pR „

È J 1 Ø É Ù Ê J Ú

È J 1 Ø Û Ë  ÌT01  T01  pR „ ÊÉ J ÚÙ Ü ÍÌ ÝÜ

È J 1 Ø Û Ë  T01 Ì1   pR „ ÊÉ J ÚÙ Ü ÍÌ ÝÜ

(3.50)

Substituting Eq. (3.50) in (3.45b), Kt t  s

(T01  T02 ) c p Ë T01 Ì1   pR „ ÍÌ

È J 1 Ø É Ù Ê J Ú

Û Ü cp ÝÜ

Wa Wisen

(3.51)

From Eqs. (3.48) and (3.51), we can write Wa

c p (T01  T02 )

È J 1 Ø Û Ë  c p Kt t  s T01 Ì1   pR „ ÊÉ J ÚÙ Ü ÍÌ ÝÜ

(3.51a)

Comparing Eqs. (3.47) and (3.51), we can say that ht t–t > ht t–s

3.6.2 Overall Isentropic Efficiency, Stage Efficiency and Comparison and Relation between Stage Efficiency and Overall Efficiency for Expansion Process Referring to Figure 3.11, we have Process 1–2¢ = isentropic expansion from p1 to p2 Process 1–2 = actual expansion from p1 to p2

(3.52)

180

Fundamentals of Turbomachinery T or h

DWisen1

pA

C

Wisen

B

DWa2

DWisen2

A

DWa1

p1

1

E

DWa3

D DWisen3

Wa

pB

p2

2

F 2¢ s

Figure 3.11

Effect of reheating on expansion.

The working medium leaves the 1st stage and enters the 2nd stage at C. Isentropic expansion in the 2nd stage is represented by CD and actual expansion by CE. The temperature difference between the ideal process (TC – TD) of 2nd stage is greater than, if the compression would have taken place as a single stage between PA to PB, i.e. process between A and B. \

(TC – TD) < (TA – TB)

Hence, the isentropic work of the 2nd stage is less. This is due to the inefficiency of the previous stage. This effect is called ‘Reheat’. Hence the isentropic work as a single stage (between p1 to p2) is greater than the sum of isentropic work of individual stages (p1 to pA, pA to pB, pB to p2). Hence, we can say that “in a multistage turbine each succeeding stage is benefitted by the inefficiency of the previous stage.” The multistage compressor as shown in Figure 3.11 is operating between p1 and p2. This is split into three stages of equal pressure ratios. Let hts-s = hto = overall isentropic expansion efficiency between pressure limits of p1 to p2 hs = stage efficiency (isentropic efficiency) (a total of three stages) = hs1 = hs2 = hs3 ps = stage pressure ratio = ps1(p1 to pA) = ps2(pA to pB) = ps3(pB to p2) p1 pA

pA pB

pB p2

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

181

pR = static-to-static overall pressure ratio between p1 to p2 (considered as a single stage) = p2/p1 pR0 = total-to-total (stagnation) overall pressure ratio between p01 to p02 (considered as a single stage) = p01/p02 DWisen = isentropic work in each stage (considered as a multistage) = DWisen1 = DWisen2 = DWisen3 Wa = actual work developed considered as a single stage (between p1 to p2) Wisen = isentropic work developed (considered as a single stage) Wa1 = Wa2 = Wa3 = actual work developed in each stage (considered as a multistage expansion) = hsDWisen Let us consider only one stage (i.e. 1st stage), i.e. pressure limit between p1 to pA. \

Ks1

'Wa1 'Wisen1

Kt1

Process 1–C Process 1–A

DWa = actual work required for one stage Similarly, Ks 2

Kt 2

'Wa 2 'Wisen2

Process C –E Process C –D

Ks 3

Kt 3

'Wa3 'Wisen3

Process E –2 Process E –F

Ks

3

Ç Ksi i 1

6'Wa 6'Wisen

'Wa1 'Wa 2 'Wa3   'Wisen1 'Wisen2 'Wisen2 Wa 6'Wisen

Sum of efficiencies of three stages

Process 1–2 6'Wisen

(3.53)

Now consider a single stage instead of three stages (i.e. 1–2 and 1–2¢) Kto

Wa Wisen

Process 1–2 Process 1–2 „

(3.54)

From Eqs. (3.53) and (3.54), we have Wa = hs SDWisen

(3.55)

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Fundamentals of Turbomachinery

Wa = htoWisen

and

(3.56)

Equating Eqs. (3.55) and (3.56), KtoWisen

Ks 6'Wisen

Kto Ks

6'Wisen Wisen

(3.57)

SDWisen > Wisen, therefore, hto > hs

(3.58)

or We know that

Thus, for the expansion process the overall isentropic turbine efficiency is greater than the stage efficiency.

3.6.3 Polytropic Efficiency or Infinitesimal Stage Efficiency (hp) of an Expansion Process A finite stage can be considered as made up of an infinite number of small stages (infinitesimal stages). Each small stage has an efficiency, hp, called the small stage efficiency or infinitesimal stage efficiency or polytropic efficiency. One stage (p1 to pA) of Figure 3.10 is considered to correlate polytropic efficiency and stage efficiency. This is shown in Figure 3.12. T or h p1 1 p T dT ¢

dT

p – dp

3

pA = pC C

A s

Figure 3.12

Infinitesimal stage efficiency (polytropic efficiency).

Let P and T be the pressure and temperature at the entry of the small stage respectively. dT and dT¢ are the actual and isentropic temperature drop respectively. Therefore, Kp

Actual temperature drop Isentropic temperature drop

dT dT „

(3.59)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

Ë ( p  dp) Û Ì p ÜÝ Í

(T  dT „) T

J

1 J

Ë dp Û Ì1  p Ü Í Ý

dT „ 1 T

183

1

J

J

Expanding the right hand side and dropping higher orders, dT „ T

1

dT „ T

(J  1) dp J p

1

or

(J  1) dp J p

(3.60)

Substituting for dT¢ from Eq. (3.59) into (3.60),

(J  1) dp J p

dT K pT dT T

or

Kp

(J  1) dp J p

Integrating the above equation between 1 and C and taking g as the constant, ln

p (J  1) ln C J p1

TC T1

Kp

Kp

TC T1 p (J  1) ln C J p1 ln

or

(3.60a)

The irreversible adiabatic (actual) expansion process can be considered a polytropic process (that is why, it is called polytropic efficiency) with index n.

TC T1 where

H

(J 1)

Ë pC Û J Ì Ü Í p1 Ý

Kp

Ë pC Û Ì Ü Í p1 Ý

HK p

Ë pC Û Ì Ü Í p1 Ý

J 1 J

Equating the indexes of Eq. (3.60b), i.e. J 1 Kp J

n 1 n

n 1 n

( pA

pC )

(3.60b)

184 \

Fundamentals of Turbomachinery

È n  1Ø È J Ø ÉÊ Ù n Ú ÉÊ J  1 ÙÚ

Kp

Ëp Û  T1 Ì C Ü Í p1 Ý

TC

J J  (J  1) K p

(3.60c)

HK p

Ëp Û (T1  TC ) T1  T1 Ì C Ü Í p1 Ý

HK p

Ë È p Ø HK p Û T1 Ì1  É C Ù Ü Ì Ê p1 Ú Ü Í Ý

(3.61)

Applying Eq. (3.61) to one stage (p1 to pA) as shown in Figure 3.12 and denoting its efficiency by hs, T1  TC È J  1Ø Û Ë  Ì È p1 Ø ÊÉ J ÚÙ Ü T1 Ì1  É Ù Ü Ì Ê pC Ú Ü Í Ý

Ks

È J 1Ø Û Ë  Ì È p1 Ø ÉÊ J ÙÚ Ü (T1  TC ) Ks T1 Ì1  É Ù Ü Ì Ê pC Ú Ü Í Ý

\

(3.62)

Equating Eqs. (3.61) and (3.62),

Ks

Èp Ø 1 É C Ù Ê p1 Ú Èp Ø 1 É 1 Ù Ê pC Ú

\

Ks

HK p

Èp Ø 1 É 1 Ù Ê pC Ú

È J  1Ø É Ù Ê J Ú

È J  1Ø É Ù Kp Ê J Ú

Èp Ø 1É 1 Ù Ê pC Ú

È J  1Ø É Ù Ê J Ú

(3.63)

HK p

1  ps

(3.64)

1  psH

For multistage expansion, hs is replaced by the overall efficiency ht s-s of the turbine and ps by the overall pressure ratio pR and pR0. \

HKp

Kt s  s

1  pR

; Kt t t 1  pRH

HKp

1  pR 0

1  pRH0

(3.65)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

185

3.6.4 Multistage Machine with Constant Stage Pressure Ratio Let there be K stages as shown in Figure 3.13. T or h

P1 1

1st

2 2¢

2nd

3 3¢

PA

e stag stag

3rd

4

PB

e

stag

PC

e

4¢

K th

e stag PK+1

(K+1) (K+1)¢ s

Figure 3.13

ps

Expansion process in the stages of a multistage turbine.

p1 pA

pA pB

pB pC

"

pK pK 1

C = stage pressure ratio

pR = overall pressure ratio (static to static) pR

p1 pK 1

psK

(3.66)

Substituting pR = psK in Eq. (3.65), \

HK p K

1  ps

Kt s  s

1  psH K

For stage 1 (Figure 3.13) apply the equation, 'T1

Ë T Û T1 Ì1  2 Ü T1 Ý Í

T1  T2

Ë È p Ø HKp Û T1 Ì1  É C Ù Ü Ì Ê p1 Ú Ü Í Ý 'T1

where

C

 HK p

T1 (1  ps

 HK p

(1  ps

)

)

CT1

Ë È p Ø [(1J ) / J ]–Kp Û Ü T1 Ì1  É 1 Ù Ì Ê pC Ú Ü Í Ý

(3.66a)

186

Fundamentals of Turbomachinery

For stage 2, DT2 = T2 – T3 = CT2 T2 = T1 – DT1 = T1 – CT1 = T1(1 – C);

\ DT2 = C(1 – C)T1

Similarly, for stage 3, DT3 = T3 – T4 = CT3 T3 = T1 – DT1 – DT2 = [T1CT1 – CT1(1 – C)] \

DT3 = C[T1 – CT1 – CT1(1 – C)] = CT1[1 – C – C(1 – C)] = CT1[(1 – C)2] DT4 = CT1[(1 – C)3] ; DTi = T1C(1 – C)i–1 DTK = T1C[(1 – C)K–1] (For Kth stage) (DT)T = total actual temperature drop in the turbine

'T1  'T2  'T3  ...  'TK

K

Ç T1C(1  C)i 1

i 1

(DT)T = [1 – (1 – Substituting C

 HK p

(1  ps

C)K]T

1

) in the above equation,  HK p K

( ' T )T

[1  (1  1  ps

( ' T )T

(1  ps

 HK p K

) ]T1

(3.67)

) T1

Equation (3.67) gives the overall temperature drop in the machine. Apply Eq. (3.62) to 1st stage of Figure 3.13. Stage 1

'T1 where

a

(T1  T2 ) T1 [1  (T2 / T1 )] KsT1 (1  psH ) (1  psH ) Ks ; H

aT1

(J  1) / J

Stage 2 'T2

T2  T3

T2Ks (1  ps H )

aT2

= a(T1 + T2 – T1) = a(T1 – DT1) = a(T1 – aT1) = aT1(1 – a) Stage 3 'T3

T3  T4

T3Ks (1  psH )

aT3

= a(T2 + T3 – T2) = a(T2 – DT2) = a(T1 + T2 – T1 + T3 – T2) = a(T1 – DT1 – DT2) = a[T1 – aT1 – aT1(1 – a)] = a[T1(1 – a) – aT1(1 – a)] = aT1(1 – a – a + a2) = aT1(1 – a)2

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

187

Stage K DTK = aT1(1 – a)K–1 (DT)T = total temperature drop (overall actual temperature drop) = DT1 + DT2 + DT3 + ... + DTK K

Ç 'T

aT1 [1  (1  a)  (1  a)2  ...  (1  a)K 1 ]

i 1

= T1[(1 – a)K – 1] Substituting a

(1  psH ) Ks in the above expression, ( 'T )T

T1 [1  [1  (1  psH ) Ks ]K ] KÛ Ë Ë È 1 Ø Û Ü Ì T1 1  Ì1  É1  H Ù Ks Ü Ì Ì Ê ps Ú ÜÝ Ü Í Í Ý K Ë Ë H Û Û  ( p 1) s T1 Ì1  Ì1  Ks Ü Ü Ì ÍÌ psH ÝÜ ÜÝ Í

(3.68)

Overall isentropic efficiency (ht s-s) between the state points 1 and K + 1 for K stages.

Kt s s

(T1  TK 1 )a (T1  T( K 1) „)isen

6'T (T1  T( K 1) „)isen

( 'T )T (T1  T( K 1) „)isen

(3.69)

Applying the isentropic process relation between the state point 1 and K + 1, T1 T( K 1)

From Eq. (3.66),

„

Ë p1 Û Ì Ü Í ( pK 1 ) Ý

Ë p1 Û Ì Ü Í pK 1 Ý

H



H

psK H

Substituting Eqs. (3.68) and (3.70) in (3.69),

Kt s  s

KÛ Ë Ë È H ps  1 Ø Û Ü Ì T1 1  Ì1  É H Ù Ks Ü Ì Ì Ê ps Ú Ü Ü Ý Ý Í Í Ë 1 Û T1 Ì1  K H Ü ps ÜÝ ÌÍ

(3.70)

188

Fundamentals of Turbomachinery

Kt s  s

\

Ë È pH  1 Ø Û 1  Ì1  É s H Ù Ks Ü ÍÌ Ê ps Ú ÝÜ È psK H  1 Ø É Ù Ê psK H Ú

K

(3.71)

3.6.5 Reheat Factor for Expansion Process (RF) Equation (3.57) can be written as RF

Kto Ks

6'Wisen Wisen

(3.72)

We know that SDWisen > Wisen, hence RF is always greater than unity.

3.7 EXAMPLES EXAMPLE 3.1 A jet of gas has the following data. Temperature = 593 K, g = 1.3, R = 469 J/kg-K, Mach number 1.2. Calculate for static and stagnation conditions (a) the velocity of sound and (b) the enthalpy. Solution: (a) Velocity of sound (static = a, stagnation = a0): J RT gc

a

\

1.3 – 469

N-m kg-m – 593 K – 1 kg-K N-s2

a = 601.29 m/s T0

Ans.

Ë (J  1) 2 Û T Ì1  M Ü 2 gc Í Ý

or

T0 = 721.088 K

\

a0

J RT0 gc

Ë 0.3 – (1.2)2 Û 593 Ì1  Ü 2 – 1 ÝÜ ÍÌ

1.3 – 469 – 721.088 – 1

663.06 m/s

Ans.

(b) Enthalpy (static = h, stagnation = h0): h

c pT

2.032 – 593 1204.97 kJ/kg

cp

J R J 1

h0

c p T0

1.3 – 0.469 0.3

Ans.

2.032 kJ/kg-K

2.032 – 721.088 1465.25 kJ/kg

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

189

EXAMPLE 3.2 A gas is stored in a reservoir at 300 K. Determine the velocity of sound in it and the stagnation enthalpy. Assume g = 1.658, and the molecular weight of the gas to be 40 kg/kg-mole. Solution: (a) Velocity of sound (a): From Eq. (3.7),

a

gc J

kg -m

R T0 M

N-s

kN-m – 300 K

– 1.658 – 8.314

2

kg - mole K – 40

1.658 – 8.314 – 300 m × N-m × 1000 40 N-s2

kg kg - mole

295.46 m/s

Ans.

(b) Stagnation enthalpy (h0): cp

J –R J 1

h0

c pT0

1.658 R – 0.658 M

0.5237

1.658 8.314 – 0.658 40

0.5237 kJ/kg-K

kJ – 300 K 157.12 kJ/kg kg K

Ans.

EXAMPLE 3.3 Air at STP and at 14 m/s is accelerated isentropically in a nozzle to 225 m/s. Find (a) the change in temperature, (b) the change in pressure, (c) the change in density, (d) the change in stagnation pressure, and (e) the change in stagnation temperature. Solution: STP air

T1 = 15 + 273 = 288 K, p1 = 101.325 kPa V2 = 225 m/s, V1 = 14 m/s

(a) Change in temperature (DT): Ws

(h1  h2 ) 

(V12  V22 ) Q 2 gc

(Eq. (3.19))

Isentropic process Q = 0, Flow through nozzle, Ws = 0 \

h1  h2

For perfect gas,

or

V22  V12 ) 2 gc

c p (T1  T2 )

(V22  V12 ) 2 gc

T1  T2

(V22  V12 ) 2 gc c p

'T

190

Fundamentals of Turbomachinery

\

(2252  142 ) 2 –1

'T

m2 s2 –

kg -m N- s

2

– 1.005

kN -m – 1000 kg -K

DT = 25.09 K

or \

Ans.

T2 = 288 – 25.09 = 262.91 K

(b) Change in pressure (Dp): J

1.4

Ë T2 Û J  1 Ë 262.91 Û 0.4 0.7269 Ì Ü Ì 288 Ü Í Ý Í T1 Ý p2 = p1 × 0.7269 = 101.325 × 0.7269 = 73.65 kPa

p2 p1 \

Dp = p1 – p2 = 101.325 – 73.65 = 27.675 kPa

Ans.

(c) Change in density (Dr): 1

1

Ë T2 Û J  1 Ë 262.91 Û 0.4 0.7962 Ì Ü Ì 288 Ü Í Ý Í T1 Ý r2 = r1 × 0.7962 = 1.226 × 0.9762 = 0.9762 kg/m3

U2 U1

Ë Ì' U1 Í

p1 RT1

101.325 0.287 – 288

Û 1.226 kg/m 3 Ü Ý

Dr = r1 – r2 = 1.226 – 0.9762 = 0.2498 kg/m3

\

Ans.

(d) Change in stagnation temperature (DT0):

From Eq. (3.11), T02

or \

T2 

V22 2 gc c p

262.91 K 

T02 = 288.1 K DT0 = T02 – T01 = 0

2252 2

m2 s2 –

kg -m N- s

2

– 1.005 – 1000

N-m kg -K

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

191

(e) Change in stagnation pressure (Dp0): J

Ë T02 Û J  1 Ì Ü Í T2 Ý

p02 p2

1.4

Ë 288 Û 0.4 Ì 262.91 Ü Í Ý

or

p02 = 101.326 kPa

\

Dp0 = p02 – p01 = 0

(Eq. (3.13))

1.3757

Ans.

EXAMPLE 3.4 The test section of a supersonic wind tunnel is having standard air (STP) at a Mach number 5. Find the condition of air in the reservoir. Solution:

Condition in the reservoir means, p0, T0 at STP, T = 288 K,

p = 101.325 kPa

(a) Stagnation temperature (temperature of the air in the reservoir) (T0): T0

Ë (J  1) 2 Û T Ì1  M Ü 2 gc Í Ý

(Eq. (3.12))

Ë 0.4(5)2 Û 288 Ì1 – Ü 1728 K 2 – 1 ÜÝ ÌÍ (b) Stagnation pressure (p0):

Ans.

1.4

J

Ë 0.4 – 52 Û 0.4 p0 101.325 Ì1  Ü 2 – 1 ÜÝ ÌÍ = 53610.02 kPa = 536.1 bar Ë (J  1) 2 Û J  1 p Ì1  M Ü 2 gc Í Ý

(Eq. (3.14)) Ans.

EXAMPLE 3.5 Adiabatic flow system of air at some point is having Mach number M = 3, total temperature 300 K, static pressure 0.6 bar, and at some other point Mach number M = 1.7. Calculate (a) the total temperature, (b) the stagnation pressure, (c) the static pressure, (d) the amount of heat transfer which caused reduction in Mach number (is this positive or negative?), and (e) the index of polytropic process. Solution:

M1 = 3, T01 = 300 K, p1 = 0.6 bar, M2 = 1.7

T02 T01

Ë (J 1 M 22 (1  J M12 )2 ÍÌ M12 (1  J M22 )2 Ë1  (J Ì Í

 1) 2 Û M2 Ü 2 Ý  1) 2 Û M1 Ü 2 Ý

Ë 0.4 – 1.72 Û (1.7)2 (1  1.4 – 32 )2 Ì1  Ü 2 Í Ý 2Û Ë 0.4 – 3 (3)2 (1  1.4 – 1.72 )2 Ì1  Ü 2 Ý Í

62.02 – 13.6 1.134 127.1592 – 5.05

192

Fundamentals of Turbomachinery

(a) The total temperature (T02): T02 = T01 × 1.314 = 300 × 1.314 = 394.08 K

Ans.

(b) The total pressure (p02): J

p02

Ë J  1 2 ÛJ 1 p2 Ì1  M2 Ü 2 Í Ý

p02 p2

0.4 Ë 2 Û 0.4 Ì1  2 – 1.7 Ü Í Ý

(Eq. (3.14))

1.4

4.94

p02 = p2 × 4.94 = 1.617 × 4.94 = 7.9885 bar

Ans.

(c) The static pressure (p2): p2 p1

\

1  J M12

1  1.4 – 32

1  J M22

1  1.4 – 1.72

2.6952

p2 = p1 × 2.6952 = 0.6 × 2.6952 = 1.617 bar

Ans.

(d) Amount of heat transfer (Q1-2): Q1-2 = cp(T02 – T01) = 1.005(394.08 – 300) = 94.55 kJ/kg

Ans.

(e) Index of polytropic process (n): J

p01

Ë (J  1) 2 Û J  1 p1 Ì1  M1 Ü 2 Í Ý 1.4

0.4 Ë Û 0.4 0.6 Ì1  – 32 Ü 2 Í Ý

22.04 bar

\ For non-flow polytropic process n

p02 p01 or

7.9885 22.04

\

ln 0.3625

\

Ë T02 Û n  1 Ì Ü Í T01 Ý

n

Ë 394.08 Û n  1 Ì 300 Ü Í Ý

n n (1.3136) 1

n

0.3625

(1.3136) n  1

n ln 1.3136 n 1

n = 0.788

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

193

EXAMPLE 3.6 Air at STP at a zero velocity expands to M = 0.8. Calculate (a) the final density and (b) the change in density. Solution:

At STP,

T0 = 15 + 273 = 288 K,

p0 RT0

U0

p0 = 101.325 kPa, g = 1.4

101.325 kN kN -m m 2 – 0.287 – 288 K kg- K

1.226 kg/m 3

(a) The final density (r): U0 U

\

1

1

Ë (J  1) 2 Û J  1 Ì1  2 M Ü Í Ý

Ë (1.4  1) Û 0.4 (0.8)2 Ü Ì1  2 Í Ý

(Eqn. (3.15))

r = 0.9072 kg/m3

Ans.

(b) The change in density (Dr): Dr = r0 – r = 1.226 – 0.9072 = 0.3188 kg/m3

Ans.

EXAMPLE 3.7 Air enters a straight axis symmetric duct at 300 K, 3.5 bar and 150 m/s and leaves it at 275 K, 2.2 bar and 270 m/s. The area of cross section at entry is 550 cm2. Assume adiabatic flow, g = 1.4, R = 287.0 J/kg-K. Calculate (a) the stagnation temperature, (b) the mass flow rate, and (c) the area of cross section at exit. Solution: J R J 1

cp

1.4 – 287 1004.5 J/kg-K 0.4

(a) Stagnation temperature (T02):

T01

T1 

V12 2c p gc

300 K 

150 2 2 – 1004.5 – 1

311.199 K

Flow is adiabatic, therefore, stagnation temperature at entry and at exit should be same, i.e., T01 = T02.

T02

T2 

V22 2c p gc

275 

270 2 2 – 1004.5 – 1

311.29 K

Ans.

(b) Mass flow rate (m): 3.5 – 100 0.287 – 300

U1

p1 RT1

m

U1 A1 V1

4.065

kg m3

4.065 kg/m 3

–

550 m 2 – 10 4 – 150 m s

33.54 kg/s

Ans.

194

Fundamentals of Turbomachinery

(c) Area of cross section at exit (A2): U1 A1 V1

U2 A2 V2

2.2 – 100 0.287 – 275

or

U2

p2 RT2

\

A2

U1 A1 V1 U2 V2

2.787 kg/m 3

4.065 – 550 – 10 4 – 150 2.787 – 270

0.04456 m 2

Ans.

EXAMPLE 3.8 An air stream flows at the rate of 1 kg/s in a duct of 10 cm diameter. The stagnation temperature is 40°C. The static pressure at one section of the duct is 0.41 bar. Calculate (a) the Mach number, (b) the velocity and (c) the stagnation pressure at this section. Solution: T0 = 40 + 273 = 313 K, p = 0.41 bar, m = 1 kg/s, d = 10 cm m = rAV;

\

p

U RT ; T

V

127.3 U

V2

m A

UV

p UR

1 S È 10 Ø 4 ÉÊ 100 ÙÚ

or

(0.9T)2 = 2009(313 – T)

\

T = 281.5 K

and

V

0.9 – T

127.3

0.41 – 10 5 0.287 – 1000 – U

127.3 – T 142.86

2J R (T0  T ) (J  1)

2

\

U

127.3 V

142.86 U

0.9T 2 – 1.4 – 0.287 – 1000 – (313  T ) 0.4

0.9 – 281.5

(Eqn. (3.12))

253.55 m/s

(a) Mach Number (M): T0 T 313 281.5

(J  1) 2 Û Ë Ì1  2 M Ü Í Ý 1

0.4 2 M ? M 2

0.75

Ans.

(b) Velocity (V): V = 255.35 m/s

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

195

(c) Stagnation pressure (p0): p0 p

1

(J  1) 2 M 2

1

0.4 – (0.75)2 2

0.4561 bar

Ans.

EXAMPLE 3.9 An air stream at 375 K has sonic velocity. Calculate (a) the velocity of sound at 375 K, (b) the velocity of sound at the stagnation condition, and (c) the stagnation enthalpy. Assume g = 1.4, R = 0.287 kJ/kg-K Solution: (a) Velocity of sound at 375 K (a): From Eq. (3.6),

J RTgc

a

\

kg -m kN -m – 1000 – 375 K – kg -K N-s2

1.4 – 0.287

a = 388.17 m/s

Ans.

(b) Velocity of sound at the stagnation condition (a0): Given data: Air jet velocity (fluid velocity) = Sonic velocity (velocity of sound) \

V = a = 388.17 m/s a2 1  V2 J 1 2

or

a02 J 1

a2 1  a2 (J  1) 2

a02 (J  1)

a2 (2  J  1) 2

a02

a2 (J  1) 2

or

a02

\

a0 = 425.22 m/s

(Here V = velocity) a2 2

Ë 2 Û Ì J  1  1Ü Í Ý

(388.17)2 – 2.4 2

Ans.

(c) Stagnation enthalpy (h0): From Eq. (3.16), h0

a02 (J  1) gc

(425.22)2 m 2 N- s2 0.4 – 1 s2 kg-m

452030.12 J/kg

Ans.

196

Fundamentals of Turbomachinery

EXAMPLE 3.10 An air stream at p = 1.00 bar, T = 350 K and velocity = 400 m/s is brought to rest, (a) adiabatically, (b) isentropically. Calculate the stagnation pressure and temperature in the two cases. Solution: (a) Stagnation temperature (T0): From Eq. (3.11),

T0

\

T

2

V 2c p gc

m2 s2 350 K  kN-m N-s2 – 100 2 – 1.005 kg-K kg-m 400 2

T0 = 429.6 K (For both adiabatic and isentropic)

Ans.

(b) Stagnation pressure (p0) for isentropic process: J

p0 p

Ë T0 Û J  1 ÌT Ü Í Ý p0 = p × 1.609

\

1.4

Ë 429.6 Û 0.4 Ì 350 Ü Í Ý

1.609

= 1 × 1.609 = 1.609 bar

Ans.

EXAMPLE 3.11 At the entry of a flow passage, the pressure, temperature and Mach number are 2.5 bar, 30°C, 1.5 respectively. If the exit Mach number is 2.5, calculate the following for adiabatic flow of a perfect gas: (a) stagnation temperature, (b) temperature and velocity of gas at exit, (c) the flow rate per m2 of the inlet cross section. Assume g = 1.4, R = 0.5 kJ/kg-K Solution:

T1 = 30 + 273 = 303 K, p1 = 2.5 bar,

M1 = 1.5, M2 = 2.5

(a) Stagnation temperature (T01): For adiabatic flow, T01 = T02 = T0 T01 T1

1

(J  1) 2 M1 2

1

(1.4  1) – 1.52 2

1.45

T01 = T1 × 1.45 = 303 × 1.45 = 439.35 K

Ans.

(b) Temperature (T2) and velocity of gas at exit (V2): U1

p1 RT1

2.5 – 10 2 kN

kg K

m 2 – 0.5 kN m – 303 K

a1

J RT1 gc

M1

V1 ? V1 a1

T02 T2

1

1.65 kg/m 3

1.4 – 0.5 – 303 – 1000

M1a1

(J  1) 2 M2 2

1

1.5 – 460.54 0.4 – 2.52 2

460.54 m/s

690.82 m/s 2.25

(Eq. (3.12))

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

\

T01 2.25

T2

39.35 195.27 K 2.25

J gc RT2

a2

197 Ans.

1.4 – 1 – 0.500 – 195.27 – 1000

369.7 m/s

V2 = M2a2 = 2.5 × 369.7 = 924.3 m/s

Ans.

È mØ (c) The flow rate per m2 of the inlet cross section É Ù : Ê A1 Ú

m = r1 A1 V1 m A1

U1 V1

1.65

kg m

3

– 690.82

m s

1139.85

kg/s

Ans.

m2

EXAMPLE 3.12 A stream of an ideal gas having cp = 1.000 kJ/kg-K, g = 1.4, velocity = 210 m/s. Calculate (a) the dynamic temperature. Also find (b) the dynamic pressure and (c) the kinetic pressure if the total temperature is 100°C and the static pressure is 1.2 bar. g = 1.4, cp = 1.0 kJ/kg-K, V1 = 210 m/s,

Solution:

T0 = 100 + 273 = 373 K, p1 = 1.2 bar (a) Dynamic temperature (T0 – T1):

(T0  T1 ) \

V12 2c p gc

210 2 2 – 1 – 1000 – 1

22.05 K

Ans.

T1 = T0 – 22.05 = 373 – 22.05 = 350.95 K

(b) Dynamic pressure (p0 – p1): For isentropic flow, J

1.4

Ë T0 Û J  1 Ë 373 Û 0.4 1.2377 Ì Ü Ì 350.95 Ü Í Ý Í T1 Ý p0 = p1 × 1.2377 = 1.2 × 1.2377 = 1.4853 bar p0 p1

\ \

(p0 – p1) = 1.4853 – 1.2 = 0.2853 bar

Ans.

(c) Kinetic pressure: U1 V12 2 gc

p1V12 RT1 2 gc 1.2 – 100

kN m2

– 210 2

26270.1578 N/m 2

m2 s2

N- s2

– 0.28

0.2627 bar

kN -m – 350.95 K – 2 – kg -m kg -K

Ans.

198

Fundamentals of Turbomachinery

EXAMPLE 3.13 A pitot tube placed in a wind tunnel gave the following readings: Static pressure 41 bar, stagnation pressure 100 bar, stagnation temperature 100°C. Calculate the acoustic velocity. Solution: p1 = 41 bar, p0 = 100 bar, T0 = 100 + 273 = 373 K Acoustic velocity (a):

\

T0 T

Ë p0 Û Ì Ü Í pÝ

T

T0 1.29

J

1

0.4

Ë 100 Û 1.4 Ì 41 Ü Í Ý

J

373 1.29

gc J RT

a

1.29

289.12 K 1 – 1000 – 1.4 – 0.287 – 289.12

= 340.83 m/s

Ans.

EXAMPLE 3.14 An air compressor has the following data: Inlet pressure 1.02 bar, Exit pressure = 1.5 bar, Inlet temperature = 300 K, Exit temperature = 340 K. Calculate (a) the isentropic compression efficiency and (b) the polytropic efficiency. Solution: p1 = 1.02 bar,

T2 T1

„

Ë p2 Û Ì Ü Í p1 Ý

J

1

p2 = 1.5 bar,

Ë 1.5 Û Ì 1.02 Ü Í Ý

J

T1 = 300 K,

0.2857

1.1165

T2 = 340 K

(Refer to Figure 3.6)

T2¢ = T1 × 1.1165 = 300 × 1.1165 = 334.95 K (a) Isentropic compression efficiency (hcs-s): From Eq. (3.29) and Figure 3.6, Kcs  s

T2  T1 T2  T1 „

334.95  300 340  300

0.87375 87.38%

Ans.

(b) Polytropic efficiency (infinitesimal stage efficiency) (hp): From Eq. (3.39) and Figure 3.8,

Kp

(J  1) p ln 2 J p1 T2 ln T1 = 88.04%

0.4 Ë 1.4 Û – ln Ì Ü 1.4 Í 1.02 Ý – 100 Ë 340 Û ln Ì Ü Í 300 Ý Ans.

Pressure difference is very small, hence, hcs-s and hp are very close, i.e. hcs-s » hp.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

199

EXAMPLE 3.15 Modify Example 3.14 in such a way that, p1 = 1.02 bar, T1 = 300 K, p2 = 3.0 bar, isentropic compression efficiency = 0.75. Calculate (a) the exit static temperature and (b) the infinitesimal stage efficiency. Solution: p1 = 1.02 bar, p2 = 3.0 bar, T1 = 300 K, pR

p2 p1

T2 T1

Ë p2 Û J  1 Ì Ü Í p1 Ý

3.0 1.02

hcs-s = 0.75,

(Eq. (3.25))

2.941

J

„

\

(2.941)0.2857

1.361

(Refer to Figure 3.6)

T2¢ = T1 × 1.361 = 300 × 1.361 = 408.298 K

(a) Exit static temperature (T2): From Eq. (3.29) and Figure 3.6, Kcs  s

\

0.75

T2  T1 T2  T1 „

T2  T1 408.298  300 144.4 0.75 0.75 T2 = T1 + 144.4 = 300 + 144.4 = 444.4 K

T2  T1

\

T2 T1

Now,

„

444.4 300

Ans.

1.48

(b) Infinitesimal efficiency (hp) From Eq. (3.39) and Figure 3.8,

Kp

(J  1) p ln 2 J p1 T ln 2 T1

0.4 – ln 2.941 1.4 – 100 ln 1.48

78.44%

Ans.

In this case, the pressure difference is high. Hence the infinitesimal stage efficiency (polytropic efficiency, hp) is higher than the isentropic compression efficiency, hcs-s. This is due to preheating of the fluid. EXAMPLE 3.16 An air compressor has six stages of equal pressure ratio 1.4. The mass flow rate is 45 kg/s. The overall isentropic compression efficiency is 84%. Entry pressure is 1 bar and T1 = 40°C. Calculate (a) the state of the air at the exit, (b) polytropic efficiency, (c) each stage efficiency, (d) power required to drive the compressor (overall efficiency of drive = 0.9). Assume g = 1.4, R = 0.287 kJ/kg-K, cp = 1.005 kJ/kg-K.

200

Fundamentals of Turbomachinery

Solution:

(a) The state of the air at the exit (pe = pK+1), (TK+1 = Te): ps = each stage pressure ratio = 1.4, K = No. of stages = 6 pR = overall pressure ratio = psK pe ? pe p1

pR

\

pR – p1

(1.4)6

7.5295

7.5295 – 1.0

pe = 7.5295 bar T( K 1) „

Ë p( K 1)„ Û Ì Ü Í p1 Ý

T1

(J 1) / J

7.52950.2857

1.78

T(K+1)¢ = T1 × 1.78 = 313 × 1.78 = 557.14 K hcs-s = overall compressor efficiency T( K 1)  T1

Ë ÌH Í

„

TK

\

1



 T1

(T( K 1) „  T1 )

TK 1

Kcs  s

 T1

J 1 J

0.4 1.4

557.14  313  313 603.64 K 0.84

( 'T )T

T1 ( pRH  1) Kcs  s

or

( ' T )T

313 (7.52950.2857  1) 0.84

\

(DT)T = TK+1 – T1 = Te – T1 = 290.75 K

\

Û 0.2857 Ü Ý

290.75 K

Te = 290.75 + T1 = 290.75 + 313 = 603.75 K

Ans.

(b) Polytropic efficiency (hp): From Eq. (3.39),

Kp

p (J  1) ln e p1 J Te ln T1

(c) Efficiency of each stage (hs): From Eq. (3.42),

0.4 7.5295 – ln 1.4 1 603.75 ln 313

0.87799 87.8%

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes J

1 J

ps

Ks

201

1

H K

ps p  1

1.40.2857  1 0.2857 1.4 0.878

0.8721 87.21%

1

(d) Power required to drive the compressor (P):

mc p ( 'T )T

P

Kco

45 – 1.005 – 290.75 0.84

15653.77 kW

EXAMPLE 3.17 Example 3.15 is modified in such a way that each stage has the same temperature rise. Calculate (a) the pressure ratio and (b) the efficiency of each stage. Solution:

DTi = pressure rise in the ith stage = pressure rise in each stage

H Kp Kp

J J 1

('T )T K

290.75 6

48.46 K

0.286 0.878

0.326 ; H

J 1 J

0.878 –

1.4 0.4

0.4 1.4

0.286

3.073 3.073

Stage No.

Ti

1 2 3 4 5 6

313.00 361.46 409.92 458.38 506.84 555.3

Ë Ti  1 Û Ì Ü Í Ti Ý [Eq. (3.39a)]

ps

1.5564 1.472 1.4137 1.3618 1.3239 1.0956

Ks

( ps0.286  1)

( ps0.326  1) (Eq. (3.42))

0.8694 0.8704 0.8711 0.8718 0.8723 0.8758

EXAMPLE 3.18 A two-stage gas turbine develops 22 MW at the shaft. The inlet temperature is 1500 K. The overall pressure ratio of the turbine is 8 and the isentropic expansion efficiency 0.9. Assume that the pressure ratio of each stage is same. Calculate (a) the pressure ratio of each stage, (b) polytropic efficiency, (c) the mass flow rate, and (d) the efficiency and power of each stage. Assume, g = 1.4, cp = 1.005 kJ/kg-K, overall drive efficiency = 0.90 Solution: Shaft power (S.P.) = 22 MW T1 = 1500 K pR = 8 ht = 0.9 = ht s–s

202

Fundamentals of Turbomachinery

T or h

ps = C (each stage) K = 2 = No. of stages

p1 1

(a) Pressure ratio of each stage (ps):

\

1st stage pA

pR

psK

ps

p1R/ K

81 / 2

2.8284

Ans.

2nd stage pB

(b) Polytropic efficiency (hp): 2

T1  T2 T1  T2

Kt s  s

2¢

„

s

Similar to Eq. (3.47), we can write

Kt s  s

'T or \

T1  T2 (J  1) Û Ë  T1 Ì J ÜÝ Í1  pR

(T1  T2 ) Kt s  s T1 (1  pR0.286 )

0.9 – 1500 – (1  8 0.286 )

DT = 605.18 K T2 = T1 – 605.18 = 1500 – 605.18 = 894.82 K

T1 T2

1500 894.82

p1 p2

8 1

Kp

T2 T1 p (J  1) ln 2 p1 J

Kp

3.5 – ln 1.6763 ln 8

1.6763

8

ln

\

= overall drive efficiency

T J ln 1 J  1 T2 p ln 1 p2

(Eq. (3.60a))

0.8695 86.95%

Ans.

(c) The mass flow rate (m): Overall drive efficiency =

S.P. P

Now

22 – 1000 24444.4 kW 0.90 P = mcpDT = m × 1.005 × 605.18 = 24444.4 kW

\

m = 32.32 kg/s

\

P

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

203

(d) The efficiency (hs) and power (P) of each stage: (1  J )

1  ps J

Ks

Kp

(1  J )

1  ps J

Pressure ratio is same in both the stages hence the efficiency is also the same. 1J Kp J

1  1.4 0.8695 4

1J J Ks1

\

1  1.4 1.4

 0.2484

 0.2857

(1  2.8284  0.2484 )

Ks 2

(1  2.8284  0.2857 )

0.8857

88.57%

Ans.

The actual temperature drop in each stage. J

T1 TA \ \

1

J

Ë p1 Û J Ì Ü Í pA Ý

ps

1

J

Kp

(2.8284)0.286 – 0.8695

1.2951

(Eq. (3.60b))

T1 1500 1158.25 K 1.2951 1.2951 DT1 = temperature drop in first stage TA

= T1 – TA = 1500 – 1158.25 = 341.75 K Similarly, J

TA T2 \ \

1

Ë pA Û J Ì Ü Í p2 Ý

Kp

(2.8284)0.286 – 0.8695

1.2951

TA 1158.25 894.33 K 1.2951 1.2951 DT2 = TA – T2 = 1158.25 – 894.33 = 263.92 K T2

The power developed in each stage P1 = power developed in first stage = mcpDT1 = 39.32 × 1.005 × 341.75 = 13504.798 kW and

P2 = mcpDT2 = 39.32 × 1.005 × 263.92 = 10429.22 kW

Ans. Ans.

EXAMPLE 3.19 Reconsider Example 3.18 in such a way that equal power is developed in both the stages. Calculate the pressure ratio and the efficiency of each stage. Also calculate the overall pressure ratio.

204

Fundamentals of Turbomachinery

Solution:

Take data from Example 3.17 as follows: hp = 0.8695, P = 23913 kW, m = 39.32 kg/s, cp = 1.005 kJ/kg-K, P1 + P2 = 23913 kW

Equal power in both the stages, i.e. P1 = P2

P1  P2 23913 11956.5 kW 2 2 P1 = mcpDT1 = mcpDT2 P2

P1

11956.5 = 39.32 × 1.005 × DT1 \

DT1 = T1 – TA = 302.5 K

\

TA = T1 – 302.57 = 1500 – 302.57 = 1197.43 K

Similarly, DT2 = TA – T2 = 302.57 \

DT2 = TA – 302.57 = 1197.43 – 302.57 = 894.86 K

p1 pA

ps1

Ë T1 Û Ì Ü Í TA Ý

J

1 J

1

–

3.5

Kp

Ë 1500 Û 0.8695 Ì 1197.43 Ü Í Ý

2.4765

Ans.

\ Similarly, 3.5

pA p2

pS 2

Ë TA Û 0.8695 Ì Ü Í T2 Ý

3.5

Ë 1197.43 Û 0.8695 Ì 894.86 Ü Í Ý

3.2298

Ans.

0.8837

Ans.

0.88765

Ans.

Efficiencies (hs2, hs2): Ks1 Ks 2

(1  ps10.2484 )

(1  2.47650.2484 )

(1  ps20.2484 )

(1  3.22980.2484 )

(1  ps102857 )

(1  ps202857 )

(1  2.47650.2857 )

(1  2.22980.2857 )

Overall pressure ratio (pR): pR = ps1 × ps2 = 2.4765 × 3.2298 = 8.0

Ans.

EXAMPLE 3.20 Air enters a compressor at a static pressure of 1.5 ata, a static temperature of 15°C and a flow velocity of 50 m/s. At the exit the static pressure is 3 ata, the static temperature 100°C and the flow velocity 100 m/s. The outlet is 2 m above the inlet. Calculate (a) the isentropic change in total enthalpy and (b) the actual change in total enthalpy. Solution: Data: p1 = 1.5 ata, T1 = 15 + 273 = 288 K, V1 = 50 m/s, p2 = 3.0 ata, T2 = 100 + 273 = 373 K, V2 = 100 m/s,

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

T or h

205

p 02¢ = p02

02 02¢

p 2 = p 2¢ 2 2¢ p01 01

p1

1 s

z2 = 2 m, z1 = 0

T02

T2 

V22 z g  2 2 gc c p gc c p

(Eq. (3.11))

N- s2 kg -K 100 2 m 2  373 K  – 2 2 s 1 – kg -m – 1.005 kN -m – 1000 2 2 m – 9.81 m N- s kg m K 1.005 s2 – kg -m kN -m – 1000

T01

T1 

V12 2 gc c p

288 K 

50 2 2 – 1 – 1.005 – 1000

377.995 K

289.24 K

We have for isentropic process,

Ë T1 Û Ì Ü Í T2 Ý „

\

T2

„

Ë p1 Û Ì Ü Í p2 Ý

J

1 J

351.08 K

0.4

Ë 1.5 Û 1.4 Ì 3.0 Ü Í Ý T2  „

V22 2 gc c p

351.108 

100 2 2 – 1.005 – 1000 – 1

356.06 K

(a) Isentropic change in total enthalpy (Dh0): 'h0

h02  h01 „

c p (T02  T01 ) „

(356.06  289.24) – 1.005

= 67.1492 kJ/kg

Ans.

(b) The actual change in total enthalpy (Dh0a): Dh0a = (h02 – h01) = cp(T02 – T01) = 1.005 × (377.995 – 289.24) = 89.199 kJ/kg

Ans.

206

Fundamentals of Turbomachinery

EXAMPLE 3.21 Air flows through an air turbine where its stagnation pressure is decreased in the ratio 5 : 1. The total-to-total efficiency is 0.8 and the air flow rate is 5 kg/s. The inlet total temperature is 280 K. Calculate (a) the actual power output, (b) the actual exit total temperature, (c) the actual exit static temperature if the exit flow velocity is 100 m/s, and (d) the total-to-static efficiency of the device. Solution: p01 p02

5 , Kt t  t 1

0.8, m

5 kg/s, T01

280 K, V2

100 m/s

p01

T or h 01

p1

1

02

p02

02¢ p2 2 2¢ s

For isentropic process,

T02 T01

„

\

Ë p02 Û Ì Ü Í p01 Ý

J

1 J

0.4

>[email protected]

0.631

T02¢ = T01 × 0.631 = 280 × 0.631 = 176.68 K

(a) Actual power output (Pa): Pa P1

Kt t t

 p (T01  T02 ) mc  p (T01  T02 ) mc „

 p (T01  T02 ) Kt t t mc

Pa

„

= 0.8 × 5 × 1.005 × (280 – 176.68) = 415 kW (b) The actual exit total temperature (T02): Pa = mcp(T01 – T02) = 415 kW or

T02

T01 

Pa mc p

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

280 

415 5 – 1.005

197.344 K

207 Ans.

(c) The actual static temperature (T2) if the exit flow velocity is 100 m/s:

or

T02

T2 

V22 2 gc c p

T2

T02 

V22 2 gc c p

197.344 

100 2 192.37 K 2 – 1.005 – 1000 – 1

Ans.

(d) The total-to-static efficiency of the device (ht t–s)

T02 or

„

T2

„

T2  „

V22 2 gc c p

T02  „

V22 2 gc c p

176.68 

Kt t  s

\

h01  h 02 h01  h2

„

100 2 2 – 1 – 1.005 – 1000

171.71 K

T01  T02 T01  T2 „

280  197.34 280  171.71

0.7633 or 76.33%

Ans.

EXAMPLE 3.22 A gas having a molecular weight of 4 and g equal to 5/3 expands isentropically in a turbine through a total pressure ratio of 5 to 1. The initial total temperature is 1000 K. Calculate the change in total enthalpy assuming that the gas is perfect. Solution:

M = 4, g = 5/3, p01/p02 = 5,

R

R M

kJ – kg mole K 4

8.314

RJ 2.0785 – 1.67 J 1 0.67 For isentropic process: cp

T01 T02

„

Ë p01 Û Ì Ü Í p02 Ý

J

T01 = 1000 K

1 kg kg mole

2.0785 kJ/kg-K

5.181 kJ/kg-K 1 J

0.67

>[email protected]

1.9073

208

Fundamentals of Turbomachinery p01

T or h 01

p1

1 p02

02 02¢

p2 2 2¢ s

or

T02

„

T01 1.9073

1000 1.9073

524.294 K

The change in total enthalpy (Dh0): 'h0

h01  h02

c p (T01  T02 )

„

„

5.181 – (1000  524.294)

Ans.

2464.157 kJ/kg

EXAMPLE 3.23 Air enters a blower at a total pressure of 1 atm, a total temperature of 30°C and a flow velocity of 55 m/s. At the exit the total temperature is 41.2°C and the flow velocity is 150 m/s. Calculate (a) the change in total pressure between the inlet and the exit of the blower and (b) the change in static pressure expressed in cm of water. Solution:

Data:

T or h

02

p01 = 1 atm

02¢

p2

T01 = 30 + 273 = 303 K V1 = 55 m/s

2¢

T02¢ = 41.2 + 273 = 314.2 K For isentropic process and perfect gas,

T1

T1 

01 p1

V12

1

2c p gc

T01 

V12 2c p gc

2

p01

V2 = 150 m/s

T01

p02

s

303 

2

55 2 – 1.005 – 1000

301.5 K

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

\

J

p01 p1

Ë T01 Û J Ì Ü Í T1 Ý

p1

p01 1.0175

1.4

Ë 303 Û 0.4 Ì 301.5 Ü Í Ý

1

1 atm 1.0175

Ë T02 Û J  1 Ì Ü Í T1 Ý

\

p02

p1 – 1.1554

„

T02 

p2 p1

Ë T2 Û J  1 Ì Ü Í T1 Ý

p2

p1 – 1.0175

„

3.5

\

„

Ë 303.01 Û Ì 301.5 Ü Í Ý

„

150 2 2 – 1 – 1.005 – 1000

314.2 

J

„

1.554

0.983 – 1.1554 1.1357 atm

V22 2c p gc

T2

„

Ë 314.2 Û Ì 301.5 Ü Í Ý

„

1.0175

0.983 atm

J

p02 p1

„

209

303.01 K

3.5

1.0175

0.983 – 1.0175 1 atm

(a) Change in total pressure between the inlet and exit of the blower ( p02 – p0): „

p02  p01 „

1.1357  1

Ans.

0.1357 atm

(b) The change in static pressure expressed in cm of water ( p02  p0 ): „

p2  p1 „

1  0.983

0.017 atm

0.17 m of H 2 O

Ans.

EXAMPLE 3.24 Air as a perfect gas undergoes an increase in total pressure of 180 mm of H2O during passage through a blower. The inlet static pressure is 1 atm, velocity 50 m/s and the inlet static temperature is 25°C. Evaluate (a) the exit total temperature if the process is isentropic. If the exit velocity is 135 m/s, find also (b) the exit static pressure and (c) the static temperature. Solution: U1

p1 RT1

101.325 0.287 – 298

p01

p1 

U1V12 2 gc

101.325 

T01

T1 

V12 2c p gc

298 K 

1.185 kg/m 3

1.185 – 50 2 2 – 1 – 1000

102.81 kN/m 2

50 2 2 – 1 – 1.005 – 1000

299.24 K

210

Fundamentals of Turbomachinery

T or h

p02

02 02¢

p2 p02¢ – p01 = 180 mm H 2O

2

2¢

p1 = 1 atm V1 = 50 m/s V2 = 135 m/s

p01

T1 = 25°C

01 p1 1 s

We know that 101.325 kPa = 1033.58 cm of H2O \

18 cm of H2O =

\

p02  p01

1.76459 kPa

180 mm of H 2 O 1.76459 kPa

„

\

101.325 – 18 1033.58

p02¢ = p01 + 1.76459 = 102.81 + 1.76459 = 104.57 kPa

(a) Exit total temperature (T02 ) if the process is isentropic: „

T02 T01

„

Ë p02 Û Ì Ü Í p01 Ý

J

„

1

0.4

Ë 104.57 Û 1.4 Ì 102.81 Ü Í Ý

J

1.005

T02¢ = T01 × 1.005 = 299.24 × 1.005 = 300.69 K

Ans.

(b) The exit static pressure (p2): For a small pressure change, we have

or

p02

p2 

U2V22 2 gc

p2

p02 

U2V22 2 gc

(Q r1 = r2)

104.57 

(c) The exit static temperature (T2): We know that \

r1 = r2 p1 RT1

p2 RT1

1.185 – 1352 2 – 1000

93.77 kPa

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

\

T2

p2 T1 p1

93.77 – 298 101.325

275.785 K

211 Ans.

EXAMPLE 3.25 A centrifugal pump rotates at 1000 rpm, having inlet diameter (eye diameter) 100 mm and outlet diameter 400 mm. The inlet and exit blade angles are 20° and 10° respectively. The blade depth is 60 mm. Assuming radial entry, i.e. inlet whirl velocity to be zero, and hydraulic efficiency of 90%, calculate (a) the volume flow rate, (b) the static and stagnation pressure rise across the impeller, (c) the power transferred to the fluid, and (d) the input power to the impeller. Solution:

Data: Speed: Eye diameter: Exit diameter: Inlet blade angle: Outlet blade angle: Blade depth: Radial entry: Hydraulic efficiency:

N = 1000 rpm d1 = 100 mm d2 = 400 mm b1 = 20° b2 = 10° b1 = b2 = 60 cm a1 = 90°, Vw1 = 0, V1 – Vf1 hH = hmax = 90%

To determine: Q, Dp, Dp0, P, Pi

u1

S d1 N 60

S – 0.1 – 1000 60

5.24 m/s

u2

S d2 N 60

S – 0.4 – 1000 60

20.94 m/s

From inlet velocity triangle, tan E1

\

V1 u1

V1 = u1 × tan b1 = 5.24 × tan 20° = 1.907 m/s

212

Fundamentals of Turbomachinery

(a) The volume flow through the impeller (Q): Q = p d1 b1 Vf 1 = p × 0.1 × 0.06 × 1.907 = 0.036 m3/s

Ans.

(b) The static pressure (Dp) and stagnation pressure (Dp0) rise across the impeller: S d1 b1 V f 1

Q

\

d1 Vf 1 d2

Vf 2

S d2 b2 V f 2

0.100 – 1.907 0.400

0.47675 m/s

Neglecting potential changes across the impeller, p2  p1 V22  V12  2g Ug

Hm

For an incompressible fluid the total pressure head difference is 'p0

Ë p2 V22 Û Ë p1 V12 Û   Ì ÜÌ Ü ÌÍ U g 2 g ÜÝ ÌÍ U g 2 g ÜÝ

p02  p01 Ug

Hm

From the exit velocity triangle,

tan E2 \

\ \

Vf 2

0.477 20.94  Vw 2

u2  Vw 2

tan 10’

Vw2 = 18.235 m/s Kmano

gH m Vw 2 u2

Kmano

0.9

gH m ; 20.94 – 18.235

\

Hm = 35.03 m

Dp0 = Hmrg = 35.03 × 1000 × 9.81 = 343644.3 N/m 2

343.64 kN/m 2

From the exit velocity triangle, V2

V f22  Vw22

0.4772  18.242

18.25 m/s

Dp = static head rise across the impeller 'p

p2  p1 Ug 35.03 

\

Hm 

V22  V12 2g

18.252  1.9072 2 – 9.81

18.25 m

p2 – p1 = 18.25 × 1000 × 9.81 = 179.0 kPa

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

213

(c) The power transmitted to the fluid (W.P.): W.P. = rgQHm 1000

kg m3

– 9.81

m s2

– 0.036

m3 – 35.03 m s

N m m = 12027.55 kg 2 s s m 12.027 kW s (d) The input power to the impeller (P):

Ans.

12027.55 N

P

EXAMPLE 3.26

W.P. KH

12.027 0.9

13.36 kW

An axial flow turbine has the following design data:

Blade speed: Mass flow rate: Gas temperature (stagnation) at inlet: Gas temperature (stagnation) at outlet: The fixed blade exit angle: Axial velocity is constant:

u1 = u2 = u = 300 m/s m = 3 kg/s T01 = 550°C T02 = 300°C a1 = 70° with respect to axial direction Vf1 = Vf 2 = 210 m/s

Calculate (a) the rotor blade gas angle, (b) the degree of reaction, (c) the blade loading coefficient, (d) the power output, and (e) the blading efficiency. Solution: (a) The rotor blade angles (b1, b2):

Vw1

Vf 1 tan D 1

210 tan 20’

576.97 m/s

W.D. = –cp(T02 – T01) = cp(T01 – T02) = 1.005 or

u(Vw 2  Vw1 ) kJ gc 1000 kg

\

Vw 2  Vw1

\

kJ (550 – 300) K kg-K

1.005(550  300) 1.005 – 250 – 1000 300

837.5 m/s

Vw2 = 837.5 – Vw1 = 837.5 – 576.97 = 260.53 m/s

214

Fundamentals of Turbomachinery

Vf

tan E1 \

210 576.97  300

Vw1  u

0.7582

b1 = 37.17°

Vf

tan E2 \

Ans.

210 260.53  300

Vw 2  u

0.3746

b2 = 20.54°

Ans.

(b) Degree of reaction (R):

\

Vr22

V f2  (Vw2  u)2

2102  (260.53  300)2

358293.9 m 2 /s2

Vr21

V f2  (Vw1  u)2

2102  (576.97  300)2

120812.4 m 2 /s2

R

Alternatively:

Vr22  Vr21

Vr22  Vr21 2u(Vw1  Vw 2 )

358293.9  120812.4 2 – 300 – (576.97  260.53)

0.473

Ans.

(Vw2  u)2  V f 2  [(Vw1  u)2  V f2 ] Vw22  u2  V f2  Vw21  u2  2Vw2 u  2Vw1u  V f2

Vr22  Vr21

(Vw22  Vw21 )  2u(Vw1  Vw 2 )

(Vw 2  Vw1 )(Vw 2  Vw1 )  2u(Vw1  Vw2 )

\

R

Vr22  Vr21 2u(Vw1  Vw 2 )

(Vw 2  Vw1 )(Vw 2  Vw1  2u) 2u (Vw1  Vw 2 )

or

R

Vw 2  Vw1 1 2u

260.53  576.97  1 0.473 2 – 300

Ans.

(c) Blade loading coefficient (y): \ \

Vf u Vf u

(cot E1  cot E 2 ) (cot D 1  cot D 2 )

W.D. mu 2 W.D. mu 2

[For derivation, see Example 2.4] \

210 (cot 37.17’  cot 20.54’) 300

2.7914

Ans.

(d) Power developed (P): P = –mDh0 = –mcp(T02 – T01) = mcp(T01 – T02)

(Eq. (2.9a))

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

215

kg kJ – 1.005 – (550  330) K s kg-K

Ans.

3

753.75 kW

(e) Blading efficiency (hb):

Vf

sin D 1

V1

2

sin D 1

210 sin 20’

614 m/s

u V1

From Eq. (2.43g) and I

Kb

Vf

; V1

2 1  2I cos D1  I

2

2

2 300 È 300 Ø 1 2 – cos 20’  É Ê 614 ÙÚ 614

2

0.8089

Ans.

Alternatively:

Kb

u (Vw1  Vw 2 ) V12 2



(Vr22

 Vr21 ) 2

300 – 837.5 2

614 358293.9  120812.4  2 2

0.8178

Ans.

EXAMPLE 3.27 In an axial flow reaction turbine, nozzle is inclined at an angle of 65° with respect to the axial direction. The axial component of the air velocity at the exit of the nozzle is 180 m/s. The degree of reaction is 50% and the blade speed is 180 m/s. Calculate (a) the rotor blade angles at inlet and outlet and (b) the degree of reaction. For the same blade speed, axial velocity and nozzle angle, the absolute velocity at the rotor exit is axial and equal to the axial velocity at the inlet. Solution: Data: Case I Machine: Nozzle inclination: Axial component: Degree of reaction: Blade speed: Case II Blade speed: Axial velocity: Nozzle angle:

Axial flow reaction turbine a1 = 65° with respect to axial direction Vf1 = Vf 2 = Va = Vf = 180 m/s R = 50% u = 180 m/s

u = 180 m/s Vf = 180 m/s a1 = 90° – 65° = 25° V2 = axial, a2 = 90°, V2 = Vf2, Vw2 = 0, V2 = Vf1 R = 50%, a1 = b2, a2 = b1; Vr1 = V2, Vr2 = V1

Vw1

Vf tan D 1

180 tan 25’

386.00 m/s

216

Fundamentals of Turbomachinery

(a) Blade angles (b1, b2):

tan E1 \

Vf

180 386  180

Vw1  u

b1 = 41.15°;

0.874

b2 = a1 = 25°

Ans.

(b) Degree of reaction (R):

Vf

80 425.92 m/s sin D 1 sin 25’ V2 = Vf1 = 180 m/s (data) V1

Vr1

V f2  (Vw1  u)2

Vr 2

u2  V22

R

(386  180)2  180 2

180 2  180 2

273.56 m/s

254.6 m/s

(Vr22  Vr21 )

(254.62  273.562 )

(V12  V22 )  (Vr22  Vr21 )

(425.922  180 2  254.62  273.562 )

 0.072

Ans.

Alternatively:

R R

(Vw 2  Vw1 ) 1 2u 386   1  0.072 2 – 180

(Vw2 = 0, Q a2 = 90°)

Ans.

Or R

(Vr22

 Vr21 )

2u(Vw 2  Vw1 )

2

(254.6  273.562 ) 2 – 180 – (386)

 0.0721

Ans.

EXAMPLE 3.28 A centrifugal blower runs at a speed of 3200 rpm, its impeller outet diameter 80 cm. The impeller blades are designed for a constant radial velocity of 60 m/s from inlet to outlet. There is no guide vanes at the inlet. If the degree of reaction is 0.6. The total-to-total efficiency is 0.8. Assume ambient pressure and temperature as 100 kPa and 25°C respectively. The mass flow rate is 2.5 kg/s. Calculate (a) the exit blade angle, (b) the power input to the blower, and (c) the exit stagnation pressure. Solution:

Data: Machine: Speed: Impeller outer diameter: Constant radial velocity: Degree of reaction: Total-to-total efficiency: Ambient pressure: Ambient temperature: Mass flow rate:

Centrifugal blower N = 3200 rpm d2 = 80 cm Vf1 = Vf2 = Vf = 60 m/s R = 0.6 ht-t = 0.8 p01 = 100 kPa T01 = 25 + 273 = 283 K m = 2.5 kg/s

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

217

There are no guide vanes at the inlet. It means that the inlet absolute velocity V1 is axial, i.e. a1 = 90°, V1 = Vf, Vw1 = 0. Usually the inlet absolute velocity V1 at the inlet is radial or axial. In case of centrifugal compressors, a certain amount of Vw1 is maintained to keep the inlet Mach number low. (a) Exit blade angle (b2):

u2

S d2 N 60

R

1 2

S – 0.8 – 3200 60

V f 2 cot E2 Ø È ÉÊ 1  ÙÚ u

(See Example 2.12)

2

60 cot E2 Ø 1È ÉÊ1  Ù 2 134.04 Ú

0.6

134.04 m/s

\ b2 = 65.92°

Ans.

(b) Power input to the blower (P):

tan E 2 \

Vf

60 (134.04  Vw 2 )

(u2  Vw 2 )

tan 65.92’

Vw2 = 107.52 m/s P

W.D. – m

2.5

m u2 Vw 2 gc

kg m m N-s2 – 134.04 – 107.52 – s s s kg-m

35.940 kW

Ans.

(c) Exit stagnation pressure (p02): P = mcp(T02 – T01) 35.940 kW

\

kg kJ – 1.005 (T02  T01 ) K s kg-K [Assume air as working medium \ cp = 1.005 kJ/kg-K] 2.5

(T02 – T01) = 14.305 K ËT Û T01 Ì 02  1Ü Í T01 Ý (T02  T01 ) „

Kt

or



t

0.8

(T02  T01 ) (T02  T01 ) „

Ë È p Ø (J 1) / J Û  1Ü T01 Ì É 02 Ù ÌÍ Ê p01 Ú ÜÝ (T02  T01 )

Ë È p Ø 0.4 /1.4 Û 283 Ì É 02 Ù  1Ü ÌÍ Ê 100 Ú ÜÝ ; ? p02 14.305

114.87 kPa

Ans.

218

Fundamentals of Turbomachinery

EXAMPLE 3.29 A single-stage axial flow blower with no inlet guide vanes, operates at 3500 rpm. The tip and hub diameters of the rotor are 20 and 12 cm respectively. The air is turned through an angle of 22° with respect to the axial direction during passage through the rotor at the arithmetic mean diameter. Degree of reaction is 0.75. Assume ST conditions and no inlet stator. Calculate (a) axial flow velocity, (b) power input and (c) blade angles. Solution:

Data: Machine: Axial flow blower Speed: N = 3500 rpm Tip diameter: d1 = 20 cm Hub diameter: d2 = 12 cm Exit air angles: a2 = 90° – 22° = 68° [The given angle a2 is with respect to axial direction, here it is changed with reference to tangential direction.] Degree of reaction: R = 0.75 Ambient pressure: p01 = 101.3 kPa Ambient temperature: T01 = 25 + 275 = 298 K To determine: Va1 = Va2 = Va, P, b1, b2

In the absence of inlet stator, inlet absolute velocity is axial, i.e. a1 = 90°, Vw1 = 0, V1 = Va u

S (d1  d2 ) N 60

S (0.2  0.12) – 3500 60 – 2

29.32 m/s

p = rRT \

U

p RT

101.3

kN m

2

A = area of flow = R

–

1 kg-K 1 – 0.287 kN-m 298 K

S 2 (d1  d22 ) 4

Va (cot E2  cot E1 ) 2u

1.184 kg/m 3

S (0.22  0.122 ) 4

0.0201 m 2

Va Ë u  Vw 2 V uÛ  Ü 1  w2 Ì Va Ý 2u Í Va 2u

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

or

0.75 1 

Vw 2 ; ? Vw 2 2 – 29.32

219

14.66 m/s

(a) Axial velocity (Va): Va Vw 2

tan D 2

Va 14.66

tan 68’ ; ? Va

36.29 m/s

Ans.

(b) Blade angles (b1, b2):

Va 36.29 u 29.32 b1 = 51.06°

tan E1 \

tan E2

\

Va u  Vw 2

1.24 Ans.

36.29 29.32  14.66

2.475

b2 = 68°

Ans.

(c) Power input (P):

– 0.0201 m 2 – 36.29

U AVa

P

m uVa (cot E1  cot E2 ) gc

0.864

\

kg

m

1.184

m

3

m s

0.864 kg/s

kg m m N-s2 – 29.32 – 36.29 (cot 51’  cot 68’) – s s s kg-m

P = 373.0 W

Ans.

EXAMPLE 3.30 A centrifugal machine has the following data: The speed of the machine is 1500 rpm, the inner diameter is 20 cm. The density of the fluid handled is 1000 kg/m3. The stage pressure rise is 200 kPa and the static pressure rise is 100 kPa. The mass flow rate is 0.5 kg/s. Calculate (a) the ratio of the inlet diameter to the exit diameter of the impeller, (b) the motor power required to drive the machine if the motor efficiency is 0.75, (c) the degree of reaction, and (d) the volume flow rate. Solution: Machine: Speed: Density of the fluid: Stage pressure rise: Static pressure rise: Mass flow rate: Motor efficiency: Inner diameter:

Centrifugal type N = 1500 rpm r = 1000 kg/m3 (Dp0)st = 200 kPa (Dp)r = 100 kPa m = 0.5 kg/s hmotor = 0.75 d1 = 20 cm

220

Fundamentals of Turbomachinery

To determine: d1/d2, Pm (motor power), R, Q ( 'p0 )st ( 'p)r ( 'p0 ) st  ( 'p)r (200  100)

or

kN m2

U [(u22  u12 )  (Vr21  Vr22 )  (V22  V12 )] 2 gc

(Eq. (2.24k))

U [(u22  u12 )  (Vr21  Vr22 )] 2 gc

(Eq. (2.24j))

U (u22  u12 ) 2 gc 1000

kg m3

100 1000 –

–

1 kN-s2 – S 2 N 2 ( d22  d12 ) m 2 – 2 kg-m 1000 – 60 – 60 s2

1 1500 2 ( d22  d12 ) – S2 – – 2 1000 3600

or

d22  d12

\

d22

\

d2 = 0.269 m

0.03242 0.03242  d12

0.03242  0.04

0.07242 m 2

(a) The ratio of inlet diameter to exit diameter (d1/d2): d1 d2

0.2 0.269

0.743

Ans.

(b) Motor power (Pm): (u22  u12 )  (Vr21  Vr22 )  (V22  V12 ) 2 gc

W.D.

u2Vw 2 gc

W.D.

( 'p0 ) st U

200

Pm

m W.D. Kmotor

0.5

kN m2

–

m3 1000 kg

0.2 kJ/kg

kg kJ 1 – 0.2 – s kg 0.75

0.133 kW

Ans.

(c) The degree of reaction (R): R

( 'p ) r ( 'p) st

100 kPa 200 kPa

0.5

Ans.

(d) Volume flow rate (Q): Q

m U

0.5

kg m3 – s 1000 kg

0.0005 m 3 /s

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

221

EXAMPLE 3.31 The power input to a centrifugal machine with radial impeller is 35 kW. The speed of the motor which is driving the machine is 3000 rpm. The efficiency of the machine is 0.85 and the transmission efficiency is 0.88. The radial velocity is 0.2u2 and is constant. Machine handles 250 m3/min at the entry. Assume that the pressure and temperature at its entry are 0.95 bar and 300 K. Calculate (a) the power required by the electric motor, (b) the stage pressure developed by the machine, (c) the impeller diameters if d2 = 2d1, (d) the blade angle at entry, (e) the impeller width at entry and exit, and (f) the degree of reaction. Solution:

Data Centrifugal machine (fan, blower, pump etc.) (Radial type, b2 = 90°) Ideal power input: Pi = 35 kW Speed: N = 3000 rpm Machine efficiency: hm/c = 0.85 Transmission efficiency: htra = 0.88 Radial velocity: Vf = Vf1 = Vf 2 = 0.2u2 Volume flow rate: Q = 250 m3/min Inlet pressure: p1 = 0.95 bar Inlet temperature: T1 = 300 K

To determine: Pm, (Dp0)st, d2, d1, b1, b1, b2, R (a) Power required by the electric motor (Pm): Ideal power Machine efficiency

Pa

actual power

Pa

35 kW 0.85

Pm

Actual power (Pa ) Transmission efficiency

41.17 kW

41.17 0.88

46.79 kW

Ans.

(b) Stage pressure developed by the machine (Dp0)st: Pi

\

( 'p0 ) st

( 'p0 ) st – m U

Pi Q

( 'p0 ) st – Q

35 kW

250 m 60 s (Dp0)st = wDH

3

35 – 60 s kN-m 250 m 3 s

(Eqs. (2.24g) and (2.24h))

8.4 kN/m 2

w = specific weight of water in kN/m3 = rg DH = water column in m

222

Fundamentals of Turbomachinery

\

'H

or

'H

( 'p0 ) st w

8.4 kN

8.4

kN m

2

–

1 kg m 1000 3 – 9.81 2 m s

m3

–

0.8563 m of H 2 O m 1000 kg 2 s DH = 85.63 cm of water column 9.81 m 2

Ans.

(c) Impeller diameters (d1, d2): (Assume fluid = Air) U

Km / c

p1 RT1

Ks

0.95 – 10 5

1.1034 kg/m 3

287 – 300 m 3

( 'p0 ) st U (W.D.) a

stage efficiency

[Radial impeller,

( 'p0 ) st U (u2 Vw 2 ) gc

( 'p0 ) st U (u2 u2 ) gc

\ u2 = Vw2]

Hence,

u22

( 'p0 )st gc U Ks

or

u22

8956.53 m 2 /s2

\

d2 = 0.6025 m

8.4

kN m2

–

1 – m3 1 kg-m – 1000 – – 1.1034 kg 0.85 kN-s2

(S d2 N )2 60

[(S – d2 – 3000)]2 60

Ans.

d2 0.6025 0.30125 m 2 2 (d) Blade angle at entry (b1):

and

Ans.

d1

tan E1 \

Vf 1 u1

0.2

u2 u1

0.2 –

d2 d1

0.2 –

0.6025 0.30125

0.4

b1 = 21.8°

Ans.

b2 = 90° (radial impellers)

0.2 – S d2 N 60 (e) Impeller width (b1, b2): Vf 1

Q \

0.2u2

S d1b1 V f 1

b1 = 0.233 m

? b1

0.2 – S – 0.6025 – 3000 60

Q S d1V f 1

18.93 m/s

250 60 – S – 0.30125 – 18.93 Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

b2

and

b1

d1 d2

0.233 –

0.30125 0.6025

0.1163 m

223 Ans.

(f) The degree of reaction (R): S d1 N S – 0.30125 – 3000 u1 60 60 u1 = 47.32 m/s u2 = 94.64 m/s

cos E1

u1 ; ? Vr1 Vf 1

u1 cos E1

47.32 cos 21.8’

50.96 m/s

Vr2 = Vf 2 = Vf1 = Vf = 18.93 m/s [(u22  u12 )  (Vr21  Vr22 )] U 2 gc

( 'p)r

(Eq. (2.24i))

(94.642  47.322 )  (50.962  18.932 ) m 2 kg N-s2 – 1.1034 2 – 3 – 2 –1 kg-m s m

or

(Dp)r = 4941.1 N/m2 = 4.941 kN/m 2

\

R

( 'p)r ( 'p0 ) st

4.941 8.4

Ans.

0.588

EXAMPLE 3.32 Water leaves a lawn sprinkler with an absolute velocity of 2.5 m/s, the diameter of the sprinkler is 0.25 m and it rotates at a speed of 150 rpm. Calculate (a) the utilization factor and (b) the degree of reaction. Solution:

Data: Refer to Figure 2.8.

Machine:

Lawn sprinklers (radial flow reaction turbine)

Exit absolute velocity: Diameter of the sprinkler: Speed:

V2 = 2.5 m/s d2 = 0.25 m = (2 × sprinkler arm) N = 150 rpm

S d2 N S – 0.25 – 150 1.964 m/s 60 60 Vr2 = u2 + V2 = 1.964 + 2.5 = 4.4634 m/s u2

224

Fundamentals of Turbomachinery

(a) Utilization factor (e):

H

2 V 1  r2 u2

2 4.4634 1 1.964

u2 Vw 2 gc

u2 V2 gc

(Eq. (2.43a))

0.6111

Ans.

Alternatively: W.D. \

W.D.

H

W.D. 

1.964 – 2.5 1

4.91 2.52 4.91  2 –1

V22 2 gc

4.91 J/kg

(Q Vw2 = V2)

0.611

Ans.

(Eq. (2.45b))

Ans.

(b) The degree of reaction (R): R

1 H

1 0.611

1.636

EXAMPLE 3.33 In a mixed flow pump with radial impellers, fluid enters axially. Impellers are designed in such a way that the exit relative velocity is equal to the inlet tangential blade speed. The inlet hub diameter is 9 cm and the impeller tip diameter 26 cm. The speed of the machine is 3000 rpm. The flow velocity is constant. Calculate (a) the dgree of reaction and (b) the work input. Solution: Data: Machine: Inlet hub diameter: Impeller tip diameter: Speed: Radial impellers: Axial inlet: Flow velocity is constant:

Mixed flow pump, Axial inlet, axial outlet d1 = 9 cm d2 = 26 cm N = 3000 rpm b2 = 90°, Vr2 = Vf 2, u2 = Vw2 a1 = 90°, V1 = Va1, Vw1 = 0 \ Va1 = Vf 2, Vr2 = u1

u1

S d1 N 60

S – 0.09 – 3000 60

14.14 m/s

u2

S d2 N 60

S – 0.26 – 3000 60

40.84 m/s

Va1 = Vf 2 = V1 = Vr2 = u1 = 14.14 m/s (given data) V2 tan E1

u22  Vr22 V1 u1

14.14 14.14

40.842  14.142 1

? E1

45’

43.22 m/s

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

225

(a) The degree of reaction (R):

R

Va (cot E1  cot E2 ) 2u

R

14.14 (cot 45’  cot 90’) 2 – 14.14

(Eq. (2.33)) Ans.

0.5

(b) The work done (W.D.): W.D.

u2 – u2 gc

u2 Vw 2 gc

(40.84)2 1

1667.91 J/kg

Ans.

EXAMPLE 3.34 The following data refers to an axial compressor. The degree of reaction is 0.65, stage efficiency 0.8, mean diameter 0.70 m, speed of the machine 1500 rpm, mass flow rate of air 55 kg/s, and mechanical efficiency 0.85. The inlet and exit temperatures of the rotor are 35°C and 60°C, respectively. Calculate (a) the actual temperature rise (stage temperature rise), (b) the pressure rise, and (c) the power input to the motor. Solution:

Data: Machine:

Axial compressor

Degree of reaction: Stage efficiency: Mean diameter: Speed: Mass flow rate: Mechanical efficiency: Inlet temperature to rotor: Exit temperature to rotor:

R = 0.65 hs = 0.8 d = 0.7 m N = 1500 rpm m = 55 kg/s hm = 0.85 TRi = 35°C = T1 TRe = 60°C

(a) Actual temperature rise (DT)s: R

\

(TRe  TRi ) ( 'T ) s

( 'T ) R ( 'T ) s

(60  35) ( 'T ) s

0.65

(DT)s = 38.46 K

Ans.

(b) Power input to the motor (Pm): W.D. Pm

mc p ( 'T ) s W.D. Km

55

kg kJ – 1.005 38.46 K s kg-K

2125.96 0.85

2501.13 kW

2125.96 kW

Ans.

226

Fundamentals of Turbomachinery

(c) The pressure rise (pR): (T3  T1 ) (T3  T1 )

pR

Ë 0.8 – 38.46 Û  1Ü Ì 308 Í Ý

„

Ë ( p / p )(J 1) / J Û  1Ü T1 Ì 3 1 38.46 ÌÍ ÜÝ

Ë (T / T ) Û T1 Ì 3 1  1Ü Í ( 'T ) s Ý

Ks

„

0.8

J /(J 1)

1.396

Ans.

EXAMPLE 3.35 Air at a temperature of 290 K, flows in a centrifugal compressor running at 20000 rpm, m = 0.8, total-to-total efficiency = 0.75, and exit diameter = 50 cm. Assume that the absolute velocities at the inlet and outlet are same. Calculate (a) the ideal exit stagnation temperature of air passing through the compressor and (b) the stage pressure ratio. Solution: Machine:

Centrifugal compressor

Inlet stagnation temperature: Speed: Total-to-total efficiency: Exit diameter: Inlet and exit absolute velocity

T01 = 290 K N = 20000 rpm, m = 0.8 ht–t = 0.75 d2 = 0.5 m V1 = V2

Radial tip vane, b2 = 90°, u2 = Vw2

Assume

S d2 N 60

u2 W.D. =

S – 0.5 – 20000 60

P u2 Vw 2 gc

P u2 u2 gc

523.6 m/s

0.8 – 523.62 1

219325 W

(a) Ideal exit temperature (T03¢): W.D. = cp(T03 – T01) 219325 = 1005(T03 – 290); Kt  t

0.75

T03  T01 T03  T01 „

\ T03 = 508.3 K T03  290 ; ? T03„ 508.3  290 „

453.7 K

Ans.

(b) Stage pressure ratio: p03 p01

Ë T03 Û Ì Ü Í T01 Ý „

J

/(J 1)

1.4 / 0.4

Ë 453.7 Û Ì 290 Ü Í Ý

4.79

Ans.

EXAMPLE 3.36 Free air delivered by a compressor is 20 kg/min. The inlet conditions are 1 bar and 20°C static. The velocity of air at the inlet is 60 m/s. The isentropic efficiency of the compressor is 0.7. The total head pressure ratio is 3. Find (a) the total head temperature at the exit and (b) the power required by the compressor if the mechanical efficiency is 95%.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes T

Solution: Data: Machine: Total head pressure ratio: Mass flow rate: Inlet pressure (static): Inlet temperature (static): Inlet absolute velocity: Isentropic efficiency: Mechanical efficiency:

T01

3

p3

3¢ p01 01

p1

1 s

60 2 2 – 1 – 1005

J

p01 p1

Ë T01 Û Ì Ü Í T1 Ý

T03 T01

Ë p03 Û Ì Ü Í p01 Ý

„

03¢

p03

V2 T1  1 2 gc c p 293 

\

Compressor p03/p01 = 3 m = 20 kg/min p1 = 1 bar T1 = 20°C V1 = 60 m/s ht–t = 0.7 hm = 0.95

03

227

/(J 1)

294.79 K 1.4 / 0.4

Ë 294.79 Û Ì 293 Ü Í Ý

(J 1) / J

? T03

„

; ? p01 1.021 bar

Ëp Û T01 Ì 03 Ü Í p01 Ý

(J 1) / J

294.79 – 30.4 /1.4

403.5 K

(a) The total head temperature (T03): T03  T01 ; ? T03  T01 T03  T01

Kt  t

\

„

403.5  294.79 0.7

155.3 K

T03 = 155.3 + T01 = 155.3 + 294.79 = 450.09 K

Ans.

(b) The power required by the compressor (P):

P \

m – W.D. Km

20 c p – (T03  T01 ) 60 Km

20 1.005 – 155.3 – 60 0.95

P = 54.76 kW

Ans.

EXAMPLE 3.37 A two-stage centrifugal compressor is delivering 500 m3 of free air per minute. The suction conditions are 1 bar and 15°C. The compression ratio and isentropic efficiency of each stage are 1.25 and 80%. Calculate the isentropic efficiency for the entire compression process. Solution:

Data: Machine: No. of stages: Volume flow rate:

Centrifugal compressor K=2 Q = 500 m3/min

228

Fundamentals of Turbomachinery

Sunction temperature: T01 = 15°C Sunction pressure: T01 = 1 bar Compressor ratio each stage: ps = 1.25 Isentropic efficiency of each stage: hs1 = hs2 = hs = 0.8 The isentropic efficiency for the entire compression process (hc t–t): J 1 J

H T

0.4 1.4

0.2857 p03

03¢ 3 03≤

II stage p02 02

I stage

02¢

p01

01 s

Kc t  t

( psK H  1)

(1.252 – 0.2857  1)

[1  ( psH  1) / Ks ]K  1

[1  (1.250.2857  1) / 0.8]2  1

0.7937

(Eq. (3.42g)) Ans.

Alternatively: (J 1) / J

T02 T01

Ë p02 Û Ì Ü Í p01 Ý

Ks1

T02  T01 T02  T01

„

; ? T02

„

0.8

„

288 – 1.250.2857

307.96  288 ; ? T02 T02  288

306.96 K 311.7 K

T03¢ = T02(p03/p02)(g –1)/g = 311.7 × (1.25)0.2857 = 332.22 K Ks 2

T03  T02 T03  T02 „

0.8

332.22  311.7 ; ? T03 T03  311.7

337.34 K

From Eq. (3.42c), 2 pR = p02/p01 = overall pressure ratio = ps

(1.25)2

1.5625

Now consider a single stage from p01 to p03, T03” T01

Ë p03 Û Ì Ü Í p01 Ý

(J 1) / J

; ? T03”

Ëp Û T01 Ì 03 Ü Í p01 Ý

0.2857

288 – (1.5625)0.2857

327.17 K

229

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

\

T03”  T01

Kc t t

T03  T01

327.17  288 337.34  288

Ans.

0.7938

EXAMPLE 3.38 Inlet (static) conditions to an air compressor are 1.02 bar and 300 K and the machine develops pressure of 1250 mm of water column and temperature of 13 K. Calculate (a) the compressor efficiency and (b) the final pressure of air if the temperature changes from 300 K to 500 K, assuming the same efficiency. Solution:

Data:

Case I

T

Machine:

Air compressor

Inlet temperature: Inlet pressure: Exit temperature: Pressure developed:

T1 p1 T2 Dp

= = = = =

2

p2

2¢

300 K 1.02 bar 313 K (p2 – p1) 1250 mm of water column

p1 1 s

Case II Inlet temperature:

T1 = 300 K ; Final temperature T2 = 500 K

Initial pressure:

p1 = 1.02 bar ; Compressor efficiency hc s–s

Dp = wDH = rgDH

1000 1000

kg m

3

kg 3

– 9.8 –

m 2

m s2

– 1.250 m

– 9.81 – 1.250 m =

m s = 12.263 kN/m2 = 0.12263 bar

kN m3

– 9.81 – 1.25 m

Dp = p2 – p1 \ p2 = Dp + p1 = 0.12263 + 1.02 = 1.143 bar (J 1) / J

Ë p2 Û Ëp Û ? T2 T1 Ì 2 Ü Ì Ü Í p1 Ý Í p1 Ý (a) The compressor efficiency (hc s–s): T2 T1

(J 1) / J

Ë 1.143 Û 300 – Ì Ü Í 1.02 Ý

„

„

Kc s  s

T2  T1 T2  T1 „

390.89  300 313  300

0.2857

309.89 K

0.761

Ans.

(b) Final pressure of air (p2): Case II Kc s  s

T2  T1 T2  T1

È p2 Ø ÉÊ p ÙÚ

Ë T2 Û Ì Ü Í T1 Ý

1

„

0.761 

T2  300 ; ? T2„ 500  300

Q /(Q 1)

„

„

J /(J 1)

? p2

ËT Û p1 Ì 2 Ü Í T1 Ý „

452.15 K

Ë 452.15 Û 1.02 – Ì Ü Í 300 Ý

3.5

4.29 bar

Ans.

230

Fundamentals of Turbomachinery

EXAMPLE 3.39 The inlet stagnation pressure and temperature are 1.05 bar and 20°C. The exit stagnation temperature is 300 K. Calculate (a) the final pressure in mm of water, (b) the isentropic enthalpy change and (c) the actual enthalpy change. Assume a total-to-total efficiency of 0.75. Solution:

T

Data:

02

p02

02¢

Machine:

Fan or blower

Inlet stagnation pressure:

p01 = 1.05 bar

Inlet stagnation temperature:

T01 = 20°C

Exit stagnation temperature:

T02 = 300 K

Total-to-total efficiency:

ht–t = 0.75

p01 01 s

To determine: p02 in mm of water column, (Dh0)isen, (Dh0)a T02  T01 T02  T01

Kt t

„

T02  293 ; ? T02„ 300  293

0.75

„

298.25 K

(a) Final pressure (DH) in terms of mm of water column: J

/(J 1)

3.5

p02 p01

Ë T02 Û Ì Ü Í T01 Ý

p02

w 'H ; 'H

p02 Ug

111.734

or

'H

111.734 kN – m – s2 – 9.81 1000 kg m

111.734 9.81

\

DH = 113.89 mm of water column

„

Ë 298.25 Û Ì 293 Ü Í Ý p02 w

? p02

1.11734 bar kN

m2

–

111.734 kN/m 2

m 3 s2 1000 kg – 9.81 m

11.389 m

Ans.

(b) Isentropic enthalpy change (Dh0)isen: (Dh0)isen = cp(T02¢ – T01) = 1.005(298.25 – 293) = 5.276 kJ/kg

Ans.

(c) Actual enthalpy change (Dh0)a: (Dh0)a = cp(T02 – T01) = 1.005 × (300 – 293) = 7.035 kJ/kg

Ans.

EXAMPLE 3.40 The overall pressure ratio through a three-stage gas turbine is 10 and the efficiency is 0.86. The inlet temperature is 1400 K. Assume the temperature rise in each stage to be the same. Calculate for each stage (a) the pressure ratio and (b) the stage efficiency. Solution:

Data: Machine: No. of stages: Overall pressure ratio:

Gas turbine K=3 pR = 10

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

231

ht s–s = 0.86 T1 = 1400 K

Efficiency: Inlet temperature: T

p1

1

p2

2

2¢

p3

3 3¢

pK

2≤

4¢ K¢

3≤ s

The assumption is that the temperature rise is the same in each stage. TK T1

TK¢

(J 1) / J

„

Ë pK Û Ì Ü Í p1 Ý = 0.518 × 1400

Ë1Û Ì 10 Ü Í Ý

0.2857

0.518

= 725.15 K

ht s–s = (T1 – TK)/(T1 – TK¢) = 0.86 = (1400 – TK)/(1400 – 725.15) \

TK = 819.63 K T1 – TK = 1400 – 819.63 = 580.37 K \ 3(DT)s = 580.37

(DT)s1 = (DT)s2 = (DT)s3 = (DT)s; \

DTs1 = T1 – T2 = 580.37/3 = 193.5 K Kp

ln (T1 / TK ) [(J  1) / J ] (ln p1 / pK )

ln (1400 / 819.63) 0.2857 – ln 10

(Eq. (3.60a))

0.815

First Stage (a) Pressure ratio (p1/p2): e = (g – 1)/g 1 / HK

(Eq. (3.60b)) J /(J 1)K p

Ë T1 Û p Ë 1400 Û Ì Ü Ì 1206.54 Ü Í Ý Í T2 Ý (b) Stage efficiency (hs1): p1 p2

Ks1

Èp Ø 1É 2Ù Ê p1 Ú Èp Ø 1É 2Ù Ê p1 Ú

HK p

J 1 / J

È 1 Ø 1É Ê 1.895 ÙÚ

1.4 /(0.4 – 0.815)

Ë 1400 Û Ì 1206.54 Ü Í Ý

1.895

Ans.

0.4 – 0.815 / 1.4

È 1 Ø 1É Ê 1.895 ÙÚ

0.4 /1.4

0.8285

(Eq. (3.63))

Ans.

232

Fundamentals of Turbomachinery

Second Stage (a) Pressure ratio (p2/p3): T2 – T3 = 193.5 \ T3 = T2 – 193.5 \

T3 = 1206.54 – 193.5 = 1012.8 K p2 p3

1 / HK p

1.4 /(0.4 – 0.815)

Ë T2 Û Ì Ü Í T3 Ý

Ë 1207.56 Û Ì 1012.8 Ü Í Ý

2.12

Ans.

(b) Stage efficiency (hs2):

Ks 2

È 1 Ø 1 É Ê 2.12 ÙÚ

(0.4 – 0.815) / 1.4

È 1 Ø 1 É Ê 2.12 ÙÚ

0.8308

0.4 /1.4

Ans.

Third Stage (a) Pressure ratio (p3/p4): T3 – TK = 193.5

\ TK = T3 – 193.5

TK = 1012.8 – 193.5 = 819.2 K p3 pK

J

Ë T3 Û Ì Ü Í TK Ý

/ K p (J 1)

1.4 /(0.815 – 0.4)

Ë 1012.8 Û Ì 812.2 Ü Í Ý

2.487

Ans.

(b) Stage efficiency (hs2):

Ks 3

È 1 Ø 1 É Ê 2.487 ÙÚ

0.815 – (0.4 /1.4)

È 1 Ø 1 É Ê 2.487 ÙÚ

0.4 /1.4

0.8341

Ans.

EXAMPLE 3.41 The flow rate through the compressor is 50 kg/s. The inlet static conditions are 1 bar and 30°C. Exit temperature from the last stage is 650 K (static). The compressor has five stages of equal pressure ratio of 1.5. Calculate (a) the exit pressure from the last stage, (b) the ideal exit temperature from the last stage, (c) the overall efficiency, (d) the polytropic efficiency, and (e) the stage efficiency. Solution:

Data: Machine: Mass flow rate: Inlet static pressure: Inlet static temperature:

Compressor m = 50 kg/s p1 = 1 bar T1 = 30°C

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

Exit temperature: No. of stages:

233

T6 = TK+1 = 650 K K=5 6¢

T

p6

6 p5 5¢

p4

5

p3

4¢ 4

p2

3¢

3

p1

2¢

2 1

s

Pressure ratio in each stage, ps = 1.5 pK 1 p1

pR

p6 p1

psK

(1.5)5

7.594

(Eq. (3.42c))

(a) Exit pressure from the last stage (pK+1): pK+1 = p6 = p1 × 7.594 = 1 × 7.594 = 7.594 bar

Ans.

(b) Ideal exit temperature from the last stage (T6¢): T6¢/T1 = (p6/p1)(g –1)/g = (7.594)0.2857 = 1.7846 \

T6¢ = T1 × 1.7846 = 303 × 1.7846 = 540.741 K

Ans.

(c) Overall compression efficiency (hc s–s): From Eq. (3.24d), Kc s  s

T6  T1 T6  T1

540.741  303 650  303

„

0.685

Ans.

Alternatively:

Kc s  s

(1.55 – 0.2857  1)

( psK H  1) Ë È pH  1 Ø Û Ì1  É s ÙÜ ÍÌ Ê Ks Ú ÝÜ

k

5

Ë È 1.50.2857  1 Ø Û Ì1  É ÙÜ 1 ÍÌ Ê 0.745 Ú ÜÝ

0.6855

Ans.

(d) Polytropic efficiency or infinitesimal efficiency (hp): Kp

[(J  1) / J ] ln ( pK 1 / p1 ) ln (TK 1 / T1 )

0.2857 – ln 7.594 ln (650 / 303)

0.759

(Eq. (3.39))

Ans.

234

Fundamentals of Turbomachinery

(e) Stage efficiency (hs): e = (g – 1)/g = 0.4/1.4 = 0.2857 Ks

(1.50.2857  1)

( psH  1) H / Kp

( ps

0.745

(1.50.2857 / 0.759  1)

 1)

(Eq. (3.42))

Ans.

EXAMPLE 3.42 Air is compressed in a compressor from 1 bar to 8 bar (static) and temperature 30°C to 700°C (static). Calculate (a) the polytropic efficiency, (b) the overall efficiency, (c) the index of actual process, (d) the number of stages assuming that pressure ratio is same in each stage, (e) the stage efficiency, (f) the change in pressure in each stage in terms of water column, and (g) the preheat factor. Solution:

Data:

T( K 1) „ T1

\

Machine:

Compressor

Inlet pressure (static): Final pressure: Inlet temperature: Final temperature: Assumption:

p1 = 1 bar pK+1 = 8 bar T1 = 30°C TK+1 = 700°C Equal pressure change in each stage.

Ë pK 1 Û Ì Ü Í p1 Ý

(J 1) / J

Ë8Û Ì1 Ü Í Ý

0.2857

1.8114

T

T(K+1)¢ = T1 × 1.8114 = 303 × 1.8114

(K+1) p K+1

(K+1)¢

4

4¢

T(K+1)¢ = 548.85 K

Kp

È J  1Ø pK 1 ÉÊ J ÙÚ ln p 1 TK 1 ln T1

0.2857 – ln ln

700 303

3

3¢

(a) Polytropic efficiency (hp) (Eq. (3.39)):

p2 2¢

8 1

0.7095

2

p1

Ans. 1 s

(b) Overall isentropic efficiency (hc s–s):

Kc s  s

T( K 1) „  T1 TK 1  T1

548.85  303 700  303

0.6193

(Eq. (3.42g))

Ans.

(c) Index of actual process (n): n

J Kp 1  J (1  K p )

1.4 – 0.7095 1  1.4(1  0.7095)

1.674195

(Eq. (3.39b))

Ans.

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

235

(d) No. of stages assuming same pressure ratio (K): Assume pressure ratio in each stage = 1.5. pR = pK+1/p1 = 8/1 = psK

1.5K

\ K = 5.1285

(Eq. (3.42c)) Ans.

Assuming K = 5, recalculate the stage pressure ratio. \

pK 1 / p1

ps5

8 /1

\ ps = 1.516

(e) Stage efficiency (hs): e = (g – 1)/g = (1.4 – 1)/1.4 = 0.2857 Ks

(1.5160.2857  1)

( psH  1) H / Kp

 1)

( ps

(Eq. (3.42))

0.6920

(1.5160.2857 / 0.7095  1)

Ans.

(f) Change in pressure in each stage in terms of water column (DH): (Dp)s = p2 – p1 = p3 – p2 = p4 – p3 = p5 – p4 = p6 – p5; \ (Dp)s = wDH 'H

( 'p) s Ug

'H

(0.516) 1 – kN – m – 9.81 1000 kg – m

p2  p1 Ug

(1.516  1) – 100

kN

1

m 1000 kg – 9.81 m m3 s2 2

0.0526 m

DH = 52.599 mm of water column

Ans.

(g) Preheat factor:

PF

Kc s  s Ks

0.6193 0.6920

0.8949

(Eq. (3.42i))

EXAMPLE 3.43 A turbine has four stages and each stage pressure ratio is 2. The inlet static temperature is 630°C. The mass flow rate is 30 kg/s. The overall efficiency is 0.8. Calculate (a) the polytropic efficiency, (b) the stage efficiency, (c) the power developed, and (d) the reheat factor. Solution:

T

Data:

Machine:

Turbine

No. of stages: Pressure ratio in each stage: Inlet static temperature: Mass flow rate: Overall efficiency:

K=4 ps = 2 T1 = 630 + 273 m = 30 kg/s ht s–s = 0.8

pR

p1 / pK 1

psK

(Eq. (3.66))

p1

1

2¢

p2 2

p3 3

p4

3¢ 4

p5

4¢ 5 5¢

s

236

Fundamentals of Turbomachinery

HK p

HK p K

1  pR

Kt s  s

1  ps

1  pRH

1  (1/ ps )

1  psH K

HK p K

(Eq. (3.65))

1  (1/ ps )H K

(a) Polytropic efficiency (hp):

0.8

1  (1 / 2)

0.2857 – K p – 4

1  (1/ 2)0.2857 – 4

; ? Kp

0.5816

(Eq. (3.66a))

Ans.

(Eq. (3.63))

Ans.

(b) Stage efficiency (hs):

Ks

HK p

1  (1 / ps )

1  (1 / ps )H

1  (1/ 2)0.5816 – 0.2857 1  (1 / 2)0.2857

0.6056

(c) Power input (P): p1 / pK 1

psK

(2) 4 K pH

Ë pK 1 Û Ì Ü Í p1 Ý

TK 1 T1

16 Ë1Û Ì 16 Ü Í Ý

(Eq. (3.66)) 0.5816 – 0.2857

0.01039

(Eq. (3.60b))

We know that P = mcp(T1 – TK+1) = mcpT1[1 – (TK+1/T1)] = mcpT1(1 – 0.01039) 30

kg kJ – 1.005 – 903 K – (0.9896) s kg-K

26942.71 kW

Ans.

(d) Reheat factor (RF): RF = ht s–s/hs = 0.8/0.6056 = 1.321

(Eq. (3.72))

Ans.

EXAMPLE 3.44 Air flows through an axial compressor and develops a total head pressure ratio 5:1 and total-to-total isentropic efficincy of 0.88. The compressor is designed for 50% reaction with inlet and outlet blade angles of 45° and 10° respectively with respect to axial direction. The inlet total head temperature is 300 K. The mean blade speed and axial velocity are constant throughout the compressor. The mean blade speed and a work done factor are 210 m/s and 0.85 respectively. Calculate (a) the polytropic efficiency, (b) the number of stages, (c) the inlet Mach number relative to rotor at the mean blade height of the first stage. Solution:

Data: Machine:

axial compressor

p03/p01 = pRO = 5/1 hc t–t = 0.88 u1 = u2 = u = 210 m/s Va1 = Va2 = Va

R = 50% b1 T01 y b2

= = = =

a2 = 90° – 45° = 45° 300 K 0.85 a1 = 90° – 10° = 80°

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

237

(a) Polytropic efficiency (hp): e = (g – 1)/g = 0.4/1.4 = 0.2857 Kc t t

(50.2857  1)

( pR 0 ) H  1 H / Kp

( pR 0 )

(Eq. (3.42b))

1

0.2857 / K p

(5

0.88; (Eq. (3.42b)) ? K p

 1)

0.9036 Ans.

(b) Number of stages (K): Va

u cot D1  cot E1

Vf

210 cot 80’  cot 45’

(Eq. (5.13))

178.52 m/s

From Eq. (5.25c), DT = T02 – T01 = T03 – T01 = [y uVa(cot a2 – cot a1)]/cpgc = [0.85 × 210 × 178.52 (cot 45° – cot 80°)]/1005 (DT)s = 26.12 K = rise in temperature in one stage T0 K 1 T01

T0K+1

H /K

Ë p0 K 1 Û p Ì Ü Í p01 Ý = T01 × 1.6634

Ë5Û Ì1 Ü Í Ý

0.2857 / 0.9036

1.6634

(Eq. (3.39a))

= 300 × 1.6634 = 499.02 K

T0K+1 – T01 = 499.02 – 300 = 199.02 K K

Temperature rise in K stages Temperature rise in one stage

199.02 26.12

7.62; K  8 stages

Ans.

(c) Mach number at first stage (M1): V1

Va sin D1

T1

T01 

Vr1

Va1 sin E1

178.52 sin 80’

V12 2c p gc

181.3 m/s

T01 

178.52 sin 45’

V12 2 – 1005 – 1

252.47 m/s

283.65 K

(Eq. (3.11))

238

Fundamentals of Turbomachinery

M1

Vr1

252.47

J RT1

0.7478

1.4 – 287 – 283.65

Ans.

EXAMPLE 3.45 Air is compressed in an axial compressor. Atmospheric temperature and pressure at inlet are 20°C and 1 bar. Exit total pressure and temperature are 4 bar and 175°C. Exit static pressure is 3.3 bar. Calculate (a) the total-to-total efficiency, (b) the polytropic efficiency and (c) the exit air velocity. Solution:

Data: Refer to Figure 5.4

Machine Axial compressor,

p01 = 1 bar, T01 = 293 K, p03 = 4 bar, T03 = 175 + 273 = 448 K, P2 = 3.3 bar

Ë p03 Û Ì Ü Í p4 Ý

T03 T01

„

(J 1) / J

Ë4Û Ì1Ü Í Ý

0.2857

1.486; ? T03

„

293 – 1.486

435.6 K

(a) Total-to-total efficiency (hc t–t): Kc t t

(T03  T01 ) (T03  T01 ) „

435.4  293 448  293

0.9187

(b) Polytropic efficiency (hp): Kp

[(J  1) / J ] ln ( p03 / p01 ) ln (T03 / T01 )

0.2857 – ln 4 ln 448 / 293

(Eq. (3.39))

0.933

Ans.

(c) Air velocity at exit (V2): T2 T03

Ë p2 Û Ì Ü Í p03 Ý

T03

T02

J

1 / J

T2 

; ? T2

V22 2c p gc

Ëp Û T03 Ì 2 Ü Í p03 Ý

424.04 

0.2857

V22 2 – 1005

Ë 3.3 Û 448 – Ì Ü Í 4 Ý

0.2857

448; ? V2

424.04 K

219.45 m/s

Ans.

EXAMPLE 3.46 Air enters the 10-stage axial compressor at the rate of 4 kg/s. Air enters the compressor at static temperature of 300 K. The pressure ratio (static) is 6.5. The exit static temperature is 550 K. The compressor is designed for symmetrical stages. The axial velocity is constant and is 120 m/s. The mean blade velocity is 175 m/s. Calculate (a) the blade angles, (b) the power given to the air, and (c) the static-to-static efficiency. Solution:

Data:

Machine = Axial compressor, m = 4 kg/s, T1 = 300 K, pK+1/p1 = 6.5, K = 10, TK+1 = 550 K, Va = 120 m/s, u = 175 m/s

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

T( K 1) „

Ë pk 1 Û Ì Ü Í p1 Ý

T1

(J 1) / J

? T( K 1) „

Ëp Û T1 Ì K 1 Ü Í p1 Ý

239

0.2857

300 – 6.50.2857

512.12 K

(a) Static-to-static efficiency (hc s–s):

Kc s  s

(T( K 1) „  T1 )

512.12  300 550  300

(TK 1  T1 )

0.8485

(K+1) T

(K+1)¢

pK+1 p4

(b) Power input (P):

4

4¢

P = mcp(TK+1 – T1) = 4 × 1.005 × (550 – 300) = 1005 kW

Ans.

(c) Blade angles (b1, b2):

p3

3

3¢

p2 2¢

2

(DT)s = temperature change per stage = (TK+1 – T1)/10

p1

1

= (550 – 300)/10 = 25 K

s

W.D. per stage = uDVw/gc = 175 × DVw/gc = cp(DT)s 1005 – 25 175

\

'Vw

or

DVw = Va (cot b1 – cot b2) = 120(cot b1 – cot b2) (Eqs. (5.16) and (5.27)) (i)

143.6 m/s

R = Va (cot b1 + cot b2)/2u 0.5 = 120(cot b1 + cot b2)/2 × 175

or

(ii)

Solving Eqs. (i) and (ii), b1 = 36.53°

and

b2 = 81.5°

Ans.

IMPORTANT EQUATIONS •

a = sonic velocity =

gc J •

M

V a

RT M

gc

Jp U

gc J RT

(3.6)

(3.7) (3.8)

240

Fundamentals of Turbomachinery

V2 2 gc

(3.10)

V2 2c p gc

(3.11)

•

h0

c pT 

•

T0

T

•

T0

Ë (J  1) 2 Û T Ì1  M Ü 2 gc Í Ý

•

p0

Ë (J  1) 2 Û J 1 p Ì1  M Ü 2 Í Ý

•

p0

p

•

U0

Ë (J  1) 2 Û J  1 U Ì1  M Ü 2 Í Ý

•

a0

J RT0

•

a0

(J  1) h0

(3.12) J

Uv2 2 gc

(3.14)

( U  constant)

(3.14a)

1

(3.15)

(3.16)

For compression process • hc t–t = isentropic compression efficiency based on total-to-total J 1

T01 [( pR 0 ) J  1] T02  T01

(3.27)

• hc s–s = isentropic compression efficiency based on static to static J 1

T1 [( pR ) J  1] T2  T1

•

•

(3.30)

hs = hs1 = hs2 = C = Each stage efficiency in a multistage compressor 'Wisen 1

'Wisen 2

'Wisen 3

'Wa1

'Wa 2

'Wa3

È 'Wisen Ø 6É Ê 'Wa ÙÚ

hco = overall isentropic compression efficiency in a finite multistage compressor

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

•

Kco

Wisen Wa

•

Kco Ks

Wisen 6 'Wisen

(3.33) (3.35a)

•

hs > hco

•

hp = polytropic or infinitesimal stage efficiency in a finite stage

p (J  1) ln C J p1 TC ln T1 •

(3.39)

hc s–s = hs = isentropic compression efficiency for one finite stage È J 1 Ø Ù J Ú

È p2 Ø ÉÊ ÉÊ p ÙÚ 1

È J  1Ø Ù J Ú

È p2 Ø ÉÊ ÉÊ p ÙÚ 1

È J  1Ø É Ù Ê J Ú

•

Kco

pR

È J  1Ø 1 É Ù Ê J Ú Kp

pR

1

(3.42)

1

Kp

1

1

(3.42b)

1

•

hco = overall isentropic efficiency in a multistage (finite stage) compressor

•

pR = overall pressure ratio = È J  1Ø É ÙK Ê J Ú

•

Kco

pR

È J  1Ø K É Ù Ê J Ú Kp

pR • •

241

pK 1 p1

psK

1

(3.42c)

(3.42d)

1

pR = pressure ratio of one stage in a multistage compressor (K stage) (DT)T = total temperature rise (actual temperature rise) Ë È J  1Ø K Û Ì pÉÊ J ÙÚ K p  1Ü T1 Í R Ý

(3.42f)

242

Fundamentals of Turbomachinery

Expansion process •

ht t–t = isentropic expansion efficiency based on total-to-total (T01  T02 ) È J 1 Ø Û Ë É Ù T01 Ì Ê J Ú Ü Í1  pR 0 Ý

•

Kt t  s

•

pR 0

(T01  T02 ) È J 1 Ø Û Ë É Ù T01 Ì Ê J ÚÜ Í1  pR „ Ý p01 p02

pR

Wa Wisen

(3.51)

p01 p01  p2 p2

„

„

•

(3.47)

p01 p02

„

•

Wa Wisen

„

ht t–t > ht t–s hs = isentropic efficiency of each stage in a multistage turbine = hs1 = hs2 = hs3 = C 'Wa1 'Wisen1

•

•

'Wa 2 'Wisen2

'Wa3 'Wisen3

Wa 6'Wisen

hto = overall isentropic turbine efficiency in a finite multistage turbine

Kto Ks

Wa Wisen

(3.54)

6 'Wisen Wisen

(3.57)

•

hto > hs

•

hp = polytropic or infinitesimal stage efficiency in a finite stage

(3.58)

TC T1 p (J  1) ln C J p1 ln

•

Ë TC Û Ì Ü Í T1 Ý

Ë pC Û Ì Ü Í p1 Ý

J

1 J

–

1

Kp

(3.60a)

Ë pC Û Ì Ü Í p1 Ý

n 1 n

(3.60b)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

• •

Kp

È n  1Ø È J Ø ÉÊ Ù n Ú ÉÊ J  1 ÚÙ

(3.60c)

ht s–s = hs = isentropic expansion efficiency for one finite stage Èp Ø 1É 1 Ù Ê pC Ú

È J 1 Ø É Ù Kp Ê J Ú

Èp Ø 1É 1 Ù Ê pC Ú

•

243

(3.63)

È J 1 Ø É Ù Ê J Ú

ht s–s = overall isentropic efficiency in a multistage (finite stage) compressor È J  1Ø É Ù Kp Ê J Ú

1  pR

(3.65)

È J  1Ø É Ù Ê J Ú

1  pR •

pR = overall pressure ratio =

p1 pK 1

( ps )K

(3.66)

È J  1Ø É Ù pK Ê Ú

•

Kt s  s

J

1  ps

È É Ê

1  ps

• •

K

(3.66a)

J  1Ø K J ÙÚ

pR = pressure ratio of one stage in a multistage turbine (K stage) (DT)T = total actual temperature drop (overall actual temperature drop) È J  1Ø Ë Û É Ù Kp K T1 Ì Ê J Ú Ü Í1  ps Ý

(3.67)

REVIEW QUESTIONS 1. 2. 3. 4. 5.

What is velocity of sound? Derive an equation for the velocity of sound for a perfect gas. Explain the term Mach number. What are subsonic, supersonic and hypersonic flows? Explain the concept of total or stagnation properties. Derive the expressions for stagnation temperature, stagnation pressure, stagnation density and stagnation enthalpy.

244

Fundamentals of Turbomachinery

6. Define total-to-total, total-to-static, static-to-static and static-to-total efficiencies for power developing and power consuming turbomachines and write the T–s diagrams. 7. What is infinitesimal stage efficiency in the expansion and compression process? Derive the corresponding equation. 8. Show that

n

J J  (J  1) K p

where n = polytropic index g = isentropic index hp = polytropic efficiency or infinitesimal stage efficiency. 9. Show that for finite turbine stages

Kt s  s

È J  1Ø É Ù Kp Ê J Ú

1  pR

(For single stage)

È J  1Ø É Ù Ê J Ú

1  pR

Kt o

È J  1Ø É ÙKpK Ê J Ú

1  pR

È J  1Ø É ÙK Ê J Ú

(For multistage, i.e. K stages)

1  pR

10. Show that for finite turbine stages, (DT)T = total actual temperature drop È J  1Ø Ë Û É Ù Kp K T1 Ì Ê J Ú Ü  1 p Í Ý R

11. Show that Kp

È n  1Ø È J Ø ÉÊ Ù n Ú ÉÊ J  1 ÙÚ

(For turbine)

Kp

È n Ø È J  1Ø ÉÊ n  1 ÙÚ ÉÊ J ÙÚ

(For compressor)

Kp

È J  1Ø p2 ÉÊ J ÙÚ ln p 1 T2 ln T1

12. Show that

(For compressor)

Thermodynamics of Fluid Flow and Thermodynamic Analysis of Compression and Expansion Processes

T2 T1 È J  1Ø p2 ÉÊ J ÙÚ ln p

245

ln

Kp

(For turbine)

13. Show that È J 1 Ø Ù Kp Ú

Ë T2 Û Ì Ü Í T1 Ý

Ë p2 ÛÉÊ J Ì Ü Í p1 Ý

Ë T2 Û Ì Ü Í T1 Ý

Ë p2 ÛÉÊ J Ì Ü Í p1 Ý

È J 1 Ø 1 ٖ Ú Kp

K 1

Ë p2 Û K Ì Ü Í p1 Ý

(For turbine)

K 1

Ë p2 Û K Ì Ü Í p1 Ý

(For compressor)

14. Show that (a)

hs > hco

where

hs = stage efficiency (summation of each stage efficiency in a multistage compressor)

(For compressor)

hco = overall isentropic efficiency (a)

hto > hs

where

hs = stage efficiency (summation of each stage efficiency in a multistage turbine)

(For turbine)

hto = overall isentropic efficiency.

EXERCISES 3.1 A gas has the following data. Temperature = 600 K, g = 1.4, R = 500 J/kg-K, Mach. number = 1.5. Calculate for the static and stagnation conditions (a) velocity of sound and (b) enthalpy. 3.2 Air at NTP and at 15 m/s is accelerated isentropically in a nozzle to 250 m/s. Find (a) the change in temperature, (b) the change in stagnation pressure, (c) the change in density, and (d) the change in stagnation temperature. 3.3 Test section of a supersonic wind tunnel is having standard air (STP) at a Mach number 6. Find the condition of air in the reservoir. 3.4 Adiabatic flow system of air at some point is having Mach number M = 4, total temperature 310 K, static pressure 0.7 bar, at some other point Mach number M = 1.8. Calculate (a) the total temperature, (b) the stagnation pressure, (c) the static pressure (d) the amount of heat transfer which caused reduction in Mach number (is this positive or negative) and (e) the index of polytropic process. 3.5 Air enters a straight axisymmetric duct at 300 K, 3.5 bar and 150 m/s and leaves it at 275 K, 2.2 bar and 270 m/s. The area of cross section at entry is 550 cm2. Assume

246

Fundamentals of Turbomachinery

adiabatic flow, g = 1.4, R = 287.0 J/kg-K. Calculate (a) the stagnation temperature, (b) the mass flow rate, and (c) the area of cross section at exit. 3.6 At the entry of a flow passage, the pressure, the temperature and Mach number are 2.5 bar, 30°C, 1.5 respectively. If the exit Mach number is 2.5, calculate the following for adiabatic flow of a perfect gas. (a) Stagnation temperature (b) Temperature and velocity of gas at exit (c) The flow rate per m2 of the inlet. 3.7 An air compressor has 6 stages of equal pressure ratio 1.5. The mass flow rate is 50 kg/s. The overall isentropic compression efficiency is 90%. Entry pressure is 1 bar, T1 = 50°C, Calculate. (a) The stage of the air at the exit (b) Polytropic efficiency (c) Each stage efficiency (d) Power required to drive the compressor (overall efficiency drive = 0.9) 3.8 A two-stage gas turbine develops 25 MW at the shaft. The inlet temperature is 1500 K. The overall pressure ratio of the turbine is 8.2 and isentropic expansion efficiency is 0.91. Assume that the pressure ratio of each stage is same. Calculate: (a) Pressure ratio of each stage (b) Polytropic efficiency (c) The mass flow rate (d) The efficiency and power of each stage. Assume g = 1.4, cp = 1.005 kJ/kg-K, overall drive efficiency = 0.93. 3.9 Air enters a compressor at a static pressure of 1.7 ata. a static temperature of 15°C and a flow velocity of 50 m/s. At the exit the static pressure is 3.5 ata. The static temperature 110°C, and the flow velocity 110 m/s. The outlet is 2.2 m above the inlet. Calculate (a) the isentropic change in total enthalpy and (b) the actual change in total enthalpy. 3.10 Air flows through an air turbine where its stagnation pressure is decreased in the ratio 5:1. The total-to-total efficiency is 0.82 and the air flow rate is 5.5 kg/s. The inlet total temperature is 280 K. Calculate (a) the actual power output, (b) the actual exit total temperature, (c) the actual exit static temperature if the exit flow velocity is 110 m/s, and (d) the total-to-static efficiency of the device.

4

Centrifugal Compressors and Pumps

CENTRIFUGAL COMPRESSORS

4.1 WORKING PRINCIPLE, COMPONENTS AND DESCRIPTION Figure 4.1 shows the main components of a centrifugal compressor. The main components are: (a) inlet pipe, (b) impeller disc and impeller (a rotor having a number of vanes or blades), (c) diffuser (with and without vanes or both), (d) casing, (e) collection chamber, and (f) outlet pipe. The gas (air or fluid) enters the impeller eye of a centrifugal compressor in an axial direction with absolute velocity V1. The gas then flows radially through the impeller passage due to centrifugal force. The impeller rotates at a very high speed (20000 to 30000 rpm). Energy is imparted to the gas by the rotating blades where it is converted into kinetic energy as it moves from radius r1 to r2, along with a small pressure rise during its radial flow in the impeller. The impeller vanes at the eye are bent to provide shockless entry. The gas leaving from the impeller blades is turned through an angle b2, and leaves with an absolute velocity V2 at an angle a2. The gas then enters the diffuser. The diffuser surrounding the impeller converts the K.E. into pressure energy. Hence, there is rise in static pressure of the gas. Gas then enters the casing, and the outlet pipe, where some more K.E. is converted into pressure energy. The clearance between the impeller blades and inner walls of the casing must be kept as small as possible to reduce leakage. Modern compressors are designed in such a way that equal pressure rise takes place in the impeller diffuser. 247

248

Fundamentals of Turbomachinery

5 3 2

3

4

4 8

2

7 9

(b)

3 17 16 2 15

1

14

10

13 11

12

5

(a)

18

7 5 6 4

(1) Driving shaft, (2) Impeller disc or hub, (3) Impeller blade, (4) Diffuser blade or diffuser, (5) Collection chamber or collector or volute, (6) Casing, (7) Inducer section, (8) Radial blade impeller, (9) Shroud, (10) Impeller eye, (11) Accelerating nozzle, (12) IGV, (13) Hub radius, rh, (14) Mean radius, rm, (15) Tip radius (impeller inlet radius) (eye tip), rt or r1, (16) Impeller exit radius (eye root), r2, (17) Diffuser exit radius, r3, (18) Outlet from the volume casing

3 2

D 1

Figure 4.1 Main components of a centrifugal compressor.

The maximum pressure obtained in a single stage compressor is in the range of 110 to 250 kPa. For very high pressures of the order of 10 bar, multistage centrifugal compressors are used.

4.2 WORK DONE AND PRESSURE RISE No work is done on the gas in the diffuser because the diffuser simply converts one type of energy into another type. Work absorbed by the gas depends upon the condition of the gas at inlet and outlet of the impeller. Assumptions for ideal energy transfer are: •

The velocity of the gas at inlet to the impeller is axial. \

a1 = 90°,

Vw1 = 0, V1 = Vf1

249

Centrifugal Compressors and Pumps

• • •

No losses occur due to friction. No energy loss or gain occurs due to heat transfer to or from the gas. The gas leaves the impeller in such a way that, u2 = Vw2.

Velocity triangles for centrifugal compressor (Figure 4.2): AC = u2 = Vw2 = without slip

ACE = without slip

BD = u2 = with slip

BDE = with slip

AB = slip = Vw2 – Vw2¢

BC = Vw2¢ = with slip

The relative velocity of gas, Vr1, at inlet makes an angle b1, with the direction of motion. The gas flows through the impeller and is turned through an angle of 90° and ideally, the outlet is in radial direction, i.e. the absolute velocity at exit, V2, is such that its whirl component is equal to u2. \

b2 = 90°,

u2 = Vw2,

Vr2 = Vf 2 = ideal case

From Euler’s energy equation, (Vw1u1  Vw 2 u2 ) gc

W.D. A

B a2

C a 2¢

V2

D b 2¢

b2 V2¢

Vf 2 = Vr 2

Vr 2¢

Tip diameter, d2

E

Vr1 b1

Hub radius (rh)

a1 = 90°, V1 = Vf 1, VW1 = 0

a1 V1 = Vf 1

Eye tip or shroud radius (r1)

w

Blade width, b (b)

(a)

Figure 4.2

(Contd.)

250

Fundamentals of Turbomachinery

Vw2¢ Vw2 = u2 b 2¢

B a 2¢

b2

Vr 2¢

Vf 2 = Vf 2¢ = Vr 2

D

C

A

a2

V2¢ V2

E

wr

ad

/s

Eye tip r2

Ey

oo er

V1 = Vf 1

t

rh

rm

Vr1 a1

b1 u1

r1

(c)

Figure 4.2

Velocity diagram for a centrifugal compressor: (a) Front view of the impeller and exit velocity triangle (shaft rotation in anticlockwise direction). (b) Side view of the impeller and inlet velocity triangle. (c) Impeller with inlet and exit velocity triangles (impeller rotation in clockwise).

Since it is a power absorbing machine, W.D.

(u2Vw 2  Vw1u1 ) gc u2Vw 2 J/kg gc

u2 u2 gc

u22 J gc kg

Vw22 J/kg gc

(Q Vw1 = 0) (Q Vw2 = u2) (4.1)

Centrifugal Compressors and Pumps

\

P

U QVw22 W gc

251 (4.2)

where r = density of fluid (kg/m3) Q = volume flow rate of the fluid (m3/s) gc = 1 kg-m/N-s2 Equation (4.1) shows that the workdone on the gas is independent of the entering velocity V1 and dimensions of the eye. It also show that the maximum work done by the impeller on the gas depends on the speed of the rotor and tip diameter of the impeller. Due to certain limitations, if the speed is limited, an increase in the pressure rise can be obtained purely by increasing the tip diameter d2, otherwise we go for multistage. However, in case of multistaging, the frontal area increases and that increases the bulkiness of the compressor. From the thermodynamic analysis, Ë Ë V 2 gz Û V 2 gz Û Q  m Ì h1  1  1 Ü W  m Ì h2  2  2 Ü gc ÜÝ gc ÜÝ 2 gc 2 gc ÍÌ ÍÌ

For the adiabatic process, Q = 0

\

h01

h1 

V12 gz1  2 gc gc

stagnation condition

h02

h2 

V22 gz2  2 gc gc

stagnation condition

W.D. = h02 – h01 = Dh0

(4.3)

For perfect gases, W.D. = cp(T02 – T01) Kc t



(h02  h01 ) (h02  h01 )

(4.4)

ÈT Ø c pT01 É 02  1Ù T Ê 01 Ú

(4.5)

„

t

W.D.

From Eqs. (4.1), (4.3) and (4.4), we have 'h0 Kct

\

h02  h01

c p (T02  T01 )

u22 gc

(4.6)

c p (T02  T01 ) „



t

T02  T01

c p (T02  T01 )

T02  T01 Kc t t „



(4.7)

252

Fundamentals of Turbomachinery

From Eqs. (4.6) and (4.7), we get (T02„  T01 )

u22 Kc t t

(4.8)

c p gc

We have for isentropic process, J

1

T02„ T01

Ë p02 Û Ì Ü Í p01 Ý

p02 p01

Ë T02„ Û J 1 Ì Ü Í T01 Ý

J

J

or

J

Ë (T02„  T01  T01 ) Û J  1 Ì Ü T01 Í Ý J

Ë (T02„  T01 ) Û J 1  1Ü Ì T01 Í Ý Substituting Eq. (4.8) in above equation, J

Ë u22 Kc t t Û J 1 Ì  1Ü ÌÍ c p gcT01 ÜÝ

pR1

p02 p01

Let

pR1

p02 be the stagnation pressure rise between the impeller inlet and the exit. p01

If

V1 = V2, then Eq. (4.9) reduces to

(4.9)

J

Let

Ë u22 Kc t t Û J 1 Ì  1Ü ÌÍ c p gc T1 ÜÝ

pR

p2 p1

pR

p2 be the static pressure rise between the impeller inlet and the exit. p1

(4.10)

Normally in centrifugal compressors, pR1 (p02/p01) is limited to 4 per stage. The losses that occur in the compressor are due to friction, leakage, shock and turbulance, etc. The adiabatic efficiency of the compressor is usually limited to 80% because of the above mentioned losses, though still less in many cases. In large compressors, the mechanical efficiency can be taken as unity, if it is not mentioned.

4.2.1 Enthalpy–Entropy Diagram In Figure 4.3, 0–1 = flow process in accelerating nozzle 1–2 = impeller; 2–3 = diffuser; 3–4 = volute casing

253

Centrifugal Compressors and Pumps

h01

V2 h1  1 2 gc

h02

V2 h2  2 2 gc

h03

V2 h3  3 2 gc

h04

h4 

T or h

p02

p3

02

2

02¢

V3 2

03² 3

03¢

2

V2 2

p2 3¢

p01

V42

p1

2 gc

2

h02 = h03 = h04

2¢ 00 0

01

2 V1

h00 = h01 h02 = h03

2 1 s

Figure 4.3

Enthalpy–entropy diagram for a centrifugal compressor.

In centrifugual compressors the static pressure rise is partly due to the action of the impeller and partly due to the action of the diffuser, however, the stagnation pressure rise is entirely due to the impeller reduced by the losses in the diffuser. Here: V1 = absolute velocity to the impeller inlet V2 = absolute velocity from the impeller exit and inlet to the diffuser V3 = absolute velocity from the diffuser exit.

4.2.2 Overall Pressure Ratio pR0 = overall pressure ratio = Kc t to

p03 p01

overall total-to-total compressor efficiency Total isentropic enthalpy rise between inlet and exit Actual enthalpy rise between the same total pressure limits (h03  h01 ) (h03  h01 ) „

or

Kc t to

(T03  T01 ) (T03  T01 ) „

ËT Û 1 T01 Ì 03  1Ü Í T01 Ý (T03  T01 ) „

(4.10a)

254

Fundamentals of Turbomachinery

Kc t to (T03  T01 )

or

Ë T03„ Û  1Ü Ì Í T01 Ý

or

T03 T01

Ë Kc t to (T03  T01 ) Û Ì1  Ü T01 Í Ý

p03 p01

Ë T03„ Û J  1 Ì Ü Í T01 Ý

„

T01

J

J

Ë Kc t to (T03  T01 ) Û J  1 Ì1  Ü T01 Í Ý

pRO

(4.11)

We know that there is no work involved during the process of diffuser, i.e. 2 to 3. Therefore, from Eq. (4.6), 'h0

h02  h 01

c p (T02  T01 )

c p (T03 T01 )

u22 gc

(4.11a)

Substituting the above relation in Eq. (4.11), we get

p03 p01

pR 0

Ë u22 Ì Kc t to gc c p Ì1  Ì T0 Ì ÍÌ

J

ÛJ 1 Ü Ü Ü Ü ÝÜ

(4.12)

Equations (4.9) and (4.12) can be used to calculate the stagnation pressure ratio between the impeller inlet to impeller exit and impeller inlet to diffuser outlet respectively. Energy balance between the impeller inlet (1–1) and the diffuser exit (3–3), Dh0 = h03 – h01 Ë V32 Û Ë V12 Û Ì h3  Ü  Ì h1  Ü 2 gc ÝÜ ÍÌ 2 gc ÝÜ ÍÌ (h3  h1 ) 

(V32  V12 ) 2 gc

The diffuser design should be such that the exit velocity from the diffuser must be equal to the inlet absolute velocity to the impeller V3 = V1 \

Dh0 = h3 – h1 = cp(T3 – T1) Kc t  to

(h03  h01 ) ( h03  h01 ) „

(4.12a)

Centrifugal Compressors and Pumps

255

Ë V32 Û Ë V12 Û Ì h3  Ü  Ì h1  Ü 2 gc ÜÝ Í 2 gc Ý ÌÍ Ë V32 Û Ë V12 Û    h h Ì 3 Ü Ì 1 Ü 2 gc ÜÝ Í 2 gc Ý ÍÌ „

Kc t to

\

Ë (V32  V12 ) Û Ì (h3  h1 )  Ü 2 gc Ì Ü Ì (V32  V12 ) Ü Ì ( h3  h1 )  Ü 2 gc ÍÌ ÝÜ „

(h3  h1 ) ( h3  h1 ) „

c p (T3  T1 ) „

c p (T3  T1 )

Kc s  s

(4.12b)

4.2.3 Limiting Inlet Velocity The tip of the inducer vane dt is very important with respect to Mach number. If the Mach number at entry to the impeller is greater than unity, then shock waves will form. Generally the eye root diameter dh is as small as possible, the limiting factor for dh being the shaft diameter and bearing arrangement (Figure 4.1).

4.3 PRESSURE COEFFICIENT (fp) Because of compressor losses as well as the exit kinetic energy, the actual pressure rise is less than the theoretical, specified by the impeller tip speed. This is expressed by a quantity called pressure coefficient (fp). Ip

Isentropic work required for actual pressure rise Isentropic work required by the impeller tip speed Isentropic work required for a static pressure change from p1 to p3 Isentropic work required by the impeller tip speed Isentropic work Euler’s work

c p (T3  T1 )

Ip

'h „ W.D.

Ip

gc c p T1 Ë T3 Û  1Ü Ì 2 u2 Í T1 Ý

„

u22 gc „

(Eq. (4.1))

(4.12c)

256

Fundamentals of Turbomachinery

Ip

Ë J 1 Û Ü gc c pT1 Ì p3 J  1 Ì Ü u22 Ì p1 Ü ÜÝ ÍÌ

(4.12d)

From Eq. (4.15), we have

V

u22 gc

c p (T03  T01 ) ; ? (T03  T01 )

V u22 gc c p

(4.12e)

From Eq. (4.12a) and (4.12c), we get IP u22 gc cP

Kc t to (T3  T1 ) Kc t to (T03  T1 )

(4.12f)

From Eqs. (4.12e) and (4.12f), we get I p u22

V u22

gc c p

gc c p

Ip

\

Kc t to

V Kc t to

(4.12g)

4.4 BLADE ANGLES AT EYE ROOT AND EYE TIP Figure 4.4 shows the inlet velocity triangles at eye root and eye tip. Assumption: Axial inlet, i.e. a1 = 90°, \ Vf1 = V1, Vw1 = 0

Figure 4.4

Inlet velocity triangles: (a) Eye root. (b) Eye tip.

uh tan E1 root

S dh N ; 60 V fh uh

;

u1 tan E1tip

S d1 N 60 Vf 1 u1

Centrifugal Compressors and Pumps

E1 root

tan 1

V fh uh

;

\

E1tip

tan 1

257

Vf 1 u1

where dh dt uh u1 b1 root b1 tip

= eye root diameter (hub diameter) = d1 = eye tip diameter (impeller inlet diameter) (tip diameter) = speed at eye root = speed at eye tip = blade angle at eye root = b1 = blade angle at eye tip.

4.5 EYE CONDITIONS FOR AN IMPELLER For a given flow rate and eye root diameter, the annular area may be made large or small. Figure 4.5 shows the three different eye conditions for a centrifugal compressor. Assume a zero whirl at entry to a centrifugal compressor, i.e. a1 = 90°,

V1 = Vf1,

Vw1 = 0

(a) For a given flow rate Q, if the eye tip diameter (d1 or dt) is large, then from continuity considerations, the axial velocity Vf1 is low (giving a low inlet absolute velocity V1) and the blade speed u1 and the relative velocity Vr1 are high (Figure 4.5(a)). (b) If the eye tip diameter is medium, the axial velocity Vf1 is medium (giving a medium absolute velocity V1) as well as the blade speed u1 and the relative velocity Vr1 are medium (Figure 4.4(b)). (c) If the eye tip diameter is low, the axial velocity Vf1 is low (giving a low absolute velocity V1) and the blade speed u1 and the relative velocity Vr1 are low (Figure 4.4(c)).

Figure 4.5

Eye conditions for an impeller and the corresponding velocity triangles.

258

Fundamentals of Turbomachinery

The above analysis shows that the relative velocity is a function of eye tip speed (eye tip diameter). We have from Figure 4.5,

\

Vr21

u12  V12

u12  V f21

Vr21

2 Ë Û 4Q Ë S dt N Û Ì 60 Ü  Ì 2 2 Ü Í Ý ÍÌ S (dt  dh ) ÝÜ

2

(4.13)

Here, dh, Q, N are fixed. Differentiating Vr1, w.r.t. dt and equating to zero will yield a value of dt for minimum Vr1. This is shown in Figure 4.6.

Figure 4.6 Variation of inlet relative velocity with eye tip diameter in case of centrifugal compressor.

4.6 INFLUENCE OF IMPELLER BLADE SHAPE Figure 4.7 shows the inlet and exit velocity triangles for different values of exit blade angle, b2, i.e. impeller shape.

Figure 4.7

Inlet and exit velocity triangles for different exit blade angles b2, and for same u2.

(a) If b2 < 90°, i.e. backward curved vanes, Vw2 is minimum, power required is minimum. (b) If b = 90°, i.e. radial curved vanes, Vw2 lies between (a) and (c), power required also lies between (a) and (c).

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(c) If b > 90°, i.e. forward curved vanes, Vw2 is maximum, power required is maximum. From the efficiency point of view, the backward curved blades are the best.

4.7 SLIP FACTOR (s) Due to the large amount of mass of air (gas) flowing through the impeller, it has certain inertia. Due to the formation of eddies, the velocity of whirl at exit reduces, and flow is turned through an angle less than 90° compared to 90° turn in an ideal condition. This effect is known as ‘slip’. This can be reduced by increasing the number of impeller vanes and reducing the clearance space. The velocity triangles for actual (with slip) and ideal (without slip) conditions are shown in Figure 4.8. (a) Backward curved vanes b2 < 90°: Ideal = velocity triangle ADC ® without slip Actual = velocity triangle ABC ® with slip Assume,

a1 = 90°,

V1 = Vf1, Vw1 = 0

W.D. = 'h0

u2 Vw 2 gc

W.D. = 'h0

V u2Vw 2 (with slip) gc

(without slip)

(b) Radial curved vanes b2 = 90°: Ideal = velocity triangle ADC ® without slip Actual = velocity triangle ABC ® with slip Assume,

a1 = 90°, Vw1 = 0, V1 = Vf1 W.D. = 'h0

u2 Vw 2 gc

W.D. = 'h0

u22 V gc

V

\

u2 u2 gc

u22 gc

(without slip) (with slip)

Vw 2 u2

(4.14)

Equations (4.1), (4.9), (4.10) and (4.12) now reduce to W.D. =

V u22 gc

'h0

(4.15) J

pR1

Ë u22 Kc t t V Û J  1 Ì1  Ü c p gc T01 Ü ÌÍ Ý

(4.16)

260

Fundamentals of Turbomachinery Slip B Slip B

V2

Vr 2¢

V2¢

V2¢

D Vr 2

Vr 2¢

Vw 2¢

b2 = 90° C

Vw 2

b 2¢

A

b 2¢ = 90° C

Vw 2¢

u2

u2

(a) Backward curved vanes, b2 < 90°

Figure 4.8

Vf 2 =Vr 2

Vf 2¢

b2

Vf 2 A

D V2

= Vw 2

(b) Radial curved vanes, b2 = 90°

Exit velocity triangles with and without slip. J

pR

Ë u22 Kc t t V Û J  1 Ì1  Ü c p gc T1 Ü ÌÍ Ý

pR 0

Ë Kc t to V u22 Û J  1 Ì1  Ü gc c pT01 Ü ÌÍ Ý

(4.16a)

J

(4.17)

4.8 POWER FACTOR (j) Some of the power supplied by the impeller is used in overcoming losses which have a breaking effect such as disc friction or windage and the power input is therefore, modified by a factor j. Equations (4.15), (4.16), (4.16a) and (4.17) can now be reduced to W.D. = 'h0

VM u22 gc

(4.18) J

pR1

Ë u22 V M Kc t t Û J  1 Ì1  Ü c p gc T01 Ü ÌÍ Ý

pR

Ë Kc t t u22 V M Û J  1 Ì1  Ü c p gcT1 Ü ÌÍ Ý

pR 0

Ë Kc t to u22 V M Û J  1 Ì1  Ü c p gc T01 Ü ÌÍ Ý

(4.19)

J

(4.19a)

J

(4.20)

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4.9 PREWHIRL AND INLET GUIDE VANES In the case of compressor, with a uniform axial component of velocity at the impeller eye, the critical region occurs at the eye tip where the blade speed and relative velocity are the highest. Hence, it is advisable to check flow condition at the critical point, i.e. at the eye tip. From Figure 4.4, it can be seen that there is a value of eye tip speed which will give the minimum relative velocity (Vr1). For a given flow rate Q, (i) If the impeller eye is large, V1 is low and the eye tip speed u is high. (ii) If the impeller eye is small, V1 is high and the eye tip speed u is low. In both the cases, Vr1 is maximum whereas Vr1 is minimum in between these two cases. Differentiating Eq. (4.13) w.r.t. Vr1 and equating to zero, we get a value of dt for minimum Vr1 (keeping dh, Q and N constant). (dt2  dh2 )

\

2 – 60 2 (4Q / S )2 (S N )2

(4.20a)

Substituting Eq. (4.20a) in (4.13), we get Vr21 Vr1

(S N )2 (dt2  d h2 ) 2 – 60 2



(S dt N )2

(S N )2

60 2

60 2

[((dt2  d h2 ) / 2)  dt2 ]

(S N )2 (3dt2  d h2 ) 60 2 – 2

(4.20b)

Flow Mach number corresponding to minimum Vr1 can be calculated by knowing the static temperature of air at inlet. With an axial entry (Figure 4.9(a)), i.e. Vw1 = 0 (a1 = 90°), absolute velocity of air entering the compressor varies from impeller hub (serial no. 13 in Figure 4.1) to impeller tip (serial no. 1 in Figure 4.1), velocity u1 reaches maximum at the tip of the impeller. Substitute this condition in Vr21 u12  V12 , Vr1 will also be large so that the inlet Mach number may go beyond unity resulting in an obstruction (choking) at the compressor inlet. Even if it is below unity, local Mach number may be more than unity resulting in choking. Mach number should be maintained less than unity, i.e. 0.9 for safer operation at the impeller eye tip. If the inlet relative velocity Vr1 is very high and if the flow rate and speed cannot be altered for efficient operation, it is possible to reduce the relative velocity by giving the fluid an initial pre-rotation or prewhirl. This can be done without altering Vf1 and hence Q. This is known as prewhirl. Prewhirl can be obtained by providing inlet guide vanes installed directly in front of the eye as shown in Figure 4.9(a). Figure 4.9 shows the prewhirl vanes and the corresponding inlet velocity triangle. From Figure 4.8(b), triangle ACE = without prewhirl; a1 = 90°, Vw1 = V1 = Vf1 W.D. = (u2Vw2 – u1Vw1)/gc = u2Vw2/gc

(4.20c)

From Figure 4.8(b), triangle ABD = with prewhirl; a1¢ = 90°, Vr1¢ < Vr1, V1¢ > V1, b1¢ > b1, a1¢ < a1.

262

Fundamentals of Turbomachinery A Prewhirl angle V1¢

(a)

u

a1¢ B

b1

a 1¢

b 1¢

C

Vr 1

Prewhirl angle

V1¢

Vr 1¢

Vr 1 Vr 1¢ b1¢

a1

C u1

Vw1¢

Vw1¢ A

Va1 = V1 = Vf 1¢

Prewhirl vanes

b1 D

E

u1¢ (b)

Va = V1 = Vf D

B (c)

Figure 4.9

Inlet guide vanes (prewhirl vanes): (a) Inlet guide vanes. (b) Inlet velocity triangle with and wihout prewhirl. (c) Inlet velocity triangle with and without prewhirl (common base and separate flow velocity).

\ \

Angle of prewhirl = 90° – a1¢ = ÐBAC (4.20d) W.D. = (u2Vw2 – u1¢Vw1¢) In Figure 4.9(c), triangle ABC = without prewhirl; triangle ADC = with prewhirl, angle of prewhirl = ÐDCB. Figure 4.8(a) clearly shows that prewhirl vanes impart a whirl component Vw1¢ (BC) to the fluid, thus reducing relative velocity (Vr1 to Vr1¢) to an acceptable value. However, the work capacity (W.D.) is reduced (W.D. according to Eq. (4.20c) < W.D. according to Eq. (4.20d)). This is because the whirl component is zero in Eq. (4.20c) whereas it is not zero in Eq. (4.20d). The disadvantage of the positive prewhirl is to be reduce the energy transfer by an amount u1¢Vw1¢/gc, however, the inlet Mach number can be maintained less than the critical value so that the flow will not be obstructed.

4.10 DIFFUSER Energy is imparted to the air (gas) by the impeller. Therefore, the absolute velocity of the gas at the impeller exit (V2) is high, which is reduced to a lower velocity, V3, in the diffuser as shown in the Figure 4.3. The amount of K.E. converted into static pressure rise (p3 – p2) in the diffuser depends on the degree of reaction and the efficiency of the diffusion process. A good diffuser must have minimum losses (p03 – p02) and maximum efficiency.

4.10.1 Vanless Diffuser It is a diffuser where the gas is diffused before it leaves the stage through volute casing. Static pressure rise takes place in the vaneless diffuser simply due to the diffusion process from a smaller diameter d2 to a larger diameter d3 as shown in Figure 4.10.

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263

Figure 4.10 Diffuser passage with diverging walls.

4.10.2 Determination of Diffuser Inlet Angle, Width and Length of the Diffuser Passages A2 = area at inlet to the vaneless diffuser = p d2b2 = 2p r2b2 A3 = pd3b3 = 2p r3b3 Assuming that the flow is frictionless in the diffuser and the angular momentum is constant, we have r2Vw2 = r3Vw3 = constant (4.21) Applying continuity equation at the entry and exit sections of the vanless diffuser, we get r2Vf 2 A2 = r3Vf 3A3 r2Vf 2(2pr2b2) = r3Vf 32pr3b2 r2Vf 2r2b2 = r3Vf 3r3b3 For a small pressure rise, r2 = r3 Suppose

\ Vf 2r2b2 = Vf 3r3b3 If b2 = b3, i.e. for a constant width (parallel wall), Vf 2r2 = Vf 3r3 V32 = absolute velocity at the exit of the vanless diffuser

V f23  Vw23 2

2

Ë r2 Û 2 Ë r2 Û 2 Ì Ü V f 2  Ì Ü Vw 2 Í r3 Ý Í r3 Ý 2

Ë r2 Û 2 2 Ì Ü (V f 2  Vw 2 ) r Í 3Ý

(4.22)

264

Fundamentals of Turbomachinery 2

Ë r2 Û 2 Ì Ü V2 Í r3 Ý From Eqs. (4.21), (4.22) and (4.23), we get Vf 3

Vw3 Vw 2

Vf 2

V3 V2

D2

D3

tan 1

(4.23)

r3 r2

(4.24)

Vf 2

tan 1

Vw 2

Vf 3 Vw3

(4.25)

Equation (4.24) clearly shows that diffusion is directly proportional to the diameter ratio (d3/d2). A large-sized vaneless diffuser is required to get a large static pressure rise. Sometime it is impracticable to have a large-sized vanless diffuser. However, for industrial applications, a large-sized vaneless diffuser is accepted. Vaneless diffuser is economical and used for a wide range of operations.

4.10.3 Width of the Impeller Channel (Figure 4.10) Energy is imparted to the gas by the rotating blades. About half of this appears as static pressure head and the rest half as kinetic energy. Therefore, the pressure at impeller outlet p2, is assumed to be equal to ( p  p1 ) p2 p1  2 (4.26) 2 Similarly, the temperature rise from the eye to the outlet may be assumed about half of the total temperature rise, and neglecting exit velocity. DT0 = total temperature rise W.D. = Dh0 = cpDT0 =

\

\

u22 ; gc

'T0

u22 gc c p

(4.27)

[Assumption: radial blades, \ b2 = 90°, Vw2 = u2] (T2 – T1) = static temperature rise between the impeller inlet and exit. = (1/2) × total temperature rise

(T2  T1 )

'T0 2

u22 2 gc c p

(4.28)

We have Q2 A2V2 V2 m 3 /s V2 = exit velocity V2 volume flow rate at exit

p2V2 \

V2

mRT2 ; mRT2 ; p2

(4.29)

p1V1

mRT1

V1

mRT1 p1

Centrifugal Compressors and Pumps

\

\

mR T2 p2 mR T1 p1

Q2 Q1

V2 V

Q2 Q1

T2 /T1 p2 /p1

1

T2 p2 T1 p1

265

p1 T2 p2 T1

(4.30)

At impeller inlet, For radial flow at inlet, Vf1 \

Q1 = A1V1 = pd1b1Vf1 = Va1 = V1 Q1 = A1V1 = pd1b1V1

Q S d1V1 For n blades on the impeller each of thickness t, we get

\

b1

b1

At impeller outlet,

Q (S d1  nt ) V1

Q S d2V2 For n blades on the impeller each of thickness t, we get b2

b2

Q (S d2  nt ) V2

(4.31)

(4.31a)

(4.31b)

(4.31c)

4.11 SURGING OF CENTRIFUGAL COMPRESSORS At low flow rates an unsteady flow and vibration called surging can occur. Consider a centrifugal compressor operating at constant speed and as per the characteristic curve shown in Figure 4.11. Let point D indicate the operating or design point of the curve. Assume the machine to be P delivering fluid to a tank or pipe system from which D it is removed for use. p2¢ E S If the demand on the system is increased, the delivered flow from the unit as well as from the machine will be increased and the operating point moves towards right of D, i.e. N, and the delivery pressure is decreased. If the demand from the system decreases the delivery pressure increases, it continues N Q up to P. If the demand from the system decreases further (left to P) the delivery pressure, p03, will Figure 4.11 Surging phenomenon of centrifugal compressor. continue to decrease causing a further drop in flow

266

Fundamentals of Turbomachinery

and a further drop in p03, and so on until a point S is reached, where the flow (volume or mass) is zero (the delivery from the machine stops). The mass flow may even become negative through the compressor (fluid tends to flow back into the machine). If the fluid is still removed from the system, then the pressure gradually drops and it will become less than the pressure indicated by the point S. When this occurs the machine starts delivering fluid but will not build up pressure rapidly enough to follow the curve from S to P. The increased flow from the machine causes the pressure to rise rapidly along E to P and the delivery ceases until the system pressure drops below S. The point P is known as the pulsation point, the highest point in the characteristic curve. The frequency and intensity of pulsation depends upon the slope of the curve, the rise at which the fluid is being removed, the nature of the fluid handled, head produced and the number of stages. Surging should be avoided since the effects of surge can cause serious damage to the machine. In small machines the effect may go unnoticed. A further increase in mass flow (away from N) sees the slope of the curve increasing until it is almost vertical at point N, where the pressure rise is zero. Theoretically, point N would be reached when all the input power is absorbed in overcoming internal friction.

CENTRIFUGAL PUMPS

4.12 INTRODUCTION In general, pump is a power absorbing device used to increase the pressure energy of the liquid and lifts liquids from a lower to a higher level at the expense of mechanical energy. In other words, it is defined as a machine, which converts mechanical energy into pressure energy. It can also be defined as a machine, which converts the mechanical energy into fluid (liquid) energy. Pumps can be broadly classified into two categories: (a) positive displacement pumps and (b) rotadynamic or dynamic pressure pumps. In this chapter we shall discuss the rotadynamic pumps, i.e. centrifugal pumps only.

4.13 CENTRIFUGAL PUMPS A turbomachine, which converts mechanical energy into pressure energy by means of centrifugal action on the liquid and raises the liquid from a lower level to a higher level is known as centrifugal pump. When a certain amount of liquid is rotated by an external energy (mechanical energy) inside the pump casing, a forced vertex is set up, which raises the pressure head of the rotating liquid purely by centrifugal action.

4.14 WORKING PRINCIPLE The first step in the operation of a centrifugal pump is priming. Priming is the operation of filling the suction pipe, casing and a portion of the delivery pipe (up to the check valve) with

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267

the liquid to be pumped in order to avoid any air bubble in the line up to the check valve. Even the presence of a small air pocket may result in no delivery of liquid from the pump. After the pump is primed, the delivery valve is still kept closed and the electric motor is started to rotate the impeller. The rotation of the impeller produces a forced vertex, whcih results in an increased pressure. When the delivery valve is opened, the liquid is made to flow in an outward radial direction with increased velocity and pressure. When the fluid is discharged to the delivery, a partial vaccum is created near the eye of the impeller causing the liquid from the sump (which is at atmospheric pressure) to rush through the suction pipe to replace the liquid that is being discharged. The casing should be so shaped that part of the kinetic energy is converted into useful pressure energy. The developed head is purely due to the whirling motion of the liquid imparted by the rotating impeller and is not due to any displacement or impact.

4.15 MAIN PARTS OF A CENTRIFUGAL PUMP Figure 4.12 illusrates the following main parts of a centrifugal pump. 1. Impeller 2. Casing 3. Suction pipe 4. Delivery pipe 5. Delivery valve or check valve or regulating valve.

Figure 4.12

Main parts of a centrifugal pump.

4.15.1 Impeller The impeller is a wheel provided with a series of backward curved vanes. It is mounted on a shaft, which is connected to an electric motor.

268

Fundamentals of Turbomachinery

4.15.2 Casing Casing in an airtight chamber surrounding the impeller. It is similar to the casing of a reaction turbine. Types of casings are: 1. Volute casing 2. Volute casing with vertex chamber 3. Diffuser casing

4.15.3 Suction Pipe, Foot Valve and a Strainer It is a pipe connected to the centre of the impeller, which is known as an eye. The lower end of the suction pipe dips into liquid in a sump from which the liquid is to be lifted up. The lower end of the suction pipe is fitted with a foot valve and strainer. The strainer will avoid the entry of any foreign matter. The foot valve is a non-return or one-way type valve, which opens only in the upward direction. To minimize the losses in the suction pipe, the diameter should be large and bends should be avoided. Care should be taken so that there is no leakage of air on the suction side.

4.15.4 Delivery Pipe The lower end of the delivery pipe is connected to the outlet of the pump and the other end is connected to the tank, where the liquid is stored or pumped from. Usually the diameter of the suction and delivery pipes must be same and its value depends upon the quantity to be lifted per unit time.

4.15.5 Delivery Valve or Check Valve or Regulating Valve A check valve or delivery valve is provided in the delivery pipe near the pump to regulate the discharge from the pump. This valve should be closed before the pump is switched on and it is opened as soon as the pressure builds up. Also, the valve should be closed before the pump is switched off so that the delivery pressure is not transmitted to the suction pipe.

4.16 CLASSIFICATION OF CENTRIFUGAL PUMPS 4.16.1 According to the Working Head Low head pump • • • •

Head limited to 15 m. Liquid may enter either from one or from both sides of the impeller depending upon the flow rate of liquid. Usually a horizontal shaft is used. Type of casing is volute with no guide vanes.

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269

Medium head pump • • •

Head between 15 m to 40 m. Liquid may enter either from one or from both sides of the impeller depending upon the flow rate of liquid. Volute casing with guide vanes.

High head pump • • • •

Head above 40 m. Usually a multistage arrangement. Shaft may be horizontal or vertical. For deep wells, vertical shafts are used.

4.16.2 According to the Type of Casing Volute casing Volute casing is of spiral form with increasing cross-sectional area towards the discharge end (Figure 4.13(a)). The increase in area of flow decreases the velocity of flow. The pressure of the liquid increases as the velocity decreases. It acts as a diffuser. In order to accommodate more and more water from tongue to eye and to maintain the mean velocity of flow constant at any cross-sectional area, the increasing cross-sectional area of casing is necessary. Loss of kinetic head due to eddy formation is avoided.

Figure 4.13

Types of casing: (a) Volute. (b) Vortex. (c) Diffuser.

Vortex casing In a vortex casing a circular chamber is introduced between the impeller and the casing (Figure 4.13(b)). The efficiency of this type of casing is more compared to the simple volute casing because loss of head due to eddies is reduced considerably.

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Fundamentals of Turbomachinery

Diffuser casing Diffusion casing consists of guide wheels with stationary vanes called diffuser (Figure 4.12(c)) provided at the outlet of impeller vanes. Water leaving the impeller enters the guide vanes without shock. The area of the fixed vanes (guide vanes or diffusion blade vanes) increases. Therefore, the velocity of flow through guide vanes reduces and consequently the pressure of water increases. The liquid leaving the guide vanes enters the collecting volute of uniform or varying cross-sectional area.

4.16.3 According to Fluid Entrance to the Impeller Single entry pump If water enters from one side of the impeller, it is called the single entry pump. See Figure 4.14(a).

Figure 4.14

Types of fluid entrance: (a) Single entry pump. (b) Double entry pump.

Double entry pump If water enters from both sides of the impeller, it is called the double entry pump. See Figure 4.14(b). This type of pump is suitable for large flow rates of liquid.

4.16.4 According to the Direction of Flow of Water through the Impeller Radial flow pump If the flow through the impeller is in the radial direction, it is known as the radial flow pump (Figure 4.15(a)). This is suitable for high head and low discharge. Almost all centrifugal pumps are designed for radial flow.

Centrifugal Compressors and Pumps

Figure 4.15

271

Types of flow direction: (a) Radial flow. (b) Mixed flow. (c) Axial flow.

Axial flow pump It is the reverse of an axial flow turbine (Figure 4.15(c)). Here, centrifugal force is not used for creating increased pressure head and, therefore, it is not a centrifugal pump. This pump is suitable for high discharge and low head.

Mixed flow pump A mixed flow is a combination of radial flow and axial flow type (Figure 4.15(b)). This pump is suitable for high discharge and low head. Mixed flow pumps are suitable for irrigation purposes.

4.16.5 According to Number of Impellers Single stage pump If a centrifugal pump consists of one impeller, then it is known as single stage pump. It is usually used for low heads. The position of the shaft may be horizontal or vertical.

Multistage pump If a centrifugal pump consists of two or more impellers, the pump is called multistage pump. The impeller may be mounted on the same shaft or on different shafts (Figure 4.16). The types of multistage pumps are: (a) To produce high head (b) To discharge a large quantity of head.

4.16.6 According to Liquid Handled Open impeller pump An open impeller pump (Figure 4.17(a)) is used when the liquid to be lifted is a mixture of liquid (75%) and solids (25%), for example, to lift slurry (mixture of sand and water).

272

Fundamentals of Turbomachinery

Delivery Impeller

Common delivery pipe

Inlet Shaft

Delivery Pump 2

Pump 1

Impeller Impeller Delivery

Guided passage

Pump 1

(a)

Figure 4.16

Figure 4.17

Impeller Pump 2 (b)

Multistage pumps: (a) Series arrangement. (b) Parallel arrangement.

Types of pumps based on liquid handled: (a) Open impeller. (b) Semi-open impeller. (c) Closed impeller.

Semi-open impeller If the vanes are covered with shroud on one side of the impeller only (Figure 4.17(b)), then it is a semi-open impeller pump, for example, a pump to lift sewage water.

Closed impeller pump If the vanes are covered with shrouds on both sides of the impeller (Figure 4.17(c)), then it is a closed impeller pump, for example, a pump to lift non-viscous liquids like water.

4.16.7 According to Specific Speed Low specific speed If the specific speed is between 10–30 rpm, then the pump is called low specific speed pump, for example, the radial flow.

Medium specific speed If the specific speed is between 30–50 rpm, then the pump is called medium specific speed pump, for example, the radial flow.

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273

High specific speed If the specific speed is between 40–450 rpm, then the pump is called the high specific speed pump, for example, the radial pump (50–80 rpm), the mixed flow pump (80–160 rpm), and the axial flow pump (100–450 rpm).

4.17 HEADS OF A CENTRIFUGAL PUMP The heads of a centrifugal pump may be expressed in two ways. 1. Static head 2. Manometric head or effective head or total head or gross head or head actually measured across the pump.

4.17.1 Static Head (HS) The static head is the sum of suction head (hs) and delivery head (hd), i.e. H S = hs + hd

(4.32)

Suction head (hs): It is the vertical height between the centre line of the centrifugal pump and top surface of the liquid. See Figure 4.12. Delivery head (hd): It is the vertical height between the centre line of the pump and the water surface in the overhead tank to which water is delivered. See Figure 4.12.

4.17.2 Manometric Head (Hm) It is defined as the head against which a centrifugal pump has to work. (a) Hm = head imparted by the impeller – head loss in the pump (HL) = work done by the impeller on the fluid per kg of liquid – losses within the pump (HL) Vw 2 u2 Û  H L J/kg Ü gc Ü Ü or Ü Vw 2 u2 Ü  HL m Ü g Ý (b) Hm = total head at outlet of the pump – total head at inlet of the pump.

(4.33)

Ë po Vo2 Û Ëp Û V2   Z o Ü  Ì i  i  Zi Ü Ì ÍÌ U g 2 g ÝÜ ÍÌ U g 2 g ÝÜ Ë pd Vd2 Û Ëp Û V2   Zd Ü  Ì s  s  Zs Ü Ì ÍÌ U g 2 g ÝÜ ÍÌ U g 2 g ÝÜ

(4.34)

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Fundamentals of Turbomachinery

po = pressure head at outlet of the pump = hd Ug

Vo2 = velocity head at outlet of the pump. 2g 2 = velocity head in delivery pipe = Vd /2 g

Zd = vertical height of the outlet of the pump from the datum. Here the suffix i or s represents the inlet conditions and o or d represents the outlet conditions i.e. (c)

Vi2 2g

Vs2 , Zi 2g

pi Ug

hs ,

Hm

hs  hd  h fs  h fd 

Zs Vd2 2g

(4.35)

hfs = frictional head loss in suction pipe hfd = frictional head loss in delivery pipe Vd = velocity of water in delivery pipe. Vs2 Vd2 and if the pressure 2g 2g gauges are at the same level, (Zs = Zd), then Eq. (4.34) reduces to

If the diameter of suction and delivery pipes are same,

Hm

pd p  s Ug Ug

hd  hs

(4.36)

4.18 EFFICIENCIES OF CENTRIFUGAL PUMP 4.18.1 Manometric Efficiency (hmano) It is the ratio of the manometric head to the head imparted by the impeller to the fluid.

Kmano

Hm Vw 2 u2 / g

Hm g Vw 2 u2

Û Ü Ü Ü Power delivered by the pump, measured in head Ü Power imparted by the impeller (head imparted by impeller) ÜÝ

Hm H m  losses

Hm Hm  HL

Water Power Power of the impeller

Head imparted by the runner (impeller) is as per the ideal velocity triangle.

(4.37)

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275

4.18.2 Mechanical Efficiency (hm) It is the ratio of the power actually delivered by the impeller to the power supplied at the shaft. Km

Mechanical energy supplied to rotor Mechanical energy supplied to shaft

Km

Vw 2 u2 / gc Vw 2 u2 / gc  Mechanical losses

Impeller power Shaft power (SP)

(4.38a)

Power at the impeller = Power actually delivered by the impeller = Work done by impeller per second.

4.18.3 Hydraulic Efficiency (hH) KH

Useful hydrodynamic energy in fluid Mechanical energy supplied to rotor

(4.38b)

4.18.4 Volumetric Efficiency (hv) Actual discharge Theoretical discharge

Kv

(4.38c)

4.18.5 Overall Efficiency (ho) It is the ratio of the power output of the pump to the power input to the pump. Ko

Water power Shaft power

WP SP

Ko

Water Power Impeller Power – Impeller Power Shaft Power

(4.39) Kmano – Km

(4.40)

4.19 WORK DONE BY THE PUMP Figure 4.18 shows the velocity diagrams at inlet and outlet of the impeller of a centrifugal pump. Centrifugal pump is a turbomachine, hence, Euler’s turbine equation can be used here. We have,

W.D. = workdone on water by the impeller = (Vw2u2 – Vw1u1)/gc J/kg

(4.41)

For maximum work done by the impeller, water should enter radially, i.e. a1 = 90°, therefore, Vw1 = 0. Therefore, Eq. (4.41) reduces to W.D.

Vw 2 u2 J/kg gc

(4.42)

276

Fundamentals of Turbomachinery

Figure 4.18

Velocity triangles for centrifugal pumps.

We have W.D.

(u22  u12 )  (V22  V12 )  (Vr21  Vr22 ) 2 gc

(4.43)

From Figure 4.18, we have Vr21

u12  V12

Vr22

V f22  (u2  Vw 2 )2

(4.44)

V f22  u22  Vw22  2u2Vw2

V22  u22  2u2Vw2

(4.45)

Substituting Eqs. (4.44) and (4.45) in (4.43), W.D. =

u22  u12  V22  V12  u12  V12  V22  u22  2 u2Vw 2 2 gc Vw2 u2 J/kg gc

(Eqs. (4.42) and (4.46) are the same) OR

Referring to Figure 4.12, we have Point i = suction inlet, 2 = impeller outlet and

1 = impeller inlet d = casing outlet

(4.46)

Centrifugal Compressors and Pumps

277

Applying Bernoull’s equation between i and 1, we get pi Vi2   Zi Ug 2g

p1 V12   Z1  Losses between i and 1 Ug 2g

(4.47)

Applying Bernoull’s equation between 1 and 2, we get (Water enters the impeller with Vr1 and leaves the impeller with Vr2.) p1 Vr21 (u2  u12 )   Z1  2 U g 2g 2g

p2 Vr22   Z 2  Losses between 1 and 2 U g 2g

(4.48)

Applying Bernoull’s equation between 2 and d, we get p2 V22   Z2 Ug 2g

Pd Vd2   Z d  Losses between 2 and d Ug 2g

(4.49)

Adding Eqs. (4.47) and (4.48) and (4.49),

pi Vi2 p V2 V2 u2 u2 p   Zi  1  r1  Z1  2  1  2  2  Z 2 Ug 2g U g 2g 2 g 2 g U g 2g p V2 p p1 V12 V2   Z1  2  r 2  Z 2  d  d  Z d  Total losses Ug 2g Ug 2g Ug 2g or

Ë pd Vd2 Û Ëp Û V2   Z d Ü  Ì i  i  Zi Ü  Total losses = Ì ÍÌ U g 2 g ÝÜ ÍÌ U g 2 g ÝÜ (u22  u12 ) (V22  V12 ) (Vr21  Vr22 )   2g 2g 2g

Hm  H L

(4.50)

(Suffix ‘i’ can be replaced by ‘S’, i.e. suction) \

W.D. = Hm + HL = work imparted by the impeller Equations (4.33) and (4.50) are same. P = power imparted by the impeller to the water = W.D. × r × Q u2 Vw 2 UQ gc

m m kg m 3 N- s2 s s m 3 s kg -m

N-m or W s

(4.51)

where Q = discharge = p d1 b1 Vf1 = p d2 b2 Vf2 (m3/s) r = density of the fluid (kg/m3) b = width of the impeller (m) Vf = flow velocity (m/s)

(4.52)

278

Fundamentals of Turbomachinery

4.20 PRESSURE RISE IN PUMP, IMPELLER AND MANOMETRIC HEAD Applying Bernolli’s equation to inlet and outlet, Energy at inlet + work done by impeller = Energy at outlet. From Eq. (4.48), V1 = Vf1, Z1 = Z2 and neglecting losses p1 Vr21 (u22  u12 )   Ug 2g 2g

p2 Vr22 ( p2  p1 )  ; Ug 2g Ug

(Vr21  Vr22 ) (u22  u12 )  2g 2g

(4.53)

Substituting Eqs. (4.44) and (4.45) in RHS of (4.53), u12  V12  V22  u22  2u2Vw 2  u22  u12 ) 2g

V12  V22  2u2Vw 2 2g

V f21  V22  2Vw 2 u2

V f21  (Vw22  V f22 )  2Vw 2 u2

2g

2g

(' V22

Vw22  V f22 )

V f21  Vw22  V f22  2Vw 2 u2 2g V f21  V f22  (u2  V f 2 cot E2 )2  2u2 (u2  V f 2 cot E2 ) 2g

(' Vw 2

u2  V f 2 cot E2 )

V f21  V f22  u22  V f22 cot 2 E 2  2u2 V f 2 cot E2  2u22  2u2V f 2 cot E 2 2g

\

( p2  p1 ) Ug

V f21  V f22  u22  V f22 cot 2 E2

V f21  u22  V f22 (1  cot 2 E2 )

2g

2g

(V f21  u22  V f22 cosec 2 E2 ) 2g

pressure rise (m)

(4.54)

4.21 MINIMUM STARTING SPEED When the pump is started, there will be no flow until the pressure difference in the impeller is large enough to overcome the manometric head. If the impeller is rotating and if there is no flow, then the water is rotating in a forced vertex. Centrifugal pressure head for no flow of water = (u22  u12 ) / 2 g Unless this pressure head is equal to or greater than the manometric head, the pump will not deliver water. By this the minimum speed can be determined.

Centrifugal Compressors and Pumps

The flow will commence only if (u22  u12 ) / 2 g • H m 2

2

2

2

i.e.

1 Ë S d2 N Û 1 Ë S d1 N Û  • Hm 2 g ÌÍ 60 ÜÝ 2 g ÌÍ 60 ÜÝ

i.e.

1 2g

Ë ÌK mano Í

279 (4.55)

gH m Û Ü Vw2 u2 Ý

Vw 2 u2 Vw 2 1 Ë S d1 N Û Ë S d2 N Û Ì 60 Ü  2 g Ì 60 Ü • Kmano g • Kmano g Í Ý Í Ý

Ë S d2 N Û Ì 60 Ü Í Ý

For minimum speed, using the equal sign, S 2N2 2g

Ë ( d22  d12 ) Û Vw 2 S d2 N Ì Ü K mano g 60 ÌÍ 3600 ÜÝ

(4.56)

60

\

N

3600 – 2Kmano Vw 2 d2 S – (d22  d12 ) – 60

minimum speed for a centrifugal pump.

4.22 MULTISTAGE PUMPS We know that, the head developed by a centrifugal pump is proportional to the diameter and speed of the impeller. However, there is a limitation for the diameter and speed of the impeller. Therefore, maximum head developed with a single impeller is up to 50 m. For more head and discharge, a multistage pump is preferred. Figure 4.16 shows the different arrangements of multistage pumps. The liquid leaves the suction pipe, enters the first impeller at inlet and is discharged at outlet with increased pressure. The liquid leaving from the first impeller enters the second impeller as shown in Figure 4.16(a). The pressure of the liquid leaving the second impeller is more than the pressure of the liquid leaving the first stage. This arrangement is called the series arrangement. Total head developed = n1 × Hm

(4.57)

Figure 4.16(b) shows parallel arrangement of the pumps. The liquid leaving each pump is discharged into a common pipe. Each pump is dipped in the same sump. In this case, the discharge obtained is more against the same head. Total discharge = n2 × Q where, n1 = number of identical impellers mounted on the same shaft, Hm = head developed by each impeller (pump), n2 = number of identical pumps arranged in parallel and Q = discharge from each pump (impeller).

4.23 CAVITATION Cavitation occurs on the suction side of the pump as lowest pressure exists just below the pump on the suction side. Due to height of installation of the pump above the sump, the pressure on

280

Fundamentals of Turbomachinery

the suction side is below the atmospheric value. The head drop across foot-valve and the frictional loss and also kinetic head all contribute to such sub-atmospheric pressures. If the pressure drops below the vapour pressure head, vaporization and bubble formations occur at the inlet to runner. Soon, energy is added, and therefore bubbles collapse. This results in water rushing to such spots. This causes mechanical failure too, cavities are formed on surfaces. This whole process is cavitation.

4.24 EXAMPLES (Centrifugal Compressors) EXAMPLE 4.1 Air at a temperature of 290 K, flows in a centrifugal compressor running at 20,000 rpm, slip factor = 0.8, ht–t = 0.80, d2 = 0.60 m. Assume that the absolute velocities at the inlet and outlet are same. Calculate (a) the temperature rise of air passing through the compressor and (b) the stage pressure ratio. Solution: Stagnation temperature at inlet: Speed: Slip factor:

T01 = 290 K N = 20,000 rpm s = 0.8 Kt to

Total-to-total efficiency: Exit diameter (impeller exit): Absolute velocity:

d2 = 0.6 m V1 = V2

To determine: T03 , pR 0

T or h

„

u2

02¢

S d2 N 60

02 03

03¢

S – 0.6 – 20000 = 628.318 m/s 60 From Eq. (4.15),

W.D. =

0.8

V u22

0.8 – 628.3182

gc

m s

2

2

3 2

2¢

N-s kg-m

01 1

= 315827.341 J/kg = 315.83 kJ/kg (a) The temperature rise of air passing through the compressor (T03¢): W.D. = Dh0 = cp(T03 – T01) (Eq. (4.12e)) or

315827.341 = 1005(T03 – 290)

\

T03 = 604.26 K Kc t  t0

\

T03

„

(T03  T01 ) (T03  T01 ) „

541.41 K

0.8

T03  290 604.26  290 „

p3

p2

3¢

2

p02 p03

(Eq. (4.10a))

p01 p1 s

281

Centrifugal Compressors and Pumps

(b) Stage pressure (pR0): We have for isentropic process J

pR 0

Ë T03 Û J  1 Ì Ü Í T01 Ý

p03 p01

„

1.4

Ë 541.41 Û 0.4 Ì 290 Ü Í Ý

8.891

Ans.

EXAMPLE 4.2 Free air delivered by a compressor is 20 kg/min. The inlet conditions are 1 bar and 20°C static. The velocity of air at the inlet is 60 m/s. The isentropic efficiency of the compressor is 0.7. The total head pressure ratio is 3. Find (a) the total head temperature at the exit and (b) the power required by the compressor if the mechanical efficiency is 95%. Solution: Mass flow rate: m = 20 kg/min Inlet static temp.: Inlet static pressure: p1 = 1 bar Ab. velocity at inlet: Total-to-total Total head efficiency: pressure ratio: K t  t0 0.7 Mechanical efficiency: hm = 0.95% We have

T01

T1  V12 / 2 gc c p 293 

02¢

\

p01

T03

Ëp Û T01 Ì 03 Ü Í p01 Ý

„

„

1.4 Ë 294.79 Û 0.4

2¢

p02 p03

p3

P2 p01 p1

01 1

s

1

J

J

3 2

Ë T01 Û J Ì Ü Ì 293 Ü Í Ý Í T1 Ý = 1.021 bar

T03

03

3¢

1

Ëp Û T01 Ì 03 Ü Í p01 Ý

02

03¢

= 294.79 K

p01 p1

p03/p01 = pR0 = 3

T or h

60 2 2 – 1 – 1005 J

T1 = 20 + 273 = 293 K V1 = 60 m/s

J

1 J

0.4

294.79 – 3 1.4

403.5 K

(a) Total head temperature (T03): Kc t to 0.7

(T03  T01 ) (T03  T01 ) „

403.5  294.79 ; T03  294.79

?

T03

450.09 K

Ans.

282

Fundamentals of Turbomachinery

(b) Power required (P):

m – c p – (t03  t01 )

m – W.D. Km

P

Km

20 1.005 – (450.09  294.79) – 60 0.95

54.76 KW

Ans.

EXAMPLE 4.3 A two-stage centrifugal compressor delivers 500 m3 of free air per min. The suction conditions are 1 bar and 15°C. The compression ratio and isentropic efficiency of each stage are 1.25 and 80% respectively. Find the isentropic efficiency for the entire compression process. Solution:

Data: Two stage centrifugal compressor

Mass of air delivered:

m

Total stagnation temperature at inlet: Total stagnation pressure at inlet:

T01 p01

Compression ratio of each stage:

pR1

„

or Now, or

T02

„

K c t  t1

0.8

Ë p02 Û Ì Ü Í p01 Ý

J

p02 p01

Kc t t1

Isentropic efficiency of each stage:

T02 T01

500 m 3 /min = 15 + 273 = 288 K = 1.0 bar

0.8 Kc t t2

1 J

T or h

288 – 1.25

0.4 1.4

306.96 K

T03¢

(T02  T01 ) (T02  T01 ) „

T02¢

306.96  288 ; T02  288

\ T02 = 311.7 K

J

1

T03” T02

Ë p03 Û Ì Ü Í p02 Ý

T03”

311.7 – 1.25 1.4

J

0.4

For the second stage, we have K c t  t2

(T03  T02 ) (T03  T02 ) „

332.22 K

p03

T03 T03²

We have for the second stage,

\

p03 p02

pR 2

1.25

T01

T02

ep tag 02 II s

e p01

tag

Is

s

Centrifugal Compressors and Pumps

or

0.8

332.22  311.7 ; T03  311.7

\

283

T03 = 337.34 K

pR0 = total pressure ratio = overall pressure ratio pR0 = (pR)2 = (1.25)2 = 1.5625 We have for the isentropic process, 1

J

1

 pR0 J

T03

 pR 0

„

or

J

T03 T01 „

J

0.4

1.5625 1.4 – T01

0.4

288 – 1.5625 1.4

327.167 K

Overall isentropic efficiency (isentropic efficiency for the entire process (K c t -to ) ): Kc t  to

(T03  T01 ) (T03  T01 ) „

327.167  288 337.34  288

0.7938

Ans.

EXAMPLE 4.4 The inlet conditions of a centrifugal compressor are 1 bar 30°C, running at 10000 rpm. It delivers a free air stream of 1.5 m3/s. The compression ratio is 5. The velocity of flow is 50 m/s and is constant. Assume that the blades are radial at outlet. The slip factor is 0.92. Calculate (a) the temperature of air at outlet, (b) the power required, (c) the impeller diameter, (d) the blade angle at inlet, and (e) the diffuser inlet angle. Assume that power factor is 1.11 and isentropic efficiency 0.90. Solution:

Data: Static inlet pressure: Static inlet temperature: Speed: Volume of air delivered: Compression ratio: Flow velocity: Radial blade at exit: Slip factor: Power factor: Isentropic efficiency:

To determine: T2, P, d2, b1, a2 (a) Static exit temperature (T2): We have for isentropic process, 1–2¢

p1 = 1 bar T1 = 30 + 273 = 303 K N = 10000 rpm v = 1.5 m3/s pR = 5 Vf1 = Vf2 = 50 m/s b2 = 90°, u2 = Vw2 s = 0.92 j = 1.11 hc s-s = 0.9

284

Fundamentals of Turbomachinery J

T or h

1

Ëp Û T2 T1 Ì 2 Ü 303  5 Í p1 Ý From Eq. (4.12b), we have J

„

Kc s  s

0.4 1.4

02¢

479.897 K

T2 = 499.55 K

p02 p03

p3

p2 2

2¢

p01 p1

01

(479.897  303) (T2  303)

\

3

3¢

„

0.9

03

03¢

(T2  T1 ) (T2  T1 )

or

02

1

s

Ans.

(b) Power (P): We have

pv

mRT

\

m

p1v RT1

100 – 1.5 kN m 3 kg- K 1 0.287 – 303 m 2 s kN - m K

1.725 kg/s

From Eq. (4.3), we have W.D. = Dh0 = mcp(T2 – T1) P 1.725

kg kJ – 1.005 (499.55  303) K s kg-K

339.05 kW

Ans.

(c) Impeller diameter (impeller outer diameter, d2): From Eq. (4.18), we have P

or

Now,

339.05

576.2

'h0

W.D.

V M u22 gc – 1000

0.92 – 1.11 – u22 ; 1 – 1000

S d2 N ; 60

\

\

u2 = 576.20 m/s

d2 = 1.1 m

(d) Blade angle at inlet (b1): Let the impeller diameter at inlet be 0.5d2 \

d1 = 0.5d2 = 0.5 × 1.1 = 0.55 m

Now,

u1

S d1 N 60

S – 0.55 – 10000 = 287.98 m/s 60

Ans.

285

Centrifugal Compressors and Pumps

From inlet velocity triangle,

Vf 1

tan E1

50 287.98

u1

0.1736 ;

\

b1 = 10.94°

Ans.

(e) Diffuser angle at inlet (a2): From exit velocity triangle,

Vf 2

tan D 2

50 0.92 – 576.20

V u2

0.0943;

?

D2

5.39’ (Vw 2

u2 )

Ans.

EXAMPLE 4.5 The following data refers to a centrifugal compressor. Calculate the compressor pressure ratio. Tip speed = 550 m/s. Static temperature at inlet = 300 K. Power factor = 1.05. Slip factor = 0.85. g = 1.4. cp = 1.005 kJ/kg-K. Isentropic efficiency = 0.82. Solution:

Data: Tip speed u2 = 550 m/s, hc s-s = 0.82

From Eq. (4.18), we have V M u22 gc – 1000

W.D. =

0.85 – 1.05 – 550 2 1 – 1000

c p (T2  T1 ) 1.005(T2  T1 )

\

(T2 – T1) = actual temperature rise = 268.64 K

\

(T2  T1 ) = isentropic temperature rise = hc s-s × (T2 – T1) „

= 0.82 × 268.64 = 220.28 K \

T2

„

T1  220.28

300  20.28

520.28 K

Compressor pressure ratio (pR):

pR

p2 p1

Ë T2 ÛJ Ì Ü Í T1 Ý „

J

1

1.4

Ë 520.28 Û 0.4 Ì 300 Ü Í Ý

6.87

Ans.

286

Fundamentals of Turbomachinery

EXAMPLE 4.6 Initial conditions of air entering a centrifugal compressor are 1 bar and 10°C static. The power input to the compressor is 450 kW. The total pressure at exit is 5 bar. The velocity of air at inlet is 150 m/s and the speed of the compressor is 20,000 rpm. The hub diameter is 12 cm. Assume isentropic efficiency as 0.8 and slip factor as 0.9. Calculate (a) the change in total temperature, (b) the impeller diameter at outlet and inlet, and (c) the mass flow rate of air. Solution:

Data: Machine: Inlet static temperature: Inlet static pressure: Power input: Total pressure at exit: Inlet velocity: Speed: Hub diameter: Isentropic efficiency: Slip factor:

Centrifugal compressor T1 = 10 + 273 = 283 K p1 = 1 bar P = 450 kW p02 = 5 bar V1 = 150 m/s N = 20,000 rpm dh = 12 cm hc t-t = 0.8 s = 0.9

To determine: (T02 – T01), d2 , d1 , m

T01

T or h

V2 T1  1 2 gc c p

02¢

02 03

03¢

T01

283 

3

2

150 2 – 1000 – 1.005 J

2 2¢

Ë T01 ÛJ Ì Ü Í T1 Ý

p 01

Ë 294.2 Û p1 – Ì Ü Í 283 Ý

100

1

kN

Ë 294.2 Û –Ì Ü 2 m Í 283 Ý

3.5

114.6 kN/m 2

We have for isentropic process 02–02¢,

T02

„

Ëp Û T01 Ì 02 Ü Í p01 Ý

J

1 J

0.4

Ë 500 Û 1.4 283 Ì Ü Í 114.6 Ý

431.1 K

(a) Actual rise in stagnation temperature (T02 – T01): Kc t t

p01 p1

01

Ì 283 Ü Í Ý 3.5

p3

p2

3¢

1.4 Ë 294.2 Û 0.4

p01 p1

1

294.2 K

p02 p03

(T02  T01 ) (T02  T01 ) „

s

Centrifugal Compressors and Pumps

\

T02  T01

T02  T01 Kc t t „

431.1  283 0.8

287 Ans.

185.13 K

(b) Mass flow rate (m  ): P = power input = W.D. = mcp(T02 – T01) 450 kW

m

(Eq. (4.3))

kg kJ – 1.005 – 185.13 K; ? m = 2.42 kg/s s kg-K

Ans.

(c) Impeller exit diameter (d2): From Eq. (4.15), we have W.D. = P

m u22 V gc

450 kW = 2.42

or \

kg s

– u22

m2 s2

– 0.9

kN- s2 kg -m – 1000

u2 = 454.55 m/s

or

454.55

S d2 N 60

S – d2 – 2000 ; \ d2 = 0.434 m 60

Ans.

(d) Impeller inlet diameter (d1): r1 = density of air at inlet =

m

2.42 \

p1 RT1

100 kN kN -m m 2 – 0.287 – – 283 K kg-K S 2 (d1  d h2 ) V f 1 U1 4

1.2312 kg/m 3

S 2 m2 kg (d1  0.122 ) m 2 – 150 2 – 2 – 1.23 3 4 s m d1 = 0.12046 m

kg s

(V1

V f 1 , D1

90’) Ans.

EXAMPLE 4.7 A centrifugal compressor runs at a speed of 550 m/s with no prewhirl. Suppose the slip is 0.95 and isentropic efficiency of compressor is 0.85. Calculate the following for standard sea level: (a) Pressure ratio, (b) the work required, and (c) the power required. Assume, mass flow rate of 25 kg/s and cp = 1.005 kJ/kg-K, and ambient temperature of 15°C.

288

Fundamentals of Turbomachinery

Solution:

Data: Speed of the impeller at exit: Slip factor: Isentropic efficiency: Mass flow rate: Stagnation temperature:

u2 = 550 m/s s = 0.95 hc t-t = 0.85 m = 25 kg/s T01 = 15 + 273 = 288 K

To determine: pR1, W.D., Power (a) Pressure ratio (pR1): J

pR1

Ë u22 Kc t t V Û J  1 Ì1  Ü c p gc T01 Ü ÌÍ Ý

p02 p01

(Eq. (4.16))

1.4

Ë 550 2 – 0.85 – 0.95 Û 0.4 1  Ì Ü ÍÌ 1.005 – 288 – 1000 ÝÜ

1.19

Ans.

(b) The work required (W.D.): From Eq. (4.15), we have W.D. = 'h0

V u22 gc

0.95 – 550 2

m 2 N- s2 s2 kg-m

28737.5 J/kg

Ans.

(c) Power required to deliver 25 kg/s (P): We have P

W.D – m

28.738

kg kJ – 25 s kg

718.45 kW

Ans.

EXAMPLE 4.8 A centrifugal compressor runs at a speed of 15000 rpm and delivers 30 kg of air per second. Exit radius is 0.35 m, relative velocity at exit is 100 m/s at an exit angle of 75°. Assume axial inlet and T01 = 300 K and p01 = 1 bar. Calculate (a) the torque, (b) the power required to drive the compressor, (c) the ideal head developed, (d) the work done, and (e) the exit total pressure. Solution:

Data:

Speed: Mass flow rate: Exit diameter:

N = 15000 rpm m 30 kg/s d2 = 0.7 m

Centrifugal Compressors and Pumps

Exit relative velocity: Exit angle Axial inlet

289

Vr2 = 100 m/s b2 = 75° a1 = 90°, Vw1 = 0, V1 = Vf1

To determine: T, P, H, W.D, p02 u2 = tip rotor velocity

S d2 N 60 From exit velocity triangle,

S – 0.7 – 15000 60

549.8 m/s

Vw2 = u2 – Vr2 cos b2 = 549.8 – 100 cos 75° = 523.896 m/s (a) Torque (T): m r2 Vw 2 gc

T

30

kg s

– 0.35 m – 523.896

m N- s2 s kg -m

5500.91 N-m

Ans.

8640.8 kW

Ans.

(b) Power (P): P

TZ

T – 2S N 60

5500.91 15000 kN-m – 2S – 1000 60 s

(c) Work done (W.D.): W.D. =

u2 Vw 2 gc

550 – 523.896 1 – 1000

288.1428 kJ/kg

(Eq. (4.1))

Ans.

(d) Ideal head (H):

H

u2 Vw 2 g

550

m m s2 – 523.896 – 9.81 m s s

29372.35 m

Ans.

(e) Exit total pressure (p02): J 1 Ë Û Ë È p Ø 0.286 Û Ì È p02 Ø J Ü Ì 02 1 1.005 300  – –  1Ü W.D. = c pT01 Ì É Ù Ü É Ù Êp Ú Ì Ê p01 Ú Ü ÌÍ 01 ÜÝ Í Ý

\

p02 p01

10.44 ; or p02

10.44 bar

288.14

Ans.

290

Fundamentals of Turbomachinery

EXAMPLE 4.9 A centrifugal compressor develops a pressure ratio of 5.0 and an air consumption of 30 kg/s. The inlet temperature and pressure are 15°C and 1 bar respectively. Isentropic efficiency 0.85. Calculate (a) the work done, (b) the total temperature, and (c) the power required. Solution:

Data: Pressure ratio:

pR1 = 5 m 30 kg/s

Mass of air: Inlet temperature: Inlet pressure: Isentropic efficiency:

T01 = 15 + 273 = 288 K p01 = 1 bar hc t-t = 85%

To determine: W.D., T02, P (a) Work done (W.D.): From Eqs. (4.3) and (4.7), we have W.D. = cp(T02 – T01) =

c pT01 Ë T02 Û  1Ü Ì Kc t t Í T01 Ý „



J 1 Ë Û c pT01 Ì È p02 Ø J Ü  1Ü Kc t t Ì ÉÊ p01 ÙÚ ÌÍ ÜÝ

\

0.4 Û 1.005 kN-m – 288 K Ë 1.4 Ì(5)  1Ü 198.8 kJ/kg kg-K – 0.85 ÌÍ ÜÝ

W.D. =

Ans.

(b) Exit total temperature (T02):

T02

T01 

W.D. cp

288 K  198.8

kJ – kg

1 1.005

kJ kg-K

485.81 K

Ans.

5964 kW

Ans.

(c) Power (P): P

W.D. – m 198.8

kJ kg – 30 kg s

EXAMPLE 4.10 A centrifugal compressor with an overall diameter of 100 cm turns at 5000 rpm. Air is supplied to the centrifugal compressor at 20°C and 1 bar. The mass flow rate of air is 25 kg/s. Isentropic pressure ratio is 2.5. Calculate (a) the isentropic efficiency, (b) the rotor power, and (c) the shaft power. Assume slip factor of 0.9.

Centrifugal Compressors and Pumps

Solution:

291

Data: Overall diameter: Speed: Inlet stagnation temperature: Inlet pressure: Mass flow rate: Isentropic pressure ratio: Slip factor:

d2 = 100 cm N = 5000 rpm T01 = 20 + 273 = 293 K p01 = 1 bar m = 25 kg/s pR1 = 1.5 s = 0.9

To determine: hc t–t, P, S.P. (a) Isentropic efficiency (hct–t):

u2

S d2 N 60

S – 1 – 5000 60

261.8 m/s J

pR1

p02 p01

Ë u22 V Kc t t Û J  1 Ì1  Ü c p gc T01 Ü ÌÍ Ý

p02 p01

Ë 261.82 – 0.9 – Kc t t Û 0.4 Ì1  Ü ÌÍ 1000 – 1.005 – 1 – 293 ÜÝ

(Eq. (4.16)) 1.4

or

pR1

1.5; ? Kc t t

0.586

Ans.

(b) Power (P): ËT Û T01 Ì 02  1Ü T Í 01 Ý (T02  T01 ) „

K c t t

\

(T02  T01 )

(T02  T01 ) (T02  T01 ) „

È J 1 Ø T01 É pR1J  1Ù ÉÊ ÙÚ Kc t t 0.4

293 – (1.51.4  1) 0.586

Now,

P

m W.D. 1.005

m – 'h0

61.41 K c p (T02  T01 ) m

kg kJ – 61.41 K – 25 s kg -K

1543 kJ/s (or kW)

Ans.

292

Fundamentals of Turbomachinery

(c) Shaft Power (S.P.): Assume mechanical efficiency, 0.97 S.P. =

P Km

Rotor power Km

1543.0 0.97

1590.7 kW

Ans.

EXAMPLE 4.11 A centrifugal compressor delivers 15 kg/m of air at 15000 rpm. The eye root and eye tip diameters are 180 mm and 310 mm respectively. The ambient conditions are 15°C and 1 bar. Calculate (a) the impeller vane angles at root and eye. Prewhirl is at angle of 20°. Inlet flow velocity is constant and is 150 m/s. Also, calculate (b) p1, T1, r1, and (c) relative Mach number. Solution:

Data:

m = 15 kg/s N = 15000 rpm dh = 0.18 m dt = d1 = 0.31 m T01 = 15 + 273 = 288 K p01 = 1 bar = 20° Vf1 = Vf2 = Vf = 150 m/s

Mass flow rate: Speed: Eye root diameter: Eye tip diameter: Ambient temperature: Ambient pressure: Prewhirl angle: Inlet flow velocity:

(a) Inlet van angles at tip and root (b1 tip, b1 root):

S d1 N 60

u1

ut

uh

S dh H 60

S – 0.31 – 15000 60

S – 0.18 – 15000 60

243.47 m/s

141.37 m/s

A Prewhirl V1¢ = Vf 1 = Vf 1¢

20° V1

B

a1 Vw1

u1

Vr 1¢ (without prewhirl)

a1¢ = 90°, Vw1¢ = 0, V1¢ = Vf 1¢

Vr 1

C

DABD = with prewhirl

b1 D b 1¢

a 1¢

DACE = without prewhirl

E

ui

Inlet velocity triangle with prewhirl

From inlet velocity triangle

V1

Vf 1 cos 20’

150 cos 20’

159.63 m/s

Centrifugal Compressors and Pumps

293

From inlet velocity triangle, Vw1 = Vf1 tan 20° = 150 × tan 20° = 54.596 m/s

tan E1tip tan E1 root

Vf 1 u1  Vw1 Vf 1 uh  Vw1

150 ; 243.47  54.6

\

150 ; 141.37  54.6

\ b1

b1

tip

root

= 38.5°

Ans.

= 59.95°

Ans.

(b) Inlet static conditions: pressure, temperature, density (p1, T1, r1):

T1

T01 

V12 2 gc c p

288 

(159.63)2 = 275.3 K 2 – 1 – 1.005 – 1000

p1

Ë T ÛJ p01 Ì 1 Ü Í T01 Ý

U1

p1 RT1

J

1

85.42

Ë 275.3 Û 100 – Ì Ü Í 288 Ý kN

m

2

–

3.5

Ans.

85.42 kN/m 2

1 kN-m 0.287 – – 288 K kg-K

(c) Relative Mach number at tip of the eye (M1

1.033 kg/m 3

tip):

a1 = acoustic velocity J RT1

332.6

N-m kg

332.6

1.4 – 287

kg-m s2

–

m kg

N-m – 275.3 K kg-K

332.6

m2 s2

= 332.6 m/s

From inlet velocity triangle, Vr1

V f21  (u1  Vw1 )2

= 241.2 m/s M1tip

Vr1 a1

241.2 332.6

0.7252

150 2  (243.47  54.6)2

Ans.

Ans.

294

Fundamentals of Turbomachinery

EXAMPLE 4.12 A centrifugal compressor running at 20000 rpm has the outer diameter of 500 mm. A 600 kW motor is used to drive the compressor. The blade angle at the impeller outlet is 26° measured from the radial direction. The flow velocity at exit is 120 m/s. Calculate (a) mass flow rate of air assuming mechanical efficiency = 0.95 and no slip, (b) eye tip and hub diameter if a radius ratio of 0.3 is selected for the impeller eye and if the velocity at inlet is 100 m/s with zero whirl, (c) what will be the overall total-to-total isentropic efficiency if an overall total pressure ratio 6.0 is required. Assume that the flow to the inlet is incompressible and the ambient air conditions are 101.325 kPa and 15°C. Solution: Speed: N = 20000 rpm Outer diameter: d2 = 500 mm Motor capacity (power): S.P. = 600 kW Blade angle at impeller outlet: b2 = 90 – 26 = 64° [Given w.r.t. radial direction, here considered w.r.t. tangential direction] The flow velocity at exit: Vf2 = 120 m/s Mechanical efficiency: hm = 0.95 No slip: s = 1 Radius ratio: rh/r1 = 0.3 Velocity at inlet: V1 = 100 m/s Zero whirl at inlet: Vw1 = 0, i.e. a1 = 90°, V1 = Vf1 Overall total-to-total pressure ratio: pR0 = 6 Ambient temperature: T01 = 15 + 273 = 288 K Ambient pressure: p01 = 101.325 kPa To determine: m , dt ( d1 ), d h , Kc t to C

B

02¢

V2

V2¢

T or h Vr 2¢

03¢ b2

E

A

b 2¢

Vw2 Exit velocity triangle

Triangle ACD = Ideal (without slip) Triangle ABD = Actual (with slip) Variables with ‘dash’ as superscript = Actual Variable without ‘dash’ as superscript = Ideal

3 2

2¢

p02 p03

p3

p2

3¢

D

F u2

Vw1¢

03

Vr 2

Vf 2

Vf 2¢

02

p1 p1

01 1

s

Centrifugal Compressors and Pumps

295

From exit velocity triangle (no slip) ideal,

u2 Vw2

S d2 N S – 0.5 – 20000 523.6 m/s 60 60 = u2 – Vf2 cos 64° = 523.6 – 120 cos 64° = 470.995 m/s

(a) Mass flow rate ( m ) : P = Euler power = m u2 Vw 2 / gc m

or

P Km

or \

kg s

– 523.6

m m N- s2 – 470.995 – s s kg -m

246612.982m W

246.613m kW

mechanical efficiency =

Euler work Shaft work

P S.P.

246.613m W 600 kW m = 2.313 kg/s

0.95

Ans.

(b) Eye tip, dt(d1) and hub diameter (dh): From continuity equation, we have,

or

m

U1 a1 V1

m

U1 S (r12  rh2 ) V1

ËÈ r Ø 2 Û U1 S rh2 Ì É 1 Ù  1Ü V1 Ì Ê rh Ú Ü Í Ý

Incompressible flow at inlet (given data) U1

U01

p01 RT01

101.325 0.287 – 288

1.226 kg/m 3

2.313 = 1.226 × p × rh2 [(3.33)2  1] – 100 \ We have

rh = 0.024397 m = 2.4397 cm dh = hub diameter = 2 × rh = 2.4397 × 2 = 4.879 cm r1 rh

3.33

Ans.

\ r1 = rh × 3.33 = 2.4397 × 3.33

\

r1 = 8.124 cm

\

d1 = eye diameter = 2 × r1 = 2 × 8.124 = 16.25 cm

Ans.

296

Fundamentals of Turbomachinery

(c) Compressor overall total-to-total efficiency (Kc t - to ) :

Ë T03 Û Ì Ü Í T01 Ý „

Ë p03 Û Ì Ü Í p01 Ý

J

1

0.4

J

>[email protected]

1.6685

From Eqs. (4.10a) and (4.11a), we have

Kc t

c p (T03  T01 )

c p (T03  T01 ) „

 p (T03  T01 ) mc

'h0

W.D.

P

„



to

m – 1.005 – T01 246.613 m

„

Ë T03 Û  1Ü Ì T Í 01 Ý „

1.005 – 288 (1.6685  1) 246.613

0.785

Ans.

EXAMPLE 4.13 Ambient condition are 288 K and 101.325 kPa. Assume zero whirl at inlet and radial blades. The tip speed of the impeller is 375 m/s. The radial velocity at exit is 30 m/s. The slip factor is 0.9. Total-to-total efficiency is 90%, the flow area at the exit of the impeller is 0.095 m2. Calculate (a) the mach number at impeller tip and (b) the mass flow rate. Solution:

Data:

Ambient temperature: (stagnation temperature) Ambient pressure: Zero whirl at inlet: Radial blades: Tip speed of the impeller: Radial velocity at exit: Slip factor: Total-to-total efficiency: Flow area at exit of the impeller: To determine: M2, m

T01 = 288 K p01 = 101.325 kPa Vw1 = 0, a1 = 90°, V1 = Vf1 b2 = 90°, u2 = Vw2 u2 = 375 m/s Vf 2 = 30 m/s s = 0.9 hc t–t = 0.9 a2 = 0.095 m2

(a) Mach number at impeller tip (M2): Triangle Triangle Triangle Triangle

ABC ADC ABC ADE V

= = = =

Actual (with slip) Ideal (without slip) Actual (with slip) Ideal (without slip)

AC AE

Vw 2 Vw 2

„

Vw 2 u2

„

Centrifugal Compressors and Pumps B

Vf 2¢ A

V2

V2

V2¢

Vr 2¢

V2¢

Vr 2

Vf 2

F Vw 2

C

E

C

u2

„

Vw2 = u2

E

Exit velocity triangles (radial blades)

V u2

Vw 2

Vf 2 = Vr 2 = Vf 2¢

b2

a2 a 2¢ Vw 2¢

A

Exit velocity triangles (non-radial blades)

or

Vr 2¢

Vf 2¢

b2¢

b2 Vw 2¢

D

D

B

297

0.9 – 375

337.5 m/s

From exit velocity triangle (radial vanes) (with slip), V f22  Vw22

V2

„

30 2  337.52

W.D. = Dh0 = cp(T02 – T01) = or

T02

T01 

V u22 c p gc

or

T02

288 

0.9 – 3752 1.005 – 1000 – 1

Now,

T2

T02 

(V2 )2 2c p gc

M2

V u2 Vw 2 gc

V u22 gc

413.93 K

„

413.93 

\

338.8 m/s

V2 a2

338.82 2 – 1.005 – 1000 – 1

V2

„

338.8

„

J RT2

356.83 K

1.4 – 287 – 356.83

0.895

 ) at impeller exit: (b) Mass flow rate ( m Applying total-to-total efficiency between the impeller inlet and the exit. Kc t t

(T02  T01 ) (T02  T01 ) „

J

p02 p01

Ë T02 Û J  1 Ì Ü Í T01 Ý „

Ë T02  T01 T01 ÛJ  Ì Ü T01 Ý Í T01 „

J

1

Ans.

298

Fundamentals of Turbomachinery

ËKc t t (T02  T01 ) ÛJ  1Ü Ì T01 Í Ý

J

1

Ë ÛJ T01 È T02 Ø ÌKc t t  1Ù  1Ü É Ú T01 Ê T01 ÌÍ ÜÝ

J

1

1.4

Ë È 413.93 Ø Û 0.4 = 3.195 Ì 0.9 ÉÊ 288  1ÙÚ  1Ü Í Ý From Figure 4.3, we have

p2 p02 \

Ë T2 ÛJ Ì Ü Í T02 Ý

J

1

Ë 356.83 Û Ì 413.93 Ü Í Ý

3.5

0.595

p2 = p02 × 0.595 = 3.195 × p01 × 0.595 = 101.325 × 3.195 × 0.595

\

p2 = 192.6 kN/m2

\

U2

\

m

p2 RT2

192.6 0.287 – 356.82

1.88 kg/m 3

U2 a2 V f 2

1.88

kg m3

– 0.095 m 2 – 30

m s

5.36 kg/s

Ans.

EXAMPLE 4.14 A centrifugal compressor has 10 radial vanes. The impeller tip diameter is 170 mm with a slip factor of 0.9. The speed of the machine is 45000 rpm. The mass flow rate of air is 0.5 kg/s with no whirl at inlet. At inlet to the impeller the mean diameter of the eye is 65 mm while the annulus height at the eye is 25 mm. The static pressure and temperature at the impeller inlet are 95 kPa and 295 K respectively. Calculate: (a) (b) (c) (d)

The theoretical power transferred The blade angle at the mean diameter at impeller inlet The stagnation temperature at impeller exit (T02) The stagnation pressure at impeller exit if the total-to-total efficiency of the impeller is 0.9.

Solution:

Data:

Number of radial vanes: Impeller tip diameter: Slip factor: Speed:

= 10 d2 = 170 mm s = 0.9 N = 45000 rpm

Centrifugal Compressors and Pumps

m

The mass flow rate:

299

0.5 kg/s

Vw1 = 0, a1 = 90°, V1 = Vf1 (r1 + rh)/2 = 32.5 r1 – rh = 25 mm T1 = 295 K p1 = 95 kPa hc t–t = 0.9

No whirl at inlet: Mean diameter of the eye: Annulus height of the eye: Inlet static temperature: Inlet static pressure: Total-to-total efficiency: To determine: P, E1m , T02 , p02 u2 = impeller tip speed

S d2 N 60

u2

S – 0.17 – 45000 60

400.6 m/s

(a) Theoretical power (P): From Eq. (4.15), we have P

m V u22 gc

0.5 – 0.9 – (400.6)2 1 – 1000

72.2 kW

We have

r1  rh 2

r1 – rh = 25; \

r1 = 45 mm

32.5

rh = 45 – 25 = 20 mm At the eye,

U1

p1 RT1

95

kN m

2

–

1 kN-m 0.287 – 295 K kg-K

1.22 kg/m 3

We have

\ or \

0.5

m

U1 a1 V f 1

kg s

1.22

kg 3

U1 a1 V1

(a1 = 90°, \ V1 = Vf1)

– S (r12  rh2 ) m 2 – V1

m s

m 0.5 = 1.22 × p × (0.0452 – 0.022) × V1 V1 = 87.3 m/s

(b) The blade angle at the mean diameter at impeller inlet (b1m): At mean radius,

u1m

S d1 N 60

p – 0.065 – 45000 60

Ans.

300

Fundamentals of Turbomachinery

\

u1m = 153.15 m/s

From inlet triangle, V1 u1 m

tan E1 m

\

87.3 153.15

b1m = 29.7°

Ans.

(c) The stagnation temperature at impeller exit (T02): h1 

h01

V12 2 gc

c pT1 

(1005 – 295)  W.D. – m

P

P  h01 m

h02

\

V12 2 gc

87.32 2 –1

m 'h0

300.28 kJ/kg m (h02  h01 )

72.2 kW kJ  300.28 kg kg 0.5 s

kJ s kJ .  300.28 kg s kg

144.4

444.69 kJ/kg

We have h02 = cpT02 \

444.69 1.005

T02

Ans.

442.47 K

(d) The stagnation pressure at impeller exit (p02): Refer Figure 4.3, we have for impeller (1–2) Kc t t

(h02  h01 ) (h02  h01 ) „

or

c p (T02  T01 )

Kc t t –

or

(T02  T01 )

129.66 1.005

or

T02

„

„

„

P m

h02  h01 P/m „

0.9 – 72.2 0.5

129.31 K

T01  129.31

T1 

V12  129.31 2c p gc

129.96 kJ/kg

Centrifugal Compressors and Pumps

295 K 

\

(87.3)2  129.31 K 2 – 1 – 1005

301

428.1 K

For isentropic process, we have J

T01

h01 cp

p01

Ë T Û J 1 p1 Ì 01 Ü Í T1 Ý

p1 298.8 K ; p01

300.28 1.005 J

95

kN

Ë 298.8 Û –Ì Ü 2 m Í 295 Ý

Ë T1 Û J  1 Ì Ü Í T01 Ý 3.5

99.35

kN m2

J

\

p02 p01

\

p02

Ë T02 Û J  1 Ë 428.1 Û3.5 3.5203 Ì Ü Ì 298.8 Ü Í Ý Í T01 Ý = p01 × 3.5203 = 99.35 × 3.5203 = 349.74 kN/m2 „

Ans.

EXAMPLE 4.15 A centrifugal compressor runs at 12000 rpm, delivers 10 kg/s of air and its operating pressure ratio is 4. The slip and power factor are 0.91 and 1.06 respectively. The isentropic compression efficiency is 0.92. The Mach number of air leaving the impeller vanes is to be unity to have shockless entry. Calculate (a) the impeller outlet diameter and (b) the axial depth of the impeller. Ambient conditions are 101.325 kPa and 288 K. Assume overall isentropic efficiency 0.83 and zero whirl at inlet. Solution:

Data: Speed: Mass of air delivered: Operating pressure ratio: Slip factor: Power factor: Isentropic compressor efficiency: Impeller exit Mach number: Stagnation temperature at inlet: Stagnation pressure at inlet:

N = 12000 rpm m = 10 kg/s pR0 = 4 s = 0.91 j = 1.06 hc t–t = 0.92 M2 = 1 T01 = 288 K p01 = 101.325 kPa Kc t to

Overall efficiency: Zero whirl at inlet:

Vw1 = 0, a1 = 90°, V1 = Vf 1

To determine: d2, b2 (a) Impeller outer diameter (d2): From Eq. (4.20), we have J

pR 0

p03 p01

0.83

Ë Kc t t u22 V M Û J  1 0 Ì1  Ü gc C pT01 Ü ÌÍ Ý

302

Fundamentals of Turbomachinery

4 \ or

Ë 0.83 – u22 – 0.91 – 1.06 Û Ì1  Ü ÌÍ 1 – 1.005 – 288 – 1000 ÜÝ

3.5

u2 = 419.2 m/s

419.2

S d2 N 60

S – d2 – 12000 ; 60

\

d2 = 0.667 m

Ans.

(b) Axial depth of the impeller (b2):

T03  T01

J 1 Ë Û T01 Ì È p03 Ø J Ü  1Ü Ì É Ù Kc t t0 Ê p01 Ú ÌÍ ÜÝ

288 0.286 (4  1) 168.84 K 0.83

\

T03 = 168.84 + T01 Kc t t

(T02  T01 ) (T02  T01 ) „

J

p02 p01

Ë T02 Û J  1 Ì Ü Í T01 Ý

J

Ë T02  T01 T01 Û J  1  Ì Ü T01 Ý Í T01 „

J

Ë Kc t t (T02  T01 ) ÛJ 1  1Ü Ì T01 Í Ý or

p02 p01

J

Ë 0.92 ÛJ 1 Ì 288 (168.84)  1Ü Í Ý

4.53

T03 = T02 = 168.84 + T01 = 168.84 + 288 = 456.84 K (T02 = T03, refer to Figure 4.3) Now, \

M2

V2

V2

J RT2

V2

20.05 T2

T2

T02 

V2 2 gc c p

1.4 – 287 – T2

456.84 

1

(20.05)2 T2 2 – 1 – 1005

380.73 K

Centrifugal Compressors and Pumps

303

Refer to Figure 4.3, Slip D

B V2 V2¢

Vr 2¢

ABC = Actual = with slip

Vr 2 = Vf 2

ADC = Ideal = without slip

Vf 2¢

A

C

Vw 2¢ u 2 = Vw 2 Exit velocity triangle J

3.5

p2 p02

Ë T2 Û J  1 Ì Ü Í T02 Ý

Ë 380.73 Û Ì 456.84 Ü Í Ý

p2 p01

p p2 – 02 p02 p01

0.528 – 4.53

0.528 2.392

p2 = p01 × 2.392 = 101.325 × 2.392 = 242.37 kPa

\

U2

p2 RT2

242.37 0.287 – 380.73

V

Vw 2 Vw 2

Vw 2 u2

„

2.218 kg/m 3

„

Vw 2 = s u2 = 0.91 × 419.2 = 381.472 m/s „

V2

„

M 2 J RT2

1 – 1.4 – 287 – 380.73

391.123 m/s

(391.123)2  (381.472)2

86.35 m/s

From exit velocity triangle, Vf 2

„

V2  Vw 2 „

„

From continuity equation, m U2 a2 V f 2

10 \

kg s

2.218

kg m

3

b2 = 0.025 m

U2 S d2 b2 V f 2

„

– S – 0.667 m – b2 m – 86.35

m s Ans.

304

Fundamentals of Turbomachinery

EXAMPLE 4.16 Flow rate of air in a rotary compressor is 6 kg/s. The air at initial conditions of 1 bar and 15°C enters the compressor with negligible velocity and leaves the compressor impeller at a pressure of 30 bar through a 310 mm diameter pipe. If the compressor is reversible and adiabatic and the mechanical efficiency of the compressor is 80%, calculate (a) the temperature of the air leaving the compressor and (b) the power required to drive the compressor. Solution:

Data: Mass flow rate of air:

m

Inlet pressure: Inlet temperature: Negligible velocity: Exit pressure: Exit diameter: Mechanical efficiency:

p1 T1 V1 p2 d2 hm

6 kg/s

= 1 bar = 15 + 273 = 288 K = 0 = 30 bar = 310 mm = 80%

To determine: T2, Power (a) Exit temperature (T2): We have for isentropic process 1–2¢

T2

Ëp Û T1 Ì 2 Ü Í p1 Ý

J

1 J

0.4 Ë 30 Û 1.4

288 Ì Ü Í1Ý

T or h 02¢

02 03

03¢

3

p3

p2

3¢ 2

761.4 K

p02 p03

2¢

p01 p1

01

(b) Power required (P): From Eq. (4.3), we have

1

s

W.D. = Dh0 = cp(T02 – T01) Neglecting inlet and exit velocities, we have

or

P

m 'h

P

6

kg s

m c p (T2  T1 ) – 1.005

kJ (761.1  288) K = 2852.79 kW kg -K

Ans.

EXAMPLE 4.17 Air is compressed in a centrifugal compressor from 27°C to 150°C and pressure from 1 bar to 3 bar. Calculate (a) the isentropic efficiency of the compressor and (b) the power developed. Assume a mass flow rate of 28 kg/min. Solution:

Data: Machine:

Centrifugal compressor

Inlet temperature:

T1 = 27 + 273 = 300 K

Centrifugal Compressors and Pumps

Final temperature: Inlet pressure: Exit pressure:

T2 = 150 + 273 = 423 K p1 = 1 bar p2 = 3 bar m

Mass flow rate: To determine: hc t–t, Power

Ë p2 Û Ì Ü Í p1 Ý

T2 T1

„

\

T2

28 kg/m T or h

(a) The isentropic efficiency (hc t–t): For isentropic process 1–2¢ J

02¢

0.4

J

3 1.4

3

p3

p01 p1

01

410.62  300 423  300

„

p02 p03

p2

2¢

(T2  T1 ) (T2  T1 )

Kc t  t

03

2

= 300 × 1.369 = 410.62 K \

02

3¢

1.369

T1 – 1.369

„

28/60 kg/s

03¢

1

305

1

0.899

s

Ans.

(b) Power developed (P): From Eq. (4.3), we have W.D. = Dh0 = cp(T02 – T01) Neglecting K.E. changes,

\

m 'h

 p (T2  T1 ) mc

P

m – W.D.

P

28 kg kJ – 1.005 – (423  300) K 60 s kg -K

57.69 kW

Ans.

EXAMPLE 4.18 A single-sided centrifugal compressor delivers 10 kg/s with a total pressure rise of 5 : 1, and operates at a speed of 15000 rpm. The inlet conditions are 1 bar and 27°C. Calculate (a) the rise in total temperature, (b) the tip speed of the impeller, (c) the tip diameter, (d) the inlet eye annulus area, and (e) the theoretical power to drive the compressor. Assume slip factor = 0.92, isentropic efficiency = 0.81, air enters axially with a velocity of 140 m/s. Solution:

Data: Mass flow rate: Total pressure rise or stagnation pressure rise:

Speed: Inlet temperature: Inlet pressure:

m 10 kg/s p02 5 p01 1 N = 15000 rpm T1 = 27 + 273 = 300 K p1 = 1 bar

pR1

306

Fundamentals of Turbomachinery

s = hc t–t = a1 = Vf1 =

Slip factor: Isentropic efficiency: Axial entry: Axial velocity:

0.92 0.81 90°, Vw1 = 0, V1 = Vf1 140 m/s

To determine: (T02 – T01), u2, d2, a, Power Inlet stagnation temperature,

T01

T1 

V12 2 gc c p

300 K 

T02

„

140 2 m 2 N - s2 kg - K 2 – 1 – 2.005 s2 kg - m N - m

Ëp Û T01 – Ì 02 Ü Í p01 Ý

J

1 J

0.4

300 – 3 1.4

309.75 K

475.15 K

(a) Rise in total temperature (T02 – T01):

\

Kc t t

(T02  T01 ) (T02  T01 )

(T02  T01 )

(T02  T01 ) Kc t t

T or h

„

02¢

02 03

03¢

3

„

2 2¢

p3

p2

3¢

475.15  300 0.81 = 216.23 K

p02 p03

p01 p1

01

Ans.

1

(b) Tip speed of the compressor (u2): From Eqs. (4.3) and (4.15), W.D. = 'h0 c p (T02  T01 )

or \

1.005

V u22 gc

s

(Assuming air leaves radially)

0.92 – u22 1 – 1000

0.92 – u22 kJ (216.23) K kg-K 1 – 1000 u2 = 486.0 m/s

Ans.

(c) Blade tip diameter (d2):

S d2 N S – d2 – 15000 60 60 d2 = 0.619 m

u2 \

486.0 Ans.

Centrifugal Compressors and Pumps

307

(d) Inlet annulus area (a1): p1v1

RT1 ; v1

RT p1

v1

specific volume

0.287

kN -m 300 K – kg-K kN 100 2 m

0.861

m3 kg

From continuity equation, we have

\

m

U1 a1 V1

a1

m U1 V1

m v1 V1

10

kg s

2

– 0.861 –

m3 1 – m kg 140 s

0.0615 m 2

Ans.

(e) Power (P): From Eq. (4.3), we have m (W.D.)

P

10

kg s

m – 'h0

– 1.005

m – c p (T02  T01 )

kJ – (216.23) K kg -K

2173.1 kW

Ans.

EXAMPLE 4.19 Air enters a centrifugal compressor axially fitted with radial vanes from a practically quiescent atmosphere of 1 bar and 15°C. The air leaves the diffuser with negligible velocity. The tip diameter of the impeller is 45 cm and the speed of the machine is 20000 rpm. Calculate the exit static temperature and pressure. Assume no losses. Solution:

Data: Axial air entry: a1 = 90°, V1 = Vf1, Vw1 = 0 Radial vanes: b2 = 90°, Vw2 = u2, V12 = Vr2 Quiescent atmosphere means, stagnation condition. Inlet temperature: T01 = 15 + 273 = 288 K Inlet pressure: p01 = 100 kN/m2 Exit air velocity from diffuser: V3 = 0 (negligible) Tip diameter of the impeller: d2 = 45 cm Speed: N = 20000 rpm No losses:

To determine: T3, p3 u2 = peripheral velocity of impeller tip \

u2

S d2 N 60

S – 0.45 – 20000 60

471.24 m/s

308

Fundamentals of Turbomachinery

From Eq. (4.1), W.D. =

u22 gc

(471.24)2 1000

222.1 kJ/kg

(radial blades, \ Vw2 = u2)

(a) Exit static temperature and pressure (T3, p3): W.D. = Dh0 = cpDT0 = cp(T03 – T01) \

'T0

T03  T01

W.D. cp

222.1 kJ kJ kg 1.005 kg -K

220.99 K

Ë V32 Û Ë V12 Û ÌT3  Ü  ÌT1  Ü 2 gc c p ÝÜ ÍÌ 2 gc c p ÝÜ ÍÌ

T3  T1

[Data: Exit velocity from the diffuser is negligible, i.e. V3 = 0, Inlet is quiesent condition \ V1 » 0] \ \

DT0 = DT = T3 – T1 = 220.99 K T3 = 220.96 + T1 = 220.99 + 288 = 508.99 K

Ans.

For isentropic process 1–3 J

\

1.4

Ë T3 ÛJ  1 Ë 508.99 Û 0.4 Ì Ü Ì 288 Ü Í Ý Í T1 Ý 2 p3 = 733.8 kN/m = 7.338 bar p3 p1

Ans.

EXAMPLE 4.20 A centrifugal supercharger (centrifugal compressor) is connected to the manifold of an engine combustion chamber. It develops a total pressure of 2.0 bar. The ambient pressure and temperature are 0.65 bar and –8°C, respectively. The mass flow rate is 1.0 kg/s. Isentropic efficiency is 0.81, and the inlet velocity into the impeller is 110 m/s. The speed of the machine is 20000 rpm. Calculate (a) the impeller diameter, (b) the annulus area for the entry, and (c) the theoretical power required to drive the impeller. Assume no prewhirl and radial discharge.

Centrifugal Compressors and Pumps

309

Solution: Total pressure: Ambient pressure: Ambient temperature:

p04 = 2.0 p01 = 0.65 bar T01 = –8 + 273 = 265 K = T1

(Q V1 = 0)

m 1.0 kg/s

Mass flow rate:

Kc t to

Isentropic efficiency: Inlet velocity into impeller: Speed: No prewhirl: Radial discharge: Ambient condition 1

0.81

V2 = 110 m/s N = 20000 rpm b3 = 90°, Vw3 = u3, Vr3 = Vf3

Entry 2

Diffuser outlet 4

Exit 3

Diffuser

p04, T04, V4 = 0 Engine manifold

V1 = 0 p01, T01 (a) T or h

T

2 bar = p4

04

04¢ 2 bar

3

04 3¢

04¢

0.65 bar 0.65 bar

01

01

s (b)

Let

State 1 2 3 4

— — — —

2

s (c)

Ambient condition Entry condition to impeller inlet Exit condition from impeller Exit condition from diffuser to engine manifold.

For isentropic process between 01 to 04¢

(Reference Figure (c))

310

Fundamentals of Turbomachinery

T04 T01

„

\

Ë p04 Û Ì Ü Í p01 Ý

J

1 J

0.4

Ë 2 Û 1.4 Ì 0.65 Ü Í Ý

1.379

T04¢ = T01 × 1.379 = 265 × 1.379 = 365.35 K Kc t to

\

(T04  T01 ) (T04  T01 ) „

365.35  265 T04  265

0.81

T04 = 388.9 K = T4 [Assumption: neglect ambient velocity and exit velocity. \ V1 = V4 = 0]

(a) Impeller diameter (d3): Applying the law of conservation between 1 and 4, we get

h1 

V12 V2  W.D. h4  4 2 gc c p 2 gc c p W.D. = h4 – h1 = cp (T4 – T1) 1.005 W.D.

\ or \

kJ – (388.9  265) K 124.51 kJ/kg kg-K

u32 gc – 1000

124.51

u3 = 352.9 m/s

(Equation (4.1), and radial blades)

S d3 N S – d3 – 20000 60 60 d3 = 0.337 m

352.9

(b) Annulus area for the entry (a2): From energy balance between 1 and 2, we have V12 2 gc

h2 

or

(h1  h2 )

V22 2 gc

or

c p (T1  T2 )

V22 2 gc

or

1.005 (265  T2 )

h1 

or

V22 2 gc

110 2 2 – 1000

T2 = 258.98 K

Ans.

Centrifugal Compressors and Pumps

311

We have J

\ \

Ë T2 ÛJ  1 Ë 258.98 Û Ì Ü Ì 265 Ü Í Ý Í T1 Ý p2 = 92.27 kPa p2 p1

3.5

0.923

We have mRT2 ;

P2 v2

or i.e.

v2 = volume flow rate

RT2 m 0.287 – 258.98 – 1 0.8055 m 3 /s p2 92.27 v2 = cross-sectional area × velocity v2 = a2V2 v2

2

m3 v2 s a2 m V2 110 s 2 a2 = 0.00732 m 0.8055

\

Ans.

(c) Theoretical power required to drive the compressor (P): P

m 'h0

m (T04  T01 )c p

= 1.0 × (388.9 – 265) × 1.005 = 124.52 kW

Ans.

EXAMPLE 4.21 The stagnation conditions of the air at the compressor intake are 1 bar, 32°C. The speed of the machine is 15,000 rpm. The intake absolute velocity is axial. Assume radial blades, exit flow velocity 150 m/s and total-to-total efficiency of the velocity triangle at the exit of the rotor to be 0.81. Compute (a) the slip coefficient and (b) the pressure ratio between the impeller inlet and the exit. Assume slip = 50 m/s and rotor diameter at exit to be 60 cm. Solution: Data: Outlet diameter: Intake stagnation temperature: Intake stagnation pressure: Speed: Intake absolute velocity is axial: Assume radial exit blades: Total-to-total efficiency: Slip:

d2 = 60 cm T01 = 32 + 273 = 305 K p01 = 1 bar N = 15,000 rpm V1 = Vf1, Vw1 = 0, a1 = 90° b2 = 90°, Vw2 = u2 hc t–t = 0.81 = 50 m/s

312

Fundamentals of Turbomachinery

To determine: s, pR1 Slip B

D V2

V2¢

Vf 2 = Vr 2

Vr 2¢

Vf 2¢

ADC = Ideal velocity triangle (without slip) ABC = Actual velocity triangle (with slip)

b2 = 90° A

C

Vw 2¢

b2¢ < 90° u 2 = Vw 2

(a) Slip coefficient (s):

S d2 N S – 0.6 – 15000 471.24 m/s Vw 2 60 60 Slip = BD = (Vw2 – V w2¢) = 50.0 u2

Vw 2

„

V

Vw 2  50

Vw 2 Vw 2

471.24  50

421.24 471.24

„

421.24 m/s

0.894

(b) Pressure ratio (pR1): From Eq. (4.16), we have

or

pR1

pR1

Ë (471.24)2 – 0.894 – 0.81 Û Ì1  Ü 1000 – 1 – 1.005 – 305 ÝÜ ÍÌ

(T02  T01 ) Kc t t

or

1

W.D. = Dh0 = cp(T02 – T01) =

or or

J

Ë u22 V Kc t t ÛJ Ì1  Ü c p gcT01 Ü ÌÍ Ý

0.894 – (471.24)2 1 – 1000 – 1.005 (T02  T01 ) (T02  T01 ) „

„

V u2 gc – 1000

197.54 K

T02  T01 197.54

0.81

T02  T01 = 197.54 × 0.81 = 160.0 K „

3.5

4.376

Centrifugal Compressors and Pumps

\

313

T02¢ = T01 + 160.0 = 305 + 160.0 = 465 K

\

Ë p02 Û Ì Ü Í p01 Ý

pR1

J

Ë T02 Û J  1 Ì Ü Í T01 Ý „

Ë 465 Û Ì 305 Ü Í Ý

3.5

Ans.

4.376

EXAMPLE 4.22 The following data refers to a centrifugal compressor. The speed of the machine is 6000 rpm, impeller tip diameter 110 cm, mass flow rate of air 22 kg/s. The inlet pressure and temperature are 1 bar and 30°C respectively, slip coefficient is 0.91, pressure coefficient 0.75 and mechanical efficiency 0.98. Calculate (a) the overall static pressure ratio, (b) the adiabatic efficiency of the impeller (based on static-to-static), (c) the static temperature of the air at the exit, and (d) the shaft power input. Solution: Speed: Impeller tip diameter:

N = 6000 rpm d2 = 110 cm

Mass flow rate of air:

m

Inlet pressure: Inlet temperature: Slip coefficient: Pressure coefficient: Mechanical efficiency:

22 kg/s

p1 = 1 bar T1 = 300 C + 273 = 303 K s = 0.91 fp = 0.75 hm = 0.98

To determine: pRO, hc s–s, T3, SP T or h

Slip

p02

02 02¢

03

03¢

3

B

p03

V2

p3 V2¢

p2

3¢ 2

Vf

p01

2¢

p1

A

01 1

s

u2 = Vw2

Ëp Û (a) Overall static pressure ratio Ì 3 = pR0 Ü : Í p1 Ý

S d2 N 60

S – 1.1 – 6000 60

Vr 2¢ Vf 2 = Vr 2

C

Vw2¢

Triangle ABC = Actual velocity triangle (with slip) Triangle ADC = Ideal velocity triangle (without slip)

u2

D

345.6 m/s

314

Fundamentals of Turbomachinery

From Eq. (4.12c), Ip

or

0.75

\

(T3  T1 ) „

\

c p (T3  T1 ) „

u22 gc

1.005(T3  T1 ) „

(345.6)2 1000 89.12 K

T3 = T1 + 89.12 K = 303 + 89.12 = 392.12 K „

T3 T1

„

Ë p3 Û Ì Ü Í p1 Ý

1

J

J

„

Ë p3 Û Ì Ü Í p1 Ý

J

1 J

J

\

pR 0

Ë T3 Û J  1 Ì Ü Í T1 Ý

Ë 392.12 Û Ì 303 Ü Í Ý

„

> pR0 @

J

1 J

3.5

2.466

Ans.

(b) The adiabatic efficiency (based on static-to-static) (hc s–s): W.D.

or

(T3  T1 )

\

Kc s  s

'h

c p (T3  T1 )

0.91 – 345.62 1 – 1000 – 1.005 (T3  T1 ) (T3  T1 ) „

V u22 gc – 1000 108.15 K

89.12 – 100 108.15

82.41%

Ans.

(c) The static temperature at the exit (T3): \

T3 = T1 + 108.15 = 303 + 108.15 = 411.15 K

Ans.

(d) The shaft power input (SP): SP =

m – W.D. Km

= 2439.99 kW

22

kg kJ 1 – 1.005 – 108.15 K – s kg-K 0.98

Ans.

Centrifugal Compressors and Pumps

315

4.25 EXAMPLES (Centrifugal Pumps) EXAMPLE 4.23 The diameters of an impeller of a centrifugal pump at inlet and outlet are 30 cm and 60 cm respectively. Determine the minimum starting speed of the pump if it works against a head of 30 m. Solution:

Data: Inlet diameter: Outlet diameter: Head:

d1 = 30 cm d2 = 60 cm Hm = 30 m

To determine: Minimum speed (N) Minimum speed (N): (u22  u12 ) 2g

or

S 2 d22 N 2 60 2 – 2 – g



S 2 d12 N 2 60 2 – 2 – g N2

\ \

(Eq. (4.55))

Hm 30

30 – 2 – 60 2 – 9.81 2

S –

(d22



30 – 2 – 9.81 – 3600

d12 )

S 2 – (0.6 2  0.32 )

N = 891.72 rpm

Ans.

EXAMPLE 4.24 A centrifugal pump delivers water against a head of 20 m at the rate of 100 lit/s at the speed of 1500 rpm. The impeller diameter is 25 cm and width at outlet is 5 cm. The manometric efficiency is 75%. Determine the vane angle at the outer periphery of the impeller. Solution:

Data: Head: Discharge: Speed: Impeller diameter at outlet: Impeller width at outlet: Manometric efficiency:

Hm = 20 m Q = 100 lit/s N = 1500 rpm d2 = 25 cm b2 = 0.5 cm hman = 0.75

To determine: b2 (a) Outlet vane angle (b2):

u2

S d2 N 60

S – 0.25 – 1500 60

Q = pd2b2Vf 2

19.54 m/s

316

Fundamentals of Turbomachinery

0.1

m3 s

K man

gH m Vw 2 u2

0.75 9.81 \

(Q a1 = 90°, Vw1 = 0)

m s2

1

– 20 m – Vw 2

m m – 19.64 s s

Vw2 = 13.32 m/s

tan E2 \

\ Vf 2 = 2.546 m/s

S – 0.25 m – 0.05 m – V f 2

Vf 2

2.546 19.64  13.32

u2  Vw 2

0.402

b2 = 21.94°

Ans.

EXAMPLE 4.25 A centrifugal pump delivers 1800 lit/min against a total height of 20 m. Its speed is 1450 rpm. Inner and outer diameters of impeller are 120 mm and 240 mm respectively and the diameter of suction and delivery pipes are both 100 mm. Determine the blade angles b1 and b2 of the impeller vane if the water enters radially. Neglect friction and other losses. Solution:

Data: Discharge: Head: Speed: Inner diameter of impeller: Outer diameter of impeller: Diameter of suction pipe: Diameter of delivery pipe: Radial water entry:

Q = 1800 lit/min Hm = 20 m N = 1450 rpm d1 = 0.12 m d2 = 0.24 m ds = 0.1 m dd = 0.1 m a1 = 90°, Vw1 = 0, Vf 1 = V1

To determine: b1 and b2, neglect all losses (a) Blade angles (b1, b2):

u1

S d1 N 60

S – 0.12 – 1450 60

9.11 m/s

u2

S d2 N 60

S – 0.24 – 1450 60

18.22 m/s

Hm

\

Vw 2 u2 ; Vw 2 g

Vw2 = 10.77 m/s

Hm g u2

20 – 9.81 18.22

Centrifugal Compressors and Pumps

\

Vs

Vf 2

Vd

tan E1

\

Q S 2 ds 4

Vf 1

S – 0.12 0.03 – 4

Q S 2 dd 4

Vf 1 u1

0.03 – 4

S – 0.12

3.82 m/s

3.82 m/s

3.82 9.11

b1 = 22.75°

tan E2

\ \

317

Vf 2 u2  Vw 2

Ans.

3.82 18.22  10.77

0.5125

b2 = 27.15°

Ans.

EXAMPLE 4.26 A centrifugal pump having outer diameter equal to two times the inner diameter and running at 1000 rpm works against a total head of 40 m. The velocity of flow through the impeller is constant and equal to 2.5 m/s. The vanes are set back at an angle of 40° at outlet. If the outer diameter of the impeller is 50 cm and width at outlet is 5 cm, determine (a) the vane angle at the inlet, (b) the work done per second by the impeller on water, and (c) the manometric efficiency. Solution:

Data: Impeller outer diameter: Speed: Head: Velocity of flow: Vane angle at outlet: Width at outlet:

To determine: E1 ,

d2 = 2d1 = 50 cm N = 1000 rpm Hm = 40 m Vf1 = Vf2 = 2.5 m/s b2 = 40° b2 = 5 cm

W.D. , Kman s

u1

S d1 N 60

S – 0.25 – 1000 60

u2

S d2 N 60

S – 0.5 – 1000 60

13.09 m/s 26.18 m/s

Q = pd2b2Vf 2 = p × 0.5 × 0.05 × 2.5 = 0.1963 m3/s

tan E2

Vf 2 u2  Vw2

2.5 26.18  Vw 2

318

Fundamentals of Turbomachinery

or

2.5 (26.18  Vw 2 )

tan 40’

\

Vw2 = 23.2 m/s

(a) Vane angle at inlet (b1):

Vf 1

tan E1 \

u1

2.5 13.09

0.191

b1 = 10.81°

Ans.

(b) Work done (W.D.): W.D. =

U QVw 2 u2 gc 119227.9088

1000

kg m3

– 0.1963

m3 m m N- s2 – 23.2 – 26.18 – s s s kg -m

N-m or W s

Ans.

(c) Manometric efficiency (hman): Kman

9.81 – 40 – 100 23.2 – 26.13

gH m Vw 2 u2

64.6%

Ans.

EXAMPLE 4.27 A centrifugal pump runs at 950 rpm. Its outer and inner diameters are 500 mm and 250 mm. The vanes are set back at 35° to the wheel rim. If the radial velocity of water through the impeller is constant at 4 m/s, find (a) the angle of vane at the inlet, (b) the velocity if water at the outlet, (c) the direction of water at the outlet and (a) the work done by the impeller per kg of water. Entry of water at inlet is radial. Solution:

Data: Speed: Impeller outer diameter: Impeller inner diameter: Vanes outlet angle: Radial velocity: Radial entry:

N d2 d1 b2 Vf 1 a1

= = = = = =

950 rpm 0.5 m 0.25 m 35° Vf 2 = 4 m/s 90°, Vw1 = 0, V1 = Vf 1

To determine: b1, V2, a2, W.D./kg

u1

S d1 N 60

S – 0.25 – 950 60

12.44 m/s

u2

S d2 N 60

S – 0.5 – 950 60

24.88 m/s

Centrifugal Compressors and Pumps

Vw 2

u2 

Vf 2

24.88 

tan E 2

4 tan 35’

319

19.17 m/s

(a) Vane angle at inlet (b1):

Vf 1

tan E1 \

4 12.44

u1

0.32150

b1 = 17.83°

Ans.

(b) Absolute velocity of water at exit (V2): V2

Vw22  V f22

19.172  42

19.53 m/s

Ans.

(c) Direction of water at outlet (a2):

tan D 2 \

Vf 2 Vw 2

4 19.17

0.2086

a2 = 11.78°

(d) Work done (W.D.): W.D. =

Vw 2 u2 gc

W.D. = 475.42

19.17

m m N-s2 – 24.8 – s s kg-m

N-m J or kg kg

Ans.

EXAMPLE 4.28 A centrifugal pump of 1.2 m diameter runs at 250 rpm and pumps 1880 lit/s, the average lift being 6 m. The angle which the vane makes at exit with the tangent to the impeller is 26° and the radial velocity of flow is 2.5 m/s. Determine the useful power and the efficiency. Find also the least speed to start pumping against a head of 6 m, the inner diameter of the impeller being 0.6 m. Solution:

Data: Outer diameter of the impeller: Speed: Discharge:

d2 = 1.2 m N = 250 rpm Q = 1880 lit/s

320

Fundamentals of Turbomachinery

Manometric head: Vane angle at exit: Radial velocity of flow: Inner diameter of the impeller:

Hm = 6 m b2 = 26° Vf 2 = 2.5 m/s d1 = 0.6 m

To determine: P, h, least speed to start

u2

S d2 N 60 Vf 2

tan E 2 \

S – 1.2 – 250 60

15.71 m/s

2.5 (15.71  Vw 2 )

u2  Vw 2

tan 26’

Vw2 = 10.58 m/s

(a) Manometric efficiency (hman): 9.81 – 6 10.58 – 15.71

gH m Vw 2 u2

K man

0.354

35.4%

Ans.

(b) Power output of the pump (WP): WP =

gUQH m gc

9.81

m s2

– 1000

= 110656.8 N-m/s (or

kg m3

– 1.88

m3 N- s2 –6m– s kg -m

W)

Ans.

(c) Power to the impeller (P): P

UQVw 2 u2 gc

1000

kg m3

– 1.88

= 312478.2 N-m/s (or

W)

m3 m m N- s2 – 10.58 – 15.71 – s s s kg -m

Ans.

Centrifugal Compressors and Pumps

321

(d) Least speed (NL): u22  u12 2g

Now,

Hm

or

S 2 N L2 (d22  d12 )

or

N L2

\

NL = 199.4 rpm

H m – 2 g – 3600

6 – 2 – 9.81 – 3600

6 – 2 – 9.81 – 3600 S 2 – (1.22  0.62 )

Ans.

EXAMPLE 4.29 A centrifugal pump impeller has outside diameter of 200 mm and rotates at 2900 rpm. The vanes are curved backward at 25° to the wheel tangent. The velocity of flow is constant at 3 m/s. Assume hydraulic efficiency as 75% and determine the head generated. Also determine the power required to run the impeller, if the breadth of the wheel at outlet is 15 mm. Neglect the effect of vane thickness, mechanical friction and leakage in the pump. Solution:

Data: Impeller outside diameter: Speed: Vane angle at exit: Velocity of flow is constant: Hydraulic efficiency: Breadth:

d2 = 200 mm N = 2900 rpm b2 = 25° Vf 1 = Vf 2 = 3 m/s hH = hman = 0.75 b2 = 0.015 m

To determine: Hm, P

u2 tan E2 \

S d2 N 60

S – 0.2 – 2900 60

Vf 2

Vf 1

u2  Vw 2

30.37  Vw 2

Vw2 = 23.94 m/s

30.37 m/s tan 25’

322

Fundamentals of Turbomachinery

(a) Manometric head (Hm): Hm g Vw 2 u2

K man

\

H m – 9.81 23.94 – 30.37

0.75

Hm = 55.55 m

Ans.

(b) Power required to run the impeller (P): Q = pd2b2Vf 2 = p × 0.2 × 0.015 × 3 = 0.0283 m3/s \

P =

gUQH m gc

9.81

m s2

= 15421.96 N-m/s

– 1000

kg m3

– 0.0283

m3 N- s2 – 55.55 m – s kg -m

(or W)

Ans.

EXAMPLE 4.30 The internal and external diameters of an impeller of a centrifugal pump which is running at 1000 rpm are 20 cm, 40 cm respectively. The discharge through pump is 0.04 m3/s and the velocity of flow is constant and equal to 2.0 m/s. The diameters of the suction and delivery pipes are 15 cm and 10 cm respectively and suction and delivery heads are 6 m (abs) and 30 m (abs) of water respectively. If the outlet vane angle is 45° and the power required to drive the pump is 16 kW, determine (a) the vane angle of the impeller at inlet, (b) the overall efficiency of the pump, and (c) the manometric efficiency of the pump. Solution:

Data: Speed: Internal diameter: External diameter: Discharge: Velocity of flow: Suction pipe diameter: Delivery pipe diameter: Suction head: Delivery head: Outlet vane angle: Power:

N = 1000 rpm d1 = 20 cm d2 = 40 cm Q = 0.04 m3/s Vf1 = Vf2 = 2 m/s ds = 15 cm dd = 10 cm hs = 6 m ab hd = 30 m ab b2 = 45° P = 16 kW

To determine: b1, ho, hman

u1

S d1 N 60

S – 0.2 – 1000 60

10.47 m/s

u2

S d2 N 60

S – 0.4 – 1000 60

20.94 m/s

Centrifugal Compressors and Pumps

\

Q = AdVd

Vs

Q As

Vd

0.04

Q Ad

0.04 S – 0.12 4

323

– 5.09 m/s

2.26 m/s

S – (0.15)2 4

(a) Vane angle at inlet (b1):

tan E1

Vf 1 u1

2.0 ; 10.47

? E1

10.48’

Ans.

(b) The overall efficiency of the pump (ho):

or

\

Hm

Ë pd Vd2 Û Ëp Û V2   Zd Ü  Ì s  s  Zs Ü Ì ÌÍ U g 2 g ÜÝ ÌÍ U g 2 g ÜÝ

Hm

Ë 5.092 Û Ë 2.262 Û    30 6 Ì Ü Ì Ü 2 – 9.81 ÝÜ ÍÌ 2 – 9.81 ÝÜ ÍÌ U gQH m 1000 U

Ko

25.06 m

1000 – 9.81 – 0.04 – 25.06 1000 – 16

0.6146

Ans.

(b) Manometric efficiency (hman):

tan E2 \

Vf 2

2 20.94  Vw 2

u2  Vw 2

tan 45’

Vw2 = 18.94 m/s Kman

gH m Vw 2 u2

9.81 – 25.06 – 100 18.94 – 20.94

61.98%

Ans.

EXAMPLE 4.31 The outer diameter of an impeller of a centrifugal pump is 40 cm and the width at outlet 5 cm. The pump is working against a total head of 15 m and is running at 800 rpm. If the vane angle at outlet is 40° and manometric efficiency is 75%, determine (a) the velocity of flow at outlet, (b) the velocity of water leaving the vane, (c) the angle made by absolute velocity at outlet with the direction of motion and (d) the discharge. Solution:

Data: Speed: Outer diameter of impeller: Width of outlet: Head:

N = 800 rpm d2 = 40 cm b2 = 5 cm Hm = 15 cm

324

Fundamentals of Turbomachinery

b2 = 40° hman = 75%

Vane angle at outlet: Manometric efficiency: To determine: Vf2, V2, a2, Q

u2

S d2 N 60

S – 0.4 – 800 60

K man

gH m Vw 2 u2

9.81 – 15 Vw 2 – 16.75

16.75 m/s 0.75

Vw2 = 11.71 m/s (a) Velocity of flow at outlet (Vf2):

tan E2 \

Vf 2

Vf 2

u2  Vw 2

16.75  11.71

tan 40’

Vf 2 = 4.23 m/s

Ans.

(b) Velocity of water leaving the vane (V2): V2

V f22  Vw22

4.232  11.712

12.45 m/s

Ans.

(c) Angle made by absolute velocity at outlet (a2):

tan D 2 \

Vf 2 Vw 2

4.23 11.71

0.36

a2 = 19.8°

Ans.

(d) Discharge (Q): Q = pd2b2Vf 2 = p × 0.4 × 0.05 × 4.23 \

Q = 0.265 m3/s

EXAMPLE 4.32 A centrifugal pump delivers 50 lit of water per second against a total head of 24 m at 1500 rpm. The velocity of flow is maintained constant at 2.4 m/s and blades are curved backward at 30° to tangent at exit. The inner diameter is half of the outer diameter, if the manometric efficiency is 80%. Find (a) the blade angle at inlet, (b) the power required to drive the pump, and (c) the torque. Assume radial entry. Solution:

Data: Discharge: Head: Speed: Velocity of flow: Vane angle at exit:

Q = 50 lit/s Hm = 24 m N = 1500 rpm Vf1 = Vf 2 = 2.4 m/s b2 = 30°

Centrifugal Compressors and Pumps

Inner diameter of blade: Manometric efficiency: Radial entry:

325

d1 = 0.5d2 hman = 80% a1 = 90°, Vw1 = 0, V1 = Vf1

To determine: b1, P, T

u2 K man

tan E 2

\ \

S d2 N 60

S – d2 – 1500 60

9.81 – 24 Vw 2 – 78.54 d2

Hm g Vw 2 u2

0.8

78.54d2 m/s

Vf 2

2.4

u2  Vw 2

78.54d2 

3.75 d2

tan 30’

d2 = 0.2464 m/s d1 = 0.1232 m/s Vw 2

3.75 d2

3.75 0.2464

15.22 m/s

S d1 N S – 0.1232 – 1500 9.677 m/s 60 60 u2 = 78.54 × 0.2464 = 19.354 m/s u1

\

(a) Vane angle at inlet (b1):

tan E1

Vf 1 u1

2.4 2.677

0.248

b1 = 13.93°

Ans.

(b) Power (P): P

UQVw 2 u2 gc

1000

= 14730 N-m/s

kg m2

– 0.05

(or W)

m3 m m N- s2 – 15.22 – 19.354 – s s s kg -m

Ans.

326

Fundamentals of Turbomachinery

(c) Torque (T): 2S N 60 P = Tw

2S – 1500 60

Z

\

T

P Z

14.730 157.1

157.1 rad/s

kN-m s s rad

(w = angular velocity)

0.09378 kN-m

Ans.

EXAMPLE 4.33 A centrifugal pump has a specific speed of 0.08 revolutions per second. The blades are forward facing on the impeller and the outlet angle is 120° to the tangent to an impeller passage with at outlet equal to one-tenth of the diameter. The discharge through the pump is 0.05 m3/s at a vertical height of 40 m. The suction and delivery pipes are each of 150 mm diameter. The pump has a combined pipe length of 50 m with a friction factor of 0.005. Other losses (pipe entry, exit, bends, etc.) are four times the velocity head in the pipes. Blades occupy 5% of the circumferential area and the hydraulic efficiency is 75%. Calculate the diameter of the pump impeller. Solution: Specific speed: Forward facing blades: Width of outlet: Discharge: Vertical height: Suction and delivery pipes diameters: Length of the pipe: Friction factor:

Ns = 0.08 rps b2 = 120° b2 = d2/10 Q = 0.05 m3/s Hs = 40 m ds = dd = 150 mm L = 50 m f = 0.005

Other losses: Blades occupy:

4Vd2 /2 g = 5% of the circumferential area hH = hman = 0.75

Hydraulic efficiency: To determine: d2

Q = discharge = VdAd = 0.05 m3/s

0.05 0.05 – 4 2.83 m/s S 2 S (0.15)2 dd 4 Total losses = pipe friction losses + other losses Vd

4 f LVd2 4Vd2  2 gdd 2g 4 – 0.005 – 50 – 2.832 4 – 2.832 = 4.35 m  2 – 9.81 – 0.15 2 – 9.81

Centrifugal Compressors and Pumps

327

Hm = total required head = Hs + total losses = 40 + 4.35 = 44.35 m NQ1 / 2

Ns

\

( gH )3 / 4

0.08

N = 34.08 rps = 2044.9 rpm

Flow area = S d2 \

d2 0.95 10

Q = flow area × flow velocity

or

0.05

or

Vf 2

\

N (0.05)1 / 2 9.81 – 44.35

S d22 – 0.95 – V f 2 10 0.1675 d22

m/s

Fluid power developed by pump Fluid power supplied to impeller

KH

K max

Vw 2

gH m Kman u2

9.81 – 44.35 0.75 – S d2 N

gH m Vw 2 u2

9.81 – 44.35 0.75 – S – d 2 – 34.08

5.42 m/s d2

The diameter of the pump impeller (d2): From outlet velocity triangle BCD, tan 60’

Vf 2 (Vw 2  u2 )

0.6175 Ë 5.42 Û d22 Ì  S d2 34.08Ü Í d2 Ý

185.5 d23  9.39d2  0.6175 0

\

d2 = 0.218 m

Ans.

EXAMPLE 4.34 A centrifugal pump has the following data: inlet diameter = 150 mm, outlet diameter = 310 mm, width of vane at inlet = 50 mm, vane angle at inlet = 0.5 radians, thickness of blades occupy 10% of the pheriphery, speed = 100 radians/second, vane angle at exit 40°. Calculate (a) the discharge (assume shockless entry), (b) the head developed by the pump, (c) the absolute velocity at exit, (d) the pressure rise in the impeller, and (e) the percentage of total work done which is converted into kinetic energy. Solution:

Data: w = angular velocity = 2p N = 2 × p × N = 100 rad/s N = 15.915 rps, 955 rpm

328

Fundamentals of Turbomachinery

u1 = p d1N = p × 0.15 × 15.915 = 7.499 m/s u2 = p d2N = p × 0.31 × 15.915 = 15.49 m/s

From inlet velocity triangle,

tan E1 \

Vf 1

Vf 1

u1

7.499

tan 28.65’

SE1 Ë Ì Radians = 180 Í

0.5 ; ? E1

Û 28.65’Ü Ý

Vf1 = 4.097 m/s Vw2 = u2 – Vf 2 cot b2 = 15.499 – 1.983 cot 40° = 13.14 m/s Q = p d2b2Vf 2 × 0.9 = p × 0.31 × 0.05 × Vf 2 × 0.9 = 0.0869 [Q Vf1 = Vf 2]

\

Vf2 = 1.983 m/s

(a) Discharge (Q): Q = p d1b1Vf 1 × 0.9 = p × 0.15 × 0.05 × 4.097 × 0.9 = 0.0869 m3/s

Ans.

(b) Head developed (Hm): For shockless entry, losses are neglected. u2Vw 2 \ Hm  losses g \

Hm

15.499 – 13.14 9.81

20.75 m

Ans.

(c) Absolute velocity at exit (V2): V2

Vw22  V f22

13.142  1.9322

(d) Pressure rise in the impeller (pR): From Eq. (5.23), pR

V f21  u22  V f22 cosec2 E2 2g

13.29 m/s

Ans.

Centrifugal Compressors and Pumps

329

4.0972  15.4992  1.98322  cosec2 40’ 2g 4.0972  15.4992  1.98322  cosec 2 40’ 2 – 9.81

12.99 m

Ans.

(e) Percentage of total work converted into KE at exit (% KE to W.D.): K.E. of water at exit =

V22 2 gc 13.292 m 2 N- s2 – 2 – 1 s2 kg-m

W.D.

N-m J or kg kg

u2 Vw 2 gc

15.499 % KE of W.D. =

88.296

m m N- s2 – 13.14 – kg-m s s

203.66 J/kg

K.E. of the water at exit – 100 W.D. 88.296 – 100 203.66

43.36%

Ans.

EXAMPLE 4.35 A centrifugal pump has the following data: Suction head (hs) = 5 m, delivery head (hd) = 35 m, head lost in suction = 50 cm, head lost in delivery 5 m, diameter of suction pipe 110 mm, diameter of delivery pipe 100 mm, total head 50 m, manometric efficiency 80%, overall efficiency 75%. Calculate (a) the power developed by the pump, (b) the power at the inlet to the impeller, (b) the power required to motor, (d) the total suction head, (e) the delivery head. Solution:

Data: Suction head: Delivery head: Head lost in suction: Head lost in delivery: Diameter of suction pipe: Diameter of delivery pipe: Total head: Manometric efficiency: Overall efficiency:

= 5 m = 35 m = 50 cm = 5 mm = 110 mm = 100 mm = 50 m hman = 80% ho = 75% hS hd hfs hfd ds dd Hm

330

Fundamentals of Turbomachinery

Vd2 2g

Hm

hs  hd  h fs  h fd 

50

5  35 

Vd2 50 5 ; ? Vd 100 2 – 9.81

Ad Vd

S 2 d d Vd 4

Q

S – 0.12 – 9.4 4

9.4 m/s

0.074 m 3 /s

(a) Water power (power of the water at the discharge end) (WP): WP =

UQgH m gc

1000

kg m3

m3 m N- s2 – 9.81 – 50 m – s kg -m s2

– 0.074

= 36297 N-m/s

Ans.

(b) Power at the inlet to the impeller (power transferred to the impeller or work done on the impeller or power of the impeller) (P): WP P WP Kman

Kman

P

36.297 0.8

45.37 kW

Ans.

(c) Power to the motor (Pm): Ko

WP ; Pm

Pm

WP Ko

Pm

36.297 0.75

48.396 kW

Ans.

(d) Total suction head (Hs):

or

S – d s2 – Vs 4 Vs = 7.71 m/s

\

Hs

Q

As Vs

hs  h fs  5  0.5 

S – (0.11)2 Vs 4

0.074

Vs2 2g

(7.71)2 2 – 9.81

8.59 m

[below atmospheric head, \ vacuum] Ans.

(e) Total delivery head (Hd): Hd

Hm  Hs 

Vd2 2g

50  8.59 

9.42 2 – 9.81

36.906 m

Ans.

Centrifugal Compressors and Pumps

331

EXAMPLE 4.36 A centrifugal pump is used to discharge water from a drainage. The length of the suction pipe is 4000 mm. The end which is dipped in drainage is 0.8 m2 and the other end of the pipe is connected to the pump inlet, i.e. the eye of the pump has an area of 0.6 m2, pressure at the eye is 60 kPa. Calculate the discharge and neglect all friction losses. Assume that the pressure acting on the surface of the drainage is 101.325 kN/m2. Solution:

Data: Suction Suction Suction Suction

pipe inlet area: pipe exit area (eye side): side pressure (eye side): head

As1 As2 ps2 Zs2

= = = =

0.8 m2 0.6 m2 60 kPa hs = 4000 mm

Q = discharge = As1Vs1 = As2Vs2 0.8Vs1 = 0.6Vs2 ; \ Vs1 = 0.75Vs2 Applying Bernaulli’s equation between suction pipe inlet and exit, pa Vs21 Z s1   U g 2g gc Vs21 101.325 – 1000  0 1000 – 9.81 2 – 9.81

ps 2 Vs22 Z s 2   Ug 2g gc 60 – 1000 1.332 Vs21  4 1000 – 9.81 2 – 9.81 2

N m N m N + = + +m ¥ kg m m kg m m kg-m m 2 ¥ 3 ¥ 2 s2 ¥ 2 m 2 ¥ 3 ¥ 2 s2 ¥ 2 m s s m s s s2 N N N Nm Nm N +m= +m+m N N N \

Vs1 = 2.12 m/s

Discharge (Q): Q = As1Vs1 = 0.8 × 2.12 = 1.694 m3/s

Ans.

EXAMPLE 4.37 Following data refers to a centrifugal pump: outer diameter of the impeller (impeller diameter) = 0.75 m, discharge = 1.5 m3/s, total head = 0.85 m, speed = 1000 rpm, width at outlet = 0.8 m. The leakage loss is 4% of the discharge, mechanical losses 8.5 kW, and hydraulic efficiency 80%. Calculate (a) the vane exit angle, (b) the power at shaft of the motor, (c) the overall efficiency, (d) the mechanical efficiency, and (e) the volumetric efficiency. Solution:

Data: Outer diameter of impeller: Discharge:

d2 = 0.75 m Q = 1.5 m3/s

332

Fundamentals of Turbomachinery

Total head: Speed: Width at outlet: Leakage loss: Mechanical loss: Hydraulic efficiency:

Hm N b2 DQ hman

u2

S d2 N 60

S – 0.75 – 1000 60

K man

gH m Vw 2 u2

9.81 – 85 39.27 – Vw 2

= = = = = =

85 m 1000 rpm 18 cm 0.04 × Q 8.5 kW 80% = hH

39.27 m/s

0.8

Vw2 = 26.54 m/s (DQ + Q) = total quantity of water entered into the pump (DQ + Q) = pd2b2Vf 2 = p × 0.75 × 0.08 × Vf 2 = 1.5 + 0.06 \

Vf 2 = 8.28 m/s

(a) Vane exit angle (b2): From exit velocity triangle,

tan E2

Vf 2

8.28 39.27  26.54

33.03’

Ans.

1000 – 1.56 – 26.54 – 39.27  8.5 1634.37 kW 1000

Ans.

u2  Vw 2

?

E2

(b) Power at the shaft (SP): SP

U (Q  'Q) Vw 2 u2  mechanical losses gc

(c) Mechanical efficiency (hm): P = power transferred to the impeller Vw 2 u2 U (Q  'Q) gc Km

P SP

1625.87 – 100 1637.37

39.27 – 26.54 – 1000 – 1.56 1000 – 1 99.48%

1625.87 kW

Ans.

(d) Volumetric efficiency (hv): Kv

Q (Q  'Q)

1.5 – 100 1.56

96.15%

Ans.

Centrifugal Compressors and Pumps

333

(e) Overall efficiency (ho): ho = hm × hman × hv = 0.8 × 0.9948 × 0.9615 = 0.7652

Ans.

EXAMPLE 4.38 A centrifugal pump with 1.2 m diameter runs at 200 rpm and pumps 1.88 m3/s, the average lift being 6 m. The angle which the vane makes at exit with the tangent to the impeller is 26° and the radial velocity is 2.5 m/s. Determine the manometric efficiency and the least speed to start the pump if the inner diameter of the impeller is 0.6 m. Solution:

Data: Impeller outer diameter: Speed: Discharge: Total head: Vane angle at exit: Radial velocity at exit: Impeller inlet diameter:

d2 = 1.2 m N = 200 rpm Q = 1.88 m3/s Hm = 6 m b2 = 26° Vf 2 = 2.5 m/s d1 = 0.6 m

To determine: hman, NL

S d2 N 60

S – 1.2 – 200 60

12.57 m/s

S d1 N 60 From exit velocity triangle,

S – 0.6 – 200 60

6.285 m/s

u2 u1

Vf 2

tan E 2 \

u2  Vw 2

2.5 12.57 – Vw 2

tan 26’

Vw2 = 7.44 m/s

(a) Manometric efficiency (hman): Kman

gH m Vw 2 u2

9.81 – 6 – 100 7.44 – 12.57

63%

(b) Minimum speed (NL): u22  u12 2g

We have 2

Ë S d2 N L Û Ë S d1 N L Û Ì 60 Ü  Ì 60 Ü Í Ý Í Ý

or

S 2 N L2 60 2

Hm

2

(d22  d12 )

2 gH m

2 – 9.81 – 6

2 – 9.81 – 6

Ans.

334

Fundamentals of Turbomachinery

NL

\

2 – 9.81 – 6 – 60 – 60 S

2

(d22



2 – 9.81 – 6 – 3000

d12 )

S 2 (1.22  0.62 )

NL = 199.39 rpm

Ans.

EXAMPLE 4.39 Following data refers to a centrifugal pump: Impeller inlet diameter: d1 = 0.4 m Impeller outlet diameter: d2 = 0.7 m Velocity of flow at exit: Vf 2 = 2.5 m/s Exit vane angle: b2 = 40° Manometric efficiency: hman = 75% Calculate the minimum speed (NL). Solution:

u2

S d2 N 60

S – 0.7 – N 60

S d1 N S – 0.4 – N 60 60 From exit velocity triangle, u1

0.037 N m/s 0.02 N m/s

Vw2 = u2 – Vf 2 cot b2 = 0.037N – 2.5 cot 45° = 0.037N – 2.5 9.81 – H m (0.037 N  2.5) 0.037 N

Kman

gH m Vw 2 u2

Hm

0.75 – 0.037 N (0.037 N  2.5) 9.81

0.75 0.00283 (0.037 N 2  2.5N )

Minimum speed (NL): We have or

u22  u12 2g (0.037 N L )2

Hm (0.02 N L )2

2 – 9.81 – 0.00283 (0.037 N L2  2.5N L )

NL = 1.51 rps or 95.36 rpm

Ans.

EXAMPLE 4.40 A three-stage centrifugal pump has impellers 50 cm diameter and 3 cm width at outlet. The thickness of the blades has reduced the circumferential area by 10%. The manometric efficiency is 90%, overall efficiency 80%, whirl velocity at exit = 20 m/s, radial velocity at outlet is 2.25 m/s, and speed 1000 rpm. Calculate (a) the head generated, (b) the discharge, (c) the exit vane angle, and (d) the shaft power.

Centrifugal Compressors and Pumps

Solution:

Data: Number of stages: Impeller outer diameter: Width at outlet: Area reduced due to thickness: Manometric efficiency: Overall efficiency: Whirl velocity at exit: Radial velocity at exit: Speed:

335

n = 3 d2 = 50 cm b2 = 3 cm = 10% hman = 0.9 ho = 0.8 Vw2 = 20 m/s Vf 2 = 2.25 m/s N = 1000 rpm

To determine: Hm, Q, b2, SP. Area of flow = p × d2 × b2 × 0.9 = p × 0.5 × 0.03 × 0.9 = 0.04241 m2

u2

S d2 N 60

S – 0.5 – 1000 60

26.18 m/s

(a) Discharge (Q): Q = area of flow × Vf 2 = 0.04241 × 2.25 = 0.09542 m3/s

Ans.

(b) Exit vane angle (b2): From exit velocity triangle,

tan E2 \

Vf 2

2.25 26.18  20.0

u2  Vw2

b2 = 20.0°

Ans.

(c) Total Head (HmT): We have or \

K man Hm

gH m Vw 2 u2 K mano Vw 2 u2 g

0.9 – 20 – 26.18 9.81

48.04 m

HmT = total head developed due to three stages = n × Hm = 3 × 48.04 = 144.11 m

(d) Shaft power (SP): Ko

P SP

Vw 2 u2 UQ gc – SP

Ans.

336

Fundamentals of Turbomachinery

Vw 2 u2 UQ gc – Ko

\

SP =

or

SP = 62.45 kW

20 – 26.18 – 1000 – 0.09542 1000 – 0.8

Ans.

EXAMPLE 4.41 A four-stage centrifugal pump has four identical impellers keyed to the same shaft. The speed of the shaft is 500 rpm. The total manometric head developed from four impellers is 50 m. The width at exit is 5 cm and the diameter at exit is 50 cm. The whirl velocity at exit is 10 m/s, and the radial flow velocity at exit is 2 m/s. Calculate (a) the discharge, (b) the exit vane angle, and (c) the manometric efficiency. Solution:

Data: Number of stages: Speed: Total manometric head: Width at exit: Diameter at exit: Whirl velocity at exit: Radial velocity at exit:

n = 4 series N = 500 rpm HmT = 50 m b2 = 5 cm d2 = 60 cm Vw2 = 10 m/s Vf 2 = 2 m/s

To determine: Q, b2, hman

u2

S d2 N 60

S – 0.6 – 500 60

15.71 m/s

(a) Discharge (Q): Q = pd2b2Vf 2 = p × 0.6 × 0.05 × 2 = 0.1885 m3/s (b) Exit vane angle (b2): From exit velocity triangle,

tan E 2

Vf 2 u2  Vw 2

2 (15.71  10)

b2 = 19.3°

Ans.

(c) Manometric efficiency (hman):

Hm K man

H mT 4 gH m Vw 2 u2

50 4

12.5 m

9.81 – 12.5 – 100 10 – 15.71

78.1%

Ans.

EXAMPLE 4.42 Prove that the manometric head of a centrifugal pump running at speed N and giving a discharge Q, may be written as Hman = AN2 + BNQ + CQ2 where A, B, C are constants.

Centrifugal Compressors and Pumps

337

Solution: From Eq. (4.33), Hm

Vw2 u2  HL g

Vw 2 u2 KV22  g 2g

(i)

V22 is the part of the head not converted into pressure head and being wasted. 2g KV22 is the part of the head and other losses not converted into pressure head. 2g

S d2 N 60

u2

K1 N

(ii)

Q = discharge = pd2b2Vf2 Vf 2

where K1 and K3 are constants. From the exit velocity triangle,

Q S d2 b2

K2Q

Vw2 = u2 – Vf 2 cot b2

(iii)

(iv)

Substituting Eqs. (ii) and (iii) in (iv), Vw 2

K1 N  K 2 Q cot E 2

Now,

V22

V f22  Vw22

\

V22

(K 2Q)2  (K1 N  K 2Q cot E2 )2

Substituting Eqs. (ii), (iii), (v) and (vi) in (i), Hm

(K1 N  K 2 Q cot E2 ) K1 N K  [(K 2 Q)2  (K 2 N  K 2Q cot E2 )2 ] g 2g

2 K1 N (K1 N  K 2Q cot E 2 )  K [(K 2Q)2  ( K1 N  K 2Q cot E 2 )2 ] 2g 2 K12 N 2  2 K1K 2 NQ cot E2  KK 22Q 2  KK12 N 2  KK 22Q 2 cot 2 E2  2KK1 NK 2Q cot E 2 2g

(v)

(vi)

338

Fundamentals of Turbomachinery 2

2K12 N 2 - KK12 N + K 2 K1 K2 NQ cot b 2 - 2K1K 2 NQ cot b 2 =

– KK22 Q2 cot 2 b 2 - KK 22Q2

2g

N 2 (2K12  KK12 ) NQ (2 K1K 2 K cot E2  2K1K 2 cot E2 )  2g 2g 

Q 2 (KK 22 cot 2 E 2  KK 22 ) 2g

(vii)

Let A C

2K1K 2 K cot E2  2 K1K 2 cot E2 Û Ü 2g Ü Ü 2 2 2 KK2 cot E2  KK2 Ü  ÜÝ 2g 2K12  KK12 ; B 2g

(viii)

Substituting Eq. (viii) in (vii), we have Hm = AN2 + BNQ + CQ2

Proved.

EXAMPLE 4.43 Following data refers to a centrifugal pump: Impeller diameter: d2 = 1.2 m Speed: N = 1450 rpm Exit blade angle: b2 = 30° Radial velocity at exit: Vf 2 = 18 m/s Calculate (a) the power required/kg, (b) the pressure head at impeller outlet, and (b) the efficiency. If a diffuser is used at the end of the impeller, the exit velocity from the impeller is reduced by 60%, then calculate (d) the pressure head at impeller outlet and (e) the efficiency. Solution:

u2

S d2 N 60

S – 1.2 – 1450 60

91.11 m/s

From exit velocity triangle, Vw2 = u2 – Vf 2 cot b2 = 91.11 – 18 cot 30° Vw2 = 59.93 m/s (a) Power required per kg (W.D.): W.D. =

Vw 2 u2 gc

59.33 – 91.11 1

5460.5 J/kg

Ans.

Centrifugal Compressors and Pumps

339

(b) Pressure head at impeller outlet (Ho): V2 = absolute velocity at exit V2

V f22  Vw22

182  59.932

62.57 m/s

Vo = Velocity head at impeller exit = Vo

or

(62.57)2 2 – 9.81

V22 2g

199.57 m

Hm = total head developed = 59.93 – 91.11 9.81

Vw 2 u2 g

556.597 m

Ho = Hm – Vo = 556.597 – 199.57 = 357.02 m

Ans.

(c) Efficiency (h): K

Ho – 100 Hm

357.02 – 100 556.597

64.14%

Ans.

(d) Pressure head at impeller with diffuser (hod), and diffuser exit velocity reduced by 60%. V2d = V2 × 0.4 = 62.5 × 0.4 = 25.03 m/s \

Vod = exit velocity head at impeller outlet with diffuser. V22d 2g

\

(25.03)2 2 – 9.81

31.93 m

hod = Hm – Vod = 556.597 – 31.93 = 524.67 m

Ans.

(e) Efficiency with diffuser (hd): Kd

hod – 100 Hm

524.67 – 100 556.597

94.26%

EXAMPLE 4.44 Following data refers to a centrifugal pump: Impeller diameter: d2 = 40 cm Exit vane angle: b2 = 30° Speed: N = 1500 rpm Velocity flow is constant: Vf 1 = Vf 2 = 2 m/s Hydraulic efficiency: hH = hman = 90%

Ans.

340

Fundamentals of Turbomachinery

Calculate (a) the pressure rise in the impeller, (b) the percentage of KE recovered in the volute casing, (c) the total head developed, and (d) the KE recovered in the casing. Solution:

S d2 N 60

u2

S – 0.4 – 1500 60

31.42 m/s

From the exit velocity triangle, Vw2 = u2 – Vf 2 cot b2 = 31.42 – 2 cot 30° = 27.96 m/s V2 = absolute velocity at exit of the impeller V f22  Vw22

22  27.962

28.03 m/s

(a) Pressure rise in the impeller (pR) (head produced by the impeller): Applying Bernoulli’s equation between the inlet and the outlet of the impeller, using suffix 1 and 2, respectively. p1 V12 Vw 2 u2   Ug 2g g

Now,

pR

p2 p  1 Ug Ug

p2 V22  Ug 2g

V12 V22 Vw 2 u2   2g 2 g g

22 (28.03)2 27.96 – 31.42   2 – 9.81 2 – 9.81 9.81

(Q V1 = Vf 1)

pR = 49.71 m (b) Total head developed (Hm): K man

\

KH

gH m Vw 2 u2

9.81 – H m 27.96 – 31.42

0.9

Hm = 80.597 m

Ans.

(c) K.E recovered in the casing (KER) KER = total head – pressure rise in the impeller = 80.597 – 49.71 = 30.887 m

Ans.

Centrifugal Compressors and Pumps

341

(d) Percentage of KE recovered in the volute casing: KE recovered in the casing KE corresponding to the exit absolute velocity 30.887

30.887 – 2 – 9.81

V22

(28.03)2

– 100

77.13%

Ans.

2g

EXAMPLE 4.45 A single-stage centrifugal pump with impeller diameter of 30 cm rotates at 2000 rpm and lifts 3 m3 of water per second to a height of 30 m with an efficiency of 75%. Find the number of stages and the diameter of each impeller of a similar multistage pump to lift 5 m3 of water/second to a height of 200 m, when rotating at 1500 rpm. Solution: First pump (single stage) Outer diameter of the impeller; d1 = 30 cm Speed: N1 = 2000 rpm Discharge: Q1 = 3 m3/s Total head: Hm1 = 30 m Efficiency: h1 = 0.75 Second pump (multistage) Discharge: Q2 = 5 m3/s Total head for n stages: HmT = 200 m Total head for single stage: Hm2 = ? Speed: N2 = 1500 rpm To determine: No. of stages (n), d2 We have the relation,

Ns or

2000 – 3 30

N 2 Q2

H m3 /14

H m3 /24

1500 – 5

3/4

\

N1 Q1

H m3 /24

Hm2 = 28.73 m

(a) Number of stages (n): HmT = nHm2 \

n

H mT Hm2

200 28.73

6.96  7

Ans.

342

Fundamentals of Turbomachinery

(b) Outer diameter of second pump (d2): d2 N 2

d1 N1

Hm2

H m1

d2 – 1500

or

0.3 – 2000

28.73

\

30

d2 = 0.39 m

Ans.

IMPORTANT EQUATIONS Centrifugal compressors u22 gc

u2Vw 2 gc

• W.D.

Vw22 gc

ËT Û • W.D. = c pT01 Ì 02  1Ü T Í 01 Ý

[' Vw1

0, Vw 2

u2 ]

c p (T02  T01 )

(4.1) (4.5)

J

•

pR1

p02 p01

pR

p2 p1

Ë u22 Kct t Û J 1  1Ü Ì ÜÝ ÍÌ c p gcT01

[ pR1

stagnation pressure rise]

(4.9)

J

•

•

pR 0

• 'h0

Ë u22 Kct t Û J 1  1Ü Ì ÌÍ c p gc T1 ÜÝ

p03 p01

[ pR

Ë u22 Ì Kct to gc c p Ì1  Ì T01 Í

(h3  h1 ) 

static pressure rise]

(4.10)

J

Û J 1 Ü Ü Ü Ý

[ pR 0

overall pressure rise]

(4.12)

(V32  V12 ) 2 gc

• Ip

Ë J 1 Û Ü gc c pT1 Ì p3 J  1Ü Ì 2 ÜÝ u2 ÌÍ p1

• Ip

VKct to

• Vr21

4Q Ë S dt N Û Ë Û ÌÍ 60 ÜÝ  Ì 2 2 Ü Í S ( dt  d h ) Ý

(I p

pressure coefficient)

(4.12d) (4.12g)

2

2

(4.13)

Centrifugal Compressors and Pumps

• W.D. = 'h0

V u2Vw 2 gc

• W.D. = 'h0

u22V gc

(V

( E2

343

slip factor) (E2  90’)

90’)

J

•

pR1

Ë u22 Kct t V Û J  1 Ì1  Ü c p gc T01 ÜÝ ÌÍ

pR

Ë u22Kct t V Û J  1 Ì1  Ü c p gc T1 ÜÝ ÌÍ

(4.16)

J

•

(4.16a) J

•

Ë Kct  to V u22 Û J  1 Ì1  Ü gc c pT01 ÝÜ ÍÌ

pR 0

• W.D. = 'h0

VM u22 gc

(M

(4.17)

power factor)

(4.18)

J

•

pR1

Ë u22 VMKct t Û J  1 Ì1  Ü c p gc T01 ÝÜ ÍÌ

pR

Ë u22 VMKct t Û J  1  1 Ì Ü c p gcT1 ÝÜ ÍÌ

(4.19)

J

•

(4.19a)

J

•

Ë u22 VMKct  to Û J  1 Ì1  Ü c p gc T01 ÝÜ ÍÌ

pR 0

(S N )2 (3dt2  d h2 )

• Vr1

Q2 Q1

• Q

(4.20b)

60 2 – 2

• W.D.

•

(4.20)

Ë u2Vw 2  u1 Vw1 Û Ì Ü gc Í Ý „

„

(with prewhirl)

T2 / p2 T1 / p1 (S d1  nt ) b1V1

(4.30) (S d2  nt ) b2V2

(4.31)

344

Fundamentals of Turbomachinery

Centrifugal pump • Hm

Vw 2 u2 J ; Hm  H L , in kg gc

• Hm

Ë pd Vd2 Û Ëp Û V2   Zd Ü  Ì s  s  Zs Ü Ì Í Ug 2g Ý Í Ug 2g Ý

• Hm

hs  hd  h fs  h fd 

• Hm = hd – hs • Kmano

(4.33)

(4.34)

Vd2 2g

(4.35)

[Zs = Zd, ds = dd]

(4.36)

Hm g Vw 2 u2

(4.38)

Water power Power of the impeller

Impeller power Shaft power (SP)

• Km • Ko

Vw 2 u2  H L , in m g

(4.38a)

Kmano – Km

(4.40)

• W.D.

Vw 2 u2 gc

(4.42)

• W.D.

Hm  HL

• Pressure rise =

(u22  u12 ) (V22  V12 ) (Vr21  Vr22 )   2g 2g 2g

p2  p1 Ug

V f21  u22  V f22 cosec 2 E2 2g

• Minimum starting speed = N

m

60 – 2 – Kmano Vw 2 d2 S (d22  d12 )

(4.50)

(4.54)

(4.56)

REVIEW QUESTIONS 1. Draw the schematic diagram of a centrifugal compressor stage indicating the names of its principle parts. 2. Name the types of impellers and draw the corresponding velocity triangles at entry and exit. 3. Explain why the radial tipped impeller is used widely in centrifugal compressors. 4. Derive an expression for the overall pressure ratio developed in the centrifugal compressor. 5. Explain the method to calculate the blade angle at the impeller eye root and at the eye tip.

Centrifugal Compressors and Pumps

345

6. Define (a) slip factor, (b) power input factor, and (c) pressure coefficient in the centrifugal compressor. 7. Explain prewhirl in centrifugal compressors. 8. Explain flow in the vaneless space of a diffuser. 9. Explain the method to calculate the propeller channel in centrifugal compressors. 10. Explain the method to determine the diffuser inlet vane angle, the width, and the length of the diffuser passage. 11. Explain with the help of a diagram, the surging of centrifugal compressor. 12. Draw the enthalpy–entropy diagram for the centrifugal compressor and briefly explain the same. 13. Derive an expression for the work done and the power in respect of the centrifugal compressor. 14. What is the difference between the isentropic efficiency based on total and static conditions. 15. What is NPSH? Explain cavitation and its effects on the performance of a centrifugal pump. 16. Explain what is priming in pumps. 17. With the help of neat diagram explain the various losses and efficiencies of a centrifugal pump. 18. Sketch the velocity triangles at the outlet of centrifugal pumps with (a) radial vanes, (b) forward vanes, and (c) backward curved vanes respectively. 19. Is it desirable to have lower or higher value of NPSH? Justify your answer with the help of the relevant equation. 20. Sketch a centrifugal pump. Indicate all the parts. Also mention the function of each part. 21. Explain the phenomenon of cavitation as it happens in the centrifugal pump. Can it be prevented? Write your answer giving reasons. 22. What are the ill effects of cavitation? 23. With a neat sketch, explain the different types of centrifugal pump casings. What is manometric head and manometric efficiency? 24. Define the terms (a) mechanical efficiency and (b) overall efficiency as applied to centrifugal pumps.

EXERCISES 4.1 A rotary compressor having a pressure ratio 10 : 1, takes in 30 kg/s of air at 15°C. Calculate the power required in kW. 4.2 A centrifugal compressor compresses the air from 1 bar, 15°C to 2 bar. It deliver 100 kg/s of air. The exit temperature is 100°C and no heat is added to the air. Calculate (a) the power required and the (b) efficiency of the compressor.

346

Fundamentals of Turbomachinery

4.3 The following data refers to a centrifugal compressor: Overall diameter of the impeller: 0.5 m Eye tip diameter: 0.3 m Eye root diameter: 0.20 m Speed: 15000 rpm Total mass flow rate: 18 kg/s Inlet total temperature: 300 K Total isentropic efficiency: 80% Power input factor: 1.05 Slip factor: 0.91

4.4

4.5

4.6

4.7

4.8

4.9

Calculate (a) the total head pressure ratio, (b) the power required to drive the compressor, and (c) the inlet angles of the vanes at the root and at the tip of the impeller eye. The stagnation conditions of the air at the compressor intake are 1 bar and 30°C. The speed of the machine is 15,000 rpm. Intake absolute velocity is axial. Assume radial exit blades, exit flow velocity of 140 m/s and total-to-total efficiency of the compressor to be 0.80. Draw the velocity triangle at the exit of the rotor and compute (a) the pressure ratio between the impeller inlet and the exit and (b) the slip coefficient. Assume, slip = 50.0 m/s and the rotor diameter at outlet to be 60 cm. The following data refers to a centrifugal compressor. Speed of the machine 6000 rpm, impeller tip diameter 100 cm, mass flow rate of air 25 kg/s, atmosphere pressure and temperature are 1 bar and 30°C respectively. Slip coefficient 0.91, pressure coefficient 0.75 and mechanical efficiency 0.98. Calculate (a) the overall pressure ratio, (b) the adiabatic efficiency of the impeller, (c) the static temperature of the air at the exit, and (d) the shaft power input. A centrifugal pump with an impeller outlet diameter of 375 mm runs at 750 rpm and delivers 35 lit/s of water. The radial velocity at the impeller exit is 2 m/s. The difference between the water levels at the overhead tank and the sump is 14.2 m including frictional losses. The total power input needed to run the pump is 6.1 kW, its mechanical and volumetric efficiencies being 0.95 and 0.96 respectively. The rotor blades are backward curved with an exit angle of 45°. Compute (a) the ideal head developed with no slip and no hydraulic losses and (b) the actual pump efficiency. A centrifugal pump is required to discharge water at the rate of 0.15 m3/s while running at 1480 rpm against a head of 30 m. The impeller diameter is 25 cm and the width at the outlet is 6 cm. The manometric efficiency is 75%. Determine the vane angle at the outlet, assuming radial entry. A centrifugal pump is running at 1000 rpm. The outlet vane angle of impeller is 45° and the velocity of flow at the outlet is 2.5 m/s. The discharge through the pump is 200 lit/s when the pump is working against the total head of 20 m. If the manometric efficiency of the pump is 80%, determine (a) the diameter of the impeller and (b) the width of impeller at the outlet. A centrifugal pump discharges 0.15 m3/s of water against a head of 12.5 m. The speed of the impeller is 600 rpm. The outer and inner diameters of the impeller are 50 cm and

Centrifugal Compressors and Pumps

347

25 cm respectively and the vanes are bent back at 35° to the tangent at exit. If the area of flow remains 0.07 m2 from inlet to outlet, determine (a) the manometric efficiency of the pump and (b) the vane angle at inlet. 4.10 A centrifugal pump running at 1450 rpm discharges 110 lit/s against a head of 23 m. If the diameter of the impeller is 25 cm and its width 5 cm, find the vane angle at the outer periphery. The manometric efficiency of the pump is 75%. 4.11 A centrifugal pump is running at 1000 rpm. The outlet vane angle of the impeller is 30° and the velocity of flow at outlet is 3 m/s. The pump is working against a total head of 30 m and the discharge through the pump is 0.3 m3/s. If the manometric efficiency is 75%, determine (a) the diameter of the impeller and (b) the width of the impeller at the outlet.

5

Axial Flow Compressors

5.1 INTRODUCTION An axial flow compressor is a power-absorbing turbomachine and also a pressure producing machine. The name itself indicates that the working fluid (air or gas) enters and leaves the compressor in an axial direction (i.e. u1 = u2 = u). Figure 5.1 shows the blade passages of an axial flow compressor, an impulse turbine, and a reaction turbine. The compressor blade row has inlet area less than that of exit. This is similar to a diffuser (subsonic). Hence, the fluid is diffused with pressure gain in the axial compressor blade row. In case of the reaction turbine blade row, the inlet area is more than that of exit and acts as a nozzle with the fluid being accelerated in the blade passages. In case of the impulse turbine blade row, both the inlet and exit area are same. u

u

1 1 2

(a)

Figure 5.1

(b)

(c)

Blade passages (shape or form): (a) Axial compressor. (b) Impulse turbine. (c) Reaction turbine. 348

Axial Flow Compressors

349

Axial compressor and reaction turbine blades are aerofoil sections whereas the impulse blade sections are formed from circular arcs (uniform across section).

5.2 DESCRIPTION AND PRINCIPLE OF OPERATION An axial compressor stage consists of a row of moving blades attached to the periphery of a rotor hub followed by a row of fixed blades fixed to the casing (walls of the outer casing). The compressor is made up of a number of such stages to give an overall pressure ratio from inlet to outlet. Figure 5.2 shows a few compressor stages with pressure and velocity distribution. All angles are referred to the tangential direction (plane of moving blades).

Figure 5.2

An axial compressor with velocity and pressure distribution.

One row of rotor blades combined with the next row of stator blades constitutes a stage; the pressure ratio developed in a stage is about 1.2. Depending upon the overall pressure ratio, the number of stages in a compressor is decided. At the inlet to the compressor, an extra row of fixed vanes, called inlet guide vanes (IGV), are fitted. These vanes are also called upstream guide vanes (UGV). These do not form part of the stage but are meant solely to guide the air at the correct angle onto the first row of moving blades. The function of IGV is to change the axial direction of the approaching flow to the desired direction (a1). Therefore, the first stage experiences additional losses arising from flow through the guide vanes. It will be seen that as the fluid moves from stage to stage, the height of the blades decreases, so that a constant axial velocity (Vf1 = Vf 2 = Vf 3 = Vf = constant) through the compressor is maintained as the density increases from stage to stage. We have m mass flow rate of fluid

350

Fundamentals of Turbomachinery

= r1a1Vf1

= r2a2Vf 2

(5.1)

m = 1st stage = 2nd stage or

m = r1 p d h1 Vf1 = r1 p d h2 Vf 2 = r1h1 = r2h2

(Q Vf1 = Vf 2)

(5.2)

If the density increases from stage to stage, then the height should decrease from stage to stage. A constant axial velocity is not a compulsion but it is convenient from the point of view of design. The blade peripheral velocities at inlet and outlet are the same. There is no flow in the radial direction. The whirl component of absolute velocity is in the direction of blade motion. Air or gas with an absolute velocity V1 and angle a1 (w.r.t. tangential direction) from the first stage diffuser (stator) is passed into the second stage rotor as shown in Figure 5.3. The rotor row has tangential velocity (peripheral velocity) u, and combining the two velocity vectors gives the relative inlet velocity, Vr1 at angle b1 (w.r.t. tangential direction). The exit velocity triangle for second stage rotor can be drawn similar to any axial machine. The absolute velocity V2 moves into the stator row (diffuser row) where the flow direction is changed to a3 with absolute velocity V3. If the following stage is the same as the preceding stage, the stage is said to be normal. For a normal stage V1 = V3, a1 = a3. Vr2 is less than Vr1, showing that diffusion of the relative velocity has taken place with some static pressure rise across the rotor blades. As the fluid flows (glides) over the rotor blades, the kinetic energy is imparted to the fluid by means of rotating blades. The absolute velocity of the fluid increases in the rotor blades (V1 increases to V2). A part of this kinetic energy is converted into pressure energy in the converging area between the rotor blades. The fluid is then passed over to the stator blades where a further pressure rise takes place and in addition to this the flow is directed to enter the next stage of moving blades with an absolute velocity V3 = V1 and at an angle a3 = a1. There is rise in pressure continuously from stage to stage (Figure 5.2). When the fluid particles flow either over rotating blades or over the fixed blades the fluid gets deflected and there is a reduction in the relative velocity due to frictional effect. It is clear that the compression takes place in the moving as well as stator blades.

5.3 STAGE VELOCITY TRIANGLE Figure 5.3 shows the velocity triangles for a compressor stage. Let subscripts 1, 2 and 3 represent fluid entry into the rotor, exit from the rotor and exit from the stator respectively. a1, a2, a3 = air angles with respect to the absolute velocity in the tangential direction b1, b2, b3 = air angles with respect to the relative velocity in the tangential direction If the following stage is the same as the preceding one, the stage is said to be normal. For a normal stage V1 = V3, a1 = a3, Vr2 < Vr1, showing that diffusion of the relative velocity has taken place with some static pressure rise across the rotor blades. From the velocity triangle at entry, Vf1 = V1 sin a1 = Vr1 sin b1

(5.3)

351

Axial Flow Compressors

Upstream guide vanes Entry velocity Vr1 triangle b1 x1 Rotor blades Vr 2 x2

Vf 1

(a)

V1 a1 Vw1

u

u b1 b2

u Vf 2

b2 u

a2 Vw2

b3

Vf 3

Vw2

(b)

Vw1 Vf 1

Vf 2

Exit velocity triangle h 2, p 2 h02, p02 Diffuser blades

V2

Vr 3

Figure 5.3

h 1, p 1 h01, p01

V1

Vr 2 V2

a2 a1

Vr1

h 3, p 3

DVw = (Vw2 –Vw1)

h03, p03

(c)

V3 a3

Velocity triangles for a compressor or stage and combined velocity diagram: (a) inlet velocity triangle. (b) Exit velocity triangle. (c) Combined velocity diagram (symmetrical blading, R = 0.5, V1 = V3, a1 = a3, Vf = constant)

Vw1 = V1 cos a1 = Vf1 cot a1

(5.4)

x1 = Vr1 cos b1 = Vf1 cot b1

(5.5)

u = x1 + Vw1

(5.6a)

u = V1 cos a1 + Vr1 cos b1

(5.6b)

= Vf1 cot a1 + Vf1 cot b1 = Vf1 (cot a1 + cot b1)

(5.6c)

From the velocity triangle at the exit, Vf2 = V2 sin a2 = Vr2 sin b2

(5.7)

Vw2 = V2 cos a2 = Vf2 cot a2

(5.8)

x2 = Vr2 cos b2 = Vf2 cot b2

(5.9)

u = X2 + Vw2

(5.10a)

= V2 cos a2 + Vr2 cos b2

(5.10b)

= Vf2 cot a2 + Vf2 cot b2

(5.10c)

= Vf2 (cot a2 + cos b2)

(5.10d)

We have that, Vf1 = Vf2 = Vf3 = Vf = constant V1 sin a1 = Vr1 sin b1 = V2 sin a2 = Vr2 sin b2

(5.11) (5.12)

352

Fundamentals of Turbomachinery

From Eqs. (5.6c) and (5.10d), we have u = flow coefficient I Vf = cot a1 + cot b1 = cot a2 + cot b2

(5.13)

From Eqs. (5.6a) and (5.10a), we have (x1 – x2) = (Vw2 – Vw1) or

(5.14)

Vf1(cot b1 – cot b2) = Vf2(cot a2 – cot a1)

(5.15)

From Eqs. (5.14) and (5.15), we have (Vw2 – Vw1) = Vf1(cot b1 – cot b2) = Vf2(cot a2 – cot a1)

(5.16)

5.4 WORK DONE Refer to Figure 5.3. From Euler’s energy equation, we have

(u2Vw 2  u1Vw1 )

W.D. =



gc u (Vw 2  Vw1 ) gc

J/kg

J/kg

(5.17)

Substituting Eq. (5.16) in (5.17), W.D. = Dh0 =

uV f (cot D 2  cot D1 ) gc



uV f (cot E1  cot E2 ) gc

J/kg

(5.18)

5.5 TEMPERATURE AND ENTROPY DIAGRAM FOR A STAGE OF AN AXIAL FLOW COMPRESSOR The flow through a stage is shown thermodynamically on the T–s diagram in Figure 5.4. Assuming adiabatic flow through the stages, h03 = h02 The W.D. equation can be written as W.D. = Dh0 = (h02 – h01) J/kg ©

(h02  h01 )  ªh2 «ª

\

(5.19)

V22 ¸

© V12 ¸ h  ¹ ª 1 ¹ gc ¹º ª« 2 gc º¹

W.D. = (h02  h01 )  (h2  h1 )

V22  V12 2 gc

(5.19a)

Axial Flow Compressors T(h)

p03

p02 02 03≤

03

03¢

p01 rel

3¢

p02 rel

01 rel

2¢

2

V3 2

3 3≤

p3

2

V2 2

02 rel

2

p01

p1

01 2

V1 2

1 s

Figure 5.4 Temperature–entropy diagram for a stage of an axial flow compressor.

From Eqs. (5.17) and (5.19a), we have u (Vw 2  Vw1 ) gc

Ë V2 V2 Û ( h2  h1 )  Ì 2  1 Ü ÍÌ 2 gc 2 gc ÝÜ

u (Vw 2  Vw1 ) Ë V22 V2 Û Ì  1 Ü gc ÍÌ 2 gc 2 gc ÝÜ

or

(h2  h1 ) 

or

(h2  h1 ) 

or

(h2  h1 ) 

2u(Vw 2  Vw1 )  (Vw22  Vw21 ) 2 gc

or

(h2  h1 ) 

2u(Vw 2  Vw1 )  (Vw 2  Vw1 ) (Vw 2  Vw1 ) 2 gc

or

(h2  h1 ) 

(Vw 2  Vw1 )[2u  (Vw 2  Vw1 )] 2 gc

or

(h2  h1 ) 

(Vw2  Vw1 )[(u  Vw 2 )  (u  Vw1 )] 2 gc

0

2u(Vw 2  Vw1 )  [(V f22  Vw22 )  (V f21  Vw21 )] 2 gc

From inlet and exit velocity triangles, (u – Vw2) = x2,

(u – Vw1) = x1,

0

0 0

0

0

353

354

Fundamentals of Turbomachinery

or

Vw2 = u – x2,

Vw1 = u – x1

\

(h2  h1 ) 

[(u  x2 )  (u  x1 )] ( x2  x1 ) 2 gc

or

(h2  h1 ) 

( x2  x1 ) ( x2  x1 ) 2 gc

or

(h2  h1 ) 

x22  x12 2 gc

Now,

x22

or

x22  x12

Vr22  V f2  Vr21  V f2

or

x22  x12

Vr22  Vr21

Vr22  V f2 ;

0

0

0

x12

(5.20)

Vr21  V f2

Substituting the above equation in (5.20), we have (h2  h1 ) 

or \

Vr22  Vr21 2 gc

0

Ë Vr22 Û Ë Vr21 Û Ì h2  Ü  Ì h1  Ü 2 gc ÜÝ ÌÍ 2 gc ÜÝ ÌÍ h01 rel = h02 rel

0

(5.21) (5.22)

Stagnation enthalpy remains the same in the relative system.

5.6 OVERALL PRESSURE RATIO PER STAGE (pR0) The energy input to the axial compressor will be absorbed in raising the pressure and velocity of the air or gas and same will be used in overcoming the various friction losses. However, the complete work input will appear as a stagnation temperature rise of the air irrespective of the isentropic efficiency. Hence, a relation can be developed between overall pressure ratio per stage, stagnation temperature rise and overall isentropic efficiency. hc t–t = Overall isentropic efficiency Ideal isentropic work input Actual work input (h03  h01 ) (h03  h01 ) „

c p (T03  T01 ) „

c p (T03  T01 )

ËT Û T01 Ì 03  1Ü Í T01 Ý (T03  T01 ) „

Axial Flow Compressors

355

We have J

p03 p01

Ë T03„ Û J  1 Ì Ü Í T01 Ý

J

Ë T03„  T01  T01 Û J  1 Ì Ü T01 Í Ý J

Ë T03„  T01 T01 Û J  1  Ì Ü T01 Ý Í T01 J

Ë Kc t to (T03  T01 ) Û J  1 pR 0 (5.23) Ì1  Ü T01 Í Ý However, the whole of the work input will appear as a stagnation temperature rise of the air (fluid) regardless of the isentropic efficiency. \ Equation (5.18) can be written as p03 p01

W.D. = Dh0 = (h02 – h01) = (h03 – h01) = cp(T03 – T01) = cpDT0

uV f (cot E1  cot E2 ) gc \

(T02  T01 )

(T03  T01 )

uV f (cot E1  cot E 2 ) gc c p

(5.24)

5.7 WORK DONE FACTOR (y) So far we have assumed Vf as constant. Actually, it is not so. Due to the tip clearance and boundary layer, there is a decrease in the energy transfer to the air (fluid). To account for this decrease in the energy transfer, a correction (empirical) factor ‘work done factor’ y (work done coefficient) is introduced. \

y = Work done factor =

Actual work absorbing capacity Ideal work absorbing capacity

Equations (5.17), (5.18) and (5.24) reduce to (5.25a), (5.25b) and (5.25c) respectively. W.D. = W.D. (T02  T01 )

\ u (Vw 2  Vw1 ) gc

(5.25a)

\ uV f (cot D 2  cot D1 )

\ uV f (cot E1  cot E2 )

gc

gc

(T03  T01 )

\ uV f (cot E1  cot E 2 ) gc c p

(5.25b) (5.25c)

356

Fundamentals of Turbomachinery

5.8 FLOW COEFFICIENT (f) It is also called compressor velocity ratio. It helps in obtaining large turning angles. Vf

I

u

5.9 PRESSURE COEFFICIENT (fp) It is defined as the ratio of the actual stagnation in enthalpy rise in the stage to the dynamic pressure equivalent of the blade velocity (i.e. kinetic energy of the fluid which has the same speed as the blades). Therefore,

'h0

Ip or

'h0 – 2 – gc

2

u2

u /2 gc

u – 'Vw – 2 – gc

Ip

2

u gc

'Vw – 2 u

If the work done factor is included, then

Ip

\ – 'Vw – 2 u

5.10 DEGREE OF REACTION (R) R= =

Static enthalpy rise in rotor Static enthalpy rise in stage (a set of rotor + stator) 'hR 'hR  'hS

'TR 'TR  'TS

(h2  h1 ) ( h3  h1 )

(h2  h1 ) Û (h2  h1 )  ( h3  h2 ) Ü Ü Ü Ü ÜÝ

DTR = temperature rise in rotor DTS = temperature rise in stator From Eq. (5.21), (h2  h1 )

(Vr21  Vr22 ) 2 gc

We have V1 = V3 and W.D. equation as (h3  h1 )

(h03  h01 )

( h02  h01 )

Substituting for (h2 – h1) and (h3 – h1) in (5.26),

u(Vw 2  Vw1 ) gc

(5.26)

Axial Flow Compressors

(Vr21  Vr22 ) u (Vw 2  Vw1 ) 2 gc gc

R

But,

Vw2

\

357

(V f2  x12 )  (V f2  x22 ) 2u (Vw 2  Vw1 )

( x12  x22 ) ( x1  x2 )( x1  x2 ) 2u (Vw 2  Vw1 ) 2u (Vw 2  Vw1 ) = (u – x2); Vw1 = (u – x1)

R

( x1  x2 )( x1  x2 ) 2u(u  x2  u  x1 )

( x1  x2 ) ( x1  x2 ) 2u ( x1  x2 )

Substituting Eqs. (5.5) and (5.9) in the above equation, we get V f (cot E1  cot E 2 )

R

Alternatively: By energy balance in rotor, h1 

V12 W 2 gc

h2 

V22 2 gc

(Q W = W.D.)

'hR

( h2  h1 ) W 

V22 2 gc

h3 

'hS

h3  h2

By energy balance in stator, h2 

(5.27)

2u

V32 2 gc

h3 

(V22  V12 ) 2 gc V12 2 gc

(Q V1 = V3)

V22  V12 2 gc

Substituting for DhR and DhS in (5.26), W R

Ë (V 2  V12 ) Û Ë (V22  V12 ) Û W Ì 2 Ü Ü  Ì Ý Í 2 gc Ý Í 2 gc

R 1

or

(V22  V12 ) 2 gc

W

(V22  V12 ) 2 gc W

(V22  V12 ) 2 gcW

Substituting for W.D. from Eq. (5.17) and

V22

(V f2  Vw22 ) and V12

R 1

(V f2  Vw21 ) in the above equation, we get

(V f2  Vw22 )  (V f2  Vw21 ) (V  Vw1 ) 2 gc u w 2 gc

358 or

Fundamentals of Turbomachinery

R 1

[(Vw 2  Vw1 ) (Vw 2  Vw1 )] (V  Vw1 ) 1  w2 2u (Vw 2  Vw1 ) 2u

Substituting Eqs. (5.4) and (5.8) in the above equation, R 1

V f (cot D 2  cot D1 ) 2u

(5.28)

Substituting R = 50% in Eqs. (5.27) and (5.28),

u Vf

cot E1  cot E 2

(5.29)

u Vf

cot D 2  cot D1

(5.30)

Comparing Eqs. (5.29) and (5.30), we can say that for R = 50%, a1 = b2,

a2 = b1, V1 = Vr2,

V2 = Vr1

5.11 COMBINED VELOCITY TRIANGLES FOR DIFFERENT VALUES OF R Case 1: If R = 50%, a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1, Vf = constant (Figure 5.5). Substituting R = 50% in Eq. (5.26), we get h2 – h1 = h3 – h2, that is, the static enthalpy and temperature increase in the rotor and stator are equal. The inlet and exit velocity triangles are equal.

Figure 5.5

R = 50%, symmetrical triangles.

Case 2: If R > 0.5 (Figure 5.6), substituting R > 0.5 in Eq. (5.26) it can be concluded that static enthalpy rise in the rotor (h2 – h1) is greater than that in the stator (h3 – h2), also the static pressure rise is greater in the rotor than that in the stator.

Axial Flow Compressors

Figure 5.6

359

R > 0.5, unsymmetrical triangles.

Case 3: If R < 0.5 (Figure 5.7), substituting R < 0.5 in Eq. (5.26) it can be concluded that static enthalpy and pressure rises are greater in the stator than those in the rotor. u

Vw 2

Vw 1

A b1 b2 Vr 2

Vf 2

V1

Vf1 V2

Vr 1

D

C a2 a1

B

DVw = (Vw 2 – Vw 1)

Figure 5.7

R < 0.5, unsymmetrical triangles.

Case 4: If V1 is axial, a1 = 90°, V1 = Vf1, Vw1 = 0 (Figure 5.8), we would have: • • • •

No guide blades. Large static pressure rise in rotor compared to that in stator. R lying between 60% and 90%. Large relative velocity, hence the stage efficiency would be less than that of the 50% reaction machine. u Vw 2 A

C

b1

a1 a2

b2 Vr 2

V1 = Vf 1 \VW1 = 0

Vf 2 V2 D

Vr1 B

DVw = (Vw 2 – Vw 1) = Vw 2

Figure 5.8 V1 axial, unsymmetrical triangles.

360

Fundamentals of Turbomachinery

Case 5: If V2 is axial, a2 = 90°, Vw2 = 0, Vf 2 = V2 (Figure 5.9), then: • • • • • • •

The absolute velocity would be minimum. Static pressure rise would occur entirely in the rotor. A small pressure drop would take place in stator. R > 100% Energy transfer per stage would be very low, therefore a bulky machine would be required for a given pressure rise. The stage efficiency would be low. Due to minimum V2, overall efficiency would be greater compared with that in other types of compressors. u Vw 1

A b1

Vf 1

b2

C a2 a1

V1

V2 = Vf 2 \VW2 = 0

Vr 1 Vr 2 B

D

DVw = (Vw 2 – Vw 1) = –Vw 1

Figure 5.9 V2 axial, unsymmetrical triangles.

5.12 Radial Equilibrium Conditions (Refer to Figure 5.10) Euler’s momentum equation in the radial direction for axisymmetrical, three-dimensional flow in cylindrical coordinates is given by VR

V2 dVr dV  Va R  w dr dx r



1 dp U dr

(5.31)

We know that absolute velocity V is resolved into three mutually perpendicular components. Therefore, VR = radial component Vw = tangential component Va = axial component r = radius dp radial pressure gradient dr Equation (5.31) is the complete radial equilibrium equation.

Axial Flow Compressors

361

Figure 5.10 Radial equilibrium of fluid element.

For steady flow conditions along concentric stream lines, the velocity, static pressure and density are constant with time and dVR = 0, i.e. radial component of the velocity is negligible, no flow occuring in the radial direction. Such a condition of the flow is known as radial equilibrium. Substituting dVR = 0 in Eq. (5.31), Vw2 r

1 dp U dr

(5.32)

Equation (5.32) will be referred to as the radial equilibrium equation. Really in the spaces between the blade rows, Vr is very much smaller than either Va or Vw and can be assumed to be negligible. Equation (5.32) gives the pressure gradient in the flow for radial equilibrium conditions. Let us derive an energy equation. The variation of enthalpy with radius is considered. We have for stagnation enthalpy, h0

h

V2 2 gc

h

1 – (VR2  Va2  Vw2 ) 2 gc

For radial equilibrium VR = 0, then h0

h

(Va2  Vw2 ) 2 gc

(Va = Vf in axial flow machines)

dV f dh0 dh dV  Vf  Vw w dr dr dr dr From the second law of thermodynamics, Tds

dh 

dp U

(5.33)

362

Fundamentals of Turbomachinery

or

dh dh dr

or

T

Tds 

dp U

ds dT 1 dp 1 dU  ds   dp dr dr U dr U 2 dr

Dropping the higher order terms, dh dr

T

ds 1 dp  dr U dr

Substituting the above equation in (5.33),

dh0 dr

T

dV f dV ds 1 dp   Vf  Vw w dr U dr dr dr

Substituting Eq. (5.32) in the above equation, dh0 dr

Assume T

ds dr

0

T

dV f dVw ds Vw2   Vf  Vw dr r dr dr

(5.34)

(constant values of entropy in radial direction)

dV f dh0 dVw Vw2 (5.34) Vf  Vw  dr dr dr r Stagnation enthalpy will increase in the axial direction whereas in the radial direction it will remain uniform. Then,

\

dh0 0 in any plane between a pair of blade rows. dr Equation (5.34) reduces to Vf

dV f dr

 Vw

dVw Vw2  dr r

(5.35)

0

or

dV f dV 1 È 2 dr Ø r 2Vw w  Vw2 2r Ù  2V f 2 É dr dr Ú dr r Ê

\

2 1 d (rVw )2 d (V f )  dr dr r2

0

0 (5.36)

5.13 AIR ANGLE DISTRIBUTION 5.13.1 Free Vertex Flow Let us determine air angle distribution with respect to blade height using the free vertex flow theory. The blade height of axial compressor varies from hub to tip.

Axial Flow Compressors

363

If Vf is maintained constant across the annulus (hub to tip), then dV f dr

0, i.e. V f

constant,

V f 1h

V f 1m

V f 1t

V f 2h

V f 2m

V f 2t

where the subscripts h = hub, m = mean, t = tip. Substituting this condition in Eq. (5.35), dVw Vw2  dr r

0

or

dVw Vw  dr r

0

or

dVw dr  Vw r

0

Vw

On integration, we get

rVw = constant

(5.37)

Thus, the whirl velocity varies inversely with radius. This is known as free vertex condition. Equation (5.37) can be written for h, m, t, as follows: rhVw1h = rmVw1m = rtVw1t = C1 = constant 1 rhVw2h = rmVw2m = rtVw2t = C2 = constant 2

Air angles with respect to blade height 1. Rotor entry Hub: From Figure 5.11(c), inlet velocity triangle, cot D1h

Vw1h Vf

C1 rhV f

cot E1h

uh  Vw1h Vf

(5.38) uh Vw1h  Vf Vf

(5.39)

uh C  1 V f rhV f

Tip: From Figure 5.11(a), inlet velocity triangle, cot D1t

Vw1t Vf

C1 rt V f

cot E1t

ut C  1 V f rt V f

(5.39a) (5.39b)

364

Fundamentals of Turbomachinery

B

B

Vr 1t

A

Vr 2t

V1t

Vf

b1t

a1t

a 2t

b2t

A

C

Vw 1t

ut

V2t

Vf

C

Vw 2t

ut (a) R > 0.5

B Vr 1m

B

a1m

b1m

A

Vr 2m

V1m

Vf

C

Vw 1h

um

V2m

Vf

b2m

A

a2m Vw 2m

um

C

(b) R = 0.5 B

B Vr 1h

A

V1h

Vf

Vr 2h

a1h

b1h uh

Vr 2h

Vr 2m

(f)

Figure 5.11

D

V1t V2t

Vr 1t C A

um (e)

B

Vr 2t

V2m

A

C

Vw 2h

B V1m

Vr 1m C

uh

a 2h

Exit velocity triangle

(c) R < 0.5

D

V2h

A

b2h uh

B V1h

Vr 1h

A

Vw 1h

Inlet velocity triangle

D

C

V2h

Vf

C ut (d)

Air angle variations with respect to blade height (radial equilibrium theory): (a) Tip. (b) Mean. (c) Hub. (d) Combined velocity triangle at tip. (e) Combined velocity triangle at mean. (f) Combined velocity triangle at hub.

Axial Flow Compressors

365

2. Rotor exit Hub: From Figure 5.11(c), exit velocity triangle, cot D 2 h

Vw 2 h Vf

C2 rh V f

cot E2 h

uh  Vw1h Vf

(5.40a) uh Vw 2 h  Vf Vf

uh C  2 V f rh V f

(5.40b)

Tip: From Figure 5.11(a), exit velocity triangle cot D 2t

Vw 2t Vf

C2 rt V f

cot E2t

ut C  2 V f rt V f

(5.41a)

(5.41b)

Specific work W.D. = Dh0 = h02 – h01 = u(Vw2 – Vw1) Applying Eq. (5.37),

Vw 2

C2 ; r

Vw1

C1 r

C Ø ÈC W.D. = Z r É 2  1 Ù Ê r r Ú

Z (C2  C1 )

C

The following points will be observed, if free vertex theory is adopted (Figure 5.11) (refer to Example 5.13). • • •

The degree of reaction increases from root to tip, Rt > 0.5, Rm = 0.5, Rr < 0.5. The relative velocities in the rotor are highest at the tip, while the velocities in the stator are highest at the root. The fluid outlet angle increases from root to tip in the rotor and decreases in the stator, giving a considerable twist to the blades.

Degree of reaction We know from Eqs. (5.38) to (5.41) that the air angles in the stage are varying along the blade height, therefore the degree of reaction must also vary.

366

Fundamentals of Turbomachinery

From Eq. (5.27), we have R

Vf 2u

(cot E1  cot E2 )

V f Ë x1 x2 Û  Ì Ü 2u ÌÍ V f V f ÜÝ

x1  x2 2u

(u  Vw1 )  (u  Vw 2 ) 2u

2u  (Vw1  Vw 2 ) 2u

We have for free vertex, r × Vw = C, then the above equation reduces to R 1

or

R 1

where

K

(C1  C2 ) 2–u–r

(C1  C2 ) 2Z r

2

1

(C1  C2 ) 2–Z

K r2

(5.42) (5.43)

Equation (5.42) shows that the stage reaction in a free vertex design increases along the blade height. Therefore, at hub,

R < 0.5

at tip,

R > 0.5

at mean,

0.5 < R < 0.5

5.13.2 Constant Reaction Design In the case of free vertex design, the tangential velocity distribution should be such that r × Vw = constant For true radial equilibrium, Vf should vary with radius but usually, Vf is kept constant. But in case of constant reaction design, the degree of reaction is considered constant from root to tip (Figure 5.12). Almost constant Mach number is obtained at different radii, as Vr1 and V2 decrease slightly from tip to root. In constant reaction design: • •

The degree of reaction remains the same from root to tip. The fluid outlet angle decreases from root to tip in the rotor and decreases in the stator giving a considerable twist to the blades (refer to Example 5.14).

367

Axial Flow Compressors

B B V1t

Vr 1t

A

a1t

b1t

V2t

Vr 2t

Vf

C

Vf a2t

b2t

A

ut

C

ut (a) B B V1m

Vr 1m

A

V2m

Vr 2m

Vf

Vf a 1m

b1m um

a 2m

b2m

A

C

um

C

(b) B Vr 1r

Vf

Vr 2r

a 1r

b1r

A

B

V1r

ur

C

A

Vf

V2r

a 2r

b 2r ur

C

(c)

Figure 5.12

Air angle variations with respect to blades height (constant reaction theory, R = C) [example R = 50%]: (a) Tip. (b) Mean. (c) Root.

5.14 EXAMPLES EXAMPLE 5.1 Air enters a three stage axial flow compressor at 1 bar and 300 K. The energy input is 25 kJ/kg per each stage. The stage efficiency is 0.86. Calculate (a) the exit static temperature, (b) the compressor efficiency and (c) the static pressure ratio. Solution: Number of stages: Inlet temperature: Inlet pressure: Energy input per stage: Stage efficiency: To determine: T4, hC, pR

=3 T1 = 300 K p1 = 1 bar W.D. = 25 kJ/kg hS = 0.86

368

Fundamentals of Turbomachinery T or h p4

4 4² 4¢

p3 3²

3¢

3 p2

2¢

p1

2

1

s

W.D. = work done in first stage = Dh = (h2 – h1) = cp(T2 – T1) or

25

\

kJ kg

1.005

kJ (T2  300) kg-K

T2 = 324.88 K hS = stage efficiency (first)

\

0.86

or

T2

„

(T2  T1 ) (T2  T1 ) „

(T2  300) (324.88  300) „

321.4 K

We have for isentropic process, J

1.4

Ë T2 Û J  1 Ë 321.4 Û 0.4 1.273 Ì Ü Ì 300 Ü Í Ý Í T1 Ý p2 = p1 × 1.273 = 1 × 1.273 = 1.273 bar

p2 p1 \

„

Similarly for the 2nd stage, 25

\

kJ kg

kJ (T3  T2 ) 1.005 – (T3  324.88) kg K

T3 = 349.76 K KS 2

\

1.005

(T3”  T2 ) (T3  T2 )

\

(T3”  324.88) (349.76  324.88)

T3² = 346.27 K J

Now,

0.86

Ë T3” Û J  1 Ì Ü Í T2 Ý p3 = p2 × 1.25 p3 p2

1.4

Ë 346.27 Û 0.4 Ì 324.88 Ü Í Ý

1.25

= 1.273 × 1.25 = 1.591 bar

Axial Flow Compressors

369

(a) Exit static temperature (T4, T4²): Similarly for the 3rd stage, 25

\ Now, \

kJ kg

1.005

kJ – (T4  T3 ) 1.005 – (T4  349.76) kg

T4 = 374.64 K KS 3

0.86

Ans.

(T4”  T3 ) (T4  T3 )

(T4”  349.96) (374.64  349.76)

T4² = 371.356 K J

Ë T4” Û J  1 Ì Ü Í T3 Ý

Ë 371.35 Û Ì 349.76 Ü Í Ý

3.5

Now,

p4 p3

\

p4 = p3 × 1.233 = 1.591 × 1.233 = 1.9617 bar

1.233

We have isentropic process 1–4¢ J

0.4

Ë p4 Û J  1 Ë 1.9584 Û 1.4 1.212 Ì Ü Ì 1 Ü Í Ý Í p1 Ý T4¢ = 300 × 1.212 = 363.5 K

T4 T1

„

\

Ans.

(b) Static pressure ratio (pR): \

pR

p4 p1

1.9617 1

Ans.

1.9617

(c) Compressor efficiency (hC) KC

(T4  T1 ) (T4  T1 ) „

(363.5  300) – 100 (374.64  300)

85.1%

Ans.

EXAMPLE 5.2 The ambient conditions at inlet are 20°C and 1 bar. At exit the total head temperature and pressure are 150°C and 3.5 bar, and static pressure at exit is 3 bar. Calculate (a) the isentropic efficiency, (b) the polytropic efficiency, and (c) the air velocity at exit. Solution: Ambient temperature: Ambient pressure: Stagnation temperature at exit: Stagnation pressure at exit: Static pressure at exit: To determine: hc t–t, hp, V2

T01 p01 T02 p02 p2

= = = = =

20°C = 293 K 1 bar 150°C + 273 = 423 K 3.5 bar 3 bar

370

Fundamentals of Turbomachinery T or h p02

02 02¢

p2 2

2¢

p01 p1

01 1 J

1

T02 T01

Ë p02 Û Ì Ü Í p01 Ý

T02

T01 – 1.431

„

\

s

„

J

Ë 3.5 Û Ì 1 Ü Í Ý

0.286

1.431

293 – 1.431

419.25 K

(a) Isentropic efficiency (hc t–t): (T02  T01 ) (T02  T01 )

Kc t  t

„

(419.25  293) – 100 (423  293)

97.11%

Ans.

(b) Polytropic efficiency (hp):

Kp

p J ln 02 J  1 p01 T ln 02 T01

0.4 ln 3.5 1.4 – 100 423 ln 293

97.48%

Ans.

(c) Air velocity at exit (V2): Applying isentropic process between 2–02,

or Now, \

J

1

T02 T01

Ë p02 Û Ì Ü Í p2 Ý

T2

T02 1.04507

T02

V2

T2 

J

Ë 3.5 Û Ì 3 Ü Í Ý

0.286

423 1.04507

1.04507

404.76 K

V22 2 gc – c p

(T02  T2 )2 gc c p (423  404.76) – 2 – 1000 – 1.005

191.5 m/s

Ans.

371

Axial Flow Compressors

EXAMPLE 5.3 An axial compressor stage has the following data. Stagnation temperature and pressure at entry are 20°C and 1 bar, and the degree of reaction is 50%. f = 0.5 dm = 35 cm N = 18,000 rpm a1 = b2 = 60° h = 5 cm j = 0.88 h c t–t = 0.85 hm = 0.96

Flow coefficient: Mean blade ring diameter: Speed: Air angles at rotor and stator exit: Blade height at entry: Work done factor: Isentropic efficiency: Mechanical efficiency:

Calculate (a) the air angles at the rotor and stator entry, (b) the mass flow rate of air, (c) the power required to drive the compressor, (d) the loading coefficient, (e) the pressure ratio developed by the stage, and (f) Mach number at the rotor entry. u

T or h

Vw 2 B

A

02

Vw1

b1 b2

02¢

C

D p2

a2

V1

Vr 2 Vf

a1

2¢

Vf V2

Vr1

01

E

2 p01 p1

F 1

DVw = (Vw 2 – Vw 1)

Solution:

um I \

p02

Vf

S dm N 60 0.5

S – 0.35 – 18000 60

Vf

Vf

um

329.87

329.87 m/s

164.93 m/s

(a) Air angles at the rotor and stator entry (b1, a2): From the inlet velocity triangle (triangle CFD) cot D1

Vw1 Vf

Vw1 164.93

cot 60’

s

372

Fundamentals of Turbomachinery

\

Vw1

95.22 m/s

Vf

tan E1

164.93 329.87  95.22

um  Vw1

0.7028

b1 = 35.1° = a2

Ans.

(b) The mass flow rate (m): V1

T1

V f2  Vw21

T01 

164.932  95.222

V12 2 gc c p – 1000 J

p01 p1

p1 U1

\

293 

190.44 m/s

190.442 2 – 1 – 1.005 – 1000

274.96 K

3.5

Ë T01 Û J  1 Ë 293 Û 1.2491 Ì Ü Ì 274.96 Ü Í Ý Í T1 Ý p01 1 bar 0.801 bar 1.2491 1.2491 p1 RT1

0.801 – 100 0.287 – 274.96

1.015 kg/m 3

m = r1 Vf (p dmh1) = 1.015 × 166.93 × p × 0.35 × 0.05 = 9.199 kg/s

(c) Power required to drive the compressor (P): From Eq. (5.18), W.D. =

M um V f (cot E1  cot E2 ) gc 0.88 – 329.87 – 164.93 – (cot 35.1’  cot 60’) 1 – 1000

\

P = W.D. × m = 40.48 × 9.199 = 372.376 kW

40.48 kJ/kg

Ans.

(d) Loading factor (s) or loading coefficient: V

W.D.

40480 J/kg

2

(329.87)2 J/kg 1

u gc

(e) The pressure ratio (pR0): W.D. = cpDT0 = 1.005(T02 – T01) or

'T0

W.D. cp

40.48 1.005

40.28 K

0.372

Ans.

Axial Flow Compressors

373

ËT Û T01 Ì 02  1Ü Í T01 Ý 'T0 „

Kc t t

Now,

or

Kc t t

or

0.85

(T02  T01 ) (T02  T01 ) „

J 1 Ë Û Ì È p02 Ø J Ü  1Ü T01 Ì É Ê p01 ÙÚ ÌÍ ÜÝ 'T0







'T0

29.3 pR0.286 1 0

40.28 pR0 = 1.472

\



1 T01 pR0.286 0

Ans.

(f) Mach number at the rotor entry (M1): Vr1

V f2  (um  Vw1 )2

164.932  (329.87  95.22)2 M1

Vr1

286.81 m/s

286.81

J gc T1

„

1.4 – 1 – 1000 – 274.96

0.4622

Ans.

EXAMPLE 5.4 The second stage of an axial compressor has a mean blade speed of 250 m/s. The axial velocity is 150 m/s. Assuming symmetrical blading, the difference between the tangents of the angles at outlet and inlet is 0.6. Adiabatic efficiency of the stage is 0.85. Air enters the stage with a total temperature of 300 K. Total stagnation pressure ratio is 4.0. The mass flow rate is 1000 kg/min. Mechanical efficiency is 99%. Calculate (a) the number of stages and (b) the power required to compress. Assume that the kinetic energy of the air entering the successive stage is negligible. Also assume that angles a and b are with respect to the axial direction. Solution:

Data: Machine: Axial compressor 2nd stage Mean blade speed: um = 25 m/s Axial velocity: Vf = 150 m/s Symmetrical blading: Difference between the tangents of the angles at outlet and inlet: = 0.6 Adiabatic efficiency: hc t–t = 0.85 Air enters the stage with total temperature: T01 = 300 K Total stagnation pressure ratio: pR0 = 4 The mass flow rate: m = 1000 kg/min Mechanical efficiency: hm = 99%

To determine: K, P

374

Fundamentals of Turbomachinery

From Eq. (5.18),

W.D. = or \

um V f (cot E1  cot E2 ) gc

kJ kg

c p 'T0

'T0

22.5 1.005

22.5

250 – 150 – (0.6) 1000

22.5 kJ/kg

22.388 K

(a) Number of stages (K):

or

p02 p01

Ë Kc t t 'T0 Û Ì1  Ü T01 Ý Í

p02 p01

1.24

J

1 J

Ë 0.85 – 22.388 Û Ì1  Ü 300 Í Ý

3.5

(Eq. (5.23))

K

\ or \

Ë p02 Û 4 Ì Ü Í p01 Ý (1.24)K = 4

(K = No. of stages)

K = 6.44 » 7 stages

Ans.

(b) Power required (P):

P

 p 'T0 K mc Km 1000 kg kJ 22.388 K – 7 – 1.005 – 60 s kg-K 0.99

\

P = 2651.51 kW

Ans.

Axial Flow Compressors

375

EXAMPLE 5.5 The mass flow rate of a multistage axial compressor is 20 kg/s of air. The stage efficiency is 0.9. The inlet conditions are 1 bar and 300 K. The stage pressure ratio is constant and the temperature rise in the first stage is 20°C. The temperature at the end of isentropic compression is 500 K. Calculate (a) the delivery pressure at the end of last stage, (b) the total pressure ratio, (c) the number of stages, and (d) the power input. Solution:

Data: m 20 kg/s hS = 0.9 T1 = 300 K p1 = 1 bar pR = constant (T2 – T1) = 20°

Mass flow rate of air: Stage efficiency: Inlet temperature: Inlet pressure: Stage pressure ratio: Temperature rise in the first stage: Temperature at the end of isentropic compression (last stage)

TK = 500 K

To determine: K (number of stages), pK, pR0, power input (a) Delivery pressure at the end of last stage (pK): KP

0.9

n 0.4 – (n  1) 1.4

n n 1

\

n (J  1) (n  1) J

KS

(Eq. (3.39b))

3.15

For isentropic relation between the inlet and the end of last stage, J

1.4

Ë pK Û Ë TK Û J  1 Ë 500 Û 0.4 Ì Ü Ì Ü Ì 300 Ü Í Ý Í p1 Ý Í T1 Ý pK = p1 × 5.97

\

5.97

= 1 × 5.97 = 5.97 bar

Ans.

(b) Total pressure ratio (pR0): pR 0

pK p1

5.97 1

5.97

(c) Number of stages (K): We have the relation for polytropic process between the inlet and the exit.

TK T1

Ë pK Û Ì Ü Í p1 Ý

n 1 n

1

5.97 3.15

1.7634

Ans.

376

Fundamentals of Turbomachinery

\

TK = T1 × 1.7634 = 300 × 1.7634 = 529 K TK = actual temperature at the end (last stage) hS1 = first stage efficiency 0.9

or

(T2  T1 ) „

\

(T2  T1 ) (T2  T1 ) „

(T2  T1 ) – 0.9

20’ – 0.9 18 K

T2 = T1 + 18 = 300 + 18 = 318 K „

pR = pressure ratio of each stage = J

p2 p1

Ë T2 Û J  1 Ì Ü Í T1 Ý „

1.4

Ë 318 Û 0.4 Ì 300 Ü Í Ý

1.23

We have the relation, Ë p2 Û Ì Ü Í p1 Ý

T or h

K

5.97

pR 0

or

(1.23)K = 5.97

or

K ln 1.23 = ln 5.97

\

p2 p1

tage

9th s

TK ¢

P

2¢ 2

m c p (TK  T1 ) 20

kg kJ – 1.005 – (529  300) K s kg-K

pK

p2

K = 8.63 » 9 stages Ans.

(d) Power input (P):

Tk

age p1 Ist st

1

s

4602.9 kW

Ans.

EXAMPLE 5.6 Air at the rate of 3 kg/s and at a temperature of 20°C enters an eight-stage axial flow compressor. The pressure ratio is 6 and the isentropic efficiency is 0.9. Assume the compressor process as adiabatic. The degree of reaction is 50% and all the stages are similar. The mean blade speed is 180 m/s and the axial velocity is 100 m/s and is constant. Calculate: (a) The polytropic efficiency (b) The power to the air compressor and change in temperature/stage (c) The air angles at entry to and exit from the rotor and the stator blades. Solution:

Data: Mass flow rate:

m

Inlet temperature: Number of stages

T1 = 20°C K=8

3 kg/s

Axial Flow Compressors

Isentropic efficiency: Degree of reaction: All the stages are similar: Mean blade speed: Axial velocity:

377

hc s–s = 0.9 R = 0.5 um = 150 m/s Vf = constant = 100 m/s

Pressure ratio:

pR

6

pK p1

To determine: hP, p, b1, b2, a1 and a2

(a) Polytropic efficiency (hP): J

pR

Kc s  s

J

1 J

1

JK p

pR \

1

60.286  1 0.286

1

(Eq. (3.42b))

0.9

6 KP  1

hP = 0.922

Ans.

Alternatively:

TK Kc s  s

\ \

„

Ëp Û T1 – Ì K Ü Í p1 Ý

J

1 J

(TK  T1 ) (TK  T1 ) „

293 – (6)0.286 (489  293) (TK  293)

489 K

0.9

TK = 510.78 K

KP

(J  1) p ln K J p1 T ln K T1

0.286 ln 6 150.78 ln 293

0.9211

(Eq. (3.39))

Ans.

378

Fundamentals of Turbomachinery

(b) Power to the air compressor (P): P

m c p 'TK 3

m c p (TK  T1 )

kg kJ – 1.005 – (510.78  293) K s kg-K

656.61 kW

Ans.

Temperature change per stage (DT):

'T

'TK K

(510.78  293) 8

27.22 K

Ans.

(c) Air angles at entry and exit of rotor and stator (b1, b2, a1, a2): From Eq. (5.18),

P 8 or \

656.61 kW 8

m – umV f (cot E1  cot E 2 )

m – W.D. 8 3

gc

kg m m (cot E1  cot E 2 ) – 180 – 100 s s s 1 – 1000

(cot b1 – cot b2) = 1.5199

(i)

From Eq. (5.27),

R or \

0.5

Vf 2um

(cot E1  cot E2 )

100 (cot E1  cot E2 ) 2 – 180

cot b1 + cot b2 = 1.8

(2)

From Eqs. (i) and (ii), b1 = 31.1°C cot b2 = 1.8 – cot b1 = 1.8 – cot 31.1° b2 = 82.03° There is 50% reaction compression, therefore, a1 = b2 = 82.03° Þ ß a2 = b1 = 31.1° à

Ans.

EXAMPLE 5.7 The speed of an axial flow compressor is 15,000 rpm. The mean diameter is 0.6 m. The axial velocity is constant and is 225 m/s. The velocity of whirl at inlet is 85 m/s. The work done is 45 kJ/kg of air. The inlet conditions are 1 bar and 300 K. Assume a stage efficiency of 0.89. Calculate (a) the fluid deflection angle, (b) the pressure ratio, (c) the degree of reaction, (d) the mass flow rate of air, and (e) the shaft power if mechanical efficiency is 0.95. The power developed is 425 kW.

Axial Flow Compressors

Solution:

Data: Speed: Mean diameter: Axial velocity: Whirl velocity at inlet: Work done: Inlet temperature: Inlet pressure: Stage efficiency:

379

N = 15,000 rpm dm = 0.6 m Va = Vf = constant = 225 m/s Vw1 = 85 m/s W.D. = 45 kJ/kg T1 = 300 K p1 = 1 bar hS = 0.89

To determine (b1 – b2), pR, R, m, SP

um = mean peripheral velocity

S dm N 60

(Vw 2  Vw1 ) um gc

W.D. = 'h or \

45

kJ kg

S – 0.6 – 15000 60

1.005

471.24 m/s

c p (T2  T1 )

kJ – (T2  300) K kg-K

T2 = 344.78 K

(a) Pressure ratio (pR): KS

\

(T2  T1 ) (T2  T1 ) „

0.89

T2  300 344.78  300 „

T2¢ = 339.85 K J

pR

p2 p1

Ë T2 Û J  1 Ì Ü Í T1 Ý „

Ë 339.85 Û Ì 300 Ü Í Ý

3.5

1.547

Ans.

380

Fundamentals of Turbomachinery

(b) Fluid deflection angle (b2 – b1): W.D.

\

45

kJ kg

(Vw 2  85) – 471.24 1 – 1000

Vw2 = 180.493 m/s

From inlet velocity triangle, cot E1

\

um  Vw1 Vf

471.24  85 225

1.72

b1 = 30.17°

Ans.

From exit triangle, cot E2

or \

um  Vw 2 Vf

471.24  180.493 225

1.292

b2 = 37.73° (b2 – b1) = 37.73° – 30.17° = 7.565°

(c) Degree of reaction (R): Vf R (cot E1  cot E2 ) 2u 225 – (cot 30.17’  cot 37.73’) 2 – 471.24

0.719

Ans.

(d) Mass flow rate of air ( m ): P 425 kW

\

m

m – W.D. m

kg kJ – 45 s kg

Ans.

9.444 kg/s

(e) Shaft power (SP): hm = mechanical efficiency = \

SP

P Km

425 kW 0.95

Power SP

447.37 kW

P SP

Ans.

EXAMPLE 5.8 Following data refers to an axial compressor. The total head pressure ratio is 4, static inlet temperature 20°C, overall total isentropic efficiency 86%. The inlet and outlet air angles from the rotor blades are 45° and 78° respectively. The blades are symmetrical. The mean blade speed of 200 m/s and axial velocity remain constant throughout the compressor. Calculate (a) the polytropic efficiency, (b) the number of stages required, and (c) the inlet Mach number relative to rotor at the mean blade height of the first stage.

Axial Flow Compressors

381

Solution: Machine: Total head pressure ratio: Static inlet temperature:

Axial compressor pR0 = 4 = p0K /p01 T1 = 20°C

Overall total isentropic efficiency:

Kc t to

0.86

b1 = 45° b2 = 78° a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1 um = 200 m/s Vf1 = constant = Va

Inlet air angle from the rotor blades: Exit air angle from the rotor blades: Blades are symmetrical: Mean blade speed: Axial velocity: To determine: hP, K, M1

(a) Polytropic efficiency (hP): J

Kc t to

J

( pR 0 )

0.86

1 J

( pR 0 )

1

JKp

1 1

40.286  1 0.286

\

4 KP  1 hP = 0.8842 = 88.42%

(b) Number of stages (K): Symmetrical blading \

a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1

From the combined velocity triangle, um = AC + CD = Vf cot b1 + Vf cot a1 um = Vf (cot b1 + cot a1) = Vf (cot 45° + cot 78°)

Ans.

382

Fundamentals of Turbomachinery

\

200 cot 45’  cot 78’

Vf

164.94 m/s

FC = Vw1 = Vf cot a1 = 164.94 cot 78° = 35.06 m/s V f2  Vw21

V1

T1 

T01 or

164.942  35.062

V12 2c p gc – 1000

293 

168.62 m/s

(168.62)2 2 – 1000 – 1.005 – 1

T01 = 307.15 K

From Eq. (5.18),

um V f (cot D 2  cot D1 )

W.D. = Dh0 = work done in each stage = 'TS

gc

200 – 168.62 – (cot 45’  cot 78’) 1 – 1000 – 1.005

DTS = 26.424 K = stagnation temperature rise in each stage Isentropic process for K stages, J

T0 K T01 \

„

1

Ë p0 K Û JKP Ì Ü Í p01 Ý

0.286

> 4 @0.8842

1.566

T0K ¢ = T01 × 1.566 = 307.15 × 1.566 = 480.94 K

\ (T0K¢ – T01) = total temperature rise considering all stages = 480.94 – 307.15 = 173.79 K K

Temperature rise in all stages Temperature rise in one stage 173.79 26.424

6.57  7 stages

\

Vr1

„

Ans.

(c) Mach number at inlet (M): From inlet velocity triangle,

sin E1

T0 K  T01 'TS

Vf Vr1 Vf sin E1

or

Vr1 M1

164.94 sin 45’

Axial Flow Compressors

383

= 0.6798

Ans.

233.26 m/s

Vr1

233.26

gc J RT1

1 – 1.4 – 0.287 – 293 – 1000

EXAMPLE 5.9 An axial flow compressor is fitted with half reaction blading, the blade inlet and outlet angles being 40° and 75° respectively. The mean diameter of a certain blade pair, is 85 cm rotating at 5500 rpm. Calculate the necessary isentropic efficiency of the stage if the pressure ratio of compression is to be 1.4, when the air inlet temperature is 25°C. Solution:

Data: Degree of reaction: Blade inlet angle: Blade outlet angle: Mean diameter: Speed: Pressure ratio: Air inlet temperature:

R = 0.5 b1 = 40° b2 = 75° dm = 85 cm N = 5500 rpm pR = 1.4 = p02/p01 T1 = 25°C

To determine: hc t–t T or h

02

p02

02 ¢ A 2¢

Vr1

B

p2 2

V1

Vf

a1

b1 C u

um

01

p1

D

Vw 1

1

S dm N 60

S – 0.85 – 5500 60

s

244.68 m/s

From Eq. (5.27),

R

Vf 2–u

p01

(cot E1  cot E2 )

Vf

i.e.

0.5

\

Vf = 167.7 m/s

2 – 244.78

(cot 40’  cot 75’)

384

Fundamentals of Turbomachinery

From inlet velocity triangle,

cot E1 \

BC Vf

BC = Vf × cot b1 = 167.7 × cot 40° = 199.86 m/s

\

Vw1 = u – BC = 244.78 – 199.86 = 44.92 m/s V1

T01 or

V f2  Vw21

T1 

V12 2 gc c p

295 

173.61 m/s

173.612 2 – 1000 – 1.005

T01 = 312.995 K

umV f (cot E1  cot E2 )

W.D. c p 'T0 \

167.72  44.92 2

'T0

gc

244.78 – 167.7 (cot 40’  cot 75’) 1 – 1005

37.33 K

(a) Isentropic efficiency (hc t–t): ËT Û T01 Ì 02  1Ü Í T01 Ý 'T0 „

(T02  T01 ) (T02  T01 )

Kc t t

„

J 1 Ë Û 298 Ì È p02 Ø J  1Ü É Ù Ì Ü 37.733 Ê p01 Ú ÌÍ ÜÝ

\

hc t–t

398 (1.4 0.286  1) 37.733 = 0.798

Ans.

EXAMPLE 5.10 At entry to an axial flow compressor, the total head pressure is 1 bar and the temperature is 300 K. The overall isentropic efficiency is 85%. It is assumed that the degree of reaction remains same for all stages and it is 50%. The total number of stages is 10. All stages contribute equal amount of work. At a particular stage the blade speed at the mean height is 210 m/s and the axial velocity is 175 m/s. Assume that the absolute air angle entering the rotor is 75° and the work done factor is 0.92. Calculate: (a) The rotor air angle at inlet (b) The overall pressure ratio

Axial Flow Compressors

(c) (d) (e) (f)

The The The The

polytropic efficiency static temperature of the air at entry to the rotor static temperature of the air leaving the first stage rotor inlet relative Mach number.

Solution:

Data: Total head pressure at inlet: Total temperature at inlet:

p01 = 1 bar T01 = 300 K Kc t to

Overall isentropic efficiency: Number of stages: Degree of reaction is same for all stages: Equal amount of work from all stages: Mean blade speed: Axial velocity: Entering absolute air angle: Work done factor: To determine: E1 ,

K pK ¢

K = 10 R = 50% um = 210 m/s Vf = constant = 175 m/s a1 = 75° y = 0.92

pK = 10th stage p7

7

7¢ 6

6¢

3¢

Vw 2

p6 p5

Vr 2

Vf

p1

2

C

b1 b2

p2

3

B

A

p3

4

4¢

u

p4

5

5¢

85%

p0 K , KP , T1 , T2 , Mr1 p01

T or h

2¢

385

E

1

Vf V2

Vr 1

Vw 1 D

V1

a2 a1

F

DVW = (Vw 2 – Vw 1)

s

(a) The rotor air angle at inlet (b1): When

R = 50%,

a1 = b2,

a2 = b1,

V1 = Vr2, V2 = Vr1

From the inlet velocity triangle AFD, Vw1 cot D1 Vf \

Vw1 = Vf × cot a1 = 175 × cot 75° = 46.89 m/s

tan E1 \

Vf u  Vw1

b1 = 47.01°

175 210  46.89

1.073 Ans.

386

Fundamentals of Turbomachinery

Èp Ø (b) The overall pressure ratio É 0 K Ù = pR0 : Ê p01 Ú

DT0 = stagnation temperature rise in the stage From Eq. (5.18), 'T0

\ V f u (cot E1  cot E 2 ) gc – 1000 – c p

0.92 – 175 – 210 – (cot 47.01’  cot 75’) 1 – 1000 – 1.005

DT0 = 22.34 K = T02 – T01 T02 = T01 + 22.34 = 300 + 22.34 = 322.34 K K = number of stages

\ \

10

Total rise in stagnation temperature in all stages Stagnation temperature rise in the stage

10

(T0 K  T01 ) 'T0

(T0 K  T01 ) 22.34

T0K – T01 = 10 × 22.34 = 223.4 K T0K = T01 + 220.34 = 300 + 220.34 = 520.34 K ËT Û T01 Ì 0 K  1Ü Í T01 Ý (T0 K  T01 ) „

Kc t to

\

T0 K T01

„

(T0 K  T01 ) (T0 K  T01 ) „

0.85

0.85 – (T0 K  T01 ) 1 T01 0.85 – 223.4 1 300

\

T0 K T01

„

1.633

From isentropic relation between 01 and 0K¢ J

p0 K p01

Ë T0 K Û J  1 Ì Ü Í T01 Ý „

= (1.633)3.5 = 5.565

Ans.

Axial Flow Compressors

387

(c) The polytropic efficiency (hP): J

1

Ë p0 K Û JKP Ì Ü Í p01 Ý

T0 K T01

0.286

520.34 (5.565) KP 300 hP = 0.8914

or \

Ans.

(d) The static temperature of the air at entry to the rotor (T1): From the inlet velocity triangle AFD,

sin D1 \

V1 T1

\

Vf V1 Vf sin D1 T01 

V12 2c p

175 sin 75’ 300 

181.17 m/s 181.172 2 – 1005

T1 = 283.7 K

Ans.

(e) The static temperature of the air leaving the first stage (T2): From the exit triangle AED,

sin D 2

Vf

\ \

V2 V2 = sin a2 × Vf = sin 47.01° × 175 V2 = 128.01 m/s

Now,

T2

T02 

V22 2 – gc – 1000 – c p

(128.01)2 2 – 1 – 1000 – 1.005 = 314.19 K 322.34 

Ans.

(f) The rotor relative inlet Mach number (Mr1):

sin E1 Vr1

\

Mr1

Vf Vr1

175 Vr1

175 sin E1 Vr1 J RT1

175 sin 47.01’

239.24 m/s

239.24 1.4 – 287 – 283.7

0.71

Ans.

388

Fundamentals of Turbomachinery

EXAMPLE 5.11 An axial flow compressor is designed on the free vertex principle. The speed of the machine is 5000 rpm. The hub-to-tip ratio is 0.6 and the work done is 25 kJ/kg. The work done factor is 0.94. The stage isentropic efficiency is 0.89. Ambient conditions are 1.0 bar and 300 K. The inlet absolute velocity is 150 m/s. Calculate the following: (a) The tip radius and the corresponding rotor air angle at exit if the Mach number relative to tip is limited to 0.92. (b) The mass flow rate (c) The stagnation pressure ratio (d) The power required (e) The rotor air angles at the root section. Solution:

Data: Inlet air angle at tip: Speed:

b1t = 25° (assumption) N = 5000 rpm

Hub-to-tip ratio: Work done factor: Stagnation isentropic efficiency: Stagnation pressure at inlet: Stagnation temperature at inlet: Inlet absolute velocity: Work done: To determine: rt , E2 t , m , p2 /p1 , power, E1r , E2r

T1 M1 Vr1

T01 

V12 2 gc c p

rh 0.6 rt y = 0.94 hc t–t = 0.89 p01 = 1 bar T01 = 300 K V1 = 150 m/s W.D. = 25 kJ/kg

150 2 2 – 1000 – 1.005

288.81 K

0.92 – 1.4 – 287 – 288.81

313.4 m/s

300 K 

Vr1 J RT1 M1 J RT1

Axial Flow Compressors

389

From inlet velocity triangle,

sin E1 \

Vf

Vf

Vr1

313.4

sin E1t

sin 25’

Vf = 132.45 m/s u

Vr21  V f2  V12  V f2

AC = AD + DC

314.42  132.452

150 2  132.452

u = u1 = 382.65 m/s = ut (a) Rotor air angles at exit corresponding to tip (b2t) and tip radius (rt): From Eq. (5.18),

W.D. c p 'T0 S Now, and or or

'T0 S (cot E1  cot E 2 )

\ uV f (cot E1  cot E2 ) gc

25 cp

(T02  T01 ) 25 – gc \ uV f

25 1.005

25 kJ/kg

24.88 K

25 – 1000 – 1 0.94 – 382.65 – 132.45

(cot b1 – cot b2) = 0.5248 cot b2 = cot b1 – 0.5248 = cot b1t – 0.5248 = cot 25° – 0.5248 = 1.6197

or

b2 = 31.7° = b2t

Ans.

J

p01 p1

Ë T01 Û J  1 Ì Ü Í T1 Ý

T or h

Ë 300 Û Ì 288.81 Ü Í Ý 02

3.5

1.14

03

03¢ p3 3

3¢ 2¢

p2

2 p01

01 1

p1

s

390

Fundamentals of Turbomachinery

or

p1

p01 1.14

1 bar 1.14

Now,

U1

p1 RT1

0.876 – 100 0.287 – 288.88

and

u1

S dt N 60

0.876 bar 1.057 kg/m 3

2S rt N 60

2 – S – rt – 5000 60 rt = radius at tip = 0.731 m

382.65 \

Ans.

(b) Mass flow rate ( m ): rr = root radius = 0.6rt = 0.6 × 0.731 = 0.4385 m

rt  rr 0.731  0.4385 2 2 h = height between tip and root = rt – rr = 0.731 – 0.4385 = 0.293 m

rm

mean radius =

m

2S rm hV f U1

0.5847 m

= 2 × p × 0.5847 × 0.293 × 132.45 × 1.057 = 150.7 kg/s

Ans.

(c) Stagnation pressure ratio (pR0): From Eq. (5.23), J

p02 p01

pR 0

Ë Kc t t (T02  T01 ) Û J  1 Ì1  Ü T01 Í Ý

0.89 – 24.88 Û Ë Ì1  Ü 300 Í Ý = 1.283

3.5

Ans.

(d) Power required (P): P

W.D. – m 150.7

m c p 'T0

kg kJ – 1.005 – 24.88 K = 3768.16 kW s kg-K

(e) The rotor air angles at the root (b1r, b2r):

ur

S dr N 60

2S rt N 60

2 – S – 0.4385 – 5000 = 229.6 m/s 60

Ans.

Axial Flow Compressors

391

Consider the same inlet triangle for root section, 229.6  V12  V f2

AD = ur  DC

229.6  150 2  132.452

= 159.19 m/s tan E1r

\

b1r

Vf

132.45 AD 159.19 = 39.76°

cot E2r

gc

cot E1r 

c p 'T0 gc ur V f \

cot 39.76’ 

\

Ans.

ur V f \ (cot E1  cot E 2 )

W.D. = c p 'To \

0.832

1.005 – 24.88 – 1000 229.6 – 132.45 – 0.94

b2r = 71.9°

Ans.

EXAMPLE 5.12 Modify Example 5.11 such that the inlet absolute velocity is axial, i.e. V1 = Vf 1 = 150 m/s. Calculate the air angle at inlet at tip (b1t). Do not consider b1t , given in Example 5.11. Solution: T1 = 288.81 K

(Example 5.11)

Vr1 = 313.4 m/s

(Example 5.11)

sin E1t tan E1t

\

ut

Vf Vr1

150 ; ? E1t 313.4

28.595’

V1 ut V1 tan E1t

150 tan 28.59’

275.23 m/s

392

Fundamentals of Turbomachinery

(a) Rotor air angles at exit corresponding to tip (b2t) and tip radius (rt): DT0S = temperature drop per stage

'T0 S

T02  T01

25 cp

25.0 1.005

cot E1  cot E 2

25.0 – gc \ ut V f

25.0 – 1 – 1000 0.94 – 275.23 – 150

(Eq. (5.18))

24.88 K

0.6442

cot b2 = cot b1 – 0.6442 = cot 28.595° – 0.6442 = 1.1903 b2 = b2t = 40.03°

Ans.

J

Ë T01 Û J  1 Ì Ü Í T1 Ý

p01 p1

Ë 305 Û Ì 288 Ü Í Ý

p1 = 0.876 bar r1 = 1.057

kg/m3

S dt N 60

ut

2S rt N 60

3.5

(Example 5.11) (Example 5.11)

2 – S – rt – 5000 60

275.23

rt = 0.526 m

Ans.

(b) Mass flow rate ( m ) : rr = 0.6rt = 0.6 × 0.526 = 0.3154 m

rm

rt  rr 2

0.526  0.3154 2

0.421 m

h = rt – rr = 0.526 – 0.3154 = 0.2106 m m

2S rm hV f U1

= 2 × p × 0.421 × 0.2106 × 150 × 1.057 = 88.33 kg/s

Ans.

(c) Stagnation pressure ratio (pR0): From Eq. (5.23), È

pR 0

p02 p01

J

Ø

Ë Kc t t (T02  T01 ) Û ÉÊ J  1ÚÙ Ì1  Ü T01 Í Ý

0.89 – 24.88 Û Ë Ì1  Ü 300 Í Ý

3.5

1.283

(Example (5.11))

Ans.

Axial Flow Compressors

393

(d) Power required (P): P

W.D. – m

m – c p 'T0

= 88.33 × 1.005 × 24.88 = 2208.6 kW

Ans.

(e) The rotor air angles at the root (b1r, b2r):

ur

tan E1r

\

S dr N 2S dr N 60 60 = 165.14 m/s V1 ur

150 165.14

0.9083

b1r = 42.25° W.D. = c p 'T0 cot E 2

cot E1 

Ans.

ur V f \ (cot E1  cot E2 ) gc c p 'T0 gc ur V f \

cot 42.55’ 

\

2 – S – 0.3154 – 5000 60

1.005 – 24.88 – 1 – 1000 165.14 – 150 – 0.94

0.0271

b2r = 88.45°

Ans.

EXAMPLE 5.13 Work input to an axial compressor is 25 kJ/kg. The work done factor is 0.95 and blade velocities at the root, mean radius and tip are 160.0, 215 and 270 m/s respectively. The axial velocity remains constant from root to tip and is 155 m/s. The degree of reaction at mean radius is 50%. Calculate: (a) The inlet and exit bar blade angles at root, mean and tip for a vertex design (b) The reaction at the root and tip. Solution:

Data: Work input: Work done factor: Blade velocity at root: at mean radius: at tip: Axial velocities, constant: Degree of reaction at mean radius:

W.D. = 25 kJ/kg y = 0.95 ur = 160 m/s um = 215 m/s ut = 270 m/s Vf = constant = 155 m/s Rm = 0.5

To determine: b1r, b2r, b1m, b2m, b1t, b2t, Rr, Rt Rm = 50%,

\ a1m = b2m,

a2m = b1m

394

Fundamentals of Turbomachinery

Velocity triangle at mean radius.

Case I:

At mean radius:

(a) Air angles at mean radius (b1m, b2m): From Eq. (5.18) and introducing the work done factor, W.D. = cp(T03 – T01) = cp(T02 – T01)

\ um V f (cot E1  cot E2 ) gc – 1000 or

(cot E1  cot E 2 )

25 – 1 – 1000 0.95 – 215 – 155

0.7897

(i)

From Eq. (5.27),

R or

(cot E1  cot E2 )

V f (cot E1  cot E2 ) 2um R – 2um Vf

0.5 – 2 – 215 155

1.3871

(ii)

From Eqs. (i) and (ii), b1m = 42.576°

Ans.

From Eq. (ii), cot b2m = 1.3871 – cot b1m = 1.3871 – cot 42.576° \

b2m = 73.4°

Ans.

This is a 50% reaction, hence a1m = b2m, Case II:

a2m = b1m

At tip radius:

(a) Air angles at tip radius (b1t, b2t, a1t, a2t): Data given: ut = 270 m/s Now we apply the free vertex condition, i.e. Vw = constant

Axial Flow Compressors

395

From inlet velocity triangle ADE (same velocity triangle but R ¹ 50%) Vw1 = Vf cot a1; (ut – Vw1) = Vf cot b1 (cot a1 + cot b1)Vf = Vw1 + ut – Vw1 = ut or

(cot D1  cot E1 )

ut Vf

270 155

= 1.742

(iii)

Similarly, (cot D 2  cot E2 )

ut Vf

270 155

= 1.742

(iv)

rmVwm = rtVwt or

rm V f cot D1m

\

cot D1t

rt V f cot D1t rm cot D1m rt

um cot D1m ut

215 – cot 73.4’ 270 a1t = 76.62°

or

Ans.

From Eq. (iii), cot b1t = 1.742 – cot a1t = 1.742 – cot 76.62° b1t = 33.62°

or

Ans.

Similarly, we have rm cot a2m = rt cot a2t \

or

cot D 2t

rm cot D 2 m rt

um cot D 2 m ut

215 – cot 42.576’ 270 a2t = 49.1°

Ans.

From Eq. (iv), cot b2t = 1.742 – cot a2t = 1.742 – cot 49.1° or

b2t = 48.79°

Ans.

396

Fundamentals of Turbomachinery

Case III:

At root radius, ur = 160 m/s

(a) Air angles at root radius (b1r, b2r, a1r, a2r): rm × Vw1m = rr × Vw1r rm cot a1m = rr cot a1r cot D1r

rm cot D1m rr um cot D1m ur

or

cot D1r

\ We have

a1r cot D1r  cot E1r

215 – cot 73.4’ 160 = 68.2°

ur Vf

160 155

Ans.

1.0323

cot b1r = 1.0323 – cot a1r

or

= 1.0323 – cot 68.2° = 0.632 \ Similarly, we have

b1r = 57.595°

Ans.

rmVw2m = rr × Vw2r rm cot a2m = rr cot a2r cot D 2r

rm cot D 2 m rr um cot D 2 m ur

\

a2r

215 – cot 42.576’ 1.4625 160 = 34.36°

Ans.

We have cot D 2r  cot E2r

or

ur Vf

160 155

1.03226

cot b2r = 1.03226 – cot a2r = 1.03226 – cot 34.36° = –0.43024

\

b2r = –66.72°

Ans.

Axial Flow Compressors

397

(b) Degree of reaction (Rr, Rt): From Eq. (5.27),

Rt

Vf 2ut

(cot E1t  cot E2t )

155 – (cot 33.62’  cot 48.79’) 2 – 270

\

Rt = 0.6831 = degree of reaction at tip

Ans.

Rr = degree of reaction at root

Vf 2 – ur

[cot 57.695’  cot ( 66.72’)]

Rr = 0.0979

Ans.

EXAMPLE 5.14 Modify Example 5.13 such that 50% reaction is assumed along the whole blade. The blade speed at root, mean radius and tip are 160, 215, 270 m/s respectively. The axial velocity is 155 m/s and is constant. Calculate the air and blade angles. Solution:

Data: Degree of reaction: Blade speed at root: at mean radius: at tip: Axial velocity:

Rm = Rr = Rt = R = 50% ur = 160 m/s um = 215 m/s ut = 270 m/s Vf = constant = 155 m/s

To determine: a1r, b1r, a2r, b2r, a1m, b1m, a2m, b2m, a1t, b1t, a2t, b2t At mean radius: a1m = b2m = 73.4°

(Example 5.13)

a2m = b1m = 42.576°

(Example 5.13)

At tip: From Eq. (5.27),

R

Vf 2 – ut

(cot E1t  cot E2t )

0.5 – 2 – 270 1.742 155 From Eq. (5.18) and introducing the work done factor

or

(cot E1t  cot E2 t )

(v)

W.D. = cp(T03 – T01)

c p (T02  T01 )

\ ut V f (cot E1t  cot E2t ) gc – 1000

398

Fundamentals of Turbomachinery

or

(cot E1t  cot E2t )

25 – 1 – 1000 270 – 0.95 – 156

0.6298

(vi)

From Eqs. (v) and (vi), we have b1t = 40.15° = a2t

Ans.

From Eq. (vi), cot b2t = cot b1t – 0.629 = cot 40.15° – 0.629 b2t = 60.9° = a1t

\

Ans.

At the root: Modifying Eq. (v) for root, (cot E1r  cot E2 r )

 – 160 – 0.5 155

1.0323

(vii)

Modifying Eq. (vi) for root, (cot E1r  cot E2r )

25 – 1 – 1000 160 – 0.95 – 155

1.0611

(viii)

From Eqs. (vii) and (viii), b1r = 43.69° = a2r

Ans.

From Eq. (viii), cot b2r = cot b1r – 1.0611 = cot 43.69° – 1.0611 \

b2r = –89.17° » 0.8244° = a1r

Ans.

EXAMPLE 5.15 An axial flow compressor stage has the following data: Air inlet stagnation temperature 300 K, relative flow angle at rotor inlet measured from the axial direction 42°, flow coefficient 0.57, relative inlet Mach number onto the rotor 0.79. If the stage is normal, calculate (a) the stagnation temperature rise, (b) the stagnation pressure ratio, if the isentropic compression efficiency is 0.89. Assume the degree of reaction to be 50%. Solution: Data: Inlet stagnation temperature: T01 = 300 K Inlet relative flow angle: b1 = 90° – 42° = 48° Given data is with respect to axial direction. Hence it is considered with respect to tangential direction. Therefore: b1 = 90° – 42° = 48° Flow coefficient f = 0.57 Relative inlet Mach number Mr1 = 0.79 Normal stage (given) Degree of reaction: R = 50% Isentropic compression efficiency: hc t–t = 0.89

Axial Flow Compressors

To determine: (T02  T01 ),

p02 p01

399

pR 0

Since the stage is normal and the degree of reaction is 50%, the velocity triangles are symmetrical.

(a) Stagnation temperature rise (T02 – T01): From Eq. (5.27), R

or \

cot E 2

Vf 2u

(cot E1  cot E2 )

2R  cot E1 I

I (cot E1  cot E 2 ) 2

2 – 0.5  cot 48’ 0.57

b2 = 49.5°

From inlet velocity triangle,

V1

Vr1

Vf

Vf

sin D1

sin 49.5’

Vf

Vf

sin E1

sin 48’

1.315 V f

1.346 V f

We have for Mach number relation, Mr1 Mr21

Vr1 J RT1 Vr21 J RT1

Vr21 È Ø V12 J R É T01  2 gc p – 1000 ÙÚ Ê

400

Fundamentals of Turbomachinery

(0.79)2

or

\

(1.346V f )2 È Ø (1.315V f )2 1.4 – 287 É 300  Ù 2 – 1 – 1.005 – 1000 Ú Ê

Vf = 192.63 m/s

From Eq. (5.18), W.D. = 'h0 (T02  T01 )

or

c p (T02  T01 )

uV f (cot E1  cot E 2 ) gc – 1000

V f2 (cot E1  cot E2 ) I gc 1000 – c p (192.63)2 (cot 48’  cot 49.5’) 0.57 – 1 – 1000 – 1.005

\

(T02 – T01) = 3.00 K

Ans.

(b) Stagnation pressure ratio (pR0) Kc t t

0.89

(T02  T01 ) (T02  T01 ) „

Ë T  1Û T01 Ì 02 Ü Í T01 Ý „

or

0.89

\

pR 0

EXAMPLE 5.16 design data:

(T02  T01 )

p02 p01

J

1

Ëp Û J 1 T01 Ì 02 Ü Í p01 Ý (T02  T01 )

0.0000002  0

„

Ans.

A 50% reaction of a first stage axial flow compressor has the following

Mass flow rate: Work done: Rotation speed: Inlet absolute velocity makes an angle: Work done factor: Mean blade speed: Inlet stagnation temperature: Inlet static temperature: Isentropic efficiency: Inlet stagnation pressure:

m 25 kg/s P = 0.25 kJ/kg N = 170 rps a1 = 75° with respect to tangent y = 0.95 um = 200 m/s T01 = 300 K T1 = 285 K hc t–t = 0.89 p01 = 1 bar

401

Axial Flow Compressors

Calculate (a) the mean radius, (b) the blade and air angles (fluid angle) at mean radius, (c) the blade height, and (d) the overall pressure ratio. Solution: To determine: rm, b1m, b2m, a1m, a2m, h, pR0 T or h 02 03¢

03²

u

Vr 2

p01

C

b1 b2

2 p1

Vf

Vf V2

E

Vr 1

DVW = Vw 2 – Vw 1

1

s

T01 or

B

A p2

01

Vw 2

p3

3

3²

3¢

p03

03

V1

T1 

(At mean radius)

V12 2 gc c p – 1000

(T01  T1 ) 2gc – c p – 1000 (300  285) – 2 – 1 – 1.005 – 1000

= 173.64 m/s From inlet velocity triangle AED,

sin D1 or

Vf V1

Vf = V1 sin a1 = 173.64 × sin 75° = 167.73 m/s

Applying continuity equation to the annulus area, We have for isentropic relation,

Ë p1 Û Ì Ü Í p01 Ý \

J

Ë T1 Û J  1 Ì Ü Í T01 Ý

Ë 285 Û Ì 300 Ü Í Ý

3.5

0.8357

p1 = p01 × 0.8357 = 1 bar × 0.8357 = 0.8357 bar U1

p1 RT1

0.8357 – 100 0.287 – 285

1.022 kg/m 3

F

Vw 1 D

V1

a2 a1

402

Fundamentals of Turbomachinery

(a) The mean radius (rm):

um

S dm N 60

\

rm

200 2 – S – 170

and

dm = 2 × rm = 2 × 0.1872 = 0.3744 m

2S rm N

200

0.1872 m

Ans.

(b) Blade angles (b1m, b2m) and air angles (a1m, a2m) at mean radius: From Eq. (5.18) and introducing the work done factor, y

'h0

W.D. or

(cot E1m  cot E2 m )

c p (T03  T01 )

\ um V f (cot E1m  cot E 2 m )

25 – 1 – 1000 0.95 – 200 – 167.73

gc – 1000 (i)

0.785

From Eq. (5.27),

R or (cot E1m  cot E 2 m )

Vf 2um

(cot E1m  cot E2 m )

0.5 – 2 – 200 167.73

(ii)

1.1924

From (i) and (ii), we get b1m = 45.32° = a2m

Ans.

From Eq. (ii), we get cot b2m = 1.1924 – cot b1m = 1.1924 – cot 45.32° \

b2m = 78.49° = a1m

Ans.

(c) The blade height (h): Applying the continuity equation to the annulus, m

U1 AV f

or

h

m U1 S dm V f

\

kg 1 kg 1 1 1 – – – – m s 1.022 m 3 S 0.3744 m 167.73 s h = 0.124 m

U1 – S dm hV f

25

(d) The overall pressure ratio (pR0): ËT Û T01 Ì 03  1Ü Í T01 Ý (T03  T01 ) „

Kc t t

(T03  T01 ) (T03  T01 ) „

0.89

Ans.

Axial Flow Compressors

or

Ë T03 Û Ì Ü Í T01 Ý

Ë 0.89 – (T03  T01 ) Û Ì Ü 1 T01 Í Ý

„

0.89 –

or

\

403

25 1 – 1 1.005 300

Ë T03 Û Ì Ü 1.0738 Í T01 Ý „

1.0738

pR 0

Ë p03 Û Ì Ü Í p01 Ý J

1 J

J

1 J

 pR 0

(1.0738)3.5

1

J

J

1.283

Ans.

EXAMPLE 5.17 An axial compressor with 50% reaction is having a flow coefficient of 0.54. Air enters the compressor at stagnation condition of 1 bar and 30°C. The total-to-total efficiency across the rotor is 0.88. The total-to-total pressure ratio across the rotor is 1.26. The pressure coefficient is 0.45 and the work done factor is 0.88. The mass flow rate is 15 kg/s. Calculate: (a) The mean rotor blade speed (b) The rotor blade angles at inlet and exit (c) The power input to the system. Solution:

Data: Degree of reaction: Flow coefficient: Inlet stagnation pressure: Inlet stagnation temperature: Total-to-total efficiency across the rotor: Total-to-total pressure ratio: Pressure coefficient: Work done factor:

R = 50% f = 0.54 p01 = 1 bar T01 = 30 + 273 = 303 K hc t–t = 0.88 p02/p01 = 1.26 fp = 0.45 y = 0.88 m 15 kg/s

Mass flow rate: To determine: b1, b1, um, P Applying the isentropic relation between 01 and 02¢

T02 T01

„

\

T02

„

Ë p02 Û Ì Ü Í p01 Ý

J

1 J

(1.26)0.286

T01 – 1.06833

= 323.7 K

1.06833

303 – 1.06833

404

Fundamentals of Turbomachinery

03

T or h

p02 p03

02 02¢

p3

03¢

3 u

p2

3¢ 2

Vw 2

Vw 1

p01

2¢

b1 p1

01

b2

1

Vr 2

Vf V2

s

Kc t  t

\

(T02  T01 ) (T02  T01 )

0.88

Vf

V1

a2 a1

Vr 1

(323.7  303) (T02  303)

„

T02 = 323.7 K Dh0 = cp(T02 – T01) = 1.005 × (326.52 – 303)

or

'h0

23.64 kJ/kg

Ip

pressure coefficient =

'h0 2

u /2 gc

(a) Mean rotor blade speed (u): 23.64 – 2 – gc – 1000

0.45

or

u2

23.64 – 2 – 1 – 1000

0.45

u2 u = 324.14 m/s

\

Ans.

(b) Blade angles at exit (b2) and inlet (b1): I

Vf

0.54

u

Vf 324.1

\ Vf = 175.01 m/s

We have from Eqs. (5.17) and (5.18), 'h0

or

23.64

kJ kg

\ u ' Vw gc – 1000 0.88 – 324.14 – 'Vw 1 – 1000

Axial Flow Compressors

\

\

405

DVw = Vw2 – Vw1 = 82.95 m/s \ u V f (cot E1  cot E 2 )

u (Vw 2  Vw1 ) \

gc – 1000

gc – 1000

cot E1  cot E2

82.95 Vf

cot E1  cot E2

R – 2u Vf

From (i) and (ii)

82.95 175.01

(Eqs. 5.25a and 5.25b) (i)

0.474

0.5 – 2 – 324.14 175.01

(Eq. 5.27)

= 1.852

(ii)

b1 = 40.69°

Ans.

From Eq. (ii), we get cot b2 = 1.852 – cot b1 = 1.852 – cot 40.69° = 0.689 b2 = 55.43°

Ans.

(c) Power (P): P

m 'h0

15

kg kJ – 23.64 354.6 kW s kg

Ans.

EXAMPLE 5.18 An axial flow compressor has mean blade velocity of 250 m/s, axial velocity of 162 m/s and it is constant. The pressure ratio is 5 : 1. The relative outlet air angle is same for each stage and it is 65°. Assume polytropic efficiency of 0.89 and 50% reaction for each stage. Calculate the number of stages. Assume stagnation temperature at inlet 290 K. Solution: Data: Ambient temperature at inlet: Mean blade velocity: Axial velocity is constant: The pressure ratio: Number of stages: Relative outlet air angle: Polytropic efficiency: Degree of reaction:

T01 = 290 K um = 250 m/s Vf = constant = 162 m/s pR0 = p0K/p01 = 5 : 1 K=? b2 = 55.43° hP = 0.89 R = 0.5, a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1

To determine: K From the exit triangle AFD, we have

cot E 2 or

AB Vf

AB = Vf × cot b2 = 162 × cot 65°

406

Fundamentals of Turbomachinery T or h 0K

pK

u

Vw 2

p5

0K ¢ 5 5¢

4

4¢

3

3¢ 2

p4

B

A

p3

Vw 1

C

b1 b2

p2

Vr 2

p1

2¢ 1

Vf

Vf

D

V1

a2 a1

V2 Vr 1 F E Combined velocity triangle for one stage corresponding to mean radius

s

= 75.54 m/s V f2  (um  AB)2

V2

162 2  (250  75.54)2

= 238.07 m/s = Vr1 From the inlet velocity triangle AED,

sin E1 \

Vf Vr1

162 238.07

0.6805

b1 = 42.88°

We have from Eq. (5.17), DT0S = temperature rise in one stage

or

um V f (cot E1  cot E2 )

'T0 S

T03  T01

'T0 S

250 – 162 – (cot 42.88’  cot 55.43’) 1 – 1000 – 1.005

Ë T0 K Û Ì Ü Í T01 Ý

Ë p Û JKP overall temperature ratio = Ì 0 K Ü Í p01 Ý

gc – 1000 – c p

J

24.61 K

1

0.286

Ë 5 Û 0.89 Ì1 Ü Í Ý

1.67731

T0K = (1.67731 × T01) = 1.67731 × 290 = 486.42 K DT0K = total temperature rise considering all the stages (K stages) (Actual) = (T0K – T01) = (486.42 – 290) = 196.42 K

Axial Flow Compressors

407

(a) Number of stages (K): K

Total temperature rise in all stages (Actual) Temperature rise in one stage) 196.42 24.61

'T0 K 'T0 S

7.98  8 stages

Ans.

EXAMPLE 5.19 An axial compressor rotates at a speed of 15000 rpm and has a mean diameter of 30 cm. The rotor blade angle at inlet is 60° with respect to axial direction and the mass flow rate is 15 kg/s. The axial velocity remains the same throughout the stage. Assume that the absolute velocity at the entry is axial. A stator downstream of the rotor directs the flow axially at the stage exit. The total-to-total overall efficiency is 0.88, and the mechanical efficiency is 0.98. Assume: Inlet conditions to be standard atmospheric, work done factor 0.86, rotor blade exit angle 60°, diffuser efficiency of 0.78. Calculate: (a) The static pressure ratio across one stage (b) The static pressure ratio across the rotor (c) The degree of reaction (d) Power input Solution: Data: Speed: N = 15000 rpm Mean diameter: dm = 0.3 m The rotor blade angle at inlet: b1 = 90° – 60° = 30° Given data is with respect to axial direction, here it is considered with respect to tangential direction. Therefore: b1 = 90° – 60° = 30° m 15 kg/s Mass flow rate: Axial velocity: Vf = constant Absolute velocity is axial at inlet: V1 = Vf1, a1 = 90°, Vw1 = 0 Kc t to

Total-to-total overall efficiency:

0.88

Diffuser efficiency: hd = 0.78 Mechanical efficiency: hm = 0.98 Work done factor: y = 0.86 Ambient pressure: p01 = 1 bar Ambient temperature: T01 = 15 + 273 = 288 K Rotor blade exit angle: b2 = 60° S dm N S – 0.3 – 15000 um 235.62 m/s 60 60 From inlet velocity triangle ACD,

tan E1

V1 u

408

Fundamentals of Turbomachinery

T or h u B

A

C a2 a1 V1 = Vf 1

b1 b2

Vf

Vw 1 = 0

Vr 2 V2

E

Vr 1

3¢

3≤

2¢

2

01

p01

Vr1

Vf

V1

u

Rotor row

V2

b2

a2 u

Stator row

Vf = Vf 3

V3 a3

or

V1 = Vf1 = u × tan b1 = 235.62 × tan 30°

\

V1 = 136.04 m/s

Now,

Vr1

V12  u2

136.042  235.622

272.1 m/s

Èp Ø (a) The static pressure ratio across the stage É 3 Ù : Ê p1 Ú

From the Eq. (5.18), 'T0

p3 p2 p1

s

a1

b1

Vf

3

D 1

Vr 2

p03

03≤

03¢

Vw 2

03

02

(T02  T01 )

M uV f (cot E1  cot E2 ) gc – 1000 – c p

Axial Flow Compressors

0.86 – 235.62 – 136.04 – (cot 30’  cot 60’) 1 – 1000 – 1.005

DT0 = 31.67 kJ/kg Kc t to

0.88

T03  T01

(T03  T01 ) 31.67

„

„

27.87 K

„

T03

(T03  T01 ) (T03  T01 )

T01  27.87

„

288  27.87

= 315.87 K T03 – T01 + 31.67 = 288 + 31.67 T02 = T03 = 319.67 K Applying isentropic relation between 01 and 03¢ J

p03 p01

p03 p01

„

p03 = p01

Ë T03 Û J  1 Ë 315.87 Û3.5 1.382 Ì Ü Ì 288 Ü Í Ý Í T01 Ý × 1.382 = 1 bar × 1.382 = 1.382 bar „

From exit velocity triangle AEC,

cot E 2 \

AB Vf

AB = Vf × cot b2 = 136.04 × cot 60° = 78.54 m/s Vw2 = u – AB = 235.62 – 78.54 = 157.08 m/s V2

T2 or

T02 

136.042  157.082

V22 2 – gc – c p

319.67 

207.8 m/s

207.82 2 – 1 – 1000 – 1.005

T2 = 298.2 K T3

or

V f2  Vw22

T03 

V32 2 – gc – 1000 – 1.005

T3 = 310.46 K

0.78

(T3”  T2 ) (T3  T2 )

\

T3”

307.76 K

136.042 2 – 1 – 1000 – 1.005

(Assumption V1 = V3)

hd = diffuser efficiency or

319.67 

(T3”  298.2) (310.46  298.2)

409

410

Fundamentals of Turbomachinery

We have the isentropic relation between 3 and 03, J

Ë T3 Û J  1 Ì Ü Í T03 Ý

p3 p03 \

Ë 10.46 Û Ì 319.67 Ü Í Ý

3.5

0.903

p3 = p03 × 0.903 = 1.382 × 0.903 = 1.248 bar

Applying the isentropic relation between 2 and 3², J

Ë T2 Û J  1 Ì Ü Í T3” Ý

p2 p3 \

Ë 298.2 Û Ì 307.76 Ü Í Ý

3.5

0.8954

p2 = p3 × 0.8954 = 1.248 × 0.8954 = 1.1175 bar

T01 

T1

V12 2 – gc – c p

288 

136.042 2 – 1 – 1000 – 1.005

T1 = 278.79 K J

Ë T1 Û J  1 Ì Ü Í T01 Ý

p1 p01

Ë 78.79 Û Ì 288 Ü Í Ý

3.5

0.8925

p1 = p01 × 0.8925 = 1 bar × 0.8925 = 0.8925 bar \

p3 p1

1.248 0.8925

Ans.

1.398

Ëp Û (b) The static pressure ratio across the rotor Ì 2 Ü : Í p1 Ý p2 p1

1.1175 1.00

1.1175 bar

Ans.

(c) The degree of ratio (R): Vf

R

2u

(cot E1  cot E 2 )

136.04 – (cot 30’  cot 60’) 2 – 235.62

0.667

Ans.

(d) Power input (SP): SP =

m 'ho Km

15

= 487.17 kW

kg kJ 1 – 1.005 – 31.67 K – s kg-K 0.98

Ans.

411

Axial Flow Compressors

EXAMPLE 5.20 An axial compressor has a mean blade velocity of 200 m/s, and 50% degree of reaction. It has 10 stages. The polytropic and stage efficiency are 0.88 and 0.85 respectively. Air angles with respect to absolute velocity at inlet and exit are 15° and 45° respectively with respect to axial direction. Workdone factor 0.85, inlet stagnation pressure and temperature are 1 bar and 15°C respectively. Calculate (a) the total pressure ratio of the first stage, (b) the overall static pressure ratio and the stagnation pressure ratio considering all the stages. Solution: Data: Degree of reaction: R = 50%, a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1 Mean blade velocity: um = 200 m/s Number of stages: K = 10 Polytropic efficiency: hP = 0.88 Stage efficiency: hS = 0.85 Inlet air angles w.r.t. absolute velocity a1 = 90° – 15° = 75° Exit air angles w.r.t. absolute velocity a2 = 90° – 45° = 45° [The given air angles are w.r.t. axial, here they are considered w.r.t. tangential direction] Work done factor: y = 0.85 Inlet stagnation temperature: T01 = 15° + 273 = 288 K Inlet stagnation pressure: P01 = 1 bar Ëp Û Ëp Û Ëp Û To determine: Ì 03 Ü , Ì K Ü , Ì 0 K Ü Í p01 Ý Í p1 Ý Í p01 Ý T or h

u

K p K K¢

5

5¢

4

4¢ 3¢ 2¢ 1

3 2

p5

B

A

p4

Vw 2

Vw 1

C

b1 b2

p3 p2

Vr 2

Vf

p1 F

tan 45’

V2

Vr 1

E

DVW = (Vw 2 – Vw 1)

s

tan E1

Vf

Vf

Vf

u  Vw1

u  V f cot D1

Vf 200  V f cot 75’

Vf = 157.74 m/s

D

V1

a2 a1

412

Fundamentals of Turbomachinery

From Eq. (5.18),

c p (T03  T01 )

\ um V f (cot E1  cot E2 ) gc – 1000 0.85 – 157.74 – 200 – (cot 45’  cot 75’) 1 – 1000

19.63 kJ/kg

Ëp Û (a) Total pressure ratio in the first stage Ì 03 Ü : Í p01 Ý

hS = stage efficiency 0.85

or

T03  T01

\

T03

„

„

(T03  T01 ) (T03  T01 ) „

T03  T01 19.63 1.005 „

16.60 K T01  16.6

288  16.6

J

p03 p01

Ë T03 Û J  1 Ì Ü Í T01 Ý

p03 p01

1.2167

„

Ë 304.6 Û Ì 288 Ü Í Ý

304.6 K

3.5

1.2167 Ans.

Ëp Û Ëp Û (b) Overall static and stagnation pressure ratio Ì K Ü , Ì 0K Ü : Í p1 Ý Í p01 Ý

(T0K – T01) = total temperature rise considering all the stage, i.e. K stages. = no. of stages × temperature rise in one stage = K × DT0 10 –

(19.63) 1.005

(' 'T0

T03  T01 )

= 195.32 K \

T0K = T01 + 195.32 = 288 + 195.32 = 483.32 K

From inlet velocity triangle AED,

sin D1

V1

Vf V1

Vf sin D1

157.74 sin 75’

163.3 m/s

Axial Flow Compressors

(163.3)2 2 – 1 – 1000 – 1.005

T1

T01 

TK

T0 K 

413

274.73 K

V12 2 – gc – c p

482.32 

(163.3)2 = 470.05 K 2 – 1 – 1000 – 1.005

[Assume V1 = VK, i.e. absolute velocity of first stage = absolute velocity of last stage.]

Ë pK Û Ì Ü Í p1 Ý

KPJ

Ë TK Û J  1 Ì Ü Í T1 Ý

Ë 470.05 Û Ì 274.73 Ü Í Ý

0.88 – 1.4 0.4

5.23

Ans.

We have Ë p0 K Û Ì Ü = overall stagnation pressure ratio considering all the stage, i.e. K stages Í p01 Ý

\

Ë p0 K Û Ì Ü Í p01 Ý

Ë p0 K Û Ì Ü Í p01 Ý „

KPJ

Ë T0 K Û J  1 Ì Ü Í T01 Ý

Ë 483.32 Û Ì 288 Ü Í Ý

3.08

4.93

Ans.

IMPORTANT EQUATIONS • If a constant axial velocity is maintained through all the stages, then r1h1 = r2h2 r1 = density of the fluid h1 = height of the blades u 1 (cot D 2  cot E2 ) • Vf I • DVw = Vw2 – Vw1 = Vf1(cot b1 – cot b2) = Vf2 (cot a2 – cot a1) • W.D. = 'h0 • W.D. =

c p (T02  T01 )

u ( 'Vw ) gc

(T03  T01 ) c p

uV f (cot D 2  cot D1 )

uV f (cot E1  cot E2 )

gc

gc

• W.D. = 'h0

h02  h01

( h3  h1 ) 

h03  h01

(V32  V12 ) 2 gc

h3  h1

(h2  h1 ) 

(5.13) (5.16) (5.17) (5.18)

V22  V12 2 gc

(assumption V3 = V1)

(5.19a)

414

Fundamentals of Turbomachinery

Ë V2 Û Ë V2 Û • Ì h2  r 2 Ü  Ì h1  r1 Ü 0 2 gc ÜÝ ÌÍ 2 gc ÜÝ ÌÍ • pR0 = overall pressure for a stage

(5.21)

J

Ë Kc t to (T03  T01 ) Û J  1 Ì1  Ü T01 Í Ý

p03 p01

(5.23)

• Introduce y (work done factor) in Eq. (5.17) given above. • f = flow coefficient =

Vf u

• fp = pressure coefficient =

'ho u2 2 gc

• R = degree of reaction • R

• R • R

'hR 'hR  'hS

Vf 2u

1

(h2  h1 ) ( h3  h1 )

(5.26)

(5.27)

(cot E1  cot E2 )

Vf 2u

(cot D 2  cot D1 )

(5.28)

If R = 50%, then a1 = b2, a2 = b1, V1 = Vr2, V2 = Vr1, i.e. blades are symmetrical. •

1 dp – U dr

Vw2 r

• rVw = constant

(radial equilibrium theory)

(5.32)

(for free vertex condition)

(5.37)

REVIEW QUESTIONS 1. Sketch the velocity diagram for an axial flow compressor and derive the relation. J

p02 p01

Ë Kc t  t u ( 'Vw ) Û J  1 Ì1  Ü gc c p T01 ÝÜ ÍÌ

2. What is vertex theory? Derive an expression for vertex flow. 3. Derive an expression in terms of entry and delivery pressure and temperature and the ratio of the specific heats, for the polytropic or small stage efficiency of a rotary compressor. 4. Discuss the relative merits and demerits of axial flow and centrifugal compressors.

Axial Flow Compressors

415

5. Derive the following terms with respect to axial compressor. (a) Work done factor, (b) Flow coefficient, (c) Pressure coefficient. 6. Obtain an expression for polytropic efficiency of an axial flow compressor in terms of ‘n’ and ‘g ’. 7. Discuss the relative merits of axial flow and centrifugal flow compressors. 8. Draw the temperature–entropy diagram for an axial flow compressor. 9. What is radial equilibrium? Derive an equation for radial equilibrium. 10. Write short notes on axial flow compressor. 11. What are the factors that affect the performance of an axial flow compressor? 12. Draw the velocity triangles at the entry and exit for the following axial compressor stages (a) R = 50%, (b) R < 0.5, (c) R > 0.5, (d) R = 1. 13. Derive an expression for the degree of reaction for an axial flow compressor and prove that if R = 50%, then (1/2)f (cot b1 + cot b2) = 0.5[1 – f (cot a1 – cot b1)]. 14. Derive the following relations in case of an axial flow compressor: 1 I (b) f (cot b1 – cot b2) = blade loading coefficient

(a) cot D1  cot E1

(c)

'p0 Uu2

cot D 2  cot E2

I (cot D 2  cot D1 )

(d) hS = stage efficiency =

'p0 \ U uV f (cot D 2  cot D1 )

15. Draw the velocity triangles for hub, mean and tip radius for (a) free vertex conditions and (b) for constant reaction conditions.

EXERCISES 5.1 An axial flow compressor has an intake capacity of 10 kg/s when supplied with air at 1.03 bar and 15°C. When tested at these inlet conditions and flow rate, it was found that the air temperature at the outlet was 150°C and the insentropic efficiency of compressor was 87%. Velocities in the inlet and outlet passages were equal to 40 m/s. Calculate: (a) The discharge pressure (b) The internal compressor power (c) The shaft power input if the mechanical bearing losses were 96.5%. 5.2 A helicopter gas turbine requires an overall pressure ratio of 8.1. This is to be obtained using a two-spool layout consisting of a four-stage axial flow compressor followed by a single-stage centrifugal compressor. The polytropic efficiency of the axial flow compressor is 94% and that of the centrifugal is 85%.

416

5.3

5.4

5.5

5.6

5.7

Fundamentals of Turbomachinery

The axial compressor has stage temperature rise of 30°C, using symmetrical stages with stator outlet angle of 20°. If the mean diameter of each stage is identical, calculate the required rotational speed. Assume a work done factor of 0.86 and constant axial velocity of 150 m/s. Also assume axial inlet as the eye of the impeller, an impeller diameter of 330 mm, a slip factor of 0.90 and power input factor of 1.04. Calculate the rotational speed required for the centrifugal compressor. Ambient conditions are 1.01 bar and 288 K. Air at the temperature of 300 K enters a ten-stage axial flow compressor at the rate of 3.0 kg/s. The pressure ratio is 6.5 and the isentropic efficiency is 85%, the compression process being adiabatic. The compressor has symmetrical stages. The axial velocity of 110 m/s is uniform across the the stage and the mean blade speed of each stage is 180 m/s. Determine the direction of air at entry to and exit from the rotor and the stator blades and also the power given to the air. cp = 1.005 kJ/kg, g = 1.4. An axial flow compressor stage where the absolute velocity at the entry is axial has an motor speed of 250 m/s. A stator downstream of the rotor directs the flow axially at the stage exit. If the rotor blade angle at the exit is 60° and the mass flow rate of air is 13.5 kg/s, Calculate: (a) The static pressure rise across the stage, assuming total-to-total efficiency of 0.86. (b) The static pressure rise across the rotor and the degree of reaction. (c) The power input assuming mechanical efficiency of 0.98. Assume the inlet air conditions to be standard atmosphere and the work input factor to be 0.86. An axial flow compressor stage draws air from with the stagnation conditions of 1 atmosphere and 35°C. Assuming 50% reaction stage with a flow coefficient of 0.52 and the ratio of DVu/u = 0.25, find the rotor blade angles at the inlet and the exit as well as the mean rotor speed. The total-to-total efficiency of the stage is 0.87. When the stage produces a total-to-total pressure ratio of 1.23, determine: (a) The pressure coefficient (b) The power input to the system. Assume the work input factor to be 0.86 and mass flow rate 12 kg/s. An axial flow compressor stage is to be designed with a flow coefficient of 0.5 and degree of reaction 50%. The mean blade speed is required to be 250 m/s. The total-to-total efficiency is 0.80. The compressor has to supply 15 kg/s of air when the inlet conditions are 1 atm. and 35°C. Draw the velocity triangle at the root, the mean section and the blade tip. The ratio of blade tip to blade root diameter is 1.5. The rotor blade is to be designed such that uDVa = constant. Find the exit static pressure and the actual power input, assuming the pressure coefficient to be 0.7 and the work factor input to be 0.86. An axial compressor provides a total head pressure ratio 4:1 with an overall total head isentropic efficiency of 80%. When the inlet total head temperature is 290 K, find the polytropic efficiency. The compressor is designed for 50% reaction with inlet and outlet air angles from rotor blades of 45° and 10° respectively. The mean blade speed and axial

Axial Flow Compressors

417

velocity are constant throughout the compressor. Assume a blade speed of 180 m/s and the work done factor to be 0.86. Find the number of stages required. What is the inlet Mach number relative to rotor at the mean blade height of the first stage? 5.8 A ten-stage axial flow compressor provides a total head pressure ratio of 5:1 with an overall total and isentropic efficiency of 87% while the inlet total head temperature is 288 K. The work is divided equally between the stages and the work done factor is 0.85. Find the air angles of a stage at the design radius where the blade speed is 215 m/s. Assume the axial velocity to be constant throughout the stage at 168 m/s and the degree of reaction to be 80%. Draw the velocity triangles and describe what happens to the air when it passes through the stage. 5.9 An axial flow compressor develops pressure ratio of 1:2 in the first stage. The inlet conditions are 1.03 bar and 315 K. The overall efficiency of the compressor including bearing losses is 83%. The axial velocity is 0.47 times the blade velocity. The velocity diagram is symmetrical and the change in velocity of whirl at the mean radius is 0.5 times the axial velocity. Estimate the blade speed required and the absolute velocity of the air leaving the stationary inlet guide vanes. 5.10 All the stages of an axial flow compressor have the following conditions at the mean diameter. Degree of reaction 50%, blade peripheral speed 200 m/s. Angles of absolute velocities at inlet to exit from the rotor measured from the axial direction are 15° and 45° respectively. A stage efficiency of 88% and work done factor of 0.86 are to be taken as constant throughout the compressor. The ambient air pressure and temperature are 1 bar and 288 K respectively. Calculate, (a) The static temperature of air at inlet to the first stage. (b) The pressure ratio of the first stage. (c) The static pressure of the air at exit, from the tength stage.

6

Steam and Gas Turbines

6.1 INTRODUCTION Steam turbine is a power-generating machine in which the pressure energy of the fluid is converted into mechanical energy. This conversion of energy is due to the dynamic action of steam flowing over the blades. These blades are mounted on the periphery of a rotating wheel in the radial direction. Today the steam turbine stands as one of the most important prime movers for power generation. It converts thermal energy into mechanical work by expanding high pressure and high temperature steam. The thermal efficiency of steam turbine is fairly high compared to steam engine. The uniform speed of steam turbine at wide loads makes it suitable for coupling it with generators, centrifugal pumps, centrifugal gas compressors, etc. The principle of energy extraction from the gas is one of gradually reducing the high pressure energy by converting it into K.E. This is accomplished by passing the gas alternatively through rows of fixed time moving blades. The K.E. of the gas is reduced in the moving blades, which are attached to the turbine hub and recovered in the fixed stationary blades attached to the casing. This necessitates a gradual density decrease as the gas moves through the turbine and the blade height therefore increases towards the low pressure end, if a constant axial flow velocity is to be maintained through the turbine. The stator row is often termed the nozzle row.

6.2 CLASSIFICATION OF STEAM TURBINES Steam turbines are classified into impulse turbines and reaction turbines (or impulse reaction). Examples of impulse turbines are De Laval, Curtis, Moore, Zoelly, Rateau, etc. Examples of reaction turbines are Parson, Ljungstrom, etc. 418

Steam and Gas Turbines

419

6.3 PRINCIPLE OF OPERATION OF STEAM TURBINES 6.3.1 Impulse Turbine Impulse or impetus means sudden tendency of action without reflexes. Figures 6.1(a), (b) and (c) show a single-stage impulse turbine. This turbine is called “Simple” impulse steam turbine since the expansion of the steam takes place in one set of the nozzles. A single-stage impulse turbine consists of a set of nozzles and moving blades. High pressure steam at boiler pressure (pb) enters the nozzle and expands to low back pressure (condenser pressure) in the nozzle. Thus, the pressure energy is converted into kinetic energy increasing the velocity of steam (Figure 6.1(d)). The rapidly moving particles of steam (high velocity steam) are then directed on to a series of blades where the kinetic energy is parly absorbed and converted into an impulse force by changing the direction of flow of steam (the blades are shaped in such a way that, there is a change in the direction of the steam flow without changing its pressure) which gives rise to a change in momentum and therefore to a force. This sets the blades in motion. The velocity of steam decreases as it flows over the blades but the pressure remains constant, i.e. the pressure Nozzle

Fixed nozzle

Disk

Nozzle

Steam

Blades Bearing Steam from boiler

Blade

Disk Blade

(a)

(b) Nozzle

Runner

V1

Velocity Pressure

Boiler pressure or pressure of steam entering

Velocity of steam leaving or exit velocity (V2) Velocity of steam entering

Condenser pressure

(c)

Figure 6.1

Impulse turbine.

420

Fundamentals of Turbomachinery

at the outlet side of the blade is equal to that at the inlet side. Such a turbine is termed impulse turbine. The processes of expansion and direction changing may occur once, or a number of times in successions. The final velocity is much higher than the inlet velocity to the nozzles in the case of the single-stage turbine. Hence, there is considerable loss in K.E.

6.3.2 Reaction Turbine High-pressure steam is directly passed on the blades which also act as nozzles. The pressure of steam continuously drops as it flows through the nozzles and the velocity increases. The steam leaving the blades will exert a reactive force in the backward direction of its flow. This reactive force sets the blades in motion. This is called pure reaction. This type of turbine is no longer used. See Figure 6.2.

Figure 6.2

Pure reaction turbine.

6.3.3 Impulse Reaction Turbine In the impulse reaction turbine, power is generated by the combination of impulse action (similar to impulse turbine) and reaction (similar to pure reaction turbine) by expanding the steam in both fixed blades (act as nozzles) and moving blades as shown in Figure 6.3. The pressure of the steam drops partially in fixed blades and partially and continuously in moving blades. The blades of impulse and impulse reaction turbine are shown in Figure 6.4. There is a drop in pressure while flowing through the asymmetrical blades, hence the relative velocity of the steam increases. Figure 6.3 shows an axial flow impulse reaction turbine having two rows of fixed (attached to the casing) and two rows of moving blades (attached to the shaft). Steam is admitted throughout the circumference (alround admission) whereas in the case of impulse turbine, steam is admitted through a set of nozzles. Steam enters the fixed row of blades, undergoes a small drop in pressure and increases in velocity, then similar to an impulse turbine, steam enters the first row of moving blades. It undergoes a change in direction and momentum, this sets up an impulse to the blades and there is a small drop in pressure too, giving rise to increase in kinetic energy. The pressure drop gives rise to reaction. Hence the name impulse reaction turbine.

Steam and Gas Turbines

421

Fixed Moving Fixed Moving

Pressure Boiler pressure or pressure of steam entering Condenser pressure Velocity of steam entering

Figure 6.3

Figure 6.4

Velocity

Velocity of steam leaving

Impulse reaction turbine.

Cross section of blades of steam turbine: (a) Impulse blade. (b) Reaction blade.

6.4 IMPULSE STAGING AND NEED FOR COMPOUNDING The exit velocity of steam from the nozzle is very high if the steam pressure drops from boiler pressure to condenser pressure in a single stage. In such a situation, the turbine speed will be of the order of 30000 rpm. Few driven machines require such a high speed. However, most driven machines run at speeds around a few thousand revolutions per munute (rpm). Atmost gear trains with large speed ratios must be used if a suitable matching between the turbine speed and the driven component speed is required. It will be seen that the velocity of steam leaving the moving bodies is high. Therefore, the exit velocity involves a loss of energy and this is termed ‘carry-over loss’ or ‘leaving loss’. Always a small velocity has to be maintained to reduce the carry-over loss. In actual De Laval turbines, the “leaving velocity” may amount to about 33% of the nozzle outlet velocity. Hence, the “leaving loss” may reach about (0.33)2 or 11% of the initial kinetic energy of the steam. There is also a danger of structural failure of the runner blades due to centrifugal stresses at very high speeds. Normally the blade speed is limited to 400 m/s. In order to overcome all these demerits, it is advisable to reduce the turbine speed by the method of compounding. Absorption of inlet energy in multiple rows of moving blades is known as compounding. If inlet energy is in K.E. form, then it is called velocity compounding. If the inlet energy is in pressure form (enthalpy form), then it is called pressure compounding.

422

Fundamentals of Turbomachinery

The characteristics of simple impulse turbine are, a set of nozzles with the high-pressure steam entering the nozzles, leaving with high steam velocity leading to high blade velocity with consequent high speed of rotation and lastly, a high carry-over loss. An example of this is the De Laval turbine. The required blade tip speeds are obtained in impulse turbines by the method termed compounding. Hence, the compounding is the method of reducing the blade speed for a given overall pressure drop. Hence, multiple rotors in series are keyed on a common shaft and the jet velocity is absorbed in stages as it flows over the blades.

6.5 METHODS OF COMPOUNDING OF STEAM TURBINE Following are the methods of compounding of steam turbines: 1. Velocity compounding 2. Pressure compounding 3. Pressure and velocity compounding

6.5.1 Velocity Compounding Nozzle

Moving

Fixed

Moving

Fixed

pboiler Exit velocity V1

Figure 6.5

pcondenser

Velocity compounded impulse turbine.

Nozzle are fitted to the stationary casing. Each fixed blade row is fitted between the moving blades. The function of the fixed blade is to direct the steam coming from the first moving row to the next moving row without appreciable change in velocity. The whole of the pressure drop occurs in one set of stationary nozzles. The kinetic energy of steam gained in the nozzle is successively absorbed by moving rows and the steam is exited from the last row with very low velocity. The steam leaves axially from the last row. Due to this, the rotor speed decreases considerably. The turbine working on this principle is called the velocity compounded turbine (Figure 6.5). An examples is that of Curtis turbine. The velocity compounded impulse turbine is also called the Curtis stage.

Steam and Gas Turbines

423

6.5.2 Pressure Compounding A pressure compounded impulse turbine is shown in Figure 6.6. A number of simple impulse stages (one set of nozzles and one set of moving blades) arranged in series are known as pressure compounding. Here the turbine is provided with rows of fixed blades which act as nozzles at the entry of each row of moving blades. The total pressure drop of the steam does not take place in the first row of nozzles but is divided among all the rows of fixed blades which act as nozzles. Each of the simple impulse turbine is named “stage“ of the turbine. This arrangement is equivalent to splitting up the whole pressure drop into a series of smaller pressure drops, hence the term “pressure compounded”. Nozzle

Moving Fixed Moving Fixed Moving

Exit velocity

pboiler

Pressure Vi

Figure 6.6

pcondenser

Pressure compounded impulse turbine.

The steam leaving the boiler enters the first row of fixed blades, i.e. nozzles in which it is partially expanded. Steam with comparatively high velocity is then passed over the first row of moving blades where almost all its kinetic energy is absorbed. This completes the expansion of steam in one stage (work is done in one stage). Steam leaving the rotor of the first stage is still having pressure energy. Now, the steam enters the stator (fixed blade) of the next stage. The fixed blades of pressure compounding act as nozzles. Therefore, steam partially expands in the fixed blades, kinetic energy increases and enters the moving blades. Here again the kinetic energy is absorbed. This completes expansion in two stages. This process continues until steam reaches the condenser pressure in the last stage and leaves axially from the last row. In each stage, a small quantity of pressure energy is converted into kinetic energy and is absorbed in that stage only. Hence, the steam velocity becomes much smaller and reduces the blade (blade tip) velocities and rotational speed.

424

Fundamentals of Turbomachinery

6.5.3 Pressure and Velocity Compounding The pressure and velocity compounding turbine is shown in Figure 6.7. This arrangement is in two stages. Each stage has a two-stage velocity compounded turbine. Consider stage I and stage II individually. They are identical to velocity compounding but both the stages taken together will mean that the pressure drop occurs in the two stages. This arrangement is very popular for its simple construction, but the efficiency is low and hence seldom used.

Figure 6.7

Pressure and velocity compounded impulse turbine.

Steam leaving the boiler with high pressure enters a set of nozzles of the first stage. The pressure drops partially in the nozzles, and steam enters the first set of rotors of first stage with very high velocity. In the first set of rotors, kinetic energy is absorbed partially with no change in pressure energy. Then, steam leaves the rotors and enters the stators (fixed blades). Here, theoretically, pressure energy and kinetic energy will remain constant. Then, steam with comparatively high kinetic energy enters the second set of rotors of first stage. Here, the remaining kinetic energy is absorbed. This completes expansion in one stage. Now, steam with a comparatively high pressure energy enters the fixed blades of the second stage which act as nozzles. Here the pressure drops once again, and is converted into kinetic energy. This kinetic energy is absorbed in the next two stages of the velocity compounded turbine. Finally, the steam will leave the last stage axially at condenser pressure with low kinetic energy. An example of this type of turbine is the Curtis turbine.

Steam and Gas Turbines

425

6.6 DIFFERENCES BETWEEN IMPULSE AND REACTION TURBINES Impulse turbine

Reaction turbine

• Complete expansion of the steam takes place

• Partial expansion of the steam takes place in

in the nozzle, hence steam is ejected with very high kinetic energy.

the nozzle (fixed blade) and further expansion takes place in the rotor blades.

• Blades are symmetrical in shape.

• Blades are non-symmetrical in shape, i.e. aerofoil section.

• No change in pressure between the ends of the

• Pressure drops from inlet to outlet of the blade,

moving blade, i.e. the pressure remains constant between the ends of the blade.

i.e. difference in pressure exists between the ends of the moving blade.

• Low efficiency, i.e. part load efficiency is

• More flattened efficiency curve, hence part

poor.

load efficiency is good.

• High speed

• Relativity low speed

• Less floor area for the same power generation,

• More floor area for the same power generation,

hence compact

hence bulky.

• Used for small power generation

• Used for medium and large power generation.

• Less stages for the same power generation

• More stages for the same power generation.

6.7 ADVANTAGES OF STEAM TURBINE OVER OTHER PRIME MOVERS •

• • • • • • •

Expansion of the working medium (steam) up to the lowest possible pressure in the steam turbine results in more work output. Hence the thermal efficiency of the turbine is high. Uniform power generation in steam turbine, hence no flywheel. High and a wide range of speeds is possible in case of steam turbine. Designed for a wide range of power generation (few kW to 1000 MW) output. No reciprocating parts, hence balancing is easy. Less wear and tear, hence internal lubrication is not required leading to prolonged life. Less noise and vibration and hence reduction in the cost of foundation. The specific steam consumption is less in case of steam turbine.

6.8 VELOCITY TRIANGLES FOR IMPULSE TURBINE Figure 6.8 shows a rotor blade with separate inlet and outlet velocity triangles, whereas Figure 6.9 shows the combined inlet and outlet velocity triangles.

426

Fundamentals of Turbomachinery

Figure 6.8 Velocity triangles for an axial flow impulse stage.

where V1 = inlet absolute velocity at the inlet of moving blade or steam velocity at the outlet from nozzles V2 = absolute velocity at the outlet of moving blade Vr1 = relative velocity of steam to moving blade at entrance Vr2 = relative velocity of steam to moving blade at exit Vw1 = tangential component of absolute velocity at entrance to the rotor or velocity of whirl at the entrance of moving blade = V1 cos a1 Vw2 = tangential component of absolute velocity at outlet or velocity of whirl at the outlet of moving blade Vf1 and Vf2 = velocity of flow at entry and exit of the moving blade = radial flow velocity (for radial turbine) = axial flow velocity (for axial turbine) a1 = angle made by the absolute velocity V1 with the plane of moving blades = nozzle or jet angle or fixed blade angle at inlet a2 = Angle made by the absolute velocity V2 with the plane of moving blades = inlet angle of fixed blade (inlet angle to fixed blade) b1 = blade angle at inlet of the rotor b2 = blade angle at exit of the rotor u = circumferential or tangential velocity of the blade

Steam and Gas Turbines

ms d h r gc K

= = = = = =

427

mass flow rate of steam kg/s mean diameter of blade drum height of blade speed ratio = u/V1 1 kg-m/N-s2 blade velocity coefficient = Vr2/Vr1

Due to expansion of steam in nozzles, steam issues with an absolute velocity V1 at an angle a1 with the plane of the moving blades. If the blade peripheral velocity is u, then the relative velocity of steam with respect to the casing is Vr1. For smooth surfaced blades, friction losses are small or zero, hence, Vr2 = Vr1.

Figure 6.9

Combined velocity diagram.

6.9 PERFORMANCE PARAMETERS OF IMPULSE TURBINE 1. Power developed (P): W.D. [Vw1  ( Vw 2 )]

u gc

(Vw1  Vw 2 )

1 Ëm m Û m Ì s  s Ü s – 1 kg-m Í Ý N-s2 W.D.

(Vw1  Vw 2 ) u kJ 1000 gc kg

m2 s2

–

u J gc kg 1 N-m J = = kg-m kg kg N- s2

(6.1)

kg (Vw1  Vw 2 ) u kJ – s 1000 – gc kg

or

P

ms

\

P

ms – (Vw1  Vw 2 ) u kJ 1000 – gc s

(6.2)

428

Fundamentals of Turbomachinery

2. Diagram or blading or rotor or vane efficiency (hb): Kb

or

Kb

Work done on blades per kg of steam Energy supplied per kg of steam

(6.3)

Power developed per kg of steam Energy supplied per kg of steam (total energy supplied per kg of steam) Work output from the blades/kg of steam K.E. of steam at inlet to the blades (rotors)/kg of steam

Total energy supplied per kg of steam, [V12  (Vr21  Vr22 )] / 2 gc

Kb

(Vw1  Vw 2 ) u / gc V12

V12 / 2 gc

(If Vr1 = Vr2)

Work done/kg of steam Actual enthalpy drop/kg of steam

/ 2 gc

2(Vw1  Vw 2 ) u

(6.4)

V12

Equations (2.43a) and (6.4) are same. Hence, the utilization factor and the blading efficiency are same. 3. Nozzle efficiency (hn): Kn

or

Kn

Actual enthalpy drop in the nozzle Isentropic enthalpy drop in the nozzle [Theoretical enthalpy drop in the nozzle, ('h) theoretical ]

( 'h)actual ( 'h)theoretical

V12 2 gc ( 'h) theoretical

(6.5)

4. Gross or stage efficiency (hS): KS

Work done per kg of steam Theoretical enthalpy drop in the nozzle per kg of steam (Vw1  Vw 2 ) u gc Work done per kg of steam Available energy at the stage ( 'h) theoretical

or

KS

\

KS

i.e.

hS = hb × hn

(6.5a)

Work done by steam/kg of steam Actual enthalpy drop/kg of steam – Actual enthalpy drop/kg of steam Theoretical enthalpy drop/kg of steam

(6.6)

5. Axial thrust (Fa):

Fa

(V f 1  V f 2 ) ms gc – 1000

kN

(6.7)

Steam and Gas Turbines

429

6. Energy lost in the blade passage due to friction (ELf): ELf

(Vr21  Vr22 ) ms kW 2 gc – 1000

(6.8)

7. Energy lost in exit (energy equivalent to exit absolute velocity) (ELe): ELe

V22 ms kW 2 gc – 1000

(6.9)

6.10 EFFECTS OF FRICTION AND BLADE ANGLES ON BLADE EFFICIENCY In an impulse turbine, the relative velocity at the outlet must be equal to the relative velocity at the inlet. However, due to friction, leakage and turbulences, the relative velocity at the outlet is slightly reduced compared to that at the inlet. Hence we can define, K = blade velocity coefficient = Vr2/Vr1

(6.10)

For impulse turbine, from Eq. (6.4) and Figure 6.9, 2(Vw1  Vw 2 ) u Kb V12 (Vw1 + Vw2) = AE + AF = FB + EB = EB + FB = Vr1 cos b1 + Vr2 cos b2

(6.10a)

Ë V cos E2 Û Vr1 cos E1 Ì1  r 2 Ü Vr1 cos E1 Ý Í = Vr1 cos b1 (1 + KC)

where

\ \

C

cos E2 cos E1

Vr1 cos b1 = (V1 cos a1 – u) (Vw1 + Vw2) = (V1 cos a1 – u) (1 + KC)

Kb

2u (V1 cos D1  u)(1  KC ) V12 È uØ 2u V1 É cos D1  Ù (1  KC ) V1 Ú Ê V12

Let \

f = speed ratio =

u V1

hb = 2f (cos a1 – f)(1 + KC)

(6.10b) (6.10c) (6.11)

430

Fundamentals of Turbomachinery

6.11 CONDITION FOR MAXIMUM EFFICIENCY Differentiating hb (Eq. (6.11)) with respect to f and equating to zero to get the condition (fopt) for maximum blade efficiency, d (Kb ) dI

\

d [2I (cos D1  I ) (1  KC )] dI

0

cos a1 – 2f = 0

or

cos D1 2

Iopt

u V1

(6.12)

This is the condition for maximum blade efficiency. Substituting

I

\ If then

cos D1 in Eq. (6.11) to get Kb max , 2

Iopt

Kb max

2 cos D1 Ë cos D1 Û cos D1  (1  KC ) Ì 2 2 ÜÝ Í

Kb max

cos2 D1 – (1  KC ) 2

K = 1, C = 1 hb max =

cos2 a

(6.13)

(equiangular blades, b1 = b2) (6.14)

1

Equations (6.14) and (2.50) are same. From Eq. (6.1), we have

or

W.D. =

(Vw1  Vw 2 ) u kJ/kg gc – 1000

W.D. =

(V1 cos D1  u)(1  KC ) u kJ/kg gc – 1000

Substituting cos a1 = 2f in the above equation to get maximum work developed per kg of steam, W.D max

(V1 2I  u)(1  KC ) u kJ/kg gc – 1000

Ë Û u ÌV1 – 2 V  u Ü (1  KC ) u Í Ý 1 gc – 1000

If

K = 1, C = 1

Steam and Gas Turbines

then,

W.D max

2u 2 kJ/kg gc – 1000

431 (6.15)

The exit absolute velocity should be minimum to get the maximum blade efficiency. This is possible only when steam discharges axially from the rotor, a2 = 90°,

i.e.

\ Vw2 = 0,

V2 = Vf 2

(Vw1 + Vw2) = Vw1 = V1 cos a1 = V1 × 2f or

Vw1 = V1 × u × 2/V1 = 2u

(6.16)

The variation of f vs. hb is shown in Figure 6.10. From Eqs. (6.1), (6.10b) and (6.10c), we get IV12 (cos D1  I ) (1  KC ) 1000 – gc

(6.16a)

From Eqs. (6.2), (6.10b) and (6.10c), we get P

ms I V12 (cos D1  I ) (1  KC ) 1000 – gc

hb or W.D. or P

W.D.

2

(6.16b)

0

cos a1

cos a1 2 f=u V1

1

Substituting f = 0 and f = 1 (means, u = V1, hence a1 = 0) Figure 6.10 Blade efficiency in Eqs. (6.11), (6.16a) and (6.16b), we get vs. speed ratio. hb = 0, W.D. = 0, P = 0 Substituting f = cos a1/2 in Eqs. (6.11), (6.16a) and (6.16b), we get hb = hbmax, W.D. = (W.D.)max, P = Pmax

6.12 MULTISTAGE IMPULSE TURBINE In order to reduce high blade tip speeds, carry-over losses, high centrifugal stresses, etc. a multistage turbine is essential. Figure 6.11 shows the combined velocity triangles for both the stages. For maximum efficiency and minimum carry-over loss, the absolute velocity of steam leaving the second row must be axial i.e. V4 = Vf4, a4 = 90°. The magnitude of absolute velocity of steam leaving the first row (V2) and entering the second row remains the same. Therefore, we have V2 = V3,

a2 = a3

W1 = work done per kg of steam in first stage u (Vw1  Vw 2 ) gc (Vr1 cos E1  Vr 2 cos E2 ) u gc

If

b1 = b2, equiangular blades,

(From Eq. (6.10a)) then

K=1

432

Fundamentals of Turbomachinery Vw 1

Vw 2 F

A a2

u a1

B

E

b2

b1

V2

Vf 2

Vr 1 Vr 2

Vf1

V1

D

C

Figure 6.11

\

b4

a3

Vf 4 = V4

a4

Vw 3

u

G

Vr 4

V3

H b3

Vr 3

I

Vf 3

Combined velocity diagram for multistage impulse turbine.

2Vr1 cos E1u gc

W1

2u (V1 cos D1  u) J/kg gc

W2 = work done per kg of steam in second stage

or

W2

u(Vw3  Vw 4 ) gc

W2

uVw3 gc

(Q a4 = 90° \ Vw4 = 0)

Vw3 = GH + HI = (Vr3 cos b3 + Vr4 cos b4) For equiangular blades, E3

E4 , K2

Vr 4 Vr 3

1

Vw3 = 2Vr3 cos b3 = 2(V3 cos a3 – u) \

W2

2u (V3 cos D 3  u) gc

Steam and Gas Turbines

If

433

a2 = a3 and V2 = V3, then we get V3 cos a3 = V2 cos a2 = Vr2 cos b1 – u

(b2 = b1, K = 1)

= Vr1 cos b1 – u = (V1 cos a1 – u) – u or \

V3 cos a3 = V1 cos a1 – 2u W2

2u (V1 cos D1  2u  u) gc 2u (V1 cos D1  3u) gc

Wt = total work done = W1 + W2 2u (V1 cos D1  u) 2u (V1 cos D1  3u)  gc gc

2u (V1 cos D1  u  V1 cos D1  3u) gc 2u (2V1 cos D1  4u) gc

or

Wt

Kb 2 S

4u (V1 cos D1  2u) gc

Wt V12 2 gc

(6.17)

blade efficiency for two stages

4u 1 (V1 cos D1  2u) – 2 gc V1 2 gc

8u (V1 cos D1  2u) V12 È 2u Ø 8u V1 É cos D1  Ù V1 Ú Ê V12

\

hb 2S = 8f (cos a1 – 2f)

(6.18)

434

Fundamentals of Turbomachinery

6.13 CONDITION FOR MAXIMUM EFFICIENCY FOR TWO-STAGE IMPULSE TURBINE Differentiating hb 2S (Eq. (6.18)) with respect to f and equating to zero to get the condition for maximum efficiency, we get

dKb 2 S dI or \

\

d [8I (cos D1  2I )] dI

0

8f cos a1 – 32f = 0

cos D1 4 Substituting Eq. (6.19) in Eq. (6.18),

(6.19)

Iopt

Kb 2 S max

8 – cos D1 4

Kb 2 S max

cos2D1

2 cos D1 Û Ë Ì cos D1  Ü 4 Í Ý

(6.20)

Substituting Eq. (6.19) in Eq. (6.17), we get Wt max

4u (V – 4I  2u) gc

(for two-stage impulse)

Û u 4u Ë  2u Ü Ì V1 4 gc ÍÌ V1 ÝÜ

8u2 gc

(6.21)

6.13.1 Utilization Factor Refer to Section 2.9.1 and Eqs. (2.38) to (2.39c), (2.42), (2.43), (2.44), (2.45), (2.47) and (2.48).

6.13.2 Degree of Reaction Refer to Section 2.5 and Eqs. (2.12), (2.13), (2.14), (2.15), (2.40) and (2.41). For maximum utilization and maximum blading efficiency. Iopt

u V1

cos D1 2

(Eqs. (6.12) and (2.51)]

[Conditions b1 = b2, K = 1, a2 = 90°, Vf1 = Vf2 = V2 = V1 sin a1, Vw2 = 0] Substituting the above conditions in Eq. (6.1), we get the maximum work done as Wmax

(Vw1  Vw 2 ) u gc

uVw1 gc

uV1 cos D1 gc

Steam and Gas Turbines

or

Wmax

IV12 cos D1 gc

2I 2V12 gc

(for single-stage impulse turbine)

435 (6.22)

6.14 ADVANTAGES AND DISADVANTAGES OF VELOCITY COMPOUNDING Advantages: • Maximum possible pressure energy is converted into kinetic energy in nozzles of first stage. No more pressure drop takes place in the next stages. This reduces the stress in the turbine housing. • Fewer number of stages are required due to large kinetic energy drop in each stage. This reduces the initial cost of the turbine. • The space required is small due to compactness. • Turbine housing is meant just to avoid mechanical accidents. Hence its need not be made strong. Disadvantages: • The friction losses are more due to the high velocity of steam. • The maximum blade efficiency and utilization factor decrease with the increase in the number of stages. • Power developed in each row keeps on decreasesing. More number of stages means that more space, material and maintenance work are required. • Specific volume of steam increases continously with stages, hence the design and fabrication of blades are complicated.

6.15 IMPULSE REACTION TURBINES OR REACTION TURBINES The velocity triangle for the reaction turbine is shown in Figure 6.12. Due to expansion of steam in blades, the relative velocity at the outlet is more than that at the inlet. All the equations of the impulse turbine hold good for the reaction turbine.

Figure 6.12

Velocity triangles for the reaction turbine.

436

Fundamentals of Turbomachinery

We have the degree of reaction R from Eq. (2.12), R

Energy transfer resulting in a change of static pressure in the rotor Total energy transfer in the stage Enthalpy drop in moving blades Enthalpy drop in the stage (work done by the steam in the stage) 'h2 'h1  'h2

(6.23)

Dh1 = enthalpy drop in fixed blade Dh2 = enthalpy drop in moving blade (Vr22  Vr21 ) 2 gc

(6.24)

Dh1 + Dh2 = work done by the turbine (each stage) u (Vw1  Vw 2 ) gc

(6.25)

Substituting Eqs. (6.24) and (6.25) in (6.23), R

(Vr22  Vr21 ) 1 – u (Vw1  Vw 2 ) 2 gc gc

(Vr22  Vr21 ) 2u (Vw1  Vw 2 )

(6.26)

Substituting u1 = u2 = u (i.e. centrifugal energy neglected) and (V12  V22 )  (u12  u22 )  (Vr22  Vr21 ) 2 gc

R

u(Vw1  Vw 2 ) in Eq. (2.13), gc

(Vr22  Vr21 ) 1 – u V ( 2 gc w1  Vw 2 ) gc (Vr22  Vr21 ) 2u (Vw1  Vw 2 )

Equations (6.26) and (6.27) are same. From Figure 6.12 and if Vf 1 = Vf 2 = Vf , we get Vr2 = Vf cosec b2 Vr1 = Vf cosec b1 Vw = Vw1 + Vw2 = Vf cot b1 + Vf cot b2

(6.27)

Steam and Gas Turbines

437

Substituting the above data in Eq. (6.27),

R

V f2 (cosec 2 E 2  cosec2 E1 ) 2u V f (cot E1  cot E 2 ) [(cot 2 E2  1)  (cot 2 E1  1)]V f 2u (cot E1  cot E 2 )

V f (cot 2 E2  cot 2 E1 ) 2u (cot E1  cot E2 ) V f (cot E2  cot E1 ) (cot E 2  cot E1 ) 2u

(cot E1  cot E2 )

Vf

(cot E 2  cot E1 ) 2u If R = 50%, then Eq. (6.28) becomes R

R

or

0.5

Vf 2u

(6.28)

(cot E 2  cot E1 )

(6.29)

u V f (cot E2  cot E1 )

From Figure 6.12, we have

\

cot D 2

Vw 2 ; Vf

\

cot E 2

(Vw 2  u) ; Vf

\

Vw 2

V f cot D 2

(Vw 2  u) V f cot E 2

(Vf cot a2 + u) = Vf cot b2 u = Vf (cot b2 – cot a2)

or

(6.30)

Similarly, u = Vf (cot a1 – cot b1)

(6.31)

Comparing Eqs. (6.29), (6.30) and (6.31), we have b2 = a1, i.e.

If

R = 50%

b1 = a2 and

Vf 1 = Vf 2 = Vf

then the moving blade and the fixed blade must have the same shape, i.e. symmetrical velocity triangles. This type of turbine is called the Parson’s reaction turbine.

438

Fundamentals of Turbomachinery

6.16 CONDITION FOR MAXIMUM EFFICIENCY FOR REACTION TURBINE Assuming R = 50% and symmetrical triangles (a1 = b2 ; a2 = b1 ; V1 = Vr2 ; V2 = Vr1), from Figure 6.12 and Eq. (6.1), we have W.D.

u (Vw1  Vw 2 ) gc u (V1 cos D1  Vr 2 cos E2  u) gc u (V1 cos D1  V1 cos D1  u) gc u (2V1 cos D1  u) gc È u uØ V1 É 2 cos D1  Ù gc V1 Ú Ê

W.D.

V12 (2I cos D1  I 2 ) gc

Ë Ì' I Í

uÛ Ü V1 Ý

(6.31a)

V12 = kinetic energy supplied to the fixed blade 2 gc

Dh2 = actual kinetic energy supplied to the moving blade = actual enthalpy drop in the moving blade From Eq. (2.41), we have 'h2

(Vr22  Vr21 ) 2 gc

R(V12  V22 ) (1  R)2 gc

(6.31b)

'h2 = theoretical enthalpy drop in rotor (moving blade) „

(Vr22  Kco Vr21 ) 2 gc Kn R = 50%

R(V12  Kco V22 ) (1  R)2 gcKn

(V12  Kco V22 ) 2 gcKn

(6.32)

hco = carry-over efficiency hn = nozzle efficiency or stator efficiency Symmetrical triangles (Vr1 = V2, Vr2 = V1, a1 = b2, a2 = b1, Vf1 = Vf2) Dh1 = actual enthalpy drop in the fixed blade (V12  V22 ) 2 gc

(6.32a)

Steam and Gas Turbines

439

'h1 = theoretical static enthalpy drop in fixed blade „

(V12  Kco V22 ) 2 gcKn

(6.33)

Equations (6.32) and (6.33) are same. 'h2

„

„

(V12  KcoV22 ) 2 gcKn

Vr21

V12  u2  2uV1 cos D1

Vr21

V12 (1  I 2  2I cos D )

'h2

or

'h1

„

'h1

„

(V12  KcoVr21 ) 2 gcKn

(Q f = u/V1)

V12  KcoV12 (1  I 2  2I cos D1 ) 2 gcKn

V12 [1  Kco (1  I 2  2I cos D1 )] 2 gcKn

\

'h2  'h1 „

„

V12 (1  Kco (1  I 2  2I cos D1 )) gcKn

(6.33a)

hS = hg = Gross stage efficiency or stage efficiency Work done by the rotor Total static enthalpy drop (Theoretical energy available in the stage)

(6.33b) Substituting Eqs. (6.31a) and (6.33a) in (6.33b), KS

Kg

V12 (2I cos D1  I 2 ) gc V12 [1  Kco (1  I 2  2I cos D1 )] gc Kn Kn (2I cos D1  I 2 ) [1  Kco (1  I 2  2I cos D1 )]

Now,

hg = hb × hn

\

Kb

(2I cos D1  I 2 ) [1  Kco (1  I 2  2I cos D1 )]

2I cos D1  I 2 1  Kco  KcoI 2  2KcoI cos D1

(6.34)

440

Fundamentals of Turbomachinery

(2I cos D1  I 2 )

(6.35)

(1  Kco )  Kco (2I cos D1  I 2 )

\

1 (1  Kco )

Kb

(2I cos D1  I 2 )

 Kco

Differentiating, hb, w.r.t. f and equating to zero to get the condition for maximum blading efficiency, dKb dI

\

d (2I cos D1  I 2 ) dI

0 ; ? 2 cos D1  2I

0

fopt = cos a1 = u/V1

(6.35a)

Equations (6.12) and (6.35a) are same. Substituting Eq. (6.35a) condition in (6.35), Kb max

(2 cos D1 cos D1  cos2D1 ) 1  Kco (1  cos2D1  2 cos2D1 ) cos2D1 1  Kco (1  cos2D1 )

Case 1 If

hco = 1, then Eq. (6.36) becomes hb max = 1

Case 2 If

(6.36b)

hco = 0 (zero carry over), then Eq. (6.36) becomes hb max = cos2 a1

Case 4 If

(6.36a)

hn = 1 and hco = 1, then Eq. (6.34) becomes hg max = 1

Case 3 If

(6.36)

(6.36c)

hn = 1, hco = 0, then Eq. (6.34) reduces to hg max = cos2 a1

(6.36d)

Assuming no entry kinetic energy to the fixed blade (V2 = 0) and R = 50%, Eqs. (6.31b) and (6.32a) become 'h1

V12 2 gc

(6.37)

Steam and Gas Turbines

'h2

\

(V12  V22 ) 2 gc

(Vr22  Vr21 ) 2 gc

V12 2 gc

441 (6.38)

e = Utilization factor Work done by the stage Energy at inlet ('h1 ) + Energy due to enthalpy drop in the rotor ('h2 )

Equation (6.31a) Equation (6.37) + Equation (6.38)

or

H

u (2V1 cos D1  u) 1 – 2 gc V1 V2  1 2 gc 2 gc u (2V1 cos D1  u) gc –

V12 gc

È uØ u V1 É 2 cos D1  Ù V1 Ú Ê V12

e = 2f cos a1 – f2

\

(6.39)

Note: We know that utilization factor e and blading efficiency hb are same. Still Eqs. (2.43g) and (6.39) are different. This is due to, V2 = 0 in Eq. (6.39) but V2 ¹ 0 in Eq. (2.43g). Diferentiating Eq. (6.39) w.r.t. f and equating to zero to get the condition for maximum utilization for Parson’s reaction turbine, dH dI

d (2I cos D1  I 2 ) dI

2 cos a1 – 2f = 0 ;

or

\

0

fopt = cos a1

(6.40)

Substituting this condition (fopt = cos a1) in Eq. (6.39), emax = (2 cos a1 cos a1 – cos2 a1) = cos2 a1 Case I If

hco = 0, then hb max = emax, i.e.

Equation (6.36c) = Equation (6.41)

(6.41)

442

Fundamentals of Turbomachinery

Case II hn = 1, hco = 1, then

If

hg max = emax, i.e. Equation (6.36d) = Equation (6.41) or see Eq. (2.43d) and (2.43g).

6.17 REHEAT FACTOR (RF) From Chapter 3, we have T or h

T or h 01

p1

p01 01 p02

02 p02 02¢ 03¢

s (a)

03²

03 s

(b)

Figure 6.13

T02„

T(h)–s diagram.

Ëp Û T01 Ì 02 Ü Í p01 Ý

J

1 J

\

J 1 Ë Û J Ü Ë Û p Ì 02 T01  T02„ T01 Ì1  Ì Ü Ü p ÌÍ Í 01 Ý ÜÝ J 1 Ë Û Ì Ë p02 Û J Ü E Ì1  Ì Ü Ü p ÌÍ Í 01 Ý ÜÝ T01 – T02 = T01bhg

\

T02 = T01 (1 – bhg)

where

p03

02¢

02

(6.42)

T01 E

(6.43)

(6.44)

(6.45)

hg = stage efficiency or gross stage efficiency. Similarly, T01  T03

„

\ \ \

T02 E

T02 – T03 = T02 hgb T03 = T02 (1 – bhg)2 W = cp hg b (T01 + T02 + T03 + T04 + ... + T0n)

(6.46) (6.47)

Steam and Gas Turbines

or

W

c p Kg E T01 [1  (1  Kg E )  (1  Kg E )2  ...  (1  Kg E )n 1 ]

443 (6.47a)

n

c pT01 [1  (1  Kg E )n ] Kg Ç WSi

(6.48)

i 1

where

WSi = total isentropic work output in single stage from p01 to p0(n+1)

Available enthalpy drop can be obtained by substituting hg = 1 in the above equation. \

WS = cpT01 [1 – (1 – b)n] n

or

WS

J 1 Ë Û Ë J Ì c pT01 1  pR Ü ; where Ì pR ÌÍ ÜÝ ÌÍ

RF

Cummulative enthalpy drop Ideal enthalpy drop in single stage from P01 to P0(n+1)

Ë p02 Û Û Ì ÜÜ Í p01 Ý ÝÜ

(6.49) (6.50)

n

Ç WSi i 1

WS

W/Kg WS

W KgWS

Ko Kg

(6.51)

where ho = overall turbine efficiency W = sum of actual enthalpy drop in each stage.

6.18 BLADE DESIGN PARAMETERS The important blade design parameters like pitch, height, and mean diameters are shown in Figure 6.14. The relations between mass flow rates, specific volume, number of blades, pitch and height of the blade, mean diameter of the turbine wheel and flow velocity are given below: Let d = mean diameter of the turbine wheel h = height or breadth of the blade t = thickness of blade n = number of blades A = area ms = mass flow rate (kg/s) vs = xvg = specific volume of steam Area = pdh (neglecting thickness of blades) A1 = (pd – nt1)h1 (considering the thickness at inlet) A2 = (pd – nt2)h2 (considering the thickness at outlet) msvs = volume flow rate of steam

444 \

Fundamentals of Turbomachinery

msvs = area × flow velocity

\

msvs1 = (p d – nt1)hVf1

\

msvs2 = (p d – nt2)hVf2

\

P = pitch of blade

(see Figure 6.14)

For most turbines it can be assumed that Vf1 = Vf2 = Vf, h1 = h2 = h and t1 = t2 = t. \

msvs = [n(P – t)h]Vf

Generally t 90°.

504

Fundamentals of Turbomachinery

Part load condition

Figure 7.8

Guide blades are opened partly, the quantity of water flowing is reduced, the speed of the runner reduces, the water entry to the runner is with shock. (a) Medium speed, inlet blade angle, b1 = 90° (b) Slow speed, inlet blade angle, b1 < 90°

Inlet and outlet velocity triangles of Francis turbine for different speeds: (a) Slow. (b) Medium. (c) Fast.

Hydraulic Turbines

505

The power relation can be obtained from the moment of momentum principle. From Eq. (7.16b), we have Fx = r aV1(Vw1 ± Vw2) where the plus symbol is used if a2 < 90°, and the minus symbol is used if a2 > 90°. Also u 1 ¹ u 2. Now,

T = Fx × R

or

T = r aV1(Vw1R1 ± Vw2R2)

and

P = Tw

or

P = r aV1(Vw1R1w1 ± Vw2R2w2)

We know that, u1 = w1R1 ; \

u2 = w2R2

P = r aV1(Vw1u1 ± Vw2u2)

(7.38)

For Francis turbine a2 = 90°, i.e., Vw2 = 0 i.e. for radial discharge or axial discharge, Eq. (7.38) reduces to P = r aV1 Vw1 u1

(7.39)

hH = hydraulic efficiency Power developed by the runner Power available at the turbine inlet

or

KH

\

KH

U aV1 (Vw1u1 “ Vw2 u2 ) U gQH (Vw1u1 “ Vw 2 u2 ) gH

(7.40)

7.10 IMPORTANT DESIGN PARAMETERS OF FRANCIS TURBINE The following important design parameters in respect of Francis turbine should be noted:

l

y = flow ratio = 0.15 to 0.35 Vf 1 2 gH

l

f = speed ratio = 0.6 to 0.8 u1 2 gH

506

Fundamentals of Turbomachinery

l

u1 = peripheral velocity at inlet or circumferential velocity at inlet

l

S D1 N m/s 60 u2 = peripheral velocity at outlet S D2 N m/s 60 D1, D2 = inlet and exit diameters of runner Vf1, Vf2 = flow velocity at inlet and exit Q = Vf 1pD1B1 = Vf 2pD2B2

(neglecting thickness of the blades)

Q = (pD1 – nt1)B1Vf1 = (pD2 – nt2)B2Vf 2 B1, B2 = width of runner at inlet and exit t1, t2 = thickness of the runner at inlet and exit n = number of runner blades H = net head or effective head on the turbine at inlet p1 V12  Ug 2g

(7.40a)

p1 = pressure at inlet Radial discharge (i) a2 = 90°, Vw2 = 0, radial discharge at outlet (ii) b1 = 90°, radial entry at inlet

l If there is no loss of energy when water flows through the vanes, then H

V22 2g

Vw1 u1 “ Vw 2 u2 g

(7.40b)

7.11 DRAFT TUBE It is a pipe of gradually increasing cross-sectional area, which converts kinetic energy of water into potential energy.

7.11.1 Types of Draft Tube Following are the important types of draft tubes (Figure 7.9): (a) (b) (c) (d)

Simple elbow tubes Conical draft tube Circular inlet and square outlet Moody’s bell-mouthed tube

Hydraulic Turbines

507

Figure 7.9 shows different type of draft tubes commonly employed. In these types, the conical type is most efficient and commonly used. For the straight divergent type conical draft tube, the centre cone angle should not exceed 8°. If this angle exceeds 8°, the water flowing through the draft tube will not remain in contact with its inner surface and hence eddies are formed and the efficiency will be reduced.

Figure 7.9

Types of draft tubes.

7.11.2 Design of Draft Tube Figure 7.10 shows a conical draft tube, where Water in 1

1

Hs

Tail race h 2

Figure 7.10

Water out 2

Conical draft tube.

508

Fundamentals of Turbomachinery

Hs h pa hf V1, V2 w

= = = = = =

vertical distance of the draft tube above the tail race vertical distance between the tail race and the bottom of the draft tube atmospheric pressure loss of energy between sections 1–1 and 2–2 velocity of water at inlet and exit of the draft tube specific weight

Applying Bernoullis equation at inlet (1–1) and outlet (2–2) and using (2–2) as datum, we get p1 V12   ( H s  h) w 2g

p2 V22   0  hf w 2g

(7.41)

We know that

p2 w \

atmospheric pressure head + h

p2 pa h w w Substituting for p2/w in Eq. (7.41), p1 V12   (H s  h ) w 2g

or

p1 w

pa V22 V12    h f  Hs w 2g 2g

(7.42)

pa V2  h  2  hf w 2g ËV 2 V 2 Û pa  Hs  Ì 1  2  h f Ü w ÌÍ 2 g 2 g ÜÝ

(7.43)

Efficiency of the draft tube (hd) Actual conversion of kinetic head into pressure head Kinetic head at the inlet of draft tube

(7.44)

Theoretical conversion of kinetic head into pressure head = (V12  V22 ) / 2 g

(7.45)

Kd

Actual conversion of kinetic head into pressure head = \

Kd

[(V12

 V22 ) / 2 g] 

[(V12  V22 ) / 2 g]  h f

V12  V22  2 gh f

V12 / 2 g

V12

hf

(7.46) (7.47)

7.11.3 Functions of a Draft Tube The functions of the draft tube are as follows: (i) A reaction turbine is required to be installed above the tail race level for ease of maintenance (repair) work. Hence, some head is lost. The draft tube recovers this head

Hydraulic Turbines

509

by reducing the pressure head at the outlet to below the atmospheric level. It increases the working head of the turbine by an amount equal to the height of the runner outlet above the tail race. This creates a negative or suction head. (ii) Exit kinetic energy of water is a necessary loss in the case of turbine. A draft tube recovers part of this exit K.E. (enables to recover the K.E.) If the draft tube is too long, the suction head will be large. The pressure at the outlet of the runner may fall below the vapour pressure of water. Vapours of water will be formed causing cavitation.

7.12 KAPLAN TURBINE Figure 7.11(b) shows two views of a Kaplan Turbine. The extension of the shaft is called boss or hub. The runner vanes are fixed around the circumference of the boss. The boss is similar to the propeller of a ship (Figure 7.11(a)). The runner is enclosed in a spiral casing. Kaplan turbine is a reaction turbine working under low head (2.5 m to 50 m) and handling a large quantity of water. The specific speed is in the range of 255 to 860 rpm. Water enters and leaves the turbine axially, otherwise it is exactly similar to the Francis turbine. This turbine is named after Kaplan, an Australian engineer.

Figure 7.11

Kaplan turbine: (a) Runner. (b) Turbine with components.

7.12.1 Design Parameters of Kaplan Turbine The following design parameters of Kaplan turbine should be noted: Q = discharge through the runner S ( Do2  Db2 ) – V f 1 4

510

Fundamentals of Turbomachinery

Do = outer diameter of the runner Db = diameter of the hub u1 = u2 = peripheral velocity at inlet and exit are equal

Vf 1

S Do N 60 = Vf 2 = velocity of flow at inlet and exit

Figure 7.12

Inlet velocity triangle: (a) For outer diameter (tip diameter). (b) For inner diameter (hub diameter).

Figure 7.13

Outlet velocity triangle: (a) For outer diameter (tip diameter). (b) For inner diameter (hub diameter).

Calculations of runner inlet angle KH

Vw1u1 gH

(b1)

(Assuming radial discharge)

\

Vw1u1 = hH gH

\

(Vw1o uo)tip dia = (Vwlb ub)hub dia

•

uo = peripheral velocity of runner (tip diameter)

•

S Do N 60 ub = peripheral velocity of runner (hub diameter) S Db N 60

Hydraulic Turbines

• •

511

Vw2o = whirl velocity at exit corresponding to tip diameter = 0 Vw2b = whirl velocity at exit corresponding to hub diameter = 0

Runner inlet angle corresponding to tip diameter (Figure 7.12(a)) tan (180  E1 )

È Vf Ø ÉÊ u  V ÙÚ o w1o

Runner inlet angle corresponding to hub diameter (Figure 7.12(b)) tan (180  E1 )

Calculations of runner outlet angle

È Vf Ø ÉÊ u  V ÙÚ b w1b (b2)

Runner outlet angle corresponding to tip diameter (Figure 7.13(a))

tan E2

Vf uo

Runner outlet angle corresponding to hub diameter (Figure 7.13(b))

tan E2

Vf ub

7.13 COMPARISON BETWEEN IMPULSE AND REACTION TURBINES Impulse turbine

Reaction turbine

1. Available energy is completely converted into kinetic energy before entry into the turbine.

1. Partially converted into kinetic energy before entry into the turbine.

2. Water directly strikes the runner.

2. First, water enters a row of fixed blades, then enters the runner vanes.

3. The pressure of the flowing water remains constant and equal to the atmospheric pressure.

3. The pressure of water reduces as it flows over the vanes. The pressure of water is usually less than the atmospheric pressure at the exit of the turbine.

4. No losses if flow is regulated depending upon the load.

4. Losses will be there during flow regulation.

5. Power developed by the change in the kinetic energy of the jet.

5. Power developed partly by the change in the kinetic energy and pressure energy.

6. Turbine can be installed above the tail race.

6. Turbine is submerged in water below the tail race level. (Contd.)

512

Fundamentals of Turbomachinery

Impulse turbine

Reaction turbine

7. Casing is to prevent splashing of water and to guide water to tail race.

7. Pressure in the turbine is not uniform, hence a casing is must.

8. Runner need not run full.

8. Runner should be completely submerged.

9. Draft tube is not necessary.

9. Draft tube is necessary.

10. Impact of jet causes rotation.

10. Reaction of pressure of water causes rotation.

11. The casing has no hydraulic function.

11. The casing has hydraulic function.

Tables 7.1 and 7.2 show the specific speed and head used for different turbines. Table 7.1

Specific speed for different types of turbine

Type of turbine

Specific speed, if P is in kW

Pelton (single jet) (two jets) (multiple jets) Francis (for heads below 370 m) Kaplan (for head below 60 m)

8 to 30 25 to 40 40 to 70 70 to 450 360 to 910

Table 7.2

Head for different types of turbines

Head of water in metres

Type of turbine

0 to 30 30 to 50 50 to 150 150 to 250 250 to 300 Above 300

Either Kaplan or Francis Either Kaplan or Francis Francis Either Kaplan or Pelton Either Francis or Pelton Pelton

7.14 EXAMPLES EXAMPLE 7.1 A Pelton wheel develops 5800 kW under a net head of 180 m at a speed of 195 rpm. Find the discharge through the turbine, the wheel diameter, the number of jets required and the specific speed. Use the following assumptions: overall efficiency 86%, D/d = 12, f = 0.45 and CV = 0.985. Solution: Power developed by the runner Head Speed

P = 5800 kW H = 180 m N = 195 rpm

Hydraulic Turbines

513

ho = 0.86 D/d = 12 CV = 0.985 f = 0.45

Overall efficiency Ratio of wheel diameter to jet diameter Coefficient of velocity Speed ratio To be calculate : Q1, D1, Nj (No. of jets), NST (a) Diameter of the wheel (D): V1

0.985 2 – 9.81 – 180

CV 2 gH

= 58.53 m/s u = u1 = u2 = velocity of wheel I 2 gH

0.45 – 2 – 9.81 – 180

= 26.74 m/s u

\

S DN 60

26.74

S – D – 195 60

D = 2.62 m

Ans.

(b) Diameter of the jet (d): D d

12

d

D 12

\

2.62 12

Ans.

0.218 m

(c) No. of jets (Nj): Ko

\ Now,

\

S.P. K.E. of water leaving the nozzle

5800 – 1000 1000 – 9.81 – Q – 180 3 Q = 3.82 m /s

0.86

S.P. U gQH

Q

3.382

NJ –

S 2 d V1 4

NJ –

S – (0.218)2 – 58.53 4

Nj = 1.974 » 2

Ans.

(d) Specific speed (NST):

N ST

N P H

5/4

195 5800 180 5 / 4

22.52 rpm

Ans.

514

Fundamentals of Turbomachinery

EXAMPLE 7.2 A Pelton wheel has a water supply rate of 5 m3/s at a head of 256 m and runs at 500 rpm. Assuming a turbine efficiency of 0.85, a coefficient of velocity for nozzle as 0.985, speed ratio of 0.46, calculate (a) the power output, (b) the specific speed, (c) the number of jets, (d) the diameter of the wheel, (e) the jet diameter, (f) the number of pelton cups, and (g) the cup dimensions. Solution:

Data: Discharge: Head: Speed: Overall efficiency: Coefficient of velocity: Speed ratio:

Q = 5 m3/s H = 256 m N = 500 rpm ho = 0.85 CV = 0.985 f = 0.46

To determine: P, NST, Nj, D, d, m, cup dimensions (a) Power (P): V1 u

Now, \

0.985 2 – 9.81 – 256

CV 2 gH I 2 gH

0.46 2 – 9.81 – 256

S DN 60 D = 1.245 m u

32.6

P

Ko –

69.8 m/s 32.6 m/s

S – D – 500 60

U gQH 1000

1000 – 9.81 – 5 – 256 – 0.85 1000

10673.3 kW

Ans.

(b) Specific speed (NST) for NJ number of jets: N ST

N P H

5/4

500 – 10673.3 (256)5 / 4

50.44 rpm

Ans.

(c) Wheel diameter (D): D = 1.245 m

Ans.

(d) Jet diameter (d): \ We know that for a single jet, the specific speed = 30 rpm \

Nj

\

Q

\

No. of Jets =

Specific speed for N j jets Specific speed for one jet

S – d 2V1 N j 4 d = 0.2135 m 5

50.44 30

1.68  2

Ans.

S – d 2 – 69.8 – 2 4

Ans.

Hydraulic Turbines

515

(e) No. of cups (m): m 15 

D 2d

15 

1.245 2 – 0.2135

17.91  18

Ans.

Þ Ñ ß Ñ à

Ans.

(f) Cup dimensions Width

B = 2.8d = 2.8 × 0.2135 = 0.5978

Length

L = 2.4d = 2.4 × 0.2135 = 0.5124

Depth

T = 0.6d = 0.6 × 0.2135 = 0.1281

EXAMPLE 7.3 The three jet Pelton is required to generate 10,000 kW under a net head of 400 m. The blade angle at outlet is 15° and the reduction in the relative velocity while passing over the blade is 5%. If the overall efficiency of the wheel is 80%, CV = 0.98, speed ratio = 0.46, then find (a) the total flow in m3/s, (b) the diameter of the jet, and (c) the force exerted by a jet on the buckets. Solution:

Data: No. of jets: Shaft power: Net head: Blade angle: Drop in relative velocity: Overall efficiency: Coefficient of velocity: Speed ratio:

Nj = 3 S.P = 10,000 kW H = 400 m b2 = 15° K = 0.95 ho = 80% CV = 0.98 f = 0.46

To determine: Q, d, Fx Vw1

V1

CV 2 gH

u I 2 gH

0.98 2 – 9.81 – 400

0.46 2 – 9.81 – 400

86.82 m/s

40.75 m/s

Vr1 = V1 – u = 86.82 – 40.75 = 46.07 m/s K

0.95

Vr 2 Vr1

\ Vr2 = 0.95Vr1

Vr2 = 0.95 × Vr1 = 0.95 × 46.07 = 43.77 m/s Vw2 = Vr2 cos b2 – u = 43.77 × cos 15° – 40.75 = 1.53 m/s (a) The total discharge (Q): Ko

S.P. U gQH

10000 – 1000 1000 – 9.81 – Q – 400

0.8 ? Q

3.186 m 3 /s

Ans.

516

Fundamentals of Turbomachinery

(b) The diameter of the jet (d): q = discharge from one nozzle = S 2 d V1 4

q 1.062

Now,

Q Nj

3.186 3

S – d 2 – 86.82 ; ? d 4

1.062 m 3 /s 0.125 m

Ans.

(c) Force in the x direction (Fx) by a jet: Fx = rq(Vw1 + Vw2) = 1000 × 1.062 × (86.82 + 1.53) = 93827.7 N

Ans.

EXAMPLE 7.4 At a power station, a Pelton turbine produces 23,100 kW power under a head of 1770 m while running at 750 rpm. Evaluate for the turbine (a) the mean diameter of the runner, (b) the number of jets and jet diameter, and (c) the number of buckets. Assume that the nozzle velocity coefficient is 0.98, speed ratio 0.46 and the overall turbine efficiency 0.86. Solution:

Data: Shaft power: Head: Speed: Speed ratio: Velocity coefficient: Overall efficiency:

S.P. = 23,100 kW H = 1770 m N = 750 rpm f = 0.46 CV = 0.98 ho = 0.85

To determine: Nj, d, D, m

If NST

750 – 23100

N S.P.

N ST

5/4

H 1770 5 / 4 = 9.93, turbine has to have ONE jet. V1

CV 2 gH

u

I 2 gH

9.93 rpm

0.98 – 2 – 9.81 – 1770 0.46 – 2 – 9.81 – 1770

182.6 m/s 85.72 m/s

(a) Wheel diameter (D): u = 85.72 = pDN/60 = p × D × 750/60;

\ D = 2.183 m

Ans.

(b) Diameter of jet (d) and number of jets (Nj): Ko

S.P. U gQH

23100 – 1000 1000 – 9.81 – Q – 1770

\

Q = 1.565 m3/s

Now,

Q 1.565

S 2 d V1 4

0.85

S – d 2 – 182.6; ? d 4

0.1044 m

Ans.

Hydraulic Turbines

Since

517

D/d > 10 Nj = 1

Ans.

(c) The number of buckets (m): m 15 

D 2d

15 

2.183 2 – 0.107

25.2  25

Preferably 24 or 26 buckets.

Ans.

EXAMPLE 7.5 A double jet Pelton wheel is required to generate 7500 kW when the available head at the base of the nozzle is 400 m. The jet is deflected through 165° and the relative velocity of the jet is reduced by 15% in passing over the buckets. Determine (a) the diameter of each jet, (b) the total flow, and (c) the force exerted by the jets in tangential direction. Assume generator efficiency of 95%, overall efficiency of 80%, blade speed ratio of 0.47, and nozzle coefficient of 0.98. Solution:

Data: Number of jets: Head: Power to generate:

Nj = 2 H = 400 m Pg = 7500 kW b2 = 180° – 165° = 15° Vr2 = 0.85Vr1, hg = 0.95, ho = 0.8, f = 0.47, CV = 0.98

To determine: d, Q, Fx S.P. Vw1 u

Vr1

Pg Kg V1

7500 0.95 CV 2 gH

7894.7 kW 0.98 2 – 9.81 – 400

86.82 m/s

I 2 gH 0.47 2 – 9.81 – 400 41.64 m/s = Vw1 – u = 86.82 – 41.64 = 45.18 m/s

Vr2 = 0.85Vr1 = 0.85 × 45.18 = 38.4 m/s Vw2 = Vr2 cos b2 – u = 38.4 × cos 15° – 41.64 = – 4.55 m/s

518

Fundamentals of Turbomachinery

\ Vw2 is –ve, hence the outlet velocity triangle changes to as given below:

(a) Total discharge (Q): Ko

7894.7 – 1000 1000 – 9.81 – Q – 400

S.P U gQH

0.8

Q = 2.515 m3/s

\

Ans.

(b) Jet diameter (d): q = discharge from each jet = S 2 d V1 4 d = 0.1357 m q 1.257

\

Q Nj

2.515 1.257 m 3 /s 2

S – d 2 – 86.82 4

Ans.

(c) Force in the x direction (Fx): Fx = r Q(Vw1 ± Vw2) = 1000 × 2.515 (86.82 – 4.55) = 1206.91 kN

Ans.

EXAMPLE 7.6 A Pelton wheel has the following data. Total output = 32 MW, gross head = 250 m, speed = 400 rpm, no. of jets = 2 per wheel, CV = 0.98, maximum overall efficiency = 80.0%, head loss = 12 m for a pipe length of 250 m. Calculate (a) the number of wheels required, (b) the jet wheel diameter and pipe diameter, (c) the hydraulic efficiency, and (d) the power not utilized and going waste. Take f = coefficient of friction = 0.02 Blade coefficient K = 0.85, b2 = 15° Solution:

Data: PT = 32 MW, 2 jets per wheel, f = 0.02 ho = 80%

Hg = 250 m, N = 400 rpm, CV = 0.98, K = 0.85, hf = 12 m for a pipe length of 250 m

We know that, H = Hg – hf = 250 – 12 = 238 m Specific speed for two jets is 26 to 40.

Hydraulic Turbines

519

Let us assume that, specific speed, NST = 30 \

400 – Pw

N Pw

N ST

30 H 2385 / 4 Pw = 4915.47 kW = Power from one wheel

or

5/4

(a) Number of wheels required (Nw): Total power Power from one wheel

Nw

32 – 103 kW 4915.47 kW

PT Pw

6.51  7 units

Ans.

(b) Jet diameter (d), wheel diameter (D) and pipe diameter (dP): V1

CV 2 gH

0.98 2 – 9.81 – 238

66.67 m/s

Condition for maximum efficiency u = V1/2 whereas in actual conditions peak occurs when u u I 0.46 V1 66.97 \

u = 30.81 m/s

Now, \

S DN 60 D = 1.471 m u

30.81

S – D – 400 60

Ans.

Nj = total number of jets = 2 × no. of wheels = 2 × 7 = 14 S.P. = power produced from each jet = =

Ko

32 – 10 3 14

S.P. U qgH

Total power No. of jets

2.3 – 10 3 kW

2.3 – 1000 – 103 1000 – q – 9.81 – 238

0.8

\

q = discharge from one jet = 2.2314 m3/s

Now,

q

or \

1.2314

S 2 d V1 4

m3 S m3 – d 2 – 66.97 s 4 s d = 0.153 m

Ans.

520

Fundamentals of Turbomachinery

Assume that water is supplied to all 14 jets by one pipe only. Therefore, discharge through the pipe (QP). QP = number of jets × discharge from each jet (q) or

QP = 14 × 1.2314 = 17.24 m3/s

Now,

QP

\

VP

S 2 d P VP 4

17.24

4 – 17.24 S – d P2

We have Darcy’s formula, VP = pipe velocity hf

4 f LVP2 2 g dP

or

12

4 – 0.02 – 250 Ë 4 – 17.24 Û Ì Ü 2 – 9.81 – d P ÍÌ S d P2 ÝÜ

or

d P5

40.90 m 5

\

dP = 2.1 m

2

Ans.

(c) Hydraulic efficiency of (hH): KH

Power produced by the wheel Energy supplied to the wheel

2(Vw1  Vw 2 ) u V12 Vw1 = V1 = 66.97 m/s,

Vw2 = Vr2 cos b2 – u = 0

Vr1 = V1 – u = 66.97 – 30.81 = 36.16 m/s Vw2 = 0.85Vr1 cos b2 – u = 0.85 × 36.16 cos 15° – 30.81 = –1.12 m/s \ Vw2 is –ve, hence the outlet velocity triangle changes to the following:

Hydraulic Turbines

\

KH

2 – 30.81 (66.97  1.12)

– 100

(66.97)2

521 Ans.

90.47%

(d) The power not utilized and going as waste (Pw): From the outlet velocity triangle, Vf 2 = Vr2 sin b2 = 0.85Vr1 sin b2 = 0.85 × 36.16 × sin 15° = 7.96 m/s V f22  Vw22

7.92 2  ( 1.12)2

\

V2 = exit velocity =

\

Hw = head wasted =

\

Pw = r QgHw = 1000 × (14 × 1.2314) × 9.81 × 3.289 = 556.24 kW

EXAMPLE 7.7

V22 2g

(8.034)2 2 – 9.81

8.03 m/s

3.289 m

Ans.

Show that in a Pelton wheel, for maximum utilization, u/V1 = 0.5 and hence,

V12 (1  cos E2 ) / 4, where

energy transfer E

V1 = velocity of jet striking the runner b2 = exit blade angle Solution: u/V1 = 0.5 \

(Given)

E = P = raV1(Vw1 + Vw2)u

W

For unit mass, P

U aV1 (Vw1  Vw 2 ) u U aV1

W/kg

From inlet velocity triangle, Vw1 = V1

Vr1 = Vr2 = V1 – u

(i)

522

Fundamentals of Turbomachinery

From outlet velocity triangle, Vw2 = (Vr2 cos b2 – u) = (V1 – u) cos b2 – u Substituting Vw1, Vw2 and u/V1 = 0.5 in Eq. (i), P = u[V1 + (V1 – u) cos b2 – u]

\

P

V1 2

Ë V1 Ø V1 Û È ÌV1  ÉÊ V1  2 ÙÚ cos E2  2 Ü Í Ý

V1 2

V1 V1 Û Ë ÌV1  2 cos E2  2 Ü Í Ý

V1 2

Ë 2V1  V1 cos E2  V1 Û Ì Ü 2 Í Ý

V12 (1  cos E2 ) 4

Ans.

EXAMPLE 7.8 Following data is available for a Pelton wheel: gross head = 100 m, onefourth of the gross head is lost in friction, mean wheel speed is 18 m/s, discharge through the nozzle is 1.0 m3/s, and the angle of deflection of the jet is 165°. Calculate (a) the power given by the water to the runner and (b) the hydraulic efficiency. Solution: Data: Hg = 100 m,

1 – H g , u = 18 m/s 4 b2 = 180° – 165° = 15° hf

Q = 1.0 m3/s, To determine: P, hH

Assume,

H

Hg  h f

V1

CV 2 gH

100 

100 4

75 m

1 – 2 – 9.81 – 75

38.36 m/s

CV = 1 Vr1 = V1 – u = 38.36 – 18 = 20.36 m/s Vr1 = Vr2

(Assume no losses over the blade)

Vw2 = (Vr2 cos b2 – u) = 20.36 cos 15° – 18 = 1.666 m/s

Hydraulic Turbines

523

(a) Power (P): P = rQ (Vw1 + Vw2)u = 1000 × 1.0 × (38.36 + 1.666) × 18 = 720.472 kW

Ans.

(b) Hydraulic efficiency (hH): KH

Alternatively: KH

P

720.472 – 2 – 1000

1 U QV12 2

1000 – 1 – (38.36)2

P W.P

– 100

97.92%

720.472 – 1000 – 100 1000 – 9.81 – 75

P U gQH

97.92%

Ans.

Ans.

EXAMPLE 7.9 In a power station, a Pelton wheel produces 15,000 kW under a head of 350 m while running at 500 rpm. Assume a turbine efficiency of 0.84, coefficient of velocity for nozzle as 0.98, speed ratio 0.46 and bucket velocity coefficient 0.86. Estimate (a) the number of jets, (b) the diameter of each jet, and (c) the tangential force exerted on the buckets if the bucket deflects the jet through 165°. Solution:

Data: S.P. = 15000 kW, ho = 0.84 K = 0.86,

H = 350 m, N = 500 rpm CV = 0.98, f = 0.46, b2 = 180° – 165° = 15°

(b) d (c) Fx

To determine: (a) Nj

0.98 – 2 – 9.81 – 350

V1

CV 2 gH

81.2 m/s

u

I 2 gH

0.46 – 2 – 9.81 – 350 = 38.11 m/s

Ko

S.P. U gQH

15000 – 1000 1000 – 9.81 – Q – 350

0.84

Q = 5.2 m3/s

\ (a) No. of jets (Nj):

N ST

N S.P. H

5/ 4

500 – 15000 (350)5 / 4

40.45 rpm

For a specific speed (NST) of 40.45, 2 jets can be used \

Nj = 2

Ans.

(b) Diameter of each jet (d): Q = total discharge = 5.2 = \

d = 0.202 m

S 2 d V1 N j 4

S – d 2 – 81.2 – 2 4

Ans.

524

Fundamentals of Turbomachinery

Vr1 = V1 – u = 81.2 – 38.11 = 43.09 V1 = Vw1 = 81.2 m/s K

Vr 2 Vr1

0.86

\ Vr2 = 0.86Vr1

Vr2 = 37.057 m/s Vw2 = (Vr2 cos b2 – u) = (37.057 cos 15° – 38.11) = –2.315 m/s \ Vw2 is –ve, hence the outlet velocity triangle changes to the following:

(c) Tangential force exerted on the bucket (Fx): Fx = rQ (Vw1 ± Vw2) u = 1000 × 5.2 × (81.2 – 2.315) = 410.202 N

Ans.

Alternatively: Fx by one jet = U q (Vw1 “ Vw 2 ) u 1000 –

U–

S 2 d V1 – (Vw1 “ Vw 2 ) u 4

S – (2.202)2 – 81.2 – (81.2  2.315) = 205.278 kN 4

Ans.

EXAMPLE 7.10 A Pelton wheel generates 3500 kW of power under a net head of 420 m. Assume generator efficiency 0.95, overall efficiency 0.8, coefficient of velocity 0.98, speed ratio 0.46, and jet ratio 10. Calculate (a) the total discharge in litres, (b) the synchronous speed at 50 Hz, and (c) the mean diameter.

Hydraulic Turbines

525

Solution: Pg = 3500 kW,

hg = 0.95,

H = 420 m, f = 0.46,

CV = 0.98,

m

D d

ho = 0.8

10

To determine: (a) Q, (b) synchronous speed if frequency = 50 Hz, (c) mean diameter V1

0.98 – 2 – 9.81 – 420

CV 2 gH

u I 2 gH

0.46 – 2 – 9.81 – 420

88.96

41.76 m/s

(a) Total discharge Q: S.P. = power developed = Ko

S.P. U gQH

0.8

Pg

3500 kW 0.95

Kg

3684.2 – 1000 1000 – 9.81 – Q – 420

\

Q = 1.118 m3/s or

Now,

Q 1.118

\

d = 0.127 m

and

D = 10d = 10 × 0.127 = 1.27 m

S 2 d V1 4

3684.2 kW

1118 lt/s

Ans.

S – d 2 – 88.96 4

(Q D/d = 10)

Ans.

(b) Synchronous speed (NS):

u \

41.76

S DN S 60

S – 1.27 – N S 60

NS = 630.6 rpm

By using this speed we calculate the number of poles using the relation, NS

or

120 f P

\

P

120 – f NS

120 – 50 630.6

P = 9.71 » 10 (next even number)

\ Final synchronous speed =

120 – f P

120 – 50 = 600 rpm 10

Ans.

EXAMPLE 7.11 A double overhung Pelton wheel unit is coupled to a 10,000 kW generator. If the generator efficiency is 0.95, calculate the power developed by each runner. Solution: A double overhung Pelton wheel means that two Pelton wheels are mounted on the same shaft along with the generator.

526

Fundamentals of Turbomachinery

P-I

P-II

G Pg = 10,000 kW

Pg = 10,000 kW,

hg = 0.95

Power developed by each runner (P): PT = total power developed =

Pg Kg

10000 0.95

= 10526.32 kW \

P

10526.32 2

Ans.

5263.2 kW

EXAMPLE 7.12 A double overhung Pelton wheel is to be designed to operate a 30,000 kW generator. The effective head is 300 m. Generator efficiency is 0.94, Pelton wheel efficiency is 0.84, speed ratio 0.46, jet coefficient velocity 0.98, jet ratio 12. Calculate (a) the size of the jet, (b) the mean diameter of the runner, (c) the synchronous speed, and (d) the specific speed of each Pelton wheel. Solution: PgT = 30,000 kW, hT = ho = 0.84,

H = 300 m

hg = 0.94

f = 0.46,

CV = 0.98,

D d

To determine: d, D, N, NST, NS PT = total power developed by the Pelton wheel = 30000 0.94

31914.89 kW

S.P. = power developed from one turbine =

V1 u

Ko

31914.89 2

15957.47 kW

CV 2 gH

0.98 – 2 – 9.81 – 300

I 2 gH

KT

0.46 – 2 – 9.81 – 300

S.P. U gQH

PT 2

75.2 m/s 35.29 m/s

15957.47 – 1000 1000 – 9.81 – Q – 300

0.84

Q = 6.455 m3/s = discharge from one Pelton wheel

PgT Kg

12

Hydraulic Turbines

527

(a) Diameter of jet (d): Q

S 2 d V1 N j 4

6.455

S 2 d – 75.2 – 1 4

Assume a single jet Pelton wheel, Nj = 1 \

d = 0.331 m

Ans.

(b) Mean bucket diameter (D): D d

\

\ D = 12 × d = 12 × 0.3373

12

D = 4.05 m

Ans.

(c) Synchronous speed (NS): u

35.29

S DN 60

S – 4.05 – N 60

N = 166.417 rpm By using speed N, calculate the number of poles, poles

120 – f N

120 – 50 (assume f 166.42

50 Hz)

= 36.05 poles » 36 (even poles) \

NS

120 – f poles

120 – 50 36

166.67 rpm

Ans.

(e) Specific speed (NST): N ST

N S S.P. H

5/4

166.67 15957.47 (300)5 / 4

16.86 rpm

Ans.

EXAMPLE 7.13 Following data refers to a Pelton wheel: gross head 500 m, water supply (penstock) diameter 1 m, length of the penstock 3.5 km, coefficient of friction, f = 0.006, jet diameter 18 cm, jet deflection angle 165°, 15% friction on the bucket, peripheral velocity of bucket is 0.46 times the absolute velocity of jet leaving the nozzle, mechanical efficiency 85%. Calculate (a) the power developed by the runner, (b) the power at the shaft, (c) the hydraulic efficiency, and (d) the overall efficiency. Solution:

Data: Hg = 500 m, f = 0.006, K = 15%,

dP = 1 m, d = 18 cm, u = 0.46V1,

LP = 3500 m b2 = 180° – 165° = 15° hm = 85%

528

Fundamentals of Turbomachinery

To determine: P, S.P., hH, ho We have continuity relation, \

QP = QN

\

S 2 d P VP 4

or

VP

S 2 d V1 4

d2 d P2

V1

(Penstock discharge = Nozzle discharge) (VP = Penstock flow velocity) d 2 – V1 12

V1 d 2

Now apply Bernoulli’s equation between head race and end of nozzle. Hg = head lost due to friction +

or or

V12 2g

500

4 f LVP2 V12  dP 2 g 2g

4 – 0.006 – 3500 – V12 d 4 V12  1 – 2 – 9.81 2 – 9.81

500

4 – 0.006 – 3500 – V12 (0.18) 4 V12  1 – 2 – 9.81 2 – 9.81

V1 = 94.95 m/s u = 0.46 × V1 = 0.46 × 94.95 = 43.68 m/s

From inlet velocity triangle, V1 = Vw1 = 94.95 m/s Vr1 = V1 – u = 94.95 – 43.68 = 51.273 m/s From outlet velocity triangle, Vr2 = 0.85Vr1 = 0.85 × 51.273 = 43.582 m/s Vw2 = (Vr2 cos b2 – u) = 43.582 cos 15° – 43.68 = –1.583 m/s Vw2 is –ve, hence the outlet velocity triangle changes to Q

S 2 d V1 4

S – (0.18)2 – 94.95 4

2.416 m 3 /s

Hydraulic Turbines

529

(a) Power developed by the turbine or runner (P): P = r Qu(Vw1 – Vw2) = 1000 × 2.416 × 43.68 × (94.95 – 1.583) = 9853.10 kW

Ans.

(b) Shaft power (S.P.): S.P. P S.P. = Phm = 9353.10 × 0.85 = 8375.14 kW Km

Ans.

(c) Hydraulic efficiency (hH): KH

P 1 U QV12 2

9853.10 – 2 – 1000 1000 – 2.416 – 94.952

– 100

90.47%

Ans.

(d) Overall efficiency (ho): ho = hm × hH = 0.85 × 0.9047 = 0.769 = 76.9%

Ans.

EXAMPLE 7.14 750 litres of water enters the Pelton turbine with a net head of 50 m. The mean bucket speed is 15 m/s, and the jet is deflected by 165°. Calculate (a) hydraulic efficiency and (b) the power produced. Solution: Q = 750 lt/s, To determine: hH, P

H = 50 m

u = 15 m/s

b2 = 180° – 165° = 15°

530

Fundamentals of Turbomachinery

Assume

CV = 0.98

\

V1

\

V1 = Vw1 = 30.695 m/s

CV 2 gH

0.98 – 2 – 9.81 – 50

30.695 m/s

Vr1 = Vr2 = V1 – u = 30.695 – 15 = 15.695 m/s Vw2 = (Vr2 cos b2 – u) = 15.695 – 15 = 0.1597 m/s (a) Hydraulic efficiency (hH):

KH

2u (Vw1  Vw 2 )

2 – 15 – (30.695  0.1597)

V12

(30.695)2

– 100

= 98.24%

Ans.

(b) Power produced (P): P = rQu(Vw1 + Vw2) = 1000 × 0.75 × 15 × (30.695 + 0.1597) = 347.45 kW

Ans.

Example 7.15 Following results were obtained during a test carried out on a Pelton wheel. Head at the base of nozzle: Discharge through the nozzle: Jet diameter: Power absorbed due to friction in Pelton wheel: Power available at shaft:

30 m 300 l/s 130 mm 4 kW 62 kW

Calculate (a) the power lost in the nozzle, (b) the head lost in the nozzle, and (c) the power lost in the runner. Solution: P = power available at nozzle = power available at shaft + power lost in nozzle and runner + power lost in mechanical resistance or

rgQH = 62 + power lost in nozzle and runner + 4

Power lost in nozzle and runner

U gQH  62  4 1000 1000 – 9.81 – 0.3 – 30  62  4 1000

S 2 d V1 4

Now,

Q

\

V1 = 22.6 m/s

S – (0.13)2 – V1 4

0.3 m 3 /s

22.29 kW

Hydraulic Turbines

531

HN = head available at nozzle outlet = V12 /2 g or

HN

(22.6)2 2 – 9.81

26.03 m

(a) Head lost in the nozzle (HL): HL = head at entry to the nozzle – head at exit from the nozzle \

HL = 30 – 26.033 = 3.967 m

Ans.

(b) Power lost in the nozzle (PL):

PL

U gQH L 1000

1000 – 9.81 – 0.3 – 3.967 1000

11.675 kW

Ans.

(c) Power lost in the runner: = power lost in the nozzle and runner – power lost in the nozzle (PL) = 22.29 – 11.675 = 10.62 kW

Ans.

EXAMPLE 7.16 An impulse turbine produces 150 kW under a head of 70 m. If the head is increased to 90 m, calculate the percentage increase in speed. Solution: P1 = 150 kW,

H1 = 70 m,

H2 = 90 m

To determine: Percentage increase in speed We have Ë N Û Ì Ü Í H Ý1

Ë N Û Ì Ü Í H Ý2 1/ 2

\

N2

ËH Û N1 Ì 2 Ü Í H1 Ý

1/ 2

Ë 90 Û N1 – Ì Ü Í 70 Ý

1.34 N1

(a) Percentage increase in speed ( N 2  N1 ) – 100 N1

(1.34 N1  N1 ) – 100 13.4% N1

Ans.

EXAMPLE 7.17 A Pelton wheel produces 7000 kW under a head of 250 m. The overall efficiency is 85% and the speed is 200 rpm. Calculate (a) the unit discharge, (b) the unit power, and (c) the unit speed. Assume peripheral coefficient = 0.46. If the head on the same turbine falls during the summer season to 150 m, calculate (d) the discharge, (e) the power, and (f) the speed for this head. Solution: Ko

\

P U gQH

7000 – 1000 1000 – 9.81 – Q – 250

Q = 3.358 m3/s

0.85

532

Fundamentals of Turbomachinery

(a) Unit discharge (Qu): Qu

Q

3.358

H

250

0.2124 m 3 /s

Ans.

(b) Unit power (Pu): Pu

000

P H

3/2

(250)3 / 2

1.711 kW

Ans.

(c) Unit speed (Nu): Nu

N

200

H

250

12.65 rpm

(d) Discharge (Q1) at 150 m head:

\

Ë Q Û Ì Ü Í HÝ

Ë Q Û Ì Ü Í H Ý1

Q1

ËH Û QÌ 1Ü ÍHÝ

1/ 2

1/ 2

Ë 150 Û 3.358 Ì Ü Í 250 Ý

2.601 m 3 /s

Ans.

(e) Power (P1) at 150 m speed:

Ë P Û Ì 3/2 Ü ÍH Ý

Ë P Û Ì 3/ 2 Ü Í H Ý1

P1

ËH Û P Ì 1Ü ÍHÝ

3/2

Ë 150 Û 7000 – Ì Ü Í 250 Ý

3/2

3253.31 kW

Ans.

(f) Speed (N1) at 150 m head: Ë N Û Ì Ü Í HÝ

Ë N Û Ì Ü Í H Ý1

N1

ËH Û N Ì 1Ü ÍHÝ

1/ 2

\

1/ 2

Ë 150 Û 200 – Ì Ü Í 250 Ý

154.92 rpm

Ans.

EXAMPLE 7.18 A turbine works under a head of 30 m at 200 rpm. The discharge is 8 m3/s. The overall efficiency is 0.9. Determine the performance of the turbine under a head of 20 m. Solution: H1 = 30 m N1 = 200 rpm Q1 = 8 m3/s To determine: H2, N2, Q2, P2

H2 = 20 m ho = 0.9

Hydraulic Turbines

Ko

P1 U gQ1 H1

533

\ P1 = horgQ1H1

P1 = 0.9 × 9.81 × 1000 × 8 × 30 = 2118.96 kW (a) Speed (N2): \

Ë N Û Ì Ü Í H Ý1

\

N2

Ë N Û Ì Ü Í H Ý2 1/ 2

ËH Û N1 Ì 2 Ü Í H1 Ý

1/ 2

Ë 20 Û 200 – Ì Ü Í 30 Ý

163.299 rpm

Ans.

(b) Discharge (Q2): Ë Q Û Ì Ü Í H Ý1

Ë Q Û Ì Ü Í H Ý2 1/ 2

Q2

ËH Û Q1 Ì 2 Ü Í H1 Ý

Ë P Û Ì 3/2 Ü Í H Ý1

Ë P Û Ì 3/2 Ü Í H Ý2

P2

ËH Û P1 Ì 2 Ü Í H1 Ý

\

1/ 2

Ë 20 Û 8–Ì Ü Í 20 Ý

6.53 m 3 /s

Ans.

(c) Power (P2):

\

3/ 2

Ë 20 Û 2118.96 – Ì Ü Í 30 Ý

3/ 2

1153.42 kW

Ans.

EXAMPLE 7.19 A Pelton turbine develops 3000 kW under a head of 400 m. The overall efficiency of the turbine is 87%. If the speed ratio is 0.48 and the coefficient of velocity is 0.96 and specific speed 18, find (a) the diameter of the turbine and (b) the diameter of the jet. Solution:

Data: Power: Head: Overall efficiency: Speed ratio: Coefficient of velocity: Specific speed:

P = 3000 kW H = 400 m ho = 0.87 f = 0.8 CV = 0.96 NST = 18

To determine: D, d V1 u

CV

2 gH

I 2 gH

0.96 – 2 – 9.81 – 400 0.48 – 2 – 9.81 – 400

85.05 m/s 42.52 m/s

534

Fundamentals of Turbomachinery

N ST

N P

18

N 3000

H 400 5 / 4 N = 587.88 rpm

\

5/4

(a) Diameter of the turbine (D): u

\

S DN 60

42.52

S – D – 587.88 60

D = 1.38 m

Ans.

(b) Diameter of the jet (d): Ko

P U gQH

3000 – 1000 1000 – 9.81 – Q – 400

\

Q = 0.879 m3/s

or

Q

S 2 d V1 4 d = 0.1147 m

\

0.879

0.87

S 2 d – 85.05 4

Ans.

EXAMPLE 7.20 A Pelton wheel has a mean bucket diameter of 1 m and is running at 1000 rpm. The net head on the Pelton wheel is 700 m. If the angle of deflection of jet is 165° and discharge through the nozzle is 0.1 m3/s, coefficient of velocity CV = 0.98, find (a) the hydraulic efficiency of the turbine, and (b) the power available at the nozzle. Solution:

Data: Mean bucket diameter: Speed: Net head: Angle of deflection of jet: Discharge: Coefficient of velocity:

D=1m N = 1000 rpm H = 700 m b2 = 180° – 165° = 15° Q = 0.1 m3/s CV = 0.98

To determine: hH, P u V1

u1

S DN 60

CV 2 gH

S – 1 – 1000 60

52.36 m/s

0.98 – 2 – 9.8 – 1000

114.85 m/s

Vr1 = V1 – u1 = 114.85 – 52.36 = 62.49 m/s (a) Hydraulic efficiency (hH): Vw2 = Vr2 cos b2 – u2 = 62.49 cos 15° – 52.36 = 8.0 m/s

Hydraulic Turbines

\

KH

535

2u (Vw1  Vw 2 ) – 100 V12 2 – 52.36(114.85  8) (114.85)2

– 100

97.53%

[Vr1 = Vr2]

Ans.

(b) Power available at the nozzle (P): P

U gQH

1000 – 9.81 – 0.1 – 1000 = 686.7 kW 1000

Ans.

EXAMPLE 7.21 A Pelton wheel has a mean bucket speed of 10 m/s with a jet of water flowing at the rate of 700 lit/s. Under a head of 30 m, the buckets deflect the jet through an angle of 160°. Calculate the power and the efficiency of the turbine. Solution:

Data: Mean bucket speed: Discharge: Head: Bucket deflects:

u = 10 m/s Q = 700 lit/s H = 30 m b2 = 180° – 160° = 20°

To determine: P and hH V1

CV 2 gH

0.98 2 – 9.81 – 30

23.77 m/s

Vr1 = V1 – u1 = 23.77 – 10 = 13.77 m/s Vw1 = V1 = 23.77 m/s

536

Fundamentals of Turbomachinery

Vr2 = Vr1 = 13.77 m/s Vw2 = Vr2 cos b2 – u = 13.77 cos 20° – 10 = 2.94 m/s (a) Power developed (P):

P

UQ(Vw1  Vw 2 ) u 1000

1000 – 0.7 – (23.77  2.94) – 10 = 186.97 kW 1000

Ans.

(b) Hydraulic efficiency or turbine efficiency (hH):

2 (Vw1  Vw 2 ) u

KH

V12 2 – 10 – (23.77  2.94) (23.77)2

– 100

94.55%

Ans.

EXAMPLE 7.22 Design a Pelton wheel to run under a head of 60 m at 200 rpm while the discharge available is 200 lit/s. Assume overall efficiency to be 85%, coefficient of velocity to be 0.98, and speed ratio 0.46. Solution: Head: Speed: Discharge: Overall efficiency: Coefficient of velocity: Speed ratio:

H = 60 m N = 200 rpm Q = 200 lit/s ho = 85% CV = 0.98 f = 0.46

To determine: Design [D, d, Z1, t, W, B] V1 u

0.98 – 2 – 9.81 – 60

CV 2 gH u1

u2

Now,

u 15.78

\

D = 1.51 m

Now,

Q

\

d = 8.7 cm

0.2

I 2 gH

S DN 60

S 2 d V1 4

33.62 m/s

0.46 2 – 9.81 – 60

15.78 m/s

S – D – 200 60

Ans. S 2 d – 33.62 4

W = width of the bucket = 4d = 4 × 8.7 = 34.8 cm L = length of the bucket = 25d = 2.5 × 8.7 = 21.75 cm t = depth of the bucket = 0.6d = 0.6 × 8.7 = 5.22 cm Z = number of buckets

Ans. Þ Ñ ß Ñà

Ans.

Hydraulic Turbines

D  15 2d

\

1.51  15 2 – 0.087

537

23.68

Z » 24 buckets

Ans.

EXAMPLE 7.23 The external and internal diameters of an inward flow reaction turbine are 2.0 m and 1.0 m respectively. The head on the turbine is 60 m. The width of the vane at inlet and outlet are same and equal to 0.25 m. The runner vanes are radial at inlet and the discharge is radial at outlet. The speed is 200 rpm and the discharge is 6 m3/s. Determine: (a) The vane angle at outlet and inlet of the runner (b) The hydraulic efficiency. Solution: Speed: External diameter: Internal diameter: Head: B1 = B2 Runner vane radial at inlet: Radial discharge:

N = 200 rpm D1 = 2.0 D2 = 1.0 H = 60 m Vf1 = Vr1 = 0.25 m b1 = 90° a2 = 90°, Vw2 = 0

To determine: b1, b2, hH

From the inlet velocity triangle,

u1

Vw1

S D1 N 60

S – 2.0 – 200 60

20.95 m/s

S D2 N S – 1.0 – 200 10.47 m/s 60 60 Q = discharge = pD1B1Vf 1 (neglecting the thickness of vane) u2

538

Fundamentals of Turbomachinery

6 = p × 2.0 × 0.25 × Vf 1

or \

Vf 1 = 3.82 m/s Q = pD2B2Vf 2

Also,

6 = p × 1.0 × 0.25 × Vf 2 \

Vf 2 = 7.64 m/s

(a) Vane angle at outlet (b2) and inlet (b1):

Vf 2

tan E 2

u2

7.64 10.47

\

b2 = 36.11°

\

b1 = 90°

0.729

(Given data)

(b) Hydraulic efficiency (hH) from Eq. (7.40): KH

Vw1u1 gH

20.95 – 20.95 – 100 9.81 – 60

Þ Ñ ß Ñà

Ans.

74.56%

Ans.

EXAMPLE 7.24 In an inward flow reaction turbine, the internal and external diameters are 1.5 m and 2.5 m respectively. The width of the wheel is 30 cm and is constant. The flow through the turbine is 5.5 m3/s and the speed is 250 rpm. The head on the turbine is 45 m and the discharge is radial. Neglecting vane thickness and friction, determine the vane angles at inlet and outlet. Solution: External diameter: Internal diameter: Speed: Discharge is radial: Neglect vane thickness,

D1 = 2.5 D2 = 1.5 m N = 250 rpm a2 = 90° i.e. t = 0

Width: Discharge: Head:

B1 = B2 = 0.3 m Q = 5.5 m3/s H = 45 m Vw2 = 0 K=1

To determine: b1 and b2

u1

S D1 N 60

S – 2.5 – 250 60

32.73 m/s

S D2 N S – 1.5 – 250 19.64 m/s 60 60 Q = 5.5 = pD1B1Vf1 = p × 2.5 × 0.3 × Vf 1 Vf 1 = 2.334 m/s Q = 5.5 = pD2B2Vf 2 = p × 1.5 × 0.3 × Vf 2 Vf 2 = 3.89 m/s u2

Now, \ Also, \

Head at inlet to the turbine = Head at outlet + Power/kg developed + Losses within turbine

Hydraulic Turbines

Neglecting losses (given data),

or

H

V22  (Vw1u1  Vw 2 u2 ) 2g

H

V22  Vw1u1 2g

or

45

or

45

\

Vw1

V f22 2g

 Vw1 u1

(Radial discharge, Vw2 = 0) (Q V2 = Vf 2)

3.892  Vw1 – 32.73 2 – 9.81 = 1.35 m/s

Vane angles at inlet and exit (b1, b2): Here, Vw1 < u1, therefore the inlet velocity triangle changes to

539

540

Fundamentals of Turbomachinery

\

tan (180  E1 )

\

Vf 1 u1  Vw1

.334 31.38

180 – b1 = 4.26°

\

b1 = 175.76°

tan E2 \

Vf 2 u2

3.89 19.64

Ans.

0.1981

b2 = 11.20°

Ans.

EXAMPLE 7.25 The following data pertains to Francis turbine. Shaft power = 1000 kW, head = 200 m, overall efficiency = 85%, speed = 540 rpm, velocity of flow at inlet = 9 m/s. The ratio of width-to-diameter of wheel at inlet = 1/10, hydraulic efficiency = 87%, area occupied by thickness of blades = 7.5%. Find (a) the area of flow, (b) the angle of entry, (c) the tangential velocity and (d) the velocity of whirl at the inlet if the discharge is radial. Solution:

Data: Net head: Overall efficiency, Speed, B D

H = 200 m ho = 85% N = 540 rpm

Shaft power = 1000 kW hH = 87% Vf 1 = 9 m/s

1 , thickness = t = 7.5% of circumferential area. 10

Radial discharge, a2 = 90°, Vw2 = 0, Vf 2 = V2

Hydraulic Turbines

S.P. W.P.

Ko

P U gQH

1000 – 1000 1000 – 9.81 – Q – 200

\

Q = 0.6 m3/s

Also,

Q = Actual area of flow × Vf 1

541

0.85

7.5 È Ø ÉÊ S D1 B1  100 S D1 B1 ÙÚ V f 1

or

0.6 = 0.925 pD1B1Vf 1 = 0.925 × p × D1 × 0.1D1 × 9

\

D1 = 0.47897 m B1 = 0.1D1 = 0.1 × 0.47897 = 0.047897 m

(a) Actual area of flow (A1):

A1

Q Vf 1

0.6 9

0.0667 m 2

Ans.

(b) Tangential velocity (u1):

u1

S D1 N 60

S – 0.47897 – 540 60

13.543 m/s

Ans.

(c) Angle at entry (a1):

or Now, \

KH

Vw1u1 gH

Vw1

0.87 – 9.81 – 200 13.543

tan D1

Vf 1 Vw1

\

9 126.04

Vw1

K H gH u1 126.04 m/s

0.0714059

a1 = 4.084°

Ans.

(d) Velocity of whirl at inlet (Vw1): Vw1 = 126.04 m/s

Ans.

EXAMPLE 7.26 A Francis turbine working under a head of 30 m has a wheel diameter of 1.2 m at the entrance and 0.6 m at the exit. The vane angle at the entrance is 90° and the guide angle is 15°. The water at the exit leaves the vanes without any tangential velocity and the velocity of flow in the runner is constant. Neglecting the effect of draft turbe and losses in the guide and runner passages, determine the speed of the wheel in rpm, and the vane angle at the exit. State whether the speed calculated is synchronous or not. If not what speed would you recommend to couple the turbine with a 50 Hz alternator. Solution:

Data: H = 30 m, b1 = 90°,

D1 = 1.2 m, a1 = 15°

D2 = 0.6 m, Vw2 = 0,

f = 50 Hz Vf1 = Vf 2

542

Fundamentals of Turbomachinery

To determine: N, b2 Vane angle at entry

b1 = 90°,

Radial discharge

a2 = 90°,

Vf 1

tan D1

u1

u1 = Vw1, Vr1 = Vf1 Vw2 = 0,

;

V2 = Vf 2

Vf 1

tan 15’

u1

;

\ u1 = 3.732Vf1

From Eq. (7.40b), H

or

30 

V22 2g

(Vw1 u1 “ Vw 2 u2 ) g

V f22

(3.732 V f 1 )2

2 – 9.81

9.81

\

Vw1 u1 g

u12 g

Vf1 = 4.51 m/s

(a) Speed of the wheel (N): u1 = 3.732Vf1 = 3.732 × 4.51 = 16.83 m/s

S D1 N 60 N = 267.857 rpm u1 16.83

\

u2

S D2 N 60

S – 1.2 – N 60 Ans.

S – 0.6 – 267.857 60

8.414 m/s

(b) Vane angle at the exit (b2):

tan E2 \

Vf 2 u2

4.51 8.285

0.544

b2 = 28.56° f = frequency = 50 Hz =

\

NS = synchronous speed =

\

NS

Speed of the turbine Synchronous speed

Ans. poles – synchronous speed 60 50 – 60 P

50 – 60 250 rpm (Assume, P = 12) 12 N = 267.857 rpm

NS = 250 rpm

\ Speed of the turbine is not synchronous It is recommended that the speed of the turbine be 250 rpm.

Þ ß à

Ans.

Hydraulic Turbines

543

EXAMPLE 7.27 A dam powerhouse is proposed to be built for which a Francis turbine is required to be designed. The design head is 16 m, and the design flow rate is 8 m3/s. The speed is to be 250 rpm. An overall efficiency of 0.9, hydraulic efficiency of 0.95, a speed ratio of 0.76 and flow ratio of 0.35 may be assumed. Obtain all the salient dimensions (outer, inner diameters, width), blade angles and guide vane angles. The inner diameter is half the outer diameter and the discharge does not have any whirl component. Neglect vane thickness. Solution:

Data: Q = 8 m3/s, hH = 0.95, Radial discharge V 2 = Vf 2

H = 16 m, ho = 0.9, y = 0.35 Vw2 = 0,

N = 250 rpm f = 0.76, a2 = 90°,

To determine: D1, D2, B1, B2, a1, b1, b2 D2 = D1/2

(t = thickness, is neglected)

(a) Wheel diameters (D1, D2): \

\

flow ratio = 0.35 =

Vf 1

Vf 1 2 – 9.81 – 16

2 gH

Vf 1 = 6.20 m/s f = speed ratio = 0.76 =

\

u1 = 13.47 m/s

Now,

u1 = 13.47 m/s =

\

D1 = 1.029 m

\

D2

1.029 2

\

u2

S D2 N 60

S D1 N 60

u1

u1

2 gH

2 – 9.81 – 16

S D1 250 60 Ans. Ans.

0.5143 m

S – 0.5143 – 250 60

6.73 m/s

(b) Vane width (B1 = B2): Q = pD1B1Vf 1 = 8 = p × 1.029 × B1 × 6.2 \

B1 = 0.399 m = B2

Ans.

(c) Inlet guide angle (a1): KH

\

Vw1

Vw1u1 Vw1 – 13.47 gH 9.81 – 16 = 11.07 m/s

0.95

544

Fundamentals of Turbomachinery

\

Vf 1

6.20 Vw1 11.07 a1 = 29.25°

tan D1

0.560 Ans.

(d) Vane angles (b1, b2): Here u1 > Vw1, therefore, the inlet velocity triangle changes from Figure (a) to (b) as shown below

Vf 1

tan (180  E1 ) Now, \

2.5833 u1  Vw1 b1 = 111.16° Q = 8 = pD2B2Vf 2 = p × 0.5143 × 0.399 × Vf 2 Vf 2 = 12.41 m/s

Ans.

Vf 2

12.41 1.8438 6.73 u2 b2 = 61.52°

tan E2 \

6.2 13.47  11.07

Ans.

(e) Power developed (P): Ko

\

P U gQH

0.9

P = 1130.112 kW

P 1000 – 9.81 – 8 – 16

Ans.

EXAMPLE 7.28 The following data is given for a Francis turbine. Net head = 70 m, speed = 600 rpm, shaft power = 370 kW, overall efficiency = 80%, hydraulic efficiency = 95%, flow ratio = 0.25, Breadth ratio = 0.1, outer diameter of the runner = 2 × inner diameter of the runner,

Hydraulic Turbines

545

the thickness of vanes occupies 10% of the circumferential area of the runner, velocity of flow is constant and discharge is radial at outlet. Determine (a) the guide blade angle, (b) the runner vane angles at inlet and outlet, (c) the diameter of the runner at inlet and outlet, and (d) the width of the wheel at inlet.

Solution: Data: Net head: H = 70 m, Speed: Shaft power: S.P. = 370 kW Overall efficiency: Hydraulic efficiency: = 95% Flow ratio: Breadth ratio: = 0.1 D1 Thickness of the vanes = 10% circumferential area.

N = 600 rpm ho = 80% y = 0.25 = 2 × D2

To determine: (a) a1, (b) b1 and b2, (c) D1, D2, (d) B1 Ko

\

0.8

S.P. W.P.

P W.P.

370 – 1000 1000 – 9.81 – Q – 70

Q = 0.674 m3/s y = flow ratio = 0.25 =

\

P U gQH Vf 1 2 gH

Vf 1 2 – 9.81 – 70

Vf1 = 9.265 m/s

(a) Inner and outer diameters (D1, D2): Q

or

10 È Ø S D1 B1 Ù V f 1 ÉÊ S D1 B1  Ú 100

È B1 ÉÊ' D 1

Ø 0.1Ù Ú

0.674 = 0.9pD1B1Vf 1 = 0.9 × p × D1 × 0.1D1 × 9.265

546

Fundamentals of Turbomachinery

\

D1 = 0.507 m

D1 2

D2

\

Ans.

0.507 2

0.2535 m

Ans.

u1

S D1 N 60

S – 0.507 – 600 60

15.93 m/s

u2

S D2 N 60

S – 0.2535 – 600 60

7.97 m/s

KH

Vw1 u1 gH

0.95

Vw1 – 15.93 9.81 – 70

Vw1 = 40.95 m/s

(b) Guide blade angle (a1):

tan D1 \

Vf 1

9.265 40.95

Vw1

0.2263

a1 = 12.75°

Ans.

(c) Runner vane angles at inlet (b1) and outlet (b2):

Vf

tan E1 \

Vw1  u1

9.265 40.95  15.93

0.3703

b1 = 20.32°

tan E2 \

Vf 2 u2

Ans.

9.265 7.97

1.162

(Q Vf1 = Vf 2 = data)

b2 = 49.29°

Ans.

(d) Width at inlet (B1): B1 D1

0.1

\ B1 = 0.1D1 = 0.1 × 0.507 = 0.0507 m

Ans.

EXAMPLE 7.29 An inward flow reaction turbine with a supply of 0.60 m3/s under a head of 15 m develops 75 kW at 400 rpm. The inner and outer diameters of the runner are 40 cm and 65 cm respectively. Water leaves the exit of the turbine at 3 m/s. Calculate (a) the theoretical hydraulic efficiency and the actual hydraulic efficiency and (b) the inlet angles. Assume radial discharge and take width to be constant. Solution: Data: Q = 0.6 m3/s, N = 400 rpm, V2 = 3 m/s

H = 15 m, D1 = 65 cm,

P = 75 kW D2 = 40 cm,

Hydraulic Turbines

547

To determine: hHT, hHA, a1, b1 and b2 Radial discharge, a2 = 90°, Vw2 = 0, Vf 2 = V2 (a) Hydraulic efficiency: actual (hHA) and theoretical (hHT):

Ko K HT

75 – 1000 1000 – 9.81 – 0.6 – 15

P U gQH

0.84947 84.95%

(Head inlet – Head outlet) Head inlet

gH V22  gc 2 gc gH gc

2 gH  V22 2 gH

(2 – 9.81 – 15)  32 2 – 9.81 – 15

hHA = hHT × hV

0.9694

96.94%

Ans.

(Assume hV = 0.96)

= 0.9694 × 0.96 = 0.931 = 93.1% (b) Angles, a1, b1, b2:

u1

S D1 N 60

S – 0.65 – 400 60

u2

S D2 N 60

S – 0.4 – 400 60

12.57 m/s

8.38 m/s

Ans.

548

Fundamentals of Turbomachinery

KHT

or \ Now, \ \

0.9694

Q

\

Vw1 – 12.57 9.81 – 15

S D1 B1 V f 1

S D2 B2 V f 2

Vf 1

D2 V2 D1

Vf 1 Vw1

0.4 – 3 0.65

1.85 11.35

(Q B1 = B2, Vf2 = V2)

1.85 m/s

0.163

a1 = 9.24°

Ans.

\ The inlet velocity triangle changes to:

tan (180  E1 )

Vf 1 u1  Vw1

1.85 12.57  11.35

b1 = 123.4°

tan E 2 \

(Q Vw2 = 0)

D1Vf 1 = D2V2

Here, Vw1 < u1

Now,

Vw1 u1 gH

Vw1 = 11.35 m/s

tan D1 \

(Vw1 u1 “ Vw 2 u2 ) gH

Vf 2 u2

V2 u2

b2 = 19.697°

Ans.

3 8.38

0.358 Ans.

EXAMPLE 7.30 A Francis turbine has the following data. Power = 13 MW, head = 200 m, specific speed = 110, overall efficiency = 0.85, hydraulic efficiency = 0.9, Ratio of width to diameter of wheel = 0.1, total thickness of the blades 0.1 times the outer diameter, inlet flow

Hydraulic Turbines

549

velocity = 0.85 times the exit flow velocity, inlet flow velocity = 10 m/s. Calculate (a) the area, (b) the guide blade angle, (c) the peripheral velocity, and (d) the velocity of whirl at inlet. Assume axial discharge and take the blade width to be constant. Solution: Data:

P = 13 MW, hH = 0.9, Vf 1 = 10 m/s Vw2 = 0 N ST

\

H = 200 m, B1/D1 = 0.1, Axial discharge Vf 2 = V2 N P H5/ 4

110

NST = 110, nt = 0.1D1, a2 = 90°,

ho = 0.85 Vf 1 = 0.85Vf 2 B1 = B2

N 13 – 103 (200)5 / 4

N = 725.6 rpm Ko

0.85

P U gQH

13 – 10 6 1000 – 9.81 – Q – 200

\

Q = 7.795 m3/s

Now,

Q = 7.795 = (pD1 – nt)B1Vf1 = (pD1 – 0.1D1)B1Vf1

or \

7.795 = (pD1 – 0.1D1) × 0.1D1 × 10 = D12 (S  0.1) – 0.1 – 10 D1 = 1.6 m

(a) Gross area (Ag) and net area (An):

An

Q Vf 1

7.795 10

0.7795 m 2

Ans.

Ag = Net area + Area occupied by the blade = pD1B1 = p × 1.6 × 0.1 × 1.6 = 0.804 m2

Ans.

550

Fundamentals of Turbomachinery

(b) Guide blade angle (a1), vane angles (b1, b2):

S D1 N S – 1.6 – 725.6 60.79 m/s 60 60 Vw1u1 Vw1 – 60.79 KH 0.9 gH 9.81 – 200 Vw1 = 29.05 m/s u1

\

tan D1 \

Vf 1 Vw1

10 29.05

0.344

a1 = 18.995°

Here,

Ans.

Vw1 < u1

Therefore, the inlet velocity triangle changes to as shown in the figure below. Vf 1 10 tan (180  E1 ) 0.3151 u1  Vw1 60.79  29.05 b1 = 162.5°

Q

or \

7.795

(S D2  nt ) E2V f 2

Ans.

(S D2  0.1D1 ) – 0.1D1 –

(S D2  0.1 – 1.6) – 0.1 – 1.6 –

Vf 1 0.85

10 0.85

D2 = 1.3691 m

u2 tan E 2

S D2 N 60 Vf 2 u2

b2 = 12.75°

S – 1.3691 – 725.6 60 È Vf 1 Ø 1 ÉÊ 0.85 ÙÚ – 52.0145

52.0145 m/s

10 1 – 0.85 52.0145

0.2262

Ans.

Hydraulic Turbines

551

(c) Peripheral velocity (u1, u2): u1 = 69.79 m/s

u2 = 52.0145 m/s

Ans.

(d) Velocity of whirl at inlet (Vw1): Vw1 = 29.05 m/s

Ans.

EXAMPLE 7.31 If the pressure at inlet and outlet of the turbine are 13 bar and 0.1 bar respectively, height of the runner inlet and outlet are 7 m and 0.5 m respectively above the tail race, calculate the loss of energy of Example 7.26. Solution:

Data:

Considering the data of Example 7.26 and p1 = 12 bar

p2 = 0.1 bar

Z1 = 7 m

Z2 = 0.5 m

To determine: Loss of energy Hi = total head at inlet to the turbine Ë p1 V12 Û   Z1 Ü Ì ÌÍ U g 2 g ÜÝ Ho = total head at outlet from the turbine Ë p2 V22 Û   Z2 Ü Ì ÌÍ U g 2 g ÜÝ From inlet velocity triangle, V1

(V f21  Vw21 )

10 2  29.052

31.15 m/s

\

Hi

13 – 10 5 31.152   7 188.97 m 1000 – 9.81 2 – 9.81

and

Ho

0.1 – 10 5 10 2   0.5 1000 – 9.81 2 – 9.81

KH

Vw1 u1 gH

\

6.62 m

Power developed = Energy transfered to the runner = hH × H = 0.9 × 200 = 180 m Head at inlet = Head at outlet + Losses + Power developed

or \

188.97 = 6.62 + Head loss + 180 m Head loss = 188.97 – 6.62 – 180 = 2.354 m

Ans.

552

Fundamentals of Turbomachinery

EXAMPLE 7.32 An inward flow radial turbine has an overall efficiency of 80%. The net head across the turbine is 6 m and the required power output is 125 kW. The runner tangential velocity is 0.96 2gH while the flow velocity is 0.36 2 gH . If the speed of the runner is 250 rpm and the hydraulic losses account for 18% of the energy available, calculate the inlet guide, vane exit angle, the inlet angle of the runner vane, the runner diameter at inlet and the height of the runner at inlet. Assume that the discharge is radial. Solution:

ho = 80%,

Data:

H = 6 m,

Vf 1 = 0.36 2 gH ,

u1

P = 125 kW,

N = 250 rpm,

Hydraulic losses = 18% of energy available, Discharge is radial, a2 = 90° \

Vw2 = 0,

Vf 2 = V2

To determine: a1, b2, b1, D1, B1 KH

\

KH

Ko \

Total head at inlet – Hydraulic loss Head at inlet

6

18 –6 100 6

S.P. P = W.P. W.P.

Q = 2.655 m3/s

H  0.18H H

0.82

125 – 1000 U gQH

125 – 1000 1000 – 9.81 – Q – 6

0.8

0.96 2 gH

Hydraulic Turbines

553

(Vw1u1  Vw 2 u2 ) gH

\

KH

or

0.82

\

Vw1 = 4.6338 m/s

0.82

Vw1 u1 gH

Vw1 – 0.96 2 – 9.81 – 6 9.81 – 6

(Q Vw2 = 0)

(a) Runner diameter at inlet (D1): u1

S D1 N 60 D1 = 0.796 m u1

\

0.96 – 2 – 9.81 – 6

0.96 2 gH

10.42

10.42 m/s

S – D1 – 250 60 Ans.

(b) Height of runner = Width of the vane (B1):

Vf 1

0.36 2 gH

0.36 – 2 – 9.81 – 6

= 3.91 m/s \

Q = 2.655 = pD1B1Vf 1 = p × 0.796 × B1 × 3.91

\

B1 = 0.272 m

Ans.

(c) Guide angle at inlet (a1), vane angle at inlet (b1):

tan D1 \

Vf 1 Vw1

3.91 4.6338

0.8438

a1 = 40.2°

Here, u1 > Vw1, therefore the inlet velocity triangle changes to:

Ans.

554

Fundamentals of Turbomachinery

tan (180  E1 ) \

Vf 1 u1  Vw1

3.91 10.42  4.634

0.676

b1 = 145.95°

Ans.

EXAMPLE 7.33 A Francis turbine has the following data. Outer peripheral speed = 35 m/s, speed = 450 rpm, water enters the runner without shock, entry flow velocity is 10 m/s, water leaves the runner without shock, without whirl, leaving absolute velocity is 7 m/s. The difference between the sum of the static and potential heads at entrance to the runner and at the exit from the runner is 65 m. If the turbine develops 13,000 kW and has a flow rate of 12 m3/s of water when the net head is 120 m, calculate: (a) (b) (c) (d) (e)

The The The The The

absolute velocity of the water at entry to the runner inlet guide vane angle inlet runner vane angle head lost in the runner inlet runner diameter.

Solution:

Data: Outer pheripheral speed: Speed: Entry flow velocity: Without whirl at exit: Exit absolute velocity: Power developed:

u1 = 35 m/s N = 450 rpm Vf1 = 10 m/s Vw2 = 0 \ a2 = 90°, V2 = Vf 2 V2 = 7 m/s P = 13000 kW

Hydraulic Turbines

Q = 12 m3/s H = 120 m = HTI = HTO

Discharge: Net head: Total head at inlet to runner: Total head at exit of runner: To determine: (a) V1, (b) a1,

555

(c) b1,

(d) The head lost in the runner, (e) D1

P = rQ(Vw1u1 ± Vw2u2) 1000

kg m

3

– 12

m3 Ë m mÛ – ÌVw1 – 35 Ü s s sÝ Í

42 – 10 4 – Vw1

kg-m m s2 s

(Q Vw2 = 0)

N-m s 13000 = (420 × Vw1) kW 42 – 10 4 – Vw1

or \

42 – 10 4 10 3

– Vw1 kW

Vw1 = 30.95 m/s

(a) Absolute inlet velocity (V1): V f21  Vw21

V1

10 2  30.952

32.53 m/s

Ans.

(b) Inlet guide vane angle (a1): Here, Vw1 < u1 \ Inlet velocity triangle changes to the figure shown below:

\ or

tan D1

Vf 1 Vw1

a1 = 17.9°

10 30.95

0.323 Ans.

556

Fundamentals of Turbomachinery

(c) Runner vane angle (b1):

tan (180  E1 ) \

Vf 1 u1  Vw1

10 35  30.95

2.469

b1 = 112.05°

Ans.

(d) The head lost in the runner (HL): HTI = total head at inlet to the runner Ë p1 V12 Û   Z1 Ü Ì ÍÌ U g 2 g ÝÜ

HTO = total head at outlet of the runner Ë p2 V22 Û   Z2 Ü Ì ÌÍ U g 2 g ÜÝ

Head transferred to the runner (HR) HR

KH H

Vw1u1 g

(30.95 – 35) 9.81

110.423 m/s

Total head at inlet to the runner = Total head at exit of the runner + Head transferred to the runner + Head lost in the runner i.e. or

\

HTI = HTO + HR + HL p1 V12   Z1 U g 2g HL

P2 V22   Z 2  110.423  H L U g 2g ( p1  p2 ) (V 2  V22 )  ( Z1  Z 2 )  1  110.423 2g Ug

Difference between the static and potential heads at inlet and outlet of the runner = 65 [given data] \

or

HL

65 

(V12  V22 )  110.423 2g

65 

(32.53)2  72  110.423 2 – 9.81

HL = 6.014 m

Ans.

Hydraulic Turbines

557

(e) Inlet runner diameter (D1):

S D1 N 35 60 D1 = 1.485 m u1

\

S – D1 – 450 60 Ans.

EXAMPLE 7.34 An inward flow reaction turbine has the following data. Speed = 250 rpm, velocity flow is constant = 4 m/s, inlet diameter = 1.2 m, width at inlet = 0.15 m, radial discharge at inlet and outlet. Calculate (a) the work done per weight, (b) the power developed by the turbine, (c) the water head on the machine, and (d) the hydraulic efficiency. Solution:

Data: Speed: Flow velocity: Inlet diameter: Width at inlet: Radial discharge: Radial inlet:

S D1 N 60

u1

N = 250 rpm Vf1 = Vf 2 = 4 m/s D1 = 1.2 m B1 = 0.15 m a2 = 90°, Vw2 = 0, Vf 2 = V2 b1 = 90°, Vw1 = u1

S – 1.2 – 250 60

15.71 m/s

(a) Work done per weight (W.D./Weight): P = rQ (Vw1u1 ± Vw2u2) or

W.D. Weight

U QVw1 u1 U Qg

kg m3

–

kg m

3

(Q Vw2 = 0 for radial discharge)

m3 m m – – s s s –

3

m m – 2 s s

N-m s N s

N-m N

558

Fundamentals of Turbomachinery

Vw1 u1 g

15.71 – 15.71 9.81

W.D. Weight

\

N-m N

[Q Vw1 = u1, radial inlet]

25.16 N-m/N

Ans.

(b) Power developed (P): Q = pD1B1Vf 1 = p × 1.2 × 0.15 × 4 = 2.262 m3/s \

P = rQVw1u1 = 1000 × 2.262 × 15.71 × 15.71 = 558.26 kW

Ans.

(c) Water head on the machine (H): Applying Bernaulli’s equation, H

V22 2g

\

Vw1u1 g Vw1 u1 V22  g 2g

H

25.16 

42 2 – 9.81

25.975 m

Ans.

(d) Hydraulic efficiency (hH): KH

Vw1 u1 gH

5.16 25.975

0.9686

96.86%

Ans.

EXAMPLE 7.35 Show that the volumetric efficiency for a Francis turbine having velocity of flow through runner remaining constant is given by the relation 1 1 tan 2 D1 2 1 tan D1 1 tan E1

(a) KH

(b) K

2 2  tan 2 D1

if the turbine has radial discharge at outlet

if the turbine has radial outlet

Solution: (a) Radial discharge at outlet: a2 = 90°,

Vw2 = 0,

Vf1 = Vf2

tan D1 tan E1

Vf 1 Vw1 Vf 1

Vf2 = V2

(data given)

? Vf 1

Vw1  u1

Vw1 tan D1

? (Vw1  u1 )

(i)

Vf 1 tan E 2

(ii)

Hydraulic Turbines

559

Substituting Vf1 from Eq. (i) in Eq. (ii), (Vw1  u1 )

Vw1 tan D1 tan E1

u1

Vw1 

V22 2g

Vw1u1 g

Vw1 tan D1 tan E1

tan D1 Ø È Vw1 É1  tan E1 ÚÙ Ê

(iii)

We have H

\

H

(for radial discharge) 2 Vw1 u1 V f  2g g

Vw1 u1 V22  2g g

Substituting Eqs. (i), (ii) in (iv), H

Vw21 Ë tan D1 Û Ë (Vw1 tan D1 )2 Û Ü Ì1  ÜÌ g Í tan E1 Ý ÌÍ 2g ÜÝ Vw21 g

or

H

KH

Ë tan D1 Û Vw21 1 tan 2 D1  Ì Ü g tan 2 E 1Ý Í

Vw21 Ë tan D1 tan 2 D1 Û  Ì1  Ü g ÍÌ tan E1 2 ÝÜ

Vw1 u1 gH

Vw1 u1 g

Ë tan D1 tan 2D1 Û  Ì1  Ü tan E1 2 Ý g Í

Vw21

(iv)

560

Fundamentals of Turbomachinery

Ë tan D1 Û Vw1Vw1 Ì1  tan E1 ÜÝ Í Ë tan D1 tan 2D1 Û Vw21 Ì1   Ü tan E1 2 Ý Í

1 1 tan 2 D1 2 1 tan D1 Ø È ÉÊ1  tan E ÚÙ 1

KH

(b) Radial entry:

b1 = 90°, Vw1 = u1

\

KH

\

KH

Proved.

1

1

1 tan 2 D1 1 2 tan D1 1 ‡

1

1 tan 2 D1 2

2

Proved.

2  tan 2 D1

EXAMPLE 7.36 Three Francis turbines have the following same data. Inlet diameter = 0.5 m, same efficiency, same head, same inlet velocity of flow = 5 m/s. The other data is: Turbine 1

Turbine 2

Turbine 3

Speed 500 rpm Inlet blade angle = 60° Inlet blade angle = 90° Calculate the speed of the remaining two turbines. Solution:

Inlet blades angle 110°

Data:

Inlet diameter Speed b1 Suffix 1, 2, 3 ®

Turbine 1

Turbine 2

Turbine 3

0.5 m 500 rpm 60°

0.5 m ? 90°

0.5 m ? 110°

Turbine 1,

Turbine 2, Turbine 3

(u1)1 = inlet tangential velocity of turbine 1

(u1 )1 (tan E1 )1 \

S D1 N1 60 Vf 1 Vw1  u1

S – 0.5 – 500 60 Vw1

(Vw1)1 = 15.977 m/s

5  13.09

13.09 m/s (tan 60’)1

Hydraulic Turbines

561

All three turbine have the same efficiency. Ë Vw1 u1 Û Ì gH Ü Í Ý1

Ë Vw2 u2 Û Ì gH Ü Í Ý2

Ë Vw3 u3 Û Ì gH Ü Í Ý3

We have the same head, H1 = H2 = H3 \

(Vw1u1)1 = (Vw1u1)2 = (Vw1u1)3

\

(Vw1u1 )2 (u12 )2

(i)

15.977 – 13.09

(Vw1u1 )1

209.13 m 2 /s2

209.13 m 2 /s2

(u1)2 = 14.46 m/s

Ë S D1 N Û Ì 60 Ü Í Ý2

Now,

(u1 )2

\

(N)2 = 552.38 rpm

Ë S – 0.5 – N Û Ì Ü 60 Í Ý2

14.46 Ans.

From Eq. (i), (Vw1u1)3 = (Vw1u1)1 = 209.13 m2/s2

Vw1

[tan (180’  110’)]3 \ \

Ë 209.13 Û Ì Ü Í u1 Ý3 Ë Vf 1 Û Ì Ü Í u1  Vw1 Ý3

Ë Û 5 Ì Ü Í (u1  Vw1 ) Ý3

(u1 – Vw1)3 = 1.82 m/s

Ë 209.13 Û Ìu1  Ü u1 Ý3 Í

1.82

? (u1 )3

15.4 m/s

(ii)

562

Fundamentals of Turbomachinery

Ë S D1 N Û Ì 60 Ü Í Ý3

Ë S – 0.5 – N Û Ì Ü 60 Í Ý3

Now,

(u1 )3

\

(N)3 = 588.24 rpm

Turbine 1

(b1 < 90°), Speed = 500 rpm

Turbine 2

(b1 = 90°), Speed = 552.38 rpm Medium

Turbine 3

(b1 > 90°), Speed = 588.24 rpm Fast

15.4

Ans. Slow

EXAMPLE 7.37 A Francis turbine operates at a speed of 1250 rpm. The net head across the turbine is 125 m and the flow rate is 0.5 m3/s. The radius of the runner is 0.5 m, the height of the runner vane at inlet is 0.03 m, the angle of the inlet guide vanes is 70° from the radial direction. Assuming absolute flow velocity to be radial at exit, calculate the torque, the power, and the hydraulic efficiency. Solution: Data:

N = 1250 rpm, R1 = 0.5 m, Vw2 = 0,

H = 125 m, B1 = 0.03 m, Vf 2 = V2

Q = 0.5 m3/s a1 = 70°, a2 = 90°,

(a) Torque (T): T = rQ (Vw1R1 ± Vw2R2) = rQ Vw1R1 = 1000 × 0.5 × 0.5 × Vw1

(Q Vw2 = 0)

or

T = 250Vw1 N-m

Now,

Q = pD1B1Vf1 = p × 2 × 0.5 × 0.03 × Vf1 = 0.5

\

Vf1 = 5.31 m/s

From the inlet velocity triangle, tan D1

\ Hence,

Vw1 Vf 1

tan 70’

Vw1 5.31

Vw1 = 14.59 T = 250 × 14.59 = 3647.5 N-m

Ans.

Hydraulic Turbines

563

(b) Power (P): 3647.5 Nm – 2 – S – 1250 60 P = 477.430 kW P

\

TZ

Ans.

(c) Hydraulic efficiency (hH):

KH

P U gQH

477.43 – 1000 1000 – 9.81 – 0.5 – 125

77.87%

Ans.

EXAMPLE 7.38 A Francis turbine produces 400 kW power when running at 750 rpm, under a head of 80 m. A model of this turbine is to be tested under a head of 5 m. Model is scaled to 1:5. Calculate the speed and power of the model. Solution:

We have Ë H Û Ì 2 2Ü Í N D Ým

Ë H Û Ì 2 2Ü Í N D Ýp

(a) Speed of the model (Nm): 2

ËD Û ËH Û N 2p Ì P Ü Ì m Ü Í Dm Ý ÍÌ H p ÜÝ Nm = 937.5 rpm

N m2

\

Ë5Û 750 2 – (5)2 Ì Ü Í 80 Ý

Ans.

(b) Power of the model (Pm): Ë P Û Ì 3 5Ü Í N D Ýp

Ë P Û Ì 3 5Ü Í N D Ým 3

ËN Û Pp Ì m Ü ÌÍ N p ÜÝ Pm = 0.16 kW

\

Pm

Ë Dm Û Ì Ü ÌÍ D p ÜÝ

5

400 –

937.5 Ë 1 Û 750 ÌÍ 5 ÜÝ

5

Ans.

EXAMPLE 7.39 A Kaplan turbine has the following data. Power developed = 5000 kW, head = 40 m, speed ratio = 2.0, flow ratio = 0.6, hub diameter = 40% of outer diameter, overall efficiency = 90%. Calculate the hub diameter, the outer diameter, the speed, and the specific speed. Solution:

Data: P = 5000 kW, H = 40 m,

speed ratio, f = 2.0,

flow ratio, y = 0.6,

Db = 0.4Do, ho = 90% \

0.6

Vf 1 2 gH

\

Vf 1

0.6 2 – 9.81 – 40 = 16.81 m/s

564

Fundamentals of Turbomachinery

Ko

S.P. W.P.

0.9

\

Q = 14.16 m3/s

Now,

Q 14.16

5000 – 1000 1000 – 9.81 – Q – 40

P U gQH

S ( Do2  Db2 ) V f 1 4

S [ Do2  (0.4 Do )2 ] – 16.81 4

\

Do = 1.13 m

Ans.

\

Db = 0.4Do = 0.4 × 1.13 = 0.453 m

Ans.

\

I

u

2

\

2 gH

S Do N 60 N = 946.96 rpm u

Now, \ and

56.03

N P

N ST

H

u

2 – 2 – 9.81 – 40 = 56.03 m/s

S – 1.13 – N 60 Ans.

946.96 5000

5/4

2 2 gH

40 5 / 4

Ans.

665.64 rpm

EXAMPLE 7.40 A Kaplan turbine runner is to be designed to develop 9000 kW. The net available head is 6 m. If the speed ratio is 2.01 and the flow ratio 0.7, overall efficiency 87%, the diameter of boss being 1/3rd of the diameter of runner, find the diameter of the runner, its speed and the specific speed of the turbine. Solution: u1

Vw 1 a1

b1

V1

Vr 1

Vr 2

Vf 1

Vf 2 = V2

b2 u2

Data: Power developed: Net head: Speed ratio: Flow ratio: Overall efficiency: Diameter of the boss: To determine: Do, N, NST

P = 9000 kW H=6m f = 2.01 y = 0.7 ho = 0.87 Db = (1/3) × Do

a 2 = 90°

Hydraulic Turbines

u

I \

2 gH

2.01

565

u 2 – 9.81 – 6

u = 21.81 m/s Vf 1

\

\

2 gH

0.7

Vf 1 = 7.595 m/s P Ko 0.87 U gQH

Vf 1 2 – 9.81 – 6

9000 – 1000 1000 – 9.81 – Q – 6

Q = 175.75 m3/s

\

(a) Runner diameter (Do):

Q 175.75 \

2 S Ë 2 Ë Do Û Û Ì Do  Ì Ü Ü V f 1 4 ÍÌ Í 3 Ý ÝÜ

S 4

Ë 2 Do2 Û Ì Do  Ü 7.595 9 ÜÝ ÍÌ

Do = 5.76 m

Ans.

(b) Speed (N):

S Do N 60 N = 72.35 rpm u

\

21.81

S – 5.76 – N 60 Ans.

(c) Specific speed (NST):

N ST

N P H

72.35 – 9000

5/ 4

65 / 4

730.92 rpm

Ans.

EXAMPLE 7.41 The hub diameter of a Kaplan turbine, working under a head of 12 m, is 0.35 times the diameter of the runner. The turbine is running at 100 rpm. If the vane angle of the extreme edge of the runner at outlet is 15° and flow ratio 0.6, find (a) the diameter of the runner and boss, and (b) the discharge through the runner. The velocity of whirl at outlet is given as zero. Solution:

Data: Head: Hub diameter: Speed: Vane angle at exit: Flow ratio: Whirl velocity at exit:

To determine: Do, Db, Q

H = 12 m Db = 0.35Do N = 100 rpm b2 = 15° y = 0.6 Vw2 = 0, Vf 2 = V2, a2 = 90°

566

Fundamentals of Turbomachinery u1

Vw 1

a1

b1

V1

Vr 1

Vr 2

Vf 1

V2 = Vf 2

b2 u2

\

\

0.6

Vf 1

a 2 = 90°

Vf 1 2 – 9.81 – 12

2 gH

Vf1 = 9.2 m/s = Vf 2

tan E2 \

Vf 2 u2

9.2 u2

tan 15’

u2 = 34.33 m/s

(a) Diameter (Do, Db):

u2

34.33

S Do N 60

S Do – 100 60

\

Do = 6.55 m

Ans.

and

Db = 0.35 × Do = 0.35 × 2.2925 m

Ans.

(b) Discharge (Q): S 2 ( Do  Db2 )V f 4 Q = 271.77 m3/s

Q

\

S – (6.552  2.32 ) – 9.2 4

Ans.

EXAMPLE 7.42 A Kaplan turbine has outer and hub diameters of 4 m and 2 m respectively. It develops 25 MW when working under a head of 20 m, with an overall efficiency of 85% and running at 150 rpm. Find (a) the discharge through the turbine, (b) the peripheral velocity at the hub and at the tip of the blade, and (c) the runner blade angle at inlet and outlet. Solution:

Data: Outer diameter: Inner diameter: Power developed: Head: Overall efficiency: Speed: To determine: Q, b1, b2, uo, ub

Do = 4 m Db = 2 m P = 25 MW H = 20 m ho = 85% N = 150 rpm

Hydraulic Turbines

567

(a) Discharge (Q): Ko

P U gQH

25 – 1000 – 1000 1000 – 9.81 – Q – 20

0.85

Q = 149.91 m3/s

\

Ans.

(b) Peripheral velocity at tip (uo) and at hub (ub): S ( Do2  Db2 )V f 4 Vf1 = 15.91 m/s Q

\

149.91

uo

S Do N 60

S – 4 – 150 60

ub

S Db N 60

S – 2 – 50 60

S 2 (4  22 ) – V f 1 4

31.42 m/s

Ans.

15.71 m/s

Ans.

(c) Inlet and outlet runner blade angles corresponding to outer diameter and hub diameter: Assume hydraulic efficiency = 100% KH

Now, or

(Vw1 u1 “ Vw 2 u2 ) gH

Vw1u1 = hHgH = 1 × 9.81 × 20 = 196.2

(Q Vw2 = 0)

We know that (Vw1 ub ) hub dia

(Vw1 uo ) tip dia

196.2

Runner inlet angle (b1): Corresponding to outer diameter or tip diameter. From Eq. (i) (suffix o means, outer), Vw1o = whirl velocity at inlet corresponding to tip dia (outer diameter) 196.2 uo

196.2 31.42

6.244 m/s

Vw2o = whirl velocity at outlet corresponding to tip dia. = 0 (radial discharge) Vw1b = whirl velocity at inlet corresponding to hub diameter From Eq. (i), (Vw1 ub ) hub

196.2

Vw1b

196.2 ub

196.2 15.71

12.49 m/s

(i)

568

Fundamentals of Turbomachinery

Applying the inlet velocity triangle corresponding to outer diameter (tip diameter), Do (Figure a),

tan (180’  E1 ) \

Vf 1 u0  Vw10

15.91 31.42  6.244

0.632

b1 = 147.7°

(ii) Ans.

Applying the inlet velocity triangle corresponding to hub diameter (inner diameter), Db (Figure b)

tan (180’  E1 ) \

Vf 1 ub  Vw1b

15.91 15.71  12.49

b1 = 101.44°

4.94

(iii) Ans.

Runner outlet angle (b2): Applying the outlet velocity triangle corresponding to outer diameter (tip diameter), Do (Figure c) V f 2 V f 1 15.91 tan E 2 0.5064 (iv) uo uo 31.42 b2 = 26.86° Applying the inlet velocity triangle corresponding to hub diameter (inner diameter), Db (Figure d)

tan E2

Vf 2

Vf 1

ub

ub

b2 = 45°

15.91 1.0127 15.71

(v) Ans.

Hydraulic Turbines

569

EXAMPLE 7.43 A Kaplan turbine working under a head of 20 m develops 11,775 kW. The outer diameter of the runner is 3.5 m and the hub diameter 1.75 m. The guide blade angle at the extreme edge of the runner is 35°. The hydraulic and overall efficiencies of the turbine are 0.88 and 0.84 respectively. If the velocity of whirl is zero at outlet, determine (a) the speed of the turbine and (b) the runner vane angles at inlet and outlet at the extreme edge of the runner. Solution:

Data: Head: Power: Outer diameter: Hub diameter:

H = 20 m P = 11775 kW Do = 3.5 m (tip diameter) Db = 1.75 m (inner diameter) a1 = 35°

hH = 0.88 ho = 0.84 a2 = 90°, Vw2 = 0,

Hydraulic efficiency: Overall efficiency: Radial discharge:

V2 = Vf 2

To determine: N, b1, b2 Ko

P U gQH

0.84

\

Q = 71.447 m3/s

Now,

Q

\

Vf = 9.9 m/s

tan D1 or

71.447

Vf Vw1

11775 – 1000 1000 – 9.81 – Q – 20

S ( Do2  Db2 ) V f 1 4

9.9 Vw1

S (3.52  1.752 ) V f 4

tan 35’ 0.70

Vw1 = 14.14 m/s = Vw1o KH

Vw1 u1 gH

0.88

14.14 – u1 9.81 – 20

(a) Speed of the turbine (N): u1 = 12.21 m/s = uo = peripheral velocity at tip diameter

570

Fundamentals of Turbomachinery

S Do N 60 N = 66.63 rpm

u1 \

S – 3.5 – N 60

12.21

Ans.

(b) Runner vane angle at inlet and outlet at the extreme edge of the runner (b1, b2 corresponding to tip diameter): Equation (ii) of Example (7.42),

tan (180’  E1 ) tan E1 \

Vf 1

(If uo > Vw1o)

uo  Vw1o Vf 1 Vw10  uo

9.9 14.14 – 12.21

5.13

(If uo < Vw1o)

b1 = 78.98°

Ans.

Equation (iv) of Example 7.42

tan E2 \

Vf

9.9 12.21

uo

0.81

b2 = 39°

Ans.

EXAMPLE 7.44 A Kaplan turbine produces 30,000 kW under a head of 9.6 m, while running at 65.2 rpm. The discharge through the turbine is 350 m3/s. The tip diameter of the runner is 7.4 m. The hub diameter is 0.432 times the tip diameter. Calculate (a) the turbine efficiency, (b) the specific speed of turbine, (c) the speed ratio (base on tip diameter), and (d) the flow ratio. Solution: Power: Head: Speed: Discharge: Tip diameter: Hub diameter:

P = 30,000 kW H = 9.6 m N = 65.2 rpm Q = 350 m3/s Do = 7.4 m Db = 0.432Do

To determine: ho, NST, f, y (a) Turbine efficiency (ho): Ko

P U gQH

30000 – 1000 1000 – 9.81 – 350 – 9.6

0.91 or 91%

Ans.

(b) Specific speed (NST):

N ST

N P H

5/4

65.2 30000 9.65 / 4

668.3 rpm

Ans.

Hydraulic Turbines

571

(c) Speed ratio (f):

uo I

S Do N 60

S – 7.4 – 62.5 60

uo

25.6 2 – 9.81 – 9.6

2 gH

25.6 m/s

Ans.

1.84

(d) Flow ratio (y):

\

S ( Do2  Db2 ) V f 4 Vf = 10 m/s

and

\

Q

Vf 2 gH

350

10 2 – 9.81 – 9.6

S [7.42  (0.432  7.4)2 ]V f 4

0.7286

Ans.

EXAMPLE 7.45 An axial flow hydraulic turbine has a net head of 25 m across it. The power developed by the turbine is 25 MW, when working under a speed of 150 rpm. The tip and hub dimension are 5 m and 2.0 m respectively. Hydraulic and overall efficiencies are 0.9 and 0.85 respectively. Calculate the inlet and outlet blade angles at the mean radius. Assume axial flow at outlet. Solution:

\

dm

mean diameter =

Ko

P U gQH

Do  Db 2

52 2

25 – 10 6 1000 – 9.81 – Q – 25

Q = 119.3 m3/s

3.5 m

0.85

572

Fundamentals of Turbomachinery

um KH

\

S NDm S – 150 – 3.5 27.49 m/s 60 60 Vw1 um Vw1 – 27.49 0.9 gH 9.81 – 25

Vw1 = 8.03 m/s S ( Do2  Db2 ) V f 4 Vf = 7.27 m/s Q

\

119.93

S 2 (5  2 2 ) V f 4

(a) Inlet and outlet blade angles at the mean radius: From the inlet velocity triangle,

Vf

tan (180’  E1 )

um  Vw1

7.27 (27.49  8.03)

0.3736

b1 = 159.5°

Ans.

From the outlet velocity triangle,

tan E2

Vf um

7.27 27.49

0.265

b2 = 14.81° EXAMPLE 7.46

Ans.

A conical draft tube has the following data: Inlet diameter: Outlet diameter: Velocity of water at outlet: Total height of draft tube: Height immersed in water: Frictional losses:

= = = = = =

1.4 m 1.8 m 3 m/s 7m 1.5 m 30% of the velocity head at outlet

Calculate the pressure head at the runner outlet and the efficiency of the draft tube. Solution: A1 = cross-sectional area at inlet S 2 S d1 – 1.42 1.5394 m 2 4 4 A2 = cross-sectional area at outlet

S 2 S d2 – 1.82 2.5447 m 2 4 4 V2 = velocity of water at outlet

Q = discharge through the tube

Hydraulic Turbines

573

1.4 m

1

1

7m pa

1.5 m

2

Tail race

2

1.8 m

= A2 × V2 = 2.5447 × 3 = 7.6341 m3/s hf = frictional losses =

30 V22 – 100 2 g

30 32 – 100 2 – 9.81

= 0.1376 m A1V1 = A2V2 2.5447 – 3 4.96 m/s 1.5394 Applying Bernaulli’s theorem between sections 1–1 and 2–2 (2–2 as datum line, hence Z2 = 0)

\

V1

A2V2 / A1

p1 V12   Z1 U g 2g

p2 V22   Z2  h f U g 2g

We know that pa  1.5 Ug

p1 Ug

p2 Ug

10.3  1.5 11.8 m

11.8 

(Q Atmospheric pressure = 10.3 m)

V22 V12   h f  Z1 2g 2 g

ËV 2 V 2 Û 11.8  7  Ì 1  2 Ü  h f ÌÍ 2 g 2 g ÜÝ 11.8  7 

\

p1 Ug

(4.962  32 )  0.1376 2 – 9.81

4.143 m (abs) or  6.157 m (gauge)

The negative sign represents below atmospheric pressure hd = efficiency of draft tube

(Eq. (7.47))

Ans.

574

Fundamentals of Turbomachinery

V12  V22  2 h f g

4.962  32  2 – 9.81 – 0.1376

V12

(4.96)2

= 52.44%

Ans.

EXAMPLE 7.47

A conical draft tube has the following data: Top diameter: = 2.2 m Pressure head at inlet: = 6 m (vaccum) Discharge velocity: = 1.5 m/s Discharge: = 24 m3/s Atmospheric pressure: = 20.3 m Losses are neglected Total height of draft tube: =5m Calculate the height of the draft tube immersed. Solution:

p1 = 6 m (vaccum) = 10.3 – 6 = 4.3 m (abs) Ug

V1 = velocity of water entering into the draft tube

Q A1

24

6.314 m/s S (2.2)2 4 Applying Bernaulli’s equation between sections 1–1 and 2–2 p1 V12   Z1 Ug 2g p1 V12  5 Ug 2g

p2 V22   Z2  h f Ug 2g pa V22   h1 Ug 2g

(Q hf = 0, data)

Hydraulic Turbines

\

h1

p1 p (V 2  V22 )  2  1 5 Ug Ug 2g 4.3  10.3 

\

575

(6.3142  1.52 ) 5 2 – 9.81

h1 = 0.9173 m

Ans.

EXAMPLE 7.48 A Kaplan turbine develops 1500 kW under a head of 6 m. The turbine is set 2.5 m above the tail race level. A vacuum gauge inserted at the turbine outlet records a section head of 3.2 m. If the turbine efficiency is 85%, what will be the efficiency of the draft tube having inlet diameter of 3 m? Solution:

Data: Power developed: Head: Height of the draft tube: Pressure at outlet of the turbine: Turbine efficiency: Inlet diameter of the draft tube:

P = 1500 kW H=6m Hs = 2.5 m p1 = 3.2 m (vaccum) ho = 85% d1 = 3 m

To determine: hd Ko

\ Now, or

P U gQH

1500 – 1000 1000 – 9.81 – Q – 6

0.85

Q = 29.98 m3/s S 2 d1 V1 4 V1 = 4.24 m/s Q

29.98

S – (3.0)2 – V1 4

Applying Bernaulli’s equation between inlet and outlet of the draft tube, p1 V12   Z1 U g 2g

p2 V22   Z2  h f U g 2g

hf = 0

(assumption)

Z2 = 0 Z1 = Hs = 2.5 m p2 Ug

pa Ug

10.3

p1 = 3.2 m (vaccum) = (+10.3 – 3.2) = 7.1 (abs) Ug

(i)

576

Fundamentals of Turbomachinery

Substituting all the data in Eq. (i), \

7.1 

V2 4.242  2.5 10.3  2  0  0 2 – 9.81 2g

or

V2 = 2.06 m/s

Substituting the above data in Eq. (7.47),

Kd

V12  V22  2 gh f

(4.242  2.062  0)

V12

(4.24)2

– 100

= 76.4%

Ans.

IMPORTANT EQUATIONS •

N

Q

P

•

(7.4); (7.7); (7.9) H3/ 2 H = Hg – hf ; H = net head ; hg = gross head ; hf = head loss

•

hN = nozzle efficiency =

•

CV = nozzle velocity coefficient =

•

Fx = force exerted by the Pelton turbine = r aV1(Vw1 ± Vw2)

•

T = torque on Pelton wheel = Fx × R

(7.17a)

•

P = power = Tw = r aV1 (Vw1 ± Vw2) u = r Q (Vw1 ± Vw2)

(7.17b)

•

KH

•

K = coefficient of friction =

•

Condition for maximum efficiency, u

•

KH max

•

KH

U aV1 (Vw1 “ Vw 2 ) u if CV = 1 U gQH

(7.27)

•

Km

Shaft power P

(7.29)

Nu

H

Qu

;

H

;

Pu

V12 2 gH

(7.14) V1 2 gH

2u (Vw1 “ Vw 2 )

(7.16) (7.17)

(7.18)

V12 Vr 2 Vr1

1  K cos E2 2

S.P. U aV1 (Vw1 “ Vw 2 ) u

(7.19)

V1 2

(7.22) (7.23)

Hydraulic Turbines

•

W.P. = (1/2)rQV12

•

Ko

•

f = speed ratio =

•

m = jet ratio = D/d ; D = pitch diameter, d = jet diameter

•

z = number of bucket =

• •

Q = discharge = aV1 Velocity triangles for Pelton wheel

S.P. W.P.

577 (7.31a)

S.P. P – P W.P.

Km – KH

(7.32)

u

(7.35)

2 gH D  15 2d

0.5m  15

(7.36) (7.37)

Francis turbine • Fx = force in x direction = raV1 (Vw1u1 ± Vw2u2); +ve if a2 < 90°, –ve, if a2 > 90° • T = Fx × R = (Vw1R1 + Vw1R2) × raV1 •

D1 = 2R1 = outer diameter

•

D2 = 2R2 = inner diameter

•

P = power = raV1 (Vw1u1 ± Vw2u2)

Usually radial discharge, a2 = 90°, Vw2 = 0, V2 = Vf 2.

â

KH

â

\

flow ratio =

I

speed ratio =

â â

(Vw1u1 “ Vw 2 u2 ) gH

Vf 1 2 gH

u1 2 gH

Q = discharge = pD1B1Vf 1 = pD2B2Vf 2 (Thickness is neglected)

578

Fundamentals of Turbomachinery

•

B1 and B2 = width or height of the blade at inlet and outlet Q = (pD2 – nt2)B2Vf 2 = (pD1 – nt1)B1Vf 1 n = no. of vanes, t1, t2 = thickness of each blade at inlet and exit H = net head or effective head =

•

p1 V12  Ug 2g

Radial entry means, b1 = 90°, Vw1 = u1 Velocity triangles—Francis turbine u1 a1

Vw 1 b1 Vr 1

V1

Vf 1

Vr 2 b2

Vf 2 = V 2

u2

a 2 = 90°

(a) Slow speed

u 1 = Vw 1 a1

b1 Vf 1 = Vr 1

V1

Vr 2 b2

Vf 2 = V 2 a 2 = 90°

u2

(b) Medium speed

Vw 1

u1

a1

V1

b1 Vr 1

Vf 2 = V 2

Vr 2 b2 u 2 (c) Fast speed

Vf 1

a 2 = 90°

Hydraulic Turbines

579

Draft tube p1 V12   z1 Ug 2g

p2 V22   z2  h f Ug 2g

pa Ug

10.3

p2 Ug

pa h Ug

REVIEW QUESTIONS 1. What is the purpose of draft tube? List any two types of draft tubes with sketches. What happens if the length of the tube is long? 2. Draw a neat sketch of a Francis turbine, and indicate the following parts; scroll casing, guide vanes, runner, draft tube, output shaft. 3. Sketch a Kaplan turbine and indicate all its parts. 4. What is the significance of unit quantities? 5. Derive the unit quantities. 6. Define the unit quantities. 7. Classify hydraulic turbines. 8. Briefly explain the working of Pelton wheel. 9. Explain the main components of a Pelton wheel. 10. Define hydraulic efficiency, volumetric efficiency, mechanical efficiency, and overall efficiency. 11. Write the velocity triangles for Francis turbine for the following three cases: (a) Slow speed, (b) Medium speed, (c) High speed. 12. Write the velocity triangles for Pelton wheel and Kaplan turbine. 13. Explain the calculation procedure for inlet and exit runner angles of Kaplan turbine corresponding to tip diameter and diameter. 14. Derive an equation to calculate the height of the draft tube immersed in water. 15. Compare the characteristics of impulse and reaction turbines.

EXERCISES 7.1 Find the speed of the runner of a Pelton wheel, if it works under a head of 600 m. Assume the coefficient of velocity and velocity ratio to be 0.98 and 0.47 respectively.

580

Fundamentals of Turbomachinery

7.2 Water is supplied to a Pelton wheel at the rate of 200 lit/s. The turbine works under a head of 180 m. The speed of the wheel is 1000 rpm and the jet is deflected back through an angle of 160°. Find the power produced and hydraulic efficiency. Assume suitable data if necessary. 7.3 A Pelton wheel is working under an effective head of 50 m. It produces 75 kW at 275 rpm. The oveall efficiency is 0.85. Take CV = 0.98. Calculate (a) the diameter of the jet, (b) the width of the buckets, (c) the depth of the buckets, and (d) the number of buckets. 7.4 A Pelton wheel is working under a head of 500 m. Jet diameter = 150 mm, speed = 350 rpm, wheel-to-jet ratio = 15. The bucket deflects the jet through an angle of 165°. Calculate (a) the force exerted, (b) the power, (c) the hydraulic and overall efficiency. Take: CV = 0.97, K = 0.91, mechanical losses as 5% of the power supplied. 7.5 The following data refer to a Pelton wheel. Head = 15 m, discharge = 18 lit/s, Power = 2 kW, jet diameter = 3 cm. Calculate the power lost in the nozzle, power wasted in the tail race and the overall efficiency. 7.6 The external and inner diameters of an inward flow reaction turbine are 2.2 m and 1.2 m respectively. The turbine is running at a speed of 200 rpm. Flow velocity is constant and equal to 5.5 m/s. Take the guide blade angle equal to 12°. Draw the velocity triangles and calculate (a) the vane angles at the inlet and the outlet and (b) the absolute velocity of water leaving the guide vane. 7.7 A radially inward flow reaction turbine, working under a head of 9 m and running at 200 rpm develops 175 kW. The speed ratio and flow ratio are 0.9 and 0.3 respectively. If the overall efficiency and hydraulic efficiency of the turbine are 80% and 85% respectively, find (a) the inner diameter of the runner, (b) the width of the wheel at the inlet, (c) the guide blade angle at the inlet and (d) the inlet vane angle. 7.8 A Francis turbine working under a head of 150 m runs at 800 rpm. The velocity of water at entry is 32 m/s. The outer and inner diameters of the runner are 1.5 and 0.75 m respectively. The outlet angle of the guide blades is 12°. Calculate the runner blade angles at inlet and outlet and hydraulic efficiency, if the discharge is axial and the velocity of flow is constant through the runner. 7.9 An inward flow reaction turbine has an overall efficiency 80%. The net head is 6 m. The power output is 130 kW. Take tangential velocity = 0.98 2 gH , flow ratio = 0.44 2 gH , and speed = 250 rpm. The hydraulic losses are 20% of energy available. Calculate the inlet guide vane exit angle, the inlet angle of the runner vane, the runner diameter at inlet and outlet. Assume radial discharge. 7.10 Following data refers to a Francis turbine: Speed = 1200 rpm, net head = 130 m, discharge = 0.7 m3/s, inner diameter = 1.3 m. Height of the runner at inlet = 0.05 m. The angle of the inlet guide vanes is set at 72°. Absolute velocity at outlet is radial. Calculate (a) the torque, (b) the power, and (c) the hydraulic efficiency.

Hydraulic Turbines

581

7.11 A Kaplan turbine has the following data. Power developed = 2 MW, head = 35 m, flow ratio = 0.6 and speed ratio = 2.1. The hub diameter is 0.35 of tip diameter. Calculate the runner diameter and the speed. Take overall efficiency = 0.85. 7.12 An axial flow turbine has the following data: Net head = 25 m, speed = 200 rpm, power = 25 MW, tip diameter = 5 m and hub diameter = 2.2 m, hydraulic efficiency = 0.9, overall efficiency = 0.85. Calculate the inlet and outlet blade angles at the mean diameter. Assume axial flow at exit. 7.13 Following data refers to a Kaplan turbine: Net head = 20 m, power developed = 15 MW, overall efficiency = 80%, runner diameter = 4.2 m, hub diameter = 2 m, specific speed = 300, Hydraulic efficiency = 90%. Calculate the inlet and exit angles of the runner blades at the tip and at the hub if the flow leaving the runner is purely axial. 7.14 Following data refers to a pelton wheel: Power = 6500 kW, head = 250 m, overall efficiency = 85%, and speed = 220 rpm. Calculate the unit discharge, unit power, and unit speed. Take speed ratio = 0.45. If the head on the same turbine falls to 125 m, calculate the discharge, power, and speed for the new head. 7.15 Find the specific speed and the type of the turbine if the power developed = 7000 kW, head = 25 m, and speed = 120 rpm. Calculate its normal speed and output under a 30 m head.

BIBLIOGRAPHY

Ganesan, V., Gas Turbines, Second Edition, Tata McGraw-Hill, New Delhi. Kadambi, V. and Manohar Prasad, An Introduction to Energy Conversion, Volume III, Turbomachinery, Wiley Eastern (1977). Kearton, W.J., Steam Turbine Theory and Practice, The English Language Book Society, London. Royce N. Brown, Compressors Selection & Sizing, Gulf Publishing Company. Saravanamuttoo, H.I.H., G.F.C. Rogers and H. Cohen, Gas Turbine Theory, Fifth Edition, Pearson Education. Shephered, D.G., Principles of Turbomachinery, Macmillan Company. Yahya, S.M., Fundamentals of Compressible Flow, Wiley Eastern, New Delhi. Yahya, S.M., Turbines, Compressors and Fans, Tata McGraw-Hill, New Delhi.

583

MODEL QUESTION PAPERS (With Answers)

PAPER 1 1(a) Ans.

Define a turbomachine. Give a comparison between turbomachines and positive displacement machines. A turbomachine is a device in which energy transfer takes place between a flowing fluid and a rotating element due to a dynamic action that results in a change of pressure and moment of the fluid. For comparison between turbomachines and positive displacement machines, refer to Section 1.6.

1(b)

Starting from the first law of thermodynamics, derive an expression for the work output of a turbomachine in terms of properties at inlet and outlet.

Ans.

Refer to Section 1.7.2.

2(a)

Starting from the impulse momentum equation, derive the Euler turbine equation.

Ans.

Refer to Section 2.1.

2(b)

A dam powerhouse is proposed to be built and for which Francis turbine is rquired to be designed. The design head is 16 m, and the design flow rate is 8 m3/s. The speed is to be 250 rpm. An overall efficiency of 0.9, hydraulic efficiency of 0.95, a speed ratio of 0.76 and flow ratio of 0.35 may be assumed. Obtain all the salient dimensions (outer diameter, inner diameter and width), blade angles and guidevane angles. The inner diameter is half the outer diameter, and discharge does not have any whirl component. Neglect vane thickness.

Ans.

Refer to Example 7.27.

3(a)

Draw a neat sketch of a Francis Turbine and indicate on it the following parts: Scroll casing, Guide vanes, Runner, Draft-Tube, Output shaft.

Ans.

Refer to Figure 7.7.

3(b)

A Pelton turbine has a water supply rate of 5 m3/s at a head of 256 m and runs at 500 rpm. Assuming a turbine efficiency of 0.85, a coefficient of velocity for nozzle as 0.985, and a speed ratio of 0.46, calculate (i) the power output, (ii) the specific speed, (iii) the number of jets, (iv) the diameter of the wheel, (v) the jet diameter, (vi) the number of Pelton cups, and (vii) the cup dimensions.

Ans.

Refer to Example 7.2.

4(a)

Define and explain manometric efficiency of a centrifugal pump. Write an expression for the same.

Ans.

Manometric head is the head against which a pump has to work. It is also defined as the difference in head imparted by the pump to the water and the loss of head in the pump. It is also the difference in head at the outlet and inlet to the pump. Hence, manometric efficiency hman is expressed as 585

586

Model Question Papers

Kmax

Manometric head Head imparted by impeller to water

Hm Vw 2 u 2 / g

4(b)

What do you mean by NPSH? Is it desirable to have a lower or a higher value of NPSH? Justify your answer with the help of relevant equations.

Ans.

It is the net head in meters of the liquid necessary to force the liquid from the suction pipe into the pump impeller. It is given by

NPSH

ps Vs2 p   v U g 2g U g

where, ps = absolute pressure at pump suction, Vs = velocity in suction pipe, pv = vapour pressure. From the expression it is clear that a lower value of NPSH is desirable.

4(b)

A centrifugal pumps runs at 950 rpm. Its outer and inner diameters are 500 mm and 250 mm. The vanes are set back at 35° to the wheel rim. If the radial velocity of water through the impeller is constant at 4 m/s, find the angle of vane at inlet, the velocity and direction of wear at outlet and the work done by the impeller per kg of water. Entry of water at inlet is radial.

Ans.

Refer to Example 4.27.

5(a)

Explain the significance of unit quantities as applied to hydraulic machines. Define and derive expressions for unit flow and unit power.

Ans.

The rate of flow, the speed and power, etc. depend on the head. For comparison, these values are reduced to unit value of head. Such reduced quantities are called unit quantities. The unit quantities of similar fluid machines will be equal. Hence, these quantities help to predict the performance of similar machines. Refer to Sections 7.3.2, 7.3.3 and 7.3.4.

5(b)

Define the specific speed of a hydraulic turbine. Derive an expression for the same. What is its utility?

Ans.

Refer to Section 1.18.7. Different types of turbines have different values of specific speed. Impulse turbines have a low value of specific speed, Francis turbine medium and Kaplan turbine high. A high specific speed makes the size of turbine and powerhouse small. For low head and high output, a high specific speed turbine should be used. Thus, based on the exisiting conditions, the type of turbine can be decided.

5(c)

A model of a centrifugal pump absorbs 5 kW at a speed of 1500 rpm, pumping water against a head of 6 m. The large prototype pump is required to pump water to a head of 30 m. The scale ratio of diameters is 4. Assuming the same efficiency and similarities, find the speed and power of the prototype and the ratio of discharges of prototype and model.

Ans.

Refer to Example 1.18.

Model Question Papers

587

6(a)

What is compounding of steam turbine? What are the advantages of compounding? Illustrate by a diagram the pressure compounding of a steam turbine.

Ans.

Compounding is a method of absorbing the inlet energy in multiple rows of moving blades. Compounding reduces the loss of energy. It also reduces the centrifugal stresses in the blades to a great extent and hence the failure of blades is reduced. It also enables to use cheaper material for the construction of blades. In pressure compounding, steam is expanded from boiler pressure to some lower pressure in the first fixed blade ring. The velocity increases. This kinetic energy is absorbed by the first moving blades ring. The steam is further expanded in the second fixed blade ring and kinetic energy is absorbed by the second moving blade ring. This is repeated. There will not be any pressure drop as the steam flows over the moving blades. Refer to Figure 6.6.

6(b)

In a two row, velocity compounded impulse steam turbine, the steam from the nozzle issues at a velocity of 600 m/s. The nozzle angle is 20° to the plane of rotation of the rows, and the moving blades have equiangular blades. The intermediate row of fixed guide blades makes the steam flow again at 20° to the second moving blade ring. The frictional losses in each row are 3%. Find: (i) the inlet and outlet angles of moving blades of each row, (ii) the inlet blade angle of the guide blade, and (iii) the power output from the first and second moving blades for unit mass flow.

Ans.

Refer to Example 6.10.

588

Model Question Papers

PAPER 2 1(a)

How are turbomachines classified? Compare turbomachines with positive displacement machines.

Ans.

Refer to Section 1.8 and Section 1.6 for the respective answers.

1(b)

Derive an expression for the angular momentum equation and arrive at Euler’s turbine equation.

Ans.

Refer to Section 2.1 and Figure 2.1.

2(a)

What is cavitation? How can it be avoided in the reaction turbine?

Ans.

When water flows in a closed conduit its velocity increases and the pressure decreases with the decrease in the cross-section for flow. In such a case the pressure may sometimes become negative. This happens in turbines when the speed of the turbine is higher. When this negative pressure falls below the vapour pressure of water, the vapour of water is formed which occupies the space in the form of bubbles known as cavities. This phenomenon is called cavitation. Cavitation may be avoided by taking the following precautions: (1) The conduit should be so designed that nowhere the pressure falls below the vapour pressure. (2) The velocity of water should not exceed the safe limit. (3) The speed of the rotor should be kept within safe limits.

2(b)

Show that the hydraulic efficiency of Pelton wheel is maximum when the peripheral wheel velocity is half the absolute velocity of the jet at inlet. Further show that: (hb)max = (1 + K cos f)/2

Ans.

Refer to Sections 7.6.1 and 7.6.2.

2(c)

The three jet pelton is required to generate 10,000 kW under a net head of 400 m. The blade angle at outlet is 15° and the reduction in the relative velocity while passing over the blade is 5%. If the overall efficiency of the wheel is 80%, CV = 0.98 and the speed ratio = 0.46, find (i) the diameter of the jet, (ii) the total flow in m3/s, and (iii) the force exerted by a jet on the buckets.

Ans.

Refer to Example 7.3.

3(a)

Define the specific speed of a turbine. Derive an expression for the specific speed of a turbine.

Ans.

Refer to Section 1.18.7.

3(b)

A Kaplan turbine working under a head of 20 m, develops 11,775 kW. The outer diameter of the runner is 3.5 m and hub diameter 1.75 m. The guide blade angle at the extreme edge of the runner is 35°. The hydraulic and overall efficiencies of the turbine are 0.88 and 0.84 respectively. If the velocity of whirl is zero at outlet, determine (i) the runner vane angles at inlet and outlet at the extreme edge of the runner and (ii) the speed of the turbine.

Ans.

Refer to Example 7.44.

Model Question Papers

589

4(a)

Obtain an expression for the minimum speed for starting a centrifugal pump.

Ans.

The centrifugal pressure head for no flow of water = (u22  u12 ) / 2 g Unless this pressure head is equal to or greater than the manometric head, the pump will not deliver water. By this the minimum speed can be determined. Refer to Section Article 4.21

4(b)

A centrifugal pump having outer diameter equal to two times the inner diameter, running at 1000 rpm, works against a total head of 40 m. The velocity of flow through the impeller is constant and equal to 2.5 m/s. The vanes are set back at an angle 40° to outlet. If the outer diameter of the impeller is 50 cm and that at outlet is 5 cm, determine (i) the vane angle at the inlet (ii) the work done by the imlpeller per second on water, and (iii) the manometric efficiency.

Ans.

Refer to Example 4.26.

5(a)

Show that the pressure rise in the impeller of a centrifugal pump when frictional and other losses in the impeller are neglected is given by V f21  u22  V f22 cosec 2I 2g

Ans.

Refer to Section 4.20.

5(b)

A single stage centrifugal pump with impeller diameter of 30 cm rotates at 2000 rpm and lifts 3 m3 of water per second to a height of 30 m, with an efficiency of 75%. Find the number of stages and the diameter of each impeller of a similar multistage pump to lift 5 m3 of water per second to a height of 200 m, when rotating at 1500 rpm.

Ans.

Refer to Example 4.45.

6(a)

Derive the condition for maximum efficiency of an impulse turbine and show that the maximum efficiency is cos2 a, where a is the angle at which the steam enters the blade.

Ans.

Refer to Section 6.11.

6(b)

A certain stage of a Parson’s turbine consists of one row of fixed blades and one row of moving blades. The details of the turbine are: Mean diameter of the blades = 680 mm. ; rpm of the turbine 3000; mass of steam passing per second = 13.5 kg, steam velocity at exit from fixed blades = 143.7 m/s, blade outlet angle = 20°. Calculate the power developed in the stage and the gross efficiency assuming carry-over coefficient to be 0.74 and the efficiency of conversion of heat energy in the blade channels to be 0.92.

Ans.

Data: d = 680 mm, N = 3000 rpm, ms = 13.5 kg/s, V1 = 143.7 m/s, a1 = 20°, carryover coefficient (carry-over efficiency) hco = 0.74, nozzle efficiency or stator efficiency, hn = 0.92.

590

Model Question Papers

u = pdN/60 = p × 0.68 × 3000/60 = 106.81 m/s I

speed ratio =

u V1

106.81 143.7

0.74

'h2  'h1 = total adiabatic enthalpy drop. „

„

(Eq. (6.33a))

V12 [1  Kco (1  I 2  2I cos D1)] / gc Kn 143.72 [1  0.74(1  0.742  2 – 0.74 – cos 20’)] 0.92

18924.6 J/kg

Wd = W.D. = V12 (2I cos D1  I 2 )

(Eq. (6.31a))

= 143.72(2 × 0.74 cos 20° – 0.742) = 17410.68 J/kg Power developed (P):

P

ms Wd 1000

13.5 kg 17410.68 kJ – s 1000 kg

235 kW

Gross stage efficiency (hg): Kg

Wd 'h1  'h2 „

„

17410.68 18924.6

0.92

(Eq. (6.33b))

7(a)

Explain the method of velocity compounding of an impulse turbine for achieving rotor speed reduction.

Ans.

In velocity compounding, steam is expanded in the nozzle from the boiler pressure to back pressure. High kinetic energy is partially absorbed by the first rotor. The steam then passes over the fixed blades which change its direction of flow. Then kinetic energy is partially absorbed by the second rotor. This process is repeated. There will not be an pressure drop as the steam flows over the moving and fixed blades. Velocity decreases as steam flows over the moving blades. There is negligible velocity drop as the steam flows over the fixed blades.

7(b)

The following particulars refers to a single impulse turbine. Mean diameter of blade ring = 2.5 m, speed = 3000 rpm, nozzle angle = 20°, ratio of blade velocity to steam 0.4, blade friction factor = 0.8 m, blade angle at exit = 3° less than that at inlet, steam flow 36,000 kg/h. Draw the velocity diagram for the moving blade and estimate (i) the power developed and (ii) the blade efficiency.

Ans.

Refer to Exarnple 6.2.

Model Question Papers

591

PAPER 3 1(a)

Define a turbomachine. How are turbomachines classified? Give examples.

Ans.

Turbomachine is a device in which energy transfer takes place between a flowing fluid and a rotating element due to a dynamic action that results in a change of pressure and momentum of the fluid. For classification of turbomachines, refer to Section 1.6

1(b)

Define and explain the following terms with reference to a work-absorbing turbomachine: (i) Mechanical efficiency (ii) Hydraulic efficiency (iii) Volumetric efficiency.

Ans.

(i) Mechanical efficiency = (ii) Hydraulic efficiency = (iii) Volume efficiency =

Mechanical energy supplied to rotor Mechanical energy supplied to shaft

Useful hydrodynamic energy in fluid Mechanical energy supplied to rotor

Actual discharge Theoretical discharge or volume swept

2(a)

Derive the Euler turbine equation. State the assumptions.

Ans.

Refer to Section 2.1 for derivation of Euler turbine equation. Assumptions: (1) Water enters and leaves the vane in a direction tangential to the vane tip at inlet and outlet. (2) There is no frictional resistance as water flows over the vane.

2(b)

For a Pelton turbine, draw the general form of velocity triangles and obtain an expression for maximum efficiency.

Ans.

Refer to Section 7.6.1 and Figure 7.6.

2(c)

At a power station, a Pelton turbine produces 23,100 kW power under a head of 1770 m, while running at 750 rpm. Evaluate for the turbine (i) the number of jets and jet diameter, (ii) the mean diameter of the runner and (iii) the number of buckets. Assume that the nozzle velocity coefficient is 0.98, the speed ratio is 0.46 and the overall turbine efficiency is 0.85.

Ans.

Refer to Example 7.4.

3(a)

Sketch a Kaplan turbine and indicate all its parts.

Ans.

Refer to Figure 7.11.

3(b)

What are the functions of a draft tube in reaction turbine? What happens if the draft tube is too long?

Ans.

Functions: (1) It increases the working head of the turbine by an amount equal to the height of the runner outlet above the tail race. Thus, it creates a negative or suction head.

592

Model Question Papers

(2) It enables to recover the kinetic energy. If the draft tube is too long, the suction head will be large. The pressure at the outlet of the runner may fall below the vapour pressure of water. Vapour of water will be formed causing cavitation.

3(c)

A Kalpan turbine produces 30,000 kW under a head of 9.6 m, while running at 65.2 rpm. The discharge through the turbine is 350 m3/s. The tip diameter of the runner is 7.4 m. The hub diameter is 0.432 times the tip diameter. Calculate (i) the turbine efficiency, (ii) the specific speed of turbine, (iii) the speed ratio (based on tip diameter) and (iv) the flow ratio.

Ans.

4(a)

Refer to Example 7.44. Sketch a centrifugal pump. Indicate all its parts. Also mention the functions of the parts.

Ans.

Refer to Figure 4.11 and Section 4.15. (1) Impeller—It transfers energy to the fluid. (2) Casing—It is an airtight chamber. The enclosed impeller causes diffusion of water. (3) The suction pipe connects the pump to the water pump. (4) The delivery pipe delivers the water.

4(b)

Explain the phenomenon of cavitation in a centrifugal pump. Can it be preventcd? Write your answers with reasons.

Ans.

Cavitation in centrifugal pump is similar to that in a reaction turbine. But this occurs on the suction side of the pump as lowest pressure exists just below the pump on the suction side. The phenomenon can be prevented by using superior materials and also parts with highly polished surface finish and also by installing the pump at an appropriate height above the pump. Also refer to Section 4.23.

4(c)

A centrifugal pump impeller has on outside diameter of 200 mm and rotates at 2900 rpm. The vanes are curved backward at 25° to tlie wheel tangent. The velocity of flow is constant at 3 m/s. Assume hydraulic efficiency as 75% and determine the head generated. Also determine the power required to run the impeller, if the breadth of the wheel at outlet is 15 mm. Neglect the effect of vane thickness, mechanical friction and leakage in the pump.

Ans.

Refer to Example 4.29.

5(a)

What are unit quantities and their uses, as applied in the context of fluid machines? Explain.

Ans.

The rate of flow, speed, power, etc. depend on the head. For comparison, these values are reduced to unit value of head. Such reduced quantities are called unit quantities. The unit quantities of similar fluid machines will be equal. Hence, these quantities help to predict the performance of similar machines.

Model Question Papers

593

5(b)

Derive an expression for the specific speed of a turbine. Explain how it guides the selection process of a turbine for a powerhouse.

Ans.

Refer to Section 1.18.7 and Table 7.1. Different types of turbines have different values of specific speed. Impulse turbines have a low value specific speed, Francis turbine medium and Kaplan turbine high. A high specific speed makes the size of turbine and powerhouse small. For low head and high output a high specific speed turbine should be used. Thus based on the existing conditions, the type of turbine can be decided.

5(c)

An axial flow pump with a rotor diameter of 30 cm, handles liquid water at the rate of 2.7 m3/min while operating at 1500 rpm. The corresponding energy input is 125 J/kg, the total-to-total efficiency being 75%. If a second geometrically similar pump with a diameter of 20 cm operates at 3000 rpm. Calculate its flow rate, power input and change in total pressure?

Ans.

Refer to Example 1.20.

6(a)

Explain with neat sketches any one type of compounding of a steam turbine.

Ans.

Refer to Section 6.5

6(b)

In a two-row curtis stage of a steam turbine, running with a mean rotor speed of 225 m/s, the steam leaves the second rotor axially. The nozzle angle is 16° and for the rotor, exit blade angles are 23° and 32° respectively for the first and second row. The stator blade exit angle is 22°. Neglecting blade friction, draw the successive velocity triangles and compute the rotor efficiency. If the nozzle efficiency is 0.91, find the dryness fraction of steam at exit, if the supplied steam is at 10 bar, 180°C.

Ans.

Refer to Example 6.12.

594

Model Question Papers

PAPER 4 1(a)

Differentiate between the following with examples: (i) (ii) (iii) (iv)

Positive displacement machine and fluid machine. Axial flow machine and radial flow machine. Impulse machine and reaction machine. Outward flow machine and inward flow machine.

Ans.

(i) In a positive displcement machine the mechanical element has usually reciprocating motion and the energy transfer is due to static action. In a fluid machine the mechanical element has pure rotary motion and the energy transfer is due to dynamic action. (ii) In an axial flow machine, fluid flows in a direction parallel to the axis of the rotor shaft. In a radial flow machine, fluid flows in the radial direction, i.e. perpendicular to the rotor shaft. (iii) In an impulse turbine all the available head is converted into kinetic energy in the nozzle before the fluid flows over the rotor. The fluid pressure will be atmospheric pressure. In a reaction turbine the fluid flows under pressure. At every instance the pressure head is converted into kinetic head. (iv) In an outward flow machine, water enters the rotor at the inner periphery and flows in a radially outward direction and is discharged at the outer periphery. In an inward flow machine, water enters the rotor at the outer periphery and flows inward and is discharged at the inner periphery.

1(b)

Define hydraulic efficiency and mechanical efficiency for a turbine and a pump.

Ans.

For turbine:

Hydraulic efficiency = Mechanical efficiency =

Mechanical energy supplied by the rotor Hydrodynamic energy available from fluid Work output at the shaft Mechanical energy supplied by the rotor

Also Refer to Section 7.6.3. For pumps: Refer to Q. No. 1(b) of Paper 3 and also refer to Sections 4.18.2 and 4.18.3.

2(a)

Derive with the help of a sketch, showing velocity diagrams, the Euler turbine equation as applied to a generalized rotor of a machine.

Ans.

Refer to Section 2.1. A double jet Pelton wheel is required to generate 7500 kW when the available head at the base of the nozzle is 400 m. The jet is deflected through 165° and the relative velocity of the jet is reduced by 15% in passing over the buckets. Determine:

2(b)

Model Question Papers

595

(i) The diameter of each jet, (ii) the total flow, and (iii) the force exerted by the jets in the tangential direction. Assume generator efficiency = 95%, overall efficiency = 80%, blade–speed ratio = 0.47, and nozzle velocity coefficient = 0.98. Ans.

Refer to Example 7.5.

3(a)

What is the purpose of draft tube? List any two types of draft tubes with sketches.

Ans.

Refer to Section 7.11.3 and Figure 7.9.

3(b)

The following data is given for a Francis turbine: Net head = 70 m, speed = 600 rpm, shaft power = 370 kW, overall efficiency = 80%, hydraulic efficiency = 95%, flow ratio = 0.25, breadth ratio = 0.1, outer diameter of the runner = 2 × inner diameter of the runner. The thickness of vanes constitutes 10% of the circumferential area of the runner, the velocity of flow is constant and the discharge is radial at outlet. Calculate: (i) The guide blade angle, (ii) the runner vane angles at inlet and outlet, (iii) the diameters of the runner at inlet and outlet, and (iv) the width of the wheel at inlet.

Ans.

Refer to Example 7.28.

4(a)

With a neat sketch, explain the different types of centrifugal pump casings. What is manometric head and manometric efficiency?

Ans.

Volute casing is of spiral form with increasing cross-sectional area towards the discharge end. Loss of kinetic head due to eddy formation is avoided. Diffusion casing consists of guide wheel with stationary vanes called diffusers. Refer to Q. No. 4(a) of Paper 3. Also refer to Figure 4.12 and Section 4.16.2.

4(b)

A centrifugal pump delivers 50 litres of water per second against a head of 24 m, running at 1500 rpm. The velocity of flow of 2.4 m/s is constant and the blades are curved at 30°. The inner diameter is half the outer diameter. If the manometric

596

Model Question Papers

efficiency is 80%, find the blade angle at inlet and the power required to drive the pump. Ans.

Refer to Example 4.32.

5(a)

Explain the terms specific speed, unit speed and unit power as applied to hydraulic turbines.

Ans.

Refer to Sections 7.3.2, 7.3.3 and 7.3.4.

5(b)

The following data were obtained from the main characteristics of a Kaplan turbine of runner diameter l m. Pu =30.695, Qu = 108.6, Nu = 63.6 Estimate the runner diameter, the discharge and the speed of a similar runner working under a head of 30 m and developing 2000 kW. Also determine the specific speed of the runner.

Ans.

Refer to Example 1.21.

5(c)

What is compounding of steam turbines? Explain with a sketch the velocity compounding.

Ans.

Refer to Q. No. 6(a) at Paper 1 and Q. No. 7(a) of Paper 2. Also refer to Sections 6.4 and 6.5.1.

6(a)

Show that the condition for maximum efficiency in a De-lava1 turbine having equiangular blades is Krm

cos D1 ; where, hrm = rotor efficiency for maximum condition and a1 = nozzle 2

angle. Ans.

Refer to Q. No. 6(a) of Paper 2. Also refer to Section 6.11 and Eqs. (6.12) and (6.13).

6(b)

A De-laval turbine has a mean rotor diameter of 0.55 m and runs at 3300 rpm. The speed ratio is 0.45 and the nozzle angle at the rotor inlet is 20°. The blade-velocity coefficient is 0.91. Assuming equiangular blades, find the rotor blade angles at the inlet and the outlet. Also find the power output. If m = 10 kg/s, find the axial thrust.

Ans.

Refer to Example 6.3.

Model Question Papers

597

PAPER 5 1(a)

State the direction of flow (axial, radial or tangential) in the rotors (i) to (viii) shown below. State with reason whether the rotor in Figure (vi) belongs to energy absorbing or energy generating turbomachine.

Ans.

(i) ® Radial, (ii) ® Axial, (iii) ® Radial, (iv) ® Radial, (v) ® Radial, (vi) ® Axial (energy absorbing), (vii) ® Tangential, (viii) ® Axial

1(b)

Describe how the head–capacity relationship of a centrifugal pump is predicted at different speeds from one test curve at a fixed speed.

Ans.

If the effects of viscosity and compressibility are neglected and the chosen characteristic dimension (d) is unit, then, p1 = capacity coefficient, p2 = head coefficient and p3 = power coefficient reduce to P Q H , and 3 respectively. N N2 N

(1) For a particular speed N1 obtain the discharge Q1 and head H1 at a particular point from one test curve at a fixed speed. (2) Calculate

Q1 H1 , . N1 N12

(3) Consider any other fixed speed N2. (4) Conditions for the operating characteristic are similar, therefore, Q1 N1

Q1 H1 and N2 N12 „

H1

„

N 22

598

Model Question Papers

(5) Calculate Q1 and H 1 at speed N2 „

„

(6) Repeat serial numbers (1) to (5) for another set of Q and H for another point (Q2, H2; Q3, H3, etc.) for N1 (7) Plot the points Q1 , H1 ; Q 2 , H 2 ; Q 3 , H 3 , etc. for N 2 „

„

„

„

„

„

(8) Join all the points for N2 to get the head–capacity relationship at a fixed speed N2.

1(c)

The quantity of water available for a hydroelectric power station is 260 m3/s under a head of 1.75 m. Assuming the speed of turbines to be 50 rpm and their efficiency to be 82.5%, find the number of turbines required given that the speed of turbines used is 890 rpm when working under a head of 1 m developing 0.75 kW.

Ans.

Total power available from a hydroelectric power station (Pt ):

Ko

0.825 – 1000 – 9.81 – 260 – 1.75 1000

Pt ; ? Pt U gQH

3282.43 kW

Specific speed of a given turbine (NST): N P

NST

H

890 – 0.75

5/4

15 / 4

770.71rpm

Power developed by a given turbine for available data, N = 50 rpm, head = 1.75 m, etc. (Pa):

N ST

N Pa H5/ 4

;

? Pa

Ë 770.71 – 1.755.4 Û Ì Ü 50 ÍÌ ÝÜ

2

962.58 kW

Number of turbines (NT):

NT

Pt Pa

3282.43 kW 962.58 kW

3.41  4

Ans.

1(d)

A centrifugal pump is required to handle water at a capacity 6.75 m3/s, a head of 125 m and a speed of 350 rpm. In designing a model of this pump the laboratory conditions impose a maximum capacity of 0.127 m3/s and a power consumption of 220 kW. If the model and prototype efficiencies are assumed to be the same, find the speed of the model and the scale ratio.

Ans.

Prototype: Q = 6.75 m3/s, H = 125 m, N = 350 rpm Model: Q = 0.127 m3/s, P = 220 kW, hm = hp Power consumption for prototype (Pp):

Pp

1000 – 9.81 – 6.75 – 25 1000 – K

8277.1875 kW K

U gQH K

Model Question Papers

599

Head required for the model (Hm): \

Pm

Ë U gQH Û Ì 1000K Ü ; ? H m Í Ým

220 – 1000K 176.58K m 1000 – 9.81 – 0.127 – 9.81

Speed of the model (Nm):

ËN Q Û Ì 3/4 Ü ÌÍ H ÜÝ m \

Nm

ËN Q Û Ì 3/4 Ü ÌÍ H ÜÝ p 350 – 6.75 1253 / 4

–

176.583 / 4K 3 / 4 0.127

3306.34K 3 / 4 rpm

Ans.

Scale ratio (Dm/Dp):

Ë H Û Ì Ü Í ND 2 Ý p 2

\

Ë Dm Û Ì Ü ÍÌ D p ÝÜ

or

Dm Dp

Ë H Û Ì Ü Í ND 2 Ý m ËN Û ËH Û ÌH Ü – ÌN Ü Í Ý p Í Ým 0.386 K 0.25 1

350 176.58 K – 125 3306.34 K3 / 4 K 0.25 2.59

0.1495 K0.25

Ans.

2(a)

Obtain expressions for energy transfer E and degree of reaction R as a function of discharge blade angle b2 for a turbomachine. Make the following assumptions u2 = 2u1 with usual notations, constant radial velocity, no inlet angular momentum and an inlet fluid angle of 45°. Sketch the nature of variation of E and R with respect to b2 as it varies from 0° to 90° and discuss the salient feature of the graphs.

Ans.

Refer to Figure 2.4, Section 2.6.1, derivation up to Eqs. (2.20) and (2.21). For salient featuers, refer to the points given below Figure 2.4.

2(b)

An inward flow turbine has 0.6 reaction. The blade speed at entry is 12 m/s and the radial velocity of flow is constant at 2.4 m3/s. The rotor diameter at entry is twice that at exit. Find the utilization factor, angles of the blades at entry and exit, assuming there is no exit whirl velocity and no friction loss. Is the utlilization factor maximum?

Ans.

Refer to Example 2.30.

3(a)

Obtain general expressions for energy transferred and degree of reaction for an axial flow stage in terms of axial component Va, blade speed u, and fluid angles.

600

Model Question Papers

Rewrite the expressions for the degree of reaction in terms of Va, u and mean fluid angles. Assume Va to be constant. Ans.

Expression for energy transfered (E or W.D.): Refer to Eqs. (2.25) to (2.28) or (2.29). Expression for degree of reaction (R): Refer to Eqs. (2.32) to (2.36a).

3(b)

A single-stage axial flow blower with no inlet guide vanes, but a row of stationary vanes downstream of rotor operates at 3600 rpm. The tip and hub diameters of the rotor are 200 mm and 125 mm respectively. The air flow rate through the stage is 0.5 kg/s. The air is turned through an angle of 20° towards the axial direction during air flow over the blade. The downstream stator redirects the flow towards the axis. Assuming standard atmospheric conditions (1 bar, 25° C), and no losses in the rotor, compute the power input and the degree of reaction.

Ans.

Machine = axial blower; Tip dia, dt = 200 mm, Speed, N = 3600 r.p.m.; Hub dia, dh = 125 mm, Air flow rate = 0.45 kg/s; p01 = 1 bar, Air turn angle = b2 – b1 = 20°; T01 = 298 K. Axial flow machine: u1 = u2 = u. The downstream stator redirects the flow towards the axis = a1 = 90°, Vw1 = 0. Area of flow (A):

S (d h2  dt2 ) / 4

A

S (0.22  0.1252 ) / 4

0.01915 m2

Density of air at inlet (r01):

U01

1.013 – 105 287.2 – 298

p01 RT01

1.185 kg/s

Mean rotor blade velocity (u):

u Vf 1 Vr 1 tan E1

S (d h  dt ) N 2 – 60 Vf 2

V1

V12  u12 V1 u

S – (0.2  0.125) – 3600 2 – 60

m U01A

0.5 1.185 – 0.01915

22.033  30.632

22.03 ; ? E1 30.63

30.63 m/s

22.03 m/s

37.73 m/s

35.72’ ; E 2  E1

20’

Model Question Papers

\

601

b2 = b1 + 20° = 35.72° + 20° = 55.72° sin E 2

V1 ; Vr 2 Vr 2

V1 sin E 2

22.03 sin 55.72’

26.7 m/s

tan E 2

V1 ; ? x x

V1 tan E 2

22.03 tan 55.72’

15.02 m/s

From figure, we know that Vw2 = u – x = 30.63 – 15.02 = 15.61 m/s V2

V12  Vw22

22.032  15.612

27.0 m/s

Power input (P): W.D. = Dho = u(Vw2 – Vw1)/gc = 30.63 × 15.61 = 478.13 J/kg \

P = W.D. × m = 478.03 J/kg × 0.5 kg/s = 239.1 W

Ans.

Degree of reaction (R): R

(Vr21  Vr22 ) (V22  V12 )  (Vr21  Vr22 )

(Eq. (2.13), u1 – u2 = u)

(37.732  26.72 ) (272  22.032 )  (37.732  26.72 )

0.745

Ans.

Alternatively: R

Va (tan E 2  tan E1) 2u (tan E1 – tan E 2 )

(Eq. (2.33a))

22.03 (tan 55.72’  tan 35.72’) – 2 – 30.63 (tan 55.72’ – tan 35.72’)

0.745

Ans.

Also, R

1

(V22  V12 ) W.D – 2 – g c

Ë (272  22.032 ) Û Ì1  Ü 478.13 – 2 – 1 ÝÜ ÍÌ

0.745

(Eq. (2.34))

Ans.

4(a)

In a multistage compressor, each suceeding stage is penalized by the “inefficiency of the previous stage” while in a multistage turbine each succeeding stage is “benefited by the inefficiency of the previous stage”. State whether these statements are true or false. With the help of T-S diagrams for multistage compression and multistage expansion, justify your answer. Define preheat factor for compressors and reheat factor for turbines.

Ans.

The statements are true. Refer to Section 3.5.2, up to Eqs. (3.35a) and (3.35b). Refer to Section 3.5.5. Refer to Section 3.6.2, up to Eq. (3.58). Refer to Section 3.6.5.

602 4(b)

Model Question Papers

Show that the overall efficiency hc of an axial flow compressor consisting of m stages of efficiency hS and stage pressure ratio pr is given by (prK H  1)

Kc

Ans.

K

Ë Û 1 H pr  1Ü  1 Ì1  Í KS Ý where, e = (g – 1)/g and g is the adiabatic index. Hence, determine the expected overall efficiency of a 16-stage air compressor with pressure ratio 6.3 and stage efficiency 89.5%. Refer to Eq. (3.42g). [6.316 – 0.286  1]

Kc

0.4721

16

Ë (6.30.286  1) Û Ì1  Ü 0.895 Ý Í

Ans.

1

5(a)

Explain with the help of actual head capacity characteristic why surging occurs in blowers. List the possible means of its alleviation.

Ans.

Refer to Section 4.11.

5(b)

A centrifugal compressor runs at 15000 rpm and produces a stagnation pressure ratio of 4 between the impeller inlet and outlet. The stagnation conditions of the air at the compressor intake are 1 atm and 25°C respectively. The absolute velocity at the compressor intake is axial. If the compressor has radial blades at the exit such that Vr2 = 135 m/s and the total-to-total efficiency of the compressor is 0.78, draw the velocity triangle at the exit of the rotor and compute the slip as well as the slip coefficient. Rotor diameter at the outlet is 58 cm.

Ans.

Outer diameter, d2 = 58 cm, intake stagnation temp. T01 = 298 K, intake stagnation pressure, p01 = 1 atm, speed N = 15000 rpm, intake ab. velocity is axial, V1 = Vf1, Vw1 = 0, a1 = 90°, radial exit blade, b2 = 90°, Vw2 = u2, pressure ratio pR1 = 4, exit flow velocity, Vf 2 = 135 m/s, hct-t = 0.78; e = (g – 1)/g = 0.286. T02 Kct t T02  T01

Ëp Û T01 Ì 02 Ü Í p01 Ý

H

298 – 40.286

443 K

(T02  T01) (T02  T01) „

T02  T01 Kct t „

443  298 0.78

186 K

u2 = pd2N/60 = p × 0.58 × 15000/60 = 456 m/s 'ho

\

Vw 2

„

W.D.

u 2Vw 2 / g c „

1004 – 186 / 456

456Vw 2

„

c p (To 2  To1)

410.5 m/s

Since the blades are radial at the exit, u2 = Vw2 (a) Slip: Slip = Vw 2  Vw 2

„

456  410.5

45.5 m/s

Ans.

Model Question Papers

603

Slip D

V¢2¢

Vf 2 = Vr 1

B V2 V¢r 2¢

A

V¢f 2¢

b2¢ < 90°

V¢w 2¢

b2¢ = 90° C

u 1 – Vw 2

(b) Slip coefficient (s): V

Vw 2 Vw 2

410.5 456

„

0.9

Alternatively:

Ë u 22 V Kct t Ì1  c p gcToi ÌÍ

pRI

J

Û Ü ÜÝ

1

(Eq. (4.16))

Ans.

3.5

6(b)

Ans.

Ë 4562 – V – 0.78 Û 4 Ì1  Ans. Ü ; ? V 0.9 1004 – 1 – 298 ÜÝ ÌÍ An axial compressor stage has the following data. Inlet conditions are 1 bar and 25°C, degree of reaction is 0.5, mean blade ring diameter is 360 mm, rotational speed is 18,000 rpm, blade height at entry is 180 mm, air angles at rotor and stator exit are 25° (with respect to axial direction), axial velocity is 180 m/s, work done factor is 0.88, stage efficiency is 85% and mechanical efficiency is 96.7%. Determine the air angles at rotor and stator entry, the mass flow rate of air, and the power required to drive the compressor. Refer to Figure 5.3. Temperature at entry T01 = 293 K, degree of reaction R = 50%, pressure at entry, p01 = 1 bar, mean blade ring diameter dm = 36 cm, speed N = 18000 rpm, blade height at entry h1 = 6 cm, air angles at rotor exit a3 = (90 – 25), air angles at stator exit a1 = (90 – 25), axial velocity Va(Vf) = 180 m/s, work done factor j = 0.88, stage efficiency hS = 85%, and hm = 96.7%. u tan D 1

S dmN 60

S – 0.36 – 1800 60

Vf ; Vw 1 Vw 1

Vf tan 65’

339.29 m/s

180 tan 65’

83.94 m/s

(a) Air angles at rotor and stator entry (b1, b2): Since the stage has 50% reaction, we have a1 = b1 = 35.18°, a1 = b2 = a3 = 65°

Ans.

604

Model Question Papers

(b) Mass flow rate (m):

U1

.0 – 105 287 – 300

p1 RT1

1.61 kg/m3

m = r1pdmh1Vf 1 = 1.161 × p × 0.36 × 0.06 × 180 = 14.18 kg/s

(Eq. 5.1)

Ans.

(c) Power required to drive (P2): W.D. = [juVf (cot b1 – cot b2)]/gc

(Eq. (5.25b))

0.88 – 339.29 – 180 – (cot 35.18’  cot 65’) 1 – 1000 P

W.D. – m Km

51.18 – 14.18 0.967

51.18 kJ/kg

750 kW

Ans.

7(a)

Why is compounding is done in steam turbines? Explain what is meant by pressure compounding with the help of a schematic diagram consisting of two steam turbine stages. Distinguish between pressure compounding and velocity compounding.

Ans.

Compounding is a method of reducing the blade speed for a given overall pressure ratio by absorbing the inlet energy in multiple rows of moving blades. Compounding reduces the loss of energy. It also reduces the centrifugal stresses in the blades to a great extent and hence, failure of blades is reduced. It also enables to use cheaper material for the construction of blades. Pressure compounding: Refer to Section 6.5.2. Also, refer to Q.6(a) of Paper 1. Velocity Compounding

Pressure Compounding

The inlet energy is in kinetic form.

The inlet energy is in pressure form (enthalpy form).

Kinetic energy of steam is absorbed by moving rows.

A number of simple impulse stages (one set of nozzles and one set of moving blades) are arranged in series.

The steam is exited from the last row with very low velocity. Steam leaves axially from the last row.

P.E. is converted into K.E. in each fixed blade that acts as a nozzle. This K.E. is absorbed in that stage only. Each simple impulse turbine is named the stage of the turbine. Steam leaves axially from the last row. Steam velocity is very low in each stage, hence, it reduces the blade speed and rotational speed.

Model Question Papers

605

7(b)

In a Curtis stage with two rows of moving blades, the rotors are both equiangular. The first rotor has angles of 29° each while the second rotor has angles of 32° each. The velocity of steam at the exit of nozzle is 530 m/s. All blade velocity coefficients are 0.9. If the absolute velocity at the stage exit should be axial, find the mean blade speed.

Ans.

Refer to Example 6.11.

8(a)

Sketch the inlet and outlet velocity triangles for a Pelton wheel. Obtain an expression for maximum hydraulic efficiency in terms of nozzle velocity coefficient, bucket velocity coefficient and exit angle of relative velocity.

Ans.

Velocity triangles: Refer to Figure 7.6 P = raV1(V1 – u)(1 + K cos b2)u

(Eq. (7.20))

W = W.D. = (V1 – u)(1 + K cos b2)u/gc We know that u

I 2gH , V1

\

W

[I 2gH (CV 2gH  I 2gH )(1  K cos E 2 )] / gc

or

W = 2gHf (CV – f) [1 + K cos b2]/gc

CV 2gH

(i)

Hydraulic Efficiency (hH): KH

or

KH

W gH gc

2gH I (CV  I )(1  K cos E 2 ) gH gc gc

(ii)

2I (CV  I )(1  K cos E 2 )

For a given turbine, if f = CV/2, then work and hydraulic efficiency become maximum, and Eqs. (i) and (ii) become Wmax KH max

ËCV2 (1  K cos E 2 ) gH Û 2g c Í Ý

(iii)

ËCV2 (1  K cos E 2 )Û 2 Í Ý

(iv)

8(b)

A Francis turbine works under a head of 260 m and develops 16.2 MW at a speed of 600 rpm. The volume flow rate through the machine is 7 m3/s. If the outside wheel diameter is 1.5 m and the axial wheel width at inlet is 135 mm, find the overall efficiency, the hydraulic efficiency and inlet angles of guide blades and rotor blades. Assume a volumetric efficiency of 0.98 and the velocity at draft tube exit to be 17.7 m/s. The whirl velocity compound at the wheel exit is zero.

Ans.

Machine = Francis turbine, head H = 260 m, power developed P = 16.2 MW, speed N = 600 rpm, volume flow rate Q = 700 l/s, outside wheel diameter = 1.5 m, axial wheel width at inlet B1 = 0.135 m, volumetric efficiency hV = 0.98, exit velocity from the draft tube V4 = 17.7 m/s, whirl velocity at exit Vw2 = 0. Refer to Figure 7.8(a)

606

Model Question Papers

Overall efficiency (ho): P W.P.

Ko

P U gQH

16.2 – 106 W 1000 – 9.81 – 7 – 260

= 0.9073

Ans.

Hydraulic efficiency (hH):

KH

Ë V2 Û 1 KV ÌH  4 Ü – 2g ÜÝ H ÌÍ

KH

Ko KV

Ë 17.72 Û 1 0.98 Ì 260  ܖ 2 – 9.81ÜÝ 260 ÌÍ

0.92

Alternatively: 0.92 0.98

= 0.939

Ans.

Inlet guide (stator) blade angle (a1):

S D1 N 60

u1

S – 1.5 – 600 60

= 47.2 m/s Now,

KH

u1Vw 1 gH

\

Vw 1

KH gH u1

(Q Vw2 = 0 as per given data) 0.939 – 9.81 – 260 47.2

50.7 m/s

Q = AVf 1 = pD1B1Vf1 \

Vf 1 tan D1

\

7 S – 1.5 – 0.315 Vf 1 Vw 1

11 50.7

11m/s

0.217

a1 = 12.2°

Ans.

Inlet rotor blade angle (b1): tan D 1

\

Vf 1 (Vw 1  u1)

b1 = 72.3°

11 (50.7  47.2)

3.143

Ans.

Model Question Papers

607

PAPER 6 1(a)

Define a turbomachine. Differentiate between a turbomachine and a positive displacement machine.

Ans.

Refer to Sections 1.4 and 1.6.

1(b)

The performance of a turbine may be assumed to depend on the mass flow rate, the initial and final pressures, the temperature drop across the turbine, outer diameter of the rotor, the initial temperature and the speed of the rotor. Using the three fundamental dimensions of length, mass and time only, find the dimensionless parameters which will express the performance of the turbine.

Ans.

Refer to Example 1.7.

2(a)

In a certain turbomachine, the blade speed at exit is twice that at inlet (u2 = 2u1), the meridian component of fluid velocity at inlet is equal to that at exit (Vm1 = Vm2) and the blade angle b1 at inlet is 45°. Show that the energy transfer per unit mass and the degree of reaction are given by

cot E 2 + 2 E =  2Vm2 (2 + cot E 2 ) and R = 4 m where, b2 = blade angle at exit. From the above expression for E/m, determine the values of b2 for which the machine acts as a compressor and the values of b2 for which the machine acts as a turbine. Assume Vw1 = 0. Ans.

Refer to Section 2.6.1 and equations (2.17) and (2.19).

2(b)

Liquid water flows at the rate of 31.5 kg/s through the rotor of a turbomachine where the inlet and outlet mean diameters are 12.5 cm and 20 cm respectively. The other data relating to the turbomachine are: speed = 6000 rpm, absolute velocity of fluid at inlet = 35 m/s and is directed axially, absolute velocity at outlet = 160 m/s and makes an angle of 30° with respect to the tangential direction. Using the mean inlet and outlet diameters, determine (i) the power in kW (ii) the torque exerted, and (iii) the stagnation enthalpy change across the rotor. Draw the velocity diagrams at inlet and exit.

Ans.

Mass flow rate m = 31.5 kg/s, inlet diameter d1 = 12.5 cm, outer diameter d2 = 20 cm, speed N = 6000 rpm, absolute velocity at inlet V1 = 35 m/s, liquid is directed axially (a1 = 90°), Vw1 = 0, absolute velocity at outlet V2 = 160 m/s, a2 = 30°. u2 = (pd2N)/60 = (p × 0.2 × 600)/60 = 6.283 m/s Power (P):

Vw2 = V2 cos a2 = 160 × cos 30° = 138.6 m/s

P = (mu2Vw2)/gc = 31.5 × 6.283 × 138.6 = 27423.84 W

Ans.

The torque exerted (T): P = Tw = Tu2/r2 = (T × 6.283)/0.1;

\ T = 436.48 N-m

Ans.

608

Model Question Papers

Stagnation enthalpy change across the rotor (Dh0): W.D. = Dh0 = (u2Vw2)/gc = 6.283 × 138.6 = 870.83 J/kg

Ans.

3(a) Show that for a turbine the maximum utilization factor is given by 2I cos D 1 1  2IR cos D1

H max

Ans.

Substituting Eq. (2.39d) in (2.7),

Ë R (V12  V22 ) Û 1 W.D. = Ì(V12  V22 )  ܖ (1  R ) ÜÝ 2gc ÌÍ W.D. =

(V12  V22 )(1  R )  R (V12  V22 ) 2(1  R )gc

(V12  V22 ) 2(1  R )gc

(V12 cos2 D1  Vf 21 )  (V22 cos2 D 2  Vf 22 ) 2(1  R ) gc

(u1Vw 1  u 2Vw 2 ) gc

(u1Vw 1  u 2Vw 2 ) gc

(i)

(ii)

If the K.E. of the fluid leaving the rotor is minimum, the maximum possible portion of the fluid energy at the rotor inlet will be utilized for conversion into work. The utilization factor will therefore be a maximum if V2 = Vf2 or a2 = 90°. Substituting the condition of maximum utilization and V2 = Vf2, a2 = 90° and Vw2 = 0 in Eq. (ii), u1Vw 1

V12 cos 2D 1  (Vf 21  Vf 22 ) 2(1  R )

u1

V12 cos2D1  (Vf 21  Vf 22 ) 2(1  R )V1 cos D 1

I

u1 V1

V12

2

cos D1 

 Vf 22 )

cos2D1 

Vf 22 V12

2(1  R ) cos D1

Vf 21

1

V12



Vf 22

V12 2(1  R ) cos D1

2V12 (1  R ) cos D1

cos2D1  sin2D 1  I

(Vf 21

[Q Vw1 = V1 cos a1]

Vf 22

V12 2(1  R ) cos D1

(iii)

Model Question Papers

609

Substituting the condition of maximum utilization, i.e. V2 = Vf 2, in Eq. (2.39f) V12  Vf 22

H max

(iv)

V12  RVf 22

From Eq. (iii), we get

V12  Vf 22 Vf 22

or

V12 2I (1  R ) cos D1 V12 [1  2I (1  R ) cos D1]

(v)

From Eq. (iv), we get

H max V12  H max RVf 22

V12  Vf 22

or

H max V12  V12

H maxRVf 22  Vf 22

Vf 22 (H maxR  1)

or

V12 (H max  1) Vf 22 (H max R  1) ; ? Vf 22

V12 (H max  1) (H maxR  1)

(vi)

Equating Eqs. (v) and (vi), we get

(H max  1)V12 (H max R  1) or

V12 [1  2I (1  R ) cos D1]

H max  1 H max R  2I (1  R ) cos D1H max R  1  2I (1  R ) cos D 1

or

H max (1  R )

 2I (1  R ) cos D1 (H max R  1)

or

H max

 2I H maxR cos D1  2I cos D1

or

H max (1  2I R cos D1)

2I cos D1 (1  2R I cos D1)

2I cos D1 ; ? H max

(vii)

Under these conditions, i.e. V2 = Vf2, a2 = 90°, Vw2 = 0, Eqs. (i) and (ii) become

W.D.

u1Vw 1 gc

u1V1 cos D1 gc

IV12 cos D1 gc

(' u1

IV1)

(viii)

For impulse stage: Substituting Eq. (2.51) in (viii),

W.Dimpulse

IV12 2I gc

2I 2V12 gc

2u12 gc

(ix)

For reaction stage: Substituting Eq. (2.50) in (viii),

W.Dreaction

IV12I gc

IV12 gc

u2 gc

(x)

610

Model Question Papers

3(b)

At a stage of an axial flow impulse turbine, the mean blade diameter is 80 cm and the speed is 3000 rpm. The absolute velocity of the fluid at inlet is 300 m/s and is inclined at 20° to the plane of the wheel. If the utilization factor is 0.85 and the relative velocity at rotor exit is equal to that at inlet, determine (i) the inlet and exit blade angles and (ii) the power output in kW for a mass flow rate of 1 kg/s. Also, sketch the inlet and exit velocity diagrams.

Ans.

Refer to Example 2.33. Explain the following terms with reference to a turbine.

4(a)

(i) Overall isentropic efficiency, (ii) Stage efficiency, (iii) Polytropic efficiency, (iv) Mechanical efficiency. Ans.

Refer to Figure 3.11. (i) Overall isentropic efficiency (hts–s): Kts s

Actual work done Ideal work done

Wa Wisen

(Eq. (3.54))

(ii) Stage efficiency (hS): KS

Actual work done in one stage Ideal work done in that stage

Wa 'Wisen

(iii) Polytropic efficiency (hp): Refer to Section 3.6.3 and Eq. (3.60a). (iv) Mechanical efficiency (hm):

Km

Power at the shaft of the turbine Power developed by the runner

S.P. P

4(b)

Obtain an expression for the overall isentropic efficiency in terms of the stage efficiency, pressure ratio per stage, the number of stages and the ratio of specific heats for a turbine.

Ans.

5(a)

Refer to Section 3.6.4, i.e. derivation up to Eq. (3.71). Explain the phenomenon of surging as applied to a centrifugal compressor.

Ans.

Refer to Section 4.11.

5(b)

The following data are suggested as a basis for the design of a single-sided centrifugal compressor. Power input factor = 1.04, slip factor = 0.9, speed = 290 rps, overall diameter of impeller = 50 cm, eye tip diameter = 30 cm, eye root diameter = 15 cm, air mass flow rate = 9 kg/s, inlet stagnation temperature = 295 K, inlet stagnation pressure = 1.1 bar, total head isentropic efficiency = 78%. Air enters the impeller in an axial direction with a velocity of 143 m/s. Assuming that the radial component of fluid velocity at impeller exit is equal to the axial component of fluid velocity at impeller inlet and the total losses are equally divided between the impeller and the diffuser,

Model Question Papers

611

determine (i) the overall pressure ratio and the (ii) blade angles at impeller eye root and at eye tip. ABC = Actual with slip, ADC = Ideal without slip, s = 0.9, j = 10.4, N = 290 rps, d2 = 50 cm, dt = 30 cm, dr = 15 cm, m = 9 kg/s, T01 = 295 K. p01 = 1.1 bar, hct–t = 0.78, V1 = Vf1, Vw1 = 0, a1 = 90°, V1 = 143 m/s, Vf2¢ = Va1 = Vf1 Slip B

D V2

V2¢

Vr 2 = Vf 2

Ans.

Vr 2¢ Vf 2¢

A

b1

C

u1

Peripheral speed (u2), u2 = pd2N = p × 0.5 × 290 = 455.5 m/s Power (P) (Eq. (4.18)),

P

(VMu 22 ) / gc

0.9 – 1.04 – 455.52

194.214 kW/kg

P = 194.214 kW/kg = cp(T02 – T01) = 1.005(T02 – 295) \

T02 = 488.2 K

Overall isentropic efficiency (hct–t) (Figure Example 4.18): Kct t

(T02  To 1) (T02  T01 )

(T02  295) ; ? T02„ (488.2  295)

„

„

445.7 K

Overall pressure ratio (pR1) (Equation (4.9)): pR 1

p 02 p01

J

ËT02 Û Ì Ü Í T01 Ý

/(J 1)

„

Ë 445.7 Û Ì 295 Ü Í Ý

3.5

4.24

Blade angle at impeller eye root (b1r): Peripheral speed at eye root (ur) ur = pdrN = p × 0.15 × 290 = 136.7 m/s tan E1r

Vfr ur

143 136.7

1.046; ? E1r

46.3’

Ans.

612

Model Question Papers

Blade angle at impeller eye tip (b1t): Peripheral speed at eye tip (ut) ut = pdtN = p × 0.3 × 290 = 273.3 m/s Vft ut

tan E1t

6(a)

Vf 1 u1

143 273.3

0.523 ? E1t

27.62’ Ans.

Draw the velocity diagrams for an axial flow compressor stage and show that the degree of reaction is given by R = Va (tan g1 + tan g2)/2u

Ans.

Refer Section 2.8.1 and Eqs. (2.30) to (2.33a) (angles considered with respect to the tangential direction).

6(b)

An axial compressor stage has blade root and mean blade velocities of 150 and 200 m/s respectively. The stage is to be designed for a stagnation temperature rise of 20°C and an axial velocity of 150 m/s at the mean blade height. The work done factor is 0.93 and the degree of reaction at the mean radius is 50%. Determine (i) the blade angles at the mean radius, (ii) the blade angles and the degree of reaction at the blade root using the free vortex flow diagram.

Ans.

Stagnation temperature rise, DT0 = 20°C, axial velocity at mean blade height, Vam = 150 m/s, = Vfm, workdone factor, j = 0.93, degree of reaction at the mean radius, Rm = 0.5, ur = 150, um = 200.

Vr 1m

Vr 2m

V1m Vfm

Vfm a 1m

b 1m um

V2m

a 2m

b 2m

Vw 1m

um

Vw 2m

(i) The blade angles at the mean radius (b1m, b2m): c p 'T 0

(cot E1m  cot E 2m ) (cos E1m  cot E 2m )

MumVfm (cot E1m  cot E 2m ) g c – 1000

20 – 1 – 1000 – 1.005 0.93 – 200 – 150 Rm 2um Vfm

0.5 – 2 – 200 150

(Eq. (5.18))

0.72 1.333

(i) (Eq. (5.27))

(ii)

Model Question Papers

613

Solving Eqs. (i) and (ii), we get b1m = 44.24°; b2m = 72.96° Ans. This is for 50% reaction. Hence, a1m = b2m, a2m = b1m (ii) The blade angles and the degree of reaction at the blade root using free vortex flow design (b1r , b2r , Rr): rmVw1m = rrVw1r ; rm cos a1m = rr cot a1r

cot D1r

rm cot D1m rr

um cot D1m ur

200 – cot 72.96’ 150

0.409 ; ? D1r

67.77

We have

cot D1r  cot E1r

ur V fm

150 150

1

cot b1r = 1 – cot 67.77° = 0.5912 ; \ b1r = 59.4° Similarly, we have rmVw2m = rrVw2r ; rm cot a2m = rr cot a2r cot D 2r

rm cot D 2m rr

Ans.

um cot 44.24’ ur

200 – cot 44.24’ 150

1.369; ? D 2r

36.14’

We have cot D 2r  cot E 2r

ur Vfm

150 150

1

cot b2r = 1 – cot a2r = 1 – cot 36.14°; Rr

Vfm (cot E1r  cot E 2r ) 2ur

\ b2r = 69.73°

Ans.

150 (cot 59.40’  cot 69.73’) = 0.111 or 11.1% 2 – 150

Ans.

7(a)

What do you mean by compounding of steam turbine? Explain with the help of a schematic diagram a two-row velocity compounding turbine stage.

Ans.

Compounding is the method of reducing the blade speed for a given overall pressure drop by absorbing the inlet energy in multiple rows of moving blades. For velocity compounding—Refer to Section 6.5.1.

614

Model Question Papers

7(b)

Dry saturated steam at 10 bar is supplied to a single rotor axial flow impulse turbine, the condenser pressure being 1 bar. The nozzle efficiency is 94% and the nozzle angle at the rotor inlet is 18° to the wheel plane. The rotor blades are equiangular and move at a speed of 450 m/s. If the blade velocity coefficient for the moving blades is 0.92, determine (i) the absolute velocity of steam entering the rotor, (ii) the power output for a mass flow rate of 1 kg, (iii) the rotor efficiency and (iv) the stage efficiency.

Ans.

pb = 10 bar, pc = 1 bar, hn = 0.94, a1 = 18°, b1 = b2, u = 450 m/s, K = 0.92. Vw 2 u a 1 b1

a2 Vf 2

h

Vw 1

V2

b2 Vr 1

Vr 2

pb

1

pc

Vf 1

V1

2¢

2

S

(i) Absolute velocity of steam (V1): (Dh)isen = h1 – h2 = 401.75 kJ/kg Kn V1

0.94

(h1  h2 ) ; h1  h2 ( 'h )isen „

„

(h1  h2 ) – 2 – gc – 1000 „

377.83 377.83 – 2 – 1 – 1000

868.3 m/s

Ans. (ii) Power output (P): Vw1 = V1 cos a1 = 869.3 × cos 18° = 826.75 m/s Vf 1

V12  Vw21

869.32  826.752

Vr 1

Vf 21  (Vw 1  u )2

268.592  (826.75  450)2

268.59 m/s

462.7 m/s

Vr2 = KVr1 = 0.92 × 462.7 = 425.7 m/s

\

tan E1

Vf 1 Vw 1  u

268.59 ; ? E1 826.75  450

cos E1

u  Vw 2 Vr 2

450  Vw 2 ; ? Vw 2 425.7

P

m – u – (Vw 1  Vw 2 ) g c – 1000

E2

35.49’

 103.63 m/s

Model Question Papers

1 – 450 – (826.75  103.36) 1 – 1000 (iii) Blade efficiency or Rotor efficiency (hb): Kb

615

325.52 kW

Ans.

0.862

Ans.

2 – (Vw 1  Vw 2 )u V12 2 – (826.75  103.56) – 450 868.32

(iv) Stage efficiency (hS): hS = hn × hb = 0.94 × 0.862 = 0.8103

Ans.

Alternatively: KS

Power developed by the turbine ('h )isen

325.52 kW 401.75 – 1

0.81

Ans.

8(a)

The supply to a single jet Pelton wheel is from a reservoir 310 m above the nozzle centre through a pipe 67.5 cm in diameter and 5.6 km long. The friction co-efficienct for pipe is 0.008. The jet has a diameter of 9 cm and its velocity co-efficient is 0.97. The blade speed ratio is 0.47 and the buckets deflect the water through 170°. The relative velocity of water is reduced by 15% in passing over the buckets. Determine the hydraulic efficiency and overall efficiency of the system if the mechanical efficiency is 88%.

Ans.

Hg = 310 m, dp = 67.5 cm, Lp = 5.6 km, f = 0.008, d = 9 cm, CV = 0.97, f = 0.47, b2 = 180° – 170° = 10°, K = 0.85, hm = 88%

Qp = QN (Penstock discharge = Nozzle discharge) S 2 d p Vp 4

S 2 d V1 ; ? Vp 4

V1d 2 d p2

Now apply Bernoulli’s equation between head race and end of nozzle. Hg = head lost due to friction + nozzle + V12 / 2g

616

Model Question Papers

Hg

310

4f LpVp2

Ë V2 Û V2  ÌH  1 Ü  1 d p 2g 2g ÜÝ 2g ÌÍ

4 – f LpV12d 4 2d p gd p4



V12

Ë' V1 Í

2gCV2

310

4 – 0.008 – 5.6 – 1000 – V12 – (0.09)4

or

310

0.004277V12  0.0542V12

\

V1 = 72.83 m/s = Vw1 H

u

5

(0.675) – 2 – 9.81

V12

72.832

2gCV2

2 – 9.81 – 0.972

I 2gh



CV 2gH ÛÝ

V12 2 – 9.81 – (0.97)2

287.33 m

0.47 2 – 9.81 – 287.33

35.3 m/s

Vr1 = V1 – u = 72.83 – 35.3 = 37.54 m/s Vr2 = 0.85 × Vr1 = 0.85 × 37.54 = 31.91 m/s cos E 2

\

u  Vw 2 Vr 2

35.3  Vw 2 31.91

cos 10’

Vw2 = –3.875 m/s

Power developed by the runner,

P

P

U Q u (Vw 1  Vw 2 ) gc – 1000 100 –

U–

S d 2u (Vw 1  Vw 2 )V1 4 gc – 1000

S – (0.09)2 – 35.3 – (72.83  3.875) – 72.83 4 9 – 1000

1127.8 kW

(a) Hydraulic efficiency (hH): P 1127.8 – 2 – 4 – 1000 1 2 1000 – S – (0.09)2 – (72.83)3 U QV1 2 (b) Overall efficiency (ho): KH

0.918

Ans.

S.P. = Shaft power = hm × P = 0.88 × 1127.8 = 992.464 kW Ko

\

S.P. W.P

ho = 0.76

S.P. U gQH

992.464 – 1000 – 4 1000 – 9.81 – S – (0.09)2 – 287.33 – 72.83

Ans.

Model Question Papers

617

8(b)

The inner and outer diameters of a Francis turbine are respectively 30 cm and 60 cm. Water enters the turbine at an angle of 20° to the wheel tangentially and leaves the turbine radially. If the velocity of flow remains constant throughout at 3 m/s and the speed of runner is 300 rpm, calculate (i) the inlet and exit blade angles and (ii) the theoretical power developed if the width of the wheel at inlet is 15 cm. Neglect the thickness of blades.

Ans.

D 1 = 60 cm, D 2 = 30 cm, Guide blade angle, a 1 = 20°, radial exit, a 2 = 90°, V w2 = 0, Vf1 = Vf 2 = Vf = 3 m/s, N = 300 r.p.m., B1 = 15 cm thickness neglected.

u1

S D1N 60

S – 0.6 – 300 60

u2

S D 2n 60

S – 0.3 – 300 60

V1

Vf 1 sin D 1

3 sin 20’

9.42 m/s 4.71m/s

8.77 m/s

Vw1 = V1 cos a1 = 8.77 × cos 20° = 8.24 m/s (i) Inlet and exit blade angles (b1, b2): tan E1

\

3 9.42  8.24

2.54

b1 = 68.5°

tan E2 \

Vf 1 u1  Vw 1

Vf u2

3 4.71

Ans.

0.637

b2 = 32.5°

Ans.

(ii) Theoretical power developed (P): Q = pD1B1Vf = p × 0.6 × 0.15 × 3 = 0.848 m3/s P

U QVw 1u1 gc 1000

1000 – 0.848 – 8.24 – 8.42 1 – 1000

65.82 kW

Ans.

618

Model Question Papers

PAPER 7 1(a)

Distinguish between a turbomachine and a positive displacement machine.

Ans.

1(b)

Refer to Section 1.6. Define specific speed of a pump. Derive an expression for the specific speed of the pump from non-dimensional numbers.

Ans.

Refer to Section 1.18.7.

1(c)

Tests on a turbine runner 1.25 m is diameter at 30 m head gave the following results: power developed = 736 kW, speed = 180 rpm and discharge = 2.7 m3/s. Find the diameter, speed and discharge of a runner to operate at 45 m head and give 1472 kW at the same efficiency. What is the specific speed of both the turbines?

Ans.

Model (m): D = 1.25, H = 30 m, P = 736 kW, N = 180 rpm, Q = 2.7 m3/s Prototype (p): D = ?, H = 45 m, P = 1472 kW, N = ?, Q = ?, hm = hp Speed of the prototype (Np): (NST)p = (NST)p ËN P Û Ì 5/4 Ü ÍÌ H ÝÜ p

ËN P Û Ì 5/4 Ü ; ÍÌ H ÝÜm

? Np

211.3 rpm

Ans.

Diameter of the prototype (Dp): Ë DN Û Ì Ü Í H Ýp

Ë DN Û Ì Ü ; Í H Ým

? Dp

1.304 m

Ans.

Discharge of the prototype (Qp): hm = hp

È p Ø ÉÊ U QgH ÙÚ p

È p Ø ÉÊ U QgH ÙÚ ; m

? Qp

3.6 m3 /s

Ans.

Specific speed of both ËÍ (N ST )m = (N ST ) p ÛÝ : (NST )m

2(a)

(NST )p

ËN P Û Ì 5/4 Ü ÍÌ H ÝÜ

180 – 736 305 / 4

69.6 rpm

Ans.

Define utilization factor of a turbine. Derive an expression relating utilization factor to the degree of reaction.

Model Question Papers

Ans.

619

Utilization factor (e ): H

(V12  V22 )  (u12  u 22 )  (Vr22  Vr21 )

(Eq. 2.39(a))

[V12  (u12  u 22 )  (Vr22  Vr21 )] (u12  u 22 )  (Vr22  Vr21)

R

(V12  V22 )  (u12  u 22 )  (Vr22  Vr21) (u12  u 22 )  (Vr22  Vr21)   ‘

(Eq. (2.13))

R [V12  V22 ]  R [(u12  u 22 )  (Vr22  Vr21)]   ‘

x

\

x  Rx

x

R (V12  V22 ); ? x (1  R ) (u12  u 22 )  (Vr22  Vr21)

or

R (V12  V22 )

R (V12 V22 ) 1 R

(i)

Substituting Eq. (i) in the utilization factor equation, (V12  V22 )  H V12 

R (V12  V22 ) 1 R

R (V12

 V22 )

(1  R )(V12  V22 )  R (V12  V22 ) V12

(1  R ) 

R (V12

 V22 )

V12  V22 V12  RV22

Ans.

1 R

2(b)

At a stage in a 50% reaction axial flow turbine running at 3000 rpm, the blade mean diameter is 685 mm. If the maximum utilization for the stage is 0.915, calculate the inlet and outlet absolute velocities for the rotor. Draw the velocity triangles and find the power output for a flow of 15 kg/s.

Ans.

Refer to Example 2.22.

3(a)

With the help of the T–s diagram, explain (i) the reheat factor of a turbine and (ii) the preheat factor of compression.

3(b)

Define: (i) Stage efficiency, (ii) Overall efficiency with respect to compression process.

Ans.

(i) Preheat factor of a compressor: Let us consider only one stage (1st stage), i.e. between the pressure limits of p1 to pA \

KS 1

'Wisen1 'Wa1

Process 1-A Process 1-C

DWa1 = actual work required for one stage

Model Question Papers

T2¢ Dw isen1

C

A

T2

p2

Dwa3 E T3

Dwa2

Dw isen3 D

Dw isen2

B T3¢

F

p3 WA pA p1

K = number of stages

T (K+1)¢ 2¢

T(K+1) 2

Dwa1

T or h

Dw isen

620

1 T1

s

Similarly, KS 2

KS

'Wisen 2 'Wa 2

'Wisen 3

; KS 3

'Wa 3

3

'Wisen1

i 1

'Wa1

Ç KSi 6 'Wisen Wa



'Wisen 2 'Wa 2



'Wisen 3 'Wa 3

6'Wisen 6'Wa

6 'Wisen Process 1-2

(i)

hcs–s = overall isentropic compression efficiency between pressure limits of p1 to p2 Wisen ; ? Wa Wa

Wisen Kcs s

(ii)

From Eqs. (i) and (ii), we get Kcs s KS

Wisen 6'Wisen

Preheat factor (PF)

Ans.

(ii) Reheat factor of a turbine: hts–s = overall isentropic expansion efficiency between the pressure limits of p1 to p2 hS = stage efficiency (there are three stages in the figure) Kts s KS 1

Wa Wisen

Process 1-2 Process 1-2„

'Wa1 ; KS 2 'Wisen1

'Wa 2 ; KS 3 'Wisen 2

(iii) 'Wa 3 'Wisen 3

Model Question Papers T or h

621

Dw isen

Dw isen2

Dwa 2

A

B

E

Dw isen3

D

Wa

pB

Dwa 3

Dw isen1

pA

C

Dw a1

p1

1

p2

2 F 2¢ s

KS

3

'Wa1 'Wa 2 'Wa 3   'Wisen1 'Wisen 2 'Wisen 3

Ç KSi

i 1

6'Wa 6'Wisen

Wa 6'Wisen

(iv)

From Eqs. (iii) and (iv), we have Kts s KS

6'Wisen Wisen

Reheat Factor (RF)

Ans.

3(c)

A 16-stage axial compressor is to have a pressure ratio of 6.3 and tests have shown that a stage efficiency of 89.5% can be obtained. The intake conditions are 288 K and 1 bar. Find (i) the overall efficiency, (ii) the polytropic efficiency, and (iii) the preheat factor.

Ans.

(i) Overall efficiency (hcs–s):

(Eq. (3.42g)) (6.30.286 – 16  1)

(ps K H  1)

Kcs s

Ë 6.3H  1Û Ì1  Ü KS Ý Í

K

1

Ë (6.30.286  1) Û Ì1  Ü 0.895 Í Ý

0.472

16

Ans.

1

(ii) Polytropic efficiency (hP):

KS

(pSH  1) H / Kp

(pS

 1)

0.895

(6.30.286  1) (6.3

0.286 / Kp

 1)

; ? Kp

0.918

Ans.

Alternatively:

Kcs s

(pSHK  1) HK / K p

pS

1

0.472

(6.30.286 – 16  1) (6.3

0.286 – 16 / Kp

 1)

; ? Kp

0.918

Ans.

622

Model Question Papers

(iii) Preheat Factor (PF): Kcs s KS

PF

0.472 0.895

0.5273

Ans.

4(a)

Discuss the phenomena of slip and cavitation in centrifugal pumps.

Ans.

Refer to Section 4.7 for slip and Section 4.3 for cavitation.

4(b)

Define manometric head of a pump and derive an expression for it.

Ans.

Refer to Section 4.17.2 and Eq. (4.33) or (4.34) or (4.35). A centrifugal pump delivers 50 l/s against a total head of 24 m when running at 1500 rpm. The velocity of flow is maintained constant at 2.4 m/s and blades are curved back at 40° to the tangent at outlet. The inner diameter is half the outer diameter. If the manometric efficiency is 80%, determine (i) the blade angle at inlet and (ii) the power required to drive the pump.

4(c)

Ans.

Discharge = 0.05 m3/s, total head Hm = 24 m, speed N = 1500 rpm, velocity of flow, Vf1 = Vf2 = 2.4 m/s, b2 = 40°, d1 = 0.5d2, manometric efficiency, hman = 0.8. Kman tan E 2

9.81 – 24 Vw 2u 2

gH m Vw 2u 2 Vf 2 u 2  Vw 2

0.8; ? Vw 2u 2

2.4 ; ? u2 294.3 u2  u2

294.3

18.7 m/s, u1

9.35 m/s

(i) Blade angle at inlet (b1): tan E1

Vf u1

2.4 ; ? E1 9.35

14.4’

Ans.

(ii) Power required to drive the pump (P): P

U QVw 2u 2 gc

1000 – 0.05 – 294.3 1000 – 1

14.715 kW

Ans.

5(a)

Derive a theoretical head capacity relationship for centrifugal pumps and compressors and explain the influence of the outlet blade angle.

Ans.

Refer to Section 2.7.2 and Figure 2.6.

5(b)

Draw the velocity triangles for an axial flow compressor and show that for an axial flow compressor having no axial thrust, the degree of reaction is given by R

Ans.

Va 2u

tan E1  tan E2 tan E1 – tan E2

6(a)

Refer to Eqs. (2.33) and (2.33a). Define radial equilibrium, obtain an expression for the same.

Ans.

Refer to Section 5.12, derivation up to Eq. (5.32).

Model Question Papers

623

6(b)

Derive an expression for a static pressure rise of an axial flow compressor in terms of fluid and blade angles.

Ans.

Refer to Figure 5.3. The main function of a compressor is to raise the static pressure of air or gas. The static pressure rise in the stage depends on the flow geometry and speed of the rotor. The total static pressure rise across the stage (rotor and stator) is the sum of the static pressure rises in the rotor (DPR) and diffuser (stator) DPD blade rows. Applying Bernoulli’s equation across the rotor blade,

p1 

UVr22 2

1 UVr21 2

p2 

( 'p )R

p2  p1

1 U (Vr21  Vr22 ) 2

U 2 (Vf  x 12  Vf 2  x 22 ) 2

U U 2 Vf (cot 2 E1  cot 2 E2 ) 2 2 Similarly, applying Bernoulli’s equation across the diffuser blade, (x 12  x 22 )

( 'p )D

p3  p2

U 2 (V2  V32 ) 2

U 2 (Vw 2  Vw23 ) 2

U 2 (Vf  Vw22  Vf 2  Vw23 ) 2

U 2 (Vf cotD2 2  Vf 2 cotD2 3 ) 2

Assuming a1 = a3, ( 'p )D

UVf (cot 2D 2  cot 2D1) 2

The stage pressure rise (Dp)S, (Dp)S = p3 – p1 = DpR + (Dp)D

7(a)

UVf 2 [(cot 2 E1  cot 2 E 2 )  (cot 2D 2  cot 2D1)] Ans. 2 What is the need for compounding in steam turbines? Discuss any two methods of compounding.

Ans.

Exit velocity of steam from a single-stage impulse turbine is very high. The turbine speed is in the order of 30,000 rpm. Gear trains are required for driven machines, if suitable speed is required. Due to high exit velocity, the carry-over loss is also high. In order to avoid all these demerits, exit velocity is required to be reduced. This can be done by compounding. Refer to Sections 6.5.1 and 6.5.2.

7(b)

An axial flow impulse steam turbine has a mean rotor diameter of 55 cm and runs at 3300 rpm. The speed ratio is 0.45 and the blade velocity coefficient is 0.91. If the nozzle angle at rotor inlet is 20°, find (a) the rotor blade angles, assuming axial exit and (b) the power output for a flow of 1 kg/s.

624

Model Question Papers

u

Ans.

V1

S dN 60

u I

S – 0.55 – 3300 60

95.04 0.45

95.04

211.2 m/s

Vw1 = V1 cos a1 = 211.2 × cos 20° = 198.5 Vf 1

V12  Vw21

211.22  198.52

72.27 m/s

(a) Rotor blade angles (b1, b2): E1 Vr 1

Vf 1 Vw 1  u

72.27 ; ? E1 198.5  95.04

Vf 21  (Vw 1  u )2

34.94’

72.272  (1985  95.04)2

Ans.

126.2 m/s

Vr2 = kVr1 = 0.91 × 126.2 = 114.84 m/s cos E2

u Vr 2

95.04 ; ? E2 114.84

34.15’

Ans.

(b) Power output for a flow of 1 kg/s (P): Vw 1um g c – 1000

P

8(a)

Ans.

198.5 – 95.04 – 1 18.87 kW 1000 – 1

Ans.

A Pelton wheel produces 15,500 kW under a head of 300 m at 500 rpm. If the overall efficiency of the wheel is 84%, find (i) the required number of jets and the diameter of each jet, (ii) the number of buckets, and (iii) the tangential force exerted, assume jet ratio = 9.5. u

I 2gH

0.45 2 – 9.8 – 350

u

S DN 60

V1

CV 2gH

S – D – 500 60

37.3 m/s

37.3; ? D

0.98 – 2 – 9.8 – 350

1.424 m

81.21m/s

Model Question Papers

625

(i) Required number of jets (Nj) and the diameter of each jet (d ): D/d = 9.5; \ Ko

p UgQH

d = 0.15 m 15500 – 1000 1000 – 9.81 – Q – 300

0.84; ? Q

6.27 m3 /s

S 2 S d V1 N j – – 0.152 – 81.21 6.27 4 4 Nj = 4.36 » 5 sets Q

Now, \

Nj –

Ans.

(ii) Number of buckets (m): D 9.5 15  2d 2 (iii) Tangential force exerted (Fx): m

p Fx

15 

19.75  20

Ans.

U Q (Vw 1  Vw 2 )u 1000 U Q (Vw 1  Vw 2 )

P – 1000 u

15500 – 1000 37.3

415549.6 kN

Ans.

8(b)

Draw the neat sketch of a Francis turbine and explain the function of draft tube. Draw the velocity triangles of Francis turbine.

Ans.

Refer to Figure 7.7, Figure 7.8 and Figure 7.11.

626

Model Question Papers

PAPER 8 1(a)

Define a turbomachine. Explain how turbomachines are classified?

Ans.

Refer to Sections 1.4 and 1.8. The thrust of a propeller is assumed to depend on the axial velocity of the fluid, the density and viscosity of the fluid, the speed of the propeller in rpm and the propeller diameter. Obtain the dimensionless parameters for the propeller.

1(b)

Ans.

The physical quantities and the corresponding dimensions for the propeller are as follows. No.

Variable Name

1 2 3 4 5 6

Propeller diameter Axial velocity of fluid Fluid viscosity Speed of the propeller Density of the fluid Thrust

Symbol

Units

Dimension

D V m N r T

m m/s kg/m-s rpm kg/m3 N-kg-m/s2

L LT–1 ML–1T–1 T–1 ML–3 MLT–2

No. of dimensionless parameters = No. of physical variables No. of fundamental dimensions = 6 – 3 = 3 Let p, p2 and p3 be these dimensionless parameters. No. of repeated variables = No. of fundamental dimensions The repeated variables therefore are: D, V and r

Now,

S1

D a1V b1Uc 1N ; S 2

S1

D a1V b 1Uc 1T

D a 2V b 2 Uc 2 P and S 3

D a 3V b 3 Uc 3T

Substituting the dimensions, L°M°T° = La1(LT–1)b1 (ML–3)c1T–1 Equating the power of L, M and T 0 = a1 + b1 – 3c1 0 = c1; T, 0 = –b1 – 1 \

c1 = 0; b1 = –1; a1 = 1 \

p1 = D1V–1r0N or p1 = DN/V

Similarly, it can be shown that, S2

P UVD

and S 3

T UV 2D 2

Note: Repeated variables can be taken in any other combination. Examples, DVm, DVT, etc., then the p values will be different as shown below:

Model Question Papers

T

p1

2

UN D

V ND

p2

P

p3

2(a)

4

UND

2

T PVD

PVD 2

ND V

V ND

UVD P

UVD P

627

T

Draw the velocity triangles at inlet and exit of a turbomachine in general and show that the energy transfer per unit mass is given by E m

1 Ë(V12  V22 )  (Vr22  Vr21 )  (u12  u 22 )Û Ý 2Í

Ans.

Refer to Section 2.2, Figure 2.2 and derivation up to Eq. (2.7).

2(b)

An upward radial flow reaction turbine has radial discharge at outlet with outlet blade angle of 45°. The radial component of absolute velocity remains constant throughout and equal to 2 gH , where g is the acceleration due to gravity and H is the constant head. The blade speed at inlet is twice that at the outlet. Express the energy transfer per unit mass and the degree of reaction in terms of a1 where a1 is the direction of the absolute velocity at inlet with respect to the blade velocity at inlet. At what value of a1 will the degree of reaction be zero and unity? What are the corresponding values of energy transfer per unit mass?

Ans.

Refer to Example 2.18.

3(a)

Show that for maximum utilization factor and for the same amount of energy transfer in an axial flow impulse turbine and an axial flow reaction turbine with 50% degree of reaction,

R

= 2 ui2

Ans.

Refer to Example 2.20.

3(b)

At a stage of an axial flow impulse turbine, the mean blade diameter is 80 cm and the speed is 300 rpm. The absolute velocity of the fluid at inlet is 300 m/s and is inclined at 20° to the plane of the wheel. If the utilization factor is 0.85 and the relative velocity at rotor exit is equal to that at inlet, determine (i) the inlet and exit blade angles and (ii) the power output for a mass flow rate of 1 kg/s.

Ans.

Refer to Example 2.33.

4(a)

Show that for a finite number of stages for compression the overall isentropic efficiency is given by

628

Model Question Papers

(prK H  1)

Kc

Ë prH  1Û 1  Ì Ü KS ÝÜ ÍÌ

K

1

where, K = no. of stages, pr = pressure ratio/stage, hS = stage efficiency. Ans.

Refer to Section 3.54, up to Eq. (3.42g).

4(b)

In a multistage axia flow air compressor, air is taken at 1 bar and 15°C. It is compressed to a final pressure of 6.4 bar. The final temperature of air is 300°C. Determine the overall isentropic efficiency of the compressor and also the polytropic efficiency. If the actual temperature rise per stage is limited to 13 K, determine the number of stages required assuming that the polytropic efficiency is equal to the stage efficiency.

Ans.

p1 = 1 bar, pK+1 = 6.4 bar, T1 = 288 K, TK+1 = 573 K, K = number of stages, e = (g – 1)/g T(K 1)„

Ëp Û T1 Ì K 1 Ü Í p1 Ý

H

È 6.4 Ø 288 – É Ê 1 ÙÚ

0.286

489.7 K

(i) Overall isentropic efficiency (hcs–s):

(T(K 1)„  1)

Kcs s

(T(K 1)  1)

489.7  1 573  1

0.7069

Ans.

(ii) Polytropic efficiency (hp): pK 1 p1 TK 1 ln T1

H ln Kp

TK 1 0.7709 or T1

0.28 – ln 6.4 573 ln 288

\

n 1 n

ln [(TK 1) / T1] ln [(pK 1) / p1]

\

Kp

J 1 n – J n 1

ln 573 / 288 ln 6.4 / 1

0.286 0.37

0.772

Ë pK 1 Û Ì Ü Í p1 Ý

n 1 n

0.37

(Eq. (3.29b))

Ans.

(iii) Number of stages (K): (DT)actual = 13 K (DT)isen = hp(DT)actual = 0.772 × 13 = 10.04 K \

T2

„

T1  ( 'T )isen

288  10.04

298.04 K

(Eq. (3.37))

Model Question Papers

p Pressure ratio for 1st stage = 2 p1 pK 1 p1

È p2 Ø ÉÊ p ÙÚ

„

È 298.04 Ø ÉÊ 288 ÙÚ

3.5

1.127

K

1

pK 1 p1 p2 ln P1

ln

\

J

È T2 Ø J 1 ÉÊ T ÙÚ 1

629

K

ln 6.4 ln 1.127

15.53  16 stages

(Eq. (3.42c))

Alternatively:

'T1

13

pR

5(a)

T2  T1

ÈT Ø T1 É 2  1Ù T Ê 1 Ú

È H Ø Kp É  1Ù T1 pS É Ù Ê Ú

È 0.286 Ø 288 É pS0.722  1Ù ; ? pS ÉÊ ÙÚ pK 1 p1

6.4 1

PSK

1.127

1.127K ; ? K

15.58  16 stage

Ans.

Applying Bernoulli’s equation between the inlet and exit of the impeller of a centrifugal pump, show that the stage pressure rise is given by p2  p1

U (Vm21  u 22  Vm2 2 cosec 2 E2 ) / 2

Ans.

Refer to Section 4.20, Eq. (4.54).

5(b)

A single-sided centrifugal air compressor running at a speed of 16,500 rpm produces a pressure ratio of 4 : 1. The hub diameter at the eye of the compressor is 16 cm. Inlet of air to the rotor is axial and equal to 120 m/s. The stagnation temperature and pressure at inlet are 25°C and 1 bar. The mass flow rate is 8.3 kg/s and the total head isentropic efficiency is 78%. The pressure coefficient is 0.7. Determine (i) the eye tie diameter, (ii) the blade angle at eye root and eye tip, (iii) the impeller tip diameter and (iv) the shaft power input to the compressor if the mechanical efficiency is 97%.

Ans.

N = 16,500 rpm, p03/p01 = 4, p01 = 1 bar, T01 = 298 K, m = 8.3 kg/s

Kct t o = 0.78, dh = 0.16 m, fp = 0.7, Va1 = 120 m/s, V2 = 120 m/s T1

T01 

V12 2c p

298 

1202 2 – 1005

290.84 K

630

Model Question Papers J

p1

U1

Ë T ÛJ 1 p01 Ì 1 Ü ÍT01 Ý

Ë 290.84 Û 1– Ì Ü Í 298 Ý 5

0.918 – 10 287 – 290.84

p1 RT1

3.5

0.918 Vr 1

1.0998 kg/m3

b1

dt2

a1 u1

(i) Eye tip diameter (dt):

m

Vf 1 = Va 1

U1SVa1 (dt2  d h2 ) / 4

4 – 8.3  0.162 ; ? dt S – 1.0998 – 120

0.325

Ans.

(ii) Impeller tip diameter (d2):

Kct t o

Now,

T03  T01

V

u2

\

Ë pH Û T01 Ì 03  1Ü ÌÍ p 01 ÜÝ ; ? T03  T01 T03  T01

V u 22 gc c p

(Eq. (4.12e))

Ip

0.7 0.78

Kct t o 456.1

185.67 K

S d 2N 60

0.897

(Eq. (4.12g))

S d 2 16500 ; ? d2 60

0.528 m

Ans.

(iii) Blade angle at eye root and eye tip (b1r, b1t): ur

E1r ut

E1t

S dr N 60

S – 0.16 – 16500 60

ÈV Ø tan1 É 1 Ù Ê ur Ú S dt N 60

tan1

120 138.23

138.23 m/s

40.96’

S – 0.325 – 16500 60

ÈV Ø tan1 É 1 Ù Ê ut Ú

tan1

120 280.78

Ans.

280.78

23.14’

Ans.

(iv) Shaft power input to the compressor (P): P = mcp(T03 – T01) = 8.3 × 1.005(185.67);

\ P = 1548.77 kW

Ans.

Model Question Papers

631

6(a)

The first stage of an axial compressor is designed with no inlet guide vanes (i.e. V1 is axial). The speed is 6000 rpm and the stagnation temperature rise is 20 K. The hub-to-tip ratio is 0.6 and the work done factor is 0.93. The isentropic efficiency of the stage is 0.89. Assuming an inlet velocity of 140 m/s and abmient conditions of 1.01 bar and 288 K, calculate (i) the tip radius and the corresponding directions of Vr1 and Vr2 if the Mach No. relative to the tip is limited to 0.95, (ii) the mass flow entering the stage, and (iii) the stage stagnation pressure ratio and power required.

Ans.

Free vortex theory, i.e. Vwr = C, a1 = 90°, Vw1 = 0, V1 = Vf1, N = 6000 rpm, DT0 = 20 K, rh/rt = 0.6, j = 0.93, hct–t = 0.89, V1 = 140 m/s, T01 = 288 K, p01 = 1.01 bar.

(i) The tip radius rt, the root radius rr and the corresponding directions of Vr1 (i.e. b1) and Vr2 (i.e. b2) if the Mach number relative to the tip is limited to 0.95.

V12 2c p

T1

T01 

Vr 1

M 1 J RT1

sin E1

V1 Vr 1

288 

1402 2 – 1005

278.25 K

0.95 1.4 – 287 – 278.25

140 317.65

u1

Vr 1 cos E1

u1

2S Nrt 60

0.4407’; ? E1

317.7 cos 26.15’

2 – S – 600 – rt 60

317.7 m/s

26.15’ 285.2 m/s

285.2; ? rt

0.454 m

Ans.

632

Model Question Papers

rh = 0.6rt = 0.6 × 0.454 = 0.2724 m rm = mean radius = (rt + rr)/2 = (0.454 + 0.2724)/2 = 0.3632 m 'T0

(cot E1  cot E2 )

u MVf (cot E1  cot E 2 ) / g c c p

1005 – 20 0.93 – 285.14 – 140

(Eq. (5.25c))

0.5414

cot b2 = cot 26.15° – 0.5414 = 1.495; \ b2 = 33.77°

Ans.

(ii) The mass flow entering the stage (m): p1

U1

p01 È T01 Ø ÉÊ T ÙÚ 1

p1 RT1

1.01

3.5

È 288 Ø ÉÊ 278.25 ÙÚ

0.8953 bar

3.5

0.8953 – 105 0.287 – 103 – 278.25

A1 = area of flow at inlet

1.121kg/m3

S 2 (dt  d h2 ) 4

S – (0.9082  0.54482 ) 4

0.4137 m2

m = r1A1Va1 = 1.121 × 0.4137 × 140 = 64.9 kg/s

Ans.

(iii) The stage stagnation pressure ratio (pR0) and power required (P): From Eq. (5.23), J

pR 0

Ë Kct t (T03  T01) Û J  1 Ì1  Ü T01 Í Ý

0.89 – 20 Û Ë Ì1  288 ÜÝ Í

3.5

1.234

P = mcpDT0 = 64.9 × 1.005 × 20 = 1304.49 kW

(5.23a)

6(b)

Explain how the free vertex flow theory is used to determine the air angles at different blade heights in an axial flow compressor.

Ans.

Refer to Section 5.13.1 and derivation up to Eq. (5.41b). Show that for a single-stage axial flow impulse turbine the rotor efficiency is given by

7(a)

Krotor Ans.

Ë cos E2 Û 2(I cos D1  I 2 ) Ì1  k b Ü cos E Ý Í

Refer to Eqs. (6.3), (6.4) and Section 6.11, derivation up to Eq. (6.11).

633

Model Question Papers

7(b)

An axial flow impulse turbine has a mean rotor diameter of 55 cm and runs at 3300 rpm. The blade speed ratio is 0.45 and the nozzle angle at the rotor inlet is 20°. The mass flow rate is 10 kg/s. Determine the power output and the axial thrust assuming that the rotor blades are equiangular.

Ans.

Machine = axial flow impulse turbine, u1 = u2 = u, d = 0.55 cm, N = 3300 rpm, r = 0.45, a1 = 20°, ms = 10 kg/s, b1 = b2, Vr1 = Vr2 (assumed) Vw 1

Vw 2

F

a2

Vf 2

V2

u a 1 b2

A

B

E b1 Vr 1

Vr 2

Vf 1

V1

D

C

u

S dN 60

V1

u U

S – 0.55 – 3300 60 95 0.45

95 m/s

211.2 m/s

Vw1 = V1 cos a1 = 211.2 cos 20 = 198.46 m/s BE = Vw1 – u = 198.46 – 95 = 130.46 m/s V12  Vw21

Vf 1 tan E1

Vf 1 BE

Vf 21  BE2

Vr 1

cos E 2

72.24 130.46

211.22  198.462 0.6983

\

72.24 m/s

b1 = 34.93° = b2

72.242  103.462

126.18 m/s

Vr 2

(u  Vw 2 ) / Vr 2

Vw2 = 126.18 cos 34.90° – 95 = 8.29 m/s sin E2

Vf 2 Vr 2

Vf 2 ; sin 34.93’ 126.18

Vf 2 Vr 2

Vf 2 ; Vf 2 126.18

72.24 m/s

(a) Power output (P): P

ums (Vw 1  Vw 2 ) g c – 1000

95 – 10 – (198.46  8.29) 1 – 1000

196.3 kW

Ans.

(b) Axial thrust (Fa): Fa

ms (Vf 1  Vf 2 ) / gc

10 – (72.24  72.24) / 1 0 N

Ans.

634

Model Question Papers

8(a)

Define unit speed, unit power and unit quantity as applied to a hydraulic turbine.

Ans.

Refer to Sections 7.3.2, 7.3.3 and 7.3.4.

8(b)

A vertical shaft inward flow reaction turbine runner develops 12,365 kW and uses 10 m3/s of water when the net head is 116 m. The runner has a diameter of 1.5 m and rotates at 430 rpm. Water enters the runner with a velocity of flow of 10 m/s and comes out of the runner and enters the draft tube with a velocity of 7 m/s. The difference between the sum of the pressure and potential heads at the exit of the runner is 60 m. Determine (i) the velocity V1 and the direction a1 of water at inlet to the runner, (ii) the blade angle at inlet b1, (iii) the loss of head in the runner, and (iv) the hydraulic efficiency.

Ans.

Refer to Example 7.33 (there are only small changes in numerical values).

Model Question Papers

635

PAPER 9 1(a)

Differentiate between a turbomachine and a positive displacement machine, giving examples.

Ans.

Refer to Section 1.6.

1(b)

Briefly explain the significance of specific speed related to fluid machines.

Ans.

Refer to Section 1.18.7.

1(c)

A partially submerged body of length L is towed in water of density r and viscosity m. Show that the total resistance R expressed by the body is given by R

Ë P Lg Û U L2V 2 f Ì – 2Ü U LV V Ý Í

where g is gravitational acceleration. Ans.

Refer to Example 1.1 (almost same).

2(a)

Derive the Euler’s turbine equation with usual notations.

Ans.

Refer to Figure 2.1 and Section 2.1.

2(b)

In a radial inward flow turbine, the runner outer diameter is 75 cm and the inner diameter is 50 cm. The runner speed is 400 rpm. Water enters the runner at a velocity of 15 m/s at an angle of 15° to wheel tangent at inlet. The flow is radial at exit with a velocity of 5 m/s. Find the blade angles at inlet and exit. Also, determine the power output for a flow rate of 1.5 m3/s, the degree of reaction and the utilization factor.

Ans:

Refer to Example 2.24 (Here Q = 1.5 m3/s; in the text, Q = 1.0 m3/s)

3(a)

With the help of inlet and outlet velocity triangles, show that the degree of reaction for an axial flow compressor is given by R = (Va/u) tan gm

Ans.

Refer to Eq. (2.33) where the angles are w.r.t. tangential direction. Follow the same procedure for this problem, taking angles w.r.t. axial direction.

3(b)

A single-stage axial blower with no inlet guide vane is running at 3600 rpm. The mean diameter of the rotor is 16 cm and the mass flow rate of air through the blower is 0.45 kg/s. In the rotor the air is turned such that the absolute velocity of air at exit makes an angle of 20° with respect to the axis. Assuming that the axial component of fluid velocity remains constant, determine the power input and the degree of reaction. Assume that the density of air is constant at 1.185 kg/m3 and the area of flow is 0.2 m2. Machine = Axial blower, N = 3600 rpm, (d1 + d2)/2 = 16 cm, m = 0.45 kg/s, a2 = 90° – 20° = 70°, Va1 = Va2, P = ?, R = ?, r1 = r2 = 1.185 kg/m3, A = 0.02 m2.

636

Model Question Papers

Mean tangential speed of the rotor (u1):

u Va1

S (d1  d 2 ) N 2 – 60 m UA

S – 16 3600 – 100 60

0.45 1.185 – 0.02

18.99 m/s

30.16 m/s V1

Power input (P): (Eq. (2.29)) P = muVa (cot a2 – cot a1)/gc P = 0.45 × 30.16 × 18.99 (cot 70° – cot 90°) = 93.80 W

Ans.

Alternatively: tan E1

Va1 u

18.99 30.16

Va 2 tan D 2

Vw 2 tan E2

Va 2 u  Vw 2

0.6296; ? E1

18.99 tan 70’

32.2’

6.91

18.99 30.16  6.91

P = muVa(cot b1 – cot b2)/gc

0.817; ? E2

39.24’

(Eq. (2.28))

= 0.45 × 30.16 × 18.99 × (cot 32.2° – cot 39.24°) = 93.80 W Ans. Also, P = mu(Vw2 – Vw1) = 0.45 × 30.16 × 6.19 = 93.7 W

Ans.

Model Question Papers

637

Degree of reaction (R): R

Va (tan E2  tan E2 ) 2u (tan E1 – tan E2 )

18.99 (tan 39.24’  tan 32.2’) 2 – 30.16 (tan 39.24’ – tan 32.2’)

0.8854

Alternatively: Use Eqs. (2.34) and (2.36).

4(a)

Define the following efficiencies for a compression process: (i) Isotropic efficiency, (ii) stage efficiency and (iii) polytropic efficiency. Show that the polytropic efficiency is given by Kp

Ans:

n J 1 – n 1 J

Refer to Figure 3.7 and Eqs. (3.33), (3.32), (3.37). Kcs s

isentropic efficiency =

Isentropic work done Actual work done

Wisen Wa

Isentropic work done in one stage Actual work done in the same stage

Ks

stage efficiency =

Kp

polytropic efficiency =

Small increment in temp. of isentropic process Small increment in temp. of actual process

Refer to Eq. (3.39b).

4(b)

Turbine A has inlet conditions of 5.6 bar, 735°C and velocity of 170 m/s. The pressure at discharge is 1 bar, the temperature 182°C and velocity 230 m/s. Turbine B consists of 8 stages, each of total head efficiency of 87% and total head pressure ratio of 1.25. The inlet conditions are same as for turbine A. The discharge velocity is 100 m/s. Which turbine delivers more shaft work?

Ans:

Turbine A: p1 = 5.6 bar, T1 = 1008 K, V1 = 170 m/s, pK+1 = 1 bar, TK+1 = 455 K, VK+1 = 230 m/s. K+1

WT

K+1

638

Model Question Papers

Ë V2 m Ìh1  1 2gc ÌÍ

Ë Û VK21 Û Ü Ü WT  m ÌhK 1  2gc ÜÝ ÜÝ ÌÍ ËV 2  VK21 Û [h1  hK 1]  Ì 1 Ü ; Assume gas as perfect gas, ÍÌ 2gc ÝÜ

WT m

c p [T1  TK 1]  [V12  VK21] / 2gc

(Eq. (3.43))

= 1005(1008 – 455) + (1702 – 2302)/2 × 1 = 543,770 J/kg Turbine B: K = 8, hS = hs(t–t) = 0.87, pS = 1.25, p1 = 5.6 bar, T1 = 1008 K, V1 = 170 m/s, VK+1 = 100 m/s. (Eq. (3.71))

K(t t )

T01

T01 T(0K 1)„

T(0K 1)„ Kt (t t )

\

J 1 Û Ë È Ø È 1 Ø J ÙÜ Ì É 1  Ì1  Ks (t t ) 1  É É ÙÜ Ê pR ÙÚ ÉÊ ÙÚ Ü Ì Í Ý J 1 Ë Û K È 1Ø J Ü Ì 1  Ì Ü ÊÉ pR ÚÙ ÌÍ ÜÝ

T1 

V12 2c p

1008 

Ë p01 Û Ì Ü ÍÌ p(0K 1) ÝÜ

J

T0 K – 0.286 pS 0.894

1 J

>ps @K

J

1 J

1022.4

0.894

1022.4 K

(Eq. (3.70))

613.58 K

1.258 – 0.286

T01  T0K 1 T01  T(0K 1)„

T0K+1 = 656.9 K; WT m

1702 2 – 1005

0.286 Ø Û Ë È È 1 Ø 1  Ì1  0.87 É 1  É ÙÜ Ù Ê 1.25 Ú ÌÍ Ê Ú ÜÝ 8 0.286 – Ë Û È 1 Ø Ì1  É Ü Ù Ê 1.25 Ú ÍÌ ÝÜ

1022.4  T0K 1 1022.4  613.58

(Eq. (3.69))

(Eq. (3.43))

c p (T01  T(0K 1) )

= 1005(1022.4 – 656.9) = 367327.5 J/kg \ Turbine A produces more work output.

Ans.

5(a) Derive an expression for pressure ratio in terms of impeller tip speed for a centrifugal compressor. Ans.

Refer to Eq. (4.9) or (4.10).

639

Model Question Papers

5(b)

A centrifugal pump delivers 50 litres of water per second against a head of 24 m running at 1500 rpm. The velocity of flow of 2.4 m/s is constant and the blades are set back at 30°. The inner diameter is half the outer diameter. If manometric efficiency is 80%, determine the blade angles and the power required to drive the pump.

Ans.

d2 = 2d1, N = 1500 rpm, Hm = 24 m, Q = 50 lit/s, Vf1 = Vf2 = 2.4 m/s, b2 = 30°, hman = 80%. gH m Vw 2u 2

Kman

(i)

Vf 2 ; Vw 2 u 2  Vw 2

tan E2

u2 

Vf 2 tan E2

(ii)

Substituting Eq. (ii) in (i), 9.81 – 24 Ë 2.4 Û Ìu 2  tan 30’ Ü – u 2 Í Ý

0.8

235.44 ; (u 2  4.16)u 2

? u2

19.38 m/s

S d 2N S – d 2 – 1500 ; ? d 2 0.247 m 60 60 d1 = d2/2 = 0.247/2 = 0.1234 m; u1 = u2/2 = 19.38/2 = 9.69 m/s u2

\

19.38

Blade angle or vane angle at inlet (b1): tan E1

Vf 1 u1

2.4 9.69

0.247; E1

13.9’

Ans.

Power required to drive the pump (P): tan E2

\

Vf 2 u 2  Vw 2

2.4 ; 19.38  Vw 2

? Vw 2

15.22 m/s

P = (rQVw2u2)/gc = 1000 × 50 × 10–3 × 15.22 × 19.38 = 14751.2 W

Ans.

640

Model Question Papers

6(a)

Briefly explain the following with respect to an axial flow compressor: (i) work done factor, (ii) free vortex flow theory.

Ans.

Work done factor—Refer to Section 5.7. Free vortex flow theory—Refer to Sections 5.12.

6(b)

The following data refer to the first stage of an axial flow compressor: m = 20 kg/s, DTs = 20 K, um = 180 m/s, j = 0.96, Rm = 0.5, Vam = 150 m/s, Nm = 150 rps, T01 = 288 K, p01 = 1 bar. Determine (a) air angles at the mean radius and (b) blade height.

Ans.

(a) Air angles (blade) at mean radius (b1, b1): DTs = jumVam (cot b1 – cot b2)/gccp (Eq. (5.25c)) 20 = 0.96 × 180 × 150 × (cot b1 – cot b2)/1005; \ (cot b1 – cot b2) = 0.7755 (i) R = Vam (cot b1 + cot b2)/2um (Eq. (5.27)) 0.5 = 150 × (cot b1 + cot b2)/2 × 180; \ (cot b1 + cot b2) = 1.2 (ii) Solving Eqs. (i) and (ii), we get b1 = 45.353°, b2 = 78.02° Since R = 0.5, \ a1 = b2 = 78.02°, a2 = b1 = 45.35° (b) Blade height (h):

A

S (do2  d h2 ) / 4 S (d 0  d h ) . (d o  d h ) / 4 S 2 2h (d o  d h ) 4 2

S hd m

2S rm h

dh

do

Model Question Papers

641

um = pdmNm = 2prmNm \

rm = 180/(p × 2 × Nm) = 180/(p × 2 × 150) = 0.191 m Vw1m = Vam/sin a1 = 150/sin 78.02° = 31.82 m/s 2 Vam  Vw21m

V1m

T01  V12m / 2c p

T1

1502  31.822

153.34 m/s

288  153.34 /(2 – 1005)

J

3.5

p1 p01

Ë T1 Û J  1 Ì Ü ÍT01 Ý

\

U1

p1 RT1

\

m = r1A . Vam = r1(2prmh)Vam

or

20 = 1.091 × 2 × p × 0.191 × h × 150;

È 276.3 Ø ÉÊ Ù 288 Ú

276.3 K

0.865 – 105 287 – 276.3

; ? p1

0.865 bar

1.091kg/m3

\

h = 0.10186 m Ans.

7(a)

What is compounding in steam turbine? Explain with a neat sketch any one of method of compounding.

Ans.

The required blade tip speed (reducing the blade speed) can be obtained in an impulse turbine by the method of compounding. This can be achieved (reduced blade speed) by absorbing inlet energy in multiple rows of moving blades and is known as compounding. If the inlet energy is K.E., then it is velocity compounding; if the inlet energy is in pressure form (enthalpy form), then it is called pressure compounding. For any one method of compounding, refer to Section ..... .

7(b)

In a curtis stage, the rotors are both equiangular. The first rotor has angles of 29° each, while the second rotor has angles of 32° each. The velocity of steam at the exit of the nozzle is 530 m/s and the blade velocity coefficients are 0.9 in the first rotor, 0.95 in the stator and in the second rotor. If the absolute velocity at the stage exit should be axial, find the mean blade speed, the rotor efficiency and the power output for a flow rate of 3.2 kg/s.

Ans.

(i) Refer to Example 6.11 (similar type). The changes are: K2 = K3 = 0.95. (ii) The rotor or blade efficiency (hb): Kb

2u [(Vw 1  Vw 2 )  [Vw 3  Vw 4 ]] V12

0.79

8(a)

Derive an expression for maximum energy transfer for a Pelton turbine.

Ans.

Refer to Section 7.6.1 up to Eq. (7.22). P

UaV1 (Vw 1 “ Vw 2 ) u

(Eq. (7.17))

642

Model Question Papers

UaV1 (Vw 1 “ Vw 2 )V1 2

È ÉÊ' u

8(b)

Ans.

V1 Ø is the condition for maximum energy transfer Ù Ú 2

UaV12 (Vw 1 “ Vw 2 ) 2 The following data is given for a Francis turbine: Net head = 70 m, speed = 600 rpm, shaft power = 368 kW, ho = 85%, hh = 95%, flow ratio = 0.25, breadth ratio = 0.1, outer diameter of the runner = 2 × inner diameter of runner, velocity of flow is constant at inlet and outlet, the thickness of vane occupies 10% of the circumferential area of the runner and discharge is radial at outlet. Determine (a) the guide blade angle, (b) the runner vane angles at inlet and exit, (c) the diameters of runner at inlet and outlet, and (d) the width of the wheel at inlet. Refer to Example 7.28.

INDEX

Acoustic velocity, 158 Adiabatic efficiency, 166, 176 Axial flow compressor, 348 blade passages, 348 degree of reaction, 356 flow coefficient, 356 overall pressure ratio, 354 pressure coefficient, 356 velocity triangles, 350 work done, 352 work done factor, 355 Axial flow pump, 271 Backward curved vanes, 71, 72 Blade discharge angle, 69 Blading efficiency, 428 condition for maximum blade efficiency, 430 Blower(s), 1, 10, 60 Buckingham’s p-theorem, 12 Capacity coefficient, 21 Compounding, 421 pressure compounding, 423 pressure and velocity compounding, 424 velocity compounding, 422 advantage and disadvantages, 435

Compressible flow, 161 Compressor(s), 1, 10, 60 analysis of, 69, 76 centrifugal, 247 eye conditions, 257 impeller shape, 258 limiting inlet velocity, 255 main components, 247, 248 overall pressure ratio, 253 pressure coefficient, 255 stagnation pressure ratio, 254 surging phenomenon, 265 velocity diagram, 250 classification of, 71

Degree of reaction, 64, 75, 76, 85, 93 for different vanes, 76 Diffuser, 262 vaneless, 262, 263 Dimensional analysis, 11 application to a fluid flow problem, 14 application to turbomachines, 19 Dimensional homogenity, 11, 13 Draft tube, 506 design of, 507 efficiency of, 508 643

644

Index

functions of, 508 types of, 506 Dynamic temperature, 163

Mach number, 18, 161 Mixed flow pump, 271 Multistage pump, 271, 279

Energy transfer, 62 effect of blade discharge angle and degree of reaction, 65, 68 Enthalpy, 8 Entropy, 8 Euler equation, 60 Euler work, 63 Expansion efficiency, 176

Nozzle efficiency, 428

Fan(s), 1, 10, 60 Flow laminar, 23 turbulent, 23 Fluid machines, 2 classification of, 2 comparison with displacement machine, 5 Forward curved vanes, 71, 73 Francis turbine, 501 components of, 502 velocity triangles, 503 Free vertex, 362 Froude number, 19

Overall isentropic efficiency, 168, 171, 179 Parson’s reaction turbine, 437 Pelton wheel, 490 components of, 492 Polytropic efficiency, 171, 182 Power coefficient, 22 Preheat factor, 176 Pressure ratio, constant stage, 174, 185 Prewhirl angle, 262 Prewhirl vanes, 261, 262 Pump(s), 60 centrifugal analysis of, 69, 76 classification of, 71, 266, 268 cavitation, 279 efficiencies of, 274 main parts, 267 minimum starting speed, 278 velocity triangles, 276

Gas, 1 Head coefficient, 22 Hydraulic turbine, 487 efficiencies of, 499 types of, 488 Impulse machine, 65, 68 Impulse momentum equation, 58 Isentropic efficiency, 166, 176 static to static, 167 static to total, 167 total to static, 167 total to total, 167 Kaplan turbine, 509 Liquid, 1

Radial curved vanes, 71, 72 Radial equilibrium, 360 Radial flow pump, 270 Reaction machine, 65, 68 Reheat factor, 442, 443 Reynold’s number, 18, 23 Single stage pump, 271 Solid, 1 Specific power, 22, 23 Specific speed, 24 of pumps, 25 of turbines, 25 Stage efficiency, 168, 179, 428 Stagnation conditions, 8 Stagnation density, 164 Stagnation enthalpy, 162 Stagnation pressure, 163 Stagnation properties, 162 Stagnation state, 162

Index

Stagnation temperature, 163 Stagnation velocity, 164 Static properties, 162 Static temperature, 163 Steady flow equation, 63 Steam turbine, 418 impulse turbines, 418, 419 reaction turbines, 418, 420 impulse reaction, 420, 421 pure reaction, 420 Turbine(s), 1, 3, 10, 60 axial flow, 85 condition for maximum utilization, 93 impulse, 86 multistage, 185, 431 pressure compounded, 423 pressure and velocity compounded, 424 radial flow, 92 reaction, 86, 93, 501 velocity compounded, 422 velocity triangles, 425, 435

645

Turbomachine(s), 1, 2, 3 parts of, 3 types of, 9

Unit quantities, 489 unit discharge, 490 unit power, 490 unit speed, 489 Utilization factor, 83, 85, 92, 93

Velocity triangles for axial flow compressors and pumps, 77 for centrifugal pumps and compressors, 70, 71 for different values of degrees of reaction, 81 for various values of blade discharge angle, 65, 66 for various values of R (axial flow type turbines), 90, 91

Weber number, 18