Power Plant Engineering

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POWER PLANT ENGINEERING

IES MASTER PUBLICATION

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Copyright © 2017, by IES MASTER Publications. No part of this booklet may be reproduced, or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior permission of IES MASTER, New Delhi. Violates are liable to be legally prosecuted.

First Edition : 2017

Typeset at : IES Master Publication, New Delhi-110016 CONTENTS

1. Reciprocating Compressor...... 1 – 22

2. Dynamic Compressor ...... 23 – 69

3. Gas Power Cycles ...... 70 – 104

4. Air Craft Propulsion ...... 105 – 127

5. Steam Power Plant ...... 128 – 220

6. Combustion ...... 221 – 241

7. Steam Generators ...... 242 – 272

8. Nuclear Power Plant ...... 273 – 290 1

INTRODUCTION

Reciprocating Basically mechanical compressors are of two types: Compressors (i) Positive Displacement Compressors: It ensures positive admission and reversal of flow is not possible. This one directional Single Stage with flow is achieved by the use of valves as the case of reciprocating Clearance Volume compressor. The flow is intermittent and work is transferred by the Clearance ratio and virtue of static force e.g. reciprocating compressor, rotatory, scroll, Clearance Volume screw compressor. (ii) Non-positive Displacement Compressor: Here, there is no Multistaging of means to prevent the reversal of flow, i.e. no valve etc. The fluid is Reciprocating subjected to flow process or work is transferred by virtue of Compressor momentum transfer process by means of blades attached to rotor. e.g. axial and centrifugal compressor, fans, blower etc. Heat Rejected in RECIPROCATING COMPRESSORS 2-Stage Compressor The p–V diagram and basic sketch of reciprocating compressor is shown in figure: • When the piston is at IDC, the volume occupied by gas is called clearance volume P V = V C 3 3 2 • Process 12 : In this process of compression, the work is done on system i.e. work is consumed in compression. • Process 23 : In this process the 4 V 1 compressed air is forced through P V VS S delivery valve to high pressure pipeline C or space. So it is isobaric delivery of air. 2 Chapter-1 : Reciprocating Compressor

• Process 34 : In this process the compressed air in

clearance volume (Vc) expands and the air/system does work Delivery valve Piston Connecting i.e. work is obtained. rod D • Process 41 : It is suction process, suction valve is open D Crank during this process and delivery valve remains closed. S Cylinder L Point ‘1’ is outer most position called as outer dead center IDC ODC (ODC). So Suction valve The stroke or swept volume or piston displacement.

 2 VVVDLP 1  3  D—Cylinder diameter, L—Stroke 4 EXPRESSION FOR SHAFT WORK

A — Single Stage No Clearance Volume: Various possible processes of compresion are P • Adiabatic compression PV  = constant P 2 (12. This is the reversible adiabatic 2 2 2 n  3 Adiabatic comp (PV = Constant) compression and the condition of reversibility is not met in reality. So the Polytropic(PV n = Constant) process is ruled out.

• Isothermal compression PV = constant Isothermal comp (PV = Constant) (12. Very slow and reversible. So again it is ruled out in actual design of P1 compressor. 4 1 • So in actual design, the process will be in a b c between these two processes i.e. a polytropic compression PVn = constant D Here in compressor, there are three processes and assuming unit flow of S compressible fluid through compressor. 1. Polytropic compression 12 and area 12bc1 denotes work consumed is, p V p V  = 1 1 2 2 n 1 2. Isobaric delivery process 2–3, area 23ab2 denotes work done in delivery, so

= – p2 V2 (done on system) 3. Isobaric suction process 41, area 41ca4 denotes suction work, so

= + p1 V1 (done by system) So net work required to run the compressor on compression cycle p V -p V = 1 1 2 2 – p V +p V n-1  2 2 1 1 n = pV-pV  ...area 1 2  3  4  1 n-1 1 1 2 2

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NOTE So the net work done is also given by

p =  2 V.dp, where pVn  const p1

p n  2 So, =  Vdp =  pV–pV1 1 2 2  p1 n 1 For centrifugal compressor (steady flow)

h1 + q = h2 + w q = 0

–w = (h2 – h1) Tds dh  Vdp, ds  0    h1 h 2  Vdp Hence the expression for reciprocating and centrifugal compressor is same mathematically. So shaft work for reciprocating compressor n w = p V -p V  s n–1 1 1 2 2

–n p2 V 2  n T2  = p1 V 1  –1  = – p1 V 1  –1  n–1 p1 V 1  n–1 T1  n–1  n p  n   p V2 –1 ws = 1 1    n–1 p1   This is work done per cycle. Negative sign indicate work done on system. For mass flow rate m kg/sec. The power input [PV = mRT]

 n 1  n p  n Power = mRT  2 1 1    n 1 p1   COMPARISON OF COMPRESSION WORK IN COMPRESSION PROCESSES

2 T f Delivery pressure 2s P

 2 P 2 2 2 2 c = 2 t c s f Polytropic with friction (n > 

n n n >  P2 <

 Isentropic (n =  ) r n = 1 P 2 1 t 1 Polytropic without friction (n <  )

Isothermal (n = 1) Work done is less P1 1 for isothermal process Suction pressure

S V

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Process 12t (Isothermal)—Work done on air = Heat transfer from air to surroundings/coolant change in internal energy is zero.

Process 12c Part of work transferred from system to surrounding/coolant as heat and rest remain as increase in internal energy.

Process 12s Adiabatic, reversible no heat transfer.

Process 12f No heat transfer but friction heat stored as internal energy. (i) The various shaded areas (in figure on right side) denotes the difference of work in various processes mentioned above.

(ii) So the most efficient process is Isothermal (12t) and highest work consuming is polytropic with friction (12f)

So the order of magnitude of work involve. w2f > w2s > w2c > w2t. Based on all these process, the various types of compressor efficiencies are defined as, (i) Isothermal efficiency:

Work done in compression with isothermal process(1–2t ) nisoth = Actual work done with polytropic process (1–2c ) (ii) Adiabatic efficiency:

Work done in compression with Adia. rev. comp. (1-2S ) nadiab = n Work done on comp. with Adia Irrev. comp 1 2f 

NOTE Adiabatic efficiency is common in use. It is clear from above discussion that the low value of index ‘n’ results in low work consumption. This is achieved by cooling the cylinder by blowing air or water jacket around the cylinder. This does not mean that always low value of ‘n’ is desirable. This is decided by end use of compressed air as discussed below: (i) Air in Gas Turbine: Here the high temperature is requirement in order to reduce heat added in combustion chamber. (Inter cooling is provided to reduce extra high temperature at extra high pressure to reduce compressor work). (ii) When air is kept stored, then isothermal compression is tried to get otherwise stored heat is lost to the surrounding. (iii) pvn = C, so, lnp + nlnv = lnc or, dp/dv = –np/v for g > n > 1, for some pressure

ratio p2/p1, isothermal needs minimum work, whereas adiabatic compression needs the maximum SINGLE STAGE WITH CLEARANCE VOLUME

We know that work done on gas with no clearance

 n 1  n p  n p V  2 1 w1 = 1 1    n 1 p1   Regd. office : F-126, (Upper Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-41013406 Mob. : 8010009955, 9711853908 E-mail: [email protected], [email protected] Mechanical Engineering (Study Package) 5

Since during expansion in clearance volume Delivery pressure (process 3-4) work is done by gas. So work obtained 3 2 n  n 1  pv = c n p3  n  p V 1 n w2 = 4 4    pv = c n 1 p4   Net work required to complete cycle— Suction w = w1 – w2 pressure n 1 n 1     4 n n n p2   n p3   1 = p1 V 1   1  p4 V 4   1 n 1 p  n 1 p  1   4   VS since P = P (delivery pressure) 2 3 VC VP P1 = P4 (Suction pressure)  n 1  n p  n p V V  2 1 So w = 1 1 4     n 1 p1  

Since (V1 – V4) is suction/inducted/free air delivery (FAD) volume-drawn in cylinder per cycle and if m is mass sucked during this cycle, then

 n 1  n p  n w mRT  2 1 1    n 1 p1   So the mass sucked is independnet of clearance volume.

NOTE In work expression we have assumed that the index of compression (12) and index

of expansion (34) are same. But it may be different as nc & ne. Most of the time they are assumed same. CLEARANCE RATIO AND CLEARANCE VOLUME

• Stroke volume/displacement volume, Vp — Volume swept by piston in one stroke. P • Clearance volume, VC  ln order to avoid strike of piston at cylinder end/head and valves a space is provided, 3 2

called clearance volume ‘VC’. • This should be as small as possible. • This is also used to accomodated the exapansion of piston as temperature of piston increases. 4 1 • Clearance ratio: It is the ratio of clearance volume to

the displacement volume. So clearance ratio, VC Stroke volume, V VV P C 3  c Clearance Suction VVV1 3 P volume volume This is also called percentage clearance. • This is expressed as percentage of cylinder volume. • It is around 3% in large compressor and around 12% in small. So it varies between 3-12%.

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Volumetric Efficiency It is the ratio of actual mass of air sucked by compressor to the mass of air which the compressor could suck if it had handled a volume of air equal to displacement volume. So

Mass of air delivered per cycle m   = = 1 vol Mass of air which could fill swept volume under intake condition per cycle PD

Here, m = mass flow rate of the gas, 1 = specific volume at inlet to the compressor Since intake conditions are atmospheric conditions and usually referred as free air. So at free air condition.

Inducted volume  VV  VV 1 4  1 4 vol = ...(1) swept volume V1 V 3  V P

Now we have to find out value of V1 and V4 interms of C and VP (or V1 – V3). Process 34

1 1 p n p n VV3  2 V4 = 3   C   p4   p1 

 V1 = Vc + VP Since clearance ratio,

V1 = Vc + Vc/C

1 n p2   V1 – V4 = Vc + Vc/C – Vc   p1 

VV1 4  v = Vp

 1  n Vc  1 p2   Vc = 1      C  Vp  C p1   Vp

1 p n = 1 C  C 2  p1 

 1  n vol 1  C 1  (P2 / P 1 ) 

Other form of this (vol) expression V   =   1 [ n n n n ] vol 1 C  C   PVPV&P.VPV1 1 2 2 1 4 2 3 V2 

V4  So vol = 1 C   C  V3  Regd. office : F-126, (Upper Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-41013406 Mob. : 8010009955, 9711853908 E-mail: [email protected], [email protected]