chapter 10 vapor and combined power cycles. 10-1 the carnot vapor cycle

CHAPTER

10

Vapor and Combined Power Cycles

10-1 The Carnot Vapor Cycle

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

FIGURE10-1T-s diagram of two Carnot vapor cycles.

TH

TC

WNET

QH

QC

H

C

H

CH

T

T

T

TQW

1

1

max

max

(1)

(2)

Operating principles

10-2 Rankine Cycle: The Ideal Cycle for Vapor Power Cycles

• Operating principles• Vapor power plants• The ideal Rankine vapor power cycle• Efficiency

– Improved efficiency - superheat

QQININ

QQOUTOUT

WWTURBINETURBINE

WWPUMPPUMP

The conventional vapor power plant

QQININ

QQOUTOUT

WWTURBINETURBINE

WWPUMPPUMP

High temperature heat addition.

High temperature heat addition.

Low temperature heat rejection

Low temperature heat rejection

Work input to compress working fluid

Work input to compress working fluid

Turbine to obtain work by expansion of working fluid.

Turbine to obtain work by expansion of working fluid.

The conventional vapor power plant


FIGURE 10-2The simple ideal Rankine cycle.

A hypothetical vapor power cycle

Assume a Carnot cycle operating between Assume a Carnot cycle operating between two fixed temperatures as shown.two fixed temperatures as shown.

TT

ss

11

22 3

4

All processes are internally reversible.All processes are internally reversible.All processes are internally reversible.All processes are internally reversible.

ss

TT

11

22

33

44

3*3*

4*4*

The ideal Rankine cycle

All processes are internally reversible.All processes are internally reversible.All processes are internally reversible.All processes are internally reversible.

ss

TT

11

22

33

44

3*3*

4*4*

Reversible constant pressure heat rejection (4 1)

Reversible constant pressure heat rejection (4 1)

Reversible constant pressure heat addition (2 3)

Reversible constant pressure heat addition (2 3)

Isentropic compression (1 2)

Isentropic compression (1 2)

Isentropic expansion to produce work (3 4) or (3* 4*)

Isentropic expansion to produce work (3 4) or (3* 4*)

The ideal Rankine cycle

The ideal Rankine cycle(h-s diagram)

hh

ss

44

33

22WOUT

QH

QC11

WIN

s

43

12

43

hhQ

hhW

hhW

H

IN

OUT

H

NET

Q

W (3)

hh

ss

44

33

22WOUT

QH

QC11

WIN

Rankine cycle efficiency

Isentropic process, s = constant

All accessible states lieto the right of the process (1 2).

p11

ss

hh

p2

2

ss11 = s = s22

CVe

e

ei

i

i

j j

jCV smsmT

Q

dt

dS

N

i

iiCSCS pekehmWQdt

tdE

1

)(

Ideal Turbine WorkIdeal Turbine Work

(4)

• Steady state.• Constant mass flow• Isentropic Expansion (s = Constant)

– Adiabatic and reversible– No entropy production

• No changes in KE and PE– Usual assumption is to neglect KE and PE

effects at inlet and outlet of turbine.

Ideal turbine work

N

i

iiCSCS pekehmWQdt

tdE

1

)(

43 hhm

WOut

(4)

(5)

Ideal turbine work

12 hhm

WIN

121 PPvm

WIN

(6)

(7)

Work of Compression

hh

ss

44

3*3*

22WOUT

QH

QC11

WIN

Increased average temperature of heat addition

Increased average temperature of heat addition

Improved Rankine cycle efficiency

2*3

124*3

hh

hhhh

0*3

h

(8)

(9)

Improved Rankine cycle efficiency

10-3 Deviation of Actual Vapor Power Cycles from Idealized Ones

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.FIGURE 10-4(a) Deviation of actual vapor power cycle from the ideal Rankine cycle. (b) The effect of pump and turbine irreversibilities on the ideal Rankine cycle.

The Improved Rankine Cycle

10-4 How Can We Increase the Efficiency of the Rankine Cycle

Raising the average temperature of heat addition


FIGURE 10-9A supercritical Rankine cycle.


FIGURE 10-10T-s diagrams of the three cycles discussed in Example 9–3.

10-5 The Ideal Reheat Rankine Cycle

Assume a Carnot cycle operating between Assume a Carnot cycle operating between two fixed temperatures as shown.two fixed temperatures as shown.

TT

ss

11

22 3

4

A hypothetical vapor power cycle

A hypothetical vapor power cycle with superheat

Superheating the working fluid raises the Superheating the working fluid raises the average temperature of heat addition.average temperature of heat addition.

TT

ss

11

22

3

4

TTH,2H,2

TTH,1H,1

A hypothetical vapor power cycle: A Rankine cycle with superheat

s

T

d

b

c

T

HT

HT

Superheating the working fluid raises the average Superheating the working fluid raises the average temperature with a reservoir at a higher temperature.temperature with a reservoir at a higher temperature.

a

The extra expansion via reheating to state “d” allows a greater enthalpy to be released between states “c” to “e”.

ss

TT

f

a

b

c

p1

p2

d

e

HT

CT

The Rankine cycle with reheat

QOUT

WWININ

aa

bb

eeff

QQHH

WWOUTOUT

QQCC

ccdd

Single stage reheat. Workproduced in both turbines.Single stage reheat. Workproduced in both turbines.

The reheat cycle

Reheat Cycle Efficiency

cdab

faedcb

dcba

afedcb

hhhh

hhhhhh

QQ

WWW

)()(

(1)


FIGURE 10-11The ideal reheat Rankine cycle.

Example 1

Example - 1: A Carnot cycle

To begin our analysis of Rankine cycle operations, consider a To begin our analysis of Rankine cycle operations, consider a steady Carnot cycle (a-b-c-d-a) with water as the working fluidsteady Carnot cycle (a-b-c-d-a) with water as the working fluidoperating between to given temperature limits as shown. The operating between to given temperature limits as shown. The given data are that the boiler pressure is 500 psi (pgiven data are that the boiler pressure is 500 psi (paa = 500 psi) and = 500 psi) and

the condenser temperature is 70the condenser temperature is 70o o F (TF (Tcc = 70 = 70oo F). Determine the F). Determine the

work outputwork output, , thermal efficiencythermal efficiency,, irreversibility irreversibility, and , and work ratiowork ratio..

b

cd

a TT11

TT22

T

s

aa bb

ccdd

T1

T2

p1

p2

Example 1 - Given data

a 500

b 500

c 530

d 530

State TR

Ppsi

hBTU/lbm

sBTU/lbm-R

x

Example 1 - Computed data

a 927 500 449 0.6487 0

b 927 500 1204 1.4634 1

c 530 0.361 774 1.4634 0.698

d 530 0.361 342 0.6487 0.288

State TR

Ppsi

hBTU/lbm

sBTU/lbm-R

x

Example 1 - Process quantities

Process HBTU/lbm

WBTU/lbm

QBTU/lbm

SBTU/lbm-R

Q/TBTU/lbm-

R

s

BTU/lbm-R

a-b 755 0 755 0.8147 0.8147 0b-c -430 430 0 0 0 0c-d -432 0 -432 -0.8147 -0.8147 0d-a 107 -107 0 0 0 0Net 0 323 323 0 0 0

To get the above process quantities, the First Law for open systemsTo get the above process quantities, the First Law for open systemshas been used assuming KE and PE effects are negligible. Thehas been used assuming KE and PE effects are negligible. Theentropy production was obtained from an entropy balance for anentropy production was obtained from an entropy balance for anopen system. Note that the cyclic heat equals the cyclic work as open system. Note that the cyclic heat equals the cyclic work as required by the First Law and that entropy production is zero as required by the First Law and that entropy production is zero as required by the Clausius Equality.required by the Clausius Equality.

Thermal efficiencyThermal efficiency

Work ratioWork ratio

428.0755

323

HQ

W

249.0430

107

Turbine

in

W

Wr

Example 1 - Thermal efficiency and back work ratio

Example 2

Example 2 - A Rankine cycle without superheat

A Rankine cycle with water as the working fluid operates betweenA Rankine cycle with water as the working fluid operates betweenthe same limits as in Example l, pthe same limits as in Example l, paa = 500 psi and a condensing = 500 psi and a condensing

temperature of 70temperature of 70oo F. Assuming all processes to be internally F. Assuming all processes to be internally reversible, determine the work output, efficiency, entropy reversible, determine the work output, efficiency, entropy production, work ratio.production, work ratio.

TT

ss

aabb

ccdd

T1

T2

pa

pd


a 927 500 39.5 0.0745 0

b 927 500 1204 1.4634 1

c 530 0.361 774 1.4634 0.698

d 530 0.361 38.0 0.0745 0

State TR

Ppsi

hBTU/lbm

sBTU/lbm-R

x


Process HBTU/lbm

WBTU/lbm

QBTU/lbm

SBTU/lbm-R

Q/TBTU/lbm-

R

s

BTU/lbm-R

a-b 1164.5 0 1164.5 1.3889 1.3889 0b-c -430 430 0 0 0 0c-d -736 0 -736 -1.3889 -1.3889 0d-a 1.5 -1.5 0 0 0 0Net 0 428.5 428.5 0 0 0

Work output =Work output = 428.5 BTU/lbm428.5 BTU/lbmThermal efficiency =Thermal efficiency = 36.8%36.8%Work ratio = Work ratio = 0.00350.0035Entropy production = Entropy production = 00

Example 3

Example 3 - Effect of irreversibility in the turbine and pump

A Rankine cycle operates between the same limits as above and A Rankine cycle operates between the same limits as above and has pump and turbine efficiencies of 80%. Determine the work has pump and turbine efficiencies of 80%. Determine the work output, efficiency, work ratio, and entropy production. Assume output, efficiency, work ratio, and entropy production. Assume the condensing and ambient temperatures to be the same.the condensing and ambient temperatures to be the same.

T1

T2

ss

aa bb

ccdd

pa

pd

c’c’

a’a’

TT


The states at a’ and c’ are determined via the First Law for an The states at a’ and c’ are determined via the First Law for an open system and the definition of isentropic efficiency for turbinesopen system and the definition of isentropic efficiency for turbinesand pumps as appropriate.and pumps as appropriate.

State TR

Ppsi

hBTU/lbm

sBTU/lbm-R

x

a 530 500 39.5 0.0745 …a’ 530 500 39.9 0.0753 …b 927 500 1204 1.4364 1c 530 0.3631 774 1.4364 0.698c’ 530 0.3631 860 1.6262 0.780d 530 0.3631 38.0 0.0745 0


Process hBTU/ lbm

WBTU/ lbm

QBTU/ lbm

sBTU/lbm-R

Q/ TBTU/lbm-R

s

BTU/lbm-R

(a’-a) (0.4) (0) (0.4) (0.0008) (0.0008) (0)a-b 1164 0 1164 1.3881 1.3881 0(b-c’) (-430) (0) (0) (0) (0) (0)b-c -344 344 0 0.1628 0 0.1628c-d -822 0 -822 -1.5517 -1.5517 0(d-a’) (1.5) (-1.5) (0) (0) (0) (0)d-a 1.9 -1.9 0 0.008 0 0.0008Net 0 342 342 0 -0..1636 +0.1636

Isentropic processes are determined prior to actual processes whereIsentropic processes are determined prior to actual processes where

irreversibility is involved.irreversibility is involved.

Work output = 342 BTU/lbm Thermal efficiency = 29.4%Work ratio = 0.0054 Entropy production = 0.1636

BTU/lbm-R

Example 3 - Comparison

Example 1Carnot

Example 2Basic Rankine

Example 3Pump and Turbine

IrreversibilityWork 323 428.5 342

Work Ratio 0.249 0.0035 0.0054Thermal Efficiency 42.8% 36.8% 29.4%Entropy Production 0 0 0.1636

Example 4

Example 4 - Superheat

An internally reversible Rankine cycle is determined by specifyingAn internally reversible Rankine cycle is determined by specifyinga maximum temperature of 800a maximum temperature of 800o o F, a quality at the turbineF, a quality at the turbinedischarge of 0.9, and a minimum condensing temperature of 70discharge of 0.9, and a minimum condensing temperature of 70ooF.F.Compare the Compare the thermal efficiencythermal efficiency with that of a with that of a Carnot cycleCarnot cycleoperating between the same temperature limits.operating between the same temperature limits.

ss

bb

aa

ccddpd

paTT

Example 4 - Given and computed data

Process HBTU/lbm

WBTU/lbm

QBTU/lbm

a-b 1391 0 1391b-c -441.5 441.5 0c-d -950 0 -950d-a 0.5 -0.5 0Net 0 441

Example 4 - Thermal efficiency

Thermal efficiencyThermal efficiency

The Carnot efficiency

316.01391

441

579.01230

53011

H

CC

T

T

Example 5

Example 5 - The reheat cycle

f

a

b

c

pa pc

ss

d

e

TT Fo800

CT

Fo70

90.0cx

90.0ex

Example 5 - Given and computed data

The state at “c” has the same as the pressure specified in Example 4.The state at “c” has the same as the pressure specified in Example 4.This determines state “b”. State “a” is determined via the usualThis determines state “b”. State “a” is determined via the usualapproximation for an incompressible liquid under going processapproximation for an incompressible liquid under going processf-a.f-a.


Process HBTU/lbm

WBTU/lbm

QBTU/lbm

a-b 1315 0 1315b-c -264 264 0c-d 336 0 336d-e -442 442 0e-f -950 0 -950f-a 5 -5 0Net 0 701 701

Work output = 701 BTU/lbmWork output = 701 BTU/lbmThermal efficiency = 701/(1315+336) = 0.424Thermal efficiency = 701/(1315+336) = 0.424Carnot efficiency = 1-(530/1260) = 0.579Carnot efficiency = 1-(530/1260) = 0.579

Example 6

0.2

0.22

0.24

0.26

0.28

0.3

300 350 400 450 500 550 600 650 700

Pupper (psia)

Variation of the (First Law) cycle efficiency with a variationof the pressure of heat addition in a basic Rankine cycle with no super heat. The condenser pressure was assumed to be 14 psia.

5623

124653

6532

214653

)()(

hhhh

hhhhhh

QQ

WWW

TT

ss

1

2

3

5

6

4

p2

pmiddle

0.2

0.22

0.24

0.26

0.28

0.3

0 100 200 300 400 500 600

Pmiddle (psia)

Pup = 400 psia

Pup = 500 psia

Pup = 600 psia

The variation of cycle efficiency of a Rankine cycle with onestage of reheat as a function of the pressure at which reheat isdone. Pup is the pressure of high temperature heat addition.

Key terms and concepts

Cycle efficiencyRankine cycle with reheat

Rankine cycle with regenerationWork ratio

10-6 The Ideal Regenerative Rankine cycle

Another technique to raise the average temperature of the heat addition process.

Overview

• Review - The reheat cylce

• The Rankine Cycle with regeneration– Open and closed feedwater heaters

• Example

The Reheat Cycle

• Reheating the expanding fluid with primary heat source is made at inter-mediate points in the expansion process.

• Net effect is to raise the average expansion temperature of the turbine without raising the temperature of the heat source.

The Reheat Cycle

The extra expansionThe extra expansionallows a greater enthapyallows a greater enthapyto be released betweento be released betweenstates 3 to 4.states 3 to 4.

Here one additionalHere one additionalreheat process hasreheat process hasbeen added.been added.

The extra expansionThe extra expansionallows a greater enthapyallows a greater enthapyto be released betweento be released betweenstates 3 to 4.states 3 to 4.

Here one additionalHere one additionalreheat process hasreheat process hasbeen added.been added.

TT

s

1

2

3

5

6

4

p2

p1

5623

124653

6532

214653

)()(

hhhh

hhhhhh

QQ

WWW

TT

s

1

2

3

5

6

4

p2

p1

The Rankine cycle with regeneration


FIGURE 10-14The first part of the heat-addition processin the boiler takesplace at relatively low temperatures.

Principle of the regenerative cycle

c

a

b

f

WWOUTOUT

WIN

QQHH

QQCCA higher feed water inletA higher feed water inlettemperature as a result oftemperature as a result ofheating from States a - b.heating from States a - b.

de

• Heating of some of the compressed liquid is done to raise the average temperature of heat addition.

• Heat is supplied after the liquid is compressed to a high pressure at State a.

Impacted processes

The T-s diagram for the regenerative cycle

TT

f

b

c

ssInternal heat transfer to feed water heater.Internal heat transfer to feed water heater.

e

d

a

• Turbines cannot be designed economically with internal heat exchangers.

• Condensation could occur in the turbine.

Not practical!

Practical considerations...

The practical regenerative cycle

5

QQ2-32-3 = 0 = 0

123

WW2,out2,out

WWIN,1IN,1

QQHH 6

WWIN,2IN,2

7

QQCC

4

WW1,out1,out

Feedwater heaterFeedwater heaterFeedwater heaterFeedwater heater

45

122,341,

762,651,

1

)1(

hhQ

hhyWandhhW

hhyWandhhW

H

ININ

OUTOUT

hh

ss

1

3

4

2

WOUT,1

QQHH

QQCC

WIN,1

5

6

7

WOUT,2WIN,2

(1 kg)

(y kg)

(1-y kg)

The h-s diagram for the regenerative cycle

Energy balances and thermal efficiency

45

12347665 )1()1(

hh

hhyhhhhyhh

hh

ss

1

3

4

2

WOUT,1

QQHH

QQCC

WIN,1

5

6

7

WOUT,2WIN,2

(1 kg)

(y kg)

(1-y kg)

Feedwater heaters

Open and Closed

Feedwater Heaters

Regeneration with an open feedwater heater

QQCC

Regeneration with Regeneration with an an open feedwateropen feedwater heater at the massheater at the massfraction rate of “y” fraction rate of “y” per unit mass of per unit mass of primary the flow primary the flow rate.rate.

Regeneration with Regeneration with an an open feedwateropen feedwater heater at the massheater at the massfraction rate of “y” fraction rate of “y” per unit mass of per unit mass of primary the flow primary the flow rate.rate.

QQHH

yy 1-y1-y

WWIN,1IN,1 WWIN,2IN,2

WWOUTOUT

yy

From the outletFrom the outletof the condenser of the condenser and first feedwaterand first feedwaterpump. (1-y kg.)pump. (1-y kg.)

From theFrom theturbine.turbine.(y kg.)(y kg.)

To the secondTo the second feedwater pump.feedwater pump.(1 kg.)(1 kg.)

1-y1-y

The open feedwater heater

Regenerative cycle with a closed feedwater heater

55

11

22 33

44

66

77 88

yy1-y1-y

yy

1-y1-y

QQHH QQCC

WWTT

Closed Closed feedwaterfeedwaterheaterheater

TrapTrap

CondenserCondenser

T-s diagram for a regenerative cycle with a closed feedwater heater

1

2

34

5

67

8

ss

TT


FIGURE 10-15The ideal regenerative Rankine cycle with an open feedwater heater.


FIGURE 10-16The ideal regenerative Rankine cycle with a closed feedwater heater.

Example - Regeneration with a single extraction and an open feedwater heater

Example

A regenerative Rankine cycle with a single extraction A regenerative Rankine cycle with a single extraction provides saturated steam at 500 psi a the turbine inlet.provides saturated steam at 500 psi a the turbine inlet.Condensation takes place at 70Condensation takes place at 70oo F. One open regenerative F. One open regenerativefeedwater heater is included, using extracted steam at a feedwater heater is included, using extracted steam at a temperature midway between the limits of the cycle. temperature midway between the limits of the cycle. All processes are assumed to be internally reversible All processes are assumed to be internally reversible except that in the regenerative heater. Neglect KE and except that in the regenerative heater. Neglect KE and PE effects, and determine the thermal efficiency, PE effects, and determine the thermal efficiency, the internal irreversibility, and the extraction pressure. the internal irreversibility, and the extraction pressure. Assume steady operation.Assume steady operation.

Example - Plant diagram

WWOUTOUT

QQCC

aa

dd

ee

cc

bbQQHH

yy 1-y1-y

WWIN,1IN,1 WWIN,2IN,2

gg ff

Example - T-s diagram

aabb

cc

ddee

ff

gg

yy

1-y1-y

11

1-y1-y

TT

ss

Example - Mass fraction at State “c”Example - Mass fraction at State “c”

The mass fraction of fluid extracted at State c is obtainedThe mass fraction of fluid extracted at State c is obtainedfrom the energy balance for an open system applied tofrom the energy balance for an open system applied tothe feedwater heater. the feedwater heater. Assume adiabatic mixingAssume adiabatic mixing..

aa bb

cc

ddee

ffgg

yy

1-y1-y

11

1-y1-y

TT

ss

gfc

gfccc

hhyyh

hmhmmhm

1

chy,

fhy,1gh

Example - Entropy balance for the feedwater heaterExample - Entropy balance for the feedwater heater

Apply the entropy balance for an open system to the Apply the entropy balance for an open system to the feedwater heater.feedwater heater.

fgcg

fcg

ssyssy

syyss

1

1

cc shy ,,

ff shy ,,1gg sh ,

Given DataSTATE T

RP

psih

BTU/ lbms

BTU/lbm-Rx

a 728.7 500 …b 500 1c 728.5d 530e 530 0f 530 …g 728.7 0

Example - Given data

Computed Data (From Tables)STATE T

RP

psih

BTU/ lbms

BTU/lbm-Rx

a 728.7 500 239.1 0.3940 …b 927 500 1204.4 1.4364 1c 728.5 41 1016 1.4364 0.835d 530 0.361 774 1.4364 0.698e 530 0.361 38 0.0745 0f 530 41 38.1 0.0745 …g 728.7 41 237.6 0.3940 0

Example - Computed data

Process QuantitiesProcess H

BTU/ lbmm mh

BTU/ lbm

WBTU/ lbm

a-b 965 1 965 0b-c -188 1 -188 188c-d -242 0.796 -193 193d-e -736 0.796 -586 0e-f 0.1 0.796 0.1 -0.1f-g 199.5 0.796 159c-g -778 0.204 -159

0

g-a 1.5 1 1.5 -1.5Net … … 0 379.4

Example - Process quantities

Process QuantitiesProcess Q

BTU/ lbmds

BTU/ lbm-Ryds

BTU/ lbm-R

Q/TBTU/lbm-R

BTU/lbm-R

a-b 965.3 1.0694 1.0694 1.0694 0b-c 0 0 0 0 0c-d 0 0 0 0 0d-e -586 -1.3889 -1.1056 -1.1056 0e-f 0 0 0 0 0f-g 0.3195 0.2543c-g

0-1.0694 -0.2180

0 0.0363

g-a 0 0 0 0 0Net 379.3 … 0 -0.0362 0.0363

Note the positive entropy production in Note the positive entropy production in the feedwater heater.the feedwater heater.Note the positive entropy production in Note the positive entropy production in the feedwater heater.the feedwater heater.

Example - Process quantities

393.0965

4.379

HQ

W

Note that regeneration has increased thermalNote that regeneration has increased thermalefficiency above that of the previous example at efficiency above that of the previous example at the expense of some work output per lbm of steam.the expense of some work output per lbm of steam.

Example - Thermal efficiency

Commercial steam-power plants

• Reheat (multiple stages)

• Regeneration (multiple extractions)

• Nearly ideal heat addition – Constant temperature boiling for water


• Heat transfer characteristics of steam and water permit external combustion systems

• Compression of condensed liquid produces a favorable work ratio.


• The Rankine cycle with reheat and regeneration is advantageous for large plants.

• Small plants do not have economies of scale– Internal combustion for heat addition.

– A different thermodynamic cycle

10-7 Second-Law Analysis of Vapor Power Cycles

10-8 Cogeneration


FIGURE 10-20A simple process-heating plant.


FIGURE 10-21An ideal cogeneration plant.


FIGURE 10-22A cogeneration plant with adjustable loads.

10-9 Combined /gas-Vapor Power Cycles


FIGURE 10-24Combined gas–steam power plant.


FIGURE 10-26Mercury–water binary vapor cycle.

chapter 10 vapor and combined power cycles. 10-1 the carnot vapor cycle

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