10-1

Chapter 10 VAPOR AND COMBINED POWER CYCLES

Carnot Vapor Cycle

10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisturecontent allowed is about 10%.

10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heattransfer processes to two-phase systems to maintain isothermal conditions severely limits the maximumtemperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisturecontent which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.

10-3E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the quality at the end of the heat rejection process, and the net work output are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We note that

and

19.3%R1.833R0.67211

R0.672F0.212R1.833F1.373

Cth,

psia14.7@sat

psia180@sat

H

L

L

H

TT

TTTT

T

14.7 psia

180 psiaqin

1

4

2

3(b) Noting that s4 = s1 = sf @ 180 psia = 0.53274 Btu/lbm·R,

0.15344441.1

31215.053274.044

fg

f

sss

x s

(c) The enthalpies before and after the heat addition process are

Btu/lbm2.111216.85190.014.346Btu/lbm14.346

22

psia180@1

fgf

f

hxhhhh

Thus,

and,Btu/lbm148.1Btu/lbm0.7661934.0

Btu/lbm0.76614.3462.1112

inthnet

12in

qw

hhq

10-2

10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 20 kPa = 60.06°C = 333.1 K, the thermalefficiency becomes

36.3%3632.0K523K333.1

11Cth,H

L

TT

T

qout

qin

1

4

2

320 kPa

(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,

Thus,

kJ/kg1092.3kJ/kg3.1715K523K333.1

kJ/kg3.1715

inout

250@in

qTT

qq

hq

H

LL

Cfg

250 C

s(c) The net work output of this cycle is

kJ/kg623.0kJ/kg3.17153632.0inthnet qw

10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the amount of heat rejected, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 250°C = 523 K and TL = Tsat @ 10 kPa = 45.81°C = 318.8 K, the thermalefficiency becomes

39.04%K523K318.811Cth,

H

L

TT

T

qout

qin

1

4

2

310 kPa

(b) The heat supplied during this cycle is simply the enthalpy of vaporization ,

Thus,

kJ/kg1045.6kJ/kg3.1715K523K318.8

kJ/kg3.1715

inout

C250@in

qTT

qq

hq

H

LL

fg

250 C

s(c) The net work output of this cycle is

kJ/kg669.7kJ/kg3.17153904.0inthnet qw

10-3

10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. Thethermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The thermal efficiency is determined from

T

1

4

2

3

th, C60 273 K350 273 K

1 1TT

L

H46.5%

(b) Note that350 C

s2 = s3 = sf + x3sfg

= 0.8313 + 0.891 7.0769 = 7.1368 kJ/kg·KThus , 60 C

s(Table A-6)MPa1.402

2

2

KkJ/kg1368.7C350

PsT

(c) The net work can be determined by calculating the enclosed area on the T-s diagram,

Thus,kJ/kg16235390.11368.760350Area

KkJ/kg5390.10769.71.08313.0

43net

44

ssTTw

sxss

LH

fgf

The Simple Rankine Cycle

10-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump,(2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heatrejection in a condenser.

10-8C Heat rejected decreases; everything else increases.

10-9C Heat rejected decreases; everything else increases.

10-10C The pump work remains the same, the moisture content decreases, everything else increases.

10-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involvefriction and pressure drops in various components and the piping, and heat loss to the surrounding mediumfrom these components and piping.

10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) thesame as the boiler inlet pressure in ideal cycles.

10-13C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam willadversely affect the turbine work output.

10-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher thanthe temperature of the cooling water.

10-4

10-15 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle and the net power output of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg58.34304.354.340kJ/kg04.3

mkPa1kJ1

kPa503000/kgm001030.0

/kgm001030.0

kJ/kg54.340

in,12

33

121in,

3kPa50@1

kPa50@1

p

p

f

f

whh

PPw

hh

v

vvT

kJ/kg3.2272

7.23048382.054.340

8382.05019.6

0912.15412.6kPa50

KkJ/kg5412.6kJ/kg3.2994

C300MPa3

44

44

34

4

3

3

3

3

fgf

fg

f

hxhh

sss

xss

P

sh

TP

3 MPa

4

2

3

150 kPa

qout

qin

s

Thus,

kJ/kg9.7188.19317.2650kJ/kg8.193154.3403.2272kJ/kg7.265058.3433.2994

outinnet

14out

23in

qqwhhqhhq

and

27.1%7.26508.193111

in

outth q

q

(b) MW25.2kJ/kg9.718kg/s35netnet wmW

10-5

10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of thesteam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg90.20109.1081.191kJ/kg09.10

mkPa1kJ1kPa10000,10/kgm00101.0

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

p

p

f

f

whh

PPw

hh

v

vv T

kJ/kg2089.72392.10.7934191.81

4996.70.64926.5995kPa10

KkJ/kg6.5995kJ/kg3375.1

C500MPa10

44

44

34

4

3

3

3

3

fgf

fg

f

hxhh

sss

xss

P

sh

TP

0.7934

10 MPa

4

2

3

110 kPa

qout

qin

s

(b)

kJ/kg4.12759.18972.3173kJ/kg9.189781.1917.2089kJ/kg2.317390.2011.3375

outinnet

14out

23in

qqwhhqhhq

and

40.2%kJ/kg3173.2kJ/kg1275.4

in

netth q

w

(c) skg164.7 /kJ/kg1275.4

kJ/s210,000

net

net

wW

m

10-6

10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality ofthe steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg68.20387.1181.191kJ/kg11.87

0.85/mkPa1

kJ1kPa1010,000/kgm0.00101

/

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

p

pp

f

f

whh

PPw

hh

v

vvT

0.874

0.793

123928119152282

kJ/kg5.2282kPa10

kJ/kg5.22827.20891.337585.01.3375

kJ/kg7.20891.23927934.081.191

44996.7

6492.05995.6kPa10

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

44

4

4

433443

43

44

44

34

4

3

3

3

3

...

hhh

xhP

hhhhhhhh

hxhh

sss

xss

P

sh

TP

fg

f

sTs

T

fgfs

fg

fss

s

s

qin

qout

2

4

s

10 MPa

4

2

3

110 kPa

(b)

kJ/kg7.10807.20904.3171kJ/kg7.209081.1915.2282kJ/kg4.317168.2031.3375

outinnet

14out

23in

qqwhhqhhq

and

34.1%kJ/kg3171.5kJ/kg1080.7

in

netth q

w

(c) kg/s194.3kJ/kg1080.7

kJ/s210,000

net

net

wWm

10-7

10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressurelimits is considered. The minimum turbine inlet temperature, the rate of heat input in the boiler, and thethermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist.2 Kinetic and potential energy changes are negligible.

T

4x4 = 0.9

2

3

1 Qout

2 psia ·

Qin

1250 psia·

Analysis (a) From the steam tables (Tables A-4E,A-5E, and A-6E),

Btu/lbm77.9775.302.94Btu/lbm75.3

ftpsia5.4039Btu1psia21250/lbmft0.01623

/lbmft01623.0

Btu/lbm02.94

in,12

33

121in,

3psia2@1

psia2@1

p

p

f

f

whh

PPw

hh

v

vv

s

F13373

3

43

3

44

44

Btu/lbm4.1693psia1250

RBtu/lbm7450.174444.19.017499.0

Btu/lbm6.10137.10219.002.94

Th

ssP

sxss

hxhh

fgf

fgf

(b) Btu/s119,67297.771693.4lbm/s7523in hhmQ

(c)

42.4%Btu/s119,672Btu/s68,96711

Btu/s68,96794.021013.6lbm/s75

in

out

14out

QQ

hhmQ

th

10-8

10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressurelimits. The minimum turbine inlet temperature, the rate of heat input in the boiler, and the thermalefficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kineticand potential energy changes are negligible. T

Qout·

2s

s

1250 psia

Qin·

4s 4x4 = 0.9

2

3

12 psia

Analysis (a) From the steam tables (Tables A-4E, A-5E,and A-6E),

Btu/lbm43.9841.402.94Btu/lbm41.4

85.0/ftpsia5.4039

Btu1psia21250/lbmft0.01623

/

/lbmft01623.0

Btu/lbm02.94

in,12

33

121in,

3psia2@1

psia2@1

p

Pp

f

f

whh

PPw

hh

v

vv

RBtu/lbm7450.174444.19.017499.0Btu/lbm6.10137.10219.002.94

44

44

fgf

fgf

sxsshxhh

The turbine inlet temperature is determined by trial and error ,

Try 1:

8171.04.9180.14396.10130.1439

Btu/lbm4.9187.10218069.002.94

8069.074444.1

17499.05826.1

Btu/lbm.R5826.1Btu/lbm0.1439

F900psia1250

43

43

44

344

3

3

3

3

sT

fgsfs

fg

f

fg

fss

hhhh

hxhh

sss

sss

x

sh

TP

Try 2:

8734.03.9436.14986.10136.1498

Btu/lbm3.9437.10218312.002.94

8312.074444.1

17499.06249.1

Btu/lbm.R6249.1Btu/lbm6.1498

F1000psia1250

43

43

44

344

3

3

3

3

sT

fgsfs

fg

f

fg

fss

hhhh

hxhh

sss

sss

x

sh

TP

By linear interpolation, at T = 0.85 we obtain T3 = 958.4 F. This is approximate. We can determine state 3exactly using EES software with these results: T3 = 955.7 F, h3 = 1472.5 Btu/lbm.

(b) Btu/s103,05598.431472.5lbm/s7523in hhmQ

(c)

33.1%Btu/s103,055Btu/s68,96911

Btu/s969,6894.021013.6lbm/s75

in

outth

14out

QQ

hhmQ

10-9

10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between thespecified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

T

kJ/kg.0327707.596.271kJ/kg5.07

mkPa1kJ1

kPa255000/kgm00102.0

/kgm000102.0

kJ/kg96.271

in,12

33

121in,

3kPa25@1

kPa25@1

p

p

f

f

whh

PPw

hh

v

vv

kJ/kg2.22765.23458545.096.271

8545.09370.6

8932.08210.6kPa25

KkJ/kg8210.6kJ/kg2.3317

C450MPa5

44

44

34

4

3

3

3

3

fgf

fg

f

hxhh

sss

xss

P

sh

TP

4

2

3

1 Q·25 kPa

out

Qin

5 MPa ·

s

The thermal efficiency is determined from

kJ/kg2.200496.2712.2276kJ/kg2.304003.2772.3317

14out

23inhhqhhq

and

Thus,24.5%96.075.03407.0

3407.02.30402.200411

gencombthoverall

in

outth q

q

(b) Then the required rate of coal supply becomes

and

tons/h150.3tons/s04174.0kg1000

ton1kJ/kg29,300

kJ/s1,222,992

kJ/s992,222,12453.0

kJ/s300,000

coal

incoal

overall

netin

CQ

m

WQ

10-10

10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as theworking fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg40.8958.082.88

kJ/kg58.0mkPa1

kJ1kPa7001400/kgm0008331.0

/kgm0008331.0

kJ/kg82.88

in,12

33

121in,

3MPa7.0@1

MPa7.0@1

p

p

f

f

whh

PPw

hh

v

vv T

kJ/kg.2026221.1769839.082.88

9839.058763.0

33230.09105.0MPa7.0

KkJ/kg9105.0

kJ/kg12.276

vaporsat.MPa4.1

44

44

34

4

MPa4.1@3

MPa4.1@33

fgf

fg

f

g

g

hxhh

sss

xss

P

ss

hhP

R-134a

1.4 MPa

4

2

3

10.7 MPa

qout

qin

s

Thus ,

kJ/kg34.1338.17372.186kJ/kg38.17382.8820.262kJ/kg72.18640.8912.276

outinnet

14out

23in

qqwhhqhhq

and

7.1%kJ/kg186.72kJ/kg13.34

in

netth q

w

(b) kW40.02kJ/kg13.34kg/s3netnet wmW

10-11

10-22 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the coolingwater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg198.8706.781.191

kJ/kg7.06mkPa1

kJ1kPa107,000/kgm0.00101

/kgm00101.0

kJ/kg.81191

in,12

33

121in,

3kPa10@1

kPa10@1

p

p

f

f

whh

PPw

hh

v

vv T

kJ/kg3.62151.23928201.081.191

8201.04996.7

6492.08000.6kPa10

KkJ/kg8000.6kJ/kg411.43

C500MPa7

44

44

34

4

3

3

3

3

fgf

fg

f

hxhh

sss

xss

P

sh

TP

7 MPa

4

2

3

110 kPa

qout

qin

s

Thus,

kJ/kg7.12508.19615.3212kJ/kg8.196181.1916.2153kJ/kg5.321287.1984.3411

outinnet

14out

23in

qqwhhqhhq

and 38.9%kJ/kg3212.5kJ/kg1250.7

in

netth q

w

(b) skg036 /.kJ/kg1250.7kJ/s45,000

net

net

wWm

(c) The rate of heat rejection to the cooling water and its temperature rise are

C8.4CkJ/kg4.18kg/s2000

kJ/s70,586)(

kJ/s,58670kJ/kg1961.8kg/s35.98

watercooling

outwatercooling

outout

cmQ

T

qmQ

10-12

10-23 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressurelimits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg9.921911.881.191kJ/kg8.11

0.87/mkPa1

kJ1kPa107,000/kgm00101.0

/

/kgm00101.0

kJ/kg91.811

in,12

33

121in,

3kPa10@1

kPa10@1

p

pp

f

f

whh

PPw

hh

v

vvT

kJ/kg1.23176.21534.341187.04.3411

kJ/kg6.21531.2392820.081.191

8201.04996.7

6492.08000.6kPa10

KkJ/kg8000.6kJ/kg.43411

C500MPa7

433443

43

44

44

34

4

3

3

3

3

sTs

T

fgfs

fg

f

hhhhhhhh

hxhh

sss

xss

P

sh

TP

qout

qin2

4

s

7 MPa

4

2

3

110 kPa

Thus,

kJ/kg2.10863.21255.3211kJ/kg3.212581.1911.2317kJ/kg5.321192.1994.3411

outinnet

14out

23in

qqwhhqhhq

and 33.8%kJ/kg3211.5kJ/kg1086.2

in

netth q

w

(b) skg4341 /.kJ/kg1086.2kJ/s45,000

net

net

wWm

(c) The rate of heat rejection to the cooling water and its temperature rise are

C10.5CkJ/kg4.18kg/s2000

kJ/s88,051)(

kJ/s,05188kJ/kg2125.3kg/s41.43

watercooling

outwatercooling

outout

cmQ

T

qmQ

10-13

10-24 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankinecycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg57.26115.1042.251kJ/kg15.10

mkPa1kJ1kPa20000,10/kgm001017.0

/kgm001017.0

kJ/kg42.251

in,12

33

121in,

3kPa20@1

kPa20@1

p

p

f

f

whh

PPw

hh

v

vv

kJ/kg3.1845

5.23576761.042.251

6761.00752.7

8320.06159.5kPa20

KkJ/kg6159.5kJ/kg5.2725

1MPa10

44

44

34

4

3

3

3

3

fgf

fg

f

hxhh

sss

xss

P

sh

xP

s

Rankinecycle

4

3

2

1

T

kJ/kg869.90.15949.2463kJ/kg0.159442.2513.1845kJ/kg9.246357.2615.2725

outinnet

14out

23in

qqwhhqhhq

0.3539.24630.159411

in

outth q

q

(b) Carnot Cycle analysis:

s

Carnotcycle

4

2 3

1

T

kJ/kg9.1093)5.2357)(3574.0(42.251

3574.00752.7

8320.03603.3kPa20

KkJ/kg3603.3kJ/kg8.1407

0C0.311

C0.311kJ/kg5.2725

1MPa10

11

11

21

1

2

2

2

32

3

3

3

3

fgf

fg

f

hxhhs

ssx

ssP

sh

xTT

Th

xP

kJ/kg565.43.7527.1317kJ/kg4.7519.10933.1845

kJ/kg7.13178.14075.2725

outinnet

14out

23in

qqwhhqhhq

0.4307.13174.75111

in

outth q

q

10-14

10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the workingfluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycleare to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Properties The specific heat of geothermal water is taken to be 4.18 kJ/kg.ºC.Analysis (a) We need properties of isobutane,which are not available in the book. However,we can obtain the properties from EES. Turbine:

kJ/kg74.689C5.179

kPa410

kJ/kg40.670kPa410

KkJ/kg5457.2kJ/kg54.761

C147kPa3250

44

4

434

4

3

3

3

3

hTP

hss

P

sh

TP

s

4

3 2

1

Geothermalwater out

Geothermalwater in

heat exchanger

pumpturbine

air-cooledcondenser

0.78840.67054.76174.68954.761

43

43

sT hh

hh

T

qin

qout

2s

4

s

3.25 MPa

4s

2

3

1410 kPa

(b) Pump:

kJ/kg82.27881.501.273kJ/kg81.5

90.0/mkPa1

kJ1kPa4103250/kgm001842.0

/

/kgm001842.0

kJ/kg01.273

in,12

33

121in,

3kPa410@1

kPa014@1

p

Pp

f

f

whh

PPw

hh

v

vv

kW941,21kJ/kg)74.68954.761kJ/kg)(305.6()( 43outT, hhmW

kW1777kJ/kg)1kJ/kg)(5.8305.6()( inp,12inP, wmhhmW

kW20,1651777941,21inP,outT,net WWW

Heat Exchanger:

kW656,162C)90160(C)kJ/kg..184(kJ/kg).9555()( outingeogeoin TTcmQ

(c) 12.4%0.124656,162165,20

in

netth Q

W

10-15

10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Themass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output fromthe turbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water forgeothermal water (Tables A-4 through A-6)

1661.02108

09.64014.990kJ/kg14.990

kPa500

kJ/kg14.9900

C230

22

12

2

11

1

fg

f

hhh

xhh

P

hxT

The mass flow rate of steam through the turbine is kg/s38.20kg/s)230)(1661.0(123 mxm

(b) Turbine:

kJ/kg7.2344)1.2392)(90.0(81.19190.0kPa10

kJ/kg3.2160kPa10

KkJ/kg8207.6kJ/kg1.2748

1kPa500

444

4

434

4

3

3

3

3

fgf

s

hxhhxP

hss

P

sh

xP

productionwell

Flashchamber

65

4

3

2condenser

1

steamturbine

separator

reinjectionwell

0.6863.21601.27487.23441.2748

43

43

sT hh

hh

(c) The power output from the turbine is

kW15,410kJ/kg)7.23448.1kJ/kg)(27438.20()( 433outT, hhmW

(d) We use saturated liquid state at the standard temperature for dead state enthalpy

kJ/kg83.1040

C250

0

0 hxT

kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE

7.6%0.0757622,203

410,15

in

outT,th E

W

10-16

10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Thetemperature of the steam at the exit of the second flash chamber, the power produced from the secondturbine, and the thermal efficiency of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

1661.0kJ/kg14.990

kPa500

kJ/kg14.9900

C230

212

2

11

1

xhh

P

hxT

kg/s80.1911661.0230kg/s38.20kg/s)230)(1661.0(

316

123

mmmmxm

Flashchamber

separator

9

8

7

Flashchamber

6

5

4

3

2condenser

1

steamturbine

separator

reinjectionwell

productionwell

kJ/kg7.234490.0kPa10

kJ/kg1.27481

kPa500

44

4

33

3

hxP

hxP

kJ/kg1.26931

kPa150

0777.0kPa150

kJ/kg09.6400

kPa500

88

8

7

7

67

7

66

6

hxP

xT

hhP

hxP

C111.35

(b) The mass flow rate at the lower stage of the turbine is kg/s.9014kg/s)80.191)(0777.0(678 mxm

The power outputs from the high and low pressure stages of the turbine are

kW15,410kJ/kg)7.23448.1kJ/kg)(27438.20()( 433outT1, hhmW

kW5191kJ/kg)7.23443.1kJ/kg)(269.9014()( 488outT2, hhmW

(c) We use saturated liquid state at the standard temperature for the dead state enthalpy

kJ/kg83.1040

C250

0

0 hxT

kW621,203kJ/kg)83.10414kg/s)(990.230()( 011in hhmE

10.1%0.101621,203

5193410,15

in

outT,th E

W

10-17

10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heatsource. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbineand the binary cycle, and the thermal efficiencies for the binary cycle and the combined plant are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

1661.0kJ/kg14.990

kPa500

kJ/kg14.9900

C230

212

2

11

1

xhh

P

hxT

kg/s80.19120.38230kg/s38.20kg/s)230)(1661.0(

316

123

mmmmxm

kJ/kg7.234490.0kPa10

kJ/kg1.27481

kPa500

44

4

33

3

hxP

hxP

kJ/kg04.3770

C90

kJ/kg09.6400

kPa500

77

7

66

6

hxT

hxP

The isobutene propertiesare obtained from EES:

/kgm001839.0kJ/kg83.270

0kPa400

kJ/kg01.691C80kPa400

kJ/kg05.755C145kPa3250

310

10

10

10

99

9

88

8

v

hxP

hTP

hTP

1

19

8

7

65

4

3

2

1

flashchamber

condenser

air-cooledcondenser

separator

BINARYCYCLE

pump

heat exchanger

isobutaneturbine

reinjectionwell

productionwell

steamturbine

kJ/kg65.27682.583.270kJ/kg.82.5

90.0/mkPa1

kJ1kPa4003250/kgm001819.0

/

in,1011

33

101110in,

p

pp

whh

PPw v

An energy balance on the heat exchanger gives

kg/s105.46isoiso

118iso766

kg276.65)kJ/-(755.05kg377.04)kJ/-09kg/s)(640.81.191(

)()(

mm

hhmhhm

(b) The power outputs from the steam turbine and the binary cycle are

kW15,410kJ/kg)7.23448.1kJ/kg)(27438.19()( 433steamT, hhmW

10-18

kW6139)kJ/kg82.5)(kg/s46.105(6753

kW6753kJ/kg)01.691.05kJ/kg)(755.46105()(

,isoisoT,binarynet,

98isoT,

inp

iso

wmWW

hhmW

(c) The thermal efficiencies of the binary cycle and the combined plant are

kW454,50kJ/kg)65.276.05kJ/kg)(755.46105()( 118isobinaryin, hhmQ

12.2%0.122454,50

6139

binaryin,

binarynet,binaryth, Q

W

kJ/kg83.1040

C250

0

0 hxT

kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE

10.6%0.106622,203

6139410,15

in

binarynet,steamT,plantth, E

WW

The Reheat Rankine Cycle

10-29C The pump work remains the same, the moisture content decreases, everything else increases.

10-30C The T-s diagram of the ideal Rankinecycle with 3 stages of reheat is shown on theside. The cycle efficiency will increase as thenumber of reheating stages increases.

10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the averagetemperature at which heat is added will be higher in this case.

1

I

2

10 s

T

3

4

5

6

7

8

II III9

10-19

10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankinecycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),

T

kJ/kg.5425912.842.251

kJ/kg8.12mkPa1

kJ1kPa208000/kgm001017.0

/kgm700101.0

kJ/kg42.251

in,12

33

121in,

3kPa20@1

kPa20@1

p

p

f

f

whh

PPw

hh

v

vv

kJ/kg2.23855.23579051.042.251

9051.00752.7

8320.02359.7kPa20

KkJ/kg2359.7kJ/kg2.3457

C500MPa3

kJ/kg1.3105MPa3

KkJ/kg7266.6kJ/kg5.3399

C500MPa8

66

66

56

6

5

5

5

5

434

4

3

3

3

3

fgf

fg

f

hxhh

sss

x

ssP

sh

TP

hss

P

sh

TP

20 kPa

8 MPa 4

3

6

2

5

1s

The turbine work output and the thermal efficiency are determined from

andkJ/kg0.34921.31052.345754.2595.3399

2.23852.34571.31055.3399

4523in

6543outT,

hhhhq

hhhhw kJ/kg1366.4

Thus,

38.9%kJ/kg3492.5kJ/kg1358.3

kJ/kg3.135812.84.1366

in

netth

in,,net

qw

www poutT

10-20

10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entryfeature of EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure turbine exit Also, the T-s diagram is to be plotted.Analysis The problem is solved using EES, and the solution is given below.

"Input Data - from diagram window"{P[6] = 20 [kPa]P[3] = 8000 [kPa] T[3] = 500 [C]P[4] = 3000 [kPa] T[5] = 500 [C]Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"}

"Pump analysis"function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''

if (x6>1) then x6$='(superheated)'if (x6<0) then x6$='(subcooled)'

end

Fluid$='Steam_IAPWS'

P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,T=T[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine""Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])

10-21

vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=QUALITY(Fluid$,h=h[6],P=P[6])"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler""Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])

"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100

200

300

400

500

600

700

s [kJ/kg-K ]

T [C

]

8000 kPa

3000 kPa

20 kPa

3

4

5

6

Ideal R ankine cycle w ith reheat

1,2

SOLUTIONEff=0.389 Eta_p=1 Eta_t=1Fluid$='Steam_IAPWS' h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg]h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg]h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg]P[1]=20 [kPa] P[2]=8000 [kPa] P[3]=8000 [kPa]P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa]Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K]s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K]s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K]s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C]T[3]=500 [C] T[4]=345.2 [C] T[5]=500 [C]T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C]v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg] v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg]W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg]W_t_lp=1072 [kJ/kg] x6s$='' x[1]=0x[6]=0.9051

10-22

10-34 A steam power plant that operates on a reheat Rankine cycle is considered. The quality (ortemperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and themass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg43.20262.1081.191

kJ/kg10.62

0.95/mkPa1

kJ1kPa1010,000/kgm0.00101

/

/kgm001010.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

p

pp

f

f

whh

PPw

hh

v

vvT

KkJ/kg7642.7kJ/kg1.3479

C500MPa1

kJ/kg0.29027.27831.337580.01.3375

kJ/kg8.2783MPa1

KkJ/kg5995.6kJ/kg75.133

C500MPa10

5

5

5

5

433443

43

434

4

3

3

3

3

sh

TP

hhhhhhhh

hss

P

sh

TP

sTs

T

ss

s

2

610 kPa

10 MPa 4s4

3

6

2

5

1s

vapordsuperheatekJ/kg8.26642.24611.347980.01.3479

kJ/kg2.24611.23929487.081.191

exitturbineat9487.04996.7

6492.07642.7kPa10

655665

65

66

66

56

6

g

sTs

T

fgsfs

fg

fss

s

s

h

hhhhhhhh

hxhh

sss

xss

P

From steam tables at 10 kPa we read T6 = 88.1 C.(b)

kJ/kg8.127662.104.1287

kJ/kg8.37490.29021.347943.2021.3375

kJ/kg4.12878.26641.34790.29021.3375

inp,outT,net

4523in

6543outT,

www

hhhhq

hhhhw

Thus the thermal efficiency is

34.1%kJ/kg3749.8kJ/kg1276.8

in

netth q

w

(c) The mass flow rate of the steam is

kg/s62.7kJ/kg1276.9kJ/s80,000

net

net

wW

m

10-23

10-35 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and themass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg90.20109.1081.191

kJ/kg10.09mkPa1

kJ1kPa01000,10/kgm00101.0

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

p

p

f

f

whh

PPw

hh

v

vvT

kJ/kg2.24611.23929487.081.191

exitturbineat4996.7

6492.07642.7kPa10

KkJ/kg7642.7kJ/kg1.3479

C500MPa1

kJ/kg8.2783MPa1

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

66

66

56

6

5

5

5

5

434

4

3

3

3

3

fgf

fg

f

hxhh

sss

x

ssP

sh

TP

hss

P

sh

TP

0.9487

10 kPa

10 MPa 4

3

6

2

5

1s

(b)

kJ/kg3.159909.104.1609

kJ/kg5.38687.27831.347990.2011.3375

kJ/kg3.16092.24611.34797.27831.3375

in,outT,net

4523in

6543outT,

pwww

hhhhq

hhhhw

Thus the thermal efficiency is

41.3%kJ/kg3868.5kJ/kg1599.3

in

netth q

w

(c) The mass flow rate of the steam is

skg050 /.kJ/kg1599.3kJ/s80,000

net

net

wW

m

10-24

10-36E A steam power plant that operates on the ideal reheat Rankine cycle is considered. The pressure at which reheating takes place, the net power output, the thermal efficiency, and the minimum mass flow rateof the cooling water required are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm11.7239.272.69

Btu/lbm2.39ftpsia5.4039

Btu1psia1800/lbmft01614.0

F69.101

/lbmft01614.0

Btu/lbm72.69

in,12

33

121in,

psia1@sat1

3psia1@sat1

psia1@sat1

p

p

whh

PPw

TT

hh

v

vvT

1 psia

800 psia4

3

6

2

5

1s

Btu/lbm0.10617.10359572.072.69

9572.084495.1

13262.08985.1psia1

RBtu/lbm8985.1Btu/lbm4.1431

F800psia23.62

pressure)reheat(the

Btu/lbm5.1178

vaporsat.

RBtu/lbm6413.1Btu/lbm0.1456

F900psia800

66

66

56

6

5

5

5

5

@sat4

@434

3

3

3

3

4

4

fgf

fg

f

ss

ssg

hxhh

sss

xss

P

sh

TP

PP

hhss

sh

TP

g

g

psia62.23

(b)

Btu/lbm3.99172.690.1061

Btu/lbm8.16365.11784.143111.720.1456

16out

4523in

hhq

hhhhq

Thus,

39.4%Btu/lbm1636.8Btu/lbm991.3

11in

outth q

q

(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of thesteam in the condenser, which is 101.7 F,

lbm/s641.0F45101.69FBtu/lbm1.0

Btu/s103.634

Btu/s10634.3Btu/s1063943.0114

outcool

44inthnetinout

TcQ

m

QWQQ

10-25

10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressurelimits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler,and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg95.20614.1581.191

kJ/kg.1415mkPa1

kJ1kPa10000,15/kgm00101.0

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@sat1

kPa10@sat1

p

p

whh

PPw

hh

v

vv

T

kJ/kg2.2817MPa161.2

kJ/kg53.3466pressurereheattheC500

KkJ/kg3988.74996.790.06492.0

kJ/kg7.23441.239290.081.191kPa10

KkJ/kg3480.6kJ/kg.83310

C500MPa15

434

4

5

5

65

5

66

66

56

6

3

3

3

3

hss

P

hP

ssT

sxss

hxhh

ssP

sh

TP

fgf

fgf

kPa2161

10 kPa

15 MPa4

3

6

2

5

1s

(b) The rate of heat supply is

kW45,038kJ/kg2.281753.346695.2068.3310kg/s124523in hhhhmQ

(c) The thermal efficiency is determined from

Thus,

42.6%kJ/s45,039kJ/s25,834

11

kJ/s,83525kJ/kg81.1917.2344kJ/s12

in

outth

16out

QQ

hhmQ

10-26

10-38 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure,the net power output, and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg3.30271.29482.335885.02.3358

?

95.0?

KkJ/kg2815.7kJ/kg2.3358

C450MPa2

kJ/kg3.30271.29485.347685.05.3476

kJ/kg1.2948MPa2

KkJ/kg6317.6kJ/kg5.3476

C550MPa5.12

655665

65

656

6

66

6

5

5

5

5

4334

43

43

434

4

3

3

3

3

sTs

T

s

sT

sT

ss

hhhhhhhh

hss

P

hxP

sh

TP

hhhh

hhhh

hss

P

sh

TP

2

4

6P = ?

12.5 MPa4s

3

T

4

5

PumpCondenser

BoilerTurbine

21

6

3

s6s

2s

5

1

The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EESfrom the above equations:

P6 = 9.73 kPa, h6 = 2463.3 kJ/kg,(b) Then,

kJ/kg59.20302.1457.189

kJ/kg14.020.90/

mkPa1kJ1kPa73.912,500/kgm0.00101

/

/kgm001010.0

kJ/kg57.189

in,12

33

121in,

3kPa10@1

kPa73.9@1

p

pp

f

f

whh

PPw

hh

v

vv

Cycle analysis:

kW10,242kg2273.7)kJ/-.8kg/s)(36037.7()(

kJ/kg7.227357.1893.3027

kJ/kg8.36033.24632.33583.30275.3476

outinnet

16out

4523in

qqmW

hhq

hhhhq

(c) The thermal efficiency is

36.9%369.0kJ/kg3603.8kJ/kg2273.7

11in

outth q

q

10-27

Regenerative Rankine Cycle

10-39C Moisture content remains the same, everything else decreases.

10-40C This is a smart idea because we waste little work potential but we save a lot from the heat input. The extracted steam has little work potential left, and most of its energy would be part of the heat rejectedanyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little workoutput.

10-41C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing.

10-42C Both cycles would have the same efficiency.

10-43C To have the same thermal efficiency as the Carnot cycle, the cycle must receive and reject heat isothermally.Thus the liquid should be brought to the saturated liquidstate at the boiler pressure isothermally, and the steam mustbe a saturated vapor at the turbine inlet. This will requirean infinite number of heat exchangers (feedwater heaters),as shown on the T-s diagram.

Boilerexit

s

TBoilerinlet

qin

qout

10-28

10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwaterheater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg81.25139.042.251

kJ/kg0.39mkPa1

kJ1kPa20400/kgm001017.0

/kgm001017.0

kJ/kg42.251

in,12

33

121in,

3kPa20@1

kPa20@1

pI

pI

f

f

whh

PPw

hh

v

vv

T

kJ/kg0.22145.23578325.042.251

8325.00752.7

8320.07219.6kPa20

kJ/kg7.26654.21339661.066.604

9661.01191.5

7765.17219.6MPa4.0

KkJ/kg7219.6kJ/kg9.3302

C450MPa6

kJ/kg73.61007.666.604

kJ/kg6.07mkPa1

kJ1kPa4006000/kgm0.001084

/kgm001084.0

kJ/kg66.604

liquidsat.MPa4.0

77

77

57

7

66

66

56

6

5

5

5

5

in,34

33

343in,

3MPa4.0@3

MPa4.0@33

fgf

fg

f

fgf

fg

f

pII

pII

f

f

hxhh

sss

xss

P

hxhh

sss

xss

P

sh

TP

whh

PPw

hhP

v

vvqout

3 y

4

1-y

0.4 MPa

20 kPa

6MPa

6

5

7

2

qin

1s

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heater. Noting that Q W ,ke pe 0

326332266

(steady)0

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

outin

systemoutin

where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,

1462.081.2517.266581.25166.604

26

23

hhhh

y

Then,

kJ/kg7.167542.2510.22141462.011kJ/kg2.269273.6109.3302

17out

45in

hhyqhhq

And kJ/kg1016.57.16752.2692outinnet qqw(b) The thermal efficiency is determined from

37.8%kJ/kg2692.2kJ/kg1675.7

11in

outth q

q

10-29

10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwaterheater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg50.25708.642.251

kJ/kg6.08mkPa1

kJ1kPa206000/kgm0.001017

/kgm001017.0

kJ/kg42.251

in,12

33

121in,

3kPa20@1

kPa20@1

pI

pI

f

f

whh

PPw

hh

v

vv

T

9

qou

3y

4

1-y

0.4 MPa

20 kPa

6MPa

6

5

7

28

qin

1

/kgm400108.0

kJ/kg66.604

liquidsat.MPa4.0

3MPa4.0@3

MPa4.0@33

f

fhhPvv

s

kJ/kg73.610

kJ/kg73.61007.666.604

kJ/kg07.6mkPa1

kJ1kPa4006000/kgm001084.0

938338

in,39

33

393in,

hPPhh

whh

PPw

pII

pII

v

v

Also, h4 = h9 = h8¸ = 610.73 kJ/kg since the two fluid streams which are being mixed have the sameenthalpy.

kJ/kg0.22145.23578325.042.251

8325.00752.7

8320.07219.6kPa20

kJ/kg7.26654.21339661.066.604

9661.01191.5

7765.17219.6MPa4.0

KkJ/kg7219.6kJ/kg9.3302

C450MPa6

77

77

57

7

66

66

56

6

5

5

5

5

fgf

fg

f

fgf

fg

f

hxhh

sss

xss

P

hxhh

sss

xss

P

sh

TP

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heater. Noting that Q ,0pekeW

3628366282

outin

(steady)0outin

1

0

hhyhhyhhmhhmhmhm

EE

EEE

eeii

system

where y is the fraction of steam extracted from the turbine ( /m m6 5 ). Solving for y,

1463.050.25773.61066.6047.2665

50.25773.610

2836

28

hhhhhh

y

Then,kJ/kg4.167542.2510.22141463.011

kJ/kg2.269273.6109.3302

17out

45in

hhyqhhq

And kJ/kg1016.84.16752.2692outinnet qqw(b) The thermal efficiency is determined from

37.8%kJ/kg2692.2kJ/kg1675.411

in

outth q

q

10-30

10-46 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwaterheaters. The net power output of the power plant and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

P III P II P I

fwh fwh I Condenser

BoilerTurbine

6

5

43

21

10

9

8

7 T

10 MPa

1 - y

85

6

3y

4

0.2 MPa

5 kPa

0.6 MPa

91 - y - z

7

10

2

1s

(a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg95.13720.075.137

kJ/kg02.0mkPa1

kJ1kPa5200/kgm0.001005

/kgm001005.0

kJ/kg75.137

in,12

33

121in,

3kPa5@1

kPa5@1

pI

pI

f

f

whh

PPw

hh

v

vv

kJ/kg10.35mkPa1

kJ1kPa60010,000/kgm0.001101

/kgm001101.0

kJ/kg38.670

liquidsat.MPa6.0

kJ/kg13.50542.071.504kJ/kg0.42

mkPa1kJ1

kPa200600/kgm0.001061

/kgm001061.0

kJ/kg71.504

liquidsat.MPa2.0

33

565in,

3MPa6.0@5

MPa6.0@55

in,34

33

343in,

3MPa2.0@3

MPa2.0@33

PPw

hhP

whh

PPw

hhP

pIII

f

f

pII

pII

f

f

v

vv

v

vv

kJ/kg7.2618

6.22019602.071.504

9602.05968.5

5302.19045.6

MPa2.0

kJ/kg8.2821MPa6.0

KkJ/kg9045.6kJ/kg8.3625

C600MPa10

kJ/kg73.68035.1038.670

99

99

79

9

878

8

7

7

7

7

in,56

fgf

fg

f

pIII

hxhh

sss

x

ssP

hss

P

sh

TP

whh

10-31

kJ/kg0.21050.24238119.075.137

8119.09176.7

4762.09045.6kPa5

1010

1010

710

10

fgf

fg

f

hxhh

sss

xss

P

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

FWH-2:

548554488

outin

(steady)0outin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

system

where y is the fraction of steam extracted from the turbine ( /m m8 5 ). Solving for y,

07133.013.5058.282113.50538.670

48

45

hhhhy

FWH-1: 329332299 11 hyhzyzhhmhmhmhmhm eeii

where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,

1373.007136.0195.1377.261895.13771.5041

29

23 yhhhh

z

Then,

kJ/kg2.13888.15560.2945kJ/kg1556.8137.752105.01373.007133.011

kJ/kg0.294573.6808.3625

outinnet

110out

67in

qqwhhzyq

hhq

and

MW30.5kW540,30kJ/kg1388.2kg/s22netnet wmW

(b) 47.1%kJ/kg2945.0kJ/kg1556.8

11in

outth q

q

10-32

10-47 [Also solved by EES on enclosed CD] A steam power plant operates on an ideal regenerativeRankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam throughthe boiler for a net power output of 250 MW and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg10.19229.081.191kJ/kg29.0

mkPa1kJ1

kPa10300/kgm0.00101

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

pI

pI

f

f

whh

PPw

hh

v

vv

kJ/kg0.27555.20479935.087.720

9935.06160.4

0457.26317.6MPa8.0

KkJ/kg6317.6kJ/kg5.3476

C550MPa5.12

kJ/kg727.83MPa12.5,

C4.170

/kgm001115.0

kJ/kg87.720

liquidsat.MPa8.0

kJ/kg52.57409.1343.561kJ/kg13.09

mkPa1kJ1

kPa30012,500/kgm0.001073

/kgm001073.0

kJ/kg43.561

liquidsat.MPa3.0

99

99

89

9

8

8

8

8

5556

MPa0.8@sat6

3MP8.0@6

MPa8.0@766

in,34

33

343in,

3MPa3.0@3

MPa3.0@33

fgf

fg

f

af

f

pII

pII

f

f

hxhh

sss

xss

P

sh

TP

hPTT

TT

hhhP

whh

PPw

hhP

vv

v

vv

1-y-zz

y

3Closed

fwh P II

P I

Openfwh

Condenser

BoilerTurbine

5

6

4

72

1

11

10

9

8

T

7

95

6

3

40.3 MPa

z10 kPa

0.8 MPa

12.5MPa

y

101 - y - z

8

11

2

1s

kJ/kg0.21001.23927977.081.191

7977.04996.7

6492.06317.kPa10

kJ/kg5.25785.21639323.043.561

9323.03200.5

6717.16317.6MPa3.0

1111

1111

811

11

1010

1010

810

10

fgf

fg

f

fgf

fg

f

hxhh

sss

x

ssP

hxhh

sss

x

ssP

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

4569455699

outin

(steady)0systemoutin 0

hhhhyhhmhhmhmhm

EE

EEE

eeii

10-33

where y is the fraction of steam extracted from the turbine ( 510 / mm ). Solving for y,

0753.087.7200.275552.57483.727

69

45

hhhhy

For the open FWH,

310273310102277

outin

(steady)0systemoutin

11

0

hzhhzyyhhmhmhmhmhmhm

EE

EEE

eeii

where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,

1381.010.1925.2578

10.19287.7200753.010.19243.561

210

2723

hhhhyhhz

Then,

kJ/kg12491.15001.2749kJ/kg1.150081.1910.21001381.00753.011

kJ/kg1.274936.7275.3476

outinnet

111out

58in

qqwhhzyq

hhq

and

kg/s200.2kJ/kg1249

kJ/s250,000

net

net

wWm

(b) 45.4%kJ/kg2749.1kJ/kg1500.1

11in

outth q

q

10-34

10-48 EES Problem 10-47 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted.Analysis The problem is solved using EES, and the solution is given below.

"Input Data"P[8] = 12500 [kPa] T[8] = 550 [C] P[9] = 800 [kPa] "P_cfwh=300 [kPa]"P[10] = P_cfwhP_cond=10 [kPa] P[11] = P_cond W_dot_net=250 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency"Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages"Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages"Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages"Eta_pump = 100/100 "Pump isentropic efficiency"

"Condenser exit pump or Pump 1 analysis"

Fluid$='Steam_IAPWS'P[1] = P[11] P[2]=P[10]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Open Feedwater Heater analysis"z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)

"Boiler condensate pump or Pump 2 analysis"P[5]=P[8]P[4] = P[5] P[3]=P[10]v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])

"Closed Feedwater Heater analysis"P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy"h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]"

10-35

T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]"s[5]=entropy(Fluid$,P=P[6],h=h[5])h[6]=enthalpy(Fluid$,P=P[6],x=0)T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]"s[6]=entropy(Fluid$,P=P[6],x=0)

"Trap analysis"P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])

"Boiler analysis"q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler"h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8])

"Turbine analysis"ss[9]=s[8]hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9])Ts[9]=temperature(Fluid$,s=ss[9],P=P[9])h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9])"Definition of turbine efficiency for high pressure stages"T[9]=temperature(Fluid$,P=P[9],h=h[9])s[9]=entropy(Fluid$,P=P[9],h=h[9])ss[10]=s[8]hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10])Ts[10]=temperature(Fluid$,s=ss[10],P=P[10])h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10])"Definition of turbine efficiency for Intermediate pressure stages"T[10]=temperature(Fluid$,P=P[10],h=h[10])s[10]=entropy(Fluid$,P=P[10],h=h[10])ss[11]=s[8]hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11])Ts[11]=temperature(Fluid$,s=ss[11],P=P[11])h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11])"Definition of turbine efficiency for low pressure stages"T[11]=temperature(Fluid$,P=P[11],h=h[11])s[11]=entropy(Fluid$,P=P[11],h=h[11])h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine"

"Condenser analysis"(1-y-z)*h[11]=q_out+(1-y-z)*h[1]"SSSF First Law for the Condenser"

"Cycle Statistics"w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_inW_dot_net = m_dot * w_net

10-36

turb turb th m [kg/s]0.7 0.7 0.3916 231.6

0.75 0.75 0.4045 224.30.8 0.8 0.4161 218

0.85 0.85 0.4267 212.60.9 0.9 0.4363 207.9

0.95 0.95 0.4452 203.81 1 0.4535 200.1

0.7 0.75 0.8 0.85 0.9 0.95 10.39

0.4

0.41

0.42

0.43

0.44

0.45

0.46

200

205

210

215

220

225

230

235

turb

th

m[kg/s]

turb = pump

0 2 4 6 8 10 120

100

200

300

400

500

600

s [kJ/kg-K]

T[C] 12500 kPa

800 kPa

300 kPa

10 kPa

Steam

1,2

3,4

5,6

7

8

910

11

10-37

10-49 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwaterheater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg61.19280.081.191kJ/kg0.80

mkPa1kJ1

kPa10800/kgm0.00101

/kgm00101.0

kJ/kg81.191

in,12

33

121in,

3kPa10@1

kPa10@1

pI

pI

f

f

whh

PPw

hh

v

vv

T

kJ/kg7.24941.23929627.081.191

9627.04996.7

6492.08692.7kPa10

KkJ/kg8692.7kJ/kg3.3481

C500MPa8.0

kJ/kg1.2812MPa8.0

KkJ/kg7585.6kJ/kg0.3502

C550MPa10

kJ/kg12.73126.1087.720kJ/kg10.26

mkPa1kJ1

kPa80010,000/kgm0.001115

/kgm001115.0

kJ/kg87.720

liquidsat.MPa8.0

88

88

78

8

7

7

7

7

656

6

5

5

5

5

in,34

33

343in,

3MPa8.0@3

MPa8.0@33

fgf

fg

f

pII

pII

f

f

hxhh

sss

xss

P

sh

TP

hss

P

sh

TP

whh

PPw

hhP

v

vv

10 MPa

1 - y

63

4

y

10 kPa

0.8 MPa 7

5

8

2

1s

6

1-y7

6

P II P I

Openfwh

Condenser

BoilerTurbine

4

32

1

8

5

y

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

326332266

outin(steady)0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,

2017.061.1921.281261.19287.720

26

23

hhhh

y

Then,

kJ/kg6.14665.18381.3305kJ/kg5.183881.1917.24942017.011

kJ/kg1.33051.28123.34812017.0112.7310.35021

outinnet

18out

6745in

qqwhhyq

hhyhhq

and kg/s54.5kJ/kg1466.1kJ/s80,000

net

net

wW

m

(b) 44.4%kJ/kg3305.1kJ/kg1466.1

in

netth q

w

10-38

10-50 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

9

10

Mixingchamber

1-y

7y

P II P I

Closedfwh

Condenser

BoilerTurbine

4

32

1

8

5

6

T

10 MPa

1 - y

6

3

410

9y

10 kPa

0.8 MPa 7

5

8

2

1s

(a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg13.73126.1087.720kJ/kg10.26

mkPa1kJ1kPa80010,000/kgm0.001115

/kgm001115.0

kJ/kg87.720

liquidsat.MPa8.0

kJ/kg90.20109.1081.191kJ/kg10.09

mkPa1kJ1kPa1010,000/kgm0.00101

/kgm00101.0

kJ/kg81.191

in,34

33

343in,

3MP8.0@3

MPa8.0@33

in,12

33

121in,

3kPa10@1

kPa10@1

pII

pII

af

f

pI

pI

f

f

whh

PPw

hhP

whh

PPw

hh

v

vv

v

vv

Also, h4 = h9 = h10 = 731.12 kJ/kg since the two fluid streams that are being mixed have the same enthalpy.

kJ/kg7.24941.23929627.081.191

9627.04996.7

6492.08692.7kPa10

KkJ/kg8692.7kJ/kg3.3481

C500MPa8.0

kJ/kg7.2812MPa8.0

KkJ/kg7585.6kJ/kg0.3502

C550MPa10

88

88

78

8

7

7

7

7

656

6

5

5

5

5

fgf

fg

f

hxhh

sss

x

ssP

sh

TP

hss

P

sh

TP

10-39

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

3629363292

outin

(steady)0systemoutin

1

0

hhyhhyhhmhhmhmhm

EE

EEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m3 4 ). Solving for y,

2019.090.20113.73187.7207.2812

90.20113.731

2936

29

hhhhhhy

Then,

kJ/kg6.14668.18375.3304kJ/kg9.183781.1917.24942019.011

kJ/kg5.33047.28123.34812019.0113.7310.35021

outinnet

18out

6745in

qqwhhyq

hhyhhq

and

kg/s54.5kJ/kg1467.1kJ/s80,000

net

net

wW

m

(b) 44.4%kJ/kg3304.5kJ/kg1837.8

11in

outth q

q

10-40

10-51E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater andtwo open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of theplant, and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

High-PTurbine

5P III

1-y-z1211

z

10

4

Openfwh II

1-y

y

P II P I

Openfwh I

Condenser

BoilerLow-PTurbine

6

32

1

8

7

9

T

9

10z

1500 81 - y5

6

3

y4 250 psia

1 psia

40 psia 140 psia

1 - y - z

11

7

12

2

1s

(a) From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm84.6912.072.69Btu/lbm0.12

ftpsia5.4039Btu1

psia140/lbmft0.01614

/lbmft01614.0

Btu/lbm72.69

in,12

33

121in,

3psia1@1

psia1@1

pI

pI

f

f

whh

PPw

hh

v

vv

Btu/lbm41.38031.409.376Btu/lbm4.31

ftpsia5.4039Btu1

psia2501500/lbmft0.01865

/lbmft01865.0Btu/lbm09.376

liquidsat.psia250

Btu/lbm81.23667.014.236

Btu/lbm0.67ftpsia5.4039

Btu1psia40250/lbmft0.01715

/lbmft01715.0Btu/lbm14.236

liquidsat.psia40

in,56

33

565in,

3psia250@5

psia250@55

i,34

33

343in,

3psia40@3

psia40@33

pIII

pIII

f

f

npII

pII

f

f

whh

PPw

hhP

whh

PPw

hhP

v

vv

v

vv

Btu/lbm5.1308psia250

RBtu/lbm6402.1Btu/lbm5.1550

F1100psia1500

878

8

7

7

7

7

hss

P

sh

TP

10-41

Btu/lbm4.1052

7.10359488.072.69

9488.084495.1

13262.08832.1

psia1

Btu/lbm0.1356psia40

RBtu/lbm8832.1Btu/lbm3.1531

F1000psia140

Btu/lbm8.1248psia140

1212

1212

1012

12

111011

11

10

10

10

10

979

9

fgf

fg

f

hxhh

sss

x

ssP

hss

P

sh

TP

hss

P

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

FWH-2:

548554488

outin

(steady)0systemoutin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m8 5 ). Solving for y,

1300.081.2365.130881.23609.376

48

45

hhhh

y

FWH-1

321133221111

outin

(steady)0systemoutin

11

0

hyhzyzhhmhmhmhmhm

EE

EEE

eeii

where z is the fraction of steam extracted from the turbine ( /m m9 5 ) at the second stage. Solving for z,

1125.01300.0184.690.135684.6914.2361

211

23 yhhhh

z

Then,

Btu/lbm4.6714.7448.1415Btu/lbm744.469.721052.41125.01300.011

Btu/lbm8.14158.12483.15311300.0141.3805.15501

outinnet

112out

91067in

qqwhhzyq

hhyhhq

and

lbm/s282.5Btu/lbm1415.8

Btu/s104 5

in

in

qQ

m

(b) MW200.1Btu1

kJ1.055Btu/lbm671.4lbm/s282.5netnet wmW

(c) 47.4%Btu/lbm1415.8Btu/lbm744.4

11in

outth q

q

10-42

10-52 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The temperature of the steam at the inlet of the closed feedwater heater, the mass flow rate of the steam extracted from the turbine for the closed feedwater heater, the net power output, and thethermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg85.26543.1442.251

kJ/kg.431488.0/)kPa2012,500)(/kgm0.001017(

/

/kgm001017.0

kJ/kg42.251

in,12

3121in,

3kPa20@1

kPa20@1

pI

ppI

f

f

whh

PPw

hh

v

vvLow-Pturbine

910

11

MixingCham.

1-y7

y

PII PI

Closedfwh Cond.

Boiler

High-Pturbine

4

32

1

8

5

6

/kgm001127.0

kJ/kg51.762

liquidsat.MPa1

3MPa1@3

MPa1@33

f

fhhPvv

kJ/kg25.77773.1451.762kJ/kg73.14

88.0/)kPa001012,500)(/kgm001127.0(

/

in,311

3

3113in,

pII

ppII

whh

PPw v

Also, h4 = h10 = h11 = 777.25 kJ/kg since the two fluid streams which are being mixed have the sameenthalpy.

kJ/kg5.32206.31855.347688.05.3476

kJ/kg6.3185MPa5

KkJ/kg6317.6kJ/kg5.3476

C550MPa5.12

655665

65

656

6

5

5

5

5

sTs

T

s

hhhhhhhh

hss

P

sh

TP

C32888

8

877887

87

878

8

7

7

7

7

kJ/kg1.3111MPa1

kJ/kg1.31111.30519.355088.09.3550

kJ/kg1.3051MPa1

KkJ/kg1238.7kJ/kg9.3550

C550MPa5

ThP

hhhhhhhh

hss

P

sh

TP

sTs

T

s

10-43

kJ/kg2.24929.23479.355088.09.3550

kJ/kg9.2347kPa20

977997

97

979

9

sTs

T

s

hhhhhhhh

hss

P

The fraction of steam extracted from the low pressure turbine for closed feedwater heater is determinedfrom the steady-flow energy balance equation applied to the feedwater heater. Noting that

,Q W ke pe 0

1788.0)51.7621.3111()85.26525.777)(1(

1 38210

yyy

hhyhhy

The corresponding mass flow rate iskg/s4.29kg/s)24)(1788.0(58 mym

(c) Then,

kJ/kg1.184042.2512.24921788.011kJ/kg7.30295.32209.355025.7775.3476

19out

6745in

hhyqhhhhq

and

kW28,550kJ/kg)1.18407.3029)(kg/s24()( outinnet qqmW

(b) The thermal efficiency is determined from

39.3%393.0kJ/kg3029.7kJ/kg1840.1

11in

outth q

q

10-44

Second-Law Analysis of Vapor Power Cycles

10-53C In the simple ideal Rankine cycle, irreversibilities occur during heat addition and heat rejectionprocesses in the boiler and the condenser, respectively, and both are due to temperature difference.Therefore, the irreversibilities can be decreased and thus the 2nd law efficiency can be increased byminimizing the temperature differences during heat transfer in the boiler and the condenser. One way ofdoing that is regeneration.

10-54 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-15 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-15,

kJ/kg8.1931kJ/kg72.2650

KkJ/kg5412.6

KkJ/kg0912.1

out

43

kPa50@21

qq

ss

sss

in

f

Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,

kJ/kg351.3

kJ/kg1068

K290kJ/kg1931.85412.60912.1K290

K1500kJ/kg2650.80912.15412.6K290

41,41041destroyed,

23,23023destroyed,

R

R

R

R

Tq

ssTx

Tq

ssTx

10-55 The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-16 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-16,

kJ/kg9.1897kJ/kg2.3173

KkJ/kg5995.6

KkJ/kg6492.0

out

in

43

kPa10@21

qq

ss

sss f

Processes 1-2 and 3-4 are isentropic. Thus, i12 = 0 and i34 = 0. Also,

kJ/kg172.3

kJ/kg1112.1

K290kJ/kg1897.9

5995.66492.0K290

K1500kJ/kg3173.2

6492.05995.6K290

41,41041destroyed,

23,23023destroyed,

R

R

R

R

Tq

ssTx

Tq

ssTx

10-45

10-56 The exergy destruction associated with the heat rejection process in Prob. 10-22 is to be determinedfor the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-22,

kJ/kg8.1961kJ/kg4.3411

KkJ/kg8000.6

KkJ/kg6492.0

out

3

43

kPa10@21

qh

ss

sss f

The exergy destruction associated with the heat rejection process is

kJ/kg178.0K290kJ/kg1961.8

8000.66492.0K29041,41041destroyed,

R

R

Tq

ssTx

The exergy of the steam at the boiler exit is simply the flow exergy,

03003

03

023

030033 2ssThhqzssThh V

where KkJ/kg2533.0kJ/kg95.71

K290@kPa100,K290@0

K290@kPa100,K290@0

f

f

ssshhh

Thus, kJ/kg1440.9KkJ/kg2532.0800.6K290kJ/kg95.714.34113

10-57 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-32 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-32,

kJ/kg8.213342.2512.2385

kJ/kg1.3521.31052.3457

kJ/kg0.314054.2595.3399KkJ/kg2359.7KkJ/kg7266.6

KkJ/kg8320.0

16out

in,45

in,23

65

43

kPa20@21

hhq

q

qssss

sss f

Processes 1-2, 3-4, and 5-6 are isentropic. Thus, i12 = i34 = i56 = 0. Also,

kJ/kg212.6

kJ/kg94.1

kJ/kg1245.0

K300kJ/kg2133.82359.78320.0K300

K1800kJ/kg352.57266.62359.7K300

K1800kJ/kg3140.08320.07266.6K300

61,61061destroyed,

45,45045destroyed,

23,23023destroyed,

R

R

R

R

R

R

Tq

ssTx

Tq

ssTx

Tq

ssTx

10-46

10-58 EES Problem 10-57 is reconsidered. The problem is to be solved by the diagram window data entryfeature of EES by including the effects of the turbine and pump efficiencies. Also, the T-s diagram is to beplotted.Analysis The problem is solved using EES, and the solution is given below.

function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''

if (x6>1) then x6$='(superheated)'if (x6<0) then x6$='(subcooled)'

end"Input Data - from diagram window"{P[6] = 20 [kPa] P[3] = 8000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"}"Data for the irreversibility calculations:"T_o = 300 [K] T_R_L = 300 [K] T_R_H = 1800 [K] "Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,T=T[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine""Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=QUALITY(Fluid$,h=h[6],P=P[6])

10-47

"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler""Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in"The irreversibilities (or exergy destruction) for each of the processes are:"q_R_23 = - (h[3] - h[2]) "Heat transfer for the high temperature reservoir to process 2-3"i_23 = T_o*(s[3] -s[2] + q_R_23/T_R_H)q_R_45 = - (h[5] - h[4]) "Heat transfer for the high temperature reservoir to process 4-5"i_45 = T_o*(s[5] -s[4] + q_R_45/T_R_H)q_R_61 = (h[6] - h[1]) "Heat transfer to the low temperature reservoir in process 6-1"i_61 = T_o*(s[1] -s[6] + q_R_61/T_R_L)i_34 = T_o*(s[4] -s[3])i_56 = T_o*(s[6] -s[5])i_12 = T_o*(s[2] -s[1])

0 .0 1 .1 2 .2 3 .3 4 .4 5 .5 6 .6 7 .7 8.8 9 .9 11 .00

100

200

300

400

500

600

700

s [kJ /kg -K ]

T [C

]

8000 kP a

300 0 kP a

20 kP a

3

4

5

6

Id ea l R an k in e cyc le w ith reh eat

1 ,2

SOLUTION

Eff=0.389Eta_p=1Eta_t=1Fluid$='Steam_IAPWS'h[1]=251.4 [kJ/kg] h[2]=259.5 [kJ/kg] h[3]=3400 [kJ/kg] h[4]=3105 [kJ/kg] h[5]=3457 [kJ/kg] h[6]=2385 [kJ/kg] hs[4]=3105 [kJ/kg] hs[6]=2385 [kJ/kg] i_12=0.012 [kJ/kg] i_23=1245.038 [kJ/kg] i_34=-0.000 [kJ/kg] i_45=94.028 [kJ/kg] i_56=0.000 [kJ/kg] i_61=212.659 [kJ/kg] P[1]=20 [kPa] P[2]=8000 [kPa]

P[3]=8000 [kPa] P[4]=3000 [kPa] P[5]=3000 [kPa] P[6]=20 [kPa] Q_in=3493 [kJ/kg] Q_out=2134 [kJ/kg] q_R_23=-3140 [kJ/kg] q_R_45=-352.5 [kJ/kg] q_R_61=2134 [kJ/kg] s[1]=0.832 [kJ/kg-K] s[2]=0.8321 [kJ/kg-K] s[3]=6.727 [kJ/kg-K] s[4]=6.727 [kJ/kg-K] s[5]=7.236 [kJ/kg-K] s[6]=7.236 [kJ/kg-K] s_s[4]=6.727 [kJ/kg-K] s_s[6]=7.236 [kJ/kg-K] T[1]=60.06 [C] T[2]=60.4 [C] T[3]=500 [C]

T[4]=345.2 [C] T[5]=500 [C] T[6]=60.06 [C] Ts[4]=345.2 [C] Ts[6]=60.06 [C] T_o=300 [K] T_R_H=1800 [K] T_R_L=300 [K] v[1]=0.001017 [m^3/kg] v[2]=0.001014 [m^3/kg] v[3]=0.04177 [m^3/kg] v[4]=0.08968 [m^3/kg] vs[6]=6.922 [m^3/kg] W_net=1359 [kJ/kg] W_p=8.117 [kJ/kg] W_p_s=8.117 [kJ/kg] W_t_hp=294.8 [kJ/kg] W_t_lp=1072 [kJ/kg] x6s$=''x[1]=0

10-31

x[6]=0.9051

10-48

10-59 The exergy destruction associated with the heat addition process and the expansion process in Prob. 10-34 are to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-34,

kJ/kg8.3749kJ/kg1.3375

kJ/kg8.2664,kPa10KkJ/kg3870.8KkJ/kg7642.7

kJ/kg0.2902,MPa1KkJ/kg8464.6KkJ/kg5995.6

KkJ/kg6492.0

in

3

666

5

444

3

kPa10@21

qh

hPss

hPss

sss f

The exergy destruction associated with the combined pumping and the heat addition processes is

kJ/kg5.1289K1600kJ/kg3749.8

8464.67642.76492.05995.6K285

15,45130destroyed

R

R

Tq

ssssTx

The exergy destruction associated with the pumping process is

Thus,kJ/kg12895.05.1289

kJ/kg53.009.1062.10

12destroyed,destroyedheatingdestroyed,

,,,12destroyed,

xxx

Pvwwwx apspap

The exergy destruction associated with the expansion process is

kJ/kg247.9KkJ/kg7642.73870.85995.68464.6K285

036,

5634034destroyed,R

R

Tq

ssssTx

The exergy of the steam at the boiler exit is determined from

03003

03

023

030033 2ssThhqz

VssThh

where

KkJ/kg1806.0kJ/kg51.50

K285@kPa100K,285@0

K285@kPa100K,285@0

f

f

ssshhh

Thus,kJ/kg1495KkJ/kg1806.05995.6K285kJ/kg51.501.33753

10-49

10-60 The exergy destruction associated with the regenerative cycle described in Prob. 10-44 is to bedetermined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-44, qin = 2692.2 kJ/kg and qout = 1675.7 kJ/kg. Then the exergy destructionassociated with this regenerative cycle is

kJ/kg1155K1500

kJ/kg2692.2K290kJ/kg1675.7

K290inout0destroyed,

HLcycle T

qTq

Tx

10-61 The exergy destruction associated with the reheating and regeneration processes described in Prob. 10-49 are to be determined for the specified source and sink temperatures.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Problem 10-49 and the steam tables,

kJ/kg6.6687.28123.3481

KkJ/kg6492.0KkJ/kg8692.7

KkJ/kg7585.6

KkJ/kg0457.22016.0

67reheat

kPa10@21

7

65

MPa8.0@3

hhq

ssss

ss

ssy

f

f

Then the exergy destruction associated with reheat and regeneration processes are

kJ/kg47.8

kJ/kg214.3

6492.02016.017585.62016.00457.2K290

1

K1800kJ/kg668.6

7585.68692.7K290

2630

0

0

surr0gen0regendestroyed,

67,670reheatdestroyed,

syyssTT

qsmsmTsTx

Tq

ssTx

iiee

R

R

10-50

10-62 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. Thepower output from the turbine, the thermal efficiency of the plant, the exergy of the geothermal liquid atthe exit of the flash chamber, and the exergy destructions and exergy efficiencies for the flash chamber, theturbine, and the entire plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) We use properties of water for geothermal water (Tables A-4, A-5, and A-6)

kJ/kg.K6841.21661.0

kJ/kg14.990kPa500

kJ/kg.K6100.2kJ/kg14.990

0C230

2

2

12

2

1

1

1

1

sx

hhP

sh

xT

kg/s38.19kg/s)230)(1661.0(123 mxm

KkJ/kg7739.7kJ/kg3.2464

95.0kPa10

KkJ/kg8207.6kJ/kg1.2748

1kPa500

4

4

4

4

3

3

3

3

sh

xP

sh

xP

KkJ/kg8604.1kJ/kg09.640

0kPa500

6

6

6

6

sh

xP

kg/s81.19119.38230316 mmm

2

productionwell

reinjectionwell

separator

steamturbine

1

Flashchamber

65

4

3

condenser

The power output from the turbine is

kW10,842kJ/kg)3.24648.1kJ/kg)(27438.19()( 433T hhmW

We use saturated liquid state at the standard temperature for dead state properties

kJ/kg3672.0kJ/kg83.104

0C25

0

0

0

0

sh

xT

kW622,203kJ/kg)83.104.14kJ/kg)(990230()( 011in hhmE

5.3%0.0532622,203

842,10

in

outT,th E

W

(b) The specific exergies at various states are kJ/kg53.216kJ/kg.K)3672.0K)(2.6100(298kJ/kg)83.104(990.14)( 010011 ssThh

kJ/kg44.194kJ/kg.K)3672.0K)(2.6841(298kJ/kg)83.104(990.14)( 020022 ssThh

kJ/kg10.719kJ/kg.K)3672.0K)(6.8207(298kJ/kg)83.104(2748.1)( 030033 ssThh

kJ/kg05.151kJ/kg.K)3672.0K)(7.7739(298kJ/kg)83.104(2464.3)( 040044 ssThh

kJ/kg97.89kJ/kg.K)3672.0K)(1.8604(298kJ/kg)83.104(640.09)( 060066 ssThh

The exergy of geothermal water at state 6 is

kW17,257kJ/kg)7kg/s)(89.9.81191(666 mX

10-51

(c) Flash chamber:

kW5080kJ/kg)44.19453kg/s)(216.230()( 211FCdest, mX

89.8%0.89853.21644.194

1

2FCII,

(d) Turbine:

kW10,854kW10,842-kJ/kg)05.15110kg/s)(719.19.38()( T433Tdest, WmX

50.0%0.500)kJ/kg05.15110kg/s)(719.19.38(

kW842,10)( 433

TTII, m

W

(e) Plant:

kW802,49kJ/kg)53kg/s)(216.230(11Plantin, mX

kW38,960842,10802,49TPlantin,Plantdest, WXX

21.8%0.2177kW802,49kW842,10

Plantin,

TPlantII, X

W

Cogeneration

10-63C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purposeto the total energy supplied. It could be unity for a plant that does not produce any power.

10-64C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity.

10-65C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference.

10-66C Cogeneration is the production of more than one useful form of energy from the same energysource. Regeneration is the transfer of heat from the working fluid at some stage to the working fluid atsome other stage.

10-52

10-67 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The net power produced and the utilizationfactor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg38.670

kJ/kg41.19260.081.191kJ/kg0.60

mkPa1kJ1

kPa10600/kgm0.00101

/kgm00101.0kJ/kg81.191

MPa0.6@3

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

f

f

f

hh

whh

PPw

hh

v

vv

7

3

24P II P I

Processheater Condenser

BoilerTurbine

5

1

8

6

Mixing chamber:

332244

outin(steady)0

systemoutin 0

hmhmhmhmhm

EEEEE

eeii

or,

kJ/kg47.31857.690.311kJ/kg6.57

mkPa1kJ1kPa6007000/kgm0.001026

/kgm001026.0

kJ/kg90.31130

38.67050.741.19250.22

inII,45

33

454inII,

3kJ/kg90.311@4

4

33224

p

p

hf

whh

PPw

mhmhmh

f

v

vv

T

KkJ/kg8000.6kJ/kg4.3411

C500MPa7

6

6

6

6sh

TP

kJ/kg6.21531.23928201.081.191

8201.04996.7

6492.08000.6kPa10

kJ/kg6.2774MPa6.0

88

88

68

8

767

7

fgf

fg

f

hxhhs

ssx

ssP

hss

PQout·

3

5

4

10 kPa

7 MPa

0.6 MPa Qproces·

Qin·

7

6

8

2

1s

Then,

kW32,8666.210077,33

kW210.6kJ/kg6.57kg/s30kJ/kg0.60kg/s22.5

kW077,33kJ/kg6.21536.2774kg/s22.5kJ/kg6.27744.3411kg/s30

inp,outT,net

inpII,4inpI,1inp,

878766outT,

WWW

wmwmW

hhmhhmW

Also, kW782,15kJ/kg38.6706.2774kg/s7.5377process hhmQ

and

52.4%788,92

782,15866,32

kW788,9247.3184.3411kg/s30

in

processnet

565in

QQW

hhmQ

u

10-53

10-68E A large food-processing plant requires steam at a relatively high pressure, which is extracted fromthe turbine of a cogeneration plant. The rate of heat transfer to the boiler and the power output of thecogeneration plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis(a) From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm13.282

Btu/lbm29.9427.002.94Btu/lbm0.27

ftpsia5.4039Btu1

psia2)80)(/lbmft0.01623(86.01

/

/lbmft01623.0Btu/lbm02.94

psia80@3

inpI,12

3

3

121inpI,

3psia2@1

psia2@1

f

p

f

f

hh

whh

PPw

hh

v

vv

7

3

24P II P I

Processheater Condenser

BoilerTurbine

5

1

8

6

Mixing chamber:

E E E

E E

m h m h m h m h m hi i e e

in out system (steady)

in out

0

4 4 2 2 3 3

0 T

or,

Btu/lbm72.17229.343.169Btu/lbm3.29

86.0/ftpsia5.4039

Btu1psia801000/lbmft0.01664

/

/lbmft01664.0

Btu/lbm43.1695

13.282229.943

in,45

33

454inpII,

3Btu/lbm43.169@4

4

33224

pII

p

hf

whh

PPw

mhmhm

h

f

v

vv8s

7Qin·

3

5

480 psia

Qproces·

2 psia

1000psia

7s

6

8

2

1s

RBtu/lbm6535.1Btu/lbm2.1506

F1000psia1000

6

6

6

6sh

TP

Btu/lbm98.9597.10218475.002.94

8475.074444.1

17499.06535.1psia2

Btu/lbm0.1209psia80

88

88

68

8

767

7

fgsfs

fg

fss

s

s

ss

s

hxhhs

ssx

ssP

hss

P

Then, Btu/s6667Btu/lbm72.1722.1506lbm/s5565in hhmQ

(b)

kW2026Btu/s1921Btu/lbm959.981209.0lbm/s3Btu/lbm1209.01506.2lbm/s586.0

878766,outT, sssTsTT hhmhhmWW

10-54

10-69 A cogeneration plant has two modes of operation. In the first mode, all the steam leaving the turbineat a relatively high pressure is routed to the process heater. In the second mode, 60 percent of the steam isrouted to the process heater and remaining is expanded to the condenser pressure. The power produced andthe rate at which process heat is supplied in the first mode, and the power produced and the rate of processheat supplied in the second mode are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg47.65038.1009.640kJ/kg10.38

mkPa1kJ1

kPa50010,000/kgm0.001093

/kgm001093.0kJ/kg09.640

kJ/kg57.26115.1042.251kJ/kg10.15

mkPa1kJ1

kPa0210,000/kgm0.001017

/kgm001017.0kJ/kg42.251

inpII,34

33

343inpII,

3MPa5.0@3

MPa5.0@3

inpI,12

33

121inpI,

3kPa20@1

kPa20@1

whh

PPw

hh

whh

PPw

hh

f

f

f

f

v

vv

v

vv

7

3

2

4

P IIP I

Processheater Condens.

BoilerTurbine

5

1

8

6

T

Mixing chamber:E E E E E

m h m h m h m h m hi i e e

in out system (steady)

in out0

5 5 2 2 4 4

0

or, kJ/kg91.4945

47.650357.2612

5

44225 m

hmhmh

34

5

7

6

8

2

1s

KkJ/kg4219.6kJ/kg4.3242

C450MPa10

6

6

6

6sh

TP

kJ/kg0.21145.23577901.042.251

7901.00752.7

8320.04219.6kPa20

kJ/kg6.25780.21089196.009.640

9196.09603.4

8604.14219.6MPa5.0

88

88

68

8

77

77

67

7

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

When the entire steam is routed to the process heater,

kW9693

kW3319

kJ/kg09.6406.2578kg/s5

kJ/kg6.25784.3242kg/s5

377process

766outT,

hhmQ

hhmW

(b) When only 60% of the steam is routed to the process heater,

kW5816

kW4248

kJ/kg09.6406.2578kg/s3

kJ/kg0.21146.2578kg/s2kJ/kg6.25784.3242kg/s5

377process

878766outT,

hhmQ

hhmhhmW

10-55

10-70 A cogeneration plant modified with regeneration is to generate power and process heat. The massflow rate of steam through the boiler for a net power output of 15 MW is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.AnalysisFrom the steam tables (Tables A-4, A-5, and A-6),

kJ/kg73.61007.666.604kJ/kg07.6

mkPa1kJ1kPa4006000/kgm0.001084

/kgm001084.0

kJ/kg66.604

kJ/kg20.19239.081.191kJ/kg0.39

mkPa1kJ1kPa10400/kgm0.00101

/kgm00101.0kJ/kg81.191

inpII,45

33

454inpII,

3MPa4.0@4

MPa4.0@943

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

whh

PPw

hhhh

whh

PPw

hh

f

f

f

f

v

vv

v

vv

fwh4

7

3

2

9

P II P I

Processheater Condenser

BoilerTurbine

5

1

8

6

T

kJ/kg7.21281.23928097.081.191

8097.04996.7

6492.07219.6kPa10

kJ/kg7.26654.21339661.066.604

9661.01191.5

7765.17219.6MPa4.0

KkJ/kg7219.6kJ/kg9.3302

C450MPa6

88

88

68

8

77

77

67

7

6

6

6

6

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

sh

TP

5

0.4 MPa

10 kPa

6 MPa

7

6

8

23,4,9

1s

Then, per kg of steam flowing through the boiler, we have

kJ/kg852.0kJ/kg7.21287.26654.0kJ/kg7.26659.3302

4.0 8776outT, hhhhw

kJ/kg8.84523.60.852

kJ/kg6.23kJ/kg6.07kJ/kg0.394.0

4.0

inp,outT,net

inpII,inpI,inp,

www

www

Thus,

kg/s17.73kJ/kg845.8

kJ/s15,000

net

net

wWm

10-56

10-71 EES Problem 10-70 is reconsidered. The effect of the extraction pressure for removing steam fromthe turbine to be used for the process heater and open feedwater heater on the required mass flow rate is tobe investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input Data"y = 0.6 "fraction of steam extracted from turbine for feedwater heater and process heater"P[6] = 6000 [kPa] T[6] = 450 [C] P_extract=400 [kPa] P[7] = P_extractP_cond=10 [kPa]P[8] = P_condW_dot_net=15 [MW]*Convert(MW, kW)Eta_turb= 100/100 "Turbine isentropic efficiency"Eta_pump = 100/100 "Pump isentropic efficiency"P[1] = P[8] P[2]=P[7]P[3]=P[7]P[4] = P[7] P[5]=P[6]P[9] = P[7]

"Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'

h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Open Feedwater Heater analysis:"z*h[7] + (1- y)*h[2] = (1- y + z)*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)

"Process heater analysis:"(y - z)*h[7] = q_process + (y - z)*h[9] "Steady-flow conservation of energy"Q_dot_process = m_dot*(y - z)*q_process"[kW]"h[9]=enthalpy(Fluid$,P=P[9],x=0)T[9]=temperature(Fluid$,P=P[9],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[9]=entropy(Fluid$,P=P[9],x=0)

"Mixing chamber at 3, 4, and 9:"(y-z)*h[9] + (1-y+z)*h[3] = 1*h[4] "Steady-flow conservation of energy"T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Condensate leaves heater as sat. liquid at P[3]"

10-57

s[4]=entropy(Fluid$,P=P[4],h=h[4])

"Boiler condensate pump or Pump 2 analysis"v4=volume(Fluid$,P=P[4],x=0)w_pump2_s=v4*(P[5]-P[4])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[4]+w_pump2= h[5] "Steady-flow conservation of energy"s[5]=entropy(Fluid$,P=P[5],h=h[5])T[5]=temperature(Fluid$,P=P[5],h=h[5])

"Boiler analysis"q_in + h[5]=h[6]"SSSF conservation of energy for the Boiler"h[6]=enthalpy(Fluid$, T=T[6], P=P[6]) s[6]=entropy(Fluid$, T=T[6], P=P[6])

"Turbine analysis"ss[7]=s[6]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h[6]-Eta_turb*(h[6]-hs[7])"Definition of turbine efficiency for high pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])ss[8]=s[7]hs[8]=enthalpy(Fluid$,s=ss[8],P=P[8])Ts[8]=temperature(Fluid$,s=ss[8],P=P[8])h[8]=h[7]-Eta_turb*(h[7]-hs[8])"Definition of turbine efficiency for low pressure stages"T[8]=temperature(Fluid$,P=P[8],h=h[8])s[8]=entropy(Fluid$,P=P[8],h=h[8])h[6] =y*h[7] + (1- y)*h[8] + w_turb "SSSF conservation of energy for turbine"

"Condenser analysis"(1- y)*h[8]=q_out+(1- y)*h[1]"SSSF First Law for the Condenser"

"Cycle Statistics"w_net=w_turb - ((1- y)*w_pump1+ w_pump2)Eta_th=w_net/q_inW_dot_net = m_dot * w_net

Pextract[kPa]

th m[kg/s]

Qprocess[kW]

100 0.3413 15.26 9508200 0.3284 16.36 9696300 0.3203 17.12 9806400 0.3142 17.74 9882500 0.3092 18.26 9939600 0.305 18.72 9984

0 2 4 6 8 10 120

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

6000 kPa

400 kPa

10 kPa

Steam

1

2 3,4,9

5

6

7

8

10-58

100 200 300 400 500 60015

15.5

16

16.5

17

17.5

18

18.5

19

Pextract [kPa]

m [

kg/s

]

100 200 300 400 500 6009500

9600

9700

9800

9900

10000

Pextract [kPa]

Qpr

oces

s [k

W]

100 200 300 400 500 6000.305

0.31

0.315

0.32

0.325

0.33

0.335

0.34

0.345

Pextract [kPa]

th

10-59

10-72E A cogeneration plant is to generate power while meeting the process steam requirements for a certain industrial application. The net power produced, the rate of process heat supply, and the utilizationfactor of this plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis(a) From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm5.1229psia120

RBtu/lbm6348.1Btu/lbm0.1408

F800psia600

Btu/lbm49.208

737

7

6543

753

3

3

3

12

F240@1

hss

P

hhhh

sssh

TP

hhhh f

7

3 54

P

Processheater

Boiler Turbine

2

1

6

kW2260Btu/s2142Btu/lbm5.12290.1408lbm/s12

755net hhmW

(b)

Btu/s19,45049.208185.1229120.14086

117766

process

hmhmhm

hmhmQ eeii T

1

3,4,52

120 psia

600 psia

76

Btu/s19,4505.1229120.1408649.20818

776611process hmhmhmhmhmQ iiee

s(c) u = 1 since all the energy is utilized.

10-60

10-73 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The mass flow rate of steam that must besupplied by the boiler, the net power produced, and the utilization factor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

7

3

24P II P I

Processheater Condenser

BoilerTurbine

5

1

8

6 T

Qout·

3

5

4

10 kPa

7 MPa

0.6 MPa Qproces·

Qin·

7

6

8

2

1s

Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg38.670

kJ/kg40.192596.081.191kJ/kg596.0

mkPa1kJ1

kPa01600/kgm0.00101

/kgm00101.0kJ/kg81.191

MPa6.0@3

inpI,12

33

121inpI,

3kPa01@1

kPa10@1

f

f

f

hh

whh

PPw

hh

v

vv

Mixing chamber:

kJ/kg90.311)1()kJ/kg))(192.4075.0(kJ/kg))(670.3825.0( 44

442233

hh

hmhmhm

kJ/kg47.318563.690.311kJ/kg.5636

mkPa1kJ1kPa0067000/kgm0.001026

/kgm001026.0

inII,45

33

454inII,

3kJ/kg90.311@4

p

p

hf

whh

PPw

f

v

vv

KkJ/kg8000.6kJ/kg4.3411

C500MPa7

6

6

6

6sh

TP

kJ/kg9.2773MPa6.0

767

7 hss

P

kJ/kg6.2153kPa10

868

8 hss

P

10-61

kg/s4.088kJ/kg38.6709.2773kJ/s8600

7

7

377process

mm

hhmQ

This is one-fourth of the mass flowing through the boiler. Thus, the mass flow rate of steam that must besupplied by the boiler becomes

kg/s16.35kg/s)4.088(44 76 mm

(b) Cycle analysis:

kW17,919115033,18

kW6.114kJ/kg6.563kg/s16.35kJ/kg0.596kg/s4.088-16.35

kW033,18kJ/kg6.21534.3411kg/s4.088-16.35kJ/kg9.27734.3411kg/s088.4

inp,outT,net

inpII,4inpI,1inp,

868767outT,

WWW

wmwmW

hhmhhmW

(c) Then,

and

52.4%524.0581,50

8600919,17

kW581,5046.3184.3411kg/s16.35

in

processnet

565in

QQW

hhmQ

u

Combined Gas-Vapor Power Cycles

10-74C The energy source of the steam is the waste energy of the exhausted combustion gases.

10-75C Because the combined gas-steam cycle takes advantage of the desirable characteristics of the gascycle at high temperature, and those of steam cycle at low temperature, and combines them. The result is a cycle that is more efficient than either cycle executed operated alone.

10-62

10-76 A combined gas-steam power cycle is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a simple ideal Rankine cycle. The mass flow rate of the steam, the net power output, andthe thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a) The analysis of gas cycle yields

kW6447100,5547,11

kW11,547K3.6791500KkJ/kg1.005kg/s14

K3.679161K1500

kW5100K3005.662KkJ/kg1.005kg/s14

kW784,11K5.6621500KkJ/kg1.005kg/s14

K5.66216K300

gas,gas,gasnet,

87air87airgas,

4.1/4.0/1

7

878

56air56airgas,

67air67airin

4.1/4.0/1

5

656

CT

pT

kk

pC

p

kk

WWW

TTcmhhmW

PP

TT

TTcmhhmW

TTcmhhmQ

PP

TT

420 K

1500 K

STEAMCYCLE

GASCYCLE

7

8

9

5

Qin·

Qout·

6

15 kPa

10 MPa

3400 C

T

42

1

300 K

s

From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg06.23613.1094.225

kJ/kg10.12mkPa1

kJ1kPa1510,000/kgm0.001014

/kgm001014.0kJ/kg94.225

inpI,12

33

121inpI,

3kPa15@1

kPa15@1

whh

PPw

hh

f

f

v

vv

kJ/kg8.20113.23727528.094.225

7528.02522.7

7549.02141.6kPa15

KkJ/kg2141.6kJ/kg0.3097

C400MPa10

44

44

34

4

3

3

3

3

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW

kg/s1.275kg/s14kJ/kg06.2360.3097

K4203.679KkJ/kg1.005

0

air23

98air

23

98

98air23

outin(steady)0

systemoutin

mhh

TTcm

hhhh

m

hhmhhmhmhm

EEEEE

ps

seeii

(b)

kW13719.121384

kW9.12kJ/kg10.12kg/s1.275

kW1384kJ/kg5.20110.3097kg/s1.275

steamp,steamT,steamnet,

steamp,

43steamT,

WWW

wmW

hhmW

ps

s

and kW781964481371gasnet,steamnet,net WWW

(c) 66.4%kW11,784

kW7819

in

netth Q

W

10-63

10-77 [Also solved by EES on enclosed CD] A 450-MW combined gas-steam power plant is considered.The topping cycle is a gas-turbine cycle and the bottoming cycle is an ideal Rankine cycle with an openfeedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustionchamber, and the thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields (Table A-17)

kJ/kg02462 K460

kJ/kg873518325450141

5450kJ/kg421515K1400

kJ/kg56354019386114

3861kJ/kg19300K300

1212

1110

11

1010

98

9

88

1011

10

89

8

.hT

.h..PPPP

.P.hT

.h..PPP

P

.P.hT

rr

r

rr

r

From the steam tables (Tables A-4, A-5, A-6),

kJ/kg01.25259.042.251kJ/kg0.59

mkPa1kJ1

kPa20600/kgm0.001017

/kgm001017.0kJ/kg42.251

inpI,12

33

121inpI,

3kPa20@1

kPa20@1

whh

PPw

hh

f

f

v

vv

1400 K

GASCYCLE

10

11

8

Qin·

3

9

4

20 kPa Qout·

0.6 MPa 460 K

STEAMCYCLE

12

8 MPa

6

5400 C

T

7

2

1

300 K

s

kJ/kg53.67815.838.670kJ/kg8.15

mkPa1kJ1kPa6008,000/kgm0.001101

/kgm001101.0kJ/kg38.670

inpI,34

33

343inpII,

3MPa6.0@3

MPa6.0@3

whh

PPw

hh

f

f

v

vv

kJ/kg2.20955.23577821.042.251

7821.00752.7

8320.03658.6kPa20

kJ/kg1.25868.20859185.038.670

9185.08285.4

9308.13658.6MPa6.0

KkJ/kg3658.6kJ/kg4.3139

C400MPa8

77

77

57

7

66

66

56

6

5

5

5

5

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

sh

TP

Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW

10-64

steamkgair /kg8.9902.46280.73553.6784.3139

0

1211

45air

1211air45

outin

(steady)0systemoutin

hhhh

mm

hhmhhmhmhm

EE

EEE

s

seeii

(b) Noting that for the open FWH, the steady-flow energy balance equation yieldsQ W ke pe 0

326336622

outin

(steady)0systemoutin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

Thus,

kJ/kg23.9562.20951.25861792.011.25864.31391

extractedsteamoffractionthe1792.001.2521.258601.25238.670

7665

26

23

hhyhhw

hhhhy

T

kJ/kg3.44419.3005.6358.73542.1515

kJ/kg56.94815.859.01792.0123.9561

891110in,gasnet,

,,in,steamnet,

hhhhwww

wwywwww

CT

IIpIpTpT

The net work output per unit mass of gas is

kJ/kg8.54956.9483.444 99.81

steamnet,99.81

gasnet,net www

and

kW720,215kJ/kg5.63542.1515kg/s818.5

kg/s7.188kJ/kg549.7

kJ/s450,000

910airin

net

netair

hhmQ

wWm

(c) 62.5%kW720,215kW450,000

in

net

QW

th

10-65

10-78 EES Problem 10-77 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input data"T[8] = 300 [K] "Gas compressor inlet"P[8] = 14.7 [kPa] "Assumed air inlet pressure""Pratio = 14" "Pressure ratio for gas compressor"T[10] = 1400 [K] "Gas turbine inlet"T[12] = 460 [K] "Gas exit temperature from Gas-to-steam heat exchanger "P[12] = P[8] "Assumed air exit pressure"W_dot_net=450 [MW] Eta_comp = 1.0Eta_gas_turb = 1.0 Eta_pump = 1.0Eta_steam_turb = 1.0P[5] = 8000 [kPa] "Steam turbine inlet"T[5] =(400+273) "[K]" "Steam turbine inlet"P[6] = 600 [kPa] "Extraction pressure for steam open feedwater heater"P[7] = 20 [kPa] "Steam condenser pressure"

"GAS POWER CYCLE ANALYSIS"

"Gas Compressor anaysis"s[8]=ENTROPY(Air,T=T[8],P=P[8])ss9=s[8] "For the ideal case the entropies are constant across the compressor"P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit"Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp >w_comp_isen"h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor,assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s"h[8]=ENTHALPY(Air,T=T[8])hs9=ENTHALPY(Air,T=Ts9)h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming:adiabatic, ke=pe=0"T[9]=temperature(Air,h=h[9])s[9]=ENTROPY(Air,T=T[9],P=P[9])

"Gas Cycle External heat exchanger analysis"h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0"h[10]=ENTHALPY(Air,T=T[10])P[10]=P[9] "Assume process 9-10 is SSSF constant pressure"Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in

"Gas Turbine analysis"s[10]=ENTROPY(Air,T=T[10],P=P[10])ss11=s[10] "For the ideal case the entropies are constant across the turbine"P[11] = P[10] /PratioTs11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit"Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen> w_gas_turb"h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0"

10-66

hs11=ENTHALPY(Air,T=Ts11)h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming:adiabatic, ke=pe=0"T[11]=temperature(Air,h=h[11])s[11]=ENTROPY(Air,T=T[11],P=P[11])

"Gas-to-Steam Heat Exchanger""SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic,W=0, ke=pe=0"m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5]h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12])

"STEAM CYCLE ANALYSIS""Steam Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'P[1] = P[7] P[2]=P[6]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"Open Feedwater Heater analysis"y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy"P[3]=P[6]h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"T[3]=temperature(Fluid$,P=P[3],x=0)s[3]=entropy(Fluid$,P=P[3],x=0)"Boiler condensate pump or Pump 2 analysis"P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam""Steam Turbine analysis"h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s[5]=entropy(Fluid$,P=P[5],T=T[5])ss6=s[5]hs6=enthalpy(Fluid$,s=ss6,P=P[6])Ts6=temperature(Fluid$,s=ss6,P=P[6])h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])ss7=s[5]hs7=enthalpy(Fluid$,s=ss7,P=P[7])Ts7=temperature(Fluid$,s=ss7,P=P[7])h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency"T[7]=temperature(Fluid$,P=P[7],h=h[7])

10-67

s[7]=entropy(Fluid$,P=P[7],h=h[7])"SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe"h[5] = w_steam_turb + y*h[6] +(1-y)*h[7]"Steam Condenser analysis"(1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass"Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out"Cycle Statistics"MassRatio_gastosteam =m_dot_gas/m_dot_steamW_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb +m_dot_steam*w_steam_turb) "Back work ratio"W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps)W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp)NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio MassRatiogastosteam

Wnetgas[kW]

Wnetsteam[kW]

th[%]

NetWorkRatiogastosteam

10 7.108 342944 107056 59.92 3.20311 7.574 349014 100986 60.65 3.45612 8.043 354353 95647 61.29 3.70513 8.519 359110 90890 61.86 3.95114 9.001 363394 86606 62.37 4.19615 9.492 367285 82715 62.83 4.4416 9.993 370849 79151 63.24 4.68517 10.51 374135 75865 63.62 4.93218 11.03 377182 72818 63.97 5.1819 11.57 380024 69976 64.28 5.43120 12.12 382687 67313 64.57 5.685

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

s [kJ/kg-K]

T [K

]

8000 kPa

600 kPa

20 kPa

Combined Gas and Steam Power Cycle

8

9

10

11

12

1,23,4

5

6

7

Steam CycleSteam Cycle

Gas CycleGas Cycle

10-68

5 9 13 17 21 2555

57.2

59.4

61.6

63.8

66

P ratio

th[%

]

5 9 14 18 232.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

Pratio

Net

Wor

kRat

ioga

stos

team

W dot,gas / Wdot,steam vs Gas Pressure Ratio

5 9 14 18 235.0

6.0

7.0

8.0

9.0

10.0

11.0

12.0

13.0

14.0

Pratio

Mas

sRat

ioga

stos

team

Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio

10-69

10-79 A 450-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycleand the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate ofair to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of thecombined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields

kJ/kg02462 K460

kJ/kg844.958735421515860421515

kJ/kg873518325450141

5450kJ/kg421515 K1400

kJ/kg709.182019300563519300

kJ/kg56354019386114

3861kJ/kg19300 K300

1212

111010111110

1110

1110

11

1010

898989

89

98

9

88

1011

10

89

8

.hT

....hhhh

hhhh

.h..PPPP

.P.hT

./.../hhhh

hhhh

.h..PPP

P

.P.hT

sTs

T

srr

r

Css

C

srr

r T

10

Qin·

GASCYCLE

511

11s9

9s

124 STEAMCYCLE 63 6s2

8Qout· 7s 71

s

From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg01.25259.042.251kJ/kg0.59

mkPa1kJ1

kPa20600/kgm0.001017

/kgm001017.0kJ/kg42.251

inpI,12

33

121inpI,

3kPa20@1

kPa20@1

whh

PPw

hh

f

f

v

vv

kJ/kg52.67815.838.670kJ/kg8.15

mkPa1kJ1

kPa6008,000/kgm0.001101

/kgm001101.0kJ/kg38.670

inpI,34

33

343inpII,

3MPa6.0@3

MPa6.0@3

whh

PPw

hh

f

f

v

vv

KkJ/kg3658.6kJ/kg4.3139

C400MPa8

5

5

5

5sh

TP

10-70

kJ/kg3.22411.20954.313986.04.3139

kJ/kg1.20955.23577820.042.251

7820.00752.7

8320.03658.6kPa20

kJ/kg3.26639.25854.313986.04.3139

kJ/kg9.25858.20859184.038.670

9184.08285.4

9308.13658.6MPa6.0

755775

75

77

77

57

7

655665

65

66

66

56

6

sTs

T

fgfs

fg

fs

sTs

T

fgsfs

fg

fss

s

hhhhhhhh

hxhhs

ssx

ssP

hhhhhhhh

hxhhs

ssx

ssP

Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW

steamkgair /kg6.42502.46295.84452.6784.3139

0

1211

45air

1211air45

outin

(steady)0systemoutin

hhhh

mm

hhmhhmhmhm

EE

EEE

s

seeii

(b) Noting that for the open FWH, the steady-flow energy balance equation yields0pekeWQ

326336622

outin(steady)0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii

Thus,

kJ/kg5.8243.22413.26631735.013.26634.313986.01

extractedsteamoffractionthe1735.001.2523.266301.25238.670

7665

26

23

hhyhhw

hhhh

y

TT

kJ/kg56.26119.3001.70995.84442.1515

kJ/kg9.81515.859.01735.015.8241

891110in,gasnet,

IIp,Ip,inp,steamnet,

hhhhwww

wwywwww

CT

TT

The net work output per unit mass of gas is

kJ/kg55.3889.81556.261 425.61

steamnet,423.61

gasnet,net www

kg/s2.1158kJ/kg388.55

kJ/s450,000

net

netair w

Wm

and kW933,850kJ/kg1.70942.1515kg/s1158.2910airin hhmQ

(c) 48.2%kW933,850kW450,000

in

net

QW

th

10-71

10-80 EES Problem 10-79 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input data"T[8] = 300 [K] "Gas compressor inlet"P[8] = 14.7 [kPa] "Assumed air inlet pressure""Pratio = 14" "Pressure ratio for gas compressor"T[10] = 1400 [K] "Gas turbine inlet"T[12] = 460 [K] "Gas exit temperature from Gas-to-steam heat exchanger "P[12] = P[8] "Assumed air exit pressure"W_dot_net=450 [MW] Eta_comp = 0.82Eta_gas_turb = 0.86 Eta_pump = 1.0Eta_steam_turb = 0.86P[5] = 8000 [kPa] "Steam turbine inlet"T[5] =(400+273) "K" "Steam turbine inlet"P[6] = 600 [kPa] "Extraction pressure for steam open feedwater heater"P[7] = 20 [kPa] "Steam condenser pressure"

"GAS POWER CYCLE ANALYSIS"

"Gas Compressor anaysis"s[8]=ENTROPY(Air,T=T[8],P=P[8])ss9=s[8] "For the ideal case the entropies are constant across the compressor"P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9])"Ts9 is the isentropic value of T[9] at compressor exit"Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp >w_comp_isen"h[8] + w_gas_comp_isen =hs9"SSSF conservation of energy for the isentropic compressor,assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s"h[8]=ENTHALPY(Air,T=T[8])hs9=ENTHALPY(Air,T=Ts9)h[8] + w_gas_comp = h[9]"SSSF conservation of energy for the actual compressor, assuming:adiabatic, ke=pe=0"T[9]=temperature(Air,h=h[9])s[9]=ENTROPY(Air,T=T[9],P=P[9])

"Gas Cycle External heat exchanger analysis"h[9] + q_in = h[10]"SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0"h[10]=ENTHALPY(Air,T=T[10])P[10]=P[9] "Assume process 9-10 is SSSF constant pressure"Q_dot_in"MW"*1000"kW/MW"=m_dot_gas*q_in

"Gas Turbine analysis"s[10]=ENTROPY(Air,T=T[10],P=P[10])ss11=s[10] "For the ideal case the entropies are constant across the turbine"P[11] = P[10] /PratioTs11=temperature(Air,s=ss11,P=P[11])"Ts11 is the isentropic value of T[11] at gas turbine exit"Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen> w_gas_turb"h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0"

10-72

hs11=ENTHALPY(Air,T=Ts11)h[10] = w_gas_turb + h[11]"SSSF conservation of energy for the actual gas turbine, assuming:adiabatic, ke=pe=0"T[11]=temperature(Air,h=h[11])s[11]=ENTROPY(Air,T=T[11],P=P[11])

"Gas-to-Steam Heat Exchanger""SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic,W=0, ke=pe=0"m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5]h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12])

"STEAM CYCLE ANALYSIS""Steam Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'P[1] = P[7] P[2]=P[6]h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"Open Feedwater Heater analysis"y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy"P[3]=P[6]h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"T[3]=temperature(Fluid$,P=P[3],x=0)s[3]=entropy(Fluid$,P=P[3],x=0)"Boiler condensate pump or Pump 2 analysis"P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam""Steam Turbine analysis"h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s[5]=entropy(Fluid$,P=P[5],T=T[5])ss6=s[5]hs6=enthalpy(Fluid$,s=ss6,P=P[6])Ts6=temperature(Fluid$,s=ss6,P=P[6])h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])ss7=s[5]hs7=enthalpy(Fluid$,s=ss7,P=P[7])Ts7=temperature(Fluid$,s=ss7,P=P[7])h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency"T[7]=temperature(Fluid$,P=P[7],h=h[7])

10-73

s[7]=entropy(Fluid$,P=P[7],h=h[7])"SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe"h[5] = w_steam_turb + y*h[6] +(1-y)*h[7]"Steam Condenser analysis"(1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass"Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out"Cycle Statistics"MassRatio_gastosteam =m_dot_gas/m_dot_steamW_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+m_dot_steam*(w_steam_turb - w_steam_pumps)"definition of the net cycle work"Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb +m_dot_steam*w_steam_turb) "Back work ratio"W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps)W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp)NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio MassRatiogastosteam

Wnetgas[kW]

Wnetsteam[kW]

th[%]

NetWorkRatiogastosteam

6 4.463 262595 187405 45.29 1.4018 5.024 279178 170822 46.66 1.634

10 5.528 289639 160361 47.42 1.80612 5.994 296760 153240 47.82 1.93714 6.433 301809 148191 47.99 2.03715 6.644 303780 146220 48.01 2.07816 6.851 305457 144543 47.99 2.11318 7.253 308093 141907 47.87 2.17120 7.642 309960 140040 47.64 2.21322 8.021 311216 138784 47.34 2.242

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.0200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

1500

1600

s [kJ/kg-K]

T [K

]

8000 kPa

600 kPa

20 kPa

Combined Gas and Steam Power Cycle

8

9

10

11

12

1,23,4

5

6

7

Steam CycleSteam Cycle

Gas CycleGas Cycle

10-74

5 9 12 16 19 2345.0

45.5

46.0

46.5

47.0

47.5

48.0

48.5

Pratio

th [

%]

Cycle Thermal Efficiency vs Gas Cycle Pressure Ratio

5 9 12 16 19 231.4

1.5

1.6

1.7

1.8

1.9

2.0

2.1

2.2

2.3

Pratio

Net

Wor

kRat

ioga

stos

team

W dot,gas / W dot,steam vs Gas Pressure Ratio

5 9 12 16 19 234.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

Pratio

Mas

sRat

ioga

stos

team

Ratio of Gas Flow Rate to Steam Flow Rate vs Gas Pressure Ratio

10-75

10-81 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressureturbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of thecombined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) We obtain the air properties fromEES. The analysis of gas cycle is as follows

kJ/kg62.475C200

kJ/kg98.87179.7638.130480.08.1304

kJ/kg79.763kPa100

kJ/kg6456.6kPa700C950

kJ/kg8.1304C950

kJ/kg21.5570/16.29047.50316.290

/

kJ/kg47.503kPa700

kJ/kg6648.5kPa100

C15kJ/kg50.288C15

1111

109910109

109

10910

10

99

9

99

787878

78

878

8

77

7

77

hT

hhhhhhhh

hss

P

sPT

hT

hhhhhhhh

hss

P

sPT

hT

sTs

T

s

Css

C

s

80.

Combustionchamber

89

Compressor Gasturbine

710

Heatexchanger11

3Steamturbine

4

6

5 Condenserpump2

1

T

From the steam tables (Tables A-4, A-5,and A-6 or from EES),

kJ/kg37.19965.781.191kJ/kg.567

80.0/mkPa1

kJ1kPa106000/kgm0.00101

/

/kgm00101.0kJ/kg81.191

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

whh

PPw

hh

p

f

f

v

vv

kJ/kg4.23661.23929091.081.191

9091.04996.7

6492.04670.7kPa10

KkJ/kg4670.7kJ/kg5.3264

C400MPa1

66

66

56

6

5

5

5

5

fgsfs

fg

fss

s hxhhs

ssx

ssP

sh

TP

6

4s

810

1 MPa

4

5

GASCYCLE

9

10s

11

7

Qin·

8s

10 kPa

STEAMCYCLE

Qout·

6 MPa3

6s2

1

950 C

15 C

s

10-76

1.6%0158.09842.011PercentageMoisture

9842.0kJ/kg5.2546kPa10

kJ/kg0.25464.23665.326480.05.3264

6

66

6

655665

65

x

xhP

hhhhhhhh

sTs

T

(b) Noting that Q for the heat exchanger, the steady-flow energy balance equationyields

0pekeW

kJ/kg0.2965)62.47598.871)(10()5.3264()37.1995.3346()15.1( 44

1110air4523

outin

hh

hhmhhmhhm

hmhm

EE

ss

eeii

Also,

sTs

T

ss

hhhhhhhh

hssP

sh

TP

433443

43

434

4

3

3

3

3 MPa1?

MPa6

The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or usingEES from the above relations. The answer is T3 = 468.0ºC. Then, the enthalpy at state 3 becomes: h3 = 3346.5 kJ/kg

(c) kW4328kJ/kg98.8718.1304kg/s10109airgasT, hhmW

kW2687kJ/kg50.28821.557kg/s1078airgasC, hhmW

kW164126874328gasC,gasT,gasnet, WWW

kW1265kJ/kg0.25465.32640.29655.3346kg/s1.156543ssteamT, hhhhmW

kW7.8kJ/kg564.7kg/s1.15ssteamP, pumpwmW

kW12567.81265steamP,steamT,steamnet, WWW

kW289712561641steamnet,gasnet,plantnet, WWW

(d) kW7476kJ/kg21.5578.1304kg/s1089airin hhmQ

38.8%0.388kW7476kW2897

in

plantnet,th Q

W

10-77

Special Topic: Binary Vapor Cycles

10-82C Binary power cycle is a cycle which is actually a combination of two cycles; one in the hightemperature region, and the other in the low temperature region. Its purpose is to increase thermalefficiency.

10-83C Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle(cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the bottomingcycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potentialenergies, and assuming the heat exchanger is well-insulated, the steady-flow energy balance relation yields

43123142

outin

(steady)0systemoutin 0

hhmhhmorhmhmhmhm

hmhm

EE

EEE

BABABA

iiee

Thus,

mm

h hh h

A

B

3 4

2 1

10-84C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low.

10-85C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization.

10-86C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.

10-78

Review Problems

10-87 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant cc can be expressed as where d are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained.

cc g s g s g g iW Q/ n an s s g,outW Q/

Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as

cctotal

in

out

in

gg

in

g,out

in

ss

g,out

out

g,out

WQ

QQ

WQ

QQ

WQ

QQ

1

1

1

where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg,out is the heat rejected from the gas cycle and supplied to the steam cycle.

Using the relations above, the expression can be expressed asg s g s

cc

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

in

out

in

out

outg,

out

in

outg,

outg,

out

in

outg,

outg,

out

in

outg,

outg,

out

in

outg,sgsg

1

111

1111

Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combinedcycle is determined to be

cc g s g s 0 4 0 30 0 40 0 30. . . . 0.58

10-88 The thermal efficiency of a combined gas-steam power plant cc can be expressed in terms of thethermal efficiencies of the gas and the steam turbine cycles as . It is to be shown that

the value ofcc g s g s

cc is greater than either of .g s or

Analysis By factoring out terms, the relation can be expressed as cc g s g s

cc g s g s g s g

Positive since <1

g

g

( )1

or cc g s g s s g s

Positive since <1

s

s

( )1

Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbinecycles alone.

10-79

10-89 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6),

kPa2780565

5

66

66

6

6

C600

KkJ/kg5488.74996.792.06492.0kJ/kg5.23921.239292.081.191

92.0kPa10

Pss

T

sxsshxhh

xP

fgf

fgf

T

600 CTSINGLE

10 kPa

25MPa 4

3

s6

2

5

1

6

7DOUBL

10 kPa

25MPa 4

3

8

2

5

1

600 C

(b) Double Reheat :

C600and

KkJ/kg3637.6C600MPa25

5

5

34

4

33

3

TPP

ssPP

sTP

xx

s

Any pressure Px selected between the limits of 25 MPa and 2.78 MPa will satisfy the requirements, and canbe used for the double reheat pressure.

10-80

10-90E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as theworking fluid is considered. The exit temperature of the geothermal water from the vaporizer, the rate of heat rejection from the working fluid in the condenser, the mass flow rate of geothermal water at thepreheater, and the thermal efficiency of the Level I cycle of this plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine),

F267.42

2

12brinebrine

F325FBtu/lbm1.03lbm/h384,286Btu/h000,790,22T

T

TTcmQ p

(b) The rate of heat rejection from the working fluid to the air in the condenser is determined from thesteady-flow energy balance on air,

MBtu/h29.7F555.84FBtu/lbm0.24lbm/h4,195,100

89airair TTcmQ p

(c) The mass flow rate of geothermal water at the preheater is determined from the steady-flow energy balance on the geothermal water,

lbm/h187,120geo

geo

inoutgeogeo

F8.2110.154FBtu/lbm1.03Btu/h000,140,11

m

m

TTcmQ p

(d) The rate of heat input is

and

, , , ,

, ,

Q Q Q

W

in vaporizer reheater

net

Btu / h

kW

22 790 000 11140 000

33 930 000

1271 200 1071

Then,

10.8%kWh1

Btu3412.14Btu/h33,930,000

kW1071

in

netth Q

W

10-81

10-91 A steam power plant operates on the simple ideal Rankine cycle. The turbine inlet temperature, thenet power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required areto be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg2501.8kJ/kg0.2574

kJ/kg79.17404.675.168kJ/kg6.04

mkPa1kJ1

kPa7.56000/kgm0.001008

C29.40/kgm001008.0

kJ/kg75.168

kPa5.7@4

kPa5.7@4

inp,12

33

121inp,

kPa7.5@sat1

3kPa5.7@1

kPa5.7@1

g

g

f

f

sshh

whh

PPw

TT

hh

v

vv T

6 MPa

4

2

3

17.5 kPa

qout

qin

s

C1089.23

3

43

3 kJ/kg2.4852MPa6Th

ssP

(b)

kJ/kg1.22723.24054.4677kJ/kg3.240575.1680.2574kJ/kg4.467779.1742.4852

outinnet

14out

23in

qqwhhqhhq

and

48.6%kJ/kg4677.4kJ/kg2272.1

in

netth q

w

Thus,

kJ/s19,428kJ/s40,0004857.0inthnet QW

(c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of thesteam in the condenser, which is 40.29 C,

kg/s194.6C1540.29CkJ/kg4.18

kJ/s20,572

kJ/s572,20428,19000,40

outcool

netinout

TcQm

WQQ

10-82

10-92 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. The thermal efficiency of the cycle and the mass flow rate of the steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg82.15207.1575.137kJ/kg15.07

mkPa1kJ1kPa515,000/kgm0.001005

/kgm001005.0kJ/kg75.137

inp,12

33

121inp,

3kPa5@1

kPa5@1

whh

PPw

hh

f

f

v

vv T

kJ/kg9.23670.24239204.075.137

9204.09176.7

4762.07642.7kPa5

KkJ/kg7642.7kJ/kg1.3479

C500MPa1

kJ/kg3.2971MPa1

KkJ/kg9781.6kJ/kg7.3434

C500MPa5

kJ/kg4.3007MPa5

KkJ/kg3480.6kJ/kg8.3310

C500MPa15

88

88

78

8

7

7

7

7

656

6

5

5

5

5

434

4

3

3

3

3

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

hssP

sh

TP

hssP

sh

TP

1 MPa 5 MPa

6

7

5 kPa

15 MPa 4

3

8

2

5

1s

Then,

kJ/kg9.18622.22301.4093kJ/kg2.223075.1379.2367

kJ/kg1.40933.29711.34794.30077.343482.1528.3310

outinnet

18out

674523in

qqwhhq

hhhhhhq

Thus,

45.5%kJ/kg4093.1kJ/kg1862.9

in

netth q

w

(b) kg/s64.4kJ/kg1862.9

kJ/s120,000

net

net

wWm

10-83

10-93 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwaterheater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle,and the irreversibility associated with the regeneration process are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

1-y6

P II P I

Openfwh

Condenser

BoilerTurbine

4

32

1

7

5

y

T

6

7qout

3 y

4

1-y0.5 MPa

10 kPa

10 MPa

6s

5

7s

2

qin

1s

(a) From the steam tables (Tables A-4, A-5, and A-6),

/kgm001093.0kJ/kg09.640

liquidsat.MPa5.0

kJ/kg33.19252.081.191kJ/kg0.52

95.0/mkPa1

kJ1kPa10500/kgm0.00101

/

/kgm00101.0kJ/kg81.191

3MPa5.0@3

MPa5.0@33

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

f

f

p

f

f

hhP

whh

PPw

hh

vv

v

vv

kJ/kg02.65193.1009.640kJ/kg10.93

95.0/mkPa1

kJ1kPa50010,000/kgm0.001093

/

inpII,34

33

343inpII,

whh

PPw pv

kJ/kg1.26540.21089554.009.640

9554.09603.4

8604.15995.6

MPa5.0

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

66

66

56

6

5

5

5

5

fgsfs

fg

fss

s

s hxhhs

ssx

ssP

sh

TP

kJ/kg3.27981.26541.337580.01.3375

655665

65sT

sT hhhh

hhhh

10-84

kJ/kg7.20891.23927934.081.191

7934.04996.7

6492.05995.6

kPa1077

77

57

7fgsfs

fg

fss

s

s hxhhs

ssx

ssP

kJ/kg8.23467.20891.337580.01.3375

755775

75sT

sT hhhh

hhhh

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that ,Q W ke pe 0

326332266

outin

(steady)0systemoutin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,

1718.033.1923.279833.19209.640

26

23

hhhh

y

Then,

kJ/kg4.9397.17841.2724kJ/kg7.178481.1918.23461718.011

kJ/kg1.272402.6511.3375

outinnet

17out

45in

qqwhhyq

hhq

and

skg7159 /.kJ/kg939.4

kJ/s150,000

net

net

wW

m

(b) The thermal efficiency is determined from

34.5%kJ/kg2724.1kJ/kg1784.7

11in

outth q

q

Also,

KkJ/kg6492.0KkJ/kg8604.1

KkJ/kg9453.6kJ/kg3.2798

MPa5.0

kPa10@12

MPa5.0@3

66

6

f

f

sssss

shP

Then the irreversibility (or exergy destruction) associated with this regeneration process is

kJ/kg39.256492.01718.019453.61718.08604.1K303

1 2630

0surr

0gen0regen syyssTT

qsmsmTsTiL

iiee

10-85

10-94 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an openfeedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

1-y6

P II P I

Openfwh

Condenser

BoilerTurbine

4

32

1

7

5

y

T

qout

3 y

4

1-y

0.5 MPa

10 kPa

10 MPa

6

5

7

2

qin

1s

(a) From the steam tables (Tables A-4, A-5, and A-6),

/kgm001093.0kJ/kg09.640

liquidsat.MPa5.0

kJ/kg30.19250.081.191

kJ/kg0.50mkPa1

kJ1kPa10500/kgm0.00101

/kgm00101.0kJ/k81.191

3MPa5.0@3

MPa5.0@33

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

f

f

f

f

hhP

whh

PPw

ghh

vv

v

vv

kJ/kg7.20891.23927934.081.191

7934.04996.7

6492.05995.6kPa10

kJ/kg1.26540.21089554.009.640

9554.09603.4

8604.15995.6MPa5.0

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

kJ/kg47.65038.1009.640

kJ/kg.3810mkPa1

kJ1kPa50010,000/kgm0.001093

77

77

57

7

66

66

56

6

5

5

5

5

inpII,34

33

343inpII,

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

sh

TP

whh

PPw v

10-86

The fraction of steam extracted is determined from the steady-flow energy equation applied to thefeedwater heaters. Noting that Q ,0pekeW

326332266

outin(steady)0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m6 3 ). Solving for y,

1819.031.1921.265431.19209.640

26

23

hhhh

y

Then,

kJ/kg0.11727.15526.2724kJ/kg7.155281.1917.20891819.011

kJ/kg6.272447.6501.3375

outinnet

17out

45in

qqwhhyq

hhq

and skg0128 /.kJ/kg1171.9

kJ/s150,000

net

net

wWm

(b) The thermal efficiency is determined from

%0.43kJ/kg2724.7kJ/kg1552.7

11in

outth q

q

Also,

KkJ/kg6492.0

KkJ/kg8604.1KkJ/kg5995.6

kPa10@12

MPa5.0@3

56

f

f

sss

ssss

Then the irreversibility (or exergy destruction) associated with this regeneration process is

kJ/kg39.06492.01819.015995.61819.08604.1K303

1 2630

0surr

0gen0regen syyssTT

qsmsmTsTiL

iiee

10-87

10-95 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. Thefraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

/kgm001101.0kJ/k38.670

liquidsat.MPa6.0

kJ/kg53.22659.094.225kJ/kg0.59

mkPa1kJ1

kPa15600/kgm0.001014

/kgm001014.0kJ/kg94.225

3MPa6.0@3

MPa6.0@33

inpI,12

33

121inpI,

3kPa15@1

kPa15@1

f

f

f

f

ghhP

whh

PPw

hh

vv

v

vv

kJ/kg8.25183.23729665.094.225

9665.02522.7

7549.07642.7kPa15

kJ/kg2.3310MPa6.0

KkJ/kg7642.7kJ/kg1.3479

C500MPa0.1

kJ/kg8.2783MPa0.1

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

kJ/kg73.68035.1038.670kJ/kg10.35

mkPa1kJ1kPa60010,000/kgm0.001101

99

99

79

9

878

8

7

7

7

7

656

6

5

5

5

5

inpII,34

33

343inpII,

fgf

fg

f

hxhhs

ssx

ssP

hssP

sh

TP

hssP

sh

TP

whh

PPw v

7

6

1-y8

P II P I

Openfwh

Condens.

BoilerTurbine

4

32

1

9

5

y

3

48

1 MPa

0.6 MPa

15 kPa

10 MPa6

5

9

2

7

1

T

s

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

328332288

outin(steady)0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m8 3 ). Solving for y,

0.14453.2262.331053.22638.670

28

23

hhhhy

(b) The thermal efficiency is determined from

kJ/kg7.196294.2258.25181440.011kJ/kg7.33898.27831.347973.6801.3375

19out

6745in

hhyqhhhhq

and 42.1%kJ/kg3389.7kJ/kg1962.7

11in

outth q

q

10-88

10-96 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

7

6

1-y8

P II P I

Openfwh

Condenser

BoilerTurbine

4

32

1

9

5

y

T

1-y3

4 866s 8sy

5

99s

2

7

1s

(a) From the steam tables (Tables A-4, A-5, and A-6),

/kgm001101.0kJ/kg38.670

liquidsat.MPa6.0

kJ/kg54.22659.094.225kJ/kg0.59

mkPa1kJ1

kPa15600/kgm0.001014

/kgm001014.0kJ/kg94.225

3MPa6.0@3

MPa6.0@33

in,12

33

121in,

3kPa15@1

kPa15@1

f

f

pI

pI

f

f

hhP

whh

PPw

hh

vv

v

vv

kJ/kg8.2783MPa0.1

KkJ/kg5995.6kJ/kg1.3375

C500MPa10

kJ/kg73.68035.1038.670kJ/kg10.35

mkPa1kJ1kPa60010,000/kgm0.001101

656

6

5

5

5

5

inpII,34

33

343inpII,

ss

s hssP

sh

TP

whh

PPw v

kJ/kg4.28788.27831.337584.01.3375

655665

65sT

sT hhhh

hhhh

10-89

kJ/kg2.33372.33101.347984.01.3479

kJ/kg2.3310MPa6.0

KkJ/kg7642.7kJ/kg1.3479

C500MPa0.1

877887

87

878

8

7

7

7

7

sTs

T

ss

s

hhhhhhhh

hssP

shP

T

kJ/kg5.26728.25181.347984.01.3479

kJ/kg8.25183.23729665.094.225

9665.02522.7

7549.07642.7kPa15

977997

97

99

99

79

9

sTs

T

fgsfs

fg

fss

s

s

hhhhhhhh

hxhhs

ssx

ssP

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

328332288

outin

(steady)0systemoutin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m8 3 ). Solving for y,

0.142753.2263.333553.22638.670

28

23

hhhhy

(b) The thermal efficiency is determined from

kJ/kg2.209794.2255.26721427.011kJ/kg1.32954.28781.347973.6801.3375

19out

6745in

hhyq

hhhhq

and

36.4%kJ/kg3295.1kJ/kg2097.2

11in

outth q

q

10-90

10-97 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for theopen feedwater heater, the thermal efficiency of the cycle, and the net power output are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis

(a) From the steam tables (Tables A-4, A-5, and A-6),

High-PTurbine

111-y-z

13

12 z

10

4

Closedfwh

5y

P II

P I

Openfwh I

Condenser

BoilerLow-PTurbine

63

21

7

8

9

T 1 MPa

10

1115 MPa

z

8

56

3

y

4

1 - y - z5 kPa 7

0.2 MPa

0.6 MPa

12

9

13

2

1s

kJ/kg95.13720.075.137kJ/kg0.20

mkPa1kJ1

kPa5200/kgm0.001005

/km001005.0kJ/kg75.137

inpI,12

33

121inpI,

3kPa5@1

kPa5@1

whh

PPw

ghh

f

f

v

vv

/kgm100110.0kJ/kg38.670

liquidsat.MPa6.0

kJ/kg41.52070.1571.504kJ/kg15.70

mkPa1kJ1kPa20015,000/kgm0.001061

/kgm001061.0kJ/kg71.504

liquidsat.MPa2.0

3MPa6.0@6

MPa6.0@766

inpII,34

33

343inpII,

3MPa2.0@3

MPa2.0@33

f

f

f

f

hhhP

whh

PPw

hhP

vv

v

vv

KkJ/kg7642.7kJ/kg1.3479

C500MPa0.1

kJ/kg8.2820MPa0.1

KkJ/kg6796.6kJ/kg1.3583

C600MPa15

kJ/kg686.23 mkPa1kJ1kPa60015,000/kgm0.00110138.670

10

10

10

10

989

9

8

8

88

33

6566556

sh

TP

hss

P

sh

TP

PPvhhTT

10-91

kJ/kg1.23680.24239205.075.137

9205.0

9176.74762.07642.7

kPa5

kJ/kg9.3000MPa2.0

kJ/kg2.3310MPa6.0

1313

1313

1013

13

121012

12

111011

11

fgf

fg

f

hxhh

sss

x

ssP

hss

P

hss

P

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to thefeedwater heaters. Noting that Q ,0pekeW

4561145561111

outin

(steady)0systemoutin 0

hhhhyhhmhhmhmhm

EE

EEE

eeii

where y is the fraction of steam extracted from the turbine ( /m m11 5 ). Solving for y,

06287.038.6702.331041.52023.686

611

45

hhhhy

For the open FWH,

31227

3312122277

outin

(steady)0systemoutin

11

0

hzhhzyyhhmhmhmhm

hmhm

EE

EEE

eeii

where z is the fraction of steam extracted from the turbine ( /m m12 5 ) at the second stage. Solving for z,

0.116595.1370.3000

95.13738.67006287.095.13771.504

212

2723

hhhhyhh

z

(b)

kJ/kg9.17244.18303.3555kJ/kg4.183075.1370.23681165.006287.011

kJ/kg3.35558.28201.347923.6861.3583

outinnet

113out

91058in

qqwhhzyq

hhhhq

and

48.5%kJ/kg3555.3kJ/kg1830.4

11in

outth q

q

(c) kW72,447kJ/kg1724.9kg/s42netnet wmW

10-92

10-98 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the steamtables (Tables A-4, A-5, and A-6),

kJ/kg8.25183.23729665.094.225

9665.02522.7

7549.07642.7

kPa15

KkJ/kg7642.7kJ/kg1.3479

C500MPa1

kJ/kg7.2843MPa1

KkJ/kg7266.6kJ/kg5.3399

C500MPa8

kJ/kg81.503

kJ/kg94.225

99

99

89

9

8

8

8

8

767

7

6

6

6

6

C120@3

12

kPa15@1

fgf

fg

f

f

f

hxhhs

ssx

ssP

shP

hssP

sh

TP

hhhhhh

T4

7

8

P II P I

Processheater

Condenser

BoilerTurbine

5

3

2

1

9

6

T

6 88 MPa

31 MPa5 7

42The mass flow rate through the process heater is

kg/s929.2kJ/kg81.5037.2843

kJ/s7,000

37

process3 hh

Qm 15 kPa

91sAlso,

or,8.25181.3479992.27.28435.3399kJ/s3000

993.2

66

986766989766

mm

hhmhhmhhmhhmWT

It yields kg/s873.36m

and kg/s881.0992.2873.3369 mmmMixing chamber:

E E E

E E

m h m h m h m h m hi i e e

in out system (steady)

in out

0

4 4 2 2 3 3

0

or, kJ/kg60.440873.3

81.503992.294.225881.0

4

332254 m

hmhmhh

Then,

kW12,020kJ/kg2843.73479.1kg/s0.881kJ/kg440.603399.5kg/s3.873

788566in hhmhhmQ

(b) The fraction of steam extracted for process heating is

77.3%kg/s3.873kg/s2.992

total

3

mm

y

10-93

10-99 A combined gas-steam power plant is considered. The topping cycle is an ideal gas-turbine cycle and the bottoming cycle is an ideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of totalheat input, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields

kJ/kg63523 K520

kJ/kg35860356545081

5450kJ/kg421515 K1400

kJ/kg125268499231118

23111kJ/kg16290 K290

1111

109

10

99

87

8

77

910

9

78

7

.hT

.h..PPPP

.P.hT

.h..PPP

P

.P.hT

rr

r

rr

r

From the steam tables (Tables A-4, A-5, and A-6),

gwhh

PPw

hh

f

f

kJ/k95.20614.1581.191

kJ/kg15.14mkPa1

kJ1kPa1015,000/kgm0.00101

/kgm00101.0

kJ/kg81.191

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

v

vv

3 MPa

4

5

1400 K

GASCYCLE

9

10

11

7

Qin·

8

10 kPa

STEAMCYCLE

Qout·

15 MPa

3

T

62

1

290 K

s

kJ/kg8.22921.23928783.081.191

8783.04996.7

6492.02355.7kPa10

KkJ/kg2359.7kJ/kg2.3457

C500MPa3

kJ/kg7.27819.17949880.03.1008

9880.05402.3

6454.21434.6MPa3

KkJ/kg1434.6kJ/kg9.3157

C450MPa15

66

66

56

6

5

5

5

5

44

44

34

4

3

3

3

3

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

hxhhs

ssx

ssP

sh

TP

Noting that for the heat exchanger, the steady-flow energy balance equation yields0pekeWQ

kg/s262.9kg/s3063.52335.86095.2069.3157

1110

23air

1110air23outin

s

seeii

mhhhhm

hhmhhmhmhmEE

(b)

kW102.80 5kW280,352kJ/kg7.27812.3457kg/s30kJ/kg12.52642.1515kg/s262.9

45reheat89airreheatairin hhmhhmQQQ

(c)

55.6%

,

kW280,352kW124,409

11

kW409124kJ/kg81.1918.2292kg/s30kJ/kg16.29063.523kg/s262.9

in

outth

16711airsteamout,airout,out

QQ

hhmhhmQQQ s

10-94

10-100 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and thebottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, therate of total heat input, and the thermal efficiency of the combined cycle are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) The analysis of gas cycle yields (Table A-17)

kJ/kg63.523K520

kJ/kg4.95835.86042.151585.042.1515

kJ/kg35.8603.565.45081

5.450kJ/kg42.1515K1400

kJ/kg1.58580.0/16.29012.52616.290

/

kJ/kg12.526849.92311.18

2311.1kJ/kg16.290K290

1111

109910109

109

109

10

99

787878

78

87

8

77

910

9

78

7

hT

hhhhhhhh

hPP

PP

PhT

hhhhhhhh

hPPPP

PhT

sTs

T

srs

r

r

Css

C

srs

r

r

s

s

T91400 K

Qin·

GASCYCLE

1010s 38

8s 3 MPa15 MPa 5

11

10 kPa

STEAMCYCLE

Qout·

44s

290 K 276s 61

s

From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg95.20614.1581.191kJ/kg15.14

mkPa1kJ1kPa1015,000/kgm0.00101

/kgm00101.0kJ/kg81.191

inpI,12

33

121inpI,

3kPa10@1

kPa10@1

whh

PPw

hh

f

f

v

vv

kJ/kg1.28387.27819.315785.09.3157

kJ/kg7.27819.17949879.03.1008

9880.05402.3

6454.21434.6MPa3

KkJ/kg1428.6kJ/kg9.3157

C450MPa15

433443

43

44

44

34

4

3

3

3

3

sTs

T

fgsfs

fg

fss

s

hhhhhhhh

hxhhs

ssx

ssP

sh

TP

10-95

kJ/kg5.24678.22922.345785.02.3457

kJ/kg8.22921.23928782.081.191

8783.04996.7

6492.02359.7kPa10

KkJ/kg2359.7kJ/kg2.3457

C500MPa3

655665

65

66

66

56

6

5

5

5

5

sTs

T

fgsfs

fg

fss

s

hhhhhhhh

hxhhs

ssx

ssP

sh

TP

Noting that Q for the heat exchanger, the steady-flow energy balance equation yields0pekeW

kg/s203.6kg/s3063.5234.95895.2069.3157

0

1110

23air

1110air23

outin

(steady)0systemoutin

s

seeii

mhhhhm

hhmhhmhmhm

EE

EEE

(b)kW207,986kJ/kg1.28382.3457kg/s30kJ/kg1.58542.1515kg/s203.6

45reheat89airreheatairin hhmhhmQQQ

(c)

44.3%kW207,986kW115,805

11

kW115,805kJ/kg81.1915.2467kg/s30kJ/kg16.29063.523kg/s203.6

in

outth

16711airsteamout,airout,out

QQ

hhmhhmQQ sQ

10-101 It is to be shown tha the exergy destruction associated with a simple ideal Rankine cycle can be expressed as

tthinqx Carnotth,destroyed , where th is efficiency of the Rankine cycle and th, Carnot is

the efficiency of the Carnot cycle operating between the same temperature limits.Analysis The exergy destruction associated with a cycle is given on a unit mass basis as

R

R

Tq

Tx 0destroyed

where the direction of qin is determined with respect to the reservoir (positive if to the reservoir andnegative if from the reservoir). For a cycle that involves heat transfer only with a source at TH and a sink atT0, the irreversibility becomes

thCthCthth

HHH

qqTT

qqqq

TTq

Tq

TqTx

,in,in

0

in

outinin

0out

in

0

out0destroyed

11

10-96

10-102 A cogeneration plant is to produce power and process heat. There are two turbines in the cycle: a high-pressure turbine and a low-pressure turbine. The temperature, pressure, and mass flow rate of steam atthe inlet of high-pressure turbine are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg1.23446.20479.278860.09.2788

kJ/kg6.20471.23927758.081.191

7758.04996.7

6492.04675.6

kPa10

KkJ/kg4675.6kJ/kg9.2788

vaporsat.MPa4.1

544554

54

55

45

45

5

MPa4.1@4

MPa4.1@44

sTs

T

fgsfs

fg

fss

s

g

g

hhhhhhhh

hxhhs

ssx

ssP

sshhP

T

3

44

55s

2

1s

and

kg/min9.107kg/s799.1kJ/kg444.8kJ/s800

kJ/kg8.4441.23449.2788

lowturb,

IIturb,turblow

54lowturb,

wW

m

hhw

Therefore ,

kJ/kg0.28439.278815.54

kJ/kg54.15kg/s18.47kJ/s1000

=kg/min11081081000

4highturb,3

43turbhigh,

,turbhighturb,

total

hwh

hhmW

w

m

I

kg/s18.47

KkJ/kg4289.61840.49908.02835.2

9908.09.1958

96.8298.2770MPa4.1

kJ/kg8.277075.0/9.27880.28430.2843

/

44

44

34

4

433443

43

fgsfs

fg

fss

s

s

Tss

T

sxssh

hhx

ssP

hhhhhhhh

Then from the tables or the software, the turbine inlet temperature and pressure becomes

C227.5MPa2

3

3

3

3KkJ/kg4289.6

kJ/kg0.2843TP

sh

10-97

10-103 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The rate of process heat, the net power produced, and the utilization factor of the plant are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the steam tables (Tables A-4, A-5, and A-6),

kJ/kg47.908

kJ/kg71.25329.242.251kJ/kg.292

88.0/mkPa1

kJ1kPa022000/kgm0.001017

/

/kgm001017.0kJ/kg42.251

MPa2@3

inpI,12

33

121inpI,

3kPa02@1

kPa20@1

f

p

f

f

hh

whh

PPw

hh

v

vv

Mixing chamber:

kJ/kg81.491kg/s)11()kJ/kg)71kg/s)(253.411(kJ/kg)47kg/s)(908.4( 44

442233

hh

hmhmhm 24P P I

Conden.

Boiler

3

Processheater

5

1

6

7Turbine

8

kJ/kg02.49921.781.491kJ/kg.217

88.0/mkPa1

kJ1kPa00208000/kgm0.001058

/

/kgm001058.0

inII,45

33

454inII,

3kJ/kg81.491@4

p

pp

hf

whh

PPw

f

v

vv

T

Qout·

3

5

4

20 kPa

8 MPa

2 MPa Qproces·

Qin·

7

6

8s 8

2

1

KkJ/kg7266.6kJ/kg5.3399

C500MPa8

6

6

6

6sh

TP

kJ/kg4.3000MPa2

767

7sh

ssP

s

kJ/kg3.30484.30005.339988.05.3399766776

76sT

sT hhhh

hhhh

kJ/kg5.2215kPa20

868

8sh

ssP

kJ/kg6.23575.22155.339988.05.3399866886

86sT

sT hhhh

hhhh

Then, kW8559kJ/kg47.9083.3048kg/s4377process hhmQ(b) Cycle analysis:

kW8603958698

kW95kJ/kg7.21kg/s11kJ/kg2.29kg/s7

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in

processnet

565in

Q

QW

hhmQ

u

10-98

10-104 EES The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to beinvestigated.Analysis The problem is solved using EES, and the solution is given below.

function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''

if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'

end

P[3] = 5000 [kPa] T[3] = 500 [C] "P[4] = 5 [kPa]"Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"

"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4]"SSSF First Law for the turbine"x4s$=x4$(x[4])

"Boiler analysis"Q_in + h[2]=h[3]"SSSF First Law for the Boiler"

"Condenser analysis"h[4]=Q_out+h[1]"SSSF First Law for the Condenser"

"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in

10-99

th P4

[kPa]Wnet

[kJ/kg]x4 Qin

[kJ/kg]Qout

[kJ/kg]0.3956 5 1302 0.8212 3292 19900.3646 15 1168 0.8581 3204 20360.3484 25 1100 0.8772 3158 20570.3371 35 1054 0.8905 3125 20720.3283 45 1018 0.9009 3100 20820.321 55 988.3 0.9096 3079 2091

0.3147 65 963.2 0.917 3061 20980.3092 75 941.5 0.9235 3045 21040.3042 85 922.1 0.9293 3031 21090.2976 100 896.5 0.9371 3012 2116

0 2 4 6 8 10 120

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

5000 kPa

5 kPa

Steam

1

3

4

2

0 20 40 60 80 100800

900

1000

1100

1200

1300

1400

P[4] [kPa]

Wne

t [k

J/kg

]

10-100

0 20 40 60 80 1000.28

0.3

0.32

0.34

0.36

0.38

0.4

P[4] [kPa]

th

0 20 40 60 80 1000.82

0.84

0.86

0.88

0.9

0.92

0.94

P[4] [kPa]

x[4]

10-101

10-105 EES The effect of the boiler pressure on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''

if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'

end

{P[3] = 20000 [kPa]}T[3] = 500 [C] P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"

"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4] "SSSF First Law for the turbine"x4s$=x4$(x[4])

"Boiler analysis"Q_in + h[2]=h[3] "SSSF First Law for the Boiler"

"Condenser analysis"h[4]=Q_out+h[1] "SSSF First Law for the Condenser"

"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in

10-102

th Wnet

[kJ/kg]x4 P3

[kPa]Qin

[kJ/kg]Qout

[kJ/kg]Wp

[kJ/kg]Wt

[kJ/kg]0.2792 919.1 0.9921 500 3292 2373 0.495 919.60.3512 1147 0.886 2667 3266 2119 2.684 11500.3752 1216 0.8462 4833 3240 2024 4.873 12210.3893 1251 0.8201 7000 3213 1962 7.062 12580.399 1270 0.8001 9167 3184 1914 9.251 12800.406 1281 0.7835 11333 3155 1874 11.44 1292

0.4114 1286 0.769 13500 3125 1840 13.63 12990.4156 1286 0.756 15667 3094 1808 15.82 13020.4188 1283 0.744 17833 3062 1780 18.01 13010.4214 1276 0.7328 20000 3029 1753 20.2 1297

0 2 4 6 8 10 120

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

20000 kPa

10 kPa

Steam

1

2

3

4

0 4000 8000 12000 16000 20000900

950

1000

1050

1100

1150

1200

1250

1300

P[3] [kPa]

Wne

t [k

J/kg

]

10-103

0 4000 8000 12000 16000 200000.26

0.28

0.3

0.32

0.34

0.36

0.38

0.4

0.42

0.44

P[3] [kPa]

th

0 4000 8000 12000 16000 200000.7

0.75

0.8

0.85

0.9

0.95

1

P[3] [kPa]

x[4]

10-104

10-106 EES The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$=''

if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)'

end

P[3] = 3000 [kPa] {T[3] = 600 [C]}P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"

"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[4] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])x[4]=quality(Fluid$,h=h[4],P=P[4])h[3] =W_t+h[4]"SSSF First Law for the turbine"x4s$=x4$(x[4])

"Boiler analysis"Q_in + h[2]=h[3]"SSSF First Law for the Boiler"

"Condenser analysis"h[4]=Q_out+h[1]"SSSF First Law for the Condenser"

"Cycle Statistics"W_net=W_t-W_pEta_th=W_net/Q_in

10-105

0 2 4 6 8 10 120

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

3000 kPa

10 kPa

Steam

1

2

3

4

T3

[C]th Wnet

[kJ/kg]x4

250 0.3241

862.8 0.752

344.4 0.3338

970.6 0.81

438.9 0.3466

1083 0.8536

533.3 0.3614

1206 0.8909

627.8 0.3774

1340 0.9244

722.2 0.3939

1485 0.955

816.7 0.4106

1639 0.9835

911.1 0.4272

1803 100

1006 0.4424

1970 100

1100 0.456 2139 100

200 300 400 500 600 700 800 900 1000 11000.32

0.34

0.36

0.38

0.4

0.42

0.44

0.46

T[3] [C]

th

10-106

200 300 400 500 600 700 800 900 1000 1100750

1050

1350

1650

1950

2250

T[3] [C]

Wne

t [k

J/kg

]

10-107

10-107 EES The effect of reheat pressure on the performance an ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''

if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)'

end

P[6] = 10 [kPa] P[3] = 15000 [kPa] T[3] = 500 [C] "P[4] = 3000 [kPa]"T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"

"Pump analysis"Fluid$='Steam_IAPWS'P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine"

"Low Pressure Turbine analysis"P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"

10-108

h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"x[6]=quality(Fluid$,h=h[6],P=P[6])

"Boiler analysis"Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler"

"Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])

"Cycle Statistics"W_net=W_t_hp+W_t_lp-W_pEta_th=W_net/Q_in

P4

[kPa]th Wnet

[kJ/kg]X6

500 0.4128 1668 0.99211833 0.4253 1611 0.91023167 0.4283 1567 0.87474500 0.4287 1528 0.85115833 0.428 1492 0.83327167 0.4268 1458 0.81848500 0.4252 1426 0.80589833 0.4233 1395 0.7947

11167 0.4212 1366 0.784712500 0.4189 1337 0.7755

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100

200

300

400

500

600

700

s [kJ/kg-K]

T[C]

8000 kPa

3000 kPa

20 kPa

3

4

5

6

Ideal Rankine cycle with reheat

1,2

10-109

0 2000 4000 6000 8000 10000 12000 140001300

1350

1400

1450

1500

1550

1600

1650

1700

P[4] [kPa]

Wnet

[kJ/kg]

0 2000 4000 6000 8000 10000 12000 140000.412

0.414

0.416

0.418

0.42

0.422

0.424

0.426

0.428

0.43

P[4] [kPa]

th

0 2000 4000 6000 8000 10000 12000 140000.775

0.8

0.825

0.85

0.875

0.9

0.925

0.95

0.975

1

P[4] [kPa]

x[6]

10-110

10-108 EES The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$=''

if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)'

end

Procedure Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6)P3=P[3]T5=T[5]h4=h[4]Q_in_reheat =0 W_t_lp = 0 R_P=(1/Pratio)^(1/(NoRHStages+1))

imax:=NoRHStages - 1 i:=0

REPEATi:=i+1

P4 = P3*R_P

P5=P4P6=P5*R_P

Fluid$='Steam_IAPWS's5=entropy(Fluid$,T=T5,P=P5)h5=enthalpy(Fluid$,T=T5,P=P5)s_s6=s5hs6=enthalpy(Fluid$,s=s_s6,P=P6)Ts6=temperature(Fluid$,s=s_s6,P=P6)vs6=volume(Fluid$,s=s_s6,P=P6)"Eta_t=(h5-h6)/(h5-hs6)""Definition of turbine efficiency"h6=h5-Eta_t*(h5-hs6)W_t_lp=W_t_lp+h5-h6"SSSF First Law for the low pressure turbine"x6=QUALITY(Fluid$,h=h6,P=P6)Q_in_reheat =Q_in_reheat + (h5 - h4) P3=P4

UNTIL (i>imax)

END

"NoRHStages = 2"P[6] = 10"kPa"P[3] = 15000"kPa"P_extract = P[6] "Select a lower limit on the reheat pressure"T[3] = 500"C"T[5] = 500"C"Eta_t = 1.0 "Turbine isentropic efficiency"Eta_p = 1.0 "Pump isentropic efficiency"Pratio = P[3]/P_extract

10-111

P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa"

Fluid$='Steam_IAPWS'

"Pump analysis"P[1] = P[6] P[2]=P[3]x[1]=0 "Sat'd liquid"h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"W_p=W_p_s/Eta_ph[2]=h[1]+W_p "SSSF First Law for the pump"v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"High Pressure Turbine analysis"h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency"T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,h=h[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine"

"Low Pressure Turbine analysis"Call Reheat(P[3],T[3],T[5],h[4],NoRHStages,Pratio,Eta_t:Q_in_reheat,W_t_lp,h6)h[6]=h6

{P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency"h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid$,h=h[6],P=P[6])W_t_lp_total = NoRHStages*W_t_lpQ_in_reheat = NoRHStages*(h[5] - h[4])}

"Boiler analysis"Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler"Q_in = Q_in_boiler+Q_in_reheat

"Condenser analysis"h[6]=Q_out+h[1]"SSSF First Law for the Condenser"T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])

10-112

x[6]=QUALITY(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])

"Cycle Statistics"W_net=W_t_hp+W_t_lp - W_p Eta_th=W_net/Q_in

th NoRHStages

Qin

[kJ/kg]Wnet

[kJ/kg]0.409

71 4085 1674

0.4122

2 4628 1908

0.4085

3 5020 2051

0.4018

4 5333 2143

0.3941

5 5600 2207

0.386 6 5838 22530.377

97 6058 2289

0.3699

8 6264 2317

0.3621

9 6461 2340

0.3546

10 6651 2358

1 2 3 4 5 6 7 8 9 101600

1700

1800

1900

2000

2100

2200

2300

2400

NoRHStages

Wne

t[k

J/kg

]

1 2 3 4 5 6 7 8 9 100.35

0.36

0.37

0.38

0.39

0.4

0.41

0.42

NoRHStages

th

10-113

1 2 3 4 5 6 7 8 9 104000

4500

5000

5500

6000

6500

7000

NoRHStages

Qin

[kJ

/kg]

10-114

10-109 EES The effect of extraction pressure on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input Data"P[5] = 15000 [kPa] T[5] = 600 [C] P_extract=1400 [kPa] P[6] = P_extract P_cond=10 [kPa] P[7] = P_cond Eta_turb= 1.0 "Turbine isentropic efficiency"Eta_pump = 1.0 "Pump isentropic efficiency"P[1] = P[7] P[2]=P[6]P[3]=P[6]P[4] = P[5]

"Condenser exit pump or Pump 1 analysis"Fluid$='Steam_IAPWS'h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"h[1]+w_pump1= h[2] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Open Feedwater Heater analysis:"y*h[6] + (1- y)*h[2] = 1*h[3] "Steady-flow conservation of energy"h[3]=enthalpy(Fluid$,P=P[3],x=0)T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]"s[3]=entropy(Fluid$,P=P[3],x=0)

"Boiler condensate pump or Pump 2 analysis"v3=volume(Fluid$,P=P[3],x=0)w_pump2_s=v3*(P[4]-P[3])"SSSF isentropic pump work assuming constant specific volume"w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency"h[3]+w_pump2= h[4] "Steady-flow conservation of energy"s[4]=entropy(Fluid$,P=P[4],h=h[4])T[4]=temperature(Fluid$,P=P[4],h=h[4])

"Boiler analysis"q_in + h[4]=h[5]"SSSF conservation of energy for the Boiler"h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) s[5]=entropy(Fluid$, T=T[5], P=P[5])

"Turbine analysis"ss[6]=s[5]hs[6]=enthalpy(Fluid$,s=ss[6],P=P[6])Ts[6]=temperature(Fluid$,s=ss[6],P=P[6])h[6]=h[5]-Eta_turb*(h[5]-hs[6])"Definition of turbine efficiency for high pressure stages"T[6]=temperature(Fluid$,P=P[6],h=h[6])s[6]=entropy(Fluid$,P=P[6],h=h[6])

10-115

ss[7]=s[6]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h[6]-Eta_turb*(h[6]-hs[7])"Definition of turbine efficiency for low pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])h[5] =y*h[6] + (1- y)*h[7] + w_turb "SSSF conservation of energy for turbine"

"Condenser analysis"(1- y)*h[7]=q_out+(1- y)*h[1]"SSSF First Law for the Condenser"

"Cycle Statistics"w_net=w_turb - ((1- y)*w_pump1+ w_pump2) Eta_th=w_net/q_in

th Pextract

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0.4456 50 1438 32270.4512 100 1421 31500.4608 500 1349 29270.4629 1000 1298 28050.4626 2000 1230 26590.4562 5000 1102 24160.4511 7000 1040 23050.4434 10000 961.4 21680.4369 12500 903.5 2068

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10-117

10-110 EES The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

Procedure Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net)

Fluid$='Steam_IAPWS'Tcond = temperature(Fluid$,P=P_cond,x=0)Tboiler = temperature(Fluid$,P=P[5],x=0)P[7] = P_cond s[5]=entropy(Fluid$, T=T[5], P=P[5]) h[5]=enthalpy(Fluid$, T=T[5], P=P[5]) h[1]=enthalpy(Fluid$, P=P[7],x=0)

P4[1] = P[5] "NOTICE THIS IS P4[i] WITH i = 1"

DELTAT_cond_boiler = Tboiler - Tcond

If NoFWH = 0 Then

"the following are h7, h2, w_net, and q_in for zero feedwater heaters, NoFWH = 0"h7=enthalpy(Fluid$, s=s[5],P=P[7])

h2=h[1]+volume(Fluid$, P=P[7],x=0)*(P[5] - P[7])/Eta_pump w_net = Eta_turb*(h[5]-h7)-(h2-h[1]) q_in = h[5] - h2

else

i=0REPEATi=i+1"The following maintains the same temperature difference between any two regeneration stages."T_FWH[i] = (NoFWH +1 - i)*DELTAT_cond_boiler/(NoFWH + 1)+Tcond"[C]"P_extract[i] = pressure(Fluid$,T=T_FWH[i],x=0)"[kPa]"P3[i]=P_extract[i]P6[i]=P_extract[i]If i > 1 then P4[i] = P6[i - 1]

UNTIL i=NoFWH

P4[NoFWH+1]=P6[NoFWH]h4[NoFWH+1]=h[1]+volume(Fluid$, P=P[7],x=0)*(P4[NoFWH+1] - P[7])/Eta_pump

i=0REPEATi=i+1

"Boiler condensate pump or the Pumps 2 between feedwater heaters analysis"h3[i]=enthalpy(Fluid$,P=P3[i],x=0)v3[i]=volume(Fluid$,P=P3[i],x=0)w_pump2_s=v3[i]*(P4[i]-P3[i])"SSSF isentropic pump work assuming constant specific volume"w_pump2[i]=w_pump2_s/Eta_pump "Definition of pump efficiency"h4[i]= w_pump2[i] +h3[i] "Steady-flow conservation of energy"s4[i]=entropy(Fluid$,P=P4[i],h=h4[i])T4[i]=temperature(Fluid$,P=P4[i],h=h4[i])Until i = NoFWH

10-118

i=0REPEATi=i+1"Open Feedwater Heater analysis:"{h2[i] = h6[i]}s5[i] = s[5]ss6[i]=s5[i]hs6[i]=enthalpy(Fluid$,s=ss6[i],P=P6[i])Ts6[i]=temperature(Fluid$,s=ss6[i],P=P6[i])h6[i]=h[5]-Eta_turb*(h[5]-hs6[i])"Definition of turbine efficiency for high pressure stages"If i=1 then y[1]=(h3[1] - h4[2])/(h6[1] - h4[2]) "Steady-flow conservation of energy for the FWH"If i > 1 then js = i -1 j = 0 sumyj = 0 REPEAT j = j+1 sumyj = sumyj + y[ j ] UNTIL j = jsy[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1])

ENDIFT3[i]=temperature(Fluid$,P=P3[i],x=0) "Condensate leaves heater as sat. liquid at P[3]"s3[i]=entropy(Fluid$,P=P3[i],x=0)

"Turbine analysis"T6[i]=temperature(Fluid$,P=P6[i],h=h6[i])s6[i]=entropy(Fluid$,P=P6[i],h=h6[i])yh6[i] = y[i]*h6[i]UNTIL i=NoFWHss[7]=s6[i]hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7])Ts[7]=temperature(Fluid$,s=ss[7],P=P[7])h[7]=h6[i]-Eta_turb*(h6[i]-hs[7])"Definition of turbine efficiency for low pressure stages"T[7]=temperature(Fluid$,P=P[7],h=h[7])s[7]=entropy(Fluid$,P=P[7],h=h[7])

sumyi = 0 sumyh6i = 0 wp2i = W_pump2[1] i=0REPEATi=i+1sumyi = sumyi + y[i]sumyh6i = sumyh6i + yh6[i]If NoFWH > 1 then wp2i = wp2i + (1- sumyi)*W_pump2[i]UNTIL i = NoFWH

"Condenser Pump---Pump_1 Analysis:"P[2] = P6 [ NoFWH] P[1] = P_cond h[1]=enthalpy(Fluid$,P=P[1],x=0) {Sat'd liquid}v1=volume(Fluid$,P=P[1],x=0)s[1]=entropy(Fluid$,P=P[1],x=0)T[1]=temperature(Fluid$,P=P[1],x=0)w_pump1_s=v1*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume"w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency"

10-119

h[2]=w_pump1+ h[1] "Steady-flow conservation of energy"s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])

"Boiler analysis"q_in = h[5] - h4[1]"SSSF conservation of energy for the Boiler"w_turb = h[5] - sumyh6i - (1- sumyi)*h[7] "SSSF conservation of energy for turbine"

"Condenser analysis"q_out=(1- sumyi)*(h[7] - h[1])"SSSF First Law for the Condenser"

"Cycle Statistics"w_net=w_turb - ((1- sumyi)*w_pump1+ wp2i)

endifEND

"Input Data"NoFWH = 2 P[5] = 15000 [kPa] T[5] = 600 [C] P_cond=5 [kPa]

Eta_turb= 1.0 "Turbine isentropic efficiency"Eta_pump = 1.0 "Pump isentropic efficiency"P[1] = P_condP[4] = P[5]

"Condenser exit pump or Pump 1 analysis"Call Reheat(NoFwh,T[5],P[5],P_cond,Eta_turb,Eta_pump:q_in,w_net)Eta_th=w_net/q_in

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[kJ/kg]0 0.4466 1532 34301 0.4806 1332 27712 0.4902 1243 25363 0.4983 1202 24114 0.5036 1175 23335 0.5073 1157 22806 0.5101 1143 22407 0.5123 1132 22108 0.5141 1124 21869 0.5155 1117 2167

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10-121

Fundamentals of Engineering (FE) Exam Problems

10-111 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturationdome between the pressure limits of 8 MPa and 20 kPa. Water changes from saturated liquid to saturatedvapor during the heat addition process. The net work output of this cycle is(a) 494 kJ/kg (b) 975 kJ/kg (c) 596 kJ/kg (d) 845 kJ/kg (e) 1148 kJ/kg

Answer (c) 596 kJ/kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

P1=8000 "kPa"P2=20 "kPa"h_fg=ENTHALPY(Steam_IAPWS,x=1,P=P1)-ENTHALPY(Steam_IAPWS,x=0,P=P1)T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1)+273T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273q_in=h_fgEta_Carnot=1-T2/T1w_net=Eta_Carnot*q_in

"Some Wrong Solutions with Common Mistakes:"W1_work = Eta1*q_in; Eta1=T2/T1 "Taking Carnot efficiency to be T2/T1"W2_work = Eta2*q_in; Eta2=1-(T2-273)/(T1-273) "Using C instead of K"W3_work = Eta_Carnot*ENTHALPY(Steam_IAPWS,x=1,P=P1) "Using h_g instead of h_fg"W4_work = Eta_Carnot*q2; q2=ENTHALPY(Steam_IAPWS,x=1,P=P2)-ENTHALPY(Steam_IAPWS,x=0,P=P2) "Using h_fg at P2"

10-112 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa, with a turbine inlet temperature of 600 C. Disregarding the pump work, the cycle efficiency is(a) 24% (b) 37% (c) 52% (d) 63% (e) 71%

Answer (b) 37%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

P1=10 "kPa"P2=3000 "kPa"P3=P2P4=P1T3=600 "C"s4=s3h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1) "kJ/kg"h2=h1+w_pumph3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)

10-122

h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)q_in=h3-h2q_out=h4-h1Eta_th=1-q_out/q_in

"Some Wrong Solutions with Common Mistakes:"W1_Eff = q_out/q_in "Using wrong relation"W2_Eff = 1-(h44-h1)/(h3-h2); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Using h_g for h4"W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_Eff = (h3-h4)/q_in "Disregarding pump work"

10-113 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600 C. The mass fraction of steam that condenses at the turbine exit is(a) 6% (b) 9% (c) 12% (d) 15% (e) 18%

Answer (c) 12%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

P1=10 "kPa"P2=5000 "kPa"P3=P2P4=P1T3=600 "C"s4=s3h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)x4=QUALITY(Steam_IAPWS,s=s4,P=P4)moisture=1-x4

"Some Wrong Solutions with Common Mistakes:"W1_moisture = x4 "Taking quality as moisture"W2_moisture = 0 "Assuming superheated vapor"

10-114 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10kPa and 10 MPa, with a turbine inlet temperature of 600 C. The rate of heat transfer in the boiler is 800 kJ/s. Disregarding the pump work, the power output of this plant is(a) 243 kW (b) 284 kW (c) 508 kW (d) 335 kW (e) 800 kW

Answer (d) 335 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

P1=10 "kPa"P2=10000 "kPa"

10-123

P3=P2P4=P1T3=600 "C"s4=s3Q_rate=800 "kJ/s"m=Q_rate/q_inh1=ENTHALPY(Steam_IAPWS,x=0,P=P1)h2=h1 "pump work is neglected""v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1)h2=h1+w_pump"h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)q_in=h3-h2W_turb=m*(h3-h4)"Some Wrong Solutions with Common Mistakes:"W1_power = Q_rate "Assuming all heat is converted to power"W3_power = Q_rate*Carnot; Carnot = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_power = m*(h3-h44); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Taking h4=h_g"

10-115 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulatedheat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Waterenters the heat exchanger at 200 C and 8 MPa and leaves at 350 C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heatexchanger becomes(a) 11 kg/s (b) 24 kg/s (c) 46 kg/s (d) 53 kg/s (e) 60 kg/s

Answer (a) 11 kg/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

m_gas=60 "kg/s"Cp=1.005 "kJ/kg.K"T3=800 "K"T4=400 "K"Q_gas=m_gas*Cp*(T3-T4)P1=8000 "kPa"T1=200 "C"P2=8000 "kPa"T2=350 "C"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)Q_steam=m_steam*(h2-h1)Q_gas=Q_steam

"Some Wrong Solutions with Common Mistakes:"m_gas*Cp*(T3 -T4)=W1_msteam*4.18*(T2-T1) "Assuming no evaporation of liquid water"m_gas*Cv*(T3 -T4)=W2_msteam*(h2-h1); Cv=0.718 "Using Cv for air instead of Cp"W3_msteam = m_gas "Taking the mass flow rates of two fluids to be equal"m_gas*Cp*(T3 -T4)=W4_msteam*(h2-h11); h11=ENTHALPY(Steam_IAPWS,x=0,P=P1) "Takingh1=hf@P1"

10-124

10-116 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, withreheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500 C, and theenthalpy of steam is 3185 kJ/kg at the exit of the high-pressure turbine, and 2247 kJ/kg at the exit of thelow-pressure turbine. Disregarding the pump work, the cycle efficiency is(a) 29% (b) 32% (c) 36% (d) 41% (e) 49%

Answer (d) 41%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

P1=10 "kPa"P2=8000 "kPa"P3=P2P4=4000 "kPa"P5=P4P6=P1T3=500 "C"T5=500 "C"s4=s3s6=s5h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)h2=h1h44=3185 "kJ/kg - for checking given data"h66=2247 "kJ/kg - for checking given data"h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3)s3=ENTROPY(Steam_IAPWS,T=T3,P=P3)h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4)h5=ENTHALPY(Steam_IAPWS,T=T5,P=P5)s5=ENTROPY(Steam_IAPWS,T=T5,P=P5)h6=ENTHALPY(Steam_IAPWS,s=s6,P=P6)q_in=(h3-h2)+(h5-h4)q_out=h6-h1Eta_th=1-q_out/q_in

"Some Wrong Solutions with Common Mistakes:"W1_Eff = q_out/q_in "Using wrong relation"W2_Eff = 1-q_out/(h3-h2) "Disregarding heat input during reheat"W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency"W4_Eff = 1-q_out/(h5-h2) "Using wrong relation for q_in"

10-125

10-117 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 0.5 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ/kg and the enthalpy of extracted steam is 2665 kJ/kg, the mass fraction of steam extracted from theturbine is (a) 4% (b) 10% (c) 16% (d) 27% (e) 12%

Answer (c) 16%

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

h_feed=252 "kJ/kg"h_extracted=2665 "kJ/kg"P3=500 "kPa"h3=ENTHALPY(Steam_IAPWS,x=0,P=P3)"Energy balance on the FWH"h3=x_ext*h_extracted+(1-x_ext)*h_feed

"Some Wrong Solutions with Common Mistakes:"W1_ext = h_feed/h_extracted "Using wrong relation"W2_ext = h3/(h_extracted-h_feed) "Using wrong relation"W3_ext = h_feed/(h_extracted-h_feed) "Using wrong relation"

10-118 Consider a steam power plant that operates on the regenerative Rankine cycle with one openfeedwater heater. The enthalpy of the steam is 3374 kJ/kg at the turbine inlet, 2797 kJ/kg at the location ofbleeding, and 2346 kJ/kg at the turbine exit. The net power output of the plant is 120 MW, and the fractionof steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rateof steam at the turbine inlet is(a) 117 kg/s (b) 126 kg/s (c) 219 kg/s (d) 288 kg/s (e) 679 kg/s

Answer (b) 126 kg/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

h_in=3374 "kJ/kg"h_out=2346 "kJ/kg"h_extracted=2797 "kJ/kg"Wnet_out=120000 "kW"x_bleed=0.172w_turb=(h_in-h_extracted)+(1-x_bleed)*(h_extracted-h_out)m=Wnet_out/w_turb"Some Wrong Solutions with Common Mistakes:"W1_mass = Wnet_out/(h_in-h_out) "Disregarding extraction of steam"W2_mass = Wnet_out/(x_bleed*(h_in-h_out)) "Assuming steam is extracted at trubine inlet"W3_mass = Wnet_out/(h_in-h_out-x_bleed*h_extracted) "Using wrong relation"

10-126

10-119 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbineinlet state the same, (select the correct statement)

(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease.(d) the moisture content at turbine exit will decrease.(e) the pump work input will decrease.

Answer (b) the amount of heat rejected will decrease.

10-120 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam issuperheated to a higher temperature, (select the correct statement)

(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease.(d) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.

Answer (d) the moisture content at turbine exit will decrease.

10-121 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle ismodified with reheating, (select the correct statement)

(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.

Answer (d) the moisture content at turbine exit will decrease.

10-122 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle ismodified with regeneration that involves one open feed water heater, (select the correct statement per unitmass of steam flowing through the boiler)

(a) the turbine work output will decrease.(b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease.(d) the quality of steam at turbine exit will decrease.(e) the amount of heat input will increase.

Answer (a) the turbine work output will decrease.

10-127

10-123 Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450 C at a rate of 20 kg/s and expands to a pressure of 0.4 MPa. At this pressure, 60% of the steam isextracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steamis used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with thefeedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in thecondenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabaticcondenser at a rate of 463 kg/s.

h1 = 191.81 h2 = 192.20 h3 = h4 = h9 = 604.66 h5 = 610.73 h6 = 3302.9 h7 = h8 = h10 = 2665.6 h11 = 2128.8

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3

2

9

P II P I

Processheater Condenser

BoilerTurbine

5

1

11

6

1. The total power output of the turbine is(a) 17.0 MW (b) 8.4 MW (c) 12.2 MW (d) 20.0 MW (e) 3.4 MW

Answer (a) 17.0 MW

2. The temperature rise of the cooling water from the river in the condenser is(a) 8.0 C (b) 5.2 C (c) 9.6 C (d) 12.9 C (e) 16.2 C

Answer (a) 8.0 C

3. The mass flow rate of steam through the process heater is(a) 1.6 kg/s (b) 3.8 kg/s (c) 5.2 kg/s (d) 7.6 kg/s (e) 10.4 kg/s

Answer (e) 10.4 kg/s

4. The rate of heat supply from the process heater per unit mass of steam passing through it is(a) 246 kJ/kg (b) 893 kJ/kg (c) 1344 kJ/kg (d) 1891 kJ/kg (e) 2060 kJ/kg

Answer (e) 2060 kJ/kg

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5. The rate of heat transfer to the steam in the boiler is (a) 26.0 MJ/s (b) 53.8 MJ/s (c) 39.5 MJ/s (d) 62.8 MJ/s (e) 125.4 MJ/s

Answer (b) 53.8 MJ/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).

Note: The solution given below also evaluates all enthalpies given on the figure.

P1=10 "kPa"P11=P1P2=400 "kPa"P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P5=6000 "kPa"P6=P5T6=450 "C"m_total=20 "kg/s"m7=0.6*m_totalm_cond=0.4*m_totalC=4.18 "kJ/kg.K"m_cooling=463 "kg/s"s7=s6s11=s6h1=ENTHALPY(Steam_IAPWS,x=0,P=P1)v1=VOLUME(Steam_IAPWS,x=0,P=P1)w_pump=v1*(P2-P1)h2=h1+w_pumph3=ENTHALPY(Steam_IAPWS,x=0,P=P3)h4=h3; h9=h3 v4=VOLUME(Steam_IAPWS,x=0,P=P4)w_pump2=v4*(P5-P4)h5=h4+w_pump2h6=ENTHALPY(Steam_IAPWS,T=T6,P=P6)s6=ENTROPY(Steam_IAPWS,T=T6,P=P6)h7=ENTHALPY(Steam_IAPWS,s=s7,P=P7)h8=h7; h10=h7 h11=ENTHALPY(Steam_IAPWS,s=s11,P=P11)W_turb=m_total*(h6-h7)+m_cond*(h7-h11)m_cooling*C*T_rise=m_cond*(h11-h1)m_cond*h2+m_feed*h10=(m_cond+m_feed)*h3m_process=m7-m_feedq_process=h8-h9Q_in=m_total*(h6-h5)

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