Chapter 10 VAPOR AND COMBINED POWER CYCLES · PDF file10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on

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 moisture content allowed is about 10%. 10-2C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content 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. The thermal efficiency, the quality at the end of the heat rejection process, 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) We note that

and

19.3%=−=−=η

=°===°==

R 1.833R 0.67211

R 0.672F0.212R 1.833F1.373

Cth,

psia14.7@sat

psia 180@sat

H

L

L

H

TT

TTTT

T

14.7 psia

180 psia qin

1

4

2

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

0.153=−

=−

=44441.1

31215.053274.044

fg

f

sss

x s

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

( )( ) Btu/lbm 2.111216.85190.014.346Btu/lbm 14.346

22

psia 180 @1

=+=+===

fgf

f

hxhhhh

Thus,

and, ( )( ) Btu/lbm 148.1===

=−=−=

Btu/lbm 0.7661934.0

Btu/lbm 0.76614.3462.1112

inthnet

12in

qw

hhq

η

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

10-2

10-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes

36.3%==−=−= 3632.0K 523K 333.1

11Cth,H

L

TT

η T

qout

qin 1

4

2

3 20 kPa

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

Thus,

( ) kJ/kg 1092.3=

===

==

kJ/kg 3.1715K 523K 333.1

kJ/kg 3.1715

inout

250 @in

qTT

qq

hq

H

LL

Cfg o

250°C

s(c) The net work output of this cycle is ( )( ) kJ/kg623.0 kJ/kg 3.17153632.0inthnet === qw η

10-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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 thermal efficiency becomes

39.04%=−=−=K 523K 318.811C th,

H

L

TT

η T

qout

qin 1

4

2

3 10 kPa

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

Thus,

( ) kJ/kg 1045.6=

===

== °

kJ/kg 3.1715K 523K 318.8

kJ/kg 3.1715

inout

C250 @in

qTT

qq

hq

H

LL

fg

250°C

s(c) The net work output of this cycle is ( )( ) kJ/kg 669.7=== kJ/kg 3.17153904.0inthnet qw η


10-3

10-6 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal 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

η th, C60 273 K350 273 K

= − = −++

=1 1TT

L

H46.5%

(b) Note that 350°C

s2 = s3 = sf + x3sfg = 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K Thus , 60°C

s (Table A-6) MPa 1.40≅

⋅=°=

22

2

KkJ/kg 1368.7C350

PsT

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

Thus,

( )( )

( )( ) ( )( ) kJ/kg 1623=−−=−−==

⋅=+=+=

5390.11368.760350Area

KkJ/kg 5390.10769.71.08313.0

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 heat rejection 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 involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping. 10-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same 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 will adversely 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 than the 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/kg 58.34304.354.340kJ/kg 04.3

mkPa 1kJ 1

kPa 503000/kgm 001030.0

/kgm 001030.0

kJ/kg 54.340

in,12

33

121in,

3kPa 50 @1

kPa 50 @1

=+=+==

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )(kJ/kg 3.2272

7.23048382.054.340

8382.05019.6

0912.15412.6kPa 50

KkJ/kg 5412.6kJ/kg 3.2994

C 300MPa 3

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/kg 9.7188.19317.2650

kJ/kg 8.193154.3403.2272kJ/kg 7.265058.3433.2994

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and

27.1%=−=−=7.26508.193111

in

outth q

qη

(b) ( )( ) MW 25.2=== kJ/kg 9.718kg/s 35netnet wmW &&


10-5

10-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the steam 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/kg 90.20109.1081.191kJ/kg 09.10

mkPa 1kJ 1kPa 10000,10/kgm 00101.0

/kgm 00101.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+==

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )( ) kJ/kg 2089.72392.10.7934191.81

4996.70.64926.5995kPa 10

KkJ/kg 6.5995kJ/kg 3375.1

C 500MPa 10

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/kg 4.12759.18972.3173kJ/kg 9.189781.1917.2089kJ/kg 2.317390.2011.3375

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and

40.2%===kJ/kg 3173.2kJ/kg 1275.4

in

netth q

wη

(c) skg 164.7 /kJ/kg 1275.4

kJ/s 210,000

net

net ===wW

m&

&


10-6

10-17 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality of the steam 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/kg 68.20387.1181.191kJ/kg 11.87

0.85/mkPa 1

kJ 1kPa 1010,000/kgm 0.00101

/

/kgm 00101.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+==

⋅−=

−=

==

==

p

pp

f

f

whh

PPw

hh

ηv

vv

T

( )( )

( )( )( )

0.874

0.793

=−

=−

=

==

=−−=−−=→

−−

=

=+=+=

=−

=−

=

==

⋅==

°==

123928119152282

kJ/kg 5.2282kPa 10

kJ/kg 5.22827.20891.337585.01.3375

kJ/kg 7.20891.23927934.081.191

44996.7

6492.05995.6kPa 10

KkJ/kg 5995.6kJ/kg 1.3375

C 500MPa 10

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

ηη

qin

qout

2

4

s

10 MPa

4

2

3

110 kPa

(b)

kJ/kg 7.10807.20904.3171kJ/kg 7.209081.1915.2282kJ/kg 4.317168.2031.3375

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and

34.1%===ηkJ/kg 3171.5kJ/kg 1080.7

in

netth q

w

(c) kg/s 194.3===kJ/kg 1080.7

kJ/s 210,000

net

net

wWm&

&


10-7

10-18E A steam power plant that operates on a simple ideal Rankine cycle between the specified pressure limits is considered. The minimum turbine inlet temperature, the 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.

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/lbm 77.9775.302.94Btu/lbm 75.3

ftpsia 5.4039Btu 1psia 21250/lbmft 0.01623

/lbmft 01623.0

Btu/lbm 02.94

in,12

33

121in,

3psia 2 @1

psia 2 @1

=+=+==

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

s

( )( )( )( )

F1337°=

=

=

=

⋅=+=+=

=+=+=

3

3

43

3

44

44

Btu/lbm 4.1693psia 1250

RBtu/lbm 7450.174444.19.017499.0

Btu/lbm 6.10137.10219.002.94

Th

ssP

sxss

hxhh

fgf

fgf

(b) ( ) ( )( ) Btu/s 119,672=−=−= 97.771693.4lbm/s 7523in hhmQ &&

(c) ( ) ( )( )

42.4%=−=−=

=−=−=

Btu/s 119,672Btu/s 68,96711

Btu/s 68,96794.021013.6lbm/s 75

in

out

14out

QQ

hhmQ

th &

&

&&

η


10-8

10-19E A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. The minimum turbine inlet temperature, the 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. T

Qout ·

2s

s

1250 psia

Qin ·

4s 4 x4 = 0.9

2

3

12 psia

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

( )( )( )

Btu/lbm 43.9841.402.94Btu/lbm 41.4

85.0/ftpsia 5.4039

Btu 1psia 21250/lbmft 0.01623

/

/lbmft 01623.0

Btu/lbm 02.94

in,12

33

121in,

3psia 2 @1

psia 2 @1

=+=+==

⋅−=

−=

==

==

p

Pp

f

f

whh

PPw

hh

ηv

vv

( )( )( )( ) RBtu/lbm 7450.174444.19.017499.0

Btu/lbm 6.10137.10219.002.94

44

44

⋅=+=+==+=+=

fgf

fgf

sxsshxhh

The turbine inlet temperature is determined by trial and error ,

Try 1:

( )( )

8171.04.9180.14396.10130.1439

Btu/lbm 4.9187.10218069.002.94

8069.074444.1

17499.05826.1

Btu/lbm.R 5826.1Btu/lbm 0.1439

F900psia 1250

43

43

44

344

3

3

3

3

=−−

=−−

=

=+=+=

=−

=−

=−

=

==

°==

sT

fgsfs

fg

f

fg

fss

hhhh

hxhh

sss

sss

x

sh

TP

η

Try 2:

( )( )

8734.03.9436.14986.10136.1498

Btu/lbm 3.9437.10218312.002.94

8312.074444.1

17499.06249.1

Btu/lbm.R 6249.1Btu/lbm 6.1498

F1000psia 1250

43

43

44

344

3

3

3

3

=−−

=−−

=

=+=+=

=−

=−

=−

=

==

°==

sT

fgsfs

fg

f

fg

fss

hhhh

hxhh

sss

sss

x

sh

TP

η

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

(b) ( ) ( )( ) Btu/s 103,055=−=−= 98.431472.5lbm/s 7523in hhmQ &&

(c) ( ) ( )( )

33.1%=−=−=

=−=−=

Btu/s 103,055Btu/s 68,96911

Btu/s 969,6894.021013.6lbm/s 75

in

outth

14out

QQ

hhmQ

&

&

&&

η


10-9

10-20 A 300-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified 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/kg 5.07

mkPa 1kJ 1

kPa 255000/kgm 00102.0

/kgm 000102.0

kJ/kg 96.271

in,12

33

121in,

3kPa 25 @1

kPa 25 @1

=+=+==

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

( )( ) kJ/kg 2.22765.23458545.096.271

8545.09370.6

8932.08210.6kPa 25

KkJ/kg 8210.6kJ/kg 2.3317

C450MPa 5

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/kg 2.200496.2712.2276kJ/kg 2.304003.2772.3317

14out

23in=−=−==−=−=

hhqhhq

and

Thus, ( )( )( ) 24.5%==××=

=−=−=

96.075.03407.0

3407.02.30402.200411

gencombthoverall

in

outth

ηηηη

ηqq

(b) Then the required rate of coal supply becomes

and

tons/h 150.3==

==

===

tons/s 04174.0kg 1000

ton1kJ/kg 29,300

kJ/s 1,222,992

kJ/s 992,222,12453.0

kJ/s 300,000

coal

incoal

overall

netin

CQ

m

WQ

&&

&&

η


10-10

10-21 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the 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 refrigerant tables (Tables A-11, A-12, and A-13),

( )

( )( )

kJ/kg 40.8958.082.88

kJ/kg 58.0mkPa 1

kJ 1kPa 7001400/kgm 0008331.0

/kgm 0008331.0

kJ/kg 82.88

in,12

33

121in,

3MPa 7.0 @1

MPa 7.0 @1

=+=+=

=

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )( ) kJ/kg .2026221.1769839.082.88

9839.058763.0

33230.09105.0MPa 7.0

KkJ/kg 9105.0

kJ/kg 12.276

vaporsat.MPa 4.1

44

44

34

4

MPa 4.1 @3

MPa 4.1 @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/kg 34.1338.17372.186

kJ/kg 38.17382.8820.262kJ/kg 72.18640.8912.276

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and

7.1%===kJ/kg 186.72kJ/kg 13.34

in

netth q

wη

(b) ( )( ) kW 40.02=== kJ/kg 13.34kg/s 3netnet 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 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/kg 198.8706.781.191

kJ/kg 7.06mkPa 1

kJ 1kPa 107,000/kgm 0.00101

/kgm 00101.0

kJ/kg .81191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+=

=

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )( ) kJ/kg 3.62151.23928201.081.191

8201.04996.7

6492.08000.6kPa 10

KkJ/kg 8000.6kJ/kg 411.43

C500MPa 7

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/kg 7.12508.19615.3212kJ/kg 8.196181.1916.2153kJ/kg 5.321287.1984.3411

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and 38.9%===kJ/kg 3212.5kJ/kg 1250.7

in

netth q

wη

(b) skg 036 /.kJ/kg 1250.7kJ/s 45,000

net

net ===wWm&

&

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

( )( )

( )( ) C8.4°=°⋅

==∆

===

CkJ/kg 4.18kg/s 2000kJ/s 70,586

)(

kJ/s ,58670kJ/kg 1961.8kg/s 35.98

watercooling

outwatercooling

outout

cmQ

T

qmQ

&

&

&&


10-12

10-23 A steam power plant operates on a simple nonideal Rankine cycle between the specified pressure limits. 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/kg 9.921911.881.191kJ/kg 8.11

0.87/mkPa 1

kJ 1kPa 107,000/kgm 00101.0

/

/kgm 00101.0

kJ/kg 91.811

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+==

⋅−=

−=

==

==

p

pp

f

f

whh

PPw

hh

ηv

vv

T

( )( )

( )( )( ) kJ/kg 1.23176.21534.341187.04.3411

kJ/kg 6.21531.2392820.081.191

8201.04996.7

6492.08000.6kPa 10

KkJ/kg 8000.6kJ/kg .43411

C500MPa 7

433443

43

44

44

34

4

3

3

3

3

=−−=−−=→

−−

=

=+=+=

=−

=−

=

==

⋅==

°==

sTs

T

fgfs

fg

f

hhhhhhhh

hxhh

sss

xss

P

sh

TP

ηη

qout

qin 2

4

s

7 MPa

4

2

3

110 kPa

Thus,

kJ/kg 2.10863.21255.3211kJ/kg 3.212581.1911.2317kJ/kg 5.321192.1994.3411

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and 33.8%===kJ/kg 3211.5kJ/kg 1086.2

in

netth q

wη

(b) skg 4341 /.kJ/kg 1086.2kJ/s 45,000

net

net ===wWm&

&

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

( )( )

( )( ) C10.5°=°⋅

==∆

===

CkJ/kg 4.18kg/s 2000kJ/s 88,051

)(

kJ/s ,05188kJ/kg 2125.3kg/s 41.43

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 Rankine cycle 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/kg 57.26115.1042.251kJ/kg 15.10

mkPa 1kJ 1kPa 20000,10/kgm 001017.0

/kgm 001017.0

kJ/kg 42.251

in,12

33

121in,

3kPa 20 @1

kPa 20 @1

=+=+==

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

( )(kJ/kg 3.1845

5.23576761.042.251

6761.00752.7

8320.06159.5kPa 20

KkJ/kg 6159.5kJ/kg 5.2725

1MPa 10

44

44

34

4

3

3

3

3

=

+=+=

=−

=−

=

==

⋅==

==

fgf

fg

f

hxhh

sss

xss

P

sh

xP

)

s

Rankine cycle

4

3

2

1

T

kJ/kg 869.9=−=−=

=−=−==−=−=

0.15949.2463kJ/kg 0.159442.2513.1845kJ/kg 9.246357.2615.2725

outinnet

14out

23in

qqwhhqhhq

0.353=−=−=9.24630.159411

in

outth q

qη

(b) Carnot Cycle analysis:

s

Carnot cycle

4

2 3

1

T

kJ/kg 9.1093)5.2357)(3574.0(42.251

3574.00752.7

8320.03603.3kPa 20

KkJ/kg 3603.3kJ/kg 8.1407

0C 0.311

C 0.311kJ/kg 5.2725

1MPa 10

11

11

21

1

2

2

2

32

3

3

3

3

=+=

+=

=−

=−

=

==

⋅==

=°==

°==

==

fgf

fg

f

hxhhs

ssx

ssP

sh

xTT

Th

xP

kJ/kg 565.4=−=−==−=−==−=−=

3.7527.1317kJ/kg 4.7519.10933.1845

kJ/kg 7.13178.14075.2725

outinnet

14out

23in

qqwhhqhhq

0.430=−=−=7.13174.75111

in

outth q

qη


10-14

10-25 A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power 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. 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/kg 74.689C5.179

kPa 410

kJ/kg 40.670kPa 410

KkJ/kg 5457.2kJ/kg 54.761

C147kPa 3250

44

4

434

4

3

3

3

3

=

°==

=

==

⋅==

°==

hTP

hss

P

sh

TP

s

4

3 2

1

Geothermal water out

Geothermal water in

heat exchanger

pumpturbine

air-cooledcondenser

0.788=−−

=−−

=40.67054.76174.68954.761

43

43

sT hh

hhη

T

qin

qout

2s

4

s

3.25 MPa

4s

2

3

1410 kPa

(b) Pump:

( )( )( )

kJ/kg 82.27881.501.273kJ/kg 81.5

90.0/mkPa 1

kJ 1kPa 4103250/kgm 001842.0

/

/kgm 001842.0

kJ/kg 01.273

in,12

33

121in,

3kPa 410 @1

kPa 014 @1

=+=+==

⋅−=

−=

==

==

p

Pp

f

f

whh

PPw

hh

ηv

vv

kW 941,21kJ/kg)74.68954.761kJ/kg)( 305.6()( 43outT, =−=−= hhmW &&

kW 1777kJ/kg) 1kJ/kg)(5.8 305.6()( inp,12inP, ===−= wmhhmW &&&

kW 20,165=−=−= 1777941,21inP,outT,net WWW &&&

Heat Exchanger:

kW 656,162C)90160(C)kJ/kg. .184(kJ/kg) .9555()( outingeogeoin =°−°=−= TTcmQ &&

(c) 12.4%0.124 ====656,162165,20

in

netth Q

W&

&η


10-15

10-26 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of steam through the turbine, the isentropic efficiency of the turbine, the power output from the 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 for geothermal water (Tables A-4 through A-6)

1661.02108

09.64014.990kJ/kg 14.990

kPa 500

kJ/kg 14.9900

C230

22

12

2

11

1

=−

=−

=

===

=

=°=

fg

f

hhh

xhh

P

hxT

The mass flow rate of steam through the turbine is kg/s 38.20=== kg/s) 230)(1661.0(123 mxm &&

(b) Turbine:

kJ/kg 7.2344)1.2392)(90.0(81.19190.0kPa 10

kJ/kg 3.2160kPa 10

KkJ/kg 8207.6kJ/kg 1.2748

1kPa 500

444

4

434

4

3

3

3

3

=+=+=

==

=

==

⋅==

==

fgf

s

hxhhxP

hss

P

sh

xP

production well

Flash chamber

65

4

3

2condenser

1

steam turbine

separator

reinjection well

0.686=−−

=−−

=3.21601.27487.23441.2748

43

43

sT hh

hhη

(c) The power output from the turbine is

kW 15,410=−=−= kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT, hhmW &&

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

kJ/kg 83.1040

C250

0

0 =

=°=

hxT

kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in =−=−= hhmE &&

7.6%==== 0.0757622,203

410,15

in

outT,th E

W&

&η


10-16

10-27 A double-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The temperature of the steam at the exit of the second flash chamber, the power produced from the second 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 for geothermal water (Tables A-4 through A-6)

1661.0kJ/kg 14.990

kPa 500

kJ/kg 14.9900

C230

212

2

11

1

=

===

=

=°=

xhh

P

hxT

kg/s 80.1911661.0230kg/s 38.20kg/s) 230)(1661.0(

316

123

=−=−====

mmmmxm

&&&

&&

Flash chamber

separator

9

8

7

Flash chamber

6

5

4

3

2condenser

1

steam turbine

separator

reinjection well

production well

kJ/kg 7.234490.0kPa 10

kJ/kg 1.27481

kPa 500

44

4

33

3

=

==

=

==

hxP

hxP

kJ/kg 1.26931

kPa 150

0777.0kPa 150

kJ/kg 09.6400

kPa 500

88

8

7

7

67

7

66

6

=

==

=°=

==

=

==

hxP

xT

hhP

hxP

C 111.35

(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

kW 15,410kJ/kg)7.23448.1kJ/kg)(274 38.20()( 433outT1, =−=−= hhmW &&

kW 5191=−=−= kJ/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/kg 83.1040

C250

0

0 =

=°=

hxT

kW 621,203kJ/kg)83.10414kg/s)(990. 230()( 011in =−=−= hhmE &&

10.1%==+

== 0.101621,203

5193410,15

in

outT,th E

W&

&η


10-17

10-28 A combined flash-binary geothermal power plant uses hot geothermal water at 230ºC as the heat source. The mass flow rate of isobutane in the binary cycle, the net power outputs from the steam turbine and the binary cycle, and the thermal efficiencies for the binary cycle and the combined 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/kg 14.990

kPa 500

kJ/kg 14.9900

C230

212

2

11

1

=

===

=

=°=

xhh

P

hxT

kg/s 80.19120.38230kg/s 38.20kg/s) 230)(1661.0(

316

123

=−=−====

mmmmxm

&&&

&&

kJ/kg 7.234490.0kPa 10

kJ/kg 1.27481

kPa 500

44

4

33

3

=

==

=

==

hxP

hxP

kJ/kg 04.3770

C90

kJ/kg 09.6400

kPa 500

77

7

66

6

=

=°=

=

==

hxT

hxP

The isobutene properties are obtained from EES:

/kgm 001839.0kJ/kg 83.270

0kPa 400

kJ/kg 01.691C80kPa 400

kJ/kg 05.755C145kPa 3250

310

10

10

10

99

9

88

8

==

==

=

°==

=

°==

v

hxP

hTP

hTP

1

19

8

7

65

4

3

2

1

flash chamber

condenser

air-cooled condenser

separator

BINARY CYCLE

pump

heat exchanger

isobutaneturbine

reinjection well

production well

steam turbine

( )( )( )

kJ/kg 65.27682.583.270kJ/kg. 82.5

90.0/mkPa 1

kJ 1kPa 4003250/kgm 001819.0

/

in,1011

33

101110in,

=+=+==

⋅−=

−=

p

pp

whh

PPw ηv

An energy balance on the heat exchanger gives

kg/s 105.46=→=

−=−

isoiso

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

kW 15,410=−=−= kJ/kg)7.23448.1kJ/kg)(274 38.19()( 433steamT, hhmW &&


10-18

kW 6139=−=−=

=−=−=

)kJ/kg 82.5)(kg/s 46.105(6753

kW 6753kJ/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

kW 454,50kJ/kg)65.276.05kJ/kg)(755 .46105()( 118isobinaryin, =−=−= hhmQ &&

12.2%==== 0.122454,50

6139

binaryin,

binarynet,binaryth, Q

W&

&η

kJ/kg 83.1040

C250

0

0 =

=°=

hxT

kW 622,203kJ/kg)83.104.14kJ/kg)(990 230()( 011in =−=−= hhmE &&

10.6%==+

=+

= 0.106622,203

6139410,15

in

binarynet,steamT,plantth, E

WW&

&&η

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 Rankine cycle with 3 stages of reheat is shown on the side. The cycle efficiency will increase as the number of reheating stages increases. 10-31C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.

1

I

2

10 s

T

3

4

5

6

7

8

II III 9


10-19

10-32 [Also solved by EES on enclosed CD] A steam power plant that operates on the ideal reheat Rankine cycle 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/kg 8.12mkPa 1

kJ 1kPa 208000/kgm 001017.0

/kgm 700101.0

kJ/kg 42.251

in,12

33

121in,

3kPa 20 @1

kPa 20 @1

=+=+=

=

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

( )( ) kJ/kg 2.23855.23579051.042.251

9051.00752.7

8320.02359.7kPa 20

KkJ/kg 2359.7kJ/kg 2.3457

C500MPa 3

kJ/kg 1.3105MPa 3

KkJ/kg 7266.6kJ/kg 5.3399

C500MPa 8

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

and ( ) ( )

( ) ( ) kJ/kg 0.34921.31052.345754.2595.3399

2.23852.34571.31055.3399

4523in

6543outT,

=−+−=−+−=

=−+−=−+−=

hhhhq

hhhhw kJ/kg 1366.4

Thus,

38.9%===

=−=−=

kJ/kg 3492.5kJ/kg 1358.3

kJ/kg 3.135812.84.1366

in

netth

in,,net

qw

www poutT

η


10-20

10-33 EES Problem 10-32 is reconsidered. The problem is to be solved by the diagram window data entry feature 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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,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_p Eff=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

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


10-22

10-34 A steam power plant that operates on a 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 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/kg 43.20262.1081.191

kJ/kg 10.62

0.95/mkPa 1

kJ 1kPa 1010,000/kgm 0.00101

/

/kgm 001010.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+=

=

⋅−=

−=

==

==

p

pp

f

f

whh

PPw

hh

ηv

vv

T

( )( )( )

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 1

kJ/kg 0.29027.27831.337580.01.3375

kJ/kg 8.2783MPa 1

KkJ/kg 5995.6kJ/kg 75.133

C500MPa 10

5

5

5

5

433443

43

434

4

3

3

3

3

⋅==

°==

=−−=−η−=→

−−

=η

=

==

⋅==

°==

sh

TP

hhhhhhhh

hss

P

sh

TP

sTs

T

ss

s

2

6 10 kPa

10 MPa 4s 4

3

6

2

5

1s

( )

( )( )

( )( )( )

( ) vapordsuperheatekJ/kg 8.26642.24611.347980.01.3479

kJ/kg 2.24611.23929487.081.191

exit turbineat 9487.04996.7

6492.07642.7kPa 10

655665

65

66

66

56

6

g

sTs

T

fgsfs

fg

fss

s

s

h

hhhhhhhh

hxhh

sss

x

ssP

>=−−=

−−=→−−

=

=+=+=

=−

=−

=

==

ηη

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

( ) ( )kJ/kg 8.127662.104.1287

kJ/kg 8.37490.29021.347943.2021.3375

kJ/kg 4.12878.26641.34790.29021.3375

inp,outT,net

4523in

6543outT,

=−=−=

=−+−=−+−=

=−+−=−+−=

www

hhhhq

hhhhw

Thus the thermal efficiency is

34.1%===kJ/kg 3749.8kJ/kg 1276.8

in

netth q

wη

(c) The mass flow rate of the steam is

kg/s 62.7===kJ/kg 1276.9kJ/s 80,000

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 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/kg 90.20109.1081.191

kJ/kg 10.09mkPa 1

kJ 1kPa 01000,10/kgm 00101.0

/kgm 00101.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+=

=

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )

( )( ) kJ/kg 2.24611.23929487.081.191

exit turbineat 4996.7

6492.07642.7kPa 10

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 1

kJ/kg 8.2783MPa 1

KkJ/kg 5995.6kJ/kg 1.3375

C500MPa 10

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/kg 3.159909.104.1609

kJ/kg 5.38687.27831.347990.2011.3375

kJ/kg 3.16092.24611.34797.27831.3375

in,outT,net

4523in

6543outT,

=−=−=

=−+−=−+−=

=−+−=−+−=

pwww

hhhhq

hhhhw

Thus the thermal efficiency is

41.3%===kJ/kg 3868.5kJ/kg 1599.3

in

netth q

wη

(c) The mass flow rate of the steam is

skg 050 /.kJ/kg 1599.3kJ/s 80,000

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 rate of 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/lbm 11.7239.272.69

Btu/lbm 2.39ftpsia 5.4039

Btu 1psia 1800/lbmft 01614.0

F69.101

/lbmft 01614.0

Btu/lbm 72.69

in,12

33

121in,

psia 1 @sat1

3psia 1 @sat1

psia 1 @sat1

=+=+=

=

⋅−=

−=

°==

==

==

p

p

whh

PPw

TT

hh

v

vv

T

1 psia

800 psia 4

3

6

2

5

1s

( )

( )( ) Btu/lbm 0.10617.10359572.072.69

9572.084495.1

13262.08985.1psia 1

RBtu/lbm 8985.1Btu/lbm 4.1431

F800psia 23.62

pressure)reheat (the

Btu/lbm 5.1178

vaporsat.

RBtu/lbm 6413.1Btu/lbm 0.1456

F900psia 800

66

66

56

6

5

5

5

5

@sat4

@434

3

3

3

3

4

4

=+=+=

=−

=−

=

==

⋅==

°==

==

==

=

⋅==

°==

=

=

fgf

fg

f

ss

ssg

hxhh

sss

x

ssP

sh

TP

PP

hhss

sh

TP

g

g

psia 62.23

(b) ( ) ( )Btu/lbm 3.99172.690.1061

Btu/lbm 8.16365.11784.143111.720.1456

16out

4523in

=−=−=

=−+−=−+−=

hhq

hhhhq

Thus,

39.4%=−=−=Btu/lbm 1636.8Btu/lbm 991.3

11in

outth q

qη

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

( ) ( )( )

( )( ) lbm/s 641.0=°−°⋅

×=

∆=

×=×−=−=−=

F45101.69FBtu/lbm 1.0Btu/s 103.634

Btu/s 10634.3Btu/s 1063943.0114

outcool

44inthnetinout

TcQ

m

QWQQ&

&

&&&& η


10-25

10-37 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits 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/kg 95.20614.1581.191

kJ/kg .1415mkPa 1

kJ 1kPa 10000,15/kgm 00101.0

/kgm 00101.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @sat1

kPa 10 @sat1

=+=+=

=

⋅−=

−=

==

==

p

p

whh

PPw

hh

v

vv

T

( )( )( )( )

( )

kJ/kg 2.2817MPa 161.2

kJ/kg 53.3466pressurereheat theC500

KkJ/kg 3988.74996.790.06492.0

kJ/kg 7.23441.239290.081.191kPa 10

KkJ/kg 3480.6kJ/kg .83310

C500MPa 15

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

kPa 2161

10 kPa

15 MPa 4

3

6

2

5

1s

(b) The rate of heat supply is

( ) ( )[ ]

( )( ) kW 45,038=−+−=−+−=

kJ/kg2.281753.346695.2068.3310kg/s 124523in hhhhmQ &&

(c) The thermal efficiency is determined from

Thus, ( ) ( )( )

42.6%=−=−=

=−=−=

kJ/s 45,039kJ/s 25,834

11

kJ/s ,83525kJ/kg81.1917.2344kJ/s 12

in

outth

16out

QQ

hhmQ

&

&

&&

η


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/kg 3.30271.29482.335885.02.3358

?

95.0?

KkJ/kg 2815.7kJ/kg 2.3358

C450MPa 2

kJ/kg 3.30271.29485.347685.05.3476

kJ/kg 1.2948MPa 2

KkJ/kg 6317.6kJ/kg 5.3476

C550MPa 5.12

655665

65

656

6

66

6

5

5

5

5

4334

43

43

434

4

3

3

3

3

=−−=−η−=→

−−

=

=

==

=

==

⋅==

°==

=−−=

−η−=→

−−

=η

=

==

⋅==

°==

sTs

T

s

sT

sT

ss

hhhhhhhh

hss

P

hxP

sh

TP

hhhh

hhhh

hss

P

sh

TP

η

2

4

6 P = ?

12.5 MPa4s

3

T

4

5

Pump Condenser

Boiler Turbine

2 1

6

3

s 6s

2s

5

1

The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above equations: P6 = 9.73 kPa, h6 = 2463.3 kJ/kg, (b) Then,

( )( )( ) ( )

kJ/kg 59.20302.1457.189

kJ/kg 14.020.90/

mkPa 1kJ 1kPa 73.912,500/kgm 0.00101

/

/kgm 001010.0

kJ/kg 57.189

in,12

33

121in,

3kPa 10 @1

kPa 73.9 @1

=+=+=

=

⋅−=

−=

==

==

p

pp

f

f

whh

PPw

hh

ηv

vv

Cycle analysis: ( ) ( )

kW 10,242==−=

=−=−=

=−+−=−+−=

kg2273.7)kJ/-.8kg/s)(3603 7.7()(

kJ/kg 7.227357.1893.3027

kJ/kg 8.36033.24632.33583.30275.3476

outinnet

16out

4523in

qqmW

hhq

hhhhq

&&

(c) The thermal efficiency is

36.9%==−=−= 369.0kJ/kg 3603.8kJ/kg 2273.7

11in

outth q

qη


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 rejected anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work output. 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 liquid state at the boiler pressure isothermally, and the steam must be a saturated vapor at the turbine inlet. This will require an infinite number of heat exchangers (feedwater heaters), as shown on the T-s diagram.

Boiler exit

s

TBoiler inlet

qin

qout


10-28

10-44 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The net work output per kg of steam 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/kg 81.25139.042.251

kJ/kg 0.39mkPa 1

kJ 1kPa 20400/kgm 001017.0

/kgm 001017.0

kJ/kg 42.251

in,12

33

121in,

3kPa 20 @1

kPa 20 @1

=+=+=

=

⋅−=

−=

==

==

pI

pI

f

f

whh

PPw

hh

v

vv

T

( ) ( )( )

( )( )

( )( ) kJ/kg 0.22145.23578325.042.251

8325.00752.7

8320.07219.6kPa 20

kJ/kg 7.26654.21339661.066.604

9661.01191.5

7765.17219.6MPa 4.0

KkJ/kg 7219.6kJ/kg 9.3302

C450MPa 6

kJ/kg 73.61007.666.604

kJ/kg 6.07mkPa 1

kJ 1kPa 4006000/kgm 0.001084

/kgm 001084.0

kJ/kg 66.604

liquidsat.MPa 4.0

77

77

57

7

66

66

56

6

5

5

5

5

in,34

33

343in,

3MPa 4.0 @3

MPa 4.0 @33

=+=+=

=−

=−

=

==

=+=+=

=−

=−

=

==

⋅==

°==

=+=+=

=

⋅−=−=

==

==

=

fgf

fg

f

fgf

fg

f

pII

pII

f

f

hxhh

sss

x

ssP

hxhh

sss

xss

P

sh

TP

whh

PPw

hhP

v

vv

qout

3 y

4

1-y

0.4 MPa

20 kPa

6 MPa

6

5

7

2

qin

1s

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater 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/kg 7.167542.2510.22141462.011kJ/kg 2.269273.6109.3302

17out

45in

=−−=−−==−=−=

hhyqhhq

And kJ/kg 1016.5=−=−= 7.16752.2692outinnet qqw (b) The thermal efficiency is determined from

37.8%=−=−=kJ/kg 2692.2kJ/kg 1675.7

11in

outth q

qη


10-29

10-45 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The net work output per kg of steam 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/kg 50.25708.642.251

kJ/kg 6.08mkPa 1

kJ 1kPa 206000/kgm 0.001017

/kgm 001017.0

kJ/kg 42.251

in,12

33

121in,

3kPa 20 @1

kPa 20 @1

=+=+=

=

⋅−=

−=

==

==

pI

pI

f

f

whh

PPw

hh

v

vv

T

9

qou

3y

4

1-y

0.4 MPa

20 kPa

6 MPa

6

5

7

28

qin

1

/kgm 400108.0

kJ/kg 66.604

liquid sat.MPa 4.0

3MPa 4.0 @3

MPa 4.0 @33

==

==

=

f

fhhPvv

s

( ) ( )( )

( ) kJ/kg 73.610

kJ/kg 73.61007.666.604

kJ/kg 07.6mkPa 1

kJ 1kPa 4006000/kgm 001084.0

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 same enthalpy.

( )( )

( )( ) kJ/kg 0.22145.23578325.042.251

8325.00752.7

8320.07219.6kPa 20

kJ/kg 7.26654.21339661.066.604

9661.01191.5

7765.17219.6MPa 4.0

KkJ/kg 7219.6kJ/kg 9.3302

C450MPa 6

77

77

57

7

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

==

=+=+=

=−

=−

=

==

⋅==

°==

fgf

fg

f

fgf

fg

f

hxhh

sss

x

ssP

hxhh

sss

x

ssP

sh

TP

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

( ) ( ) ( )( ) ( 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 =−+−

−=

−+−−

=hhhh

hhy

Then, ( )( ) ( )( ) kJ/kg 4.167542.2510.22141463.011kJ/kg 2.269273.6109.3302

17out

45in

=−−=−−==−=−=

hhyqhhq

And kJ/kg 1016.8=−=−= 4.16752.2692outinnet qqw (b) The thermal efficiency is determined from

37.8%=−=−=kJ/kg 2692.2kJ/kg 1675.411

in

outth q

qη


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