Thermo 7e sm_chap10-1

10-1

Solutions Manual for

Thermodynamics: An Engineering Approach Seventh Edition

Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011

Chapter 10 VAPOR AND COMBINED POWER CYCLES

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PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.

. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

10-2

Carnot Vapor Cycle

10-1C 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.

10-2E 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

15.5%==−=−=

=°===°==

1553.0R 861R 2.727

11

R 2.727F2.267R 861F401

Cth,

psia40@sat

psia 025@sat

H

L

L

H

TT

TTTT

η

T

40 psia

250 psia

3

2

4

1qin

(b) Noting that s4 = s1 = sf @ 250 psia = 0.56784 Btu/lbm·R,

0.137=−

=−

=2845.1

3921.056784.044

fg

f

sss

x s

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

( )( ) Btu/lbm 3.116047.82595.009.376Btu/lbm 09.376

22

psia 025 @1

=+=+===

fgf

f

hxhhhh

Thus,

and ( )( ) Btu/lbm 122===

=−=−=

Btu/lbm 2.7841553.0

Btu/lbm 2.78409.3763.1160

inthnet

12in

qw

hhq

η



10-3

10-3 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.


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

20 kPa3

2

4

1qin

qout

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

250°C

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

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

( )( ) kJ/kg623.0 kJ/kg 3.17153632.0inthnet === qw η

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.


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

10 kPa 3

2

4

1 qin

qout

(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

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



10-4

10-5 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.


Analysis (a) The thermal efficiency is determined from

η th, C60 273 K350 273 K

= − = −++

=1 1T

TL

H

46.5% T

3

2

4

1 (b) Note that 350°Cs2 = s3 = sf + x3sfg

= 0.8313 + 0.891 × 7.0769 = 7.1368 kJ/kg·K

Thus, 60°C

(Table A-6) MPa 1.40≅⎭⎬⎫

⋅=°=

22

2

KkJ/kg 1368.7C350

PsT s

(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



10-5

The Simple Rankine Cycle

10-6C 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.

10-7C Heat rejected decreases; everything else increases.

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

10-9C The pump work remains the same, the moisture content decreases, everything else increases.

10-10C 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.

10-11C 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.

10-12C We would reject this proposal because wturb = h1 - h2 - qout, and any heat loss from the steam will adversely affect the turbine work output.

10-13C 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.



10-6

10-14 A simple ideal Rankine cycle with R-134a as the working fluid operates between the specified pressure limits. The mass flow rate of R-134a for a given power production and the thermal efficiency of the cycle are to be determined.


Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg 89.6495.094.63kJ/kg 95.0 mkPa 1

kJ 1 kPa)4001600)(/kgm 0007907.0()(

/kgm 0007907.0

kJ/kg 94.63

inp,12

33

121inp,

3MPa 0.4 @1

MPa 4.0 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

qin

qout

0.4 MPa

1

3

2

4

1.6 MPa

T

s

kJ/kg 21.273 MPa 4.0

KkJ/kg 9875.0kJ/kg 07.305

C80MPa 6.1

434

4

3

3

3

3

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

hss

P

sh

TP

Thus,

kJ/kg 91.3027.20918.240

kJ/kg 27.20994.6321.273kJ/kg 18.24089.6407.305

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

The mass flow rate of the refrigerant and the thermal efficiency of the cycle are then

kg/s 24.26===kJ/kg 91.30kJ/s 750

net

net

wW

m&

&

0.129=−=−=18.24027.20911

in

outth q

qη



10-7

10-15 A simple ideal Rankine cycle with R-134a as the working fluid is considered. The turbine inlet temperature, the cycle thermal efficiency, and the back-work ratio of the cycle are to be determined.


Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg 21.6678.043.65kJ/kg 78.0 mkPa 1

kJ 1 kPa)89.4141400)(/kgm 0007930.0()(

/kgm 0007930.0

kJ/kg 43.65

kPa 89.414

inp,12

33

121inp,

3C10 @1

C10 @1

C10 @sat 1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

==

°

°

°

whh

PPw

hh

PP

f

f

v

vv

qin

qout

10°C

1

3

2

4

1.4 MPa

T

s

KkJ/kg 91295.0)67356.0)(98.0(25286.0kJ/kg 35.252)73.190)(98.0(43.65

98.0

C10

44

44

4

4

⋅=+=+==+=+=

⎭⎬⎫

=°=

fgf

fgf

sxsshxhh

xT

C53.0°==

⎭⎬⎫

⋅===

3

3

43

3 kJ/kg 91.276

KkJ/kg 91295.0kPa 1400

Th

ssP

Thus,

kJ/kg 92.18643.6535.252kJ/kg 70.21021.6691.276

14out

23in

=−=−==−=−=

hhqhhq

The thermal efficiency of the cycle is

0.113=−=−=70.21092.18611

in

outth q

qη

The back-work ratio is determined from

0.0318=−

=−

==kJ/kg )35.25291.276(

kJ/kg 78.0

43

inP,

outT,

inP,

hhw

Ww

rbw



10-8

10-16 A simple ideal Rankine cycle with water as the working fluid is considered. The work output from the turbine, the heat addition in the boiler, and the thermal efficiency of the cycle are to be determined.


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

kJ/kg 18.17665.853.167kJ/kg 65.8 mkPa 1

kJ 1 kPa)385.78588)(/kgm 001008.0()(

/kgm 001008.0

kJ/kg 53.167

kPa 8588

kPa 385.7

inp,12

33

121inp,

3C40 @1

C40 @1

C300 @sat 2

C04 @sat 1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

==

==

°

°

°

°

whh

PPw

hh

PP

PP

f

f

v

vv

qin

qout

40°C

1

3 2

4

300°C

s

T

kJ/kg 1.1775)0.2406)(6681.0(53.167

6681.06832.7

5724.07059.5

C40

KkJ/kg 7059.5kJ/kg 6.2749

1

C300

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

=°=

⋅==

⎭⎬⎫

=°=

fgf

fg

f

hxhhs

ssx

ssT

sh

xT

Thus,

kJ/kg 6.160753.1671.1775

18.1766.27491.17756.2749

14out

23in

43outT,

=−=−==−=−=

=−=−=

hhqhhqhhw

kJ/kg 2573.4kJ/kg 974.5


0.375=−=−=4.25736.160711

in

outth q

qη



10-9

10-17E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.


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

Btu/lbm 81.11140.240.109Btu/lbm 40.2

ftpsia 5.404Btu 1 psia)3800)(/lbmft 01630.0(

)(

/lbmft 01630.0

Btu/lbm 40.109

inp,12

33

121inp,

3psia 3 @1

psia 3 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 24.975)8.1012)(8549.0(40.109

8549.06849.1

2009.06413.1

psia 3

RBtu/lbm 6413.1Btu/lbm 0.1456

F900psia 800

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

qin

qout

3 psia 1

3

2

4

800 psia

T

s

Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from

lbm/s 450.3kJ 1

Btu 0.94782)Btu/lbm24.975(1456.0

kJ/s 1750)(43

outT,43outT, =⎟

⎠⎞

⎜⎝⎛

−=

−=⎯→⎯−=

hhW

mhhmW&

&&&

The rates of heat addition and rejection are

Btu/s 2987Btu/s 4637

=−=−=

=−=−=

Btu/lbm)40.109.24lbm/s)(975 450.3()(

Btu/lbm)81.1110.6lbm/s)(145 450.3()(

14out

23in

hhmQ

hhmQ&&

&&

and the thermal efficiency of the cycle is

35.6%==−=−= 3559.04637298711

in

outth Q

Q&

&η



10-10

10-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The turbine inlet temperature and the thermal efficiency of the cycle are to be determined.


qin

qout

5 psia 1

3

2

4

2500 psia

T

s


Btu/lbm 76.13758.718.130Btu/lbm 58.7


)(

/lbmft 01641.0

Btu/lbm 18.130

inp,12

33

121inp,

3psia 5 @1

psia 5 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

RBtu/lbm 52203.1)60894.1)(80.0(23488.0Btu/lbm 58.930)5.1000)(80.0(18.130

80.0psia 5

44

44

4

4

⋅=+=+==+=+=

⎭⎬⎫

==

fgf

fgf

sxsshxhh

xP

F989.2°==

⎭⎬⎫

⋅===

3

3

43

3 Btu/lbm 8.1450

RBtu/lbm 52203.1psia 2500

Th

ssP

Thus,

Btu/lbm 4.80018.13058.930Btu/lbm 1313.076.1378.1450

14out

23in

=−=−==−=−=

hhqhhq


0.390=−=−=0.13134.80011

in

outth q

qη



10-11

10-19E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined.



Btu/lbm 18.7746.772.69Btu/lbm 46.7


)(

/lbmft 01614.0

Btu/lbm 72.69

inp,12

33

121inp,

3psia 6 @1

psia 1 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

Btu/lbm 70.787)7.1035)(6932.0(72.69

6932.084495.1

13262.04116.1

psia 1


F800psia 2500

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fs

hxhhs

ssx

ssP

sh

TP

qin

qout

1 psia 1

3

2

4

2500 psia

4s

s

T

kJ/kg 13.839)70.7870.1302)(90.0(0.1302)( 4s3T3443

43T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη

Thus,

Btu/lbm 39.45541.7698.1224

Btu/lbm 41.76972.6913.839Btu/lbm 8.122418.771302.0

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

The mass flow rate of steam in the cycle is determined from

lbm/s 2.081=⎟⎠⎞

⎜⎝⎛==⎯→⎯=

kJ 1Btu 0.94782

Btu/lbm 455.39kJ/s 1000

net

netnetnet w

WmwmW

&&&&

The power output from the turbine and the rate of heat addition are

Btu/s 2549

kW 1016

===

=⎟⎠⎞

⎜⎝⎛−=−=

Btu/lbm) 4.8lbm/s)(122 081.2(

Btu 0.94782kJ 1Btu/lbm)13.8392.0lbm/s)(130 081.2()(

inin

43outT,

qmQ

hhmW

&&

&&


0.3718=⎟⎠⎞

⎜⎝⎛==

kJ 1Btu 0.94782

Btu/s 2549kJ/s 1000

in

netth Q

W&

&η

preparation. If you are a student using this Manual, you are using it without permission.

10-12

10-20E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be determined.



Btu/lbm 18.7746.772.69Btu/lbm 46.7


)(

/lbmft 01614.0

Btu/lbm 72.69

inp,12

33

121inp,

3psia 6 @1

psia 1 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv


Btu/lbm 70.787)7.1035)(6932.0(72.69

6932.084495.1

13262.04116.1

psia 1


F800psia 2500

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fs

hxhhs

ssx

ssP

sh

TP

qin

qout

1 psia 1

3

2

4

2500 psia

4s

s

T

kJ/kg 13.839)70.7870.1302)(90.0(0.1302)( 4s3T3443

43T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη


lbm/s 2.048kJ 1

Btu 0.94782Btu/lbm 839.13)(1302.0

kJ/s 1000)(43

net43net =⎟

⎠⎞

⎜⎝⎛

−=

−=⎯→⎯−=

hhW

mhhmW&

&&&

The rate of heat addition is

Btu/s 2508Btu 0.94782

kJ 1Btu/lbm)18.772.0lbm/s)(130 048.2()( 23in =⎟⎠⎞

⎜⎝⎛−=−= hhmQ &&


0.3779kJ 1

Btu 0.94782Btu/s 2508kJ/s 1000

in

netth =⎟

⎠⎞

⎜⎝⎛==

QW&

&η

The thermal efficiency in the previous problem was determined to be 0.3718. The error in the thermal efficiency caused by neglecting the pump work is then

1.64%=×−

= 1003718.0

3718.03779.0Error


10-13

10-21 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.


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


( )( ) 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

qin

qout

10 kPa 1

3

2

4

7 MPa

T

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

&

&

&&


10-14

10-22 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.



( )( )( ) ( )

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

10 kPa 1

3

2

4

7 MPa

s

qout

qin2

4

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

&

&

&&



10-15

10-23 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The rate of heat addition in the boiler, the power input to the pumps, the net power, and the thermal efficiency of the cycle are to be determined.



/kgm 001026.0kJ/kg 03.314

C753.63.813.6

kPa 503

C75 @1

C75 @ 1

kPa 50 @sat 1

1

==

=≅

⎭⎬⎫

°=−=−==

°

°

f

fhhTT

Pvv

kJ/kg 10.6 mkPa 1kJ 1 kPa)506000)(/kgm 001026.0(

)(

33

121inp,

=⎟⎠

⎞⎜⎝

⎛

⋅−=

−= PPw v


kJ/kg 13.32010.603.314inp,12 =+=+= whh

kJ/kg 4.2336)7.2304)(8660.0(54.340

8660.05019.6

0912.17219.6

kPa 50

KkJ/kg 7219.6kJ/kg 9.3302

C450kPa 6000

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fs

hxhhs

ssx

ssP

sh

TP

qin

qout

50 kPa 1

3

2

4

6 MPa

4s

s

T

kJ/kg 4.2394)4.23369.3302)(94.0(9.3302)( 4s3T3443

43T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη

Thus,

kW 18,050

kW 122

kW 59,660

=−=−=

===

=−=−=

=−=−=

122170,18

kJ/kg) kg/s)(6.10 20(

kW 18,170kJ/kg)4.2394.9kg/s)(3302 20()(

kJ/kg)13.320.9kg/s)(3302 20()(

inP,outT,net

inP,inP,

43outT,

23in

WWW

wmW

hhmW

hhmQ

&&&

&&

&&

&&

and

0.3025===660,59050,18

in

netth Q

W&

&η


10-16

10-24 The change in the thermal efficiency of the cycle in Prob. 10-23 due to a pressure drop in the boiler is to be determined.

Analysis We use the following EES routine to obtain the solution.

"Given" P[2]=6000 [kPa] DELTAP=50 [kPa] P[3]=6000-DELTAP [kPa] T[3]=450 [C] P[4]=50 [kPa] Eta_T=0.94 DELTAT_subcool=6.3 [C] T[1]=temperature(Fluid$, P=P[1], x=x[1])-DELTAT_subcool m_dot=20 [kg/s] "Analysis" Fluid$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) v[1]=volume(Fluid$, P=P[1], T=T[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in Solution DELTAP=50 [kPa] DELTAT_subcool=6.3 [C] Eta_T=0.94 Eta_th=0.3022 Fluid$='steam_iapws' h[1]=314.11 [kJ/kg] h[2]=320.21 [kJ/kg] h[3]=3303.64 [kJ/kg] h[4]=2396.01 [kJ/kg] h_s[4]=2338.1 [kJ/kg] m_dot=20 [kg/s] P[1]=50 P[2]=6000 P[3]=5950 P[4]=50 q_in=2983.4 [kJ/kg] q_out=2081.9 [kJ/kg] s[3]=6.7265 [kJ/kg-K] s[4]=6.7265 [kJ/kg-K] T[1]=75.02 [C] T[3]=450 [C] v[1]=0.001026 [m^3/kg] w_net=901.5 [kJ/kg] w_p_in=6.104 [kJ/kg] x[1]=0

Discussion The thermal efficiency without a pressure drop was obtained to be 0.3025.



10-17

10-25 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined.


Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6),

( )( )( )

kJ/kg 64.34510.554.340kJ/kg 10.5

mkPa 1kJ 1kPa 505000/kgm 001030.0

/kgm 001030.0

kJ/kg 54.340

in,12

33

121in,

3kPa 50 @1

kPa 05 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

1

2

3

4

Rankine cycle

s

( )(kJ/kg 2.2071

7.23047509.054.340

7509.05019.6

09120.19737.5kPa 50

KkJ/kg 9737.5kJ/kg 2.2794

1MPa 5

44

44

34

4

3

3

3

3

=

+=+=

=−

=−

=⎭⎬⎫

==

⋅==

⎭⎬⎫

==

fgf

fg

f

hxhh

sss

xss

P

sh

xP

)

kJ/kg 717.9=−=−=

=−=−==−=−=

7.17306.2448kJ/kg 7.173054.3402.2071kJ/kg 6.244864.3452.2794

outinnet

14out

23in

qqwhhqhhq

29.3%==−=−= 2932.06.24487.173011

in

outth q

qη

(b) Carnot Cycle analysis:

T

1

2 3

4

Carnot cycle

skJ/kg 05.989)7.2304)(2814.0(54.340

2814.05019.6

0912.19207.2kPa 50

KkJ/kg 9207.2kJ/kg 5.1154

0C9.263

C9.263kJ/kg 2.2794

1MPa 5

11

11

21

1

2

2

2

32

3

3

3

3

=+=

+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

=°==

°==

⎭⎬⎫

==

fgf

fg

f

hxhhs

ssx

ssP

sh

xTT

Th

xP

kJ/kg 557.5=−=−==−=−==−=−=

2.10827.1639kJ/kg 2.108254.3402.2071kJ/kg 7.16395.11542.2794

outinnet

14out

23in

qqwhhqhhq

34.0%==−=−= 3400.07.16392.108211

in

outth q

qη



10-18

10-26 A 120-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.



( )( )( )

kJ/kg .0523511.994.225kJ/kg .119

mkPa 1kJ 1kPa 519000/kgm 001014.0

/kgm 0010140.0

kJ/kg 94.225

in,12

33

121in,

3kPa 51 @1

kPa 15 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

p

p

f

f

whh

PPw

hh

v

vv

T

( )( ) kJ/kg 8.22084.23728358.094.225

8358.02522.7

7549.08164.6kPa 15

KkJ/kg 8164.6kJ/kg 0.3512

C550MPa 9

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhh

sss

xss

P

sh

TP

Qin

Qout

15 kPa 1

3

2

4

9 MPa

·

·

s

The thermal efficiency is determined from

kJ/kg 9.198294.2258.2208kJ/kg 9.327605.2350.3512

14out

23in=−=−==−=−=

hhqhhq

and

Thus, ( )( )( ) 28.4%===××=

=−=−=

2843.096.075.03949.0

3949.09.32769.198211

gencombthoverall

in

outth

ηηηη

ηqq

(b) Then the required rate of coal supply becomes

and

tons/h 51.9====

===

kg/s 404.14kJ/kg 29,300

kJ/s 422,050

kJ/s 050,4222843.0

kJ/s 120,000

coal

incoal

overall

netin

CQ

m

WQ

&&

&&

η



10-19

10-27 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.


Analysis (a) We use properties of water for geothermal water (Tables A-4 through A-6)

1661.0

210809.64014.990

kJ/kg 14.990kPa 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

2

reinjection well

separator

steam turbine

1

Flash chamber

65

4

3

condenser

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


10-20

10-28 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.



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

&&&

&&

productionwell

reinjectionwell

separator

steam turbine

1

3

condenser

4

5

6

Flash chamber

Flash chamber

9

separator

8

7kJ/kg 7.234490.0kPa 10

kJ/kg 1.27481

kPa 500

44

4

33

3

=⎭⎬⎫

==

=⎭⎬⎫

==

hxP

hxP

2

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



10-21

10-29 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.



1661.0

kJ/kg 14.990kPa 500

kJ/kg 14.9900

C230

212

2

11

1

=⎭⎬⎫

===

=⎭⎬⎫

=°=

xhh

P

hxT

kg/s 80.19120.38230

kg/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

steam turbine

isobutaneturbine

heat exchanger

pump

BINARY CYCLE

separator

air-cooled condenser

condenser

flash chamber

1

91

8

7

5

2

6

3

4

1reinjection well

production well

kJ/kg 04.377

0C90

kJ/kg 09.6400

kPa 500

77

7

66

6

=⎭⎬⎫

=°=

=⎭⎬⎫

==

hxT

hxP

The isobutane 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

( )( )( )

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 &&

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

&&&

&&



10-22

(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


10.6%==+

=+

= 0.106622,203

6139410,15

in

binarynet,steamT,plantth, E

WW&

&&η



10-23

The Reheat Rankine Cycle

10-30C The pump work remains the same, the moisture content decreases, everything else increases.

10-31C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 10-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansion in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the low-pressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%.

0 20 40 60 80 100 120 140 160 180200

300

400

500

600

700

800

900

s [kJ/kmol-K]

T [K

]

6000 kPa 4000 kPa

500 kPa

20 kPa 0.2 0.4 0.6 0.8

SteamIAPWS

1

2

3

4

5

6

7

10-32C 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.



10-24

10-33 An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.



kJ/kg 88.19907.881.191kJ/kg 07.8 mkPa 1

kJ 1 kPa)108000)(/kgm 001010.0()(

/kgm 001010.0

kJ/kg 81.191

inp,12

33

121inp,

3kPa 10 @1

kPa 10 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

T

kJ/kg 4.2636)0.2108)(9470.0(09.640

9470.09603.4

8604.15579.6

kPa 500

KkJ/kg 5579.6kJ/kg 3.3273

C045kPa 8000

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

1

5

2

6

3

4

500 kPa

10 kPa

8 MPa

s

kJ/kg 9.2564)1.2392)(9921.0(81.191

9921.04996.7

6492.00893.8

kPa 10

KkJ/kg 0893.8kJ/kg 5.3484

C500kPa 500

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Thus,

kJ/kg 5.15481.23735.3921

kJ/kg 1.237381.1919.2564kJ/kg 5.39214.26365.348488.1993.3273)()(

outinnet

16out

4523in

=−=−==−=−=

=−+−=−+−=

qqwhhq

hhhhq


kg/s 3.229===⎯→⎯−=kJ/kg 1548.5kJ/s 5000)(

net

net43net w

WmhhmW

&&&&


0.395=−=−=5.39211.237311

in

outth q

qη



10-25

10-34 An ideal reheat steam Rankine cycle produces 2000 kW power. The mass flow rate of the steam, the rate of heat transfer in the reheater, the power used by the pumps, and the thermal efficiency of the cycle are to be determined.


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

kJ/kg 05.43354.1551.417kJ/kg 54.15 mkPa 1

kJ 1 kPa)10015000)(/kgm 001043.0()(

/kgm 001043.0

kJ/kg 51.417

inp,12

33

121inp,

3kPa 100 @1

kPa 100 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

T

kJ/kg 3.2703)8.1889)(9497.0(47.908

9497.08923.3

4467.21434.6

kPa 2000

KkJ/kg 1434.6kJ/kg 9.3157

C045

kPa 000,15

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

1

5

2

6

4

2 MPa

100 kPa

15 MPa 3

s

kJ/kg 0.2648)5.22257)(9880.0(51.417

9880.00562.6

3028.12866.7

kPa 100

KkJ/kg 2866.7kJ/kg 2.3358

C450kPa 2000

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Thus,

kJ/kg 2.11495.22308.379

kJ/kg 5.223051.4170.2648kJ/kg 8.33793.27032.335805.4339.3157)()(

outinnet

16out

4523in

=−=−==−=−=

=−+−=−+−=

qqwhhq

hhhhq

The power produced by the cycle is

kW 2000=== )kJ/kg 1149.2)(kg/s 1.74(netnet wmW &&

The rate of heat transfer in the rehetaer is

kW 27

kW 1140===

=−=−=

kJ/kg) 4kg/s)(15.5 740.1(

kJ/kg )3.2703.2kg/s)(3358 740.1()(

inP,inP,

45reheater

wmW

hhmQ&&

&&


0.340=−=−=8.33795.223011

in

outth q

qη



10-26

10-35 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.



( )( )( )

kJ/kg .5025708.642.251

kJ/kg .086mkPa 1

kJ 1kPa 206000/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

T

( )( ) kJ/kg 7.23495.23578900.042.251

8900.00752.7

8320.01292.7kPa 20

KkJ/kg 1292.7kJ/kg 4.3248

C400MPa 2

kJ/kg 0.2901MPa 2

KkJ/kg 5432.6kJ/kg 3.3178

C400MPa 6

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

1

5

2

6

3

4

20 kPa

6 MPa

s

The turbine work output and the thermal efficiency are determined from

and ( ) ( )

( ) ( ) kJ/kg 32680.29014.324850.2573.3178

7.23494.32480.29013.3178

4523in

6543outT,

=−+−=−+−=

=−+−=−+−=

hhhhq

hhhhw kJ/kg 1176

Thus,

35.8%====

=−=−=

358.0kJ/kg 3268kJ/kg 1170

kJ/kg 117008.61176

in

netth

in,,net

qw

www poutT

η



10-27

10-36 Problem 10-35 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] = 6000 [kPa] T[3] = 400 [C] P[4] = 2000 [kPa] T[5] = 400 [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]) 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"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"


10-28

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 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

6000 kPa

2000 kPa

20 kPa

SteamIAPWS

1,2

3

4

5

6

Ideal Rankine cycle with reheat

SOLUTION Eff=0.358 Eta_p=1 Eta_t=1 Fluid$='Steam_IAPWS' Q_in=3268 [kJ/kg] Q_out=2098 [kJ/kg] W_net=1170 [kJ/kg] W_p=6.083 [kJ/kg] W_p_s=6.083 [kJ/kg] W_t_hp=277.2 [kJ/kg] W_t_lp=898.7 [kJ/kg] x6s$=''



10-29

10-37E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined.


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

Btu/lbm 06.16381.125.161Btu/lbm 81.1


)(

/lbmft 01659.0

Btu/lbm 25.161

inp,12

33

121inp,

3psia 10 @1

psia 10 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

T

Btu/lbm 5.1187)33.843)(9865.0(46.355

9865.000219.1

54379.05325.1

psia 200


F600psia 600

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

1

5

2

6

4

200 psia

10 psia

600 psia 3

s

Btu/lbm 0.1071)82.981)(9266.0(25.161

9266.050391.1

28362.06771.1

psia 10


F600psia 200

66

46

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Thus,

Btu/lbm 0.3528.9097.1261

Btu/lbm 7.90925.1610.1071Btu/lbm 7.12615.11873.132206.1631289.9)()(

outinnet

16out

4523in

=−=−==−=−=

=−+−=−+−=

qqwhhq

hhhhq


lbm/s 47.13kJ 1

Btu 0.94782Btu/lbm 352.0

kJ/s 5000

net

netnetnet =⎟

⎠⎞

⎜⎝⎛==⎯→⎯=

wW

mwmW&

&&&


Btu/s 12,250Btu/s 16,995

===

===

Btu/lbm) .7lbm/s)(909 47.13(

Btu/lbm) 1.7lbm/s)(126 47.13(

outout

inin

qmQ

qmQ&&

&&


0.2790=⎟⎠⎞

⎜⎝⎛==

kJ 1Btu 0.94782

Btu/s 16,990kJ/s 5000

in

netth Q

W&

&η



10-30

10-38E An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined for a reheat pressure of 100 psia.


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

Btu/lbm 06.16381.125.161Btu/lbm 81.1


)(

/lbmft 01659.0

Btu/lbm 25.161

inp,12

33

121inp,

3psia 6 @1

psia 10 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

T

Btu/lbm 9.1131)99.888)(9374.0(51.298

9374.012888.1

47427.05325.1

psia 100


F600psia 600

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

1

5

2

6

4

100 psia

10 psia

600 psia 3

s

Btu/lbm 2.1124)82.981)(9808.0(25.161

9808.050391.1

28362.07586.1

psia 10


F600

psia 100

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

Thus,

Btu/lbm 5.3619.9624.1324

Btu/lbm 9.96225.1612.1124Btu/lbm 4.13249.11314.132907.1631289.9)()(

outinnet

16out

4523in

=−=−==−=−=

=−+−=−+−=

qqwhhq

hhhhq


lbm/s 11.13kJ 1

Btu 0.94782Btu/lbm 361.5

kJ/s 5000

net

netnetnet =⎟

⎠⎞

⎜⎝⎛==⎯→⎯=

wW

mwmW&

&&&


Btu/s 12,620Btu/s 17,360

===

===

Btu/lbm) .9lbm/s)(962 11.13(

Btu/lbm) 4.4lbm/s)(132 11.13(

outout

inin

qmQ

qmQ&&

&&


0.2729=⎟⎠⎞

⎜⎝⎛==

kJ 1Btu 0.94782

Btu/s 17,360kJ/s 5000

in

netth Q

W&

&η

Discussion The thermal efficiency for 200 psia reheat pressure was determined in the previous problem to be 0.2790. Thus, operating the reheater at 100 psia causes a slight decrease in the thermal efficiency.



10-31

10-39 An ideal reheat Rankine with water as the working fluid is considered. The temperatures at the inlet of both turbines, and the thermal efficiency of the cycle are to be determined.


T



kJ/kg 87.19806.781.191kJ/kg 06.7 mkPa 1

kJ 1 kPa)107000)(/kgm 001010.0()(

/kgm 001010.0

kJ/kg 81.191

inp,12

33

121inp,

3kPa 10 @1

kPa 10 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛⋅

−=

−=

==

==

whh

PPw

hh

f

f

v

vv

C373.3°==

⎭⎬⎫

==

⋅=+=+==+=+=

⎭⎬⎫

==

3

3

43

3

44

44

4

4

kJ/kg 5.3085

kPa 7000

KkJ/kg 3385.6)6160.4)(93.0(0457.2kJ/kg 0.2625)5.2047)(93.0(87.720

93.0

kPa 800

Th

ssP

sxsshxhh

xP

fgf

fgf

1

5

2

6

3

4

800 kPa

10 kPa

7 MPa

s

C416.2°==

⎭⎬⎫

==

⋅=+=+==+=+=

⎭⎬⎫

==

5

5

65

5

66

66

6

6

kJ/kg 0.3302

kPa 800

KkJ/kg 6239.7)4996.7)(93.0(6492.0kJ/kg 4.2416)1.2392)(93.0(81.191

90.0kPa 10

Th

ssP

sxsshxhh

xP

fgf

fgf

Thus,

kJ/kg 6.222481.1914.2416

kJ/kg 6.35630.26250.330287.1985.3085)()(

16out

4523in

=−=−==−+−=−+−=

hhqhhhhq

and

37.6%==−=−= 3757.06.35636.222411

in

outth q

qη


10-32

10-40 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.



T

( )( )( )

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

( )( )( )( )

( )

kJ/kg 2.2817MPa 15.2

kJ/kg 61.3466pressurereheat 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 2150

1

5

2

6

3

4

10 kPa

15

s

(b) The rate of heat supply is

( ) ( )[ ]

( )( )kW 45,039=

−+−=−+−=

kJ/kg2.281761.346695.2068.3310kg/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



10-33

10-41 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.



( )( )( )

( ) ( )( ) 3) (Eq. 2.335885.02.3358

2) (Eq. ?

1) (Eq. 95.0

?

KkJ/kg 2815.7kJ/kg 2.3358

C450 MPa2

kJ/kg 3.30271.29485.347685.05.3476

kJ/kg 1.2948 MPa2

KkJ/kg 6317.6kJ/kg 5.3476

C550 MPa5.12

6655665

65

656

6

66

6

5

5

5

5

4334

43

43

434

4

3

3

3

3

ssTs

T

s

sT

sT

ss

hhhhhhhhh

hss

P

hxP

sh

TP

hhhh

hhhh

hss

P

sh

TP

−−=−−=⎯→⎯−−

=

=⎭⎬⎫

==

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

=−−=

−−=→

−−

=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

ηη

η

η

1

5

2s

6s

3

4s 12.5 MPa

P = ? 6

4

2

T

3

1 2

Turbine Boiler

CondenserPump

5

4

6

s

The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above three 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 7.227357.1893.2463

kJ/kg 8.36033.24632.335859.2035.3476

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



10-34

Regenerative Rankine Cycle

10-42C Moisture content remains the same, everything else decreases.

10-43C 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.

10-44C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing.

10-45C Both cycles would have the same efficiency.

10-46C 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



10-35

10-47E Feedwater is heated by steam in a feedwater heater of a regenerative Rankine cycle. The ratio of the bleed steam mass flow rate to the inlet feedwater mass flow rate is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Tables A-4E through A-6E or EES),

h1 ≅ hf @ 110°F = 78.02 Btu/lbm

Btu/lbm 2.1167 F250

psia 202

2

2 =⎭⎬⎫

°==

hTP

h3 ≅ hf @ 225°F = 193.32 Btu/lbm

Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance:


321

outin

(steady) 0systemoutin

0

mmmmm

mmm

&&&

&&

&&&

=+=

=∆=−

Energy balance:

)(0)peke (since

0

3212211

332211

outin

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

outin

hmmhmhmWQhmhmhm

EE

EEE

&&&&

&&&&&

&&

444 3444 21&

43421&&

+=+≅∆≅∆===+

=

=∆=−

Water 110°F 20 psia

3 2

1

Steam 20 psia 250°F

Water 20 psia 225°F

Solving for the bleed steam mass flow rate to the inlet feedwater mass flow rate, and substituting gives

0.118=−−

=−−

=kJ/kg )2.116732.193(kJ/kg )32.19302.78(

23

31

1

2

hhhh

mm&

&


10-36

10-48 In a regenerative Rankine cycle, the closed feedwater heater with a pump as shown in the figure is arranged so that the water at state 5 is mixed with the water at state 2 to form a feedwater which is a saturated liquid. The amount of bleed steam required to heat 1 kg of feedwater is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

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

1

7000 kPasat. liq.

Feedwater 7000 kPa 260°C

6

5

3

Steam 6000 kPa 325°C

4

2

kJ/kg 5.1267 0

kPa 7000

kJ/kg 5.2969 C325kPa 6000

kJ/kg 8.1134 C260kPa 7000

kPa 7000 @ 66

6

33

3

C260 @ 11

1

==⎭⎬⎫

==

=⎭⎬⎫

°==

=≅⎭⎬⎫

°==

°

f

f

hhxP

hTP

hhTP

Analysis We take the entire unit as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

631inP,33311

66inP,33311

outin

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

outin

)(

0

hmmwmhmhmhmwmhmhm

EE

EEE

&&&&&

&&&&

&&

444 3444 21&

43421&&

+=++

=++=

=∆=−

Solving this for , 3m&

kg/s 0.0779=+−

−=

+−−

=319.15.12675.2969

8.11345.1267kg/s) 1()( inP,63

1613 whh

hhmm &&

where

kJ/kg 319.1

mkPa 1kPa 1kPa )60007000)(/kgm 001319.0(

)()(

33

45kPa 6000 @ 454inP,

=⎟⎠

⎞⎜⎝

⎛⋅

−=

−=−= PPPPw fvv



10-37

10-49E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat rejected in the condenser 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-4E, A-5E, and A-6E),

Btu/lbm 29.13011.018.130Btu/lbm 11.0


)(

/lbmft 01641.0

Btu/lbm 18.130

inpI,12

33

121inPI,

3psia 5 @1

psia 5 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

1

qin

2

7

1-y 6

qout

500 psia

5 psia

y 40 psia

4

3

5

s

T

Btu/lbm 60.23746.114.236Btu/lbm 46.1


)(

/lbmft 01715.0

Btu/lbm 14.236

inpII,34

33

343inPII,

3psia 40 @3

psia 40 @3

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 63.953)5.1000)(8230.0(18.130

8230.060894.1

23488.05590.1

psia 5

Btu/lbm 4.1084)69.933)(9085.0(14.236

9085.028448.1

39213.05590.1

psia 40


F600psia 500

77

47

57

7

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

sh

TP

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

326332266

(steady) 0

1)1(

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

outin

systemoutin

=−+⎯→⎯=+⎯→⎯=

=

=∆=−

∑∑ &&&&&

&&

&&&

where y is the fraction of steam extracted from the turbine ( 36 / mm &&= ). Solving for y,

1109.029.1304.108429.13014.236

26

23 =−−

=−−

=hhhh

y

Then,

Btu/lbm 732.1Btu/lbm 1.57

Btu/lbm 330.5

=−−=−−=

=+=+=

=−−+−=−−+−=

)18.13063.953)(1109.01())(1(

46.111.0)63.9534.1084)(1109.01(4.10846.1298))(1(

17out

inPII,inPI,inP,

7665outT,

hhyq

wwwhhyhhw

Also, Btu/lbm 106160.2376.129845in =−=−= hhq

3100.01061

1.73211in

outth =−=−=

qq

η



10-38

10-50E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The change in thermal efficiency when the steam supplied to the open feedwater heater is at 60 psia rather than 40 psia is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

Btu/lbm 35.13017.018.130Btu/lbm 17.0


)(

/lbmft 01641.0

Btu/lbm 18.130

inp,12

33

121inPI,

3psia 5 @1

psia 5 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

1

qin

2

7

1-y 6

qout

500 psia

5 psia

y 60 psia

4

3

5

s

T

Btu/lbm 62.26342.120.262Btu/lbm 42.1


)(

/lbmft 01738.0

Btu/lbm 20.262

inp,34

33

343inPII,

3psia 60 @3

psia 60 @3

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 63.953)5.1000)(8230.0(18.130

8230.060894.1

23488.05590.1

psia 5

Btu/lbm 7.1113)61.915)(9300.0(20.262

9300.021697.1

42728.05590.1

psia 60


F600psia 500

77

47

57

7

66

46

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

fgf

fg

f

hxhhs

ssx

ssP

hxhhs

ssx

ssP

sh

TP

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

326332266

(steady) 0

1)1(

0

hhyyhhmhmhmhmhm

EE

EEE

eeii

outin

systemoutin

=−+⎯→⎯=+⎯→⎯=

=

=∆=−

∑∑ &&&&&

&&

&&&


1341.035.1307.111335.13020.262

26

23 =−−

=−−

=hhhh

y

Then,

Btu/lbm 713.0)18.13063.953)(1341.01())(1(Btu/lbm 103562.2636.1298

17out

45in

=−−=−−==−=−=

hhyqhhq

and 0.3111=−=−=1035

0.71311in

outth q

qη

When the reheater pressure is increased from 40 psia to 60 psia, the thermal efficiency increases from 0.3100 to 0.3111, which is an increase of 0.35%.



10-39

10-51E The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES.

Analysis The EES program used to solve this problem as well as the solutions are given below.

P[5]=500 [psia] T[5]=600 [F] P[6]=60 [psia] P[7]=5 [psia] x[3]=0 "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia-ft^3, Btu) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out Eta_th=1-q_out/q_in



10-40

P 6

[kPa] ηth

10 0.3011 20 0.3065 30 0.3088 40 0.3100 50 0.3107 60 0.3111 70 0.31126 80 0.31129 90 0.3112 100 0.3111 110 0.3109 120 0.3107 130 0.3104 140 0.3101 150 0.3098 160 0.3095 170 0.3091 180 0.3087 190 0.3084 200 0.3080

0 40 80 120 160 2000.3

0.302

0.304

0.306

0.308

0.31

0.312

Bleed pressure [kPa]

ηth



10-41

10-52 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net power output of the power plant and the thermal efficiency of the cycle are to be determined.


Analysis

P III P II P I

fwh II fwh I Condenser

Boiler Turbine

6

5

4 3

2 1

10

9

7

8

T

1

2

10

1 - y

8

91 - y - z

0.6 MPa

5 kPa

0.2 MPa

4y

3

10 MPa

5

6

7

s

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

( ) ( )( )kJ/kg 95.13720.075.137

kJ/kg 02.0mkPa 1

kJ 1kPa 5200/kgm 0.001005

/kgm 001005.0

kJ/kg 75.137

in,12

33

121in,

3kPa 5 @1

kPa 5 @1

=+=+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

==

==

pI

pI

f

f

whh

PPw

hh

v

vv

( ) ( )( )

( ) ( )( )kJ/kg 10.35

mkPa 1kJ 1

kPa 60010,000/kgm 0.001101

/kgm 001101.0

kJ/kg 38.670

liquidsat.MPa 6.0

kJ/kg 13.50542.071.504kJ/kg 0.42

mkPa 1kJ 1

kPa 200600/kgm 0.001061

/kgm 001061.0

kJ/kg 71.504

liquidsat.MPa 2.0

33

565in,

3MPa 6.0 @5

MPa 6.0 @55

in,34

33

343in,

3MPa 2.0 @3

MPa 2.0 @33

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

==

==

⎭⎬⎫=

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

==

==

⎭⎬⎫=

PPw

hhP

whh

PPw

hhP

pIII

f

f

pII

pII

f

f

v

vv

v

vv

kJ/kg 8.2821MPa 6.0

KkJ/kg 9045.6kJ/kg 8.3625

C600MPa 10

kJ/kg 73.68035.1038.670

878

8

7

7

7

7

in,56

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

=+=+=

hss

P

sh

TP

whh pIII

( )( ) kJ/kg 7.26186.22019602.071.504

9602.05968.5

5302.19045.6MPa 2.0

99

99

79

9

=+=+=

=−

=−

=

⎭⎬⎫

==

fgf

fg

f

hxhh

sss

xss

P



10-42

( )( ) kJ/kg 0.21050.24238119.075.137

8119.09176.7

4762.09045.6kPa 5

1010

1010

710

10

=+=+=

=−

=−

=

⎭⎬⎫

==

fgf

fg

f

hxhh

sss

x

ssP

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

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/kg 2.13888.15560.2945

kJ/kg 1556.8137.752105.01373.007133.011kJ/kg 0.294573.6808.3625

outinnet

110out

67in

=−=−==−−−=−−−=

=−=−=

qqwhhzyq

hhq

and

( )( ) MW 30.5≅=== kW 540,30kJ/kg 1388.2kg/s 22netnet wmW &&

(b) The thermal efficiency is

47.1%=−=−=kJ/kg 2945.0kJ/kg 1556.8

11in

outth q

qη



10-43

10-53 An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat added in the boiler 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/kg 45.25403.342.251kJ/kg 03.3 mkPa 1

kJ 1 kPa)203000)(/kgm 001017.0()(

/kgm 001017.0

kJ/kg 42.251

inp,12

33

121inp,

3kPa 20 @1

kPa 20 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

kJ/kg 7.2221)5.2357)(8357.0(42.251

8357.00752.7

8320.07450.6

kPa 20

kJ/kg 9.2851 kPa 1000

KkJ/kg 7450.6kJ/kg 1.3116

C350kPa 3000

66

66

46

6

545

5

4

4

4

4

=+=+=

=−

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

hss

P

sh

TP

5

Condenser

12 3

4 Turbine

Boiler

Closed fwh

8 7

Pump

6

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.

1

qin

2

6

1-y 5

3 MPa

20 kPa

qout

y 3 1 MPa 7

8

4

s

T

kJ/kg 53.763 C9.209

kPa 3000kJ/kg 51.762

C9.179kJ/kg 51.762

0

kPa 1000

373

3

78

7

7

7

7

=⎭⎬⎫

°===

==

°==

⎭⎬⎫

==

hTT

Phh

Th

xP

An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( ) for closed feedwater heater:

45 / mm &&=

7325

77332255

11 yhhhyhhmhmhmhm

hmhm eeii

+=++=+

=∑∑&&&&

&&

Rearranging,

2437.051.7629.285145.25453.763

75

23 =−−

=−−

=hhhh

y

Then,

kJ/kg 2353

kJ/kg 3.03kJ/kg 740.9

=−=−=

=

=−−+−=−−+−=

53.7631.3116

)7.22219.2851)(2437.01(9.28511.3116))(1(

34in

inP,

6554outT,

hhqw

hhyhhw

Also, kJ/kg 8.73703.39.740inP,outT,net =−=−= www

3136.02353

8.737

in

netth ===

qw

η



10-44

10-54 Problem 10-53 is reconsidered. The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES.


"Given" P[4]=3000 [kPa] T[4]=350 [C] P[5]=600 [kPa] P[6]=20 [kPa] P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in



10-45

P 6 [kPa]

ηth

100 0.32380 110 0.32424 120 0.32460 130 0.32490 140 0.32514 150 0.32534 160 0.32550 170 0.32563 180 0.32573 190 0.32580 200 0.32585 210 0.32588 220 0.32590 230 0.32589 240 0.32588 250 0.32585 260 0.32581 270 0.32576 280 0.32570 290 0.32563

100 140 180 220 260 3000.3238

0.324

0.3243

0.3245

0.3248

0.325

0.3253

0.3255

0.3258

0.326

Bleed pressure [kPa]

ηth



10-46

10-55 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is 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 or EES),

kJ/kg 45.25403.342.251kJ/kg 03.3 mkPa 1

kJ 1 kPa)203000)(/kgm 001017.0()(

/kgm 001017.0

kJ/kg 42.251

inp,12

33

121inp,

3kPa 20 @1

kPa 20 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛⋅

−=

−=

==

==

whh

PPw

hh

f

f

v

vv

kJ/kg 7.2221)5.2357)(8357.0(42.251

8357.00752.7

8320.07450.6

kPa 20

kJ/kg 9.2851 kPa 1000

KkJ/kg 7450.6kJ/kg 1.3116

C350kPa 3000

66

66

46

6

545

5

4

4

4

4

=+=+=

=−

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fss

s

ss

hxhhs

ssx

ssP

hss

P

sh

TP

5 6

12 3

4 Turbine

Boiler

Condenser

Closed fwh

7

Pump

kJ/kg 3.2878)9.28511.3116)(90.0(1.3116)( 5s4T4554

54T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη

kJ/kg 1.2311)7.22211.3116)(90.0(1.3116)( 6s4T4664

64T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη


1

qin

2

6

4

5s

3 MPa

20 kPa

3 1 MPa 7

1-y

y

qout

5

6s

s

T

kJ/kg 53.763 C9.209

kPa 3000

C9.179kJ/kg 51.762

0

kPa 1000

373

3

7

7

7

7

=⎭⎬⎫

°===

°==

⎭⎬⎫

==

hTT

P

Th

xP


45 / mm &&=

7325

77332255

11 yhhhyhhmhmhmhm

hmhm eeii

+=++=+

=∑∑&&&&

&&

Rearranging,

2406.051.7623.287845.25453.763

75

23 =−−

=−−

=hhhh

y

Then,

kJ/kg 235353.7631.3116kJ/kg 3.03

kJ/kg 668.5)1.23113.2878)(2406.01(3.28781.3116))(1(

34in

inP,

6554outT,

=−=−=

=

=−−+−=−−+−=

hhqw

hhyhhw

Also, kJ/kg 5.66503.35.668inP,outT,net =−=−= www

28.3%==== 0.28292353

5.665

in

netth q

wη



10-47

10-56 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is to be determined.


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

5

Condenser

12 3

4 Turbine

Boiler

Closed fwh

7

Pump

6

1

qin

2

6

4

5s

3 MPa

20 kPa

1 MPa

1-y 7

y 3

qout

5

6s

s

T

When the liquid enters the pump 10°C cooler than a saturated liquid at the condenser pressure, the enthalpies become

/kgm 001012.0

kJ/kg 34.209

C501006.6010kPa 20

3C50 @ 1

C50 @ 1

kPa 20 @sat 1

1

=≅

=≅

⎭⎬⎫

°≅−=−==

°

°

f

fhhTT

Pvv

kJ/kg 02.3 mkPa 1

kJ 1 kPa)203000)(/kgm 001012.0()(

33

121inp,

=⎟⎠

⎞⎜⎝

⎛

⋅−=

−= PPw v

kJ/kg 36.21202.334.209inp,12 =+=+= whh

kJ/kg 7.2221)5.2357)(8357.0(42.251

8357.00752.7

8320.07450.6

kPa 20

kJ/kg 9.2851 kPa 1000

KkJ/kg 7450.6kJ/kg 1.3116

C350kPa 3000

66

66

46

6

545

5

4

4

4

4

=+=+=

=−

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fss

s

ss

hxhhs

ssx

ssP

hss

P

sh

TP

kJ/kg 3.2878)9.28511.3116)(90.0(1.3116)( 5s4T4554

54T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη

kJ/kg 1.2311)7.22211.3116)(90.0(1.3116)( 6s4T4664

64T =−−=−−=⎯→⎯

−−

= hhhhhhhh

sηη


kJ/kg 53.763 C9.209

kPa 3000

C9.179kJ/kg 51.762

0

kPa 1000

373

3

7

7

7

7

=⎭⎬⎫

°===

°==

⎭⎬⎫

==

hTT

P

Th

xP



10-48


45 / mm &&=

7325

77332255

11 yhhhyhhmhmhmhm

hmhm eeii

+=++=+

=∑∑&&&&

&&

Rearranging,

2605.051.7623.287836.21253.763

75

23 =−−

=−−

=hhhh

y

Then,

kJ/kg 235353.7631.3116

kJ/kg 3.03kJ/kg 657.2)1.23113.2878)(2605.01(3.28781.3116))(1(

34in

inP,

6554outT,

=−=−=

=

=−−+−=−−+−=

hhqw

hhyhhw

Also,

kJ/kg 2.65403.32.657inP,outT,net =−=−= www

0.2781===2353

2.654

in

netth q

wη



10-49

10-57 The effect of pressure drop and non-isentropic turbine on the rate of heat input is to be determined for a given power plant.


"Given" P[3]=3000 [kPa] DELTAP_boiler=10 [kPa] P[4]=P[3]-DELTAP_boiler T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] eta_T=0.90 P[2]=P[3] P[7]=P[5] P[1]=P[6] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in



10-50

Solution with 10 kPa pressure drop in the boiler: DELTAP_boiler=10 [kPa] eta_T=0.9 Eta_th=0.2827 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=2990 [kPa] q_in=2352.8 [kJ/kg] w_net=665.1 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] w_T_out=668.1 [kJ/kg] y=0.2405 Solution without any pressure drop in the boiler: DELTAP_boiler=0 [kPa] eta_T=1 Eta_th=0.3136 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=3000 [kPa] q_in=2352.5 [kJ/kg] w_net=737.8 [kJ/kg] w_p_in=3.031 [m^3-kPa/kg] w_T_out=740.9 [kJ/kg] y=0.2437



10-51

10-58 A steam power plant operates on an ideal regenerative Rankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam through the boiler for a net power output of 400 MW and the thermal efficiency of the cycle are to be determined.


1-y-z

Condenser

Turbine

1

z 11

8

9

10

2 7

4

6

5

Open fwh

P I

P II

3

y

Closedfwh

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

( )( )( )

kJ/kg 40.19260.081.191kJ/kg 60.0

mkPa 1kJ 1

kPa 10600/kgm 0.00101

/kgm 00101.0

kJ/kg 81.191

in,12

33

121in,

3kPa 10 @1

kPa 10 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

pI

pI

f

f

whh

PPw

hh

v

vv

( )

( )( )

kJ/kg 73.68035.1038.670kJ/kg 10.35

mkPa 1kJ 1

kPa 00610000/kgm 0.001101

/kgm 001101.0

kJ/kg 38.670

liquid sat.MPa 6.0

in,34

33

343in,

3MPa 3.0 @3

MPa 3.0 @33

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−===

==

⎭⎬⎫=

pII

pII

f

f

whh

PPw

hhP

v

vv

T

1

2

11

9

1 - y - z 10

10 MPa

1.2 MPa

10 kPa

0.6 MPa

4 y

37 z

5 6

8

s

kJ/kg 5.2974MPa 2.1

KkJ/kg 9045.6kJ/kg 8.3625

C600MPa 10

kJ/kg 798.33 MPa 10 ,

C0.188

kJ/kg 33.798

liquid sat.MPa 2.1

989

9

8

8

8

8

5556

MPa .21 @sat6

MPa 2.1 @766

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

=→==

°==

===

⎭⎬⎫=

hss

P

sh

TP

hPTT

TT

hhhP f

( )( ) kJ/kg 0.21871.23928341.081.191

8341.04996.7

6492.09045.kPa 10

kJ/kg 9.2820MPa 6.0

1111

1111

811

11

10810

10

=+=+=

=−6

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

fgf

fg

f

hxhh

sss

x

ssP

hss

P


( ) ( ) ( ) ( 4569455699

outin

(steady) 0systemoutin 0

hhhhyhhmhhmhmhm

EE

EEE

eeii −=−⎯→⎯−=−⎯→⎯=

=

=∆=−

∑∑ &&&&

&&

&&&

)


05404.033.7985.297473.68033.798

69

45 =−−

=−−

=hhhh

y



10-52

For the open FWH,


)

( ) ( 310273310102277

outin


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,

( ) ( ) ( )( ) 1694.0

40.1929.282040.19233.79805404.040.19238.670

210

2723 =−

−−−=

−−−−

=hh

hhyhhz

Then,

( )( ) ( )( )kJ/kg 127815492827

kJ/kg 154981.1910.21871694.005404.011kJ/kg 282733.7988.3625

outinnet

111out

58in

=−=−==−−−=−−−=

=−=−=

qqwhhzyq

hhq

and

kg/s 313.0===kJ/kg 1278

kJ/s 400,000

net

net

wW

m&

&

(b) 45.2%==−=−= 452.0kJ/kg 2827kJ/kg 154911

in

outth q

qη


10-53

10-59 Problem 10-58 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.


"Input Data" P[8] = 10000 [kPa] T[8] = 600 [C] P[9] = 1200 [kPa] P_cfwh=600 [kPa] P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=400 [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]" 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]" PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.


10-54

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_in W_dot_net = m_dot * w_net

ηturb ηth m [kg/s] 0.7

0.75 0.8

0.85 0.9

0.95 1

0.3834 0.397

0.4096 0.4212 0.4321 0.4423 0.452

369 356.3 345.4 335.8 327.4 319.8 313

ηpump ηth m [kg/s] 0.7

0.75 0.8

0.85 0.9

0.95 1

0.4509 0.4511 0.4513 0.4515 0.4517 0.4519 0.452

313.8 313.6 313.4 313.3 313.2 313.1 313



10-55

Pcfwh [kPa] y z

100 200 300 400 500 600 700 800 900

1000

0.1702 0.1301 0.1041

0.08421 0.06794 0.05404 0.04182 0.03088 0.02094 0.01179

0.05289 0.09634 0.1226 0.1418 0.1569 0.1694 0.1801 0.1895 0.1979 0.2054

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

]

10000 kPa

1200 kPa 600 kPa

10 kPa

SteamIAPWS

1,2

3,45,6

7

8

9

10

11

0.7 0.75 0.8 0.85 0.9 0.95 10.38

0.39

0.4

0.41

0.42

0.43

0.44

0.45

0.46

ηturb

ηth



10-56

0.7 0.75 0.8 0.85 0.9 0.95 1310

320

330

340

350

360

370

η turb

m [

kg/s

]

0.7 0.75 0.8 0.85 0.9 0.95 10.4508

0.451

0.4512

0.4514

0.4516

0.4518

0.452

0.4522

312.9

313

313.1

313.2

313.3

313.4

313.5

313.6

313.7

313.8

ηpump

ηth

m [

kg/s

]

100 200 300 400 500 600 700 800 900 10000

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.22

Pcfwh [kPa]

y

z



10-57

10-60 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 the thermal efficiency are to be determined.



( )

kJ/kg 65.260

223.942.251

kJ/kg .22988.01)kPa 200080)(/kgm 0.001017(

/

/kgm 001017.0

kJ/kg 42.251

in,12

3

121in,

3kPa 20 @1

kPa 20 @1

=

+=

+=

=

−=

−=

==

==

pI

ppI

f

f

whh

PPw

hh

ηv

vv


/kgm 001127.0

kJ/kg 51.762

liquid sat.MPa 1

3MPa 1 @3

MPa 1 @33

==

==

⎭⎬⎫=

f

fhhPvv

( )

kJ/kg 48.77197.851.762kJ/kg 97.8

88.0/)kPa 00108000)(/kgm 001127.0(

/

in,311

3

3113in,

=+=+==

−=

−=

pII

ppII

whh

PPw ηv

6

5

8

1 2

3

High-P turbine

Boiler

Cond. Closed

fwh

PI PII

10

11

MixingCham.

y

7 1-y9

Low-P turbine

4

Also, h4 = h10 = h11 = 771.48 kJ/kg since the two fluid streams which are being mixed have the same enthalpy.

( )( )( ) kJ/kg 1.31407.31045.339988.05.3399

kJ/kg 7.3104MPa 3

KkJ/kg 7266.6kJ/kg 5.3399

C500MPa 8

655665

65

656

6

5

5

5

5

=−−=−−=⎯→⎯

−−

=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

sTs

T

s

hhhhhhhh

hss

P

sh

TP

ηη

( )( )( )

C349.9°=⎭⎬⎫

==

=−−=−−=⎯→⎯

−−

=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

88

8

877887

87

878

8

7

7

7

7

kJ/kg 9.3157MPa 1

kJ/kg 9.31571.31172.345788.02.3457

kJ/kg 1.3117MPa 1

KkJ/kg 2359.7kJ/kg 2.3457

C500MPa 3

ThP

hhhhhhhh

hss

P

sh

TP

sTs

T

s

ηη


10-58

( )( )( ) kJ/kg 9.25132.23852.345788.02.3457

kJ/kg 2.2385kPa 20

977997

97

979

9

=−−=

−−=⎯→⎯−−

=

=⎭⎬⎫

==

sTs

T

s

hhhhhhhh

hss

P

ηη

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

( )( ) ( )1758.0)51.7629.3157()65.26048.771)(1(

1 38210

=⎯→⎯−=−−

−=−−

yyy

hhyhhy

The corresponding mass flow rate is

kg/s 2.637=== kg/s) 15)(1758.0(58 mym &&

(c) Then,

( )( ) ( )( ) kJ/kg 8.186442.2519.25131758.011kJ/kg 2.29451.31402.345748.7715.3399

19out

6745in

=−−=−−==−+−=−+−=

hhyqhhhhq

and

kW 16,206=−=−= kJ/kg)8.18648.2945)(kg/s 15()( outinnet qqmW &&

(b) The thermal efficiency is determined from

36.7%==−=−= 3668.0kJ/kg 2945.8kJ/kg 1864.811

in

outth q

qη



10-59

10-61 A Rankine steam cycle modified with two closed feedwater heaters is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass extracted for the closed feedwater heater z and the cooling water flow rate are to be determined. Also, the net power output 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 (b) Using the data from the problem statement, the enthalpies at various states are

( )( )( )

kJ/kg 1.2561.5251kJ/kg 1.5

mkPa 1kJ 1

kPa 205000/kgm 0.00102

/kgm 00102.0kJ/kg 251

inpI,12

33

121inpI,

3kPa 20 @ 1

kPa 20 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

1

2

9

s

5

6 245 kPa

20 kPa

1.4 MPa

8 4

31-y-z

1-y

y

5 MPa

11

121-y-z

y

10 z

y+z

8

7

T

Also,

operation) valve(throttle

kJ/kg 533

1112

kPa 245 @ 113

hhhhh f

====

operation) valve(throttle

kJ/kg 830

910

kPa 1400 @ 94

hhhhh f

====

An energy balance on the closed feedwater heater gives 1131072 )(11 hzyhyhzhh ++=++

where z is the fraction of steam extracted from the low-pressure turbine. Solving for z,

0.1017=−

−+−=

−−(+−(

=5332918

)830533)(1153.0()1.256533())

117

101123

hhhhyhh

z

(c) An energy balance on the condenser gives

1112128822

2111212188

)( hmhmhmhhmhmhmhmhmhm

www

wwww

&&&&

&&&&&

−+=−+=++

Solving for the mass flow rate of cooling water, and substituting with correct units,

[ ]

[ ]

kg/s 2158=

−++−−=

∆−++−−

=

)10)(18.4()251(1)533)(1017.01153.0()2477)(1017.01153.01()50(

1)()1( 11285

wpww Tc

hhzyhzymm

&&

(d) The work output from the turbines is

kJ/kg 1271

)2477)(1017.01153.01()2918)(1017.0()3406)(1153.0(3900)1( 8765outT,

=−−−−−=

−−−−−= hzyzhyhhw

The net work output from the cycle is

kJ/kg 9.12651.51271

inP,outT,net

=−=

−= www

The net power output is

MW 63.3==== kW 300,63kJ/kg) .9kg/s)(1265 50(netnet wmW &&

The rate of heat input in the boiler is

kW 500,153kJ/kg )8303900)(kg/s 50()( 45in =−=−= hhmQ &&

The thermal efficiency is then

41.2%==== 412.0kW 500,153kW 300,63

in

netth Q

W&

&η



10-60

Second-Law Analysis of Vapor Power Cycles

10-62C In the simple ideal Rankine cycle, irreversibilities occur during heat addition and heat rejection processes 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 by minimizing the temperature differences during heat transfer in the boiler and the condenser. One way of doing that is regeneration.

10-63E The exergy destructions associated with each of the processes of the Rankine cycle described in Prob. 10-17E are to be determined for the specified source and sink temperatures.


Analysis From Problem 10-17E,

qin

qout

3 psia 1

3

2

4

800 psia

T

s

Btu/lbm 8.86540.10924.975Btu/lbm 2.134481.1110.1456

RBtu/lbm 6413.1

RBtu/lbm 2009.0

14out

23in

43

psia 3 @ 21

=−=−==−=−=

⋅==

⋅===

hhqhhq

ss

sss f

The exergy destruction during a process of a stream from an inlet state to exit state is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie

Application of this equation for each process of the cycle gives

Btu/lbm 146

Btu/lbm 377

=⎟⎠⎞

⎜⎝⎛ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

=⎟⎠⎞

⎜⎝⎛ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

R 500Btu/lbm 8.8656413.12009.0)R 500(

R 1960Btu/lbm 2.13442009.06413.1)R 500(

sink

out41041 destroyed,

source

in23023 destroyed,

Tq

ssTx

Tq

ssTx

Processes 1-2 and 3-4 are isentropic, and thus

00

=

=

34 destroyed,

12 destroyed,

x

x



10-61

10-64 The exergy destruction associated with the heat rejection process in Prob. 10-21 is to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined.


Analysis From Problem 10-21,

kJ/kg 8.1961kJ/kg 4.3411

KkJ/kg 8000.6

KkJ/kg 6492.0

out

3

43

kPa10@21

==

⋅==

⋅===

qh

ss

sss f

The exergy destruction associated with the heat rejection process is

( ) kJ/kg 178.0=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−=

K 290kJ/kg 1961.8

8000.66492.0K 29041,41041destroyed,

R

R

Tq

ssTx

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

( ) ( )( ) ( )03003

03

023

030033 2ssThhqzssThh

−−−=++−−−=

Vψ

where

( )( ) KkJ/kg 2533.0

kJ/kg 95.71

K 290 @ kPa 100,K 290@0

K 290 @ kPa 100 ,K 290@0

⋅=≅==≅=

f

f

ssshhh

Thus,

( ) ( )( ) kJ/kg 1440.9=⋅−−−=ψ KkJ/kg 2532.0800.6K 290kJ/kg 95.714.34113



10-62

10-65 The component of the ideal reheat Rankine cycle described in Prob. 10-33 with the largest exergy destruction is to be identified.


Analysis From Prob. 10-33,

kJ/kg 1.2373kJ/kg 1.8484.26364.3485kJ/kg 4.307388.1993.3273

KkJ/kg 0893.8KkJ/kg 5579.6

KkJ/kg 6492.0

out

455-4 in,

233-2 in,

65

43

kPa 10 @ 21

=

=−=−=

=−=−=⋅==⋅==

⋅===

qhhqhhq

ssss

sss f T

1

5

2

6

3

4

500 kPa

10 kPa

8 MPa

s


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


kJ/kg 267.6

kJ/kg 161.6

kJ/kg 687.1

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⎟⎠

⎞⎜⎝

⎛ −−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⎟⎠

⎞⎜⎝

⎛ −−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

K 283kJ/kg 1.23730893.86492.0)K 283(

K 883kJ/kg 1.8485579.60893.8)K 283(

K 883kJ/kg 4.30736492.05579.6)K 283(

sink

out61061 destroyed,

source

5-4 in,45045 destroyed,

source

3-2 in,23023 destroyed,

Tq

ssTx

Tq

ssTx

Tq

ssTx

Processes 1-2, 3-4, and 5-6 are isentropic, and thus,

000

=

=

=

56 destroyed,

34 destroyed,

12 destroyed,

x

x

x

The greatest exergy destruction occurs during heat addition process 2-3.



10-63

10-66 The exergy destructions associated with each of the processes of the reheat Rankine cycle described in Prob. 10-35 are to be determined for the specified source and sink temperatures.



kJ/kg 3.209842.2517.2349

kJ/kg 3.3470.29014.3248

kJ/kg 8.292050.2573.3178KkJ/kg 1292.7KkJ/kg 5432.6

KkJ/kg 8320.0

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/kg 240.6

kJ/kg 104.6

kJ/kg 1110

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−=

=⎟⎠

⎞⎜⎝

⎛ −+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+−=

K 295kJ/kg 2098.31292.78320.0K 295

K 1500kJ/kg 347.35432.61292.7K 295

K 1500kJ/kg .829208320.05432.6K 295

61,61061destroyed,

45,45045destroyed,

23,23023destroyed,

R

R

R

R

R

R

Tq

ssTx

Tq

ssTx

Tq

ssTx



10-64

10-67 Problem 10-66 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. Also, the T-s diagram is to be plotted.


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] = 6000 [kPa] T[3] = 400 [C] P[4] = 2000 [kPa] T[5] = 400 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency"} "Data for the irreversibility calculations:" T_o = 295 [K] T_R_L = 295 [K] T_R_H = 1500 [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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,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]) 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"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine"


10-65

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 "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 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

6000 kPa

2000 kPa

20 kPa

SteamIAPWS

1,2

3

4

5

6

Ideal Rankine cycle with reheat

SOLUTION Eff=0.358 Eta_p=1 Eta_t=1 Fluid$='Steam_IAPWS' i_12=0.007 [kJ/kg] i_23=1110.378 [kJ/kg] i_34=-0.000 [kJ/kg] i_45=104.554 [kJ/kg] i_56=0.000 [kJ/kg] i_61=240.601 [kJ/kg] Q_in=3268 [kJ/kg] Q_out=2098 [kJ/kg] q_R_23=-2921 [kJ/kg] q_R_45=-347.3 [kJ/kg] q_R_61=2098 [kJ/kg] T_o=295 [K] T_R_H=1500 [K] T_R_L=295 [K] W_net=1170 [kJ/kg] W_p=6.083 [kJ/kg] W_p_s=6.083 [kJ/kg] W_t_hp=277.2 [kJ/kg] W_t_lp=898.7 [kJ/kg]



10-66

10-68E The component of the ideal regenerative Rankine cycle described in Prob. 10-49E with the largest exergy destruction is to be identified.


Analysis From Prob. 10-49E,

0.1109Btu/lbm 1.732

kJ/kg 10616.2376.1298RBtu/lbm 5590.1

RBtu/lbm 39213.0

RBtu/lbm 23488.0

out

45in

765

psia 40 @ 43

psia 5 @ 21

==

=−=−=⋅===

⋅===

⋅===

yq

hhqsss

sss

sss

f

f

1

qin

2

7

1-y 6

qout

500 psia

5 psia

y 40 psia

4

3

5

s

T


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


Btu/lbm 43.6

Btu/lbm 168.9

=⎟⎠⎞

⎜⎝⎛ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

=⎟⎠⎞

⎜⎝⎛ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

R 520Btu/lbm 1.7325590.123488.0)R 520(

R 1260Btu/lbm 106139213.05590.1)R 520(

sink

out71071 destroyed,

source

in45045 destroyed,

Tq

ssTx

Tq

ssTx

For open feedwater heater, we have

[ ][ ]

Btu/lbm 5.4=−−−=

−−−=

)23488.0)(1109.01()5590.1)(1109.0(39213.0)R 520(

)1( 2630FWH destroyed, syyssTx

Processes 1-2, 3-4, 5-6, and 6-7 are isentropic, and thus

00

=

=

34 destroyed,

12 destroyed,

x

x

00

=

=

67 destroyed,

56 destroyed,

x

x

The greatest exergy destruction occurs during heat addition process 4-5.



10-67

10-69 A single-flash geothermal power plant uses hot geothermal water at 230ºC as the heat source. The power output from the turbine, the thermal efficiency of the plant, the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions and exergy efficiencies for the flash chamber, the turbine, and the entire plant are to be determined.


Analysis (a) We use properties of water for geothermal water (Tables A-4, A-5, and A-6)

kJ/kg.K 6841.21661.0

kJ/kg 14.990kPa 500

kJ/kg.K 6100.2kJ/kg 14.990

0C230

2

2

12

2

1

1

1

1

==

⎭⎬⎫

===

==

⎭⎬⎫

=°=

sx

hhP

sh

xT


kg/s 38.19kg/s) 230)(1661.0(

123

=== mxm &&

KkJ/kg 7739.7kJ/kg 3.2464

95.0kPa 10

KkJ/kg 8207.6kJ/kg 1.2748

1kPa 500

4

4

4

4

3

3

3

3

⋅==

⎭⎬⎫

==

⋅==

⎭⎬⎫

==

sh

xP

sh

xP

KkJ/kg 8604.1kJ/kg 09.640

0kPa 500

6

6

6

6

⋅==

⎭⎬⎫

==

sh

xP

2

productionwell

reinjectionwell

separator

steam turbine

1

Flash chamber

65

4

3

condenser

kg/s 81.19119.38230316 =−=−= mmm &&&

The power output from the turbine is

kW 10,842=−=−= kJ/kg)3.24648.1kJ/kg)(274 38.19()( 433T hhmW &&

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

kJ/kg 3672.0kJ/kg 83.104

0C25

0

0

0

0

==

⎭⎬⎫

=°=

sh

xT


5.3%==== 0.0532622,203

842,10

in

outT,th E

W&

&η

(b) The specific exergies at various states are

kJ/kg 53.216kJ/kg.K)3672.0K)(2.6100 (298kJ/kg)83.104(990.14)( 010011 =−−−=−−−= ssThhψ





The exergy of geothermal water at state 6 is

kW 17,257=== kJ/kg) 7kg/s)(89.9 .81191(666 ψmX &&


10-68

(c) Flash chamber:

kW 5080=−=−= kJ/kg)44.19453kg/s)(216. 230()( 211FC dest, ψψmX &&

89.8%==== 0.89853.21644.194

1

2FCII, ψ

ψη

(d) Turbine:

kW 10,854=−=−−= kW 10,842-kJ/kg)05.15110kg/s)(719. 19.38()( T433Tdest, WmX &&& ψψ

50.0%==−

=−

= 0.500)kJ/kg05.15110kg/s)(719. 19.38(

kW 842,10)( 433

TTII, ψψ

ηm

W&

&

(e) Plant:

kW 802,49kJ/kg) 53kg/s)(216. 230(11Plantin, === ψmX &&

kW 38,960=−=−= 842,10802,49TPlantin,Plantdest, WXX &&&

21.8%==== 0.2177kW 802,49kW 842,10

Plantin,

TPlantII, X

W&

&η



10-69

Cogeneration

10-70C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purpose to the total energy supplied. It could be unity for a plant that does not produce any power.

10-71C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity.

10-72C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference.

10-73C Cogeneration is the production of more than one useful form of energy from the same energy source. Regeneration is the transfer of heat from the working fluid at some stage to the working fluid at some other stage.



10-70

10-74 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. 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/kg 38.670

kJ/kg 41.19260.081.191kJ/kg 0.60

mkPa 1kJ 1

kPa 10600/kgm 0.00101

/kgm 00101.0kJ/kg 81.191

MPa 0.6 @ 3

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

==

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

f

f

f

hh

whh

PPw

hh

v

vv

P II P I

Condenser

Boiler

6

7 Turbine

Process heater

3

2 4

5

1

8

Mixing chamber:

332244

outin(steady) 0

systemoutin 0

hmhmhmhmhm

EEEEE

eeii &&&&&

&&&&&

+=⎯→⎯=

=⎯→⎯=∆=−

∑∑

or, ( )( ) ( )( )

( )( )( )

kJ/kg 47.31857.690.311kJ/kg 6.57

mkPa 1kJ 1kPa 6007000/kgm 0.001026

/kgm 001026.0

kJ/kg 90.31130

38.67050.741.19250.22

inII,45

33

454inII,

3kJ/kg 90.311 @ 4

4

33224

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

=≅

=+

=+

=

=

p

p

hf

whh

PPw

mhmhmh

f

v

vv&

&&

T


KkJ/kg 8000.6

kJ/kg 4.3411C500

MPa 76

6

6

6⋅=

=

⎭⎬⎫

°==

sh

TP

( )( ) kJ/kg 6.21531.23928201.081.191

8201.04996.7

6492.08000.6kPa 10

kJ/kg 6.2774MPa 6.0

88

88

68

8

767

7

=+=+=

=−

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

fgf

fg

f

hxhhs

ssx

ssP

hss

P

1

2

8

7

7 MPa

10 kPa

0.6 MPa 4

5

3

Qin·

Qout·

Qproces·

6

s

Then,

( ) ( )( )( ) ( )( )

( )( ) ( )( )

kW 32,866=−=−=

=+=+=

=−+−=−+−=

6.210077,33

kW 210.6kJ/kg 6.57kg/s 30kJ/kg 0.60kg/s 22.5

kW 077,33kJ/kg6.21536.2774kg/s 22.5kJ/kg6.27744.3411kg/s 30

inp,outT,net

inpII,4inpI,1inp,

878766outT,

WWW

wmwmW

hhmhhmW

&&&

&&&

&&&

Also, ( ) ( )( ) kW 782,15kJ/kg 38.6706.2774kg/s 7.5377process =−=−= hhmQ &&

( ) ( )( ) kW 788,9247.3184.3411kg/s 30565in =−=−= hhmQ &&

and 52.4%=+

=+

=788,92

782,15866,32

in

processnet

Q

QWu &

&&ε


10-71

10-75E A large food-processing plant requires steam at a relatively high pressure, which is extracted from the turbine of a cogeneration plant. The rate of heat transfer to the boiler and the power output of the cogeneration plant are to be determined.


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

( )

Btu/lbm 92.324

Btu/lbm 50.9448.002.94Btu/lbm 0.48

ftpsia 5.4039Btu 1

psia2)014)(/lbmft 0.01623(86.01

/

/lbmft 01623.0Btu/lbm 02.94

psia 140 @ 3

inpI,12

3

3

121inpI,

3psia 2 @ 1

psia 2 @ 1

==

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅×

−=

−=

==

==

f

p

f

f

hh

whh

PPw

hh

ηv

vv

P II P I

Condenser

Boiler

6

7 Turbine

Process heater

3

2 4

5

1

8

Mixing chamber: T


& & &

& &

& & & & &

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,

( )( ) ( )( )

( )( )( ) ( )

Btu/lbm 39.13133.207.129Btu/lbm .332

86.0/ftpsia 5.4039

Btu 1psia 014800/lbmft 0.01640

/

/lbmft 01640.0

Btu/lbm 07.12910

92.3245.150.945.8

in,45

33

454inpII,

3Btu/lbm 07.129 @ 4

4

33224

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

=≅

=+

=+

=

=

pII

p

hf

whh

PPw

mhmhm

h

f

ηv

vv

&

&&

1

2

8

6

7s

800psia

2 psia

140 psia 4

5

3

Qin·

Qprocess·

8s

7

s

RBtu/lbm 6812.1

Btu/lbm 2.1512F1000

psia 8006

6

6

6⋅=

=

⎭⎬⎫

°==

sh

TP

( )( ) Btu/lbm 21.9767.10218634.002.94

8634.074444.1

17499.06812.1psia 2

Btu/lbm 5.1287psia 140

88

88

68

8

767

7

=+=+=

=−

=−

=

⎭⎬⎫

==

=⎭⎬⎫

==

fgsfs

fg

fss

s

s

ss

s

hxhhs

ssx

ssP

hss

P

Then, ( ) ( )( ) Btu/s 13,810=−=−= Btu/lbm39.1312.1512lbm/s 10565in hhmQ &&

(b) ( ) ( )[ ]( ) ( )( ) ( )( )[ ]

kW 4440==−+−=

−+−==

Btu/s 4208Btu/lbm 976.211287.5lbm/s 1.5Btu/lbm 1287.51512.2lbm/s 0186.0

878766,outT, sssTsTT hhmhhmWW &&&& ηη


10-72

10-76 A cogeneration plant has two modes of operation. In the first mode, all the steam leaving the turbine at a relatively high pressure is routed to the process heater. In the second mode, 60 percent of the steam is routed to the process heater and remaining is expanded to the condenser pressure. The power produced and the rate at which process heat is supplied in the first mode, and the power produced and the rate of process heat supplied in the second mode are to be determined.



( )( )( )

( )( )( )

kJ/kg 47.65038.1009.640kJ/kg 10.38

mkPa 1kJ 1

kPa 50010,000/kgm 0.001093

/kgm 001093.0kJ/kg 09.640

kJ/kg 57.26115.1042.251kJ/kg 10.15

mkPa 1kJ 1

kPa 0210,000/kgm 0.001017

/kgm 001017.0kJ/kg 42.251

inpII,34

33

343inpII,

3MPa 5.0 @ 3

MPa 5.0 @ 3

inpI,12

33

121inpI,

3kPa 20 @ 1

kPa 20 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

whh

PPw

hh

f

f

f

f

v

vv

v

vv

2 PI

Condens.

6 Turbine

5

Boiler

Process heater

PII

4

3

7

1

8

T

1

2

8

7

5

4 3

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

5 5 2 2 4 4

0

s

or, ( )( ) ( )( ) kJ/kg 91.4945

47.650357.2612

5

44225 =

+=

+=

mhmhm

h&

&&

KkJ/kg 4219.6kJ/kg 4.3242

C450MPa 10

6

6

6

6⋅=

=

⎭⎬⎫

°==

sh

TP

( )( )

( )( ) kJ/kg 0.21145.23577901.042.251

7901.00752.7

8320.04219.6kPa 20

kJ/kg 6.25780.21089196.009.640

9196.09603.4

8604.14219.6MPa 5.0

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,

( ) ( )( )( ) ( )( ) kW 9693

kW 3319

=−=−=

=−=−=

kJ/kg09.6406.2578kg/s 5

kJ/kg6.25784.3242kg/s 5

377process

766outT,

hhmQ

hhmW

&&

&&

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

( ) ( )

( )( ) ( )( )( ) ( )( ) kW 5816

kW 4248

=−=−=

=−+−=−+−=

kJ/kg 09.6406.2578kg/s 3

kJ/kg 0.21146.2578kg/s 2kJ/kg 6.25784.3242kg/s 5

377process

878766outT,

hhmQ

hhmhhmW

&&

&&&



10-73

10-77 A cogeneration plant modified with regeneration is to generate power and process heat. The mass flow rate of steam through the boiler for a net power output of 25 MW is to be determined.




( )( )( )

( )( )( )

kJ/kg 02.86757.844.858kJ/kg 57.8

mkPa 1kJ 1kPa 4009000/kgm 0.001159

/kgm 001159.0

kJ/kg 44.858

kJ/kg 41.19361.181.191kJ/kg .611

mkPa 1kJ 1kPa 101600/kgm 0.00101

/kgm 00101.0kJ/kg 81.191

inpII,45

33

454inpII,

3MPa 6.1 @ 4

MPa 6.1 @ 943

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

====

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hhhh

whh

PPw

hh

f

f

f

f

v

vv

v

vv

( )( )

( )( ) kJ/kg 2.19901.23927518.081.191

7518.04996.7

6492.02876.6kPa 10

kJ/kg 0.27304.19349675.044.858

9675.00765.4

3435.22876.6MPa 6.1

KkJ/kg 2876.6kJ/kg 8.3118

C400MPa 9

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

1

2

8

T

6

7

10 kPa

9 MPa

1.6 MPa

5

3,4,9

Boiler

6

7 Turbine

8

1

5 Condenser

Process heater

PI PII

9

2

3

fwh

4

s

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

( ) ( )( ) ( )(

kJ/kg 7.869kJ/kg 2.19900.273035.01kJ/kg 0.27308.3118

)1( 8776outT,

=−−+−=

−−+−= hhyhhw)

( )( ) ( )

kJ/kg 1.86062.97.869

kJ/kg .629kJ/kg 8.57kJ/kg 1.6135.01

)1(

inp,outT,net

inpII,inpI,inp,

=−=−=

=+−=

+−=

www

wwyw

Thus,

kg/s 29.1===kJ/kg 860.1

kJ/s 25,000

net

net

wW

m&

&


10-74

10-78 Problem 10-77 is reconsidered. The effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate is to be investigated.


"Input Data" y = 0.35 "fraction of steam extracted from turbine for feedwater heater and process heater" P[6] = 9000 [kPa] T[6] = 400 [C] P_extract=1600 [kPa] P[7] = P_extract P_cond=10 [kPa] P[8] = P_cond W_dot_net=25 [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]" 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"



10-75

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_in W_dot_net = m_dot * w_net

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]

9000 kPa

1600 kPa

10 kPa

SteamIAPWS

1,2

3,4,5,9

6

7

8

Pextract [kPa]

ηth m [kg/s]

Qprocess

[kW] 200 400 600 800

1000 1200 1400 1600 1800 2000

0.3778 0.3776 0.3781 0.3787 0.3794 0.3802 0.3811 0.3819 0.3828 0.3837

25.4 26.43 27.11 27.63 28.07 28.44 28.77 29.07 29.34 29.59

2770 2137 1745 1459 1235 1053 900.7 770.9 659

561.8



10-76

200 400 600 800 1000 1200 1400 1600 1800 20000.377

0.378

0.379

0.38

0.381

0.382

0.383

0.384

Pextract [kPa]

ηth

200 400 600 800 1000 1200 1400 1600 1800 200025

26

27

28

29

30

500

1000

1500

2000

2500

3000

Pextract [kPa]

m [

kg/s

]

Qpr

oces

s [k

W]



10-77

10-79E 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 utilization factor of this plant are to be determined.


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

P

Boiler

2

6

1

Turbine

Process heater

3 4 5

7

Btu/lbm 5.1229psia 120


F800psia 600

Btu/lbm 49.208

737

7

6543

753

3

3

3

12

F240 @ 1

=⎭⎬⎫

==

===

⋅====

⎭⎬⎫

°==

≅=≅ °

hss

P

hhhh

sssh

TP

hhhh f

( )( )( )

kW 2260==−=

−=

Btu/s 2142Btu/lbm 5.12290.1408lbm/s 12

755net hhmW &&

T

1 6

7

600 psia

120 psia

2

3,4,5 (b)

( )( ) ( )( ) ( )( )Btu/s 19,450=

−+=−−+=

−= ∑∑

49.208185.1229120.14086117766

process

hmhmhm

hmhmQ eeii

&&&

&&&

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

s



10-78

10-80 A Rankine steam cycle modified for a closed feedwater heater and a process heater is considered. The T-s diagram for the ideal cycle is to be sketched; the mass flow rate of the cooling water; and the utilization efficiency of the plant are to be determined.


Analysis (b) Using the data from the problem statement, the enthalpies at various states are

( )( )( )

kJ/kg 9.2011.1081.191kJ/kg 1.10

mkPa 1kJ 1

kPa 1010000/kgm 0.00101

/kgm 00101.0kJ/kg 81.191

inpI,12

33

121inpI,

3kPa 20 @ 1

kPa 10 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

1

2

8

s

4

5 700 kPa

10 kPa

2 MPa

3

1-y-z

1-y

y

10 MPa

10

111-y-z

9 z

y

7

6

T

kJ/kg 00.697kJ/kg 47.908

kPa 700 @ 1110

kPa 2000 @ 983

===

====

f

f

hhhhhhh

An energy balance on the closed feedwater heater gives

3495.0

47.90829309.20147.908

)(

85

23

2385

=−−

=−−

=

−=−

hhhh

y

hhhhy

The process heat is expressed as

kg/s 60.3=°°⋅

−=

∆−

=

∆=−=

C)40(C)kJ/kg 18.4(kJ/kg )00.6972714(kg/s) 100(05.0)(

)(

106

106process

wpw

wpw

Tchhmz

m

TcmhhmzQ&

&

&&&

(c) The net power output is determined from

[ ]

kW 970,94kJ/kg) 1.10(kJ/kg )20893374)(05.03495.01(kJ/kg )27143374(05.0kJ/kg )29303374(3495.0

)kg/s 100(

))(1()()( 746454

PTnet

=

⎥⎦

⎤⎢⎣

⎡−−−−+−+−

=

−−−−+−+−=−=

PwhhzyhhzhhymWWW

&

&&&


kW 296,550kJ/kg )47.9083874)(kg/s 100()( 34in =−=−= hhmQ &&

The rate of process heat is

kW 10,085kJ/kg )00.6972714(kg/s) 100(05.0)(05.0 106process =−=−= hhmQ &&

The utilization efficiency of this cogeneration plant is

35.4%==+

=+

= 354.0kW 550,296

kW )085,10970,94(

in

processnet

Q

QWu &

&&ε



10-79

Combined Gas-Vapor Power Cycles

10-81C The energy source of the steam is the waste energy of the exhausted combustion gases.

10-82C Because the combined gas-steam cycle takes advantage of the desirable characteristics of the gas cycle 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.



10-80

10-83 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 open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, 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 (Table A-17)

( )( )

( )

kJ/kg 02462 K460

kJ/kg 873518325450141

5450kJ/kg 421515K 1400

kJ/kg 56354019386114

3861kJ/kg 19300K 300

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/kg 01.25259.042.251kJ/kg 0.59

mkPa 1kJ 1

kPa 20600/kgm 0.001017

/kgm 001017.0kJ/kg 42.251

inpI,12

33

121inpI,

3kPa 20 @ 1

kPa 20 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

1

2

7

1400 K

300 K 6

8 MPa

20 kPa 0.6 MPa

4

9

3

Qout·

Qin·

460 K STEAM CYCLE

GAS CYCLE

10

11

12

8

5 400°C

T

s

( )

( )( )

kJ/kg 53.67815.838.670kJ/kg 8.15

mkPa 1kJ 1kPa 6008,000/kgm 0.001101

/kgm 001101.0kJ/kg 38.670

inpI,34

33

343inpII,

3MPa 6.0 @ 3

MPa 6.0 @ 3

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

( )( )

( )( ) kJ/kg 2.20955.23577821.042.251

7821.00752.7

8320.03658.6kPa 20

kJ/kg 1.25868.20859185.038.670

9185.08285.4

9308.13658.6MPa 6.0

KkJ/kg 3658.6kJ/kg 4.3139

C400MPa 8

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 for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&


10-81

( ) ( )

steam kg air / kg8.99 02.46280.73553.6784.3139

0

1211

45air

1211air45

outin


=−−

=−−

=

−=−⎯→⎯=

=

=∆=−

∑∑

hhhh

mm

hhmhhmhmhm

EE

EEE

s

seeii

&

&

&&&&

&&

&&&

(b) Noting that for the open FWH, the steady-flow energy balance equation yields & &Q W ke pe≅ ≅ ≅ ≅∆ ∆ 0

( ) ( ) 326336622

outin


11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii =−+⎯→⎯=+⎯→⎯=

=

=∆=−

∑∑ &&&&&

&&

&&&

Thus,

( )

( )( )( )( ) kJ/kg 23.9562.20951.25861792.011.25864.3139

1

extracted steam offraction the 1792.001.2521.258601.25238.670

7665

26

23

=−−+−=−−+−=

=−−

=−−

=

hhyhhw

hhhhy

T

( )( )( )

( ) ( )( ) kJ/kg 3.44419.3005.6358.73542.1515

kJ/kg 56.94815.859.01792.0123.9561

891110in,gasnet,

,,in,steamnet,

=−−−=−−−=−=

=−−−=−−−=−=

hhhhwww

wwywwww

CT

IIpIpTpT

The net work output per unit mass of gas is

( ) kJ/kg 8.54956.9483.444 99.81

steamnet,99.81

gasnet,net =+=+= www

and

( ) ( )( ) kW 720,215=−=−=

===

kJ/kg 5.63542.1515kg/s 818.5

kg/s 7.188kJ/kg 549.7

kJ/s 450,000

910airin

net

netair

hhmQ

wWm

&&

&&

(c) 62.5%===kW 720,215kW 450,000

in

net

QW

ηth &

&



10-82

10-84 Problem 10-83 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated.


"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 = 1.0 Eta_gas_turb = 1.0 Eta_pump = 1.0 Eta_steam_turb = 1.0 P[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] /Pratio Ts11=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" 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])

PROPRIETARY MATERIAL

preparation. If you are a student using this Manual, you are using it without permission. . © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

10-83

"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]) 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_steam W_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"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"


10-84

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 MassRatio gastosteam

Wnetgas[kW]

Wnetsteam[kW]

ηth[%]

NetWorkRatio gastosteam

10 7.108 342944 107056 59.92 3.203 11 7.574 349014 100986 60.65 3.456 12 8.043 354353 95647 61.29 3.705 13 8.519 359110 90890 61.86 3.951 14 9.001 363394 86606 62.37 4.196 15 9.492 367285 82715 62.83 4.44 16 9.993 370849 79151 63.24 4.685 17 10.51 374135 75865 63.62 4.932 18 11.03 377182 72818 63.97 5.18 19 11.57 380024 69976 64.28 5.431 20 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

Com bined Gas and Steam Pow er Cycle

8

9

10

11

12

1,23,4

5

6

7

Steam CycleSteam Cycle

Gas CycleGas Cycle



10-85

5 9 13 17 21 2555

57 .2

59 .4

61 .6

63 .8

66

P ra tio

ηth

[%

]

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 / W dot,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

sto

stea

m

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



10-86

10-85 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air 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-2a).

Analysis Working around the topping cycle gives the following results:


K 8.530K)(8) 293( 0.4/1.4/)1(

5

656 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

K 8.57285.0

2938.530293

)()(

5656

56

56

56

56

=−

+=

−+=⎯→⎯

−

−=

−−

=

C

s

p

spsC

TTTT

TTcTTc

hhhh

η

η

K 0.75881K) 1373(

0.4/1.4/)1(

7

878 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

K 5.819)0.7581373)(90.0(1373

)()()(

877887

87

87

87

=−−=

−−=⎯→⎯−

−=

−−

= sTsp

p

sT TTTT

TTcTTc

hhhh

ηη

1

2

4s 4

3

6 MPa

20 kPa

6s

1373 K

293 K

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

320°C6

8

T

s

K 548.6 C6.275kPa 6000 @sat 9 =°== TT

Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6):

kJ/kg 5.25708.642.251kJ/kg 08.6 mkPa 1

kJ 1 kPa)206000)(/kgm 001017.0()(

/kgm 001017.0

kJ/kg 42.251

inp,12

33

121inp,

3kPa 20 @1

kPa 20 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

kJ/kg 8.2035 kPa 20

KkJ/kg 1871.6kJ/kg 6.2953

C320kPa 6000

434

4

3

3

3

3

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

shss

P

sh

TP

kJ/kg 6.2127)8.20356.2953)(90.0(6.2953

)( 433443

43

=−−=

−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη


10-87

The net work outputs from each cycle are

kJ/kg 2.275K)2937.5725.8191373)(KkJ/kg 1.005(

)()( 5687

inC,outT,cycle gas net,

=+−−⋅=

−−−=

−=

TTcTTc

www

pp

kJ/kg 9.81908.6)6.21276.2953(

)( inP,43

inP,outT,cycle steam net,

=−−=

−−=

−=

whh

www


aap

wwpa mm-hh

TTcm-hhmTTcm &&&&& 1010.0

5.2576.2953)6.5485.819)(005.1()(

)()(23

982398 =

−−

=−

=⎯→⎯=−

That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is

kg/s 279.3=×+×

==air kJ/kg )9.8191010.02.2751(

kJ/s 000,100

net

net

wW

ma

&&



10-88

10-86 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).

1

2

4s 4

3

6 MPa

20 kPa

6s

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

1373 K

320°C

6a6

8

293 K

T

Analysis With an ideal regenerator, the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit. Representing this state by “6a”

K 5.81986 == TT a

The rate of heat addition in the cycle is

kW 370,155K )5.8191373(C)kJ/kg 005.1(kg/s) 3.279(

)( 67in

=−°⋅=

−= apa TTcmQ &&

The thermal efficiency of the cycle is then

0.6436===kW 370,155kW 000,100

in

netth Q

W&

&η

s

Without the regenerator, the rate of heat addition and the thermal efficiency are

kW 640,224K )7.5721373(C)kJ/kg 005.1(kg/s) 3.279()( 67in =−°⋅=−= TTcmQ pa&&

0.4452===kW 640,224kW 000,100

in

netth Q

W&

&η

The change in the thermal efficiency due to using the ideal regenerator is

0.1984=−=∆ 4452.06436.0thη



10-89

10-87 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob. 10-86 is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. T



kg/s 21.28)3.279(1010.01010.0kJ/kg 2.1876kJ/kg 1.2696

kJ/kg 3.804)(KkJ/kg 4627.6KkJ/kg 1871.6

KkJ/kg 8320.0K 293

K 5.819

K 1373

14out

23in,23

67in,67

4

3

kPa 20 @ 21

sink

8cycle steam source,

cycle gas source,

====−=

=−=

=−=⋅=⋅=

⋅====

==

=

aw

p

f

mmhhqhhq

TTcqss

sssT

TT

T

&&

( ) kW 2278)1871.64627.6)(K 293(kg/s) 21.28(

process) c(isentropi 0

34034 destroyed,

12 destroyed,

=−=−=

=

ssTmX

X

w&&

&

1

2

4s 4

3

6 MPa

20 kPa

6s

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

320°C

86

1373 K

293 K

s

kW 8665K 293kJ/kg 2.1876

1871.68320.0)K 293(kg/s) 21.28(

sink

out41041 destroyed,

=⎟⎠

⎞⎜⎝

⎛ +−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

Tq

ssTmX w&&

kW 11260

)8320.01871.6)(293)(21.28(5.8196.548ln)005.1()293)(3.279(

)(ln 2308

90230890exchangerheat destroyed,

=

−+⎥⎦⎤

⎢⎣⎡=

−+⎟⎟⎠

⎞⎜⎜⎝

⎛=∆+∆= ssTm

TT

cTmsTmsTmX apawa &&&&&

kW 6280)8ln()287.0(293

572.7(1.005)ln)293)(3.279(lnln5

6

5

6056 destroyed, =⎥⎦

⎤⎢⎣⎡ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

PP

RTT

cTmX pa&&

kW 23,970=⎥⎦⎤

⎢⎣⎡ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

13733.804

572.71373(1.005)ln)293)(3.279(ln

source

in

6

7067 destroyed, T

qTT

cTmX pa&&

kW 639681ln)287.0(

1373819.5(1.005)ln)293)(3.279(lnln

7

8

7

8078 destroyed, =⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

PP

RTT

cTmX pa&&

The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle.


10-90

10-88 A 280-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle are to be determined.


Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields T

( )( )

( )( ) ( )

1

2

7s

5

6s

4

9s

3

Qout·

Qin·

8

12

11s

10

GAS CYCLE

STEAM CYCLE 6

7

11

9

( )

( )( )(

kJ/kg 26.421K 420

kJ/kg .4067418.59507.116186.007.1161

kJ/kg 18.59519.151.167111

1.167kJ/kg 07.1161K 1100

kJ/kg .7466082.0/19.30084.59519.300

/

kJ/kg 84.59525.15386.111

386.1kJ/kg 19.300K 300

1212

111010111110

1110

1110

11

1010

898989

89

98

9

88

1011

10

89

8

=⎯→⎯=

=−−=

−−=⎯→⎯−−

=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==⎯→⎯=

=−+=

−+=⎯→⎯−−

=

=⎯→⎯===

==⎯→⎯=

hT

hhhhhhhh

hPPP

P

PhT

hhhhhhhh

hPPP

P

PhT

sTs

T

srr

r

Css

C

srr

r

ηη

ηη

s )

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

( )

( )( )

kJ/kg 60.19280.081.191kJ/kg 0.80

mkPa 1kJ 1

kPa 01800/kgm 0.00101

/kgm 00101.0kJ/kg 81.191

inpI,12

33

121inpI,

3kPa 01 @ 1

kPa 10 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

( )

( )( )

kJ/kg 55.72568.487.720kJ/kg .684

mkPa 1kJ 1kPa 0085000/kgm 0.001115

/kgm 001115.0kJ/kg 87.720

inpI,34

33

343inpII,

3MPa 8.0 @ 3

MPa 8.0 @ 3

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

KkJ/kg 4516.6kJ/kg 3.3069

C350MPa 5

5

5

5

5⋅=

=

⎭⎬⎫

°==

sh

TP



10-91

( )( )

( ) ( )( )

( )( )

( ) ( )( ) kJ/kg 3.21865.20423.306986.03.3069

kJ/kg 5.20421.23927737.081.191

7737.04996.7

6492.04516.6kPa 10

kJ/kg 3.27301.26753.306986.03.3069

kJ/kg 1.26758.20859545.087.720

9545.06160.4

0457.24516.6MPa 8.0

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

ηη

ηη


( ) ( )

steam kg air / kg 9.259=−−

=−−

=

−=−⎯→⎯=

=

=∆=−

∑∑

26.42140.67455.7253.3069

0

1211

45air

1211air45

outin


hhhh

mm

hhmhhmhmhm

EE

EEE

s

seeii

&

&

&&&&

&&

&&&

(b) Noting that for the open FWH, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&

( ) ( ) 326336622

outin(steady) 0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii =−+⎯→⎯=+⎯→⎯=

=→=∆=−

∑∑ &&&&&

&&&&&

Thus,

( )

( )( )[ ]( ) ( )( )[ ] kJ/kg 8.7693.21863.27302082.013.27303.306986.0

1

extracted steam offraction the 2082.060.1923.273060.19287.720

7665

26

23

=−−+−=−−+−=

=−−

=−−

=

hhyhhw

hhhh

y

TT η

( )( )( )

( ) ( )( ) kJ/kg 12.12619.30074.66040.67407.1161

kJ/kg 5.76468.480.02082.018.7691

891110in,gasnet,

IIp,Ip,inp,steamnet,

=−−−=−−−=−=

=−−−=−−−=−=

hhhhwww

wwywwww

CT

TT

The net work output per unit mass of gas is

( ) kJ/kg 69.2085.764259.9112.126

425.61

steamnet,gasnet,net =+=+= www

kg/s 1341.7kJ/kg 208.69

kJ/s 280,000

net

netair ===

wW

m&

&

and ( ) ( )( ) kW 671,300=−=−= kJ/kg 74.66007.1161kg/s 1341.7910airin hhmQ &&

(c) 41.7%==== 4171.0kW 671,300kW 280,000

in

net

QW

th &

&η



10-92

10-89 Problem 10-88 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated.


"Input data" T[8] = 300 [K] "Gas compressor inlet" P[8] = 100 [kPa] "Assumed air inlet pressure" "Pratio = 11" "Pressure ratio for gas compressor" T[10] = 1100 [K] "Gas turbine inlet" T[12] = 420 [K] "Gas exit temperature from Gas-to-steam heat exchanger " P[12] = P[8] "Assumed air exit pressure" W_dot_net=280 [MW] Eta_comp = 0.82 Eta_gas_turb = 0.86 Eta_pump = 1.0 Eta_steam_turb = 0.86 P[5] = 5000 [kPa] "Steam turbine inlet" T[5] =(350+273.15) "K" "Steam turbine inlet" P[6] = 800 [kPa] "Extraction pressure for steam open feedwater heater" P[7] = 10 [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] /Pratio Ts11=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" 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])

PROPRIETARY MATERIAL

preparation. If you are a student using this Manual, you are using it without permission. . © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course

10-93

"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]) 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_steam W_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"

PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent"


10-94

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

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

]

5000 kPa

800 kPa

10 kPa

Combined Gas and Steam Power CyclePratio MassRatiogastosteam ηth[%]

10 11 12 13 14 15 16 17 18 19 20

8.775 9.262 9.743 10.22 10.7

11.17 11.64 12.12 12.59 13.07 13.55

42.03 41.67 41.22 40.68 40.08 39.4 38.66 37.86 36.99 36.07 35.08

8

9

10

11

121,2

3,4

5

6

7

Steam CycleSteam Cycle

Gas CycleGas Cycle

10 12 14 16 18 208

9

10

11

12

13

14

Pratio

Mas

sRat

ioga

stos

team

10 12 14 16 18 2035

36

37

38

39

40

41

42

43

Pratio

ηth

[%

]



10-95

10-90 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressure turbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of the combined cycle are to be determined.


Analysis (a) We obtain the air properties from EES. The analysis of gas cycle is as follows

( )( ) ( )

( )( )(

kJ/kg 62.475C200

kJ/kg 98.87179.7638.130480.08.1304

kJ/kg 79.763kPa 100

kJ/kg 6456.6kPa 700C950

kJ/kg 8.1304C950

kJ/kg 21.55780.0/16.29047.50316.290

/

kJ/kg 47.503kPa 700

kJ/kg 6648.5kPa 100

C15kJ/kg 50.288C15

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

ηη

ηη

)


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

( )( )( )

kJ/kg 37.19965.781.191kJ/kg .567

80.0/mkPa 1

kJ 1kPa 106000/kgm 0.00101

/

/kgm 00101.0kJ/kg 81.191

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

p

f

f

ηv

vv

( )( ) kJ/kg 4.23661.23929091.081.191

9091.04996.7

6492.04670.7kPa 10

KkJ/kg 4670.7kJ/kg 5.3264

C400MPa 1

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fss

s hxhhs

ssx

ssP

sh

TP

1

2

6ss

T

3

6 MPa

10 kPa

8s

Qout·

Qin·

7

11

10s

9

GAS CYCLE

STEAM CYCLE

950°C

15°C

5

4

1 MPa

10

8

4s

6

3

1 2

Steam turbine

Gas turbine

Condenserpump

5

4

Combustion chamber

11

Compressor

Heat exchanger

10

9 8

7

6


10-96

( )( )(

1.6%==−=−=

=⎭⎬⎫

==

=−−=

−−=⎯→⎯−−

=

0158.09842.011Percentage Moisture

9842.0kJ/kg 5.2546kPa 10

kJ/kg 0.25464.23665.326480.05.3264

6

66

6

655665

65

x

xhP

hhhhhhhh

sTs

T ηη)

(b) Noting that for the heat exchanger, the steady-flow energy balance equation yields 0∆pe∆ke ≅≅≅≅WQ &&

( ) ( ) ( )

[ ] kJ/kg 0.2965)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 MPa 1 ?MPa 6

−−=⎯→⎯−−

=

=⎭⎬⎫

==

==

⎭⎬⎫

==

ηη

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

(c) ( ) ( )( ) kW 4328kJ/kg 98.8718.1304kg/s 10109airgasT, =−=−= hhmW &&

( ) ( )( ) kW 2687kJ/kg 50.28821.557kg/s 1078airgasC, =−=−= hhmW &&

kW 164126874328gasC,gasT,gasnet, =−=−= WWW &&&

( ) ( )( ) kW 1265kJ/kg 0.25465.32640.29655.3346kg/s 1.156543ssteamT, =−+−=−+−= hhhhmW &&

( )( ) kW 7.8kJ/kg 564.7kg/s 1.15ssteamP, === pumpwmW &&

kW 12567.81265steamP,steamT,steamnet, =−=−= WWW &&&

kW 2897=+=+= 12561641steamnet,gasnet,plantnet, WWW &&&

(d) ( ) ( )( ) kW 7476kJ/kg 21.5578.1304kg/s 1089airin =−=−= hhmQ &&

38.8%==== 0.388kW 7476kW 2897

in

plantnet,th Q

W&

&η



10-97

Special Topic: Binary Vapor Cycles

10-91C 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.

10-92C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency.

10-93C 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.

10-94C 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.

10-95 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 bottoming cycle (cycle B) enters at state 3 and leaves at state 4. Neglecting any changes in kinetic and potential energies, 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


10-98

Review Problems

10-96 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The thermal efficiency of the cycle is to be compared when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters as a subcooled liquid.

determined power produced by the turbine and consumed by the pump are to be determined.



kJ/kg 67.34613.654.340kJ/kg 13.6 mkPa 1

kJ 1 kPa)506000)(/kgm 001030.0()(

/kgm 001030.0

kJ/kg 54.340

inp,12

33

121inp,

3kPa 20 @1

kPa 50 @1

=+=+==

⎟⎠

⎞⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

kJ/kg 0.2495)7.2304)(9348.0(54.340

9348.05019.6

0912.11693.7

kPa 50

KkJ/kg 1693.7kJ/kg 8.3658

C600kPa 6000

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

qin

qout

50 kPa 1

3

2

4

6 MPa

T

s

Thus,

kJ/kg 5.215454.3400.2495kJ/kg 1.331267.3468.3658

14out

23in

=−=−==−=−=

hhqhhq


0.3495=−=−=1.33125.215411

in

outth q

qη

When the liquid enters the pump 11.3°C cooler than a saturated liquid at the condenser pressure, the enthalpies become

/kgm 001023.0

kJ/kg 07.293

C703.113.813.11kPa 50

3C70 @ 1

C70 @ 1

kPa 50 @sat 1

1

=≅

=≅

⎭⎬⎫

°=−=−==

°

°

f

fhhTT

Pvv

kJ/kg 09.6 mkPa 1

kJ 1 kPa)506000)(/kgm 001023.0()(

33

121inp,

=⎟⎠

⎞⎜⎝

⎛

⋅−=

−= PPw v

kJ/kg 16.29909.607.293inp,12 =+=+= whh

Then,

kJ/kg 9.220109.2930.2495kJ/kg 6.335916.2998.3658

14out

23in

=−=−==−=−=

hhqhhq

0.3446=−=−=6.33599.220111

in

outth q

qη

The thermal efficiency slightly decreases as a result of subcooling at the pump inlet.



10-99

10-97E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as the working 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 the preheater, and the thermal efficiency of the Level I cycle of this plant are to be determined.


Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine),


)

)

)

( )( )( )(

F267.4°=°−°⋅=−

−=

2

2

12brinebrine

F325FBtu/lbm 1.03lbm/h 384,286Btu/h 000,790,22T

T

TTcmQ p&&

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

( )( )( )(

MBtu/h 29.7=°−°⋅=

−=

F555.84FBtu/lbm 0.24lbm/h 4,195,10089airair 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/h 187,120=

°−°⋅=−

−=

geo

geo

inoutgeogeo

F8.2110.154FBtu/lbm 1.03Btu/h 000,140,11

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%=⎟⎟⎠

⎞⎜⎜⎝

⎛==

kWh 1Btu 3412.14

Btu/h 33,930,000kW 1071

in

netth Q

W&

&η


10-100

10-98 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.



( )( )( )

kJ/kg 37.29919.1018.289kJ/kg .1910

mkPa 1kJ 1

kPa 3010,000/kgm 0.001022

/kgm 001022.0kJ/kg 18.289

inp,12

33

121inp,

3kPa 30 @ 1

kPa 30 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

T

( )( ) kJ/kg 1.25573.23359711.027.289

9711.08234.6

9441.05706.7kPa 30

KkJ/kg 5706.7kJ/kg 4.3578

C550MPa 2

kJ/kg 1.3321MPa 2

KkJ/kg 2335.7kJ/kg 7.3559

C550MPa 4

kJ/kg 9.3204MPa 4

KkJ/kg 7561.6kJ/kg 9.3500

C550MPa 10

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

5

2

8

3

4

7

6

2 MPa 4 MPa

30 kPa

10 MPa

s

Then,

( ) ( ) ( )

kJ/kg 8.15459.22677.3813kJ/kg 9.226718.2891.2557

kJ/kg 7.38131.33214.35789.32047.355937.2999.3500

outinnet

18out

674523in

=−=−==−=−=

=−+−+−=−+−+−=

qqwhhq

hhhhhhq

Thus,

40.5%==== 4053.0kJ/kg 3813.7kJ/kg 1545.8

in

netth q

wη

(b) The mass flow rate of the steam is then

kg/s 48.5===kJ/kg 1545.8kJ/s 75,000

net

net

wW

m&

&



10-101

10-99 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.


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

( )( )( )( )

kPa 2780=⎭⎬⎫

=°=

⋅=+=+==+=+=

⎭⎬⎫

==

565

5

66

66

6

6

C600

KkJ/kg 5488.74996.792.06492.0kJ/kg 5.23921.239292.081.191

92.0kPa 10

Pss

T

sxsshxhh

xP

fgf

fgf

T

1

5

2

6

s

3

425

MPa

10 kPa

SINGLE600°C

1

5

2

8

3

4

7

6

DOUBLE

10 kPa

25 MPa

s

T600°C

(b) Double Reheat:

C600 and

KkJ/kg 3637.6C600MPa 25

5

5

34

4

33

3

°==

==

⋅=⎭⎬⎫

°==

TPP

ssPP

sTP

xx

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



10-102

10-100 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwater 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.


Analysis

T

y

P II P I

Open fwh

Condenser

Boiler

4

3 2

1

6

Turbine

7

1-y

5

1

qin

2

7s 7

5

6s

10 MPa

10 kPa

0.5 MPa

1-y

4

y 3

qout

6

s


( )( )( ) ( )

/kgm 001093.0kJ/kg 09.640

liquidsat.MPa 5.0

kJ/kg 33.19252.081.191kJ/kg 0.52

95.0/mkPa 1

kJ 1kPa 10500/kgm 0.00101

/

/kgm 00101.0kJ/kg 81.191

3MPa 5.0 @ 3

MPa 5.0 @ 33

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

====

⎭⎬⎫=

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

η−=

==

==

f

f

p

f

f

hhP

whh

PPw

hh

vv

v

vv

( )( )( ) ( )

kJ/kg 02.65193.1009.640kJ/kg 10.93

95.0/mkPa 1

kJ 1kPa 50010,000/kgm 0.001093

/

inpII,34

33

343inpII,

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

η−=

whh

PPw pv

( )(kJ/kg 1.2654

0.21089554.009.640

9554.09603.4

8604.15995.6

MPa 5.0

KkJ/kg 5995.6kJ/kg 1.3375

C500MPa 10

66

66

56

6

5

5

5

5

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgsfs

fg

fss

s

s hxhhs

ssx

ssP

sh

TP

)

( )( )(

kJ/kg 3.27981.26541.337580.01.3375

655665

65

=−−=

−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη)



10-103

( )(kJ/kg 7.2089

1.23927934.081.191

7934.04996.7

6492.05995.6

kPa 1077

77

57

7

=+=+=

=−

=−

=

⎭⎬⎫

==

fgsfs

fg

fss

s

s hxhhs

ssx

ssP )

( )( )(

kJ/kg 8.23467.20891.337580.01.3375

755775

75

=−−=

−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη)

)

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

( ) ( 326332266

outin


11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii =−+⎯→⎯=+⎯→⎯=

=

=∆=−

∑∑ &&&&&

&&

&&&


1718.033.1923.279833.19209.640

26

23 =−−

=−−

=hhhh

y

Then, ( )( ) ( )( )

kJ/kg 4.9397.17841.2724kJ/kg 7.178481.1918.23461718.011

kJ/kg 1.272402.6511.3375

outinnet

17out

45in

=−=−==−−=−−=

=−=−=

qqwhhyq

hhq

and

skg 7159 /.kJ/kg 939.4

kJ/s 150,000

net

net ===wW

m&

&


34.5%=−=−=kJ/kg 2724.1kJ/kg 1784.7

11in

outth q

qη

Also,

KkJ/kg 6492.0

KkJ/kg 8604.1

KkJ/kg 9453.6kJ/kg 3.2798

MPa 5.0

kPa 10 @ 12

MPa 5.0 @ 3

66

6

⋅===⋅==

⋅=⎭⎬⎫

==

f

f

sssss

shP

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

( )[ ]

( ) ( )( ) ( )( )[ ]kJ/kg 39.25=

−−−=

−−−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−== ∑∑

6492.01718.019453.61718.08604.1K 303

1 2630

0surr

0gen0regen syyssTT

qsmsmTsTi

Liiee



10-104

10-101 An 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an open feedwater 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.


Analysis

T

1

qin

2

7

1-y

5

6

qout

10 MPa

10 kPa

0.5 MPa

4

y 3

s

y

P II P I

Open fwh

Condenser

Boiler

4

3 2

1

6

Turbine

7 1-y

5


( )( )( )

/kgm 001093.0kJ/kg 09.640

liquidsat.MPa 5.0

kJ/kg 30.19250.081.191

kJ/kg 0.50mkPa 1

kJ 1kPa 10500/kgm 0.00101

/kgm 00101.0kJ/k 81.191

3MPa 5.0 @ 3

MPa 5.0 @ 33

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

====

⎭⎬⎫=

=+=+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

f

f

f

f

hhP

whh

PPw

ghh

vv

v

vv

( )( )( )

( )( )

( )( ) kJ/kg 7.20891.23927934.081.191

7934.04996.7

6492.05995.6kPa 10

kJ/kg 1.26540.21089554.009.640

9554.09603.4

8604.15995.6MPa 5.0

KkJ/kg 5995.6kJ/kg 1.3375

C500MPa 10

kJ/kg 47.65038.1009.640

kJ/kg .3810mkPa 1

kJ 1kPa 50010,000/kgm 0.001093

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

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

( ) ( 326332266

outin(steady) 0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii =−+⎯→⎯=+⎯→⎯=

=→=∆=−

∑∑ &&&&&

&&&&&

)




10-105

1819.031.1921.265431.19209.640

26

23 =−−

=−−

=hhhh

y

Then, ( )( ) ( )( )

kJ/kg 0.11727.15526.2724kJ/kg 7.155281.1917.20891819.011

kJ/kg 6.272447.6501.3375

outinnet

17out

45in

=−=−==−−=−−=

=−=−=

qqwhhyq

hhq

and skg 0128 /.kJ/kg 1171.9

kJ/s 150,000

net

net ===wWm&

&


%0.43=−=−=kJ/kg 2724.7kJ/kg 1552.7

11in

outth q

qη

Also,

KkJ/kg 6492.0

KkJ/kg 8604.1KkJ/kg 5995.6

kPa 10 @ 12

MPa 5.0 @ 3

56

⋅===

⋅==⋅==

f

f

sss

ssss

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

( )[ ]

( ) ( )( ) ( )( )[ ]kJ/kg 39.0=

−−−=

−−−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−== ∑∑

6492.01819.015995.61819.08604.1K 303

1 2630

0surr

0gen0regen syyssTT

qsmsmTsTi

Liiee



10-106

10-102 An ideal 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.



( )( )( )

/kgm 001101.0kJ/k 38.670

liquid sat.MPa 6.0

kJ/kg 53.22659.094.225kJ/kg 0.59

mkPa 1kJ 1

kPa 15600/kgm 0.001014

/kgm 001014.0kJ/kg 94.225

3MPa 6.0 @ 3

MPa 6.0 @ 33

inpI,12

33

121inpI,

3kPa 15 @ 1

kPa 15 @ 1

====

⎭⎬⎫=

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

f

f

f

f

ghhP

whh

PPw

hh

vv

v

vv

( )( )( )

( )( ) kJ/kg 8.25183.23729665.094.225

9665.02522.7

7549.07642.7kPa 15

kJ/kg 2.3310MPa 6.0

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 0.1

kJ/kg 8.2783MPa 0.1

KkJ/kg 5995.6kJ/kg 1.3375

C500MPa 10

kJ/kg 73.68035.1038.670kJ/kg 10.35

mkPa 1kJ 1kPa 60010,000/kgm 0.001101

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

y

Condens.

5

1 2

3

4

Turbine Boiler

Open fwh

P I P II

7

6

8 9

1-y

1

7

2

9

5

6 8

1 MPa

15 kPa

10 MPa

0.6 MPa

4

3

T

s


( ) ( 328332288

outin(steady) 0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii =−+⎯→⎯=+⎯→⎯=

=→=∆=−

∑∑ &&&&&

&&&&&

)


0.144=−−

=−−

=53.2262.331053.22638.670

28

23

hhhhy


( ) ( ) ( ) ( )( )( ) ( )( ) kJ/kg 7.196294.2258.25181440.011

kJ/kg 7.33898.27831.347973.6801.3375

19out

6745in

=−−=−−==−+−=−+−=

hhyqhhhhq

and 42.1%=−=−=kJ/kg 3389.7kJ/kg 1962.7

11in

outth q

qη



10-107

10-103 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.


Analysis

T


y

P II P I

Open fwh

Condenser4

3 2

1

5

8

Boiler

7

6 Turbine

9 1-y

1

7

2

9

5

6 8 4

1-y

9s

6s 8sy

3

s


( )( )( )

/kgm 001101.0kJ/kg 38.670

liquid sat.MPa 6.0

kJ/kg 54.22659.094.225kJ/kg 0.59

mkPa 1kJ 1

kPa 15600/kgm 0.001014

/kgm 001014.0kJ/kg 94.225

3MPa 6.0 @3

MPa 6.0 @33

in,12

33

121in,

3kPa 15 @1

kPa 15 @1

====

⎭⎬⎫=

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

f

f

pI

pI

f

f

hhP

whh

PPw

hh

vv

v

vv

( )( )( )

kJ/kg 8.2783MPa 0.1

KkJ/kg 5995.6kJ/kg 1.3375

C500MPa 10

kJ/kg 73.68035.1038.670kJ/kg 10.35

mkPa 1kJ 1kPa 60010,000/kgm 0.001101

656

6

5

5

5

5

inpII,34

33

343inpII,

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

ss

s hssP

sh

TP

whh

PPw v

( )( )(

kJ/kg 4.28788.27831.337584.01.3375

655665

65

=−−=

−η−=⎯→⎯−−

=η sTs

T hhhhhhhh

)

( ) ( )(kJ/kg 2.3337

2.33101.347984.01.3479

kJ/kg 2.3310MPa 6.0

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 0.1

877887

87

878

8

7

7

7

7

=−−=−η−=⎯→⎯

−−

=η

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

sTs

T

ss

s

hhhhhhhh

hssP

sh

TP

)


10-108

( )( )

( ) ( )(kJ/kg 5.2672

8.25181.347984.01.3479

kJ/kg 8.25183.23729665.094.225

9665.02522.7

7549.07642.7kPa 15

977997

97

99

99

79

9

=−−=−−=⎯→⎯

−−

=

=+=+=

=−

=−

=

⎭⎬⎫

==

sTs

T

fgsfs

fg

fss

s

s

hhhhhhhh

hxhhs

ssx

ssP

ηη )

)


( ) ( 328332288

outin


11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii =−+⎯→⎯=+⎯→⎯=

=

=∆=−

∑∑ &&&&&

&&

&&&


0.1427=−−

=−−

=53.2263.333553.22638.670

28

23

hhhhy


( ) ( )( ) ( )( )( ) ( )( ) kJ/kg 2.209794.2255.26721427.011

kJ/kg 1.32954.28781.347973.6801.3375

19out

6745in

=−−=−−==−+−=

−+−=

hhyq

hhhhq

and

36.4%=−=−=kJ/kg 3295.1kJ/kg 2097.2

11in

outth q

qη



10-109

10-104 A steam power plant operating on the ideal reheat-regenerative Rankine cycle with three feedwater heaters is considered. Various items for this system per unit of mass flow rate through the boiler are to be determined.


16

n

13

1

Low-P Turbine

Boiler

Condenser

High-P Turbine

14

2

19

15m

z

P IV

P I

Closed FWH I

12

11 10 9

8

6 34 5

7

17

ClosedFWH II

OpenFWH P II

yx

P III

18

Analysis The compression processes in the pumps and the expansion processes in the turbines are isentropic. Also, the state of water at the inlet of pumps is saturated liquid. Then, from the steam tables (Tables A-4, A-5, and A-6),

kJ/kg 8.1011kJ/kg 3.1008kJ/kg 86.885kJ/kg 46.884kJ/kg 28.419kJ/kg 51.417kJ/kg 84.168kJ/kg 75.168

10

9

6

5

4

3

2

1

========

hhhhhhhh

kJ/kg 7.2454kJ/kg 5.2891kJ/kg 3.3481kJ/kg 0.2871kJ/kg 6.3063kJ/kg 5.3204kJ/kg 1.3423

19

18

17

16

15

14

13

=======

hhhhhhh

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. Then,

kJ/kg 8.1008 C9.233

kPa 3000

kJ/kg 91.884 C1.207

kPa 1800

11911

11

757

7

=⎭⎬⎫

°===

=⎭⎬⎫

°===

hTT

P

hTT

P

Enthalpies at other states and the fractions of steam extracted from the turbines can be determined from mass and energy balances on cycle components as follows:

Mass Balances:

znm

zyx=+=++ 1

Open feedwater heater:

3218 zhnhmh =+

Closed feedwater heater-II:

57154 yhzhyhzh +=+



10-110

Closed feedwater heater-I:

911148 )()( xhhzyxhhzy ++=++

Mixing chamber after closed feedwater heater II:

867 )( hzyyhzh +=+

Mixing chamber after closed feedwater heater I:

121110 1)( hhzyxh =++

Substituting the values and solving the above equations simultaneously using EES, we obtain

0.708820.07124

0.16670.05334

=======

nmzyx

hh

78000.0

kJ/kg 0.1009kJ/kg 08.885

12

8

Note that these values may also be obtained by a hand solution by using the equations above with some rearrangements and substitutions. Other results of the cycle are

43.9%

kJ/kg 1620kJ/kg 2890

kJ/kg 769.6kJ/kg 502.3

==−=−=

=−==−+−=

=−+−=

=−+−+−=

4394.02890162011

)()(

)()()()()(

in

outth

119out

16171213in

19171817LPout,T,

161315131413HPout,T,

qq

hhnqhhzhhq

hhnhhmwhhzhhyhhxw

η



10-111

10-105 The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined using EES.


"Given" P_boiler=6000 [kPa] P_cfwh1=3000 [kPa] P_cfwh2=1800 [kPa] P_reheat=800 [kPa] "P_ofwh=100 [kPa]" P_condenser=7.5 [kPa] T_turbine=500 [C] "Analysis" Fluid$='steam_iapws' "turbines" h[13]=enthalpy(Fluid$, P=P_boiler, T=T_turbine) s[13]=entropy(Fluid$, P=P_boiler, T=T_turbine) h[14]=enthalpy(Fluid$, P=P_cfwh1, s=s[13]) h[15]=enthalpy(Fluid$, P=P_cfwh2, s=s[13]) h[16]=enthalpy(Fluid$, P=P_reheat, s=s[13]) h[17]=enthalpy(Fluid$, P=P_reheat, T=T_turbine) s[17]=entropy(Fluid$, P=P_reheat, T=T_turbine) h[18]=enthalpy(Fluid$, P=P_ofwh, s=s[17]) h[19]=enthalpy(Fluid$, P=P_condenser, s=s[17]) "pump I" h[1]=enthalpy(Fluid$, P=P_condenser, x=0) v[1]=volume(Fluid$, P=P_condenser, x=0) w_pI_in=v[1]*(P_ofwh-P_condenser) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P_ofwh, x=0) v[3]=volume(Fluid$, P=P_ofwh, x=0) w_pII_in=v[3]*(P_cfwh2-P_ofwh) h[4]=h[3]+w_pII_in "pump III" h[5]=enthalpy(Fluid$, P=P_cfwh2, x=0) T[5]=temperature(Fluid$, P=P_cfwh2, x=0) v[5]=volume(Fluid$, P=P_cfwh2, x=0) w_pIII_in=v[5]*(P_cfwh1-P_cfwh2) h[6]=h[5]+w_pIII_in "pump IV" h[9]=enthalpy(Fluid$, P=P_cfwh1, x=0) T[9]=temperature(Fluid$, P=P_cfwh1, x=0) v[9]=volume(Fluid$, P=P_cfwh1, x=0) w_p4_in=v[5]*(P_boiler-P_cfwh1) h[10]=h[9]+w_p4_in "Mass balances" x+y+z=1 m+n=z



10-112

"Open feedwater heater" m*h[18]+n*h[2]=z*h[3] "closed feedwater heater 2" T[7]=T[5] h[7]=enthalpy(Fluid$, P=P_cfwh1, T=T[7]) z*h[4]+y*h[15]=z*h[7]+y*h[5] "closed feedwater heater 1" T[11]=T[9] h[11]=enthalpy(Fluid$, P=P_boiler, T=T[11]) (y+z)*h[8]+x*h[14]=(y+z)*h[11]+x*h[9] "Mixing chamber after closed feedwater heater 2" z*h[7]+y*h[6]=(y+z)*h[8] "Mixing chamber after closed feedwater heater 1" x*h[10]+(y+z)*h[11]=1*h[12] "cycle" w_T_out_high=x*(h[13]-h[14])+y*(h[13]-h[15])+z*(h[13]-h[16]) w_T_out_low=m*(h[17]-h[18])+n*(h[17]-h[19]) q_in=h[13]-h[12]+z*(h[17]-h[16]) q_out=n*(h[19]-h[1]) Eta_th=1-q_out/q_in

P open fwh

[kPa] ηth

10 20 30 40 50 60 70 80 90

100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290 300

0.429828 0.433780 0.435764 0.436978 0.437790 0.438359 0.438768 0.439065 0.439280 0.439432 0.439536 0.439602 0.439638 0.439647 0.439636 0.439608 0.439565 0.439509 0.439442 0.439367 0.439283 0.439192 0.439095 0.438993 0.438887 0.438776 0.438662 0.438544 0.438424 0.438301

0 50 100 150 200 250 3000.428

0.43

0.432

0.434

0.436

0.438

0.44

Pofwh [kPa]

ηth



10-113

10-106 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 at the inlet of high-pressure turbine are to be determined.



( )( )

( )( )(

kJ/kg 1.23446.20479.278860.09.2788

kJ/kg 6.20471.23927758.081.191

7758.04996.7

6492.04675.6

kPa 10

KkJ/kg 4675.6kJ/kg 9.2788

vapor sat.MPa 4.1

544554

54

55

45

45

5

MPa 4.1 @ 4

MPa 4.1 @ 44

=−−=

−−=⎯→⎯−−

=

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅====

⎭⎬⎫=

sTs

T

fgsfs

fg

fss

s

g

g

hhhhhhhh

hxhhs

ssx

ssP

sshhP

ηη)

1

2

5s

T

5

44

3

and

kg/min 9.107kg/s 799.1

kJ/kg 444.8kJ/s 800

kJ/kg 8.4441.23449.2788

lowturb,

IIturb,turblow

54lowturb,

====

=−=−=

wW

m

hhw&

&

Therefore ,

kJ/kg 0.28439.278815.54

kJ/kg 54.15kg/s 18.47kJ/s 1000

=kg/min 11081081000

4highturb,3

43turbhigh,

,turbhighturb,

total

=+=+=

−====

=+=

hwh

hhmW

w

m

I

&

&

& kg/s 18.47

( )( ) ( )

( )( ) KkJ/kg 4289.61840.49908.02835.2

9908.09.1958

96.8298.2770MPa 4.1

kJ/kg 8.277075.0/9.27880.28430.2843

/

44

44

34

4

433443

43

⋅=+=+=

=−

=−

=

⎭⎬⎫

==

=−−=

−−=⎯→⎯−−

=

fgsfs

fg

fss

s

s

Tss

T

sxssh

hhx

ssP

hhhhhhhh

ηη

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

C227.5MPa 2

°==

⎭⎬⎫

⋅==

3

3

3

3KkJ/kg 4289.6

kJ/kg 0.2843TP

sh



10-114

10-107 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine 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.


1

2

8s 8

T

6

7

8 MPa

20 kPa

2 MPa 4

5

3

Qout·

Qin·

Qprocess·

Conden.

6 Turbine

Processhe

ater

I 3

7

4 2 P I P I

Boiler

5

1

8

s


( )( )( )

kJ/kg 47.908

kJ/kg 71.25329.242.251kJ/kg .292

88.0/mkPa 1

kJ 1kPa 022000/kgm 0.001017

/

/kgm 001017.0kJ/kg 42.251

MPa 2 @ 3

inpI,12

33

121inpI,

3kPa 02 @ 1

kPa 20 @ 1

==

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

f

p

f

f

hh

whh

PPw

hh

ηv

vv

Mixing chamber:

kJ/kg 81.491kg/s) 11()kJ/kg) 71kg/s)(253. 411(kJ/kg) 47kg/s)(908. 4( 44

442233

=⎯→⎯=−+

=+

hh

hmhmhm &&&

( )( )( )

kJ/kg 02.49921.781.491kJ/kg .217

88.0/mkPa 1

kJ 1kPa 00208000/kgm 0.001058

/

/kgm 001058.0

inII,45

33

454inII,

3kJ/kg 81.491 @ 4

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

=≅ =

p

pp

hf

whh

PPw

f

ηv

vv

KkJ/kg 7266.6kJ/kg 5.3399

C500MPa 8

6

6

6

6⋅=

=

⎭⎬⎫

°==

sh

TP

kJ/kg 4.3000MPa 2

767

7 =⎭⎬⎫

==

shss

P

( ) ( )( ) kJ/kg 3.30484.30005.339988.05.3399766776

76 =−−=−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη

kJ/kg 5.2215kPa 20

868

8 =⎭⎬⎫

==

shss

P



10-115

( ) ( )( ) kJ/kg 6.23575.22155.339988.05.3399866886

86 =−−=−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη

Then,

( ) ( )( ) kW 8559=−=−= kJ/kg 47.9083.3048kg/s 4377process hhmQ &&

(b) Cycle analysis:

( ) ( )( )( ) ( )( )

( )( ) ( )( )

kW 8603=−=−=

=+=+=

=−+−=

−+−=

958698

kW 95kJ/kg 7.21kg/s 11kJ/kg 2.29kg/s 7

kW 8698kJ/kg6.23575.3399kg/s 7kJ/kg3.30485.3399kg/s 4

inp,outT,net

inpII,4inpI,1inp,

868767outT,

WWW

wmwmW

hhmhhmW

&&&

&&&

&&&

(c) Then,

and ( ) ( )( )

53.8%==+

=+

=

=−=−=

538.0905,31

85598603

kW 905,3102.4995.3399kg/s 11

in

processnet

565in

Q

QW

hhmQ

u &

&&

&&

ε



10-116

10-108E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable for Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).

Analysis Working around the topping cycle gives the following results:

R 1043R)(10) 540( 0.4/1.4/)1(

5

656 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT


R 109990.0

5401043540

)()(

5656

56

56

56

56

=−

+=

−+=⎯→⎯

−

−=

−−

=

C

s

p

spsC

TTTT

TTcTTc

hhhh

η

η

R 1326101R) 2560(

0.4/1.4/)1(

7

878 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

R 1449)13262560)(90.0(2560

)()()(

877887

87

87

87

=−−=

−−=⎯→⎯−

−=

−−

= sTsp

p

sT TTTT

TTcTTc

hhhh

ηη 1

2

4s 4

3

800 psia

5 psia

6s

2560 R

540 R

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

600°F 6

8

T

s

R 102850R 3.97850psia 800 @sat 9 =+=+= TT

Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E):

Btu/lbm 59.13241.218.130Btu/lbm 41.2


)(

/lbmft 01641.0

Btu/lbm 18.130

inp,12

33

121inp,

3psia 5 @1

psia 5 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 6.908 psia 5


F600

psia 800

434

4

3

3

3

3

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

shss

P

sh

TP

Btu/lbm 7.926)6.9089.1270)(95.0(9.1270

)( 433443

43

=−−=

−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη


Btu/lbm 5.132R)540109914492560)(RBtu/lbm 240.0(

)()( 5687


=+−−⋅=

−−−=

−=

TTcTTc

www

pp


10-117

Btu/lbm 8.34141.2)7.9269.1270(

)( inP,43


=−−=

−−=

−=

whh

www


aap

wwpa mm-hh

TTcm-hhmTTcm &&&&& 08876.0

59.1329.1270)10281449)(240.0()(

)()(23

982398 =

−−

=−

=⎯→⎯=−

That is, 1 lbm of exhaust gases can heat only 0.08876 lbm of water. Then the heat input, the heat output and the thermal efficiency are

Btu/lbm 8.187Btu/lbm )18.1307.926(08876.0R)5401028)(RBtu/lbm 240.0(1

)()(

Btu/lbm 6.350R)10992560)(RBtu/lbm 240.0()(

1459out

67in

=−×+−⋅×=

−+−=

=−⋅=−=

hhmm

TTcmm

q

TTcmm

q

a

wp

a

a

pa

a

&

&

&

&

&

&

0.4643=−=−=6.3508.18711

in

outth q

qη



10-118

10-109E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined.



1

2

4s 4

3

800 psia

10 psia

6s

2560 R

540 R

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

600°F 6

8

TAnalysis Working around the topping cycle gives the following results:

R 1043R)(10) 540( 0.4/1.4/)1(

5

656 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT

R 109990.0

5401043540

)()(

5656

56

56

56

56

=−

+=

−+=⎯→⎯

−

−=

−−

=

C

s

p

spsC

TTTT

TTcTTc

hhhh

η

η

R 1326101R) 2560(

0.4/1.4/)1(

7

878 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− kk

s PP

TT s

R 1449)13262560)(90.0(2560

)()()(

877887

87

87

87

=−−=

−−=⎯→⎯−

−=

−−

= sTsp

p

sT TTTT

TTcTTc

hhhh

ηη

R 102850R 3.97850psia 800 @sat 9 =+=+= TT

Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E):

Btu/lbm 7.16343.225.161Btu/lbm 43.2


)(

/lbmft 01659.0

Btu/lbm 25.161

inp,12

33

121inp,

3psia 10 @1

psia 10 @1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

Btu/lbm 6.946 psia 10


F600

psia 800

434

4

3

3

3

3

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

shss

P

sh

TP

Btu/lbm 8.962)6.9469.1270)(95.0(9.1270

)( 433443

43

=−−=

−−=⎯→⎯−−

= sTs

T hhhhhhhh

ηη




10-119

Btu/lbm 5.132R)540109914492560)(RBtu/lbm 240.0(

)()( 5687


=+−−⋅=

−−−=

−=

TTcTTc

www

pp

Btu/lbm 7.30543.2)8.9629.1270(

)( inP,43


=−−=

−−=

−=

whh

www


aap

wwpa mmhh

TTcmhhmTTcm &&&&& 09126.0

7.1639.1270)10281449)(240.0()(

)()(23

982398 =

−−

=−

−=⎯→⎯−=−

That is, 1 lbm of exhaust gases can heat only 0.09126 lbm of water. Then the heat input, the heat output and the thermal efficiency are

Btu/lbm 3.190Btu/lbm )25.1618.962(09126.0R)5401028)(RBtu/lbm 240.0(1

)()(

Btu/lbm 6.350R)10992560)(RBtu/lbm 240.0()(

1459out

67in

=−×+−⋅×=

−+−=

=−⋅=−=

hhmm

TTcmm

q

TTcmm

q

a

wp

a

a

pa

a

&

&

&

&

&

&

0.4573=−=−=6.3503.19011

in

outth q

qη

When the condenser pressure is increased from 5 psia to 10 psia, the thermal efficiency is decreased from 0.4643 to 0.4573.



10-120

10-110E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The cycle supplies a specified rate of heat to the buildings during winter. The mass flow rate of air and the net power output from the cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable to Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.



Analysis The mass flow rate of water is

lbm/h 2495Btu/lbm 161.25)-(962.8

Btu/h 102 6

14

buildings =×

=−

=hh

Qmw

&&

The mass flow rate of air is then

lbm/h 27,340===0.09126

249509126.0

wa

mm

&&

The power outputs from each cycle are

kW 1062Btu/h 3412.14

kW 1R)540109914492560)(RBtu/lbm 240.0(lbm/h) 340,27(

)()(

)(

5687


=

⎟⎠⎞

⎜⎝⎛+−−⋅=

−−−=

−=

TTcmTTcm

wwmW

papa

a

&&

&&

1

2

4s 4

3

800 psia

10 psia

6s

2560 R

540 R

Qout·

Qin·

5

9

8s

7

GAS CYCLE

STEAM CYCLE

600°F6

8

T

s

kW 224Btu/h 3412.14

kW 1)43.28.9629.1270(lbm/h) 2495(

)(

)(

inP,43


=

⎟⎠⎞

⎜⎝⎛−−=

−−=

−=

whhm

wwmW

a

a

&

&&

The net electricity production by this cycle is then

kW 1286=+= 2241062netW&


10-121

10-111 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 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. 3 Air is an ideal gas with variable specific heats. Analysis (a) The analysis of gas cycle yields


( )( )

( )

kJ/kg 63.523K 520

kJ/kg 38.76854.375.450121

5.450kJ/kg 42.1515K 1400

kJ/kg 18.63066.185546.112

5546.1kJ/kg 24.310K 310

1111

109

10

99

87

8

77

910

9

78

7

=⎯→⎯=

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==⎯→⎯=

=⎯→⎯===

==⎯→⎯=

hT

hPPP

P

PhT

hPPP

P

PhT

rr

r

rr

r


( ) ( )( )

gwhh

PPw

hh

f

f

kJ/k 42.20462.1281.191

kJ/kg .6212mkPa 1

kJ 1kPa 1012,500/kgm 0.00101

/kgm 00101.0

kJ/kg 81.191

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

=+=+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=−=

==

==

v

vv

1

26

312.5 MPa

10 kPa

8

Qout·

Qin·

7

11

10

9

GAS CYCLE

STEAM CYCLE

1400 K

310 K

5

4

2.5 MPa

T

s

( )( ) kJ/kg 8.23651.23929089.081.191

9089.04996.7

6492.04653.7kPa 10

KkJ/kg 4653.7kJ/kg 4.3574

C550MPa 5.2

kJ/kg 6.2909MPa 5.2

KkJ/kg 4651.6kJ/kg 6.3343

C500MPa 5.12

66

66

56

6

5

5

5

5

434

4

3

3

3

3

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

hssP

sh

TP


( ) ( )

( ) kg/s 153.9=−−

=−−

=

−=−⎯→⎯=⎯→⎯= ∑∑kg/s 12

63.52338.76842.2046.3343

1110

23air

1110air23outin

s

seeii

mhhhh

m

hhmhhmhmhmEE

&&

&&&&&&

(b) The rate of total heat input is ( ) ( )

( )( ) ( )( )

kW 101.44 5×≅=

−+−=−+−=+=

kW 200,144kJ/kg6.29094.3574kg/s 12kJ/kg 18.63042.1515kg/s 153.9

45reheat89airreheatairin hhmhhmQQQ &&&&&

(c) The rate of heat rejection and the thermal efficiency are then ( ) ( )

( )( ) ( )( )

59.1%

,

==−=−=

=−+−=

−+−=+=

5913.0kW 144,200kW 58,930

11

kW 93058kJ/kg 81.1918.2365kg/s 12kJ/kg 24.31063.523kg/s 153.9

in

outth

16711airsteamout,airout,out

QQ

hhmhhmQQQ s

&

&

&&&&&

η


10-122

10-112 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle are to be determined.



Analysis (a) The analysis of gas cycle yields (Table A-17)

( )( )

( )( ) ( )

( )

( )( )( )

kJ/kg 63.523K 520

kJ/kg 4.95835.86042.151585.042.1515

kJ/kg 35.8603.565.45081

5.450kJ/kg 42.1515K 1400

kJ/kg 1.58580.0/16.29012.52616.290

/

kJ/kg 12.526849.92311.18

2311.1kJ/kg 16.290K 290

1111

109910109

109

109

10

99

787878

78

87

8

77

910

9

78

7

=⎯→⎯=

=−−=

−η−=⎯→⎯−−

=η

=⎯→⎯=⎟⎠⎞

⎜⎝⎛==

==⎯→⎯=

=−+=

η−+=⎯→⎯−−

=η

=⎯→⎯===

==⎯→⎯=

hT

hhhhhhhh

hPP

PP

PhT

hhhhhhhh

hPPPP

PhT

sTs

T

srs

r

r

Css

C

srs

r

r

s

s

1

26s

3

15 MPa

10 kPa

8

Qout·

Qin·

7

11

10s

9

GAS CYCLE

STEAM CYCLE

1400 K

290 K

5

4

3 MPa

6

4

81

T

s


( )

( )( )

kJ/kg 95.20614.1581.191kJ/kg 15.14

mkPa 1kJ 1kPa 1015,000/kgm 0.00101

/kgm 00101.0kJ/kg 81.191

inpI,12

33

121inpI,

3kPa 10 @ 1

kPa 10 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

( )( )

( )( )( )

kJ/kg 1.28387.27819.315785.09.3157

kJ/kg 7.27819.17949879.03.1008

9880.05402.3

6454.21434.6MPa 3

KkJ/kg 1428.6kJ/kg 9.3157

C450MPa 15

433443

43

44

44

34

4

3

3

3

3

=−−=

−η−=⎯→⎯−−

=η

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

sTs

T

fgsfs

fg

fss

s

hhhhhhhh

hxhhs

ssx

ssP

sh

TP


10-123

( )( )

( )( )( )

kJ/kg 5.24678.22922.345785.02.3457

kJ/kg 8.22921.23928782.081.191

8783.04996.7

6492.02359.7kPa 10

KkJ/kg 2359.7kJ/kg 2.3457

C500MPa 3

655665

65

66

66

56

6

5

5

5

5

=−−=

−η−=⎯→⎯−−

=η

=+=+=

=−

=−

=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

sTs

T

fgsfs

fg

fss

s

hhhhhhhh

hxhhs

ssx

ssP

sh

TP


( ) ( )

( ) kg/s 203.6=−−

=−−

=

−=−⎯→⎯=

=

=∆=−

∑∑kg/s 30

63.5234.95895.2069.3157

0

1110

23air

1110air23

outin


s

seeii

mhhhhm

hhmhhmhmhm

EE

EEE

&&

&&&&

&&

&&&

(b) ( ) ( )( )( ) ( )( )

kW 207,986=−+−=

−+−=+=kJ/kg 1.28382.3457kg/s 30kJ/kg 1.58542.1515kg/s 203.6

45reheat89airreheatairin hhmhhmQQQ &&&&&

(c) ( ) ( )( )( ) ( )( )

44.3%=−=−=

=−+−=

−+−=+=

kW 207,986kW 115,805

11

kW 115,805kJ/kg 81.1915.2467kg/s 30kJ/kg 16.29063.523kg/s 203.6

in

outth

16711airsteamout,airout,out

QQ

hhmhhmQQQ s

&

&

&&&&&

η



10-124

10-113 A Rankine steam cycle modified with two closed feedwater heaters and one open feed water heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass extracted for the open feedwater heater y and the cooling water flow temperature rise are to be determined. Also, the rate of heat rejected in the condenser and the thermal efficiency of the plant are to be determined.


Analysis (b) Using the data from the problem statement, the enthalpies at various states are

1

2

12

s

7

8 140 kPa

20 kPa

1910 kPa

4

3

56

5 MPa

9

11

14

15

1-y-z-w 13

z

w

w

y

10

w+z

kJ/kg 898

kJ/kg 676

kJ/kg 458

kJ/kg 251

kPa 1910@ 126

kPa 062@ 4

kPa 140 @ 14315

kPa 02@ 1

===

==

====

==

f

f

f

f

hhh

hh

hhhh

hh

T

620 kPa

An energy balance on the open feedwater heater gives

439 1)1( hhyyh =−+

where z is the fraction of steam extracted from the low-pressure turbine. Solving for z,

0.08086=−−

=−−

=4583154458676

39

34

hhhh

y

(c) An energy balance on the condenser gives

[ ] wpwwwww Tcmhhmhyhzwhzywm ∆=−=−−++−−− &&& )()1()()1( 22115117

Solving for the temperature rise of cooling water, and substituting with correct units,

[ ]

[ ]

C9.95°=

−−++−−−=

−−++−−−=∆

)18.4)(4200()251)(08086.01()458)(0655.00830.0()2478)(0655.008086.00830.01()100(

)1()()1( 115117

pwww cm

hyhzwhzywmT

&

&

(d) The rate of heat rejected in the condenser is

kW 174,700=°°⋅=∆= C)95.9)(CkJ/kg kg/s)(4.18 4200(out wpww TcmQ &&


kW 200,300kJ/kg )8983900)(kg/s 100()( 67in =−=−= hhmQ &&

The thermal efficiency is then

41.8%==−=−= 418.0kW 200,300kW 700,17411

in

outth Q

Q&

&η



10-125

10-114 A Rankine steam cycle modified for reheat, two closed feedwater heaters and a process heater is considered. The T-s diagram for the ideal cycle is to be sketched. The fraction of mass, w, that is extracted for the closed feedwater heater is to be determined. Also, the mass flow rate through the boiler, the rate of process heat supplied, and the utilization efficiency of this cogeneration plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (b) Using the data from the problem statement, the enthalpies at various states are

( )( )( )

kJ/kg 5.2561.54.251kJ/kg 1.5

mkPa 1kJ 1

kPa 205000/kgm 0.00102

/kgm 00102.0kJ/kg 4.251

inpI,12

33

121inpI,

3kPa 20 @ 1

kPa 20 @ 1

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

whh

PPw

hh

f

f

v

vv

12

14

s 17 11

5

6

3

245 kPa

20 kPa

1.4 MPa

7 4

5 MPa

9 12

1315

z

1-y-z-w

y 8

w 16 10

1.2 MPa

150 kPa

T

kJ/kg 467

kJ/kg 830kJ/kg 533

kPa 150 @ 163

kPa 1400 @ 14154

kPa 245 @ 1213

=======

===

f

f

f

hhhhhhh

hhh

An energy balance on the closed feedwater heater gives 1631513102 )(11 hwzyhyhzhwhh +++=+++

where w is the fraction of steam extracted from the low-pressure turbine. Solving for z,

0.0620=−

−−++−=

−−−++−(

=

4673023)830)(1160.0()533)(15.0()467)(15.01160.0()5.256467(

)()

1610

15131623

hhyhzhhzyhh

w

(c) The work output from the turbines is

kJ/kg 6.1381)2620)(0620.015.01160.01()3023)(0620.0()3154)(15.0()3692)(1160.01(

)3349)(1160.01()3400)(1160.0(3894)1()1()1( 111098765outT,

=−−−−−−−+

−−−=

−−−−−−−+−−−= hwzywhzhhyhyyhhw

The net work output from the cycle is kJ/kg 5.13761.56.1381inP,outT,net =−=−= www

The mass flow rate through the boiler is

kg/s 217.9===kJ/kg 1376.5

kW 000,300

net

net

wW

m&

&


kW 700,733

kJ/kg )33493692)(kg/s 9.217)(1160.01(kJ/kg )8303894)(kg/s 9.217()()1()( 7845in

=−−+−=

−−+−= hhmyhhmQ &&&

The rate of process heat and the utilization efficiency of this cogeneration plant are

kW 85,670=−=−= kJ/kg )5333154)(kg/s 9.217)(15.0()( 129process hhmzQ &&

52.6%==+

=+

= 526.0kW 700,733

kW )670,85000,300(

in

processnet

Q

QWu &

&&ε



10-126

10-115 The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to be investigated.


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] = 10000 [kPa] T[3] = 550 [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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" s[2]=entropy(Fluid$,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_p Eta_th=W_net/Q_in



10-127

P4

[kPa] ηth Wnet

[kJ/kg] 5

15 25 35 45 55 65 75 85

100

0.4268 0.3987 0.3841 0.3739 0.3659 0.3594 0.3537 0.3488 0.3443 0.3385

1432 1302 1237 1192 1157 1129 1105 1084 1065 1040

0 2 4 6 8 10 120

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

10 MPa

5 kPa

SteamIAPWS

1,2

3

4

0 20 40 60 80 1000.32

0.34

0.36

0.38

0.4

0.42

0.44

P[4] [kPa]

ηth

0 20 40 60 80 1001000

1050

1100

1150

1200

1250

1300

1350

1400

1450

P[4] [kPa]

Wne

t [k

J/kg

]



10-128

10-116 The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated.


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$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p h[2]=h[1]+W_p "SSSF First Law for the pump" s[2]=entropy(Fluid$,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_p Eta_th=W_net/Q_in



10-129

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] ηth 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]

ηth

200 300 400 500 600 700 800 900 1000 1100750

1050

1350

1650

1950

2250

T[3] [C]

Wne

t [k

J/kg

]



10-130

10-117 The effect of number of reheat stages on the performance an ideal Rankine cycle is to be investigated.


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 REPEAT i:=i+1 P4 = P3*R_P P5=P4 P6=P5*R_P Fluid$='Steam_IAPWS' s5=entropy(Fluid$,T=T5,P=P5) h5=enthalpy(Fluid$,T=T5,P=P5) s_s6=s5 hs6=enthalpy(Fluid$,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$,h=h6,P=P6) Q_in_reheat =Q_in_reheat + (h5 - h4) P3=P4 UNTIL (i>imax) END "NoRHStages = 2" P[6] = 10"kPa" P[3] = 15000"kPa" P_extract = P[6] "Select a lower limit on the reheat pressure" T[3] = 500"C" T[5] = 500"C" Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" Pratio = P[3]/P_extract P[4] = P[3]*(1/Pratio)^(1/(NoRHStages+1))"kPa" Fluid$='Steam_IAPWS'



10-131

"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_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$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,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$,h=h[6],P=P[6]) W_t_lp_total = NoRHStages*W_t_lp Q_in_reheat = NoRHStages*(h[5] - h[4])} "Boiler analysis" Q_in_boiler + h[2]=h[3]"SSSF First Law for the Boiler" Q_in = Q_in_boiler+Q_in_reheat "Condenser analysis" h[6]=Q_out+h[1]"SSSF First Law for the Condenser" T[6]=temperature(Fluid$,h=h[6],P=P[6]) s[6]=entropy(Fluid$,h=h[6],P=P[6]) x[6]=QUALITY(Fluid$,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



10-132

ηth NoRH

Stages Qin

[kJ/kg] Wnet

[kJ/kg] 0.4097 1 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 2253

0.3779 7 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

ηth

1 2 3 4 5 6 7 8 9 104000

4500

5000

5500

6000

6500

7000

NoRHStages

Qin

[kJ

/kg]



10-133

10-118 The effect of number of regeneration stages on the performance an ideal regenerative Rankine cycle with one open feedwater heater is to be investigated.


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=0 REPEAT i=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=0 REPEAT i=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 i=0 REPEAT



10-134

i=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 = js y[i] =(1- sumyj)*(h3[i] - h4[i+1])/(h6[i] - h4[i+1]) ENDIF T3[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=NoFWH ss[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=0 REPEAT i=i+1 sumyi = 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" 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" PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission.


10-135

"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) endif END "Input Data" NoFWH = 2 P[5] = 10000 [kPa] T[5] = 500 [C] P_cond=10 [kPa] Eta_turb= 1.0 "Turbine isentropic efficiency" Eta_pump = 1.0 "Pump isentropic efficiency" P[1] = P_cond P[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

No FWH

ηth wnet [kJ/kg]

qin [kJ/kg]

0 1 2 3 4 5 6 7 8 9

10

0.4019 0.4311 0.4401 0.4469 0.4513 0.4544 0.4567 0.4585 0.4599 0.4611 0.462

1275 1125 1061 1031 1013 1000 990.5 983.3 977.7 973.1 969.4

3173 2609 2411 2307 2243 2200 2169 2145 2126 2111 2098

0 1 2 3 4 5 6 7 8 9 100.4

0.41

0.42

0.43

0.44

0.45

0.46

0.47

NoFwh

ηth

0 1 2 3 4 5 6 7 8 9 10950

1000

1050

1100

1150

1200

1250

1300

NoFwh

wne

t [k

J/kg

]

0 1 2 3 4 5 6 7 8 9 10

2000

2200

2400

2600

2800

3000

3200

NoFwh

q in

[kJ/

kg]



10-136

10-119 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as η η η η ηcc g s g s= + − where ηg g i=W Q/ n and ηs s g,out=W Q/ are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained.

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

η

η

η

cctotal

in

out

in

gg

in

g,out

in

ss

g,out

out

g,out

= = −

= = −

= = −

W

Q

Q

Q

W

Q

Q

Q

W

Q

Q

Q

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 η η η ηg s g s+ − can be expressed as

cc

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

QQ

η

ηηηη

=

−=

−++−−+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛−−⎟

⎟⎠

⎞⎜⎜⎝

⎛−+⎟

⎟⎠

⎞⎜⎜⎝

⎛−=−+

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 combined cycle is determined to be

η η η η ηcc g s g s= + − = + − × =0 4 0 30 0 40 0 30. . . . 0.58

10-120 The thermal efficiency of a combined gas-steam power plant ηcc can be expressed in terms of the thermal efficiencies of the gas and the steam turbine cycles as η η η η ηcc g s g s= + − . It is to be shown that the value of ηcc is greater

than either of η ηg s or .

Analysis By factoring out terms, the relation η η η η ηcc g s g s= + − can be expressed as

η η η η η η η η η

η

cc g s g s g s g

Positive since <1

g

g

= + − = + − >( )11 24 34

or η η η η η η η η η

η

cc g s g s s g s

Positive since <1

s

s

= + − = + − >( )11 24 34

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



10-137


b10-121 It is to e shown that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as ( )thinqx ηη −= Carnotth,destroyed , where ηth is efficiency of the Rankine cycle and ηth, 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 and negative if from the reservoir). For a cycle that involves heat transfer only with a source at TH and a sink at T0, the irreversibility becomes

( ) ( )[ ] ( )thCthCthth

HHH

qqTT

qqqq

TTq

Tq

TqTx

ηηηη −=−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

,in,in

0

in

outinin

0out

in

0

out0destroyed

11


10-138

Fundamentals of Engineering (FE) Exam Problems

10-122 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet 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.

10-123 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated 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.

10-124 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified 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.

10-125 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with regeneration that involves one open feed water heater, (select the correct statement per unit mass 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.



10-139

10-126 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 3 MPa and 10 kPa. Water changes from saturated liquid to saturated vapor during the heat addition process. The net work output of this cycle is

(a) 666 kJ/kg (b) 888 kJ/kg (c) 1040 kJ/kg (d) 1130 kJ/kg (e) 1440 kJ/kg

Answer (a) 666 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 numerical values).

P1=33000 "kPa" P2=10 "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)+273 T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273 q_in=h_fg Eta_Carnot=1-T2/T1 w_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"



10-140

10-127 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%


P1=10 "kPa" P2=3000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h1=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_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-h2 q_out=h4-h1 Eta_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"



10-141

10-128 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%


P1=10 "kPa" P2=5000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 h3=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"



10-142

10-129 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 5 kPa and 10 MPa, with a turbine inlet temperature of 600°C. The rate of heat transfer in the boiler is 300 kJ/s. Disregarding the pump work, the power output of this plant is

(a) 93 kW (b) 118 kW (c) 190 kW (d) 216 kW (e) 300 kW

Answer (b) 118 kW


P1=10 "kPa" P2=5000 "kPa" P3=P2 P4=P1 T3=600 "C" s4=s3 Q_rate=300 "kJ/s" m=Q_rate/q_in h1=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-h2 W_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"



10-143

10-130 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg/s and leave at 400 K. Water enters 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 heat exchanger 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


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) "Taking h1=hf@P1"



10-144

10-131 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500°C, and the enthalpy of steam is 3185 kJ/kg at the exit of the high-pressure turbine, and 2247 kJ/kg at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is

(a) 29% (b) 32% (c) 36% (d) 41% (e) 49%

Answer (d) 41%


P1=10 "kPa" P2=8000 "kPa" P3=P2 P4=4000 "kPa" P5=P4 P6=P1 T3=500 "C" T5=500 "C" s4=s3 s6=s5 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 h44=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-h1 Eta_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"



10-145

10-132 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 2 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ/kg and the enthalpy of extracted steam is 2810 kJ/kg, the mass fraction of steam extracted from the turbine is

(a) 10% (b) 14% (c) 26% (d) 36% (e) 50%

Answer (c) 26%


h_feed=252 "kJ/kg" h_extracted=2810 "kJ/kg" P3=2000 "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"

10-133 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater. The enthalpy of the steam is 3374 kJ/kg at the turbine inlet, 2797 kJ/kg at the location of bleeding, and 2346 kJ/kg at the turbine exit. The net power output of the plant is 120 MW, and the fraction of steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rate of 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


h_in=3374 "kJ/kg" h_out=2346 "kJ/kg" h_extracted=2797 "kJ/kg" Wnet_out=120000 "kW" x_bleed=0.172 w_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"



10-146

10-134 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 is extracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steam is 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 the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in the condenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabatic condenser at a rate of 463 kg/s.

Condenser

Boiler

6

7 10

8 Turbine

5 Process heater

P IP II

9

2

3 fwh

4 1

11

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

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

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


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



10-147

P1=10 "kPa" P11=P1 P2=400 "kPa" P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P5=6000 "kPa" P6=P5 T6=450 "C" m_total=20 "kg/s" m7=0.6*m_total m_cond=0.4*m_total C=4.18 "kJ/kg.K" m_cooling=463 "kg/s" s7=s6 s11=s6 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump h3=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_pump2 h6=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)*h3 m_process=m7-m_feed q_process=h8-h9 Q_in=m_total*(h6-h5)

10-135 ··· 10-142 Design and Essay Problems



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