e e i i - به نام یگانه مهندس گیتی ... · pdf file10-79 10-89 a steam power...

10-77

Special Topic: Binary Vapor Cycles 10-82C Binary power cycle is a cycle which is actually a combination of two cycles; one in the high temperature region, and the other in the low temperature region. Its purpose is to increase thermal efficiency. 10-83C Consider the heat exchanger of a binary power cycle. The working fluid of the topping cycle (cycle A) enters the heat exchanger at state 1 and leaves at state 2. The working fluid of the 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

10-84C Steam is not an ideal fluid for vapor power cycles because its critical temperature is low, its saturation dome resembles an inverted V, and its condenser pressure is too low. 10-85C Because mercury has a high critical temperature, relatively low critical pressure, but a very low condenser pressure. It is also toxic, expensive, and has a low enthalpy of vaporization. 10-86C In binary vapor power cycles, both cycles are vapor cycles. In the combined gas-steam power cycle, one of the cycles is a gas cycle.

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

10-78

Review Problems 10-87 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as where d are the thermal efficiencies of

the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. η η η η ηcc g s g s= + − ηg g i=W Q/ n an ηs s g,out=W Q/

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

= = −

= = −

= = −

WQ

QQ

WQ

QQ

WQ

QQ

1

1

1

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

Using the relations above, the expression can be expressed as η η η ηg s g+ − s

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

10-88 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 . It is to be shown that

the value of

η η η ηcc g s g s= + − η

ηcc is greater than either of . η ηg s or

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

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.


10-79

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

( )( )( )( )

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

600°CTSINGLE

10 kPa

25 MPa 4

3

s 6

2

5

1

6

7DOUBL

10 kPa

25 MPa 4

3

8

2

5

1

600°C (b) Double Reheat :

C600 and

KkJ/kg 3637.6C600MPa 25

5

5

34

4

33

3

°==

==

⋅=

°==

TPP

ssPP

sTP

xx

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


10-80

10-90E A geothermal power plant operating on the simple Rankine cycle using an organic fluid as 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. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The exit temperature of geothermal water from the vaporizer is determined from the steady-flow energy balance on the geothermal water (brine),

( )( )( )(

F267.4°=°−°⋅=−

−=

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


10-81

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

( )( )( )

kJ/kg 2501.8kJ/kg 0.2574

kJ/kg 79.17404.675.168kJ/kg 6.04

mkPa 1kJ 1

kPa 7.56000/kgm 0.001008

C29.40/kgm 001008.0

kJ/kg 75.168

kPa 5.7 @ 4

kPa 5.7 @ 4

inp,12

33

121inp,

kPa 7.5 @ sat1

3kPa 5.7 @ 1

kPa 5.7 @ 1

====

=+=+==

⋅−=

−=

°======

g

g

f

f

sshh

whh

PPw

TT

hh

v

vv

T

6 MPa

4

2

3

17.5 kPa

qout

qin

s

C1089.2°==

==

3

3

43

3 kJ/kg 2.4852MPa 6Th

ssP

(b)

kJ/kg 1.22723.24054.4677kJ/kg 3.240575.1680.2574kJ/kg 4.467779.1742.4852

outinnet

14out

23in

=−=−==−=−==−=−=

qqwhhqhhq

and

48.6%===kJ/kg 4677.4kJ/kg 2272.1

in

netth q

wη

Thus,

( )( ) kJ/s 19,428=== kJ/s 40,0004857.0inthnet QW && η

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

( )( ) kg/s 194.6=°−°⋅

=∆

=

=−=−=

C1540.29CkJ/kg 4.18kJ/s 20,572

kJ/s 572,20428,19000,40

outcool

netinout

TcQm

WQQ

&&

&&&


10-82

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

( )

( )( )

kJ/kg 82.15207.1575.137kJ/kg 15.07

mkPa 1kJ 1kPa 515,000/kgm 0.001005

/kgm 001005.0kJ/kg 75.137

inp,12

33

121inp,

3kPa 5 @ 1

kPa 5 @ 1

=+=+==

⋅−=

−=

====

whh

PPw

hh

f

f

v

vv

T

( )( ) kJ/kg 9.23670.24239204.075.137

9204.09176.7

4762.07642.7kPa 5

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 1

kJ/kg 3.2971MPa 1

KkJ/kg 9781.6kJ/kg 7.3434

C500MPa 5

kJ/kg 4.3007MPa 5

KkJ/kg 3480.6kJ/kg 8.3310

C500MPa 15

88

88

78

8

7

7

7

7

656

6

5

5

5

5

434

4

3

3

3

3

=+=+=

=−

=−

=

==

⋅==

°==

=

==

⋅==

°==

=

==

⋅==

°==

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

hssP

sh

TP

hssP

sh

TP

1 MPa 5 MPa

6

7

5 kPa

15 MPa 4

3

8

2

5

1s

Then,

( ) ( ) ( )

kJ/kg 9.18622.22301.4093kJ/kg 2.223075.1379.2367

kJ/kg 1.40933.29711.34794.30077.343482.1528.3310

outinnet

18out

674523in

=−=−==−=−=

=−+−+−=−+−+−=

qqwhhq

hhhhhhq

Thus,

45.5%===kJ/kg 4093.1kJ/kg 1862.9

in

netth q

wη

(b) kg/s 64.4===kJ/kg 1862.9

kJ/s 120,000

net

net

wWm&

&


10-83

10-93 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. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

1-y 6

P II P I

Open fwh

Condenser

Boiler Turbine

4

3 2

1

7

5

y

T

6

7 qout

3 y

4

1-y 0.5 MPa

10 kPa

10 MPa

6s

5

7s

2

qin

1

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

( )( )( ) ( )

/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)


10-84

( )(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

(steady) 0systemoutin

11

0

hhyyhhmhmhmhmhm

EE

EEE

eeii =−+→=+→=

=

=∆=−

∑∑ &&&&&

&&

&&&

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

1718.033.1923.279833.19209.640

26

23 =−−

=−−

=hhhh

y

Then, ( )( ) ( )( )

kJ/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&

&

(b) The thermal efficiency is determined from

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

qsmsmTsTiL

iiee


10-85

10-94 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. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

1-y 6

P II P I

Open fwh

Condenser

Boiler Turbine

4

3 2

1

7

5

y

T

qout

3 y

4

1-y

0.5 MPa

10 kPa

10 MPa

6

5

7

2

qin

1


( )( )( )

/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


10-86

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

( ) ( 326332266

outin(steady) 0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii =−+→=+→=

=→=∆=−

∑∑ &&&&&

&&&&&

)


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&

&

(b) The thermal efficiency is determined from

%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

qsmsmTsTiL

iiee


10-87

10-95 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. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

( )( )( )

/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

7

6

1-y 8

P II P I

Openfwh

Condens.

BoilerTurbine

4

3 2

1

9

5

y

3

48

1 MPa

0.6 MPa

15 kPa

10 MPa6

5

9

2

7

1

T

s

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

( ) ( 328332288

outin(steady) 0

systemoutin

11

0

hhyyhhmhmhmhmhm

EEEEE

eeii =−+→=+→=

=→=∆=−

∑∑ &&&&&

&&&&&

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

0.144=−−

=−−

=53.2262.331053.22638.670

28

23

hhhhy

(b) The thermal efficiency is determined from ( ) ( ) ( ) ( )

( )( ) ( )( ) kJ/kg 7.196294.2258.25181440.011kJ/kg 7.33898.27831.347973.6801.3375

19out

6745in

=−−=−−==−+−=−+−=

hhyqhhhhq

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

11in

outth q

qη


10-88

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

7

6

1-y 8

P II P I

Open fwh

Condenser

Boiler Turbine

4

3 2

1

9

5

y

T

1-y3

4 8 6 6s 8s y

5

9 9s

2

7

1


( )( )( )

/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

)


10-89

( ) ( )(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

shP

)

T

( )( )

( ) ( )(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

(b) The thermal efficiency is determined from ( ) ( )

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


10-90

10-97 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for the open feedwater heater, the thermal efficiency of the cycle, and the net power output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

High-P Turbine

11 1-y-z

13

12 z

10

4

Closed fwh

5 y

P II

P I

Open fwh I

Condenser

Boiler Low-P Turbine

6 3

2 1

7

8

9

T 1 MPa

10

11 15 MPa

z

8

56

3

y

4

1 - y - z 5 kPa 7

0.2 MPa

0.6 MPa

12

9

13

2

1s

( )

( )( )

kJ/kg 95.13720.075.137kJ/kg 0.20

mkPa 1kJ 1

kPa 5200/kgm 0.001005

/km 001005.0kJ/kg 75.137

inpI,12

33

121inpI,

3kPa 5 @ 1

kPa 5 @ 1

=+=+==

⋅−=

−=

==

==

whh

PPw

ghh

f

f

v

vv

( )( )( )

/kgm 100110.0kJ/kg 38.670

liquidsat.MPa 6.0

kJ/kg 41.52070.1571.504kJ/kg 15.70

mkPa 1kJ 1kPa 20015,000/kgm 0.001061

/kgm 001061.0kJ/kg 71.504

liquidsat.MPa 2.0

3MPa 6.0 @ 6

MPa 6.0 @ 766

inpII,34

33

343inpII,

3MPa 2.0 @ 3

MPa 2.0 @ 33

==

===

=

=+=+==

⋅−=

−=

==

==

=

f

f

f

f

hhhP

whh

PPw

hhP

vv

v

vv

( )( )( )

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 0.1

kJ/kg 8.2820MPa 0.1

KkJ/kg 6796.6kJ/kg 1.3583

C600MPa 15

kJ/kg 686.23 mkPa 1kJ 1kPa 60015,000/kgm 0.00110138.670

10

10

10

10

989

9

8

8

88

33

6566556

⋅==

°==

=

==

⋅==

°==

=

⋅−+=

−+=→=

sh

TP

hss

P

sh

TP

PPvhhTT


10-91

( )( )kJ/kg 1.2368

0.24239205.075.137

9205.0

9176.74762.07642.7

kPa 5

kJ/kg 9.3000MPa 2.0

kJ/kg 2.3310MPa 6.0

1313

1313

1013

13

121012

12

111011

11

=+=

+==

−=

−=

==

=

==

=

==

fgf

fg

f

hxhh

sss

x

ssP

hss

P

hss

P


( ) ( ) ( ) ( 4561145561111

outin

(steady) 0systemoutin 0

hhhhyhhmhhmhmhm

EE

EEE

eeii −=−→−=−→=

=

=∆=−

∑∑ &&&&

&&

&&&

)


06287.038.6702.331041.52023.686

611

45 =−−

=−−

=hhhhy

For the open FWH,

( ) ( ) 31227

3312122277

outin


11

0

hzhhzyyhhmhmhmhm

hmhm

EE

EEE

eeii

=+−−+=++

=

=

=∆=−

∑∑&&&&

&&

&&

&&&

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

( ) ( ) ( )( ) 0.1165=

−−−−

=−

−−−=

95.1370.300095.13738.67006287.095.13771.504

212

2723

hhhhyhh

z

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

kJ/kg 9.17244.18303.3555kJ/kg 4.183075.1370.23681165.006287.011

kJ/kg 3.35558.28201.347923.6861.3583

outinnet

113out

91058in

=−=−==−−−=−−−=

=−+−=−+−=

qqwhhzyq

hhhhq

and

48.5%=−=−=kJ/kg 3555.3kJ/kg 1830.4

11in

outth q

qη

(c) ( )( ) kW 72,447=== kJ/kg 1724.9kg/s 42netnet wmW &&


10-92

10-98 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies 7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

( )(kJ/kg 8.2518

3.23729665.094.225

9665.02522.7

7549.07642.7

kPa 15

KkJ/kg 7642.7kJ/kg 1.3479

C500MPa 1

kJ/kg 7.2843MPa 1

KkJ/kg 7266.6kJ/kg 5.3399

C500MPa 8

kJ/kg 81.503

kJ/kg 94.225

99

99

89

9

8

8

8

8

767

7

6

6

6

6

C 120 @ 3

12

kPa 15 @ 1

=+=+=

=−

=−

=

==

⋅==

°==

=

==

⋅==

°==

==≅

==

°

fgf

fg

f

f

f

hxhhs

ssx

ssP

shP

hssP

sh

TP

hhhhhh

T 4

7

8

P II P I

Process heater

Condenser

BoilerTurbine

5

3

2

1

9

6

)

T

6 8 8 MPa

31 MPa5 7

42The mass flow rate through the process heater is

( ) kg/s 929.2kJ/kg 81.5037.2843

kJ/s 7,000

37

process3 =

−=

−=

hhQ

m&

& 15 kPa9 1

s Also,

or, ( ) ( ) ( ) ( )( )

( ) ( )( )8.25181.3479992.27.28435.3399kJ/s 3000

993.2

66

986766989766

−−+−=

−−+−=−+−=

mm

hhmhhmhhmhhmWT

&&

&&&&&

It yields kg/s 873.36 =m&

and kg/s 881.0992.2873.3369 =−=−= mmm &&& Mixing chamber:

& & &

& &

& & & &

E E E

E E

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

in out system (steady)

in out

− = =

=

= → = +∑ ∑

∆ 0

4 4 2 2 3 3

0

&

or, ( )( ) ( )( ) kJ/kg 60.440873.3

81.503992.294.225881.0

4

332254 =

+=

+=≅

mhmhmhh

&

&&

Then,

( ) ( )( )( ) ( )(

kW 12,020=−+−=

−+−=kJ/kg 2843.73479.1kg/s 0.881kJ/kg 440.603399.5kg/s 3.873

788566in hhmhhmQ &&&

)

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

77.3%===kg/s 3.873kg/s 2.992

total

3

mm

y&

&


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