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1. A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined.

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

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

h1 = h f @ 10 kPa = 191.81 kJ/kg v1 = v f @ 10 kPa = 0.00101 m 3 /kg

w p,in = v1 (P2 - P1 )

( ) =

0.00101

m 3 /kg

(7,000 -10

kPa

)

1

1 kJ kPa m

3

= 7.06 kJ/kg

3 7 MPa

2

qin

h2 = h1 + w p,in = 191.81+ 7.06 = 198.87 kJ/kg

P 3 T3

= =

7 MPa 500?C

h3 s3

= =

3411.4 6.8000

kJ/kg kJ/kg

K

10 kPa

1

qout

4

s

P4 s4

= 10 = s3

kPa

x4

=

s4 - s f s fg

= 6.8000 - 0.6492 = 0.8201 7.4996

h4 = h f + x4h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg

Thus,

qin = h3 - h2 = 3411.4 - 198.87 = 3212.5 kJ/kg qout = h4 - h1 = 2153.6 - 191.81 = 1961.8 kJ/kg wnet = qin - qout = 3212.5 - 1961.8 = 1250.7 kJ/kg

and

th

=

wnet qin

= 1250.7 kJ/kg 3212.5 kJ/kg

= 38.9%

(b)

m& = W&net = 45,000 kJ/s = 36.0 kg/s

wnet 1250.7 kJ/kg

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

Q& out = m& qout = (35.98 kg/s)(1961.8 kJ/kg) = 70,586 kJ/s

Tcoolingwater

=

Q& out (m& c) coolingwater

=

(2000

70,586 kJ/s

kg/s)(4.18 kJ/kg

?C)

=

8.4?C

2. A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and the thermal efficiency of the cycle are to be determined.

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

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

h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3/kg

T

( ) wp,in

= v1(P2 - P1 )

= 0.001017 m3/kg = 6.08 kJ/kg

(6000

-

20

kPa

)

1

1 kJ kPa m

3

6 MPa

3 5 4

h2 = h1 + wp,in = 251.42 + 6.08 = 257.50 kJ/kg

2

P 3 T3

= =

6 MPa 400?C

h3 s3

= =

3178.3 6.5432

kJ/kg kJ/kg

K

P4 s4

= 2 MPa = s3

h4

=

2901.0

kJ/kg

20 kPa

1

6

s

P5 T5

= =

2 MPa 400?C

h5 s5

= =

3248.4 7.1292

kJ/kg kJ/kg

K

P6 s6

= =

20 s5

kPa

x6 h6

= =

s6 - s f = 7.1292 - 0.8320 = 0.8900

s fg

7.0752

h f + x6h fg = 251.42 + (0.8900)(2357.5)

=

2349.7

kJ/kg

The turbine work output and the thermal efficiency are determined from

wT,out = (h3 - h4 ) + (h5 - h6 ) = 3178.3 - 2901.0 + 3248.4 - 2349.7 = 1176 kJ/kg

and

qin = (h3 - h2 ) + (h5 - h4 ) = 3178.3 - 257.50 + 3248.4 - 2901.0 = 3268 kJ/kg

Thus,

wnet = wT ,out - w p,in = 1176 - 6.08 = 1170 kJ/kg

th

=

wnet qin

= 1170 kJ/kg 3268 kJ/kg

= 0.358 = 35.8%

3. A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined.

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

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

h1 = hsat@ 10 kPa = 191.81 kJ/kg

T

v 1 = v sat@ 10 kPa = 0.00101 m 3 /kg

( ) w p,in

= v1 (P2 - P1 ) = 0.00101 m3/kg (15,000 -10

= 15.14 kJ/kg

kPa) 1

1 kJ kPa m3

15 2

3 5 4

h2 = h1 + w p,in = 191.81+15.14 = 206.95 kJ/kg

10 kPa

P3 T3

= 15 MPa = 500?C

h3 s3

= =

3310.8 6.3480

kJ/kg kJ/kg K

1

6

s

P6 s6

= 10 = s5

kPa

h6

s6

= hf = sf

+ x6 h fg + x6 s fg

= 191.81+ (0.90)(2392.1) = 2344.7 kJ/kg = 0.6492 + (0.90)(7.4996) = 7.3988 kJ/kg K

T5 s5

= =

500?C s6

P5 h5

= =

2150 kPa (the reheat

3466.61 kJ/kg

pressure)

P4 s4

= =

2.15 s3

MPa

h4

=

2817.2

kJ/kg

(b) The rate of heat supply is

Q&in = m& [(h3 - h2 )+ (h5 - h4 )] = (12 kg/s)(3310.8 - 206.95 + 3466.61- 2817.2)kJ/kg

= 45,039 kW

(c) The thermal efficiency is determined from

Thus,

Q& out = m& (h6 - h1 ) = (12 kJ/s)(2344.7 -191.81)kJ/kg = 25,835 kJ/s

th

=

1

-

Q& out Q& in

= 1- 25,834 kJ/s 45,039 kJ/s

= 42.6%

4. 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.

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

Boiler 6 fwh II

P III 5

7 Turbine

8

10 9

fwh I 4

3 P II

2 P I

Condenser 1

6 10 MPa

5 0.6 MPa 4

y 3 0.2 MPa 2

5 kPa 1

7

8 1-y 9 1 - y - z 10

s

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

h1 = h f @ 5 kPa = 137.75 kJ/kg

( ) v1

w pI ,in h2

= = =

v f @ 5 kPa = 0.001005 m 3 /kg

v1 (P2 - P1 ) = 0.001005 m3/kg (200 - 5 kPa

h1 + w pI ,in = 137.75 + 0.20 = 137.95 kJ/kg

)

1

1 kJ kPa m3

=

0.20

kJ/kg

P3 = 0.2 MPa sat.liquid

h3 v 3

= =

h v

f @ 0.2 MPa f @ 0.2 MPa

= 504.71 kJ/kg = 0.001061 m3/kg

( ) w pII,in

= v 3 (P4 - P3 ) =

= 0.42 kJ/kg

0.001061

m3 /kg

(600

-

200

kPa

)

1

1 kJ kPa m

3

h4 = h3 + w pII ,in = 504.71+ 0.42 = 505.13 kJ/kg

P5 = 0.6 MPa sat.liquid

h5 v 5

= =

h f @ 0.6 MPa v f @ 0.6 MPa

= 670.38 kJ/kg = 0.001101 m3/kg

( ) w pIII,in

= v 5 (P6 - P5 ) =

= 10.35 kJ/kg

0.001101

m3/kg (10,000 - 600

kPa

)

1

1 kJ kPa m

3

h6 = h5 + w pIII,in = 670.38 +10.35 = 680.73 kJ/kg

P7 T7

= 10 MPa = 600?C

h7 s7

= 3625.8 = 6.9045

kJ/kg kJ/kg K

P8 s8

= =

0.6 s7

MPa

h8

=

2821.8

kJ/kg

P 9 s9

= =

0.2 s7

MPa

x9 h9

= =

s9 - s f = 6.9045 -1.5302 = 0.9602

s fg

5.5968

h f + x9h fg = 504.71+ (0.9602)(2201.6) =

2618.7

kJ/kg

5. 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),

h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m 3 /kg

wp,in

= =

v1 (P2 - P1 (0.001017

) m

3

/kg)(3000

-

20)kPa

1 kJ

= 3.03 kJ/kg

1 kPa m3

h2 = h1 + wp,in = 251.42 + 3.03 = 254.45 kJ/kg

Boiler

4 Turbine

5 6

P4 = 3000 kPa T4 = 350?C

h4 = 3116.1 kJ/kg s4 = 6.7450 kJ/kg K

P5 = 1000 kPa s5 = s4

h5 = 2851.9 kJ/kg

Closed

fwh 3

P6 = 20 kPa s6 = s4

x6

=

s6 - s f s fg

= 6.7450 - 0.8320 = 0.8357 7.0752

h6 = h f + x6 h fg = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg

Condenser

7

8

1 2 Pump

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.

P7 = 1000 kPa x7 = 0

h7 = 762.51 kJ/kg T7 = 179.9?C

h8 = h7 = 762.51 kJ/kg

P3 T3

= =

3000 kPa T7 = 209.9?C

h3 = 763.53 kJ/kg

An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m& 5 / m& 4 )

for closed feedwater heater:

m& i hi = m& e he

T

4

3 MPa

qin

3 1 MPa y

27

5

8 20 kPa 1-y

1

qout 6

s

m& 5 h5 + m& 2 h2 = m& 3h3 + m& 7 h7 yh5 +1h2 = 1h3 + yh7

Rearranging,

y = h3 - h2 = 763.53 - 254.45 = 0.2437 h5 - h7 2851.9 - 762.51

Then,

wT,out = h4 - h5 + (1- y)(h5 - h6 ) = 3116.1- 2851.9 + (1- 0.2437)(2851.9 - 2221.7) = 740.9 kJ/kg

wP,in = 3.03 kJ/kg

qin = h4 - h3 = 3116.1- 763.53 = 2353 kJ/kg

Also, wnet = wT,out - wP,in = 740.9 - 3.03 = 737.8 kJ/kg

th

= wnet qin

= 737.8 = 0.3136 2353

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