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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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