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ENGR 2213 Thermodynamics. F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma. T. 3. 2. 1. 4. S. 2. 3. T. 5. 2. Boiler. 3. Boiler. 3. 4. Pump. 4. 2. Pump. Turbine. 1. 5. 1. 6. 6. Condenser. 1. 4. S. Condenser. - PowerPoint PPT Presentation
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ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics
F. C. Lai
School of Aerospace and Mechanical
Engineering
University of Oklahoma
Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles
T
S
12
3
4
Turbine
Boiler
Condenser
Pump
1
2
3
4
Boiler
Condenser
Pump
1
2
3
6
4
5
T
S
1
2
3
4
5
6
Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles
net out
in in
w q1
q q 6 1
3 2 5 4
h h1
(h h ) (h h )
qin = (h3 – h2) + (h5 – h4)
qout = h6 – h1
p 2 1
t 3 4 5 6
w h hback work ratio
w (h h ) (h h )
T
S
1
2
3
4
5
6wt = (h3 – h4) + (h5 – h6)
wp = h2 – h1 = v(p2 – p1)
Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles
The reheat process in general does not significantlychange the cycle efficiency.
The sole purpose of the reheat cycle is to reducethe moisture content of the steam at the final stagesof the expansion process.
Example 1Example 1
Consider a steam power plant operating on the ideal reheat Rankine cycle. The steam entersthe turbine at 15 MPa and 600 ºC and is condensedin the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4%, determine(a) the pressure at which the steam should be reheated.(b) the thermal efficiency of this cycle.
Example 1 (continued)Example 1 (continued)
State 6: saturated mixture at p1 = 10 kPa, x6 = 0.896
State 5: superheated vapor at T5 = 600 ºC
s5 = s6 = 7.370 kJ/kg·K
h6 = hf + x6hfg = 191.83 + 0.896(2392.8) = 2335.8 kJ/kgs6 = sf + x6sfg = 0.6493 + 0.896(7.5009) = 7.370 kJ/kg·K
Table A-6, p5 = 4 MPa
Example 1 (continued)Example 1 (continued)
State 2: compressed liquid at p2 = 15 MPa
wp = v(p2 – p1) = (0.001008)(15000-10) = 15.11 kJ/kgh2 = h1 + wp = 191.83 + 15.11 = 206.94 kJ/kg
Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg
State 1: saturated liquid at p1 = 10 kPa
Example 1 (continued)Example 1 (continued)
State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC
Table A-6 h3 = 3582.3 kJ/kg s3 = 6.6776 kJ/kg·K
State 4: superheated vapor at p4 = 4 MPa
s4 = s3 = 6.6776 kJ/kg·KTable A-6 h4 = 3154.3 kJ/kg T4 = 375.5 ºC
Example 1 (continued)Example 1 (continued)
qin = (h3 – h2) +(h5 – h4) = (3582.3 – 206.94) + (3674.4 – 3154.3) = 3895.46 kJ/kg
qout = h6 – h1 = 2335.8 – 191.83 = 2143.97 kJ/kg
out
in
q 2143.971 1 0.45
q 3895.46
0.43 (without reheat)
Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles
1
2
3
7
45
6
P 1
Turbine
Boiler
Condenser
P 2 FWH
T
S
1
2 3
4
5
7
6
Turbine
Condenser
Pump
1
2
3
4
Boiler T
S
12
3
4
Open Feedwater Heater
Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles
net
in
w
q 5 6 6 7 2 1 4 3
5 4
(h h ) (1 y)(h h ) (1 y)(h h ) (h h )
(h h )
qin = h5 – h4
qout = (1 – y)(h7 – h1)
wt = (h5 – h6) + (1 – y)(h6 – h7)
wp1 = h2 – h1 = v1(p2 – p1) T
S
1
2 3
4
5
7
6wp2 = h4 – h3 = v3(p4 – p3)
1-yywp = (1 – y)wp1 + wp2
Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles
net out
in in
w q1
q q
7 1
5 4
(1 y)(h h )1
(h h )
T
S
1
2 3
4
5
7
61-yy
1
2
3
7
45
6
P 1
Turbine
Boiler
Condenser
P 2 FWHy
1-y
yh6 + (1-y)h2 = h3
3 2
6 2
h hy
h h
Example 2Example 2
Consider a steam power plant operating on the ideal regenerative Rankine cycle using open feedwater heater. The steam enters the turbineat 15 MPa and 600 ºC and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the feedwater heater. Determine(a) the fraction of steam extracted from the turbine.(b) the thermal efficiency of this cycle.
Example 2 (continued)Example 2 (continued)
State 2: compressed liquid at p2 = 1.2 MPa
wp1 = v(p2 – p1) = (0.001008)(1200-10) = 1.20 kJ/kgh2 = h1 + wp1 = 191.83 + 1.2 = 193.03 kJ/kg
Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg
State 1: saturated liquid at p1 = 10 kPa
Example 2 (continued)Example 2 (continued)
State 3: saturated liquid at p3 = 1.2 MPa
Table A-5 h3 = 798.65 kJ/kg v3 = 0.001139 m3/kg
State 4: compressed liquid at p4 = 15 MPawp2 = v3(p4 – p3) = (0.001139)(15000-1200) = 15.72 kJ/kg
h4 = h3 + wp2 = 798.65 + 15.72 = 814.37 kJ/kg
Example 2 (continued)Example 2 (continued)
State 5: superheated vapor at p5 = 15 MPa and T5 = 600 ºC
State 6: p6 = 1.2 MPa
s6 = s5 = 6.6776 kJ/kg·KTable A-6, h6 = 2859.5 kJ/kg
Table A-6 h5 = 3582.3 kJ/kg s5 = 6.6776 kJ/kg·K
State 7: p7 = 10 kPas7 = s6 = s5 = 6.6776 kJ/kg·K
Example 2 (continued)Example 2 (continued)
7 f7
fg
s s 6.6776 0.6493x 0.804
s 7.5009
State 7: saturated mixture at p4 = 10 kPa
h7 = hf + x7hfg
= 191.83 + 0.804(2392.8) = 2115.6 kJ/kg3 2
6 2
h h 798.65 193.03y 0.227
h h 2859.5 193.03
Example 2 (continued)Example 2 (continued)
qin = h5 – h4 = 3582.3 – 814.37 = 2767.93 kJ/kg
qout = (1 – y)(h7 – h1) = (1 – 0.227)(2115.6 – 191.83) = 1487.1 kJ/kg
out
in
q 1487.11 1 0.463
q 2767.93
0.43 (without reheat)