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Modern Engineering Thermodynamics

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13.9 Rankine Cycle with Reheat 479<br />

EXAMPLE 13.7<br />

The first Rankine cycle steam turbine prime mover with reheat used in the United States was at the Crawford Avenue power<br />

station of the Commonwealth Edison Company of Chicago, Illinois, which went into operation in September 1924. The primary<br />

steam was at 700.°F, 600. psia, with reheat to 700.°F at 100. psia. The isentropic efficiencies of the first and second turbine<br />

stages and the boiler feed pump were 84.0, 80.0, and 61.0%, respectively. The condenser pressure was 1.00 psia with<br />

saturated liquid being produced at its outlet.<br />

Determine<br />

a. The Rankine cycle thermal efficiency of the plant with reheat.<br />

b. The Rankine cycle thermal efficiency of the plant without reheat (assume a turbine isentropic efficiency of 82.0% for this<br />

calculation).<br />

Solution<br />

a. Using the notation of Figure 13.27, the monitoring station data for this problem are<br />

where the following calculations have been used:<br />

Then,<br />

and<br />

Station 1<br />

Station 2s<br />

p 1 = 600: psia<br />

p 2s = p 2 = 100: psia<br />

T 1 = 700:°F<br />

s 2s = s 1 = 1:5874 Btu/ðlbm . RÞ<br />

h 1 = 135:6 Btu/lbm x 2s = 0:9856<br />

s 1 = 1:5874 Btu/ðlbm . RÞ h 2s = 1175:0 Btu/lbm<br />

Station 3<br />

Station 4s<br />

p 3 = p 2s = 100: pisa p 4s = p 4 = 1:00 psia<br />

T 3 = 700:°F<br />

s 4s = s 3 = 1:8035 Btu/ðlbm . RÞ<br />

h 3 = 1379:2 Btu/lbm x 4s = 0:9054<br />

s 3 = 1:8035 Btu/ðlbm . RÞ h 4s = 1007:7 Btu/lbm<br />

Station 5<br />

Station 6s<br />

p 5 = 1:00 psia<br />

p 6s = p 6 = 600: psia<br />

x 5 = 0:00<br />

s 6s = s 5 = 0:1326 Btu/ðlbm . RÞ<br />

h 5 = 69:7 Btu/lbm h 6s = 72:5 Btu/lbm<br />

s 5 = 0:1326 Btu/ðlbm . RÞ<br />

x 2s = s 2s − s f 2<br />

= s 1 − s f 2 1:5874 − 0:4745<br />

= = 0:9856<br />

s fg2 s fg2 1:1291<br />

h 2s = h f 2 + x 2s ðh fg2 Þ = 298:6 + ð0:9856Þð889:2Þ = 1175:0 Btu/lbm<br />

x 4s = s 3 − s f 4 1:8035 − 0:1326<br />

= = 0:9054<br />

s fg4 1:8455<br />

h 4s = h f 4 + x 4s h fg4<br />

<br />

= 69:7 + ð0:9054Þð1036:0Þ = 1007:7 Btu/lbm:<br />

Since v 5 = v f (1.0 psia) = 0.01614 ft 3 /lbm, Eqs. (13.16) and (13.17) can now be used to give<br />

and<br />

Then, finally, Eq. (13.15) yields<br />

h 2 = 1350:6 − ð0:840Þð1350:6 − 1175:0Þ = 1203:1 Btu/lbm<br />

h 6 = 69:7 + ð0:01614Þð600: − 1:00Þð144/778:16Þ/ð0:610Þ<br />

= 69:7 + 2:93 = 72:6 Btu=lbm<br />

ð1350:6 − 1175:0Þð0:840Þ + ð1379:2 − 1007:7Þð0:800Þ − 2:93<br />

η T =<br />

ð1350:6 − 72:6Þ + ð1379:2 − 1203:1Þ<br />

= 0:304 = 30:4%<br />

(Continued )

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