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Modern Engineering Thermodynamics

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484 CHAPTER 13: Vapor and Gas Power Cycles<br />

EXAMPLE 13.8 (Continued )<br />

1<br />

Boiler<br />

8s<br />

Boiler<br />

feed<br />

pump<br />

7<br />

Reheater<br />

#1<br />

#2<br />

2s<br />

4s<br />

3<br />

5<br />

Turbine #1<br />

Turbine #2<br />

Turbine #3<br />

Condenser<br />

6s<br />

(a) System schematic<br />

1200<br />

1000<br />

1<br />

3 5<br />

1<br />

3<br />

5<br />

T (°F)<br />

72.8<br />

7, 8s<br />

2s<br />

4s<br />

6s<br />

s<br />

(b) System T–s diagram<br />

h<br />

2s 4s<br />

CP<br />

7, 8s<br />

s<br />

(c) System h–s diagram<br />

6s<br />

FIGURE 13.34<br />

Example 13.8.<br />

The thermodynamic data at the various points around this cycle are as follows:<br />

Station 1—Turbine 1 inlet<br />

Station 2s—Turbine 1 exit<br />

p 1 = 5000: psia<br />

p 2s = 1000: psia<br />

T 1 = 1200:°F<br />

s 2s = s 1 = 1:5068 Btu/lbm . R<br />

h 1 = 1530:8 Btu/lbm<br />

h 2s = 1316:9 Btu/lbm<br />

s 1 = 1:5068 Btu/lbm . R<br />

ðby interpolation in Table C:3aÞ<br />

Station 3—Turbine 2 inlet<br />

Station 4s—Turbine 2 exit<br />

p 3 = 1000: psia<br />

p 4s = 300: psia<br />

T 3 = 1000:°F<br />

s 4s = s 3 = 1:6532 Btu/lbm . R<br />

h 3 = 1505:9 Btu/lbm<br />

h 4s = 1343:8 Btu/lbm<br />

s 3 = 1:6532 Btu/lbm . R<br />

ðby interpolation in Table C:3aÞ<br />

Station 5—Turbine 3 inlet<br />

Station 6s—Turbine 3 exit<br />

p 5 = 300: psia<br />

p 6s = 0:400 psia<br />

T 5 = 1000:°F<br />

s 6s = s 5 = 1:7966 Btu/lbm . R<br />

h 5 = 1526:4 Btu/lbm x 6s = ð1:7966 – 0:0799Þ/1:9762 = 0:8687<br />

s 5 = 1:7966 Btu/lbm . R<br />

h 6s = 40:9 + 0:8687ð1052:4Þ<br />

= 955:1 Btu/lbm<br />

Station 7—Condenser exit<br />

Station 8s—Boiler inlet<br />

p 7 = 0:400 psia<br />

p 8s = p 1 = 5000: psia<br />

x 7 = 0:00 s 8s = s 7<br />

h 7 = h f ð0:4 psiaÞ = 40:9 Btu/lbm h 8s = h 7 + v 7 ðp 8s – p 7 Þ = 40:9 + 0:01606ð5000: – 0:400Þð144/778:16Þ<br />

v 7 = v f ð0:4 psiaÞ = 0:01606 ft 3 /lbm = 55:76 Btu/lbm

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