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Modern Engineering Thermodynamics

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13.7 Rankine Cycle with Regeneration 473<br />

pressure), and pump 2 brings the pressure the rest of the way up from 80.0 to 200. psia. Then, v 3 = v f (1.00 psia) =<br />

0.01614 ft 3 /lbm and v 6 = v f (80.0 psia) = 0.01757 ft 3 /lbm. The additional monitoring station data needed are<br />

Station 4s<br />

p 4s = p 4 = 80:0 psia<br />

s 4s = s 3 = 0:1326 Btu/ðlbm . RÞ<br />

Station 5s<br />

p 5s = p 4 = 80:0 psia<br />

h 4s = h 3 + v 3 ðp 4 – p 3 Þ x 5s = 0:9358<br />

s 5s = s 1 = 1:5466 Btu/ðlbm . RÞ<br />

= 69:7 + ð0:01614Þð80:0 – 1:00Þð144/778:16Þ h 5s = 1125:7 Btu/lbm<br />

= 69:7 + 0:236 = 69:9 Btu/lbm<br />

Station 6<br />

p 6 = 80:0 psia<br />

x 6 = 0:00<br />

Station 7s<br />

p 7s = p 7 = 200: psia<br />

s 7s = s 6 = 0:4535 Btu/ðlbm . RÞ<br />

h 6 = 282:2 Btu/lbm h 7s = h 6 + v 6 ðp 7 – p 6 Þ<br />

s 6 = 0:4535 Btu/ðlbm . RÞ<br />

= 282:2 + ð0:01757Þð200: – 80:0Þð144/778Þ<br />

= 282:2 + 0:390 = 282:6 Btu/lbm<br />

where, at station 5s, wedeterminethatx 5s = (1.5466 – 0.4535)/1.1681 = 0.9358 and h 5s = 282.2 + (0.9358)(901.4) =<br />

1125.7 Btu/lbm. Equation (13.11) now gives the value of y as<br />

y = h 6 − h 4s 282:6 − 69:9<br />

=<br />

h 5s − h 4s 1125:7 − 69:9 = 0:201<br />

then, the isentropic thermal efficiency of the cycle is given by Eq. (13.12b) with all the η s = 1.0 as<br />

<br />

ðη T Þ Rankine cycle<br />

= 1 − h <br />

<br />

<br />

2s − h 3<br />

ð1 − yÞ = 1 −<br />

863:5 − 69:7<br />

ð1 − 0:201Þ = 0:308 = 30:8%<br />

h 1 − h 7s<br />

1199:3 − 282:6<br />

with 1 regenerator<br />

By altering the value of y (the amount of steam bled from the turbine) and recomputing the value of h 7s , it becomes clear<br />

that the cycle thermal efficiency is maximized at 31.2% when y = 0.138. This is shown in Figure 13.23.<br />

32.0<br />

0.30<br />

Cycle thermal efficiency<br />

Cycle thermal efficiency (%)<br />

31.0<br />

30.0<br />

Regenerator mass fraction<br />

0.20<br />

0.10<br />

Regenerator mass fraction, y<br />

FIGURE 13.23<br />

Example 13.6, plot of solution.<br />

29.0<br />

0 40 80 120 160 200 0.00<br />

Regenerator pressure, psia<br />

Exercises<br />

16. The operator of the plant described in Example 13.6 changes the boiler outlet state from dry saturated steam at 200. psia<br />

to dry saturated steam at 400. psia. Determine the mass fraction of regeneration steam that is now required to produce<br />

saturated liquid water at 80.0 psia at the end of the open loop regenerator. Assume all the other variables remain<br />

unchanged. Answer: y = 0.210<br />

17. If the interstage steam pressure in Example 13.6 is increased from 80.0 to 150. psia, determine the new isentropic<br />

Rankine cycle thermal efficiency. Assume all the other variables remain unchanged. Answer: (η T ) isentropic Rankine = 31.0%.<br />

18. The power plant engineer mentioned in Exercise 16 just informed us that the first and second stage prime mover<br />

isentropic efficiencies are 75.0% and 68.0%, respectively, and that the boiler feed pump isentropic efficiency is 83.0%.<br />

Now determine the actual Rankine cycle thermal efficiency of this system with the 80.0 psia open loop regenerator.<br />

Answer: (η T ) Rankine = 25.8%.

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