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Modern Engineering Thermodynamics

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

Q B + Q R<br />

Primary<br />

boiler<br />

Reheater<br />

1<br />

6<br />

3<br />

2<br />

Stage<br />

1<br />

Stage<br />

2<br />

W T<br />

T<br />

2s<br />

1<br />

3<br />

Boiler<br />

feed<br />

pump<br />

W P<br />

Condenser<br />

5<br />

Q C<br />

(a) Equipment schematic<br />

4<br />

5,6s 4s<br />

s<br />

(b) Thermodynamic state diagram<br />

FIGURE 13.27<br />

A Rankine cycle power plant with reheat.<br />

The thermal efficiency of the Rankine cycle power plant with reheat shown in Figure 13.27 can be computed<br />

from the general thermal efficiency definition as<br />

ðη T Þ Rankine<br />

cyc1e with<br />

one reheat unit<br />

W<br />

= _ pm − j _W p j ð<br />

= h 1 − h 2 Þ+ ðh 3 − h 4 Þ− ðh 6 − h 5 Þ<br />

_Q B + _Q R<br />

ðh 1 − h 6 Þ+ ðh 3 − h 2 Þ<br />

and, if the liquid condensate is considered to be incompressible, then Eq. (13.7b) can be used to give<br />

ðη T Þ Rankine<br />

cyc1e with<br />

one reheat unit<br />

ð<br />

= h 1 − h 2 Þ+ ðh 3 − h 4 Þ− v 5 ðp 6 − p 5<br />

ðh 1 − h 6 Þ+ ðh 3 − h 2 Þ<br />

where (η s ) p is the isentropic efficiency of the boiler feed pump. Using Eq. (13.4a), we can introduce the isentropic<br />

efficiencies of the two turbine stages (η s ) pm1 and (η s ) pm2 as<br />

Þ/ ðη s<br />

Þ p<br />

and<br />

ðη s Þ pm1<br />

= h 1 − h 2<br />

(13.13)<br />

h 1 − h 2s<br />

Finally, the thermal efficiency of the cycle can be written as<br />

ðη s Þ pm2 = h 3 − h 4<br />

(13.14)<br />

h 3 − h 4s<br />

ðη T Þ Rankine<br />

cycle with<br />

one reheat unit<br />

ð<br />

= h 1 − h 2s<br />

Þðη s Þ pm1<br />

+ ðh 3 − h 4s Þðη s Þ pm2<br />

− v 5 ðp 6 − p 5<br />

ðh 1 − h 6 Þ+ ðh 3 − h 2 Þ<br />

Þ/ ðη s Þ p<br />

(13.15)<br />

where the values of h 2 and h 6 in the denominator of this equation are calculated from Eqs. (13.13) and<br />

(13.7b) as<br />

and<br />

h 2 = h 1 − ðh 1 − h 2s Þðη s Þ pm1<br />

(13.16)<br />

h 6 = h 5 + v 5 ðp 6 − p 5 Þ/ ðη s Þ p<br />

(13.17)

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