Question 4.199E: For the same steam power plant as shown in Fig. P4.118 and P...

For the same steam power plant as shown in Fig. P4.118 and Problem 4.198E determine the rate of heat transfer in the economizer which is a low temperature heat exchanger and the steam generator. Determine also the flow rate of cooling water through the condenser, if the cooling water increases from 55 to 75 F in the condenser.

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Condenser:    A _{7}=\pi D _{7}^{2} / 4=0.0491   ft ^{2}, v _{7}=0.01617   ft ^{3} / lbm

 

V _{7}=\dot{ m } v _{7} / A _{7}=\frac{200000 \times 0.01617}{3600 \times 0.0491}=18   ft / s

 

q=78.02-1028.7+\frac{18^{2}-600^{2}}{2 \times 25037}=-957.9   Btu / lbm

 

\dot{ Q }_{ COND }=200000(-957.9)=-1.916 \times 10^{8}   Btu / h

 

Economizer  V _{3} \approx V _{2}    since liquid v is constant:    v_{3} \approx v_{2} \text { and } A_{3}=A_{2},

 

q = h _{3}- h _{2}=323.0-85.3=237.7   Btu / lbm

 

\dot{ Q }_{ ECON }=200000(237.7)= 4 . 7 5 \times 1 0 ^{7}   Btu / h

 

Generator:    A _{4}=\pi D _{4}^{2} / 4=0.349   ft ^{2}, \quad v _{4}=0.9595   ft ^{3} / lbm

 

V _{4}=\dot{ m } v _{4} / A _{4}=\frac{200000 \times 0.9505}{3600 \times 0.349}=151   ft / s

 

A _{3}=\pi D _{3}^{2} / 4=0.349 ft ^{2}, \quad v _{3}=0.0491   ft ^{3} / lbm

 

V _{3}=\dot{ m } v _{3} / A _{3}=\frac{200000 \times 0.0179}{3600 \times 0.0491}=20   ft / s,

 

q =1467.8-323.0+\frac{151^{2}-20^{2}}{2 \times 25037}=1145.2   Btu / lbm

 

\dot{Q}_{ GEN }=200000 \times(1145.2)= 2 . 2 9 1 \times 1 0 ^{8}   Btu / h

 

 

 

 

 

 

 

 

 

 

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