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At steady state, a power cycle having a thermal efficiency of 38% generates 100 MW of electricity while discharging energy by heat transfer to cooling water at an average temperature of 70 F. The average temperature of the steam passing through the boiler is 900 F.

Determine the rate at which is discharged to the cooling water, in BTU/h

User OscarWyck
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1 Answer

5 votes

Answer:

Heat rejected = 556,724,951 Btu/hr

Step-by-step explanation:

given data

thermal efficiency = 38%

electricity = 100 MW

average temperature = 70 F

average temperature = 900 F

solution

Heat input = Output ÷ efficiency ........................1

Heat input = 100 ÷ 0.38

Heat input = 263.16 MW

and

Heat rejected = Heat input - Output ........................2

Heat rejected = 263.16 - 100

Heat rejected = 163.16 MW

Heat rejected = 556,724,951 Btu/hr

User Skamsie
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