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A steam power plant receives heat from a furnace at a rate of 300 GJ/h. Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 16 GJ/h. If the waste heat is transferred to the cooling water at a rate of 140 GJ/h, determine (a) net power output and (b) the thermal efficiency of this power plant.

User Cylindric
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1 Answer

4 votes

Answer:

Net Power output 144 GJ/h


\eta = 51.41 %

Step-by-step explanation:

Given data:

Heat received = 300 GJ/h


= (300* 10^9)/(3600) J/s


= 77.78 * 10^6 W

Heat lost
Q_2 = 16 GJ/h

Heat to the water
= Q_3 = 140 GJ/h

Net Power output
= Q_1 - Q_2 -Q_3

= 300 - 16 - 140

= 144 GJ/h


=(144* 10^9)/(3600) = 40* 10^8 J/s

thermal efficiency


\eta = (net\ power)/(heat\ supplied)


= (40* 10^6)/( 77.78 * 10^6)

= 0.514 = 51.41 %

User Fengtao Ding
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