SSC JE Mechanical Engineering 2014 Set 1 Paper

Question : 131 of 200

Marks: +1, -0

A reversible heat engine working at the rate of 100 kW has an efficiency of 20%. The magnitudes of heat transfer rate from the source and to the sink in kW would be, respectively,

200, 100
300, 200
500, 400
1000, 900

Validate

Solution:

Efficiency =

{\Work \Output }/{ \Heat \Supplied}

;

0.20 = 100 / {HS}

→

HS = 100 / 0.2 = 500 \KW

And, Work Output = Heat Supplied (from source) – Heat Rejected (to sink) ;
100 = 500 – HR ; HR = 400 KW

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