SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 109 of 200

Marks: +1, -0

A reversible heat engine receives 240 kJ/s heat from a source at 627°C and rejects heat to a sink at 300 K. The available power is:

Solution:

T_H = 627°C = 627 + 273 = 900 K

,T_L = 300 K, Q_\in = 240

kJ/s

\η ={W_{\text "out "}}/{Q_{H}}={T_{H}-T_{L}}/{T_{H}}

{W_{\text "out "}}/{240}={900-300}/{900} \

⇒{W_{\text "out "}}/{240}={2}/{3} W_{\text "out "}=

160 kJ/s

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