A heat engine receives heat at the rate of 1500 kJ/min and gives an output of 8.2 kW, then the thermal efficiency is-

Question

Question

This question was previously asked in

NPCIL SA/ST ME GJ held on 07/11/2019 Shift-1

Answer 1

Concept:

Thermal efficiency of heat engine

\({\eta _{thermal}} = \frac{{Power\;Output}}{{Heat\;Input}}\)

Calculation:

Given:

Heat input = 1500 kJ/min = 25 kW, Power output = 8.2 kW

F1 Krupalu 2.12.20 Pallavi D8

\({\eta _{thermal}} = \frac{{8.2}}{{25}} = 0.328 = 32.8\;\% \)