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A Carnot engine works between the source and the sink with efficiency 40%. How much temperature of the sink be lowered keeping the source temperature constant so that it's efficiency increases by 10% |
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Answer» 1/6 of sink temperature should be lowered keeping the source temperature constant to INCREASE the efficiency by 10 %. • Let tsink = temperature of sink and, tsource = temperature of source • According to the question, .40 = 1 – (tsink / tsource) (tsink / tsource) = .60 ………(i) • Let decrease in temperature be X. Then, .50 = 1 – [(tsink - x) / tsource] (tsink - x) / tsource = .50 ……..(ii) • On DIVIDING (i) by (ii) and SOLVING, x = (1/6) tsink
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