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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% |
Answer» <html><body><p><strong>1/6 of sink temperature should be lowered keeping the source temperature constant to <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> the efficiency by 10 %.</strong></p><p><strong>• Let tsink = temperature of sink</strong></p><p><strong>and, tsource = temperature of source</strong></p><p><strong>• According to the question,</strong></p><p><strong>.40 = 1 – (tsink / tsource)</strong></p><p><strong>(tsink / tsource) = .60 ………(i)</strong></p><p><strong>• Let decrease in temperature be <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>. Then,</strong></p><p><strong>.50 = 1 – [(tsink - x) / tsource]</strong></p><p><strong>(tsink - x) / tsource = .50 ……..(ii)</strong></p><p><strong>• On <a href="https://interviewquestions.tuteehub.com/tag/dividing-957391" style="font-weight:bold;" target="_blank" title="Click to know more about DIVIDING">DIVIDING</a> (i) by (ii) and <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a>,</strong></p><p><strong>x = (1/6) tsink</strong></p><p><strong></strong></p></body></html> | |