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ACarnotenginewhoseefficiencyis 45%takesheat fromasourcemaintainedat atemperatureof 327^(@)C.Tohave an engineof efficiency60%whatmustbe theintaketemperaturefor thesameexhaust(sink)temperature |
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Answer» Solution :Efficiency of Carnot engine `(eta_(1))=45%=0.45` Initial intake temperature `(T_(1))=327^(@)C=600K` NEW efficiency `(eta_(2))=60%=0.6` Efficiency of Carnot engine is given by `ETA=1-(T_(2))/(T_(1))` `T_(1)` is temperature of source , `T_(2)` is temperature of sink `"1st Case :"T_(2)=(1-eta)T_(1)=(1-0.45)xx600 rArr T_(2)=330K` `"2nd Case :"(T_(2))/(T_(1))=1-eta` `T_(1)=(T_(2))/(1-eta)=(330)/(1-0.6)=(330)/(0.4)` `T_(1)=825K=825-273, T_(1)=552^(@)C` |
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