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> Efficiency of a heat engine whose sink is at temperature of300K is 40%. To increase the efficiency to 60%, keeping thesink temperature constant, the source temperature must beincreased by |
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Answer» Quite an easy question to answer...EFFICIENCY of heat ENGINE N = (T2 - T1)/T2where T2 is source TEMPERATURE and T1 is the sink temperature....For n = 0.40.4 = (T2 - T1)/T2T1/T2 = 0.6As T1 = 300 K...... T2 = 300/0.6 = 500KCase 2: for efficiency n to be increased 50% of its ORIGINAL efficiency , i.e. n = 0.4( 1+ 0.5) = 0.6 = 60%Source temperature can be calculated by the same formula n = (T2 - T1)/T20.6 = (T2 - 300)/T2 ………( sink temperature is constant)T2 = 300/0.4 = 750KIncrease in source temperatureDel T = (750 - 500) = 250 K....That's your answer....The source temperature should be increased by 250 K so as to increase the efficiency of heat engine by 50 % of its original efficiency. |
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