Heat received by the air is given by Q=macv(T2′–T1′)
Say Q=600, ⟹600=1×0.81(T2′–400)
T2′=3×0.81600+400=646.9K say 647K
Available energy with the source (1200–290)×1200600=455kJ
Change in entropy of the air =macvlogeT1T2=3×0.81×loge400647=1.168kJ/K
Unavailability of the air =290×1.168=338.72kJ
Available energy with the air =600–338.72=261.28kJ
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