Solution;
Given;
T1=1000F=810.928K
T2=100F=310.928K
P=10hp=7.457kW
(a)
Thermal efficiency;
η=1−THTC
η=1−810.928310.928=0.61658
η=0.61658 or 61.658 %
(b)
Heat supplied;
η=QiP
Q=ηP=0.616587.457=12.09kJ/s
(c)
Change in entropy;
Heat rejected;
Qo=Qi−P
Qo=12.09−7.457=4.633kJ/s
Δs=T2Qo=310.2984.633×103=14.98J/sK
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