One kilogram of ammonia initially at 8.0 bar and 50°c undergoes a process to 4.5 bar, 20°c while being rapidly expanded in a piston–cylinder assembly. heat transfer between the ammonia and its surroundings occurs at an average temperature of 40°c. the work done by the ammonia is 40 kj. kinetic and potential energy effects can be ignored. determine the heat transfer, in kj, and the entropy production, in kj/k.

Heat transfer (Q) is 76.79KJ

Entropy production (S) is 0.245KJ/K

Explanation:

P1=8bar, P2=4.5bar, T1= 50°C=323K, T2=20°C=293K, T3=40°C=313K, Moles of ammonia (kmol)= 1/17=0.059kmol, Cv=R/0.4 = 8.314/0.4=20.785KJ/kmolK

Change in internal energy (∆U) = Cv(T1 - T2) = 20.785(323 - 293)=20.785×30=623.55KJ/kmol×0.059kmol=36.79KJ

Heat transfer (Q) = ∆U + W = 36.79KJ + 40KJ = 76.79KJ

Entropy production (S) = Q/T3 = 76.79/313 = 0.245KJ/K

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answer:

ya

explanation:

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