Answer:
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