1kg of ammonia is contained in a piston/cylinder, as saturated liquid at −20°C. Heat is added at 100°C until a final condition of 70°C, 800 kPa is reached. Assuming the process is reversible, find the entropy generation.
Answer: D
P1 = 190.08 kPa, v1 = 0.001504 m^3/kg, u1 = 88.76 kJ/kg, s1 = 0.3657 kJ/kg K
v2 = 0.199 m^3/kg, u2 = 1438.3 kJ/kg, s2 = 5.5513 kJ/kg K
1W2 =(1/2)(190.08 + 800)1(0.1990 – 0.001504) = 97.768 kJ
1Q2 = m(u2 – u1) + 1W2 = 1(1438.3 – 88.76) + 97.768 = 1447.3 kJ
1S2(gen) = m(s2 – s1) – 1Q2/T(res) = 1(5.5513 – 0.3657) – (1447.3/373.15)
= 1.307 kJ/K.