Question

A piston cylinder has 2.5 kg ammonia at 50 kPa, -20°C. Now it is heated to 50°C at constant pressure through the bottom of the cylinder from external hot gas at 200°C. Find the heat transfer to the ammonia and the total entropy generation.

Answer 1

C.V. Ammonia plus space out to the hot gas.

Substitute the work into the energy equation and solve for the heat transfer

1Q2 = m(h2 - h1) = 2.5 (1583.5 – 1434.6) = 372.25 kJ

1S2 gen = m(s2 – s1) - 1Q2/Tgas

= 2.5 (6.8379 – 6.3187) – 372.25/473.15

= 0.511 kJ/K

Remark: This is an internally reversible- externally irreversible process. The * is generated in the space between the 200oC gas and the ammonia. If there are any ∆T in the ammonia then it is also internally irreversible.