Question

Two kilograms of air undergo a polytropic process (n = 1.3) from 600 K and 200 kPa to 900 K. Considering the system and surroundings as an isolated system, find the entropy change of the air and the entropy production.

Answers: -0.1938 kJ/K, 0.284 (Engineering Thermodynamics)

Answer 1

Fo polytropic process,

T2 / T1 = (P2/P1)^(1-1/n)

900 / 600 = (P2 / 200)^(1-1/1.3)

P2 = 1159 kPa

Entropy change of air = m [Cp ln (T2 / T1) - R ln (P2 / P1)]

= 2 * [1.005 ln (900 / 600) - 0.287 ln (1159 / 200)]

= - 0.1938 kJ/K

Heat transfer in polytropic process .....where = 1.4 is the ideal gas specific heat ratio

Putting values, we get Q = - 143.5 kJ

Thus heat transfered to surroundings = 143.5 kJ

Entropy increase of surroundings = Q / T_surr..........taking surrounding temperature to be 300 K we get

= 143.5 / 300

= 0.4775 kJ/K

Entropy production = - 0.1938 + 0.4775

= 0.284 kJ/K