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**

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