Nitrogen cooled: entropy generation

Reading of the problem. "Energy generation" is taken as entropy generation. From the figure, the gas column is initially 0.5 m + 0.5 m = 1.0 m high, and the stops are at mid-height (0.5 m). The piston first moves down at constant pressure (250 kPa) until it reaches the stops, and then cooling continues at constant volume. The cross-sectional area is not needed.

Given: m = 1 kg, P₁ = 250 kPa, T₁ = 500 + 273.15 = 773.15 K, T₃ = 40 °C = 313.15 K, T₀ = 20 °C = 293.15 K, R = 0.297, cv = 0.743, cp = cv + R = 1.040 kJ/kg·K.

Process path

• 1-2: constant pressure, until V₂ = V₁/2. So T₂ = T₁/2 = 386.6 K (113.4 °C).
• 2-3: constant volume at the stops; T falls from 386.6 K to 313.15 K, and the pressure falls.

 P (kPa)
 250 |   2<----------1
     |   |
     |   |  2-3: constant V
 202 |   3
     +------------------> V
        V1/2         V1

 T (K)
 773 |               1
     |            /
     |         /
 387 |   2  /
     |   |
 313 |   3
     +------------------> V
        V1/2         V1

On T-V, 1-2 is a straight line (T ∝ V at constant P) and 2-3 is vertical.

States

V₁ = mRT₁/P₁ = 1 × 0.297 × 773.15/250 = 0.9185 m³; V₂ = V₃ = 0.4593 m³ P₃ = P₁ T₃/T₂ = 250 × 313.15/386.6 = 202.5 kPa

Heat lost

W₁₂ = P₁(V₂ − V₁) = 250 × (0.4593 − 0.9185) = −114.8 kJ; W₂₃ = 0 Q = ΔU + W = m cv (T₃ − T₁) + W = 0.743 × (313.15 − 773.15) − 114.8 Q = −341.8 − 114.8 = −456.6 kJ (heat lost = 456.6 kJ)

Entropy change of nitrogen

ΔS = m [cp ln(T₃/T₁) − R ln(P₃/P₁)] = 1.040 × ln(0.4050) − 0.297 × ln(0.8100) = 1.040 × (−0.9038) − 0.297 × (−0.2107) = −0.9399 + 0.0626 = −0.8774 kJ/K

Entropy change of surroundings (at 20 °C)

ΔS_surr = −Q/T₀ = 456.6/293.15 = +1.5575 kJ/K

Entropy generation

S_gen = ΔS_N₂ + ΔS_surr = −0.8774 + 1.5575

Answer: S_gen ≈ 0.68 kJ/K (positive; the process is irreversible because of heat transfer across a finite temperature difference).