--- title: "Nitrogen cooled: entropy generation" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 5 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:nitrogen-cooling-entropy-generation" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/nitrogen-cooling-entropy-generation site: BCT Engineering --- # Nitrogen cooled: entropy generation Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics - Asked in: 2070 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/nitrogen-cooling-entropy-generation - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5 ## Question A piston cylinder device shown in figure below contains 1 kg of Nitrogen initially at a pressure of 250 kPa and a temperature of 500°C. Heat is lost from the system till its temperature reaches 40°C. Sketch the process on P-V and T-V diagrams and determine the energy generation. Assume that surrounding is at 20°C. Take R = 297 J/kgK, Cv = 743 J/kgK. [Figure: cylinder containing N2 with a piston resting on stops; dimensions 0.5 m and 0.5 m marked on the cylinder height.] ## Short answer The entropy generation is 0.68 kJ/K. The process is a constant-pressure cooling until the piston hits stops, then constant-volume cooling, with heat loss causing irreversibility. ## Answer **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. ```text 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the process path for the nitrogen? A. Constant volume then constant pressure B. Isothermal throughout C. Constant pressure down to the stops (V halves), then constant volume D. Isentropic compression Answer: C. The piston falls at 250 kPa until it rests on the stops at mid-height, then cooling is at constant volume. (Key point: Process path) 2. T2 at the stops is: A. T1/2 = 386.6 K B. 313.15 K C. 2T1 D. T1 − 40 K Answer: A. At constant P, T ∝ V, and V2 = V1/2. (Key point: Temperature at stops) 3. Entropy generation is calculated as: A. ΔS_N2 only B. ΔS_N2 + ΔS_surr, with ΔS_surr = −Q/T0 C. ΔS_surr − ΔS_N2 only D. Q/T1 Answer: B. The nitrogen plus surroundings form an isolated system. (Key point: Entropy generation formula) 4. S_gen ≈ 0.68 kJ/K and positive because: A. the nitrogen entropy rises B. work is done by the gas C. pressure stays constant D. heat transfers across a finite temperature difference Answer: D. The gas cools through a finite ΔT to surroundings at 20 °C, which is irreversible. 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