--- title: "Three-process gas cycle: net work, heat, type" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:three-process-cycle-pv-sketch" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/three-process-cycle-pv-sketch site: BCT Engineering --- # Three-process gas cycle: net work, heat, type Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2073 Magh, 8 marks; 2069 Poush, 8 marks (asked 2 times) - 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/three-process-cycle-pv-sketch - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question A Gas undergoes a thermodynamic cycle consisting of three process, Process 1-2 constant Pressure, P = 1.4 bars, v1 = 0.028 m³, w12 = 10.5 kJ Process 2-3, compression with Pv = constant, U3 = U2 Process 3-1, constant volume, U1-U3 = -26.4 kJ There are no significant change in kinetic and potential energy a) Sketch the system on a P-v diagram. b) Calculate Net work for a cycle, in kJ. c) Calculate the heat transfer of Process 1-2 in kJ d) Is this a Power cycle or a Refrigerator cycle? ## Short answer Net work = 10.5 (1-2) - 18.78 (2-3) = -8.28 kJ. Process 1-2 heat transfer is 36.9 kJ (added). Because net work is negative and cycle runs anticlockwise on P-v, it is a refrigerator cycle. ## Answer **Given:** 1-2: constant P = 1.4 bar = 140 kPa, V₁ = 0.028 m³, W₁₂ = 10.5 kJ; 2-3: PV = constant (isothermal) with U₃ = U₂; 3-1: constant volume, U₁ − U₃ = −26.4 kJ. **Step 1: Volumes** W₁₂ = P (V₂ − V₁) ⇒ V₂ = V₁ + W₁₂/P = 0.028 + 10.5/140 = 0.103 m³ V₃ = V₁ = 0.028 m³ (3-1 is at constant volume) P₃ = P₂ V₂/V₃ = 140 × 0.103/0.028 = 515 kPa #### a) P-V diagram ```text P (kPa) 515 |3 ||\ || \ PV = C (2-3) || \ || `-._ 140 |1--------2 +---------------- V (m³) 0.028 0.103 ``` 1 → 2 is a horizontal line to the right, 2 → 3 is a hyperbola going up and to the left, and 3 → 1 is a vertical line down. The cycle is traversed **anticlockwise**. #### b) Net work W₂₃ = P₂V₂ ln(V₃/V₂) = 140 × 0.103 × ln(0.028/0.103) = 14.42 × (−1.3025) = −18.78 kJ W₃₁ = 0 (constant volume) W_net = 10.5 − 18.78 + 0 = **−8.28 kJ** #### c) Heat transfer in process 1-2 U₁ − U₃ = −26.4 ⇒ U₃ − U₁ = 26.4 kJ; with U₂ = U₃, U₂ − U₁ = 26.4 kJ Q₁₂ = ΔU₁₂ + W₁₂ = 26.4 + 10.5 = **36.9 kJ** (added) (Check: Q₂₃ = W₂₃ = −18.78 kJ, Q₃₁ = −26.4 kJ; ΣQ = 36.9 − 18.78 − 26.4 = −8.28 kJ = W_net.) #### d) Type of cycle The net work is negative (work is supplied to the system) and the cycle runs anticlockwise on the P-V diagram, so it is a **refrigerator (or heat pump) cycle**, not a power cycle. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. V2 is found from: A. W12 = P(V2 - V1), giving V2 = 0.103 m³ B. V2 = V1 C. V2 = W12/P1 = 0.075 m³ D. From PV = C at state 3 Answer: A. V2 = 0.028 + 10.5/140 = 0.103 m³. (Key point: Volume V2) 2. Net cycle work is: A. +8.28 kJ B. -8.28 kJ (10.5 - 18.78 + 0) C. +29.3 kJ D. 0, because it is a cycle Answer: B. W23 = P2V2 ln(V3/V2) = -18.78 kJ; W31 = 0. (Key point: Net work) 3. Q12 is: A. 26.4 - 10.5 = 15.9 kJ B. 10.5 kJ C. ΔU12 + W12 = 26.4 + 10.5 = 36.9 kJ D. -36.9 kJ Answer: C. U2 = U3 and U3 - U1 = 26.4, so ΔU12 = 26.4 kJ. (Key point: Heat in 1-2) 4. The cycle is a refrigerator (heat pump) because: A. Net work is positive B. It runs clockwise C. Heat input equals heat rejected D. Net work is negative and it runs anticlockwise on the P-V diagram Answer: D. Work is supplied to the system. (Key point: Cycle type) ## Related in Fundamental of Thermodynamics & Heat Transfer - [Three-process gas cycle: net work and heat](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Athree-process-cycle-pv-const-800kpa) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [Gas work: constant P, constant T, PV^1.35](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Agas-work-three-processes-150kpa) — Fundamental of Thermodynamics & Heat Transfer, chapter 2 - [Thermodynamic process; constant pressure heating diagrams](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Aprocess-sketch-constant-pressure-heating) — Fundamental of Thermodynamics & Heat Transfer, chapter 1