--- title: "Three-process gas cycle: net work and heat" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:three-process-cycle-pv-const-800kpa" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/three-process-cycle-pv-const-800kpa site: BCT Engineering --- # Three-process gas cycle: net work and heat Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2068 Magh, 8 marks - Video: [Finding Net heat and Net work when some TD processes data is given](https://www.youtube.com/watch?v=MtyYVTyZ6m8) — Mr.YantriK - 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-const-800kpa - 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 the following three processes: Process 1-2: expansion with PV = constant, P1 = 800 kPa, U2 = U1 Process 2-3: constant volume V2 = V3 = 2 m³, U3 - U2 = 300 kJ Process 3-1: constant pressure, W31 = -1200 kJ (a) Sketch the process on P-V and T-V diagrams. (b) Calculate the net work for the cycle. (c) Calculate the net heat for the cycle. (d) Calculate the heat transfer for process 1-2. (e) Calculate the heat transfer for process 3-1. (f) Is this power cycle or a refrigeration cycle? ## Short answer Process 1-2 isothermal (U2=U1). V1=0.5 m³, P2=200 kPa. W12 = P1V1 ln(V2/V1)=554.5 kJ. W31=−1200 kJ → net work=−645.5 kJ, net heat=−645.5 kJ. Q12=554.5 kJ, Q31=ΔU31+W31=−300−1200=−1500 kJ. Negative net work, anticlockwise P-V → refrigeration cycle. ## Answer Take the gas as ideal. Then U₂ = U₁ means T₂ = T₁, so process 1-2 (PV = constant) is isothermal. Process 3-1 is at constant pressure, P₃ = P₁ = 800 kPa. #### (a) Sketch First find the volumes. Process 3-1: W₃₁ = P(V₁ − V₃) = −1200 kJ, so V₁ − V₃ = −1200/800 = −1.5 m³. With V₃ = V₂ = 2 m³: V₁ = 2 − 1.5 = 0.5 m³ P₂ = P₁V₁/V₂ = 800 × 0.5/2 = 200 kPa States: 1 (0.5 m³, 800 kPa), 2 (2 m³, 200 kPa), 3 (2 m³, 800 kPa). Since T ∝ PV, T₃ = 4T₁. ```text P-V diagram P (kPa) 800 | 1<-----------3 | \ ^ | \ PV = C | | `-. | 200 | `------2 +------------------> V (m³) 0.5 2 T-V diagram T 4T1 | 3 | / ^ | / | | / | T1 | 1---------->2 +------------------> V (m³) 0.5 2 1->2 isothermal expansion; 2->3 heating at constant V; 3->1 compression at constant P (cooling; T ∝ V, so a straight line on T-V) ``` The path 1→2→3→1 runs anticlockwise on the P-V diagram. #### (b) Net work W₁₂ = P₁V₁ ln(V₂/V₁) = 800 × 0.5 × ln 4 = 400 × 1.3863 = 554.5 kJ W₂₃ = 0 (constant volume) W₃₁ = −1200 kJ **Net work = 554.5 + 0 − 1200 = −645.5 kJ** (work is done on the gas). #### (c) Net heat For a cycle, ΣQ = ΣW, so **net heat = −645.5 kJ** (net heat is rejected). #### (d) Heat for process 1-2 ΔU₁₂ = 0, so Q₁₂ = W₁₂ = **554.5 kJ** (added). #### (e) Heat for process 3-1 Q₂₃ = ΔU₂₃ + W₂₃ = 300 + 0 = 300 kJ. For the cycle ΔU = 0: ΔU₃₁ = −(ΔU₁₂ + ΔU₂₃) = −300 kJ. Q₃₁ = ΔU₃₁ + W₃₁ = −300 − 1200 = **−1500 kJ** (rejected). Check: 554.5 + 300 − 1500 = −645.5 kJ ✓. #### (f) Type of cycle The net work is negative (work input) and the net heat is negative, with the cycle running anticlockwise on the P-V diagram. So it is a **refrigeration (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. Why is process 1-2 (PV = constant, U2 = U1) taken as isothermal? A. For an ideal gas U depends only on T, so T2 = T1, consistent with PV = constant B. Because the volume is constant C. Because no heat is transferred D. Because pressure is constant Answer: A. For an ideal gas U2 = U1 means T2 = T1, which agrees with PV = constant. (Key point: Nature of process 1-2) 2. How is V1 found from process 3-1? A. From W31 = P(V1 − V3): V1 = 2 − 1200/800 = 0.5 m³ B. From V1 = V2 C. From PV = constant with P3 D. From V1 = 1200/800 = 1.5 m³ Answer: A. Constant pressure 800 kPa gives V1 − V3 = −1.5 m³; with V3 = 2 m³, V1 = 0.5 m³. (Key point: Finding V1) 3. Work for the isothermal process 1-2 is: A. P1(V2 − V1) = 1200 kJ B. 0, as ΔU = 0 C. P1V1 ln(V2/V1) = 554.5 kJ D. P1V1 ln(V1/V2) = −554.5 kJ Answer: C. For PV = constant, W = P1V1 ln(V2/V1) = 800 × 0.5 × ln4 = 554.5 kJ. (Key point: Work of process 1-2) 4. Q31 is obtained using ΔU31 = −300 kJ and W31 = −1200 kJ. Its value is: A. −900 kJ B. +900 kJ C. −1500 kJ D. −1200 kJ Answer: C. Q31 = ΔU31 + W31 = −300 − 1200 = −1500 kJ (heat rejected). 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