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A piston cylinder device, whose piston is resting on a set of stops initially, contains 3 kg of air at 200 kPa and 27oC. The mass of the piston is such that a pressure of 400 kPa is required to move it. Heat is now transferred to the air until its volume doubles. Determine the work done by the air, total heat transferred to the air and entropy generated during the process. Also show the process on a P-v diagram. Solution: Step 1: Problem Statement System: Mass contained within cylinder piston arrangement. It is a closed system. Working fluid: Air Process: expansion; initially at constant volume then at constant pressure Known: mass of air, initial pressure and temperature, final pressure, relation between initial and final volume Find: work transfer, heat transfer, entropy generated Step 2: Schematic Step 3: Assumptions and Approximations i. ii. iii. iv. v. vi. vii.

Leak-proof piston implying the system is closed. Cylinder walls are rigid implying change of boundary is only possible by movement of piston Rigid piston implying no work done in deformation of piston Frictional effects are negligible implying quasi-static process Changes in kinetic, potential, electrical, electromagnetic and other form of energies (except internal energies) are negligible. Air acts as an ideal gas with variable specific heats The system acts only with a thermal source at the temperature equal to final temperature of system

Step 4: Physical Laws Mass balance: not required (closed system) Energy balance: Ein – Eout = ΔEsys For closed system: [ΣQin + ΣWin] - [ΣQout + ΣWout] = ΔE

[Qin + 0] – [0 + Wout] = ΔU + negligible forms Qin – Wout = ΔU Qin – ∫P dV = m (u2 – u1) ………… (eqn 1) Entropy balance: Sin – Sout + Sgen = ΔSsys For closed system: Σ∫(δQin/T)b - Σ∫(δQout/T)b + Sgen = ΔSsys (b in suffix stands for boundary) ∫(δQin/T)source + Sgen = ΔSsys For extended system Qin/Tsource + Sgen = m (s2 – s1) ………… (eqn 2) Step 5: Properties Working fluid: air State 1: P1 = 200 kPa T1 = 27+273 = 300K v1 = RT1 / P1 = (287) (300) / (200 x 103) = 0.4305 m3/kg u1 = 214.07 kJ/kg so1 = 1.70203 kJ/kg K State 2: P2 = 200 kPa v2 = 2v1= 0.861 m3/kg T2 = P2v2 / R = (400 x 103) (0.861) / 287 = 1200 K u2 = 933.3 kJ/kg so2 = 3.17888 kJ/kg K Step 6: Calculations Wout = ∫P dV = ∫isochoricP dV + ∫isobaricP dV = 0 + P2 m (v2 – v1) = (3) (400x103) (0.861 – 0.4305) = 516.6 k J From eqn 1 Qin = m (u2 – u1) + Wout = 3 (933.3 – 214.07) + 515.6 = 2674.29 kJ From eqn 2 Sgen = m (s2 – s1) + Qin / Tsource = m [ (so2 – so1) – R ln(P2/P1)] + Qin / T2 (assumption vi & vii used) =3 [(3.1788 – 1.70203) – 287 ln(400x103/200x103)] – 2674.29/1200 = 1.6049 kJ/kg K...