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Unformatted text preview: Ψ 1 = 576.32 Btu/lbm Ψ 2 = h 2h oT o (s 2s o ) Ψ 2 = 1201.2 –129.06 – 540*(1.7933 – .60078) Ψ 2 = 428.1792 Btu/lbm Therefore, Change in availability = ψ  ψ = ₁ ₂ 576.32  428.1792 = 148.1408 Btu/lbm Solution to Problem 2: The Carnot Cycle: Data Given: Pa = 3 MPa ; Pc = 100 KPa ; Tc = 20 C = 20 + 273 = 293 K ; Compression ratio = 15 = Vc/Va (a) Thermal efficiency = Now, (b) Work output = thermal efficiency * Qh Since process A to B is an isothermal process, hence PV = constant Therefore, Now, process B to C is an adiabatic process, hence, And γ = 1.4 ( for air) Hence, 6.96 = 326.3 J/kgK Work done = 326.3 * 0.5 *0.5 = 81.75 KJ/kg (c) MEP = work done/ ( Vc – Va) Therefore, Vc = 15Va To calculate Va, we use the gas equation, Therefore, = 0.056 * 10^(3) Vc Va = 14Va MEP = Work done / 14Va = 156.5 kPa...
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 Spring '11
 urvashi
 Adiabatic process, Heat engine, Isothermal process, Isobaric process, Steam Table

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