Solution 1 and 2

Solution 1 and 2 - 1 = 576.32 Btu/lbm 2 = h 2-h o-T o (s...

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Solution to Problem 1: (a) Since the actual work done here is against atmosphere hence it is an isobaric process and Therefore, Work done = p (∆V) = 14.7 * (0.65 – 0.1) = 8.085 (b) We know, Q = W + change n internal energy Hence, ∆U = Q – W Q is given to be .50 BTU Once, we have the steam table in british units we can get ∆U which will further give us U1 and U2. Also, we will then get h1 and s1 which will help us in calculating ψ1 and ψ2. Change in availability during process = ψ1 – ψ2 Now that I have the steam table in british units, the solution is as follows… State 1: From steam table for 500 F v 1 = 3.4412 ft 3 /lbm u 1 = 1171.4 Btu/lbm h 1 = 1273.2 Btu/lbm s 1 = 1.6522 Btu/lbm – R State 2: Hence; Therefore, T2 = 320 F From steam table for 320 F v = 22.980 ft 3 /lbm u = 1161.1 Btu/lbm h = 1201.2 Btu/lbm s = 1.7933 Btu/lbm – R The change of availability solution formula is ψ - ψ
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Ψ 1 = h 1 -h o -T o (s 1 -s o ) Ψ 1 = 1273.2 –129.06 – 540*(1.6522 -.60078)
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Unformatted text preview: 1 = 576.32 Btu/lbm 2 = h 2-h o-T o (s 2-s o ) 2 = 1201.2 129.06 540*(1.7933 .60078) 2 = 428.1792 Btu/lbm Therefore, Change in availability = - = 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 = 1.4 ( for air) Hence, 6.96 = -326.3 J/kg-K 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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Solution 1 and 2 - 1 = 576.32 Btu/lbm 2 = h 2-h o-T o (s...

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