This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1 = 576.32 Btu/lbm 2 = h 2h oT o (s 2s 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/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...
View Full
Document
 Spring '11
 urvashi

Click to edit the document details