MECH 310 Problems Chapter 5 Solutions, Fall 2011

1 MECH 310 Thermodynamics I October 30, 2011 American University of Beirut, Fall 2011 Handout # 3 Chapter 05 Solved Problems Problem 5.30 a. R-410a P = 500 kPa, h = 300 kJ/kg Table B.4.1: h > hg = > superheated vapor, look in section 500 kPa and interpolate T = 0 + 20 × (300 − 287.84)/(306.18 − 287.84) = 20 × 0.66303 = 13.26°C, v = 0.05651 + 0.66303 × (0.06231-0.05651) = 0.06036 m3/kg, u = 259.59 + 0.66303 × (275.02 – 259.59) = 269.82 kJ/kg b. R-410a T = 10°C, u = 200 kJ/kg Table B.4.1: u < ug = 255.9 kJ/kg => L+V mixture, P = 1085.7 kPa x = (u – uf)/ufg = (200 - 72.24)/183.66 = 0.6956, v = 0.000886 + 0.6956 × 0.02295 = 0.01685 m3/kg, h = 73.21 + 0.6956 × 208.57 = 218.3 kJ/kg c. R-134a T = 40°C, h = 400 kJ/kg Table B.5.1: h < hg => two-phase L + V, look in B.5.1 at 40°C: x = (h – hf)/hfg = (400 − 256.5)/163.3 = 0.87875, P = Psat = 1017 kPa, v = 0.000 873 + 0.87875 × 0.01915 = 0.0177 m3/kg u = 255.7 + 0.87875 × 143.8 = 382.1 kJ/kg Problem 5.36 C.V. Water. This is a control mass Energy Eq.: m (u2 − u1 ) = Q12 − W12 Process : V = constant W12 = ∫ P dV = 0 State 1: T, x1 from Table B.1.1 v1 = vf + x1 vfg = 0.00106 + 0.25 × 0.8908 = 0.22376 m3/kg u1 = uf + x1 ufg = 503.48 + 0.25 × 2025.76 = 1009.92 kJ/kg T2, v2 = v1< vg2 = 0.50885 m3/kg so two-phase x2 = (v2 - vf2)/vfg2 = (0.22376 - 0.00108)/0.50777 = 0.43855 u2 = uf2 + x2 ufg2 = 588.72 + x2 ×1961.3 = 1448.84 kJ/kg From the energy equation Q12 = m(u2 − u1) = 2 ( 1448.84 – 1009.92 ) = 877.8 kJ 2 Problem 5.37 Problem 5.39 3 Problem 5.66 Problem 5.67 4