ME 24-221 THERMODYNAMICS – I Solutions to extra problem set from Chapters 5, 6 and 7. Fall 2000 October 30, 2000 J. Y. Murthy 5.61 Saturated, x = 1%, water at 25°C is contained in a hollow spherical aluminum vesselwith inside diameter of 0.5 m and a 1-cm thick wall. The vessel is heated until thewater inside is saturated vapor. Considering the vessel and water together as a controlmass, calculate the heat transfer for the process.C.V. Vessel and water. This is a control mass of constant volume.m = m ;U - U = Q -W = QState 1: v = 0.001003 + 0.01 x 43.359 = 0.4346 m /kg u = 104.88 + 0.01 x 2304.9 = 127.9 kJ/kgState 2: x = 1 and constant volume so v = v = V/mvg T2 = v = 0.4346 => T = 146.1°C; u = uG2 = 2555.9VINSIDE = π6 (0.5) = 0.06545 m ; mHO = 0.065450.4346 = 0.1506 kgVAl = π6()(0.52) - (0.5) = 0.00817 m mAl = ρAlVAl = 2700 x 0.00817 = 22.065 kgQ = U - U = mHO(u - u)HO + mAlCV Al(T - T) = 0.1506(2555.9 - 127.9) + 22.065 x 0.9(146.1 - 25) = 2770.6 kJ 5.63 A rigid insulated tank is separated into two rooms by a stiff plate. Room A of 0.5 m contains air at 250 kPa, 300 K and room B of 1 m has air at 150 kPa, 1000 K. Theplate is removed and the air comes to a uniform state without any heat transfer. Findthe final pressure and temperature.C.V. Total tank. Control mass of constant volume.Mass and volume: m = m$ + m%; V = V$ + V% = 1.5 m Energy Eq.: m u – m$uA – m%uB = Q – W = 0Ideal gas at 1: m$ = P$V$/RT$ = 250 × 0.5/(0.287 × 300) = 1.452 kgu $= 214.364 kJ/kg from Table A.7Ideal gas at 2: m% = P%V%/RT %= 150 × 1/(0.287 × 1000) = 0.523 kgu %= 759.189 kJ/kg from Table A.7m = m$ + m% = 1.975 kgu = (m$uA + m%uB)/m = (1.452 × 214.364 + 0.523 × 759.189)/1.975 = 358.64 kJ/kg => Table A.7 T = 498.4 KP = m RT /V = 1.975 × 0.287 × 498.4/1.5 = 188.3 kPa5.71 Two containers are filled with air, one a rigid tank A, and the other a piston/cylinder Bthat is connected to A by a line and valve, as shown in Fig. P5.71. The initial conditionsare: mA = 2 kg, TA = 600 K, PA = 500 kPa and VB = 0.5 m3, TB = 27°C, PB = 200kPa. The piston in B is loaded with the outside atmosphere and the piston mass in thestandard gravitational field. The valve is now opened, and the air comes to a uniformcondition in both volumes. Assuming no heat transfer, find the initial mass in B, thevolume of tank A, the final pressure and temperature and the work, W.Cont.: m = m = mA1 + mB1Energy: mu - mA1uA1 - mB1uB1 = -W ; W = PB1(V - V)System: P% = const = PB1 = P ; Substance: PV = mRTmB1 = PB1VB1/RTB1 = 1.161 kg ; V$ = mA1RTA1/PA1 = 0.6888 m3P = PB1 = 200 kPa ; A.7: uA1 = 434.8, uB1 = 214.09 kJ/kgmu + PV = mA1uA1 + mB1uB1 + PB1V = mh = 1355.92 kJ⇒ h = 428.95 kJ/kg ⇒ T = 427.7 K ⇒ V = mtotRT/P = 1.94 m W = 200 × (1.94 - 1.1888) = 150.25 kJ 5.83 Water at 150°C, quality 50% is contained in a cylinder/piston arrangement with initialvolume 0.05 m3. The loading of the piston is such that the inside pressure is linearwith the square root of volume as P = 100 + CV 0.5 kPa. Now heat is transferred to thecylinder to a final pressure of 600 kPa. Find the heat transfer in the process.Continuty: m2 = m1 Energy: m(u2 − u1) = 1Q2 − 1W2State 1: v = 0.1969, u = 1595.6 kJ/kg ⇒ m = V/v = 0.254 kgProcess equation ⇒ P - 100 = CV1/2 so(V/V)1/2 = (P - 100)/(P - 100)V = V x P - 100P - 100 = 0.05 x 500475.8 - 100 = 0.0885W = ⌡⌠PdV = ⌡⌠(100 + CV1/2)dV = 100x(V - V) + 23 C(V1.5 - V1.5) = 100(V - V)(1 - 2/3) + (2/3)(PV - PV)W = 100 (0.0885-0.05)/3 + 2 (600 x 0.0885-475.8 x 0.05)/3 = 20.82 kJState 2: P, v = V/m = 0.3484 ⇒ u = 2631.9 kJ/kg, T ≅ 196°CQ = 0.254 x (2631.9 - 1595.6) + 20.82 = 284 kJPV21P = 100 + C V1/21005.85 A closed cylinder is divided into two rooms by a frictionless piston held in place by apin, as shown in Fig. P5.85. Room A has 10 L air at 100 kPa, 30°C, and room B has300 L saturated water vapor at 30°C. The pin is pulled, releasing the piston, and bothrooms come to equilibrium at 30°C and as the water is compressed it becomes two-phase. Considering a control mass of the air and water, determine the work done bythe system and the heat transfer to the cylinder.P = PG HO at 30°C = PA2 = PB2 = 4.246 kPaAir, I.G.:PA1VA1 = m$R$T = PA2VA2 = PG HO at 30°CVA2→ VA2 = 100 x 0.014.246 m = 0.2355 m VB2 = VA1 + VB1 - VA2 = 0.30 + 0.01 - 0.2355 = 0.0745 m m% = VB1vB1 = 0.332.89 = 9.121x10-3 kg => vB2 = 8.166 m /kg8.166 = 0.001004 + xB2 x (32.89 - 0.001) ⇒ xB2 = 0.2483System A+B: W = 0; ∆U$ = 0 ( IG & ∆T = 0 ) uB2 = 125.78 + 0.2483 x 2290.8 = 694.5, uB1 = 2416.6 kJ/kgQ = 9.121x10-3(694.5 - 2416.6) = -15.7 kJ 6.34 A large SSSF expansion engine has two low velocity flows of water entering.High pressure steam enters at point 1 with 2.0 kg/s at 2 MPa, 500°C and 0.5 kg/scooling water at 120 kPa, 30°C enters at point 2. A single flow exits at point 3with 150 kPa, 80% quality, through a 0.15 m diameter exhaust pipe. There is aheat loss of 300 kW. Find the exhaust velocity and the power output of theengine.C.V. : Engine (SSSF)Constant rates of flow, Q.loss and W.State 1: Table B.1.3: h1 = 3467.6State 2: Table B.1.1: h2 = 125.77h3 = 467.1 + 0.8 × 2226.5 = 2248.3 kJ/kg123EngineQ.lossW.v3 = 0.00105 + 0.8 × 1.15825 = 0.92765 m3/kgContinuity : m.1+ m.2 = m.3 = 2 + 0.5= 2.5 kg/s = (AV/v) = (π/4)D2V/vEnergy Eq. : m.1h1 + m.2h2 = m.3(h3 + 0.5 V2) + Q.loss + W.V = m.3v3 / [(π/4)D2] = 2.5 × 0.92765/(0.7854 × 0.152 ) = 131.2 m/s0.5 V2 = 0.5 ×131.2×131.2/1000 = 8.6 kJ/kg ( remember units factor 1000)W. = 2 ×3467.6 + 0.5 ×125.77 – 2.5 (2248.3 + 8.6) – 300 = 1056 kW6.49 A 25-L tank, shown in Fig. P6.49, that is initially evacuated is connected by avalve to an air supply line flowing air at 20°C, 800 kPa. The valve is opened, andair flows into the tank until the pressure reaches 600 kPa.Determine the finaltemperature and …