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SecondLawSolns

# SecondLawSolns - ME 201 Thermodynamics ME 201...

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ME 201 Thermodynamics ME 201 Thermodynamics Solutions to Second Law Practice Problems 1. Ideally, which fluid can do more work: air at 600 psia and 600°F or steam at 600 psia and 600°F Solution: The maximum work a substance can do is given by its availablity. We will assume that we have a closed system so that ) s - (s T - u - u = o o o ψ We take the dead state to be at STP or 25°C and 100 kPa or 76.4 ° F and 14.7 psia. Then using the appropriate table we have Btu/lb 139.48 = 14.7 600 0.06855ln - 0.5995 - 0.7649 (537) - 91.53 - 183.30 = m air ψ and ( 29 Btu/lb .84 361 = 0.08215 - 1.5320 (537) - 44.09 - 1184.5 = m steam ψ So the steam can do more work 2. A heat pump provides 30,000 Btu/hr to maintain a dwelling at 68°F on a day when the outside temperature is 35°F. The power input to the pump is 1 hp. If electricity costs 8 cents per kilowatt-hour, compare the actual operating cost per day with the minimum theoretical operating cost per day. 1

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ME 201 Thermodynamics Solution: We sketch our device interactions The cost is given by (0.08)W = Cost net For the actual cost we have \$1.43 = hr/day) kW/hp)(24 0.7457 (0.08)(1)( = (Cost) act To calculate the minimum cost we will allow the heat pump to operate as a Carnot cycle, so that 16 = 528 495 1 1 = T T 1 1 = COP H L Carnot - - Then the minimum possible power input is ( 29 kW 0.5495 = Btu/hr 1875 = 16 30,000 = COP Q = W Carnot H min net 2 High Temperature Heat Reservoir at T H Low Temperature Heat Reservoir at T L Heat Pump Q L Q H W net Dwelling Outside
ME 201 Thermodynamics and the minimum cost is \$1.06 = hr/day) 495)(24 (0.08)(0.5 = (Cost) min 3. A cylinder/piston system contains water at 200 kPa, 200°C with a volume of 20 liters. The piston is moved slowly, compressing the water to a pressure of 800 kPa. The process is polytropic with a polytropic exponent of 1. Assuming that the room temperature is 20°C, show that this process does not violate the second law. Solution: To determine if this violate the 2nd law we will want to calculate the entropy change of the universe and compare it to zero. We have ( 29 T Q - + ) s - m(s = S surr sys 1 2 universe We now work this as a first law problem Working Fluid: Water(compressible) System: Closed System Process: Polytropic with n=1.0 State 1 State 2 T 1 = 200°C T 2 = 214.7°C P 1 = 200 kPa P 2 = 800 kPa u 1 = 2654.4 kJ/kg u 2 = 2655.5kJ/kg V 1 = 0.020 m 3 V 2 = 0.005 m 3 v 1 = 1.0803 m 3 /kg v 2 = 0.2703 m 3 /kg s 1 = 7.5066 kJ/(kg K) s 2 = 6.8811 kJ/(kg K) phase: sup.vap. phase: sup.vap. italicized values from tables, bold values are calculated Initial State: Fixed Final State: Unknown W sh = 0 Q = ????

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