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# HW22s - find P 1 = 132.82 kPa h 1 = 238.41 kJ/kg s 1 =...

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Spring 2006 1 ME 201 Thermodynamics Homework 22 Solution 1. An ideal vapor –compression refrigeration cycle with refrigerant 134a as the working fluid operates with an evaporator temperature of –20 ° C and a condenser pressure of 1.2 MPa. For a refrigerant flow rate of 3 kg/min determine: a. COP for the cycle b. Cooling load in kW for the cycle c. Carnot cycle COP for the given conditions Solution: We begin by nodding our layout. Evaporator Compressor Condenser Valve 1 2 3 4 setting up our table, and entering our operating information. Node T ( ° C) P (MPa) Fluid Phase h (kJ/kg) s kJ/(kg K) & m kg/s 1 -20 0.13282 sat.vap. 238.41 0.94564 0.05 2 55.5 1.2 sup.vap. 284.48 0.94564 0.05 3 46.29 1.2 sat.liq. 117.77 0.42441 0.05 4 -20 0.13282 2 phase x = 0.433 117.77 0.05 italicized values from tables, bold values calculated

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ME 201 Thermodynamics Spring 1999 2 We begin by noting that the state is fixed at nodes 1 and 3, so going to the tables we
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Unformatted text preview: find P 1 = 132.82 kPa h 1 = 238.41 kJ/kg s 1 = 0.94564 kJ/(kg ⋅ K) T 3 = 46.29 ° C h 3 = 117.77 kJ/kg s 3 = 0.42441 kJ/(kg ⋅ K) Now traversing the cycle 1-2 Compressor (isentropic) s 2 = s 1 = 0.94564 kJ/(kg ⋅ K) Then going to the superheat tables at 1.2 MPa we find T 2 =55.5 ° C h 3 = 284.48 kJ/kg 3-4 Valve (isenthalpic) h 4 = h 3 = 117.77 kJ/kg Then going to the saturation tables at -20 ° C we find we have a two phase mixture with quality 0.433 25.49-238.41 25.49-117.77 h h h h x f g f 4 4 = =--= Now calculating our cycle information ( 29 kW 032 . 6 117.77)-.41 (0.05)(238 h h m Q 4 1 evap = =-= & & which is the cooling load. The COP is given by 2.62 238.41-284.48 117.77-238.41 h-h h-h W Q COP 1 2 4 1 comp evap = = = = The Carnot cycle COP is given by 3.82 1 20-273 29 . 46 273 1 = 1 T T 1 = COP L H Carnot =-+-...
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