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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 12 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 34 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 25.49238.41 25.49117.77 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 2.62 238.41284.48 117.77238.41 hh hh W Q COP 1 2 4 1 comp evap = = = = The Carnot cycle COP is given by 3.82 1 20273 29 . 46 273 1 = 1 T T 1 = COP L H Carnot =+...
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.
 Spring '06
 SOMERTON

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