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EML3100_Soln7_revised

# EML3100_Soln7_revised - i =-25°C and x i = 1 h i = h g =...

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EML 3100: SOLUTION 7

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9.122 A compressor in a commercial refrigerator receives R-410a at -25°C, x = 1. The exit is at 2000 kPa, 80°C. Neglect kinetic energies and find the isentropic compressor efficiency. Solution: Assume no heat transfer, steady state, single inlet and exit flow. State 1: The point ‘i’ or inlet which is defined by by T i = -25°C and x i = 1 State 2: The point ‘e’ or exit which is defined by T e = 80°C and P e = 2000 kPa To find the isentropic compressor efficiency, we also need a point ‘e s ’, which will define the exit state for an isentropic compressor. This point is given by s= s i and P=P e .

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Energy equation 6.13: h i + 0 = w C + h e From Table B.4.1 At T
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Unformatted text preview: i = -25°C and x i = 1, h i = h g = 269.77 kJ/kg and s i =s g = 1.0893 kJ/kg From Table B.4.2, At T e = 80°C and P e = 2000 kPa, h e = 343.22 kJ/kg , s e = 1.1537 Therefore w c_actual = h i – h e = 269.77 - 343.22 = -73.45 kJ/kg To find isentropic work, we need to find the point ‘e s ’ where s = 1.0893 From Table B.4.2, s(T=80°C) = 1.1537 and s(T=60°C) = 1.878. Therefore interpolating, h es = 320.62 + (343.22 – 320.62)*(1.0893-1.0878)/(1.1537-1.0878) = 321.134 kJ/kg w c_s = h i – h es = 269.77 – 321.134 = -51.36 kJ/kg Therefore isentropic efficiency is given by η = w c_s / w C_actual = (-51.34) /(-73.45) = 0.6989 or 69.89%...
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