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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)*(1.08931.0878)/(1.15371.0878) = 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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This note was uploaded on 09/06/2011 for the course EML 3100 taught by Professor Sherif during the Spring '08 term at University of Florida.
 Spring '08
 Sherif
 Heat Transfer

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