This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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%...
View
Full Document
 Spring '08
 Sherif
 Thermodynamics, Energy, Kinetic Energy, Heat Transfer, TI, isentropic compressor efficiency

Click to edit the document details