Refrigerant134a enters a compressor operating at steady stat

Refrigerant-134a enters a compressor operating at steady state with a rate of 7 kg/s. The inlet state is saturated vapor at 160 kPa and the inlet pipe diameter is 8 cm. At the exit, the pressure is 1.2 MPa, the temperature is 60 degree C and the exit pipe diameter is 10 cm. Heat transfer from the compressor to its surroundings occurs at a rate of 5.5 kJ per kg of refrigerant flowing. Determine the power consumed by the compressor, in kW.

Solution

at the exit we have 1.2 MPa and 60°C, so we can know the properties using the tables of r134a

v=0.018404 m^3/kg

h=289.64 KJ/kg

s=0.9614 KJ/kg K

so, as a steady state, the enthropy doesnt change from enter and exit..

s1=s=0.9614 KJ/kg K, andwe know the preassure at the enter, wich is 160 kPa

so the other properties are

h1=228.55 KJ/kg

x1=0.59

v1=0.124032 m^3/kg

5.5 KJ/kg is energy lost so

starting with 7 kg/s the flow of volume is

v1*m1=7*0.124032= 0.868224 m^3/s

So finally the balance of energy is

E+h1=h2+lost

E=mas flow* ( h2-h1+lost) = 7*(289.64-228.55+5.5) = 466.13 kW!!!