Refrigerant 134a enters an air conditioner compressor at 4 bar, 20°C, and is compressed at steady state to 12 bar, 80°C. The volumetric flow rate of the refrigerant entering is 8 m3/min. The work input to the compressor is 120 kJ per kg of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the magnitude of the heat transfer rate from the compressor, in kW.

The magnitude of the heat transfer rate from the compressor is 87.05 kW

Explanation:

Initial pressure of refrigerant = 4 bar

Final pressure of refrigerant = 12 bar

From steam table,

Internal energy at 4 bar (U1) = 2554 kJ/kg

Internal energy at 12 bar (U2) = 2588 kJ/kg

Change in internal energy (∆U) = U2 - U1 = 2588 - 2554 = 34 kJ

Work input (W) = 120 kJ/kg

Quantity of heat transfer (Q) = ∆U + W = 34 + 120 = 154 kJ/kg

Volumetric flow rate of refrigerant = 8 m^3/min = 8/60 = 0.133 m^3/s

Density of refrigerant = 4.25 kg/m^3

Mass flow rate = density × volumetric flow rate = 4.25 kg/m^3 × 0.133 m^3/s = 0.56525 kg/s

Q = 154 kJ/kg × 0.56525 kg/s = 87.05 kJ/s = 87.05 kW

answer: 31, 5556. 952 seconds