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 5.5 m3/min. The work input to the compressor is 82.5 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 45.40 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) = 82.5 kJ/kg

Quantity of heat transfer (Q)

= ∆U + W

= 34 + 82.5

= 116.5 kJ/kg

Volumetric flow rate of refrigerant = 5.5 m^3/min

= 5.5/60

= 0.0917 m^3/s

Density of refrigerant = 4.25 kg/m^3

Mass flow rate = density × volumetric flow rate

= 4.25 kg/m^3 × 0.0917 m^3/s

= 0.38973 kg/s

Q = 116.5 kJ/kg × 0.38973 kg/s

= 45.40 kJ/s

= 45.40 kW

The magnitude of the heat transfer rate from the compressor is 45.40 kW

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ill do it for 100, also this isnt 80 its 40