The vapor compression heat pump operates at steady state using Refrigerant 22, w

Question

The vapor compression heat pump operates at steady state using Refrigerant 22, whose flow rate is 6kg/min. The heat pump maintains the interior of a house at 25 C, while drawing heat from the surrounding at a temperature of 10 C. The refrigerant enters the compressor as saturated vapor at -10 C, exits the compressor at 1 MPa and 50 C, and flows through the condenser at constant pressure existing as saturated liquid. The compressor operates adiabatically. Calculate (a) the power required by the compressor in kW (b) the isentropic compressor efficiency (c) the rate of heat transferred to the interior of the house in kW

Explanation / Answer

From R22 properties,

At T1 = -10 C and quality x1 = 1 (sat. vapor), we get h1 = 246 kJ/kg, s1 = 0.943 kJ/kg-K

At P2 = 1 MPa and T2 = 50 C, we get h2 = 280 kJ/kg

At P3 = 1 MPa and x3 = 0 (sat.liquid), we get h3 = 73.4 kJ/kg

During throtling, h4 = h3 = 73.4 deg C

a)

Compressor ppower = m(h2 - h1)

= 6*(280 - 246)

= 204 kJ/min = 3.4 kW

b)

For isentropic compression, s2_is = s1 = 0.913 kJ/kg-K

At P2 = 1 MPa and s2_is = 0.913 kJ/kg-K, we get h2_is = 263 kJ/kg

Efficiency = (h2_is - h1) / (h2 - h1)

= (263 - 246) / (280 - 246)

= 0.5 or 50 %

c)

Rate of heat transfer to house = m*(h2 - h3)

= 6*(280 - 73.4)

= 1239.6 kJ/min = 20.66 kW

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