Problem 5. Consider a residential heat pump system as shown in the figure below.

Question

Problem 5. Consider a residential heat pump system as shown in the figure below. The working fluid is refrigerant 134a. The states are given by (1) T10°C, P 1.8 bar, (2) B, 8 bar, h2 270 kJ/kg, (3) T3 -30 °C,A 8 bar, (4) T, -12 oC. The mass flow rate is 4.6 kg/min. The power into the compressor is 2.5 kW. Presuming the changes of kinetic energy and potential energy are negligible, now determine the rate of heat transfer from compressor to surrounding, the rate of heat transfer from condenser to surrounding, and the coefficient of performance for the heat pump. Inside air Outside air Evaporator Condenser 3 Expansion valve Compressor 2.

Explanation / Answer

Solution :-

(A) At T1 = -10 deg C and P1 = 1.8 bar we get h1 = 245 kJ/kg

h2 - h1 = 270 - 245 = 25 kJ/kg

m*(h2 - h1) = 4.6*25 / 60 = 1.9 kW

Heat transfer from compressor = 2.5 - 1.9 = 0.6 kW

(B) At T3, P3 we get h3 = 93.6 kJ/kg

Heat rejection from condensor = m*(h2 - h3) = 4.6*(270 - 93.6)/60 = 13.5 kW

(C) COP = (Work + Heat rejection) / work

= (2.5 + 13.5) / 2.5

= 6.4

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