Refrigerant-134a enters the condenser of a residential heat pump at 800 kPA and 35oC at a rate of 0.018 kg/s and leaves at 800 kPa as a saturated liquid. If the compressor consumes 1.2 kW of power, determine

(a) the COP of the heat pump

(b) the rate of heat absorbtion from the outside air.

(A) COP = 2.64

(B) rate of heat absorption = 1.9637 kW

Explanation:

mass flow rate (m) = 0.018 kg/s

work input (Win) = 1.2kW

inlet pressure (P1) = 800kPa

inlet temperature (T1) = 35 degree Celsius

h1 = 271.24 KJ/Kg

outlet pressure (P2) = 800 kPa

outlet temperature (T2) = ?

entalphy (h2) = 95.48 KJ/Kg

The entalphies are gotten from tables for refrigerant 134a at the temperatures and pressures above

(A) COP = Qh : Win

where Qh = m (h1 - h2) from the energy balance equation

Qh = 0.018 (271.24 - 95.48) = 3.1637 kW

COP = 3.1637 : 1.2 = 2.64

(B) rate of heat absorption = Qh - Win

= 3.1637 - 1.2 = 1.9637 kW