Question

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.

Answer 1

Answer:

(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