Answer:
a. The coefficient of power = 2.6364
b. The rate of heat absorption from the outside air is 1.96368kW
Explanation:
Given
First we need to get the enthalpy of R-34a.
When T = 35°C and P = 800kPa;
h1 = 271.24kj/kg
When x2 = 0 and P = 800kPa;
h1 = 95.48kj/kg
To calculate the COP, first we need to calculate the energy balance.
This is given as
Q = m(h1 - h2)
Where m = 0.018kg
Q = 0.018(271.24 - 95.48)
Q = 3.16368Kw
COP is then calculated as Q/W
Where W = Power consumption of the compressor = 1.2kW
COP = 3.16368Kw/1.2Kw
COP = 2.6364
Hence, the coefficient of power = 2.6364
b. The rate of heat absorption from the outside air is calculated as ∆Heat Rate
∆Heat Rate = Q - W
Where Q = Energy Balance = 3.16368Kw
W = Power consumption of the compressor = 1.2kW
∆Heat Rate = 3.16368Kw - 1.2kW
∆Heat Rate = 1.96368kW
Hence, The rate of heat absorption from the outside air is 1.96368kW
