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
Answer:
A) COP = 2.59
B)The rate of heat absorption from the outside air = 1.8935 Kw
Explanation:
P1 = 800 KPa
T1 = 35°C
At this condition, from the R-134 table i attached, the enthalpy is given as;
h1 = 267.34 Kj/Kg
Now,
P2 = 800 KPa
Thus, from the same table, h2 = 95.48 Kj/Kg
A) Energy balance is given as;
Q_h = m'(h1 - h2)
Where m' is mass flow rate = 0.018 kg/s
Q_h = 0.018(267.4 - 95.48) = 3.0935 Kw
Ce-efficient of performance of the refrigerator is given as;
COP = Q_h/W_in
W_in is the power consumed by the compressor = 1.2 Kw
Thus, COP = 3.0935/1.2 = 2.59
B) The rate of heat absorption from the outside air is given as;
W_l = Q_h - W_in = 3.0935 - 1.2 = 1.8935 Kw
