7–53 Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 35°C 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 and (b) the rate of heat absorption from the outside air.

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

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

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