Question

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.

Answer 1

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 2

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