Question

a refrigerator uses refrigerant-134a as the working fluid and operates on the vapor-compression refrigeration cycle. the evaporator and condenser pressures are 200 kpa and 1400 kpa, respectively. the isentropic efficiency of the compressor is 88 percent. If the refrigerant enters the compressor at a rate of 0.025 kg/s, determine the rate of cooling provided by the evaporator, the power input required to the compressor, and the coefficient of performance of the refrigerator.

Answer 1

Answer:

I. 3.316 kW

II. 1.218 kW

III. 2.72

Explanation:

At state 1, the enthalpy and entropy are determined using the given data from A-13.

At P1 = 200kpa and T1 = 0,

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2, the isentropic enthalpy is determined at P2 = 1400kpa and s1 = s2 by interpolation. Thus

h2(s) = 295.95 kJ/kg

The actual enthalpy is then gotten by

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

h3 = h4 = 120.43 kJ/kg

Heating load is determined from energy balance, thus,

Q'l = m'(h1 - h4)

Q'l = 0.025(253.07 - 120.43)

Q'l = 0.025 * 132.64

Q'l = 3.316 kW

Power is determined by using

W' = m'(h2 - h1)

W'= 0.025(301.8 - 253.07)

W'= 0.025 * 48.73

W'= 1.218 kW

The Coefficient Of Performance is Q'l / W'

COP = 3.316/1.218

COP = 2.72

Answer 2

Answer/Explanation:

The enthalpy and entropy of state 1 is obtained from the property table(A-13).

Entropy is denoted as s, and enthalpy as h. P1 and P2 are pressures of state 1 and state 2 respectively.

For state 1

P1 = 200kpa

The efficiency of the compressor, n = 0.88%

Refrigerant mass rate = 0.025kg/s

h1 = 253.07 kJ/kg

s1 = 0.9699 kJ/kgK

At state 2

P2 = 1400kpa

s1 = s2 by interpolation.

Isentropic enthalpy h2(s) = 295.95 kJ/kg.

The actual enthalpy and the isentropic enthalpy is related by:

h2 = h1 + [h2(s) - h1]/n

h2 = 253.07 + [295.95 - 253.07]/0.88

h2 = 253.07 + 48.73

h2 = 301.8 kJ/kg

Power = mass rate(h2 - h1)

Power = 0.025(301.8 - 253.07)

Power = 0.025 x 48.73

Power = 1.218 kW

The Coefficient Of Performance is given by :

Heating load/Power

To calculate the heating load.

Note:

h3 = h4 = 120.43 kJ/kg

Heating load = mass rate(h1 - h4)

Heating load = 0.025(253.07 - 120.43)

Heating load = 0.025 x 132.64

Heating load = 3.316 kW

Coefficient of Performance = 3.316/1.218

Coefficient of performance = 2.7225