Answer :
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/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