Modern Engineering Thermodynamics

14.18 Second Law Analysis of Refrigeration Cycles 581
Note that the designer computes pressure losses of 75.0 kPa in the condenser and 27.0 kPa in the evaporator. The refrigerant
mass flow rate is 0.500 kg/s, the local environmental temperature is T 0 = 25.0°C, and the isentropic efficiency of the
compressor is 70.0%. Determine
a. The irreversibility rate of each component in the system and the total irreversibility rate of the system.
b. The system COP and ε (the first and second law efficiencies).
c. Select the component with the highest irreversibility rate and suggest ways to improve it.
Solution
Using Figure 14.36 as the illustration for this example, the properties at the four stations can be found in Tables C.7e, C.7f,
and C.8d as
Station 1 Station 2s Station 3 Station 4h
Compressor inlet Compressor outlet Condenser outlet Expansion valve outlet
x 1 = 1:00 p 2s = 800: kPa x 3 = 0:0 h 4h = h 3
T 1 = −20:0°C s 2s = s 1 p 3 = 725 kPa p 4h = 160 kPa
= 0:9332 kJ/kg.K h 3 = 87:46 kJ/kg h 4h = 87:46 kJ/kg
h 1 = 235:31 kJ/kg h 2s = 271:10 kJ/kg s 3 = 0:3257 kJ/kg.K x 4h = 0:280
s 1 = 0:9332 kJ/kg.K T 2s = 39:8°C T 3 = 27:9°C s 4h = 0:3449 kJ/kg.K
p 1 = 132:99 kPa
T 4h = −15:6°C
Conditions at stations 2s, 3, and 4h are determined by interpolation in Table C.7f or with a computer program containing the
appropriate properties. Also, note that the condenser outlet temperature T 3 is about 3°C above the environmental temperature
of T 0 = 25°C, which is an appropriate temperature drop across the wall of the condenser. A temperature drop of this magnitude
also occurs in the evaporator, so the temperature of the refrigerated space is about −12.6°C instead of −15.6°C.
We can now determine the actual conditions at the outlet of the compressor from the two properties h 2 = (h 2s − h 1 )/(η s ) comp +
h 1 = 288.03 kJ/kg and p 2 = p 2s = 800 kPa. Interpolation in Table C.7f in Thermodynamic Tables to accompany Modern Engineering
Thermodynamics (or through the use of an appropriate computer program) gives the following additional properties at this state:
s 2 = 0:9814 kJ=kg . KandT 2 = 54:97°C
The condenser and evaporator heat transfer rates and the compressor work rate are
Now we can calculate the desired quantities.
_Q condenser = _m ref ðh 3 − h 2 Þ = ð0:5kg/sÞð87:46 − 288:03 kJ/kgÞ = −100:3 kJ/s
_Q evaporator = _m ref ðh 1 − h 4h Þ = ð0:5kg/sÞð235:31 − 87:46 kJ/kgÞ = 73:9 kJ/s
_Q compressor = _m ref ðh 2 − h 1 Þ = ð0:5kg/sÞð288:03 − 235:31 kJ/kgÞ = 26:36 kJ/s
a. The irreversibility rate of the compressor is given by
_I adiabatic
= _m ref T 0 ðs 2 − s 1 Þ
compressor
= 0:500 kJ
ð25:0 + 273:15 KÞ 0:9814 − 0:9317 kJ
s
kg.K
= 7:41 kJ = 7:41 kW
s
Q
_I condenser = T 0 _m ref ðs 3 − s 2 Þ − _
condenser
T 0
_I expansion
valve
= ð25:0 + 273:15 KÞ 0:500 kg
0:3257 − 0:9814 kJ
s
kg.K
= 2:55 kJ = 2:55 kW
s
= _m ref T 0 ðs 4h − s 3 Þ
−
−100:3 kJ/s
25:0 + 273:15 K
= 0:500 kg
ð25:0 + 273:15 KÞ 0:3449 − 0:3257 kJ
s
kg.K
= 2:86 kJ
s = 2:86 kW (Continued )