Modern Engineering Thermodynamics

566 CHAPTER 14: Vapor and Gas Refrigeration Cycles
EXAMPLE 14.9 (Continued )
4h
3
Condenser
Q refrig. = 422 kJ/h
Refrigeration evaporator
5
2s
Compressor
1
Freezer evaporator
6h
Q freezer = 422 kJ/h
2s
T
3
4h
5
6h
1
s
FIGURE 14.24
Example 14.9.
Solution
The properties of the refrigerant at the states shown in Figure 14.24 are as follows (see Tables C.7e, C.7f, and C.8d for the
numerical values):
Station 1—Compressor inlet
Station 2s—Compressor outlet
x 1 = 1:00
s 2s = s 1 = 0:9315 kJ/kg.K
T 1 = −18:0°C
p 2s = p 3 = p sat ð30:0°CÞ = 0:770 MPa
h 1 = 236:53 kJ/kg
h 2s = 271:0 kJ/kg
s 1 = 0:9315 kJ/kg.K
ðinterpolation in Table C:7fÞ
Station 3—Condenser outlet
Station 4h—Refrigerator evaporator inlet
x 3 = 0:00 h 4h = h 3
T 3 = 30:0°C
T 4h = 4:00°C
p 3 = p sat ð30:0°CÞ = 0:770 MPa
No further information is needed:
h 3 = h f ð30:0°CÞ = 91:49 kJ/kg
Station 5—Refrigerator evaporator outlet Station 6h—Freezer evaporator inlet
x 5 = ? h 6h = h 5 = ?
T 5 = T 4h = 4:00°C
T 6h = −18:0°C
h 5 = ?
No further information is needed:
The isentropic efficiency of the compressor is 80.0%, or (η s ) comp = 0.800.
a. The COP for this system is the ratio of the total heat rate removed from the refrigeration plus freezer compartments
divided by the power input to the compressor, or
COP =
but since h 5 = h 6h here, this equation reduces to
COP =
_ Q R + _Q F
_W C
= ðh 5 − h 4h Þ + ðh 1 − h 6h Þ
ðh 2s − h 1 Þ/ðη s Þ comp
h 1 − h 4h
=
236:53 − 91:49
ðh 2s − h 1 Þ/ðη s Þ comp
ð271:0 − 236:53Þ/0:800 = 3:37
so that the system COP does not depend on states 5 or 6h.
b. Since h 5 = h 6h , we can write the refrigeration and freezer compartment cooling rates as
_Q R = _m ref ðh 5 − h 4h Þ and _Q F = _m ref ðh 1 − h 6h Þ = _m ref ðh 1 − h 5 Þ. Solving these two equations for _m ref gives
_m ref =
Q _ R + _Q F
h 1 − h 4h
=
ð422 + 422 kJ/hÞð1 h/60minÞ
236:53 − 91:49 kJ/kg
= 0:0970 kg/min