Modern Engineering Thermodynamics

14.5 Vapor-Compression Refrigeration Cycle 545
The thermodynamic data for stations 1, 2s, and 3 remain unchanged from part b, but the isenthalpic throttling valve
changes the data of station 4s to 4h as follows.
where we have calculated
Station 4h
T 4h = T 1 = −15:0°C
h 4h = h 3 = 68:67 kJ/kg
x 4h = 0:1910
s 4h = 0:27081 kJ/ðkg.KÞ
and
Finally,
x 4h = h 4h − h f 4
=
68:67 − 27:34
= 0:1906
h fg4 216:79
s 4h = s f 4 + x 4h ðs fg4 Þ = 0:11075 + ð0:1906Þð0:83977Þ = 0:27081 kJ/ ðkg.KÞ
Q
COP isentropie
= _ L
_W
vapor-compression cycle c
= h 1 − h 4h
h 2s − h 1
231:0 − 68:67
=
256:5 − 231:0 = 6:37
ðwith throttling valveÞ
Exercises
7. Determine the pressure of the R-22 in the evaporator in Example 14.3. Answer: p evaporator = p 1 = p 4 = p sat (R-22 at −15.0°C) =
295.7 kPa.
8. If ammonia were used in the refrigeration system described in Example 14.3, determine the condenser pressure if all the
other variables remain unchanged. Answer: p condenser = p 2 = p 3 = p sat (ammonia at 20.0°C) = 857.12 kPa.
9. The head of your Engineering Department has decided to use R-134a instead of R-22 in the refrigeration system in
Example 14.3. Assuming all the other variables remain unchanged, determine the new operating pressure in the
evaporator. Answer: p evaporator = p 1 = p 4 = p sat (R-134a at −15.0°C) = 164 kPa.
The decrease in COP from 7.38 to 6.37 (13.7%) in the previous example is not normally sufficient to justify the
increased expense of manufacturing, installing, and maintaining a turbine or other prime mover between
the condenser and the evaporator in small- and medium-size systems. Also, the working fluid in this part of the
cycle contains a mixture of liquid and vapor, and it is difficult to find any prime mover that operates efficiently
and reliably with this type of two-phase fluid. Throttling expansion valves, on the other hand, are very inexpensive
and reliable under these conditions.
By introducing the isentropic efficiency of the compressor (η s ) c , the general formula for the actual thermal efficiency
(COP) of a reversed Rankine cycle can be written as
COP vapor-compression cycle
R/AC
= _ Q L
_W c
=
h 1 − h 4h
(14.6)
ðh 2s − h 1 Þ/ ðη s Þ c
and
COP vapor-compression cycle
HP
= j _Q H j
_W c
=
h 2 − h 3
(14.7)
ðh 2s − h 1 Þ/ ðη s Þ c
where h 2 = h 1 + ðh 2s − h 1 Þ/ ðη s Þ c :
Because throttling processes are ideally isenthalpic, a pressure-enthalpy diagram is often used to describe vapor
refrigeration cycles, as shown in Figure 14.12. Process 1 to 2s in this figure involves the compression of a liquidvapor
mixture. This is technically more difficult than compressing either a pure vapor or a pure liquid. A
method of eliminating this problem is to superheat the vapor, as shown in Figure 14.12b.