Modern Engineering Thermodynamics

546 CHAPTER 14: Vapor and Gas Refrigeration Cycles
T
3
2s
p
3 2s
4h 1
4h
1
s
h
(a) Vapor-compression isentropic refrigeration cycle
with isenthalpic throttling
2s
T
3
1
p
3 2s
4h
4h
1
s
(b) Same as (a) except with superheat
h
FIGURE 14.12
T–s and p–h diagrams for an isentropic vapor-compression cycle.
EXAMPLE 14.4
Repeat part c of Example 14.3 requiring that the evaporator outlet be a saturated vapor at −15.0°C and introduce a compressor
isentropic efficiency of 75.0%.
Solution
First, draw a sketch of the system (Figure 14.13). Since the evaporator outlet is a saturated vapor, the compressor outlet is a
superheated vapor, as shown in the figure.
The thermodynamic data for the four monitoring stations are (see Example 14.3 for details)
Then, from Eq. (14.7),
Station 1
Station 2s
T 1 = −15:0°C
p 2s = p 2 = 909:9 kPa
x 1 = 1:00
s 2s = s 1 = 0:95052 kJ/ ðkg.KÞ
h 1 = 244:13 kJ/kg h 2s = 271:92 kJ/kgðfrom interpolation in Table C:10bÞ
s 1 = 0:95052 kJ/ ðkg.KÞ T 2s = 39:3°C
Station 3
Station 4h
T 3 = 20:0°C
T 4h = T 1 = −15:0°C
x 3 = 0:00
h 4h = h 3 = 68:67 kJ/kg
h 3 = 68:67 kJ/kg x 4h = 0:1910
s 3 = 0:25899 kJ/ ðkg.KÞ s 4h = 0:27088 kJ/ ðkg.KÞ
Q
COP vapor-compression
= _ L
=
cycle
j _W j
R/AC
c
=
h 1 − h 4h
ðh 2s − h 1 Þ/ ðη s
244:13 − 68:67
ð271:92 − 244:13Þ/0:750 = 4:74
Þ c