Modern Engineering Thermodynamics

544 CHAPTER 14: Vapor and Gas Refrigeration Cycles
EXAMPLE 14.3 (Continued )
b. From Table C.9b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, the thermodynamic data at
the monitoring stations shown in the schematic are
where we have calculated
Station 1
T 1 = − 15:0°C
Station 2s
T 2s = 20:0°C
s 1 = s 2s = 0:89973 kJ/ ðkg.KÞ x 2s = 1:00
x 1 = 0:9395
h 1 = 231:0 kJ/kg
Station 3
T 3 = 20:0°C
x 3 = 0:00
h 2s = 256:5 kJ/kg
s 2s = 0:89973 kJ/ ðkg.KÞ
p 2s = 909:9 kPa
Station 4s
T 4s = T 1 = −15:0°C
h 3 = 68:67 kJ/kg x 4s = 0:1765
s 3 = 0:25899 kJ/ðkg.KÞ
p 3 = p 2 = 909:9 kPa
s 4s = s 3 = 0:25899 kJ/ðkg.KÞ
h 4s = 65:6 kJ/kg
and
x 1 = s 1 − s f 1
s
= s 2s − s f 1
fgx1 s
=
0:89973 − 0:11075
= 0:9395
fg1 0:83977
h 1 = h f 1 + x 1 ðh fg1 Þ = 27:33 + ð0:9395Þð216:79Þ = 231:0 kJ/kg
x 4s = s 3 − s f 4 0:25899 − 0:11075
s
= = 0:1765
fg4 0:83977
h 4s = h f 4 + x 4s ðh fg4 Þ = 27:33 + ð0:1765Þð216:79Þ = 65:59 kJ/kg
Then,
COP isentropic
=
vapor-compression
cycle ðwith expansion
turbineÞ
=
_Q L
h
=
1 − h 4s
_W c − _W t ðh 2s − h 1 Þ − ðh 3 − h 4s Þ
231:0 − 65:59
ð256:5 − 231:0Þ − ð68:67 − 65:59Þ = 7:38
which is identical to the Carnot efficiency of part a, as it should be, because the Rankine and Carnot cycles are identical
in this case (see Figure 14.10).
c. When the isentropic turbine is replaced by an adiabatic, aergonic throttling valve, the process from station 3 to station 4
becomes isenthalpic rather than isentropic, as shown in Figure 14.11.
3 2
4
Throttling
valve
Q H
Condenser
W C
Q L
3
2s
Evaporator
1
T
4h 1
s
FIGURE 14.11
Example 14.3, Solution, part c.