Modern Engineering Thermodynamics

288 CHAPTER 9: Second Law Open System Applications
EXAMPLE 9.4 (Continued )
a. The MERB for this aergonic device with negligible change in kinetic
and potential energy is
System boundary
_Q − 0 = _m ðh 2 − h 1 + 0Þ
1 2
and, from Table C.7a for R-134a, we find
0.100 lbm/s of
refrigerant-134a
and
h 1 = h f ð100°FÞ = 44:23 Btu/lbm
s 1 = s f ð100°FÞ = 0:0898 Btu/ ðlbm.RÞ
h 2 = h f ð20:0°FÞ+ x 2 h fg ð20:0°FÞ
= 17:74 + ð0:530Þð86:87Þ = 63:78 Btu/lbm
s 2 = s f ð20:0°FÞ+ x 2 s fg ð20:0°FÞ
= 0:0393 + ð0:530Þð0:2206 − 0:0393Þ = 0:1353 Btu/ ðlbm.RÞ
Q
FIGURE 9.4
Example 9.4.
Then, the MERB gives
_Q = ð0:100 lbm/sÞð63:78 − 44:23 Btu/lbmÞ = 1:955 Btu/s
and Eq. (9.13) gives
_S P = _m ðs 2 − s 1 Þ − _ Q
−
T b
1:955 Btu/s
60:0 + 459:67 R
= ð0:100 lbm/sÞ½0:1353 − 0:0898 Btu/ ðlbm.RÞŠ
= 7:88 × 10 −4 Btu/ ðs.RÞ
= ½7:88 × 10 −4 Btu/ ðs.RÞŠð778:17 ft.lbf/BtuÞ = 0:613 ft.lbf/ ðs.RÞ
b. The MERB for this device as an adiabatic, aergonic, negligible change in kinetic and potential energy system is
or
0 − 0 = _m ðh 2 − h 1 + 0Þ
h 2 = h 1
Now, station 2 is fixed by the pair of properties T = 20.0°F andh 2 = h 1 = 44.23 Btu/lbm. Consequently, the quality at
station 2 cannot be 53.0% but is instead
then,
Then, with _Q = 0, Eq. (9.13) gives
x 2 = h 2 − h f 2 44:23 − 17:74
= = 0:3049 = 30:5%
h fg2 86:87
s 2 = 0:0393 + 0:3049ð0:2206 − 0:0393Þ = 0:0946 Btu/ ðlbm.RÞ
_S P = _mðs 2 − s 1 Þ − _ Q
T b
= ð0:100 lbm/sÞ½0:0946 − 0:0898 Btu/ ðlbm.RÞŠ− 0
= 4:80 × 10 −4 Btu/ ðs.RÞ = ½4:80 × 10 −4 Btu/ ðs.RÞŠð778:17 ft.lbf/BtuÞ
= 0:374 ft.lbf/ ðs.RÞ
c. The percentage decrease in _S P brought about by adding the insulation is
7:88 × 10 −4 Btu/ ðs.RÞ− 4:80 × 10 −4 Btu/ ðs.RÞ
7:88 × 10 −4 × 100 = 39:1%
Btu/ ðs.RÞ
Note that there is a substantial decrease in the entropy production rate of Example 9.4 due to simply insulating the valve.
This results from the elimination of the entropy generated by the heat transfer present in the uninsulated valve.