Modern Engineering Thermodynamics

286 CHAPTER 9: Second Law Open System Applications
EXAMPLE 9.2 (Continued )
The modified energy rate balance equation for this system is
"
_Q
↘
0
− _W
↘
0
+ _m
h 1 − h 2 + ðV 2 1 − V2 2 Þ/2g c + gðZ 1 − Z 2 Þ/g c
⎵
Assuming liquid water is incompressible with a constant specific heat under these conditions and using the incompressible
liquid auxiliary equation for enthalpy (Eq. (6.19)) allows the preceding modifies energy rate balance equation to be
written as
h 2 − h 1 = cT ð 2 − T 1 Þ+ vp ð 2 − p 1 Þ = V1 2 − V2 2 /2gc
0
#
= 0
or
T 2 = T 1 + v ð
c p 1 − p 2 Þ− V2 2 −
V2 1
2cg c
where v = v f (at 50.0°F) = 0.01602 ft 3 /lbm (from Table C.1a of Thermodynamic Tables to accompany Modern Engineering
Thermodynamics). From the data given in the problem statement, we can compute V 1 and V 2 as
and
V 1 =
_m = 4 _mv
ρA 1 πD 2 =
1
V 2 = 4 _mv
πD 2 2
40:2000 ð lbm/s Þ 0:01602 ft3 /lbm 144 in: 2 /ft 2
πð1:000 in: Þ 2 = 0:5874 ft/s
2
D
= V 1
1 = ð0:5874Þ
D 2
1:000
0:2500
2
= 9:399 ft/s
Then,
"
T 2 = ð50:00 + 459:67 RÞ+ 0:01602 ft 3 ð95:00 − 14:70 lbf/in:2Þ 144 in: 2 /ft 2 #
/lbm
½1:0 Btu/ ðlbm.RÞŠð778:17 ft.lbf/BtuÞ
ð9:399Þ 2 − ð0:5874Þ 2 ft 2 /s 2
−
21:0 ½ Btu/ ðlbm.RÞŠ½32:174 lbm.ft/ ðlbf.s 2 ÞŠð778:17 ft.lbf/BtuÞ
= 509:9 − 0:0018 R = 509:9R
and Eq. (9.14) gives 3 _S P = _mc ln T 2
= ð0:2000 lbm/sÞ½1:000 Btu/ ðlbm.RÞŠln 509:9
T 1 509:7
= ½7:846 × 10 −5 Btu/ ðs.RÞŠð778:17 ft.lbf/BtuÞ = 0:0611 ft.lbf/ ðs.RÞ
Notice that the kinetic energy term in this example (0.0018 R) provides a negligible contribution to the exit temperature
and the vast majority of the entropy production results from the pressure loss across the nozzle. The increase in velocity
across the nozzle as converted pressure energy does decrease the entropy production rate slightly (but only by less than
1% in this case). Therefore, this nozzle is quite inefficient at converting pressure energy into kinetic energy. Its efficiency
could be improved, however, by making the nozzle outlet diameter smaller, so that the outlet velocity is substantially
increased.
Exercises
4. Determine the entropy production rate in Example 9.2 if the mass flow rate is increased from 0.2000 to 0.8000 lbm/s.
Recalculate the values of V 1 , V 2 ,andT 2 , then keep the values of all the remaining variables the same as they are in
Example 9.2. Answer: _S P = 0:288 ft.lbf/ ðs.RÞ:
5. Determine the entropy production rate in Example 9.2 if the water hose has been lying in the sun and the water
temperature is 80.00°F rather than 50.00°F. Using v = v sat (80°F), recalculate V 1 , V 2 ,andT 2 , then keep the values of all
the remaining variables the same as they are in Example 9.2. Answer: _S P = 0:068 ft.lbf/ ðs.RÞ:
6. The exit diameter on the nozzle in Example 9.2 is reduced to 0.1250 in. Determine the new entropy production rate
for this system. Keep the values of all the variables except V 2 and T 2 the same as they are in Example 9.2.
Answer: _S P = 0:064 ft.lbf/ ðs.RÞ:
2 To get a meaningful answer, this problem needs to be specified to four significant figures.
3 Note that, without carrying four significant figures in this example, the logarithm is zero.