Modern Engineering Thermodynamics

296 CHAPTER 9: Second Law Open System Applications
EXAMPLE 9.6 (Continued )
Then, from Eq. (9.34),
_S P
mixing
ðmaxÞ
= ð0:600 lbm/s
Þ½1:0 Btu/ ðlbm.RÞŠln ½1 + 0:486ð0:18ÞŠð1:18Þ −0:486
= 2:056 × 10 −3
Btu/ ðs.RÞ
ð 778:17 ft .lbf/BtuÞ = 1:60 ft.lbf/ ðs.RÞ
This example illustrates that mixing identical materials at almost the same absolute temperatures with equal mass flow rates
produces nearly the maximum possible entropy production rate. Less entropy is produced if the mixing fraction is either
y < 0.5 or y > 0.5 when T 1 ≈ T 2 .
Exercises
16. If the hot water mass flow rate is 0.500 lbm/s and the cold water mass flow rate is 0.300 lbm/s in Example 9.6,
determine the entropy production rate of the mixing process. Keep the values of all the variables except T M and y the
same as they are in Example 9.6. Answer: _S P
mixing = 1:90 ft .lbf/ ðs.RÞ.
17. Recalculate parts a and b in Example 9.6 when the hot water temperature is reduced from 140.°F to 120.°F.
Keep the values of all the variables except T M and y c the same as they are in Example 9.6. Answer:
(a) _S P
mixing = 0:966 ft .lbf/ ðs.RÞ, (b) y c = 0.489 and _S P
mixing max = 0:966 ft .lbf/ ðs.RÞ.
18. If the cold water temperature is increased from 50.0°F to 60.0°F, recalculate parts a and b in Example 9.6.
Keep the values of all the variables except T M and y c the same as they are in Example 9.6. Answer:
(a) _S P
mixing = 1:196 f .lbf/ ðs.RÞ, (b) y c = 0.488, and _S P
mixing max = 1:196 ft .lbf/ ðs.RÞ.
9.8 SHAFT WORK MACHINES
A shaft work machine is defined in Chapter 6 as any device that has work (or power) input or output through a
rotating or reciprocating shaft. These devices are normally steady state, steady flow, single-inlet, single-outlet systems,
and their resulting modified energy rate balance equation is
_Q actual − _W actual = _m h 2 − h 1 + V2 2 −
V2 1 /2gc + ðg/g c ÞðZ 2 − Z 1 Þ
(6.30)
Assuming an isothermal system boundary, the modified entropy rate balance equation is
or
ð _Q /T b Þ actual + _m ðs 1 − s 2 Þ+ _S P = 0
_Q actual = _mT b ðs 2 − s 1 Þ− T b
_S P
Using this expression for _Q in the preceding equation and solving for
_W actual produces
_W actual = _m ðh 1 − T b s 1 Þ− ðh 2 − T b s 2 Þ+ V1 2 − V2 2 /2gc + ðg/g c ÞðZ 1 − Z 2 Þ − Tb _S P (9.35)
and this can be rearranged to provide an expression for the entropy production rate of a shaft work machine as
Entropy production in a shaft work machine
= _m
ðh 1 − T b s 1 Þ− ðh 2 − T b s 2 Þ+ V2 1 − V2 2
+ gZ
ð 1 − Z 2 Þ
−
T b
2g c g c
_S P shaft work
machine
_ W actual
T b
(9.36)
For a reversible process, _S P = 0, and Eq. (9.35) reduces to
_W rev = _m ðh 2 − T b s 2 Þ− ðh 1 − T b s 1 Þ+ V2 2 −
V2 1 /2gc + ðg/g c ÞðZ 2 − Z 1 Þ
(9.37)
and since, from Chapter 7,
by comparing Eqs. (9.35) and (9.37), we see that
_W actual = _W rev + _W irr
_W irr = T b
_S P