Modern Engineering Thermodynamics

9.7 Mixing 293
Now, _m air is given and _m water can be found from the modified energy rate balance equation by combining Eq. (6.19) with
(p in ) water =(p out ) water and Eq. (6.22) with Eq. (9.23) as
Our calculations begin with this last equation:
Then, the entropy production rate equation (a) gives
nh
i
_m water = _m air c p
air /c water ðT in − T out Þ air / ðT out − T in
Þ water
_m water = ð0:2kg/sÞ 1:004 kJ/ ð kg
.KÞ
ð90:0
− 75:0KÞ
= 0:036 kg/s
4:186 kJ/ ðkg.KÞ
ð40:0 − 20:0KÞ
_S P = ð0:200 kg/sÞ½1004 J/ ðkg.KÞŠln 75:0 + 273:15
90:0 + 273:15
+ ð0:036 kg/sÞ½4186 J/ ðkg.KÞŠln
40:0 + 273:15
20:0 + 273:15
o
= 1:48 W/K
Notice that the entropy production rate in the previous example is independent of whether the heat exchanger is parallel
or counterflow, since the same amount of heat transfer occurs in each case. We would see a difference if we had included
the effect of the viscous pressure drop in the entropy production rate equation. The counterflow arrangement requires less
heat transfer area and therefore produces a smaller pressure drop than the parallel flow arrangement. Then, the counterflow
heat exchanger has a smaller entropy production rate than a parallel flow heat exchanger with the same _Q, U(ΔT ) H ,and
(ΔT ) C values.
Exercises
13. Determine the entropy production rate in Example 9.5 if the air mass flow rate is increased from 0.200 kg/s
to 0.500 kg/s. Keep the values of all the other variables except _m water the same as they are in Example 9.5.
Answer: _S P = 3:67 W/K.
14. If the cooling water mass flow rate in Example 9.5 is decreased so that it exits the heat exchanger at 60.0°C rather than
40.0°C, then determine the new entropy production rate for the heat exchanger. Keep the values of all the variables
except _m water the same as they are in Example 9.5. Answer: _S P = 1:16 W/K.
15. Suppose the air in Example 9.5 is to be cooled to 35.0°C rather than 75.0°C. Determine the new entropy production
rate for this system. Keep the values of all the variables except _m water the same as they are in Example 9.5.
Answer: _S P = 3:47 W/K.
9.7 MIXING
A mixer normally has two or more inlet flow streams but only one outlet flow stream. Often mixers are used
simply to mix different chemical species to produce a final product. When all the entering fluids have the same
composition but are at different temperatures, the mixer becomes a type of simple heat exchanger.
Consider the dual-inlet, single-exit mixer shown in Figure 9.10. The steady state, steady flow energy rate balance
equation (neglecting any change in kinetic and potential energy) is
_Q − _W + _m 1 h 1 + _m 2 h 2 − _m 3 h 3 = 0
and the similar mass rate balance equation is
_m 1 + _m 2 − _m 3 = 0
Combining these two equations and introducing the
mass fraction y as
_m 1 / _m 3 = y (9.27)
1
2
W
3
or
_m 2 / _m 3 = 1 − y
where y is always bound by 0 ≤ y ≤ 1, which gives
_Q − _W + _m 3 ½yh ð 1 − h 2 Þ+ ðh 2 − h 3 ÞŠ = 0
FIGURE 9.10
A simple mixing system.
Q
System boundary