Modern Engineering Thermodynamics

9.7 Mixing 295
Critical value of y, y c
1.0
0.8
0.6
0.4
0.2
0.0
10 −3 10 −2 10 −1 10 0 10 1 10 2 10 3
Temperature ratio, T 1 /T 2
FIGURE 9.11
The critical mass fraction required to produce a maximum entropy
production rate in adiabatic, aergonic, isobaric (i.e., viscousless)
mixing of identical incompressible liquids or ideal gases with
constant specific heats (note that y c = 1.0 at T 1 /T 2 = 0, and
y c → 0asT 1 /T 2 → ∞).
Maximum mixing S P /m c
4.0
3.0
2.0
1.0
Unavailable
region
0.0
10 −3 10 −2 10 −1 10 0 10 1 10 2 10 3
Temperature ratio, T 1 /T 2
FIGURE 9.12
The maximum entropy production rate vs. temperature ratio for a
simple mixing process.
EXAMPLE 9.6
A bathroom shower is set with equal hot and cold mass flow rates of 0.300 lbm/s. The hot water is at 140.°F and the cold
water is 50.0°F. Determine
a. The shower mixture temperature and its entropy production rate.
b. The critical mass fraction y c and the value of the maximum entropy production rate.
Solution
First, draw a sketch of the system (Figure 9.13).
The unknowns are the shower mixture temperature and its
entropy production rate, the critical mass fraction y c , and the
value of the maximum entropy production rate. The material
is liquid water.
m H = 0.300 lbm/s
T H = 140.0°F
Shower head
Assume the water is incompressible with a constant specific
heat and the mixing takes place adiabatically, aergonically,
and isobarically.
a. From Eq. (9.27), since _m H = _m C and _m M = _m H + _m C ,
y = _m H / _m M = 1/2 = 0:500 and, with T 1 = T H , T 2 = T C ,
and T 3 = T M ,
T 1
= T H 140: + 459:67
=
T 2 T C 50:0 + 459:67 = 1:18
FIGURE 9.13
Example 9.6.
m C = 0.300 lbm/s
T C = 50.0°F
T M = ?
S P = ?
y c = ?
(S P ) max = ?
Then, from Eq. (9.30), we have
T 3 = T M = T C + yT ð H − T C Þ
= 50:0 + 0:500ð140: − 50:0Þ = 95:0°F
and, from Eq. (9.32), we have (using _m 3 = _m M = _m H + _m C = 0:600 lbm/s)
n
o
_S P = 0:600 lbm/s
mixing ð Þ ½ 1:0 Btu/ ð lbm .RÞŠln ½1 + 0:5ð0:18ÞŠð1:18Þ −0:5
= ½2:051 × 10 −3 Btu/ ðs.RÞŠð778:17 ft.lbf/BtuÞ
= 1:60 ft.lbf/ ðs.RÞ
b. From Eq. (9.33), we have
y c = ð1 − T H/T C Þ + ln ðT H /T C Þ ð1 − 1:18Þ + ln ð1:18Þ
= = 0:486
ð1 − T H /T C Þ ln ðT H /T C Þ ð1 − 1:18Þln ð1:18Þ
(Continued )