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Final

Ch 12Exam


Department
Chemical Engineering
Course Code
CHE 3101
Professor
All
Study Guide
Final

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Solutions, Chapter 12
______________________________________________________________________
12.1 Following Ex. 12.1, we read the appropriate values from Fig 12.2, as shown below
for the values for various values of Qg/Ql and compute and then plot
Qg/Ql Holdup
Ratio
(1-e)/e e
ρ
avg/
ρ
water minus Δ P/
g
ρ
water
F/g
ρ
wate
r
0 1 infinite 0 1 1 0
10 3.5 0.35 0.741 0.259 0.5 0.929
20 4.4 0.22 0.820 0.180 0.59 2.272
30 5 0.167 0.857 0.143 0.72 4.040
40 5.6 0.14 0.877 0.123 0.86 6.003
0
2
4
6
8
10
0 5 10 15 20 25 30 35 40
Qg
/
Q
l
10
ρ
avg
/
ρ
water
F
/
g
ρ
water
We use 10
ρ
avg /
ρ
water to get both values on a common scale, and
F
/
g
ρ
water to get a
dimensionless representation.
______________________________________________________________________
12.2 Using the table prepared in Prob 12.1, we can compute
Qg/Ql e Vavg l , ft/s V
avg g, ft/s V
slip, ft/s
0 0.00 2.06 0.00
10 0.74 7.95 27.81 19.86
20 0.82 11.42 50.26 38.84
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 12, page 1

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

30 0.86 14.42 72.10 57.68
40 0.88 16.77 93.94 77.16
and plot
0
10
20
30
40
50
60
70
80
0 5 10 15 20 25 30 35 40
Qg
/
Q
l
A little algebra shows that Vslip =Q
l
Atube
Qg
Ql
1
ε
1
1
ε
For the values shown, in which
0.88
ε
0.74 the term in parentheses is practically proportional to Q so that the
plot is practically linear.
g/Q
l
______________________________________________________________________
12.3* For liquid only Vliq = 2.06 f
t
s For air only the velocity is 10 times as large. For
plastic tubing we use
ε
0.00006 in (See Table 6.3) Then for water R =16, 300, f =
0.0068 and . The corresponding values for air are
10,900, 0.0076 and 0.0076.
dP /dx()
friciton =0.0091 psi / ft
Fliquid =ΔP
ρ
friction
=ΔP
ΔxΔx
ρ
g
g=gΔz0.0091 lbf
in2ft
32.2 lbm
lbf
ft
s2
144 in2
ft2
32.2 ft
s2
62.3 lbm
ft3
=0.021 gΔz
Fgas =ΔP
ρ
friction
=ΔP
ΔxΔx
ρ
g
g=gΔz0.0076 lbf
in2ft
32.2 lbm
lbf
ft
s2
144 in2
ft 2
32.2 ft
s2
0.075 lbm
ft3
=2.3 gΔz
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 12, page 2
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