This

**preview**shows pages 1-2. to view the full**6 pages of the document.**Solutions, Chapter 12

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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.

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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

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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

Δx⋅Δx

ρ

⋅g

g=gΔz⋅0.0091 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

Δx⋅Δx

ρ

⋅g

g=gΔz⋅0.0076 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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