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**preview**shows page 1. to view the full**5 pages of the document.**Solutions, Chapter 11

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11.1

f

pipe ⋅Rpipe =16

See the discussion below Eq. 11.10, which shows that

f

pipe

=

3

f

pm and that

Rpipe =2

3

Rpm .

Substituting these, we find

fpmRpm =16 ⋅3Rpm

2Rpipe

⋅3

f

pm

fpipe

=72

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11.2 From Eq. 11.14 we see that in laminar flow in porous media

F∝Vs⋅Δ

x

If both

Vs and Δx are multiplied by 2 then their product must be multiplied by 2.

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11.3 The Ergun Eq. (Eq. 11.17) shows f as the sum of that for laminar and for turbulent

flow. The ratio of the laminar term to the turbulent one is

lamina

r

term

turbulent term =150

Rpm /1.75=85.7

Rpm

For = 0.1 this ratio is 857, and the turbulent term is 0.3 % ≈ 0 of the sum. Increasing

the by steps of 10 makes the ratio become 85.7, 8.57, 0.857, 0.0857, 0.00857. Thus

at = 10,000, the laminar term is 0.14% ≈ 0 of the sum.

Rpm

Rpm

Rpm

I know of no theoretical justification for adding the two terms to get a predictive

curve in the transition region. However it represents experimental data satisfactorily and

is widely used.

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11.4

RP.M. =DpVS

ρ

μ

(1 −

ε

)=

0.0025 ft ⋅2f

t

s⋅62.3 lb

m

ft3

6.73⋅10−4lbm

ft s 1−0.033

()

=690

This is at the high end of the transition region, practically all turbulent.

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11.5* k=Vs

μ

Δ

x

ρ

g−Δz()

=

0.0124 f

t

s

⎛

⎝

⎜

⎞

⎠

⎟ ⋅ 1.002cp()⋅1ft()

62.3 lbm

ft2

⎛

⎝

⎜ ⎞

⎠

⎟ ⋅32.2 ft

s2

⎛

⎝

⎜ ⎞

⎠

⎟ ⋅1.25 ft

()

⋅6.72 ⋅10−4lbm

ft s cp ⋅darcy

1.06 ⋅10−11 ft =315 darcies

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 11, page 1

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This is a large permeability. Most aquifers and petroleum reservoir rocks have

permeabilities of one darcy or less.

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11.6 Here one can argue whether to apply B.E. inside the bed, and use the interstitial

velocity or apply it to the bed outlet, and use the superficial velocity. Doing the latter we

find for Ex. 11.1

Vs

2

2

F=

Vs

2

2

g−Δz()

=

0.0124 ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

2

32.2 ft

s2

⎛

⎝

⎜ ⎞

⎠

⎟ 1.25 ft

()

=1.9 ⋅10−6

and for Ex. 11.2

Vs

2

2

F=

Vs

2

2

ΔP

ρ

=

2ft

s

⎛

⎝

⎜

⎞

⎠

⎟

2

2⋅62.3 lbm

ft3

701 lbf

in2

⎛

⎝

⎜ ⎞

⎠

⎟ ⋅32.2 lbf ft

lbm s2⋅144 in2

ft2

=1.9⋅10−6

If one used interstitial, one would find the above answers multiplied by (1/0.33)2 = 9.1.

The small values here should convince the students that in normal porous media flows we

can ignore the kinetic energy terms.

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11.7* To use Fig. 11.5 we need the capillary number. Ca =

k

σ

Δ

P

Δ

x

Here Δ

P

Δ

x

=g

ρ

and

k=

V

s

μ

Δ

x

ρ

g(−Δz) combining these we have

Ca =1

σ

⋅Vs

μ

Δx

ρ

g(−Δz)⋅

ρ

g=Vs

μ

Δx

σ

Δz=

0.0124 f

t

s

⎛

⎝

⎜

⎞

⎠

⎟ ⋅ 2.09 ⋅10−5lbf s

ft 2

⎛

⎝

⎜

⎞

⎠

⎟ ⋅ 1ft

()

4.15 ⋅10−4lbf

in

⎛

⎝

⎜ ⎞

⎠

⎟ ⋅1.25 ft

()

⋅12 in

1ft

=4.17⋅10−5

Then from Fig. 11.5 we can read the residual saturation as about 0.16

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11.8 If the bed were filled completely with water, the value on the abscissa would be 12

inches/ft. Thus in the "flooded" condition at E, the packing's void spaces are only 50 to

60% full of water.

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11.9* (a) At the low end of the curve,

ρ

Vs = 5 lbm/ft2hr and

Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 11, page 2

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