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Midterm Test ADV. TRANSPORT II 2012

Note: Please answer four exam problems by answering step-by-step questions.

Problem 1 Concentration profile of a membrane reactor

A membrane reactor (shown in figure) uses a homogeneous catalyst which cannot pass through an ultra-filtration membrane. Reagents flow continuously toward the membrane, but the catalyst is injected only at the start of the experiment. Volume averaged velocities of reagents and products are v0. The amount of the injected catalyst per membrane area is M/A.

Questions:

(1) Write down governing equation and two boundary conditions

(2) Derive the concentration profile of the catalyst

Prompt: it is a steady state convective diffusion problem; use must define correctly the boundary conditions, one of which reflects the impermeability of the membrane to the catalyst, and the second one accounts for the conservation of catalyst mass.

Solution

(1) The basic equations are

0 = - dn1dz (1)

Thus n1 = constant for all z. In addition,

n1 = -D dc1dz + c1v0 (2)

B. C. (1) z = 0, n1 = 0 (at the membrane) (2) 𝑐!𝑑𝑧

!!! = !

! (the catalyst injected per membrane area)

(2) From B. C. (1), we have n1 = 0 for all z, thus integrating Eq. (2), we have ln(c1/c10) = v0z/D => c1 = c10ev

0z/D Applying B. C. (2),

MA =

c10Dv0 => c10 =

Mv0

AD

∴ c1 = Mv0

AD ev0z/D

[Comments] If you define “+z” towards left, the boundaries of integration in B.C.2 should be 0 to ∞. Also, the final concentration will have a minus sign.

+z

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Problem 2 Controlled release of pheromones

Controlled release is important in agriculture, especially for insect control. One common example involves the pheromones, sex attractants released by insects. If you mix this attractant with an insecticide, you can wipe out all of one sex of a particular insect pest. A device for releasing one pheromone is shown schematically below. The solid pheromone evaporates with the rate of

𝑟! = 6 ∗ 10!!" 1− 1.10 ∗ 10!𝑐𝑚!

𝑚𝑜𝑙 𝑐!𝑚𝑜𝑙𝑠𝑒𝑐

where c1 is the concentration in the vapor. The permeability (DH) of this material through the polymer is 1.92*10-12 cm2/sec. The concentration of pheromone outside of the device is essentially zero.

Questions:

(1) Write down the mass balance equation for the amount of pheromone vapor in the chamber. Assume steady state

(2) What is the concentration of pheromone in the vapor?

(3) How fast is the pheromone released by the device?

Solution

(1) From a balance on the device we have:

101 AjrdtdcV −=

Since we are interested in the steady state case, the time derivative is 0 and we find that the rate of sublimation must equal the rate of transport through the membrane. Substituting for r0 and j1 we have:

( )0101.11106 11

3717

10 −=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⋅−⋅⇒= − cADH

smolc

molcmAjr

l

(2) Solving for c1 we have:

pheromone vapor well mixed by free convection

polymeric diffusion barrier of thickness 0.08 cm and area 1.6 cm2

Impermeable holder

solid pheromone

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( )scm

cmscmcm

smol

scmADH

smol

c

310

2122

17

310

17

1

106.608.0

1092.16.1

106

106.6

106

−

−

−

−

−

⋅+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

⋅=

⋅+

⋅=

38106.8cmmol−⋅=

(3) Solving for r0 or Aj1 gives:

( )smol

cmmol

cmscmcm

cADHAj 183

8

2122

11 103.3106.808.0

1092.16.1−−

−

⋅=⎟⎠

⎞⎜⎝

⎛ ⋅⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

==

[comments] Following the description in the problem “the solid evaporation rate is r0,” the mass balance equation should be the evaporation rate equals to diffusion flux times area. However, the evaporation rate assigned here (from textbook) may not be reasonable. The higher concentration

should bring a larger evaporation rate. The mass balance should be 10 Ajr =− . You didn’t lose point in both approaches.

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Problem 3 Permeation of Benzoic acid through cellulose acetate

Water containing 0.1 M benzoic acid flows at 0.1 cm/s through a 1-cm-diameter rigid tube of cellulose acetate, the walls of which are permeable to small electrolyte molecules. These walls are 0.01 centimeter thin; solutes within the walls diffuse as through water. The tube is immersed in a large well-stirred water bath. We want to know the benzoic acid concentration after 2 meters.

Questions:

(1) Review problem 2 in homework set 6. Write down the information you need.

(2) Calculate the overall mass transfer coefficient, including the mass transfer coefficient you found in (1) and the mass transfer coefficient through the tube wall.

(3) Formulate mass balance as in HW6 problem 2. Derive the concentration profile and calculate the value of concentration after 2 meters.

Solution

(1) k = 5.97×10-5 cm/s, D=1*10-5 cm2/s (2) The total mass transfer coefficient is

𝑘! =1𝑘 +

𝑙𝐷

!!

= 11

5.97 ∗ 10!! +0.0110!!

= 5.63 ∗ 10!!𝑐𝑚𝑠

(3) Mass balance over a section of the pipe: πR2vΔc = -kT(2πRΔz)(c - cext)

or - dcdz =

2kRv (c - cext)

subject to c = 0.1M at z = 0, and cext = 0

ð ln⎝⎜⎛⎠⎟⎞cc0

= - 2kTLRv

∴ c = c0 exp(-2kTL/Rv) = 0.1 exp(-2*5.63×10-5*200/0.5/0.1) = 0.064 M

[comments] Many of you simply use the concentration profile from problem 8.2 on p. 270, but it is wrong. Although the mass balance equations are similar, boundary conditions differ and so as the derived concentration profile.

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Problem 4 Oxygen concentration in a burrow

Gopher’s burrow is modeled as a long circular pipe of length LP and radius R, which is located underground at the distance LS to the soil surface (which is much larger than R). The pipe is parallel to the surface. The burrow entrance is connected with open air through a vertical hole. The gopher spends time mainly at its favorite spot at the very far end of the burrow. The gopher breathes and consumes oxygen at a steady rate M. Oxygen diffused into the burrow from the open air through the entrance and through the soil layer of thickness Ls on top of the burrow.

Assuming (1) no resistance to diffusion through the vertical holes from the open air to the channel entrances; (2) diffusion passes through the soil layer only through the top side (roof) of the burrow so that we can neglect diffusion through the other half side of the pipe wall; (3) concentration of oxygen in the burrow cross-section is uniform.

There are two diffusion coefficients in this problem: DA - in air (for diffusion inside the burrow along the axis) and DS - in soil (for diffusion through the soil layer into the burrow though its roof). Mass transport problem in the burrow layer can be considered as one-dimensional along the burrow axis Mass transport problem in the soil layer can be considered as one-dimensional in vertical direction.

(1) Write down two boundary conditions at z equals to zero and Lp.

(2) Write down the flux along the pipe and the flux through the soil.

(3) Write down the overall mass balance equation and combine it with your equations in (2), assuming a steady state.

(4) Solve the equation for the concentration profile of oxygen in the burrow.

Solution.

(1) Boundary conditions:

𝑧 = 0, 𝑐 = 𝑐!

𝑧 = 𝐿! ,𝑀 = 𝜋𝑅! 𝑗! !!!!

*jD is defined as the overall flux

(2) Flux: Along the pipe (denoted as p) and from the soil on the top of the pipe (denoted as s)

𝑗! = −𝐷!𝑑𝑐𝑑𝑧

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𝑗! = −𝐷!𝑐 − 𝑐!𝐿!

(3) Mass balance

𝜋𝑅! 𝑗! !!!" − 𝑗! ! =12 (2𝜋𝑅𝑑𝑧)𝑗!

𝑑𝑗!𝑑𝑧 =

1𝑅 𝑗!

𝐷!𝑑!𝑐𝑑𝑧! =

1𝑅𝐷!

𝑐 − 𝑐!𝐿!

(4) Parameters 𝑐∗ = 𝑐 − 𝑐! and 𝛼! = !!!!!!!

The equation becomes  𝑑!𝑐∗

𝑑𝑧! = 𝛼!𝑐∗

𝑐∗ = 𝐴𝑒!!" + 𝐵𝑒!"

B.C.

𝐴 = −𝐵

𝑀𝜋𝑅! = 𝐷!

𝑑𝑐∗

𝑑𝑧 = 𝐷!𝐵𝛼(𝑒!!!! + 𝑒!!!)

𝐵 =𝑀𝜋𝑅!

1𝐷!𝛼(𝑒!!!! + 𝑒!!!)

Finally,

𝑐 = 𝑐! −𝑀

𝐷!𝛼𝜋𝑅!(𝑒!!!! − 𝑒!!!)(𝑒!!!! + 𝑒!!!)

where

𝛼 =𝐷!

𝑅𝐿!𝐷!