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.
+z
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
dn1
0=dz
(1)
Thus n1 = constant for all z. In addition,
dc1
n1 = -D
+ c1v0
dz
(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
0
ln(c1/c10) = v0z/D => c1 = c10ev z/D
Applying B. C. (2),
M c10D
Mv0
= 0 => c10 =
A
v
AD
0
Mv v0z/D
∴ c1 =
e
AD
[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.
1
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!
𝑚𝑜𝑙
𝑠𝑒𝑐
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.
Impermeable
holder
pheromone vapor well mixed by
free convection
solid pheromone
polymeric diffusion barrier of
thickness 0.08 cm and area 1.6 cm2
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:
V
dc1
= r0 − Aj1
dt
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:
r0 = Aj1 ⇒ 6 ⋅ 10
−17
3
⎡ ⎛
⎤ mol ADH
7 cm ⎞
⎟⎟c1 ⎥
(c1 − 0)
=
⎢1 − ⎜⎜1.1 ⋅ 10
mol ⎠ ⎦ s
l
⎣ ⎝
(2) Solving for c1 we have:
2
c1 =
6 ⋅ 10 −17
mol
s
mol
mol
s
= 8.6 ⋅ 10 −8
2
cm 3
cm ⎞
⎟⎟
s ⎠
cm 3
+ 6.6 ⋅ 10 −10
s
6 ⋅ 10 −17
ADH
cm 3
+ 6.6 ⋅ 10 −10

s
=
(1.6 cm )⎛⎜⎜1.92 ⋅10
2
−12
⎝
0.08 cm
(3) Solving for r0 or Aj1 gives:
(1.6 cm )⎛⎜⎜1.92 ⋅10
2
Aj1 =
ADH
c1 =

⎝
0.08 cm
−12
cm 2 ⎞
⎟
s ⎟⎠ ⎛
mol
−8 mol ⎞
= 3.3 ⋅ 10 −18
⎜ 8.6 ⋅ 10
3 ⎟
cm ⎠
s
⎝
[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 − r0 = Aj1 . You didn’t lose
point in both approaches.
3
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
1 𝑙 !! 1
𝑘! =
+
=
𝑘 𝐷
𝑐𝑚
!!
=
5.63
∗
10
1
0.01
𝑠
+
5.97 ∗ 10!! 10!!
(3) Mass balance over a section of the pipe:
πR2vΔc = -kT(2πRΔz)(c - cext)
dc 2k
or - =
(c - cext)
dz Rv
subject to c = 0.1M at z = 0, and cext = 0
2kTL
⎛ c ⎞
ð ln⎜ ⎟ = c
Rv
⎝ 0⎠
∴ 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.
4
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!
𝑑𝑐
𝑑𝑧
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𝑗! = −𝐷!
𝑐 − 𝑐!
𝐿!
(3) Mass balance
𝜋𝑅! 𝑗!
!!!"
− 𝑗!
!
1
= (2𝜋𝑅𝑑𝑧)𝑗!
2
𝑑𝑗! 1
= 𝑗!
𝑑𝑧
𝑅
𝐷!
𝑑 ! 𝑐 1 𝑐 − 𝑐!
= 𝐷
𝑑𝑧 ! 𝑅 ! 𝐿!
!!
(4) Parameters 𝑐 ∗ = 𝑐 − 𝑐! and 𝛼 ! = !!
! !!
The equation becomes 𝑑! 𝑐 ∗
= 𝛼!𝑐∗
𝑑𝑧 !
𝑐 ∗ = 𝐴𝑒 !!" + 𝐵𝑒 !"
B.C.
𝐴 = −𝐵
𝑀
𝑑𝑐 ∗
=
𝐷
= 𝐷! 𝐵𝛼(𝑒 !!!! + 𝑒 !!! )
!
𝜋𝑅!
𝑑𝑧
𝐵=
𝑀
1
𝜋𝑅! 𝐷! 𝛼(𝑒 !!!! + 𝑒 !!! )
Finally,
𝑐 = 𝑐! −
𝑀 (𝑒 !!!! − 𝑒 !!! )
𝐷! 𝛼𝜋𝑅! (𝑒 !!!! + 𝑒 !!! )
where
𝛼=
𝐷!
𝑅𝐿! 𝐷!
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