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Tutorial/HW Week
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1. Dept of Chemical and Biomolecular Engineering CN2125 Heat and Mass Transfer Dr. Praveen Linga, E5-02-23, 6601-1487
(Radiation, Mass Transfer)
2. Tutorial/HW Week #9 WWWR Chapters 26-27 ID Chapter 14 Tutorial #9
WWWR #26.17, 26.27, 27.6, 27.22 & ID #14.33
To be discussed during the week of Mar 19 - 23, 2012.
By either volunteer or class list. Homework #9
(self practice)
WWWR # 26.22, 27.16 & ID #14.39
3. Molecular Diffusion General differential equation
One-dimensional mass transfer without reaction
4. Unimolecular Diffusion Diffusivity of gas can be measured in an Arnold diffusion cell
5. Assuming
Steady state, no reaction, and diffusion in z-direction only
We get
And since B is a stagnant gas,
6. Thus, for constant molar flux of A, when NB,z = 0,
with boundary conditions:
at z = z1, yA = yA1
at z = z2, yA = yA2
Integrating and solving for NA,z
7. since the log-mean average of B is
we get
This is a steady-state diffusion of one gas through a second stagnant gas;
8. For film theory, we assume laminar film of constant thickness d,
then, z2 – z1 = d
and
But we know
So, the film coefficient is then
9. To determine concentration profile,
if isothermal and isobaric,
integrated twice, we get
10. with boundary conditions:
at z = z1, yA = yA1
at z = z2, yA = yA2
So, the concentration profile is:
11. Example 1
16. Pseudo-Steady-State Diffusion When there is a slow depletion of source or sink for mass transfer
Consider the Arnold diffusion cell, when liquid is evaporated, the surface moves,
at any instant, molar flux is
17. Molar flux is also the amount of A leaving
Under pseudo-steady-state conditions,
which integrated from t=0 to t=t, z=zt0 to z=zt becomes
18. Example 2
26. Equimolar Counterdiffusion Flux of one gaseous component is equal to but in the opposite direction of the second gaseous component
Again, for steady-state, no reaction, in the z-direction,
the molar flux is
27. In equimolar counterdiffusion, NA,z = -NB,z
Integrated at z = z1, cA = cA1 and at z = z2, cA = cA2 to:
Or in terms of partial pressure,
28. The concentration profile is described by
Integrated twice to
With boundary conditions at z = z1, cA = cA1and at z = z2, cA = cA2
becomes a linear concentration profile:
29. Systems with Reaction When there is diffusion of a species together with its disappearance/appearance through a chemical reaction
Homogeneous reaction occurs throughout a phase uniformly
Heterogeneous reaction occurs at the boundary or in a restricted region of a phase
30. Diffusion with heterogeneous first order reaction with varying area:
With both diffusion and reaction, the process can be diffusion controlled or reaction controlled.
Example: burning of coal particles
steady state, one-dimensional, heterogeneous
31. 3C (s) + 2.5 O2 (g) ? 2 CO2 (g) + CO (g)
Along diffusion path, RO2 = 0, then the general mass transfer equation reduces from
to
For oxygen,
32. From the stoichiometry of the reaction,
We simplify Fick’s equation in terms of oxygen only,
which reduces to
33. The boundary conditions are:
at r = R, yO2 = 0 and at r = ?, yO2 = 0.21,
Integrating the equation to:
The oxygen transferred across the cross-sectional area is then:
34. Using a pseudo-steady-state approach to calculate carbon mass-transfer
output rate of carbon:
accumulation rate of carbon:
input rate of carbon = 0
Thus, the carbon balance is
35. Rearranging and integrating from
t = 0 to t = ?, R = Ri to R = Rf, we get
For heterogeneous reactions, the reaction rate is
36. If the reaction is only
C (s) + O2 (g) ? CO2 (g)
and if the reaction is not instantaneous, then
for a first-order reaction, at the surface,
then,
37. Combining diffusion with reaction process, we get
38. Example 3
42. Diffusion with homogeneous first-order reaction:
Example: a layer of absorbing liquid, with surface film of composition A and thickness ?,
assume concentration of A is small in the film,
and the reaction of A is
43. Assuming one-direction, steady-state, the mass transfer equation reduces from
to
with the general solution
44. With the boundary conditions:
at z = 0, cA = cA0 and at z = ?, cA = 0,
At the liquid surface, flux is calculated by differentiating the above and evaluating at z=0,
45. Thus,
Comparing to absorption without reaction, the second term is called the Hatta number.
As reaction rate increases, the bottom term approaches 1.0, thus
46. Comparing with
we see that kc is proportional to DAB to ½ power.
This is the Penetration Theory model, where a molecule will disappear by reaction after absorption of a short distance.
47. Example 4
55. 2- and 3-Dimensional Systems Most real systems are two- and three-dimensional
Analytical solution to general differential equation with the boundary conditions
Requires partial differential equations and complex variable theories.
56. Unsteady-State Diffusion Transient diffusion, when concentration at a given point changes with time
Partial differential equations, complex processes and solutions
Solutions for simple geometries and boundary conditions
57. Fick’s second law of diffusion
1-dimensional, no bulk contribution, no reaction
Solution has 2 standard forms, by Laplace transforms or by separation of variables
58. Transient diffusion in semi-infinite medium
uniform initial concentration CAo
constant surface concentration CAs
Initial condition, t = 0, CA(z,0) = CAo for all z
First boundary condition:
at z = 0, cA(0,t) = CAs for t > 0
Second boundary condition:
at z = ?, cA(?,t) = CAo for all t
Using Laplace transform, making the boundary conditions homogeneous
59. Thus, the P.D.E. becomes:
with
?(z,0) = 0
?(0,t) = cAs – cAo
?(?,t) = 0
Laplace transformation yields
which becomes an O.D.E.
60. Transformed boundary conditions:
General analytical solution:
With the boundary conditions, reduces to
The inverse Laplace transform is then
61. As dimensionless concentration change,
With respect to initial concentration
With respect to surface concentration
The error function
is generally defined by
62. The error is approximated by
If ? ? 0.5
If ? ? 1
For the diffusive flux into semi-infinite medium, differentiating with chain rule to the error function
and finally,
63. Transient diffusion in a finite medium, with negligible surface resistance
Initial concentration cAo subjected to sudden change which brings the surface concentration cAs
For example, diffusion of molecules through a solid slab of uniform thickness
As diffusion is slow, the concentration profile satisfy the P.D.E.
64. Initial and boundary conditions of
cA = cAo at t = 0 for 0 ? z ? L
cA = cAs at z = 0 for t > 0
cA = cAs at z = L for t > 0
Simplify by dimensionless concentration change
Changing the P.D.E. to
Y = Yo at t = 0 for 0 ? z ? L
Y = 0 at z = 0 for t > 0
Y = 0 at z = L for t > 0
65. Assuming a product solution,
Y(z,t) = T(t) Z(z)
The partial derivatives will be
Substitute into P.D.E.
divide by DAB, T, Z to
66. Separating the variables to equal -?2, the general solutions are
Thus, the product solution is:
For n = 1, 2, 3…,
67. The complete solution is:
where L = sheet thickness and
If the sheet has uniform initial concentration,
for n = 1, 3, 5…
And the flux at z and t is
68. Example 1
75. Example 2
83. Concentration-Time charts
86. Example 3
