Chapter 3 Diffusion in concentrated solutions page 3-1
Chapter 3 Diffusion in concentrated solutions
1. Diffusion flux halfway up a capillary
The concentration profile is
1 – y1l
z/l
2. Gas-oil cracking
The reaction is A 3B
Basic equations:
dz (1)
Thus n1 = constant for all z. In addition,
dz + c v (2)
Because of the reaction, 3n1 = –n2, thus cv0 = n1 + n2 = -2n1
Substitute into Eq. (2), –Dc dy1
dz = n(1 + 2y)
1 1
B. C. z = 0 y1 = 0 (due to reaction)
z = l y1 = y1l (the bulk)
Integration yields ln(1 + 2y1) = -2n1z/Dc
2l ln(1 + 2y) (3)
When y1l is small, ln(1 + 2y1l) 2y1l, thus this equation reduces to
l c (diffusing through a thin film)
For evaporation through stagnant solvent, Eq. 3.3-11 shows that (after changing coordinates)
l ln(1 – y)
Chapter 3 Diffusion in concentrated solutions page 3-2
which is very similar to Eq. (3), the solution of this problem.
3. Interface movement of evaporating benzene
1
and is known from Eq. (3.3-38)
-1
= 3.4710-4 g/cm2-s
At 60 °C, the density of benzene is 0.83g/cm3
4. Concentration profile in a membrane reactor
The basic equations are
dz (1)
Thus n1 = constant for all z. In addition,
dz + c v (2)
B. C. z = 0 n1 = 0 (at the membrane)
c dz = M/A (the catalyst injected per membrane area)
From B. C. 1, we have n1 = 0 for all z, thus integrating Eq. (2), we have
Applying B. C. 2, – M
A = c10D
v0 => c10 = – Mv0
AD (v < 0 because of the coordinates)
0
5. Diffusion Forms of diffusion coefficient
From Table 3.2-1,
similarly, v2 – v0 = –Dlnc2
Chapter 3 Diffusion in concentrated solutions page 3-3
By comparing (1) and (1), we thus have D = D*.
6. Fick’s law vs. Maxwell-Stefan equation
(a) make a steady mass balance on a shell
r4
rr
0d
d
r
r2j1
combine with Fick’s law
to give
dr r2dc1
dr
subject to
rRin c1c1sat
To solve, integrate twice to find
where A and B are integration constants.
Thus
c1sat
1
r1
rout
1
Rin
1
Rout
But
Chapter 3 Diffusion in concentrated solutions page 3-4
j1rRout Ddc1
dr rRout
Dc1sat
Rout
1
Rout
Rin
1
(b) The mass balance for the Maxwell-Stefan equation is unchanged. Now assuming ideal
solutions with constant total concentration c, we find from Eq. 3.2-6 or 3.2-7,
c1c1c2
cD v2v1
CD c1n2c2n1
1
C
Dc1j2c2j1
The solution is dilute and the polymer isn’t moving much so c1 and j2 are near zero. Thus c2
is about equal to c, and
This is the same as (a) so the solution is unchanged.
(c) We now must replace Fick’s law and the Maxwell-Stefan relation with
cD
12
‘v2v1
cD
13
‘v3v1
Details of this replacement are in Chapter 7.
Chapter 3 Diffusion in concentrated solutions page 3-5
8. Mass balance assumptions
(a) Steady flow and unsteady mass transfer in a tube.
(c) The terms are
accumulation = (axial flow in – out)
+ (radial diffusion in – out)
+ (axial diffusion in – out).
Radial, diffusion-induced flow is sensibly neglected.
9. Porous solid desiccant
2
0
0.5
1
3
02468
t
1/2
When using a semi-infinite slab model, from Eq.
The absorbed amount
M = 0
t N1Adt = 2 Dt A/V1
As observed on the right figure, this model does
not fit the experimental data. Instead, as the
problem states, we assume that the slab contains
cylindrical pores in which water can diffuse fast.
The flux out of these pores is given by Eq. 5.80
in Crank’s book (p. 87):
D
where a is the radius of the pore. Integration gives
t j1dt = M0 + A
4Dt
As the figure shows, these data fit very well to this equation. Thus,
c = 0.1759 g/min = 0.00126 mol/s
The surface area of the slab is
At 45 °C, p(sat) = 0.0946 atm,
Chapter 3 Diffusion in concentrated solutions page 3-6
D =
0.00126
2*840*7.2510-7
2
= 3.37 cm /s
2
10. Leaching of copper
Using Ex. 3.5-3, c1
t = Deff
r2
r r2 c1
r
subject to t = 0, all r, c1 = c10
The solution of this problem is (From Carslaw and Jaeger)
and the total amount leached out should be
2 0
t n1|r = R0dt = 0.8 4R0
3
From Crank, p. 93, => – Dt
11. Concentration profile in a slab
The mass transfer equation is (from Eq. (C) in Table 3.4-2)
Using Laplace transformation, these equations become
where = (c1 – c10)/c10, = z/b, and = tD/b2. Thus
The solution of Eq. (4) is
Chapter 3 Diffusion in concentrated solutions page 3-7
The final form of the solution is
(-1)n
12. Graham’s experiment
The basic equation is
bc c1(0, t) = 0 (a simplified condition) (2)
This is the semi-infinite slab, and the solution is Eq. 2.3-18
The concentration variation in the large beaker is thus
V Dt
13. Wool dyeing
This is a cylindrical diffusion problem, known as “Diffusion from a stirred solution of limited
volume”. Thus from section 5.3.3 in Crank’s book, the total amount of dyestuff in the wool
fiber, M(t), is
n=1
4(1 + )
where M is M at infinite t,
where V is dye bath’s volume, and H is the partition coefficient, and qn are the positive roots of
Chapter 3 Diffusion in concentrated solutions page 3-8
The dye concentration in the bath is
Therefore, by plotting (1 – C1/C10) vs t1/2, the initial slope is 4(1 + 1/)D/R2.
14. Truffles
This is a point source diffusion problem. The basic equation is
t = D
r2
r r2 c1
r
subject to c1(r,0) = 0
The solution can be found in Carslaw and Jaeger (p. 261) as
4Dr erfc r
4Dt
To meet the boundary condition of the ground surface, define
Finally, the flux is j1|z=0 = –D c1
z|z=0 =
J0d
r0
– erfc(r0/ 4Dt)
r2
0 + exp(-r2
0/ 4Dt)
r0Dt
15. Two immiscible substances in contact
This is a two semi-infinite slab problem.
In slab A, c1A
t = DA2c1A
z2
In slab B, c1B
t = DB2c1B
z2
These are subject to t = 0, z > 0, c1A = c1A0,
The solution of this problem is given in Carslaw and Jaeger, p. 88:
16. Flux away from a dissolving drop
For a dissolving drop, the basic equation is
For the pseudo-steady state case where the drop has constant size r, c1