NPTEL – Chemical – Mass Transfer Operation 1
MODULE 2: DIFFUSION
LECTURE NO. 3
2.3 Diffusion through variable cross-sectional area
2.3.1 Diffusion through a conduit of non-uniform cross-sectional
area
Consider a component A is diffusing at steady state through a circular conduit
which is tapered uniformly as shown in Figure 2.3. At point 1 the radius is r 1 and
at point 2 it is r2. At position Z in the conduit, A is diffusing through stagnant, nondiffusing B.
Figure 2.3: Schematic of diffusion of A through a uniformly tapered tube
At position Z the flux can be written as
NA =
N A -DAB
dPA
=
2
πr
RT (1- PA /P) dz
(2.23)
Using the geometry as shown, the variable radius r can be related to position z in
the path as follows:
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NPTEL – Chemical – Mass Transfer Operation 1
 r -r 
r =  2 1  z + r1
 z 2 - z1 
(2.24)
The Equation (2.24) is then substituted in the flux Equation to eliminate r and
then the Equation is integrated and obtained as:
NA
π
z2

z1
dz
-D
= AB
2
RT
 r2 - r1 

z
+
r


1
 z 2 - z1 

PA2
dPA
 1- P /P
PA1
A
Or
N A  1 1  DAB P P  PA2
  
ln
4  r1 r2 
RT
P  PA1
(2.25)
2.3.2 Evaporation of water from metal tube
Suppose water in the bottom of a narrow metal tube is held at constant
temperature T. The total pressure of air (assumed dry) is P and the temperature
is T. Water evaporates and diffuses through the air. At a given time t, the level is
Z meter from the top as shown in Figure 2.4. As diffusion proceeds, the level
drops slowly. At any time t, the steady state Equation holds, but the path length is
Z.
Figure 2.4: Schematic of evaporation in metal tube
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Thus the steady state Equation becomes as follows where NA and Z are
variables:
DAB P
( PA1  PA2 )
RT Z PBM
NA 
(2.26)
Where
PBM 
PB2  PB1
ln( PB2 / PB1 )

PA1  PA2
ln[( P  PA2 ) /( P  PA1 )]
(2.27)
Assuming a cross sectional area of 1 m2, the level drops dZ meter in dt sec, and
ρA(dZ.1)/MA is the kmol of A that has been left and diffused. Then
N A .1 
 A (dZ .1)
(2.28)
M A .dt
Substituting the Equation (2.28) in Equation (2.26) and integrating, one gets
A
ZF
MA
 Z dZ 
Z0
tF 
DAB P ( PA1  PA2 ) tF
RT PBM
 dt
(2.29)
0
 A ( Z F2  Z 02 ) RT PBM
2 M A DAB P ( PA1  PA2 )
(2.30)
The Equation (2.30) represents the time tF for the level to drop from a starting
point of Zo meter at t = 0 to ZF at t = tF.
2.3.3 Diffusion from a sphere
There are lots of examples where diffusion can take place through the spherical
shape bodies. Some examples are:

Evaporation of a drop of liquid

The evaporation of a ball of naphthalene

The diffusion of nutrients to a sphere-like microorganism in a liquid
Assume a constant number of moles ÑA of A from a sphere (area = 4r2) through
stagnant B as shown in Figure 2.5.
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NPTEL – Chemical – Mass Transfer Operation 1
Figure 2.5: Schematic of diffusion from a sphere
From the Fick’s law of diffusion, the rate of diffusion can be expressed as:
 P  -D dP
N A 1- A  = AB A
 Ptotal  RT dr
NA
where N A =
4πr 2
-RTN A dr
dPA

 Ptotal
2 NA
PA  -D AB dPA
4πD AB
Ptotal  PA
where
N Ar =
=
Ptotal  RT dr
4πr 2
N A dr
dPA
 RT N A dr
dPA
 Ptotal
2
 Ptotal
Ptotal  PA
2
AB r
4DAB r
( Ptotal  PA )
r2
-RTN A  1 
P
Integrating
with
limits
oftotal
PA2 P
atA )r2PA2and PA1 at r1 gives:
 Ptotal
  ln(P


A1
4πDAB  r  r1
 Ptotal  PA2 
-RTN A  1 1 


    ln 
4πDAB Ptotal  r1 r2 
 Ptotal  PA1 
 P  PA2  PA1  PA2
RTN A
As r1 <<
r2, then
1/r2 ≈ 0: 
ln  total
PBM
4πr1D AB Ptotal
 Ptotal  PA1 
N A D AB Ptotal (PA1  PA2 )


 N A1 , the flux at the surface
4πr12
RTPBM r1
(2.31)
(2.32)
(2.33)
(2.34)
(2.35)
This Equation can be simplified if PA1 is small compared to P (a dilute gas
phase), PBM ≈ P. Also setting 2r1 = D1, diameter, CA1 = PA1/RT
N A1 =
DAB
(CA1 - CA2 )
D1
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(2.36)
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