Chemistry Reference
In-Depth Information
6.2
Fick's Laws
The concept of mass transfer had not been totally understood by 1850. In the
1830s, a chemist named Thomas Graham published several sets of data to
show that there exists a relationship between mass flux and concentration with-
out introducing any mathematical model to correlate the data
[8]
. Later, in
1855, based on the idea that Fourier used for analyzing heat conduction prob-
lems, Adolf Fick proposed his famous model of mass transfer using Thomas
Graham's data:
D
dC
dx
J
52
(6-1)
m
4
is the concentration gradient
of the diffusive component in the mass transfer direction (i.e., positive x direction
in the context of the above equation), and D m
2
is the diffusive mass flux,
d
dx
kg
m
2
s
kg
where J
is the Fickian diffusion coef-
ficient.
Equation (6-1)
is known as Fick's first law of mass transfer that is the
basis for many engineering mass transfer calculations. According to
Eq. (6-1)
, the
driving force of mass transfer is the gradient of concentration and the minus sign
indicates that mass is transferred from regions with higher concentrations to those
with lower concentrations. In
Section 6.4
, it will be shown that the actual driving
force for mass transfer is the gradient of chemical potential and Fick's first law is
only exact in the case of ideal mixtures in which different components interact
with each other with similar intermolecular interactions.
Fick's first law does not include any information about the time variation
of the concentration in the system. To analyze the dynamics (time variation) of
a mass transfer process, the mass conservation law is applied.
Figure 6.1
shows
a volume element in a one-dimensional diffusion process in which there
exist no gradients of concentrations in the y and z directions. The thickness of
the element is 2
=
s
Δ
x and the cross-sectional area, which is normal to the mass
transfer direction, is A. When there is no chemical reaction taking place in the
volume element,
the law of conservation of mass yields
the following
equation:
input mass
2
output mass
5
accumulation of mass in the volume element
dm
dt
5
dC
dt
J
ð
x
2Δ
x
Þ 3
A
2
J
ð
x
1Δ
x
Þ 3
A
5
A
3
2
Δ
x
3
(6-2)
Dividing both sides of
Eq. (6-2)
by the volume of the volume element
(i.e., 2
Δ
x
3
A) results in the following equation:
J
ð
x
2Δ
x
Þ 2
J
ð
x
1Δ
x
Þ
dC
dt
5
(6-3)
Δ
2
x
