Page 587 - Bird R.B. Transport phenomena

§18.8 Diffusion in a Three-Component Gas System 567

have common asymptotes for large and small Л and do not differ from one another very
much for intermediate values of Л. Thus Fig. 18.7-3 provides a justification for the use of
Eq. 18.7-16 to estimate rj A for nonspherical particles.

§18„8 DIFFUSION IN A THREE-COMPONENT GAS SYSTEM
Up to this point the systems we have discussed have been binary systems, or ones that
could be approximated as two-component systems. To illustrate the setting up of multi-
component diffusion problems for gases, we rework the initial evaporation problem of
§18.2 when liquid water (species 1) is evaporating into air, regarded as a binary mixture
of nitrogen (2) and oxygen (3) at 1 atm and 352K. We take the air-water interface to be at
z = 0 and the top end of the diffusion tube to be at z = L. We consider the vapor pressure
of water to be known, so that x x is known at z = 0 (that is, x 10 = 341/760 = 0.449), and
the mole fractions of all three gases are known at z = L: x u = 0.10, x 2L = 0.75, x 3L = 0.15.
The diffusion tube has a length L = 11.2 cm.
The conservation of mass leads, as in §18.2, to the following expressions:
dN
- y ^ = 0 a = 1,2,3 (18.8-1)
dz
From this it may be concluded that the molar fluxes of the three species are all constants
at steady state. Since species 2 and 3 are not moving, we conclude that N and N are
2z 3z
both zero.
Next we need the expressions for the molar fluxes from Eq. 17.9-1. Since x + x +
x 2
x 3 = 1, we need only two of the three available equations, and we select the equations for
species 2 and 3. Since N 2z = 0 and N 3z = 0, these equations simplify considerably:

d X l N u dX3 N u
- г- ~ г П8 8? Ъ
—:— — —zr— X 2 , —.— — ~^z— %1 VlO.o-Z, o)
dz c% 2 l dz c% 3 5
Note that the diffusivity 2) з does not appear here, because there is no relative motion of
2
species 2 and 3. These equations can be integrated from an arbitrary height z to the top of
the tube at L, to give for constant c2)
aj3
Гр^Г г (18.8-4,5)
*2 а
X2 C% 2 J Z ) *3 C%J
Integration then gives
il = e JJMlZl>) ; |1 ex f-^|^) (18.8-6,7)
=
P
*2L \ C2) 1 2 / *3L \ C2) 1 3 /
and the mole fraction profile of water vapor in the diffusion column will be

(18.8-8)

When we apply the boundary condition at z = 0, we get
/ \T.T\
(18.8-9)
which is a transcendental equation for N .
lz