chemical

1. 3.4 Simple Diffusion models

System in which diffusion take place can be modeled by
starting with one of the material balance equation appearing
in either table 3-1 and 3-2
For table 3-1

2. A . Continuity equations of species A
∂ΡA∂t+(∂nAx∂x+∂nAy∂y+∂nAz∂z) = řA
Rectangular coordinates
Cylindrical coordinates
∂ΡA∂t+(1r*∂₍rnAr₎∂r+1r*∂nAѲ∂Ѳ+∂nAz∂z) = řA
(
(A
(
(B

3. spherical coordinates
∂ΡA∂t+(1r.r*∂₍r.rnAr₎∂r+1rsinѲ*∂nAѲsinѲ∂Ѳ+1rsinѲ*∂nAφ∂φ) = řA
B . Continuity equations of species A for constant ρ and Dab*
Rectangular coordinates
∂ΡA∂t+(Űx.∂ρA∂x+Űy.∂ρA∂y+Űz.∂ρA∂z)=
DAB(∂∂ρAρx.x+∂∂ρA∂y.y+∂∂ρA∂z.z)+řA
(
(C
(
(D

4. Cylindrical coordinates
∂ΡA∂t+(Űr.∂ρA∂r+ŰѲ.1r.∂ρA∂Ѳ+Űz.∂ρA∂z)=
DAB(1r.∂∂r₍r.∂rA∂r₎+1r.r.∂∂ρA∂Ѳ.Ѳ+∂∂ρA∂z.
z)+řA
spherical coordinates
∂ΡA∂t+(Űr.∂ρA∂r+ŰѲ.1r.∂ρA∂Ѳ+Űφ1rsinѲ.∂
ρA∂φ)=
DAB(1r.r.∂∂r₍r.r∂ρA∂r₎+1r.r.sinѲ*∂∂Ѳ₍sinѲ*∂
ρA∂Ѳ₎+1r.rsin2Ѳ*∂∂ρA∂φ.φ)+řA
(
(E
(F)

5. A. Molar flux of species A in various coordinates system
Rectangular coordinates
∂CA∂t+(∂NAx∂x+∂NAy∂y+∂NAz∂z) = ŘA
Cylindrical coordinates
∂CA∂t+(1r*∂₍rNAr₎∂r+1r*∂NAѲ∂Ѳ+∂NAz∂z) = ŘA
For table 3-2
(
(A
(B)

6. spherical coordinates
∂CA∂t+(1r.r*∂₍r.rNAr₎∂r+1rsinѲ*∂NAѲsinѲ∂Ѳ+1rsinѲ*∂NAφ∂φ) = ŘA
B . Continuity equations of species A for constant ρ and Dab*
Rectangular coordinates
∂CA∂t+(Űx.∂CA∂x+Űy.∂CA∂y+Űz.∂CA∂z)=
DAB(∂∂CA∂x.x+∂∂CA∂y.y+∂∂CA∂z.z)+ŘA
Cylindrical coordinates
∂CA∂t+(Űr.∂CA∂r+ŰѲ.1r.∂CA∂Ѳ+Űz.∂CA∂z)=
DAB
(1r.∂∂r₍r.∂rA∂r₎+1r.r.∂∂CA∂Ѳ.Ѳ+∂∂CA∂zz)+Ř
A
(
(C
(
(D
(
(E

7. spherical coordinates
∂CA∂t+(Űr.∂CA∂r+ŰѲ.1r.∂CA∂Ѳ+Űφ1rsinѲ.∂
CA∂φ)=
DAB(1r.r.∂∂r₍r.r∂CA∂r₎+1r.r.sinѲ*∂∂Ѳ₍sinѲ*∂
CA∂Ѳ₎+1r.rsin2Ѳ*∂∂CA∂φ.φ)+ŘA
(
(F
And eliminating the terms that do not apply to the system to be
modeled
The other method for modeling diffusing system is the shell balance
method made popular by bird et al . (1960)
Since this method requires that a material balance be made over a
differential element in the direction of mass transfer
It is usually employed to model one –dimensional system

8. For system of complicated geometry the elimination of terms from the
general material balance is probably the better approach
In this section both methods will be used to formulate models for one-
dimensional diffusion system in rectangular ,cylindrical and spherical
coordinates
The reduction of the material balance equation and the shell balance
method will be demonstrated for the case in which a single
component is transferred

9. Reducing the general material
balance
Let us consider the steady-state evaporation of a liquid from a small
diameter tube through a stationary gas film
As shown in figure 3-3 if we
assume
D
Liquid A
D 1
∆Z
Z = Z1 CA
=
CA1
FIGURE 3-3
evaporation through a
stagnant film
Z = Z2 CA
=
CA2
Gas B

10. We have six
assumption
•
1- that d1<<d2 then the level of the liquid in the tube will remaine
constant
•
2- for the gas film to remain stationary, gas B must be insoluble in
liquid A or component B must be at is solubility limit in A
•
3- there is no chemical reaction
•
4- at constant temperature and pressure (constant total
concentration )
•
5- the gas mixture is ideal

11. •
6- mass is transferred only in the Z direction
∂NAX∂X and ∂NAY∂Y equal to zero
At steady state conditions ∂CA∂t= 0
Thus we have ∂NAZ∂Z=0 (3-35)
But eq. (1-34) for the gas phase in the Z direction is
NBZ)
+
NAZ =-CDAB *dyAdz+yA(NAZ
If component B is stagnant then substituting NBZ =0
into the equation above and rearranging gives

12. NAZ=
-CDAB
dz
dyA
(3-37)
1-YA
By substituting eq. (3-36) into eq. (3-35) and assuming that the total
molar concentration and diffusivity are constant we obtain
d
dz
=0
1
1-YA
)
(
(3-36)
In order to complete the model boundary condition giving the
concentration of component A at the limit of the diffusion path
(the top of the tube and the gas – liquid interface )

13. By integrating eq. (3-37) once gives
YA=YA1 AT Z=Z1
(3-40)
YA=YA2 AT Z=Z2
(3-38)
That the concentration of A at the gas-liquid interface is the
concentration of A in the vapor that is in equilibrium with the
liguid
As shown in figure 3-3 the boundary condition
are
1
1-YA
dyA
dz =k1
(3-39)
By integrating eq. (3-40) once gives
Ln (1-yA)=k1z+k2 (3-41)

14. the molar flux , NAZ , at the liquid surface can be obtained by
evaluating eq. (3-36) at z = z1
ln
Z-z1
1-YA1
=
1-YA
CD AB
ln
1-YA2
1-YA1
(3-42)
After substituting the boundary condition given by eq. (3-38) and (3-39)
into eq. (3-41) ,the constant of integration ,k1 and k2, can be evaluated
The concentration profile for species A over the lenth of the tube as
NAZ|z=z1 =
Z-z1
1-YA
dyA
dz
|z=z1 (3-43)

15. Differentiate eq. ( 3-42) and evaluated the result at z = z1 the molar flux
at the gas-liquid interface is represented by
NAZ|z=z1
=
CD AB
Z2-z1
ln
1-YA2
1-YA1
(3-44)
NAZ|z=z1 =
CD AB
Z2-z1
ln
YB2
YB1
(3-45)

16. The flux can be equated to a mass transfer coefficient and a linear driving
force by defining the logarithmic mean mole fraction as
(YB)M =
YB2-YB1
(
Ln(YB2YB1
(3-46)
thus
NAZ|z=z1 =
CD AB
(z2-z1)(YB)M
(YB2-YB1) (3-47)
Therefore, a mass transfer coefficient can be defined in terms of the
diffusion coefficient and the length of the diffusion path by the
expression

17. ky
=
CD AB
(z2-z1)(YB)M
(3-49)
The equation above is similar to the boundary condition given in eq. (3-
33)
NAZ|z=z1 = Ky(YB2-YB1 )
(3-48)
Subsequently , the flux is given in terms of amass transfer coefficient and
a linear driving force as

18. If the gas mixture in the tube is ideal , the flux can be written in terms of
partial pressures at z1 and z2 since the total pressure is constant
P = P A1+PB1 = PA2+PB2
(3-51)
PA =YAP PB =YBP
(3-50)
and
thus for a binary system the molar flux can be expressed
as
NAZ|z=z1 =
PDAB
RT(Z2-Z1)(PB)M
(PB2-PB1) (3-52)