1. Diffusion of gases A and B Plus Convection
Just as momentum and energy (heat) transfer have two mechanisms for transport-molecular and convective, so does mass transfer. However, there are convective fluxes in mass transfer, even on a molecular level. The reason for this is that in mass transfer, whenever there is a driving force, there is always a net movement of the mass of a particular species which results in a bulk motion of molecules. Of course, there can also be convective mass transport due to macroscopic fluid motion. In this chapter the focus is on molecular mass transfer.
The mass (or molar) flux of a given species is a vector quantity denoting the amount of the particular species, in either mass or molar units, that passes per given increment of time through a unit area normal to the vector. The flux of species defined with reference to fixed spatial coordinates, NA is
(1)
This could be written in terms of diffusion velocity of A, (i.e.,  A - ) and average velocity of mixture, , as
(2)
Geankoplis chapter 6 more details
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2. (3) By definition Therefore, equation (2) becomes
For systems containing two components A and B,
(3)
The first term on the right hand side of this equation is diffusional molar flux of A ( or molecular diffusion ), and the second term is flux due to bulk motion (convection term).
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3. Diffusion of gases A and B Plus Convection
For diffusion in only the Z direction, the Fick’s rate equation is
where D AB is diffusivity or diffusion coefficient for component A diffusing through component B, and dCA / dZ is the concentration gradient in the Z-direction.
A more general flux relation could be written as
(4)
using this expression, Equation (3) could be written as
(5)
(Total flux)
= (molecular diffusion) + (convection)
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4. Absolute Molar and Mass Fluxes of Species A
Absolute Fluxes
Absolute Molar and Mass Fluxes of Species A
in a Binary Mixture of Species A and B
Molar Flux of Species A:
Mass Flux of Species A:
Special Case of Stationary Medium:
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5. Then Proof D AB = D BA?
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6. For a binary system containing A and B, from Equation (5),
Flux Relations diffusion
(6)
(7) or
-Can re written for B
(8)
(9)
Therefore Equation (9) becomes,
D AB = D BA ­ (10)
This leads to the conclusion that diffusivity of A in B is equal to diffusivity of B in A.
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7. Diffusivity, D AB
Fick’s law proportionality, D AB, is known as mass diffusivity (simply as diffusivity) or as the diffusion coefficient.
Diffusivity is normally reported in cm2 / sec; the SI unit being m2 / sec.
Diffusivity depends on pressure, temperature, and composition of the system.
In table, some values of DAB are given for a few gas, liquid, and solid systems.
Diffusivities of gases at low density are almost composition independent, increase with the temperature and vary inversely with pressure. Liquid and solid diffusivities are strongly concentration dependent and increase with temperature.
General range of values of diffusivity: Gases : X 10 – X m2 / sec.
Liquids : – m2 / sec.
Solids : X 10 – X m2 / sec
In the absence of experimental data, semi theoretical expressions have been developed which give approximation, sometimes as valid as experimental values, due to the difficulties encountered in experimental measurements.
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8. Diffusivity of gasses 1-Diffusion for binary gasses mixtures
Pressure dependence of diffusivity is given by
(for moderate ranges of pressures, upto 25 atm).And temperature dependency is according to
1-Diffusion for binary gasses mixtures
Lennard Jones Equation (Integral collision )
Is a collision integral based on the Lennard Jones potential
MA and MB are Molecular weight of A and B
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9. Diffusion for binary gasses mixtures (continue)
Calculation procedure
1-Find
2-From Table A find for A and B
(in K )
(in A0 )And corresponding
(in A0 ) And corresponding
(in K )
3-Estimate
and
4-Use values of step 3 to estimate
5-Use values of step 4 and Appendix A3.4 to find
6-Use the formula to find
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16. Gilliland method
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17. 2-Multi-component gasses
Diffusivity of a component in a mixture of components can be calculated using the diffusivities for the various binary pairs involved in the mixture. The relation given by Wilke is
Where D 1-mixture is the diffusivity for component 1 in the gas mixture; D 1-n is the diffusivity for the binary pair, component 1 diffusing through component n; and
is the mole fraction of component n in the gas mixture evaluated on a component –1 – free basis, that is
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18. Example. Determine the diffusivity of stagnant mixture Co 2 (1), O 2 (2) and N 2 (3) in a gas mixture having the composition:
Co2 : 28.5 %, O2 : 15%, N 2 : 56.5%,
The gas mixture is at 273 k and 1.2 * 10 5 Pa. The binary diffusivity values are given as: (at 273 K)
D 12 P = m 2 Pa/sec
D 13 P = m 2 Pa/sec
D 23 P = m 2 Pa/sec
Calculations:
Diffusivity of Co 2 in mixture
where
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19. Where Dr Mustafa Nasser Therefore = 1.93 m 2.Pa/sec
Since P = 1.2 * 10 5 Pa,
Diffusivity of O 2 in the mixture,
Where
(mole fraction on-2 free bans).
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20. and
and
D 21 P = D 12 P = m 2.Pa/sec
Therefore
= m 2.Pa/sec
By Similar calculations Diffusivity of N 2 in the mixture can be calculated, and is found to be, D 3m = * 10 –5 m 2/sec.
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21. Diffusion of gases A and B Plus Convection: Equimolar counter diffusion
A physical situation which is encountered in the distillation of two constituents whose molar latent heats of vaporization are essentially equal, stipulates that the flux of one gaseous component is equal to but acting in the opposite direction from the other gaseous component; that is, NA = - NB.
The molar flux NA, for a binary system at constant temperature and pressure is described by or
(1)
with the substitution of NB = - NA, Equation (1) becomes,
(2)
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22. For steady state diffusion Equation
For steady state diffusion Equation. (2) may be integrated, using the boundary conditions: at z = z CA = CA and z = z CA = CA2
Giving,
from which
(3)
For ideal gases,
Therefore Equation. (3) becomes
This is the equation of molar flux for steady-state equimolar counter diffusion.
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23. Example on equimolar counter diffusion
Figure 6.2-1
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24. For B find first pB1 and pB2
Negative why?
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25. Steady State Diffusion
Diffusion through a stagnant gas film
The case of diffusion of A through stagnant or nondiffusing B at steady state often occurs. In this case one boundary at end of the diffusion path is impermeable to component B, so it cannot pass trough. One example shown in figure (below) is an evaporation of pure liquid such as Benzene (A) at the bottom of narrow tube, where a large amount of inert or nondiffusing air (B) is passed over the top. The benzene vapour (A) diffuses through air (B) in the tube. The boundary of the at liquid surface at point 1 is impermeable to air, since air is insoluble in benzene liquid. Hence, air (B) cannot diffuse into or way from surface . at point 2 the partial pressure pA2 = 0 since large volume of air is passing by.
Another example is shown in the same fig. occurs in the adsorption of NH3 (A) vapour which is in air (B) by water. The water surface is impermeable to air, since air is only very slight soluble in water.
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26. From the general equation
Since N B = 0,
Rearranging,
This equation may be integrated between the two boundary conditions:
at z = z YA = YA1
And at z = z YA = yA2
Assuming the diffusivity is to be independent of concentration, and realizing that NA is constant along the diffusion path, by integrating the above equation we obtain
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27. The above Final form can be given
The log mean average concentration of component B is defined as
Since ,
Final form
For an ideal gas
The above Final form can be given
, and for mixture of ideal gases
Velocity vA=NACA
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28. where or
This is the equation of molar flux for steady state diffusion of one gas through a second stagnant gas.
Many mass-transfer operations involve the diffusion of one gas component through another non-diffusing component; absorption and humidification are typical operations defined by these equation.
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29. Please note for the above cases: for Two components
We use diffusion coefficient as DAB , for multi-components we use DA,mixture
This is same
as before
This diffusion of gas through a mixture of stagnant gas film
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32. Example: Methane diffuses at steady state through a tube containing helium. At point 1 the partial pressure of methane is p A1 = 55 kPa and at point 2, 0.03 m apart P A2 = 15 KPa. The total pressure is kPa, and the temperature is 298 K. At this pressure and temperature, the value of diffusivity is 6.75 * 10 –5 m 2/sec.
1)Calculate the flux of CH 4 at steady state for equimolar counter diffusion
2)Calculate the partial pressure at a point 0.02 m apart from point 1.
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33. Calculation:
For steady state equimolar counter diffusion, molar flux is given by
And from (1), partial pressure at 0.02 m from point 1 is:
p A = kPa
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