Ministry of Higher

Education and
Scientific Research

Advanced Mass Transfer

Master Course

By
Prof. Dr. Ahmed Daham Wiheeb

2020 - 2021
Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

17.2 Temperature and Pressure Dependence of Diffusivities

For binary gas mixtures at low pressure, DAB is inversely proportional to the pressure,
increases with increasing temperature, and is almost independent of the composition for a given
gas pair.

Where:

𝒄𝒎𝟐
𝑫𝑨𝑩 in ( )
𝒔

𝒑 in (𝒂𝒕𝒎)

𝑻 in (𝐊)

The dimensionless constants (a and b) for nonpolar gas pairs excluding helium and hydrogen
are:

𝒂 = 𝟐. 𝟕𝟒𝟓 × 𝟏𝟎−𝟒

𝒃 = 𝟏. 𝟖𝟐𝟑

While the dimensionless constants (a and b) for pairs consisting of H2O and a nonpolar gas
are:

𝒂 = 𝟑. 𝟔𝟒𝟎𝟓 × 𝟏𝟎−𝟒

𝒃 = 𝟐. 𝟑𝟑𝟒

At high pressures, and in the liquid state, the behavior of 𝑫𝑨𝑩 is more complicated. A
corresponding-states plot of the self-diffusivity 𝑫𝑨𝑨∗ for nonpolar substances is given in Fig.
17.2-1. The ordinate is (𝒄 𝑫𝑨𝑨∗ ) at pressure p and temperature T, divided by (𝒄 𝑫𝑨𝑨∗ )𝒄 at the
critical point. This quantity is plotted as a function of the reduced pressure (pr = p/pc), and the
reduced temperature (Tr = T/Tc). Because of the similarity of species A and the labeled
species A* ,the critical properties are all taken as those of species A.

From Fig. 17.2-1 we see that (𝒄 𝑫𝑨𝑨∗ ) increases strongly with temperature, especially for
liquids. At each temperature (𝒄 𝑫𝑨𝑨∗ ) decreases toward zero with increasing pressure. With
decreasing pressure (𝒄 𝑫𝑨𝑨∗ ) increases toward a low-pressure limit.

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 Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

The quantity (𝒄 𝑫𝑨𝑨∗ )𝒄 may be estimated by one of the following three methods:

i. Given (𝒄 𝑫𝑨𝑨∗ ) at a known temperature and pressure, one can read the reduced self-
diffusivity (𝒄 𝑫𝑨𝑨∗ )𝒓 from the chart and get:

(𝒄 𝑫𝑨𝑨∗ )
(𝒄 𝑫𝑨𝑨∗ )𝒄 =
(𝒄 𝑫𝑨𝑨∗ )𝒓

ii. One can predict a value of (𝒄 𝑫𝑨𝑨∗ ) in the low-density region by the methods
given in section 17.3 and then proceed as in (i).

iii. One can use the empirical formula (see Problem 17A.9):

This equation, like Eq. 17.2-1, should not be used for helium or hydrogen isotopes.

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

Where:

𝒄 : is the concentration in (gmol/cm3).

𝒄𝒎𝟐
𝑫𝑨𝑨∗ : is the self diffusivity in ( )
𝒔

𝒑𝒄 ∶ is the crectical pressure in (atm)

𝑻𝒄 ∶ is the crectical temperature in (K)

(𝒄 𝑫𝑨𝑨∗ )𝒓 = (𝝆 𝑫𝑨𝑨∗ )𝒓 for Ar, Kr, Xe, and CH4.

We turn now to the binary diffusion of chemically dissimilar species. In the absence of other
information it is suggested that Fig. 17.2-1 may be used for crude estimation of (𝒄 𝑫𝑨𝑩 ) with
𝒑𝒄𝑨 and 𝑻𝒄𝑨 replaced everywhere by √𝒑𝒄𝑨 𝒑𝒄𝑩 and √𝑻𝒄𝑨 𝑻𝒄𝑩 respectively (see Problem
17A.9 for the basis for this empiricism). The ordinate of the plot is then interpreted as:

(𝒄 𝑫𝑨𝑩 )
(𝒄 𝑫𝑨𝑩 )𝒓 =
(𝒄 𝑫𝑨𝑩 )𝒄

and 17.2-2 is replaced by:

Example 17.2-1: Estimation of Diffusivity at Low Density

Solution:

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

Example 17.2-2: Estimation of Self-Diffusivity at High Density

Solution:

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

Example 17.2-3: Estimation of Binary Diffusivity at High Density

Solution:

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

17.3 Theory of Diffusion in Gases at Low Density

The Chapman-Enskog kinetic theory should be used to find the mass diffusivity for
binary mixtures of nonpolar gases. The corresponding formula for ( 𝑫𝑨𝑩 ) is:

Where:

𝒄𝒎𝟐
𝑫𝑨𝑩 : is the mass diffusivity in ( )
𝒔

𝒑 ∶ is the pressure in (atm)

𝑻 ∶ is the temperature in (K)

𝝈𝑨𝑩 and 𝜺𝑨𝑩 : are the Lennard-Jones potential parameters

𝟏
𝝈𝑨𝑩 ∶ is the average collision diamete in (Å) → 𝝈𝑨𝑩 = (𝝈𝑨 + 𝝈𝑩 )
𝟐

𝞨𝑫,𝑨𝑩 ∶ is the collisional integral for diffusion

𝐊𝑻
and is a function of the dimensionless temperature ( )
𝜺𝑨𝑩
𝐊𝑻
* We can find 𝞨𝑫,𝑨𝑩 from Table E.2 (p. 866) as a function of (𝜺 )
𝑨𝑩

𝜺𝑨𝑩 ∶ is the maximum attractive energy between the molecules .

𝜺𝑨𝑩 = √𝜺𝑨 𝜺𝑩

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

Example 17.3-1: Computation of Mass Diffusivity for Low Density Gases

Solution:

17.7 Mass and Molar Transport by Convection

• Mass and Molar Concentrations

Mass concentration (𝝆𝜶 ) ∶ is the mass of species 𝛂 per unit volume of solution.

Molar concentration (𝑪𝜶 ) ∶ is the number of moles of species 𝛂 per unit volume of solution.
𝝆𝜶
𝑪𝜶 =
𝑴𝜶

Mass fraction (𝒘𝜶 ) ∶ is the mass of species 𝛂 per total mass of all species per unit
volume of solution .

𝝆𝜶
𝒘𝜶 =
𝝆
𝑁

𝝆 = ∑ 𝝆𝜶
𝜶=1

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

Mole fraction (𝒙𝜶 ) ∶ is the mole of species 𝛂 per total moles of all species per unit
volume of solution .

𝑪𝜶
𝒙𝜶 =
𝑪
𝑁

𝑪 = ∑ 𝑪𝜶
𝜶=1

See Table 17.7-1 (p. 534) Notation of Concentration.

• Mass Average and Molar Average Velocity

𝐕𝜶 = Velocity of species 𝜶
∶ is the average of all the velocities of molecules of species 𝜶
within a small volume.

Then, for a mixture of N species, the local mass average velocity V is defined as:

Where:

𝝆 𝐯 = is the local rate at which mass passes through a unit cross section placed
perpendicular to the velocity 𝐯.

Where:

𝐯 ∗ = is a local molar average velocity

𝒄 𝐯 ∗ = is the local rate at which moles pass through a unit cross section placed
perpendicular to the molar velocity 𝐯 ∗
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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

See Table 17.7-2 (p. 535) Notation for Velocities in Multicomponent Systems.

• Molecular Mass and Molar Fluxes

Molecular mass flux of species 𝜶 : is the flow of mass of 𝜶 a through a unit area per unit
time (kg/m2 . s).

𝒋𝜶 = 𝝆𝜶 (𝑽𝜶 − 𝑉)

That is, we include only the velocity of species 𝜶 relative to the mass average velocity v.

Molecular molar flux of species 𝜶 : is the number of moles of species 𝜶 flowing through
a unit area per unit time (kmol/m2 . s).

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ )

That is, we include only the velocity of species 𝜶 relative to the molar average velocity 𝐯 ∗ .

Fick's (first) law of diffusion describes how the mass of species A in a binary mixture is
transported by means of molecular motions. This law can also be expressed in molar units.
Hence we have the pair of relations for binary systems:

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

(𝐯𝜶 − 𝐯) sometimes referred to as diffusion velocities

(𝐯𝑨 − 𝐯 ∗ )

• Summary of Mass and Molar Fluxes

𝒘𝑨 = Mass fraction of 𝐴

𝒘𝑩 = Mass fraction of 𝐵

𝒙𝑨 = Mole fraction of 𝐴

𝒙𝑩 = Mole fraction of 𝐵

𝑴𝑨 = Molecular weight of 𝐴

𝑴𝑩 = Molecular weight of 𝐵
𝝆
𝑴 = Average molecular weight =
𝑪
𝒙𝑨 𝑴𝑨 𝒙𝑨 𝑴𝑨
𝒘𝑨 = =
𝒙𝑨 𝑴𝑨 + 𝒙𝑩 𝑴𝑩 𝑴

𝐯𝑨 = Velocity of species A (𝑚⁄𝑠)

𝐯 = Mass average velocity (𝑚⁄𝑠)

𝐯 ∗ = Molar average velocity (𝑚⁄𝑠)

𝝆𝑨 = Density of species A (mass of A ⁄unit volume of mixture), (𝑘𝑔⁄𝑚3 )

𝑪𝑨 = Molar concentration of species A (𝑚𝑜𝑙𝑒𝑠⁄𝑚3 )

𝑪 = Total molar concentration (𝑚𝑜𝑙𝑒𝑠 ⁄𝑚3 ) , (molar density of solution)

𝒋𝑨 = Mass flux of A , (𝑘𝑔⁄𝑚2 . 𝑠), with respect to the mass average velocity 𝐯

𝒋𝑨 = 𝝆𝑨 (𝑽𝑨 − 𝑉) = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝒋𝑩 = Mass flux of B , (𝑘𝑔⁄𝑚2 . 𝑠)

𝑱∗𝑨 = Molar flux of A , (𝑚𝑜𝑙𝑒⁄𝑚2 . 𝑠), with respect to the molar average velocity 𝐯 ∗

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

(𝑽𝑨 − 𝑉): Diffusion velocity of species A with respect to mass average velocity 𝐯
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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

(𝐯𝑨 − 𝐯 ∗ ): Diffusion velocity of species A with respect to molar average velocity 𝐯 ∗

𝝆𝑨
𝒄𝑨 =
𝑴𝑨
𝝆𝑨 𝒄𝑨
𝒘𝑨 = , 𝒙𝑨 =
𝝆 𝒄
𝑁

𝝆 = ∑ 𝝆𝜶
𝜶=1

𝑁
𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝐯 = ∑ 𝒘𝜶 𝐯𝜶 , 𝐯=
𝝆
𝜶=1

𝑁
∗
𝐯 = ∑ 𝒙𝜶 𝐯𝜶
𝜶=1

𝐯 − 𝐯 = ∑ 𝒘𝜶 (𝐯𝜶 − 𝐯 ∗ )
∗

𝜶=1

= 𝒘𝑨 (𝐯𝑨 − 𝐯 ∗ ) + 𝒘𝑩 (𝐯𝑩 − 𝐯 ∗ )

𝑁
∗
𝐕 − 𝑉 = ∑ 𝒙𝜶 (𝑽𝜶 − 𝑉)
𝜶=1

= 𝒙𝑨 (𝐕𝑨 − 𝑉) + 𝒙𝑩 (𝐕𝑩 − 𝑉)

𝝆𝑨 = 𝒘𝑨 𝝆 & 𝒄𝑨 = 𝒙𝑨 𝒄

𝝆𝑩 = 𝒘𝑩 𝝆 & 𝒄𝑩 = 𝒙𝑩 𝒄

𝐧𝑨 = 𝝆𝑨 𝐯𝑨 = Mass flux of species 𝑨 with respect to stationary axes.

𝐧𝑩 = 𝝆𝑩 𝐯𝑩 = Mass flux of species 𝑩 with respect to stationary axes.

𝐍𝑨 = 𝒄𝑨 𝐯𝑨 = Molar flux of species 𝑨 with respect to stationary axes.

𝐍𝑩 = 𝒄𝑩 𝐯𝑩 = Molar flux of species 𝑩 with respect to stationary axes.

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

• We add together the molecular mass flux vector and the convective mass flux vector to
get the combined mass flux vector, and similarly for the combined molar flux vector:

𝐧𝑨 = 𝒋𝑨 + 𝝆𝑨 𝐯

𝐍𝑨 = 𝑱𝑨 + 𝒄𝑨 𝐯 ∗

• Relation between mass and molar fluxes:

𝐧𝑨 = 𝑴𝑨 𝐍𝑨

• Relation between mass fraction and molar fraction gradients:

1
𝑴𝑨 𝑴𝑩 𝞩 𝒘𝑨
𝞩 𝒙𝑨 =
𝒘 𝒘 2
(𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

𝑴𝑨 𝑴𝑩 𝞩 𝒙𝑨
𝞩 𝒘𝑨 =
(𝒙𝑨 𝑴𝑨 + 𝒙𝑩 𝑴𝑩 )2

Example: Gas mixture of O2 and N2 with ( 20% of O2 & 80% of N2). Given that;

𝑉𝑂2 = 1 𝑚⁄𝑠 𝑎𝑛𝑑 𝑉𝑁2 = 2 𝑚⁄𝑠

𝑝 = 1 𝑎𝑡𝑚 𝑎𝑛𝑑 𝑇 = 300 K

𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑂2 = 32 𝑘𝑔⁄𝑘𝑚𝑜𝑙

𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑁2 = 28 𝑘𝑔⁄𝑘𝑚𝑜𝑙

Find:

1. Mass average velocity, 𝑉 ( 𝑚⁄𝑠).

2. Molar average velocity, 𝐯 ∗ ( 𝑚⁄𝑠).
3. Diffusion velocity of N2 relative to, 𝑉.
4. Diffusion velocity of O2 relative to, 𝐯 ∗ .
5. Mass flux of A and B with respect to stationary axes, 𝑉 and 𝐯 ∗ .
6. Molar flux of A and B with respect to stationary axes, 𝑉 and 𝐯 ∗ .

Solution:

𝐿𝑒𝑡: 𝑶𝟐 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑨 & 𝑵𝟐 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑩

𝑥𝐴 = 0.2 & 𝑥𝐵 = 0.8

𝑥𝐴 𝑀𝐴 𝑥𝐴 𝑀𝐴
𝑤𝐴 = =
𝑥𝐴 𝑀𝐴 + 𝑥𝐵 𝑀𝐵 𝑀
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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑀 = 𝑥𝐴 𝑀𝐴 + 𝑥𝐵 𝑀𝐵 = 0.2 ∗ 32 + 0.8 ∗ 28 = 28.8

0.2 ∗ 32
𝑤𝐴 = = 0.222
28.8

For binary mixture:

𝑤𝐵 = 1 − 𝑤𝐴 = 1 − 0.222 = 0.778

𝑥𝐵 𝑀𝐵 0.8 ∗ 28
𝑜𝑟, 𝑤𝐵 = = = 0.778
𝑀 28.8

1. Mass average velocity , 𝑉 ( 𝑚⁄𝑠).

𝜌𝐴 𝑉𝐴 + 𝜌𝐵 𝑉𝐵
𝑉= = 𝑤𝐴 𝑉𝐴 + 𝑤𝐵 𝑉𝐵 = 0.222 ∗ 1 + 0.778 ∗ 2 = 1.778 𝑚⁄𝑠
𝜌

2. Molar average velocity, 𝐯 ∗ ( 𝑚𝑜𝑙𝑒⁄𝑠).

𝑐𝐴 v𝐴 + 𝑐𝐵 v𝐵
𝐯∗ = = 𝑥𝐴 v𝐴 + 𝑥𝐵 v𝐵 = 0.2 ∗ 1 + 0.8 ∗ 2 = 1.8 𝑚⁄𝑠
𝑐

3. Diffusion velocity of N2 relative to, 𝑉.

= 𝑉𝐵 – 𝑉
= 2 − 1.778 = + 0.222 𝑚⁄𝑠

4. Diffusion velocity of O2 relative to, 𝐯 ∗ .

= v𝐴 – v ∗
= 1 − 1.8 = − 0.8 𝑚⁄𝑠

𝑝𝑀 10.1.3 ∗ 28.8
𝜌= = = 1.17 𝑘𝑔⁄𝑚3
𝑅𝑇 8.314 ∗ 300

𝜌 1.17
𝑐= = = 0.0406 𝑘𝑚𝑜𝑙 ⁄𝑚3
𝑀 28.8

𝜌𝐴 = 𝑤𝐴 𝜌 = 0.222 ∗ 1.17 = 0.26 𝑘𝑔⁄𝑚3

𝜌𝐵 = 𝑤𝐵 𝜌 = 0.778 ∗ 1.17 = 0.91 𝑘𝑔⁄𝑚3

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑜𝑟, 𝜌𝐵 = 𝜌 − 𝜌𝐴 = 1.17 − 0.26 = 0.91 𝑘𝑔⁄𝑚3

𝑐𝐴 = 𝑥𝐴 𝑐 = 0.2 ∗ 0.0406 = 0.00812 𝑘𝑚𝑜𝑙 ⁄𝑚3

𝑐𝐵 = 𝑐 − 𝑐𝐴 = 0.04067 − 0.00812 = 0.03248 𝑘𝑔⁄𝑚3

5. Mass flux of A and B with respect to stationary axes, 𝑉 and 𝐯 ∗ .

𝐧𝑨 = 𝝆𝑨 𝐯𝑨 = Mass flux of species 𝑨 with respect to stationary axes.

𝑘𝑔
= 0.26 ∗ 1 = 0.26
𝑚2 . 𝑠

𝐧𝑩 = 𝝆𝑩 𝐯𝑩 = Mass flux of species 𝑩 with respect to stationary axes.

𝑘𝑔
= 0.91 ∗ 2 = 1.82
𝑚2 . 𝑠

𝒋𝑨 = 𝝆𝑨 (𝐯𝑨 − 𝐯) = Mass flux of species 𝑨 with respect to 𝐯.

𝑘𝑔
= 0.26 ∗ (1 − 1.778) = −0.20228
𝑚2 . 𝑠

𝒋𝑩 = 𝝆𝑩 (𝐯𝑩 − 𝐯) = Mass flux of species 𝑩 with respect to 𝐯.

𝑘𝑔
= 0.91 ∗ (2 − 1.778) = 0.202
𝑚2 . 𝑠

𝒋∗𝑨 = 𝝆𝑨 (𝐯𝑨 − 𝐯 ∗ ) = Mass flux of species 𝑨 with respect to 𝐯 ∗ .

𝑘𝑔
= 0.26 ∗ (1 − 1.8) = −0.208
𝑚2 . 𝑠

𝒋∗𝑩 = 𝝆𝑩 (𝐯𝑩 − 𝐯 ∗ ) = Mass flux of species 𝑩 with respect to 𝐯 ∗ .

𝑘𝑔
= 0.91 ∗ (2 − 1.8) = 0.182
𝑚2 . 𝑠

6. Molar flux of A and B with respect to stationary axes, 𝑉 and 𝐯 ∗ .

𝐍𝑨 = 𝒄𝑨 𝐯𝑨 = Molar flux of species 𝑨 with respect to stationary axes.

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑘𝑚𝑜𝑙
= 0.00812 ∗ 1 = 0.00812
𝑚2 . 𝑠

𝐧𝑨 0.26 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝐍𝑨 = = = 0.00812
𝑴𝑨 32 𝑚2 . 𝑠

𝐍𝑩 = 𝒄𝑩 𝐯𝑩 = Molar flux of species 𝑩 with respect to stationary axes.

𝑘𝑚𝑜𝑙
= 0.03248 ∗ 2 = 0.06496
𝑚2 . 𝑠

𝐧𝑩 1.82 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝐍𝑩 = = = 0.0065
𝑴𝑩 28 𝑚2 . 𝑠

𝑱𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯) = Molar flux of species 𝑨 with respect to 𝐯.

𝑘𝑚𝑜𝑙
= 0.00182 ∗ (1 − 1.778) = −0.006317
𝑚2 . 𝑠

𝒋𝑨 −0.20228 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝑱𝑨 = = = −0.006317
𝑴𝑨 32 𝑚2 . 𝑠

𝑱𝑩 = 𝒄𝑩 (𝐯𝑩 − 𝐯) = Molar flux of species 𝑩 with respect to 𝐯.

𝑘𝑚𝑜𝑙
= 0.03248 ∗ (2 − 1.778) = 0.00721
𝑚2 . 𝑠

𝒋𝑩 0.202 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝑱𝑩 = = = 0.00721
𝑴𝑩 28 𝑚2 . 𝑠

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) = Molar flux of species 𝑨 with respect to 𝐯 ∗ .

𝑘𝑚𝑜𝑙
= 0.00812 ∗ (1 − 1.8) = −0.0065
𝑚2 . 𝑠
𝒋∗𝑨 −0.202 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝑱∗𝑨 = = = −0.0065
𝑴𝑨 32 𝑚2 . 𝑠

𝑱∗𝑩 = 𝒄𝑩 (𝐯𝑩 − 𝐯 ∗ ) = Molar flux of species 𝑩 with respect to 𝐯 ∗ .

𝑘𝑚𝑜𝑙
= 0.03248 ∗ (2 − 1.8) = 0.0065
𝑚2 . 𝑠
𝒋∗𝑩 −0.182 𝑘𝑚𝑜𝑙
𝑜𝑟, 𝑱∗𝑩 = = = 0.0065
𝑴𝑩 28 𝑚2 . 𝑠
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𝟏. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑱𝑨 + 𝑱𝑩 = 𝒄(𝐯 ∗ − 𝐯).

𝑱𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯) & 𝑱𝑩 = 𝒄𝑩 (𝐯𝑩 − 𝐯)

𝑱𝑩 + 𝑱𝑩 = 𝒄𝑨 (𝐯𝑨 − 𝐯) + 𝒄𝑩 (𝐯𝑩 − 𝐯)
= 𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯 + 𝒄𝑩 𝐯𝑩 − 𝒄𝑩 𝐯
= 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 − 𝒄𝑨 𝐯 − 𝒄𝑩 𝐯 = 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 – 𝐯 (𝒄𝑨 + 𝒄𝑩 ) … … … (1)
𝒘𝒉𝒆𝒓𝒆: 𝒄𝑨 + 𝒄𝑩 = 𝒄 & 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 = 𝒄 𝐯 ∗
Then Eq.(1) becomes:
𝑱𝑨 + 𝑱𝑩 = 𝒄 𝐯 ∗ – 𝐯 𝒄 → 𝑱𝑨 + 𝑱𝑩 = 𝒄( 𝐯 ∗ – 𝐯)

𝟐. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝐯 − 𝐯 ∗ = 𝒘𝑨 (𝐯𝑨 − 𝐯 ∗ ) + 𝒘𝑩 (𝐯𝑩 − 𝐯 ∗ ).

𝝆𝑨 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 𝝆𝑩 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩
𝐯 − 𝐯∗ = (𝐯𝑨 − )+ (𝐯𝑩 − )
𝝆 𝒄 𝝆 𝒄
𝝆𝑨 𝐯𝑨 𝝆𝑨 𝒄𝑨 𝐯𝑨 𝝆𝑨 𝒄𝑩 𝐯𝑩 𝝆𝑩 𝐯𝑩 𝝆𝑩 𝒄𝑨 𝐯𝑨 𝝆𝑩 𝒄𝑩 𝐯𝑩
= − − + − −
𝝆 𝝆𝒄 𝝆𝒄 𝝆 𝝆𝒄 𝝆𝒄
𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 𝝆𝑨 𝒄𝑨 𝐯𝑨 + 𝝆𝑨 𝒄𝑩 𝐯𝑩 + 𝝆𝑩 𝒄𝑨 𝐯𝑨 + 𝝆𝑩 𝒄𝑩 𝐯𝑩
= −
𝝆 𝝆𝒄
𝝆𝑨 (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) + 𝝆𝑩 (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 )
=𝐯−
𝝆𝒄
(𝝆𝑨 + 𝝆𝑩 ) (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) 𝝆 (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 )
=𝐯− =𝐯−
𝝆𝒄 𝝆𝒄

(𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) ∑ 𝒄𝒊 𝐯𝒊
=𝐯− =𝐯−
𝒄 𝒄
= 𝐯 − 𝐯∗

𝟑. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝐯 ∗ − 𝐯 = 𝒙𝑨 (𝐯𝑨 − 𝐯) + 𝒙𝑩 (𝐯𝑩 − 𝐯).

𝒄𝑨 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 𝒄𝑩 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝐯∗ − 𝐯 = (𝐯𝑨 − )+ (𝐯𝑩 − )
𝒄 𝝆 𝒄 𝝆
𝒄𝑨 𝐯𝑨 𝒄𝑨 𝝆𝑨 𝐯𝑨 𝒄𝑨 𝝆𝑩 𝐯𝑩 𝒄𝑩 𝐯𝑩 𝒄𝑩 𝝆𝑨 𝐯𝑨 𝒄𝑩 𝝆𝑩 𝐯𝑩
= − − + − −
𝒄 𝝆𝒄 𝝆𝒄 𝒄 𝝆𝒄 𝝆𝒄
𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 𝒄𝑨 𝝆𝑨 𝐯𝑨 + 𝒄𝑨 𝝆𝑩 𝐯𝑩 + 𝒄𝑩 𝝆𝑨 𝐯𝑨 + 𝒄𝑩 𝝆𝑩 𝐯𝑩
= −
𝒄 𝝆𝒄
𝒄𝑨 (𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 ) + 𝒄𝑩 (𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 )
= 𝐯∗ −
𝝆𝒄

(𝒄𝑨 + 𝒄𝑩 ) (𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 ) 𝒄 (𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 )

= 𝐯∗ − = 𝐯∗ −
𝝆𝒄 𝝆𝒄
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(𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 )
= 𝐯∗ −
𝝆
∑ 𝝆𝒊 𝐯𝒊
= 𝐯∗ − = 𝐯∗ − 𝐯
𝝆

𝟒. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝒋𝑨 + 𝒋𝑩 = 𝟎

𝒋𝑨 = 𝝆𝑨 (𝐯𝑨 − 𝐯) & 𝒋𝑩 = 𝝆𝑩 (𝐯𝑩 − 𝐯)

𝒋𝑨 + 𝒋𝑩 = 𝝆𝑨 (𝐯𝑨 − 𝐯) + 𝝆𝑩 (𝐯𝑩 − 𝐯)
= 𝝆𝑨 𝐯𝑨 − 𝝆𝑨 𝐯 + 𝝆𝑩 𝐯𝑩 − 𝝆𝑩 𝐯

= 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 − 𝝆𝑨 𝐯 − 𝝆𝑩 𝐯

𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝒘𝒉𝒆𝒓𝒆: 𝐯= → 𝝆 𝐯 = 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝝆
→ 𝒋𝑨 + 𝒋𝑩 = 𝝆 𝐯 − 𝝆𝑨 𝐯 − 𝝆𝑩 𝐯 = 𝝆 𝐯 − 𝐯 (𝝆𝑨 + 𝝆𝑩 ) = 𝝆 𝐯 − 𝝆 𝐯 = 0

𝟓. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑱∗𝑨 + 𝑱∗𝑩 = 𝟎

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) & 𝑱∗𝑩 = 𝒄𝑩 (𝐯𝑩 − 𝐯 ∗ )

𝑱∗𝑨 + 𝑱∗𝑩 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) + 𝒄𝑩 (𝐯𝑩 − 𝐯 ∗ )

= 𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯 ∗ + 𝒄𝑩 𝐯𝑩 − 𝒄𝑩 𝐯 ∗
= 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 − 𝒄𝑨 𝐯 ∗ − 𝒄𝑩 𝐯 ∗
= 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 − (𝒄𝑨 + 𝒄𝑩 ) 𝐯 ∗

𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩
𝒘𝒉𝒆𝒓𝒆: 𝐯∗ = → 𝝆 𝐯 ∗ = 𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩
𝒄
→ 𝑱∗𝑨 + 𝑱∗𝑩 = 𝒄 𝐯 ∗ − 𝒄 𝐯 ∗ = 0

𝟔. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑫𝑨 𝑩 = 𝑫𝑩𝑨

𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … . (1)

𝑱∗𝑩 = −𝒄 𝑫𝒃𝑨 𝞩 𝒙𝑩 → −𝑱∗𝑩 = 𝒄 𝑫𝒃𝑨 𝞩 𝒙𝑩 . . . … … … . (2)

𝑱∗𝑨 + 𝑱∗𝑩 = 𝟎 → 𝑱∗𝑨 = −𝑱∗𝑩

𝐄𝐪𝐮𝐚𝐭𝐢𝐧𝐠 𝐄𝐪. (𝟏)𝐚𝐧𝐝 𝐄𝐪. (𝟐) → −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 = 𝒄 𝑫𝒃𝑨 𝞩 𝒙𝑩 … … … … . (3)

− 𝑫𝑨𝑩 𝞩 𝒙𝑨 = 𝑫𝑩𝑨 𝞩 𝒙𝑩 … … … … . (4)

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𝒙𝑨 + 𝒙𝑩 = 𝟏 → 𝒙𝑨 = 𝟏 − 𝒙𝑩 → 𝒅𝒙𝑨 = −𝒅𝒙𝑩

→ 𝞩 𝒙𝑨 = −𝞩 𝒙𝑩 … … … … … . . (5)

𝐒𝐮𝐬𝐢𝐭𝐢𝐮𝐭𝐢𝐧𝐠 𝐄𝐪. (𝟓) 𝐢𝐧𝐭𝐨 𝐄𝐪. (𝟒) 𝐭𝐨 𝐠𝐞𝐭; 𝑫𝑨𝑩 = 𝑫𝑩𝑨

𝑴𝑩
𝟕. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑱𝑨 = 𝑱∗𝑨
𝑴
𝑱∗𝑨 𝑴 𝑱∗𝑨 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) 𝐯𝑨 − 𝐯 ∗
= → = =
𝑱𝑨 𝑴𝑩 𝑱𝑨 𝒄𝑨 (𝐯𝑨 − 𝐯) 𝐯𝑨 − 𝐯
𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝒘𝒉𝒆𝒓𝒆: 𝐯∗ = & 𝐯=
𝒄 𝝆
𝒄 𝐯 +𝒄 𝐯 𝐯𝑨 𝒄 − 𝒄𝑨 𝐯𝑨 − 𝒄𝑩 𝐯𝑩
𝑱∗𝑨 𝐯𝑨 − 𝑨 𝑨 𝒄 𝑩 𝑩
= = 𝐯 𝝆 − 𝝆 𝒄𝐯 − 𝝆 𝐯
𝑱𝑨 𝐯 − 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 𝑨 𝑨 𝑨 𝑩 𝑩
𝑨 𝝆 𝝆

𝒘𝒉𝒆𝒓𝒆: 𝒄 = 𝒄𝑨 + 𝒄𝑩 & 𝜌 = 𝝆𝑨 + 𝝆𝑩

𝑱∗𝑨 𝐯𝑨 (𝒄𝑨 + 𝒄𝑩 ) − 𝒄𝑨 𝐯𝑨 − 𝒄𝑩 𝐯𝑩 𝝆 𝒄𝑩 𝐯𝑨 − 𝒄𝑩 𝐯𝑩
= = 𝑴
𝑱𝑨 𝐯𝑨 (𝝆𝑨 + 𝝆𝑩 ) − 𝝆𝑨 𝐯𝑨 − 𝝆𝑩 𝐯𝑩 𝒄 𝝆𝑩 𝐯𝑨 − 𝝆𝑩 𝐯𝑩

𝒄𝑩 (𝐯𝑨 − 𝐯𝑩 ) 𝒄𝑩 𝑴 𝑴
= 𝑴= 𝑴= 𝝆 =
𝝆𝑩 (𝐯𝑨 − 𝐯𝑩 ) 𝝆𝑩 𝑩 𝑴𝑩
𝒄𝑩
𝑱∗𝑨 𝑴
∴ =
𝑱𝑨 𝑴𝑩

𝟖. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝒋𝑨 = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝑴
𝑱∗𝑨 = 𝑱
𝑴𝑩 𝑨
𝒋𝑨
𝒘𝒉𝒆𝒓𝒆: 𝑱𝑨 =
𝑴𝑨
𝑴
→ 𝑱∗𝑨 = 𝒋 … … … … … … … … . . (1)
𝑴𝑩 𝑴𝑨 𝑨

𝒘𝒉𝒆𝒓𝒆: 𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

𝒅𝒙𝑨
→ 𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 … … … … … … … … . . (2)
𝒅𝒛
𝟏
𝑴𝑨 𝑴𝑩 𝞩 𝒘𝑨
𝐟𝐫𝐨𝐦 𝐄𝐪. 𝐏́ 𝐢𝐧 𝐓𝐚𝐛𝐥𝐞 𝟏𝟕. 𝟕 − 𝟏 (𝐩. 𝟓𝟑𝟒), 𝞩 𝒙𝑨 =
𝒘 𝒘 2
(𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

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𝒅𝒘𝑨 𝒅𝒘𝑨 𝑴2
𝒅𝒙𝑨 = = = 𝒅𝒘𝑨
𝒘 𝒘 2 𝟏 2 𝑴𝑨 𝑴𝑩
𝑴𝑨 𝑴𝑩 (𝑴𝑨 + 𝑴𝑩 ) 𝑴𝑨 𝑴𝑩 (𝑴)
𝑨 𝑩

𝑴2
𝒅𝒙𝑨 = 𝒅𝒘𝑨 … … … … … … … … … (3)
𝑴𝑨 𝑴𝑩

Substituting Eq.(2) & Eq.(3) into Eq.(1) to get:

𝑴2 𝑴
−𝒄 𝑫𝑨𝑩 𝞩𝒘𝑨 = 𝒋
𝑴𝑨 𝑴𝑩 𝑴𝑩 𝑴𝑨 𝑨

→ 𝒋𝑨 = −(𝒄 𝑴) 𝑫𝑨𝑩 𝞩 𝒘𝑨 = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝒅𝒘𝑨
𝟗. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝒅𝒙𝑨 =
𝒘 𝒘 2
𝑴𝑨 𝑴𝑩 (𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

𝒘𝑨 𝒘 𝒘 𝒅𝒘 𝒘 𝒅𝒘 𝒅𝒘
(𝑴𝑨 + 𝑴𝑩 ) 𝑴 𝑨 − 𝑴𝑨 ( 𝑴 𝑨 + 𝑴 𝑩 )
𝑴𝑨 𝑨 𝑩 𝑨 𝑨 𝑨 𝑩
𝒙𝑨 = 𝒘 𝒘𝑩 → 𝒅𝒙𝑨 = 𝟐
𝑨 𝒘 𝒘
𝑴𝑨 + 𝑴𝑩 (𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

𝒘𝑨 𝒅𝒘𝑨 𝒘𝑩 𝒅𝒘𝑨 𝒘𝑨 𝒅𝒘𝑨 𝒘𝑨 𝒅𝒘𝑩

𝑴𝑨 𝑴𝑨 + 𝑴𝑩 𝑴𝑨 − 𝑴𝑨 𝑴𝑨 − 𝑴𝑨 𝑴𝑩
𝒅𝒙𝑨 =
𝒘 𝒘 𝟐
(𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

𝒘𝑩 𝒅𝒘𝑨 𝒘𝑨 𝒅𝒘𝑩
𝑴𝑩 𝑴𝑨 − 𝑴𝑨 𝑴𝑩
𝒅𝒙𝑨 = … … … … … … … … … . (1)
𝒘 𝒘 𝟐
(𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

𝒃𝒖𝒕: 𝒘𝑩 = 𝟏 − 𝒘𝑨 → 𝒅𝒘𝑩 = −𝒅𝒘𝑨 … … … … … … … … … (2)

Substituting Eq.(2) into Eq.(1) to get:

𝒘𝑩 𝒅𝒘𝑨 𝒘𝑨 𝒅𝒘𝑨 (𝒘𝑨 + 𝒘𝑩 ) 𝒅𝒘𝑨

𝑴𝑩 𝑴𝑨 + 𝑴𝑨 𝑴𝑩 𝑴𝑨 𝑴𝑩
𝒅𝒙𝑨 = 𝟐 =
𝒘 𝒘 𝒘 𝒘 𝟐
(𝑴𝑨 + 𝑴𝑩 ) (𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩 𝑨 𝑩

𝒃𝒖𝒕: 𝒘𝑨 + 𝒘𝑩 = 𝟏

𝒅𝒘𝑨
→ 𝒅𝒙𝑨 =
𝒘 𝒘 2
𝑴𝑨 𝑴𝑩 (𝑴𝑨 + 𝑴𝑩 )
𝑨 𝑩

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑴𝑩
𝟏𝟎. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑱𝑨 = 𝑵𝑨 + 𝒘𝑨 (𝑵𝑨 + 𝑵 )
𝑴𝑨 𝑩

𝑴𝑩
𝑱𝑨 = 𝑵𝑨 + 𝒘𝑨 (𝑵𝑨 + 𝑵 )
𝑴𝑨 𝑩
1
𝝆𝑨 𝑴𝑩 𝑴𝑨
= 𝒄𝑨 𝐯𝑨 + (𝒄𝑨 𝐯𝑨 + 𝒄 𝐯 ) ×
𝝆 𝑴𝑨 𝑩 𝑩 1
𝑴𝑨
𝑴
𝝆𝑨 𝒄𝑨 𝐯𝑨 + 𝑴𝑩 𝒄𝑩 𝐯𝑩
𝑨
= 𝒄𝑨 𝐯𝑨 + ( )
𝝆 𝑴𝑨 1
𝑴𝑨

𝑴
𝑴𝑨 (𝒄𝑨 𝐯𝑨 + 𝑴𝑩 𝒄𝑩 𝐯𝑩 )
𝑨
= 𝒄𝑨 𝐯𝑨 + 𝒄𝑨 ( )
𝝆

𝑴𝑨 𝒄𝑨 𝐯𝑨 + 𝑴𝑩 𝒄𝑩 𝐯𝑩 𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
= 𝒄𝑨 𝐯𝑨 + 𝒄𝑨 ( ) = 𝒄𝑨 𝐯𝑨 + 𝒄𝑨 ( )
𝝆 𝝆

= 𝒄𝑨 𝐯𝑨 + 𝒄𝑨 𝐯 = 𝒄𝑨 (𝐯𝑨 + 𝐯)

𝑱𝑨 = 𝒄𝑨 (𝐯𝑨 + 𝐯)

𝟏𝟏. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝐧𝑨 − 𝒘𝑨 (𝐧𝑨 + 𝐧𝑩 ) = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … … . . (1)

𝑴
𝑱∗𝑨 = 𝒋 … … … … … . . (2)
𝑴𝑨 𝑴𝑩 𝑨

𝑴𝟐
𝞩 𝒙𝑨 = 𝞩 𝒘𝑨 … … … … … … . (3)
𝑴𝑨 𝑴𝑩

Substituting Eq.(2) & Eq.(3) into Eq.(1) to get:

𝑴 𝑴𝟐
𝒋𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒘𝑨
𝑴𝑨 𝑴𝑩 𝑴𝑨 𝑴𝑩

𝒋𝑨 = −𝒄 𝑴 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝒋𝑨 = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨 … … … … … … . (4)

𝒘𝒉𝒆𝒓𝒆: 𝒋𝑨 = 𝝆𝑨 (𝐯𝑨 − 𝐯)

→ 𝒋𝑨 = 𝝆𝑨 𝐯𝑨 − 𝝆𝑨 𝐯

𝒘𝒉𝒆𝒓𝒆: 𝐧𝑨 = 𝝆𝑨 𝐯𝑨

→ 𝒋𝑨 = 𝐧𝑨 − 𝝆𝑨 𝐯

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
𝒘𝒉𝒆𝒓𝒆: 𝐯=
𝝆
𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩
→ 𝒋𝑨 = 𝐧𝑨 − 𝝆𝑨 ( )
𝝆
𝝆𝑨
→ 𝒋𝑨 = 𝐧 𝑨 − (𝝆𝑨 𝐯𝑨 + 𝝆𝑩 𝐯𝑩 )
𝝆
𝝆𝑨
→ 𝒋𝑨 = 𝐧 𝑨 − (𝐧𝑨 + 𝐧𝑩 )
𝝆
→ 𝒋𝑨 = 𝐧𝑨 − 𝒘𝑨 (𝐧𝑨 + 𝐧𝑩 ) … … … … … … … . (5)

Substituting Eq.(5) into Eq.(4) to get:

𝐧𝑨 − 𝒘𝑨 (𝐧𝑨 + 𝐧𝑩 ) = −𝝆 𝑫𝑨𝑩 𝞩 𝒘𝑨

𝟏𝟐. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑵𝑨 − 𝒙𝑨 (𝑵𝑨 + 𝑵𝑩 ) = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … … . (1)

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ ) = 𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯 ∗
𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩
𝑱∗𝑨 = 𝒄𝑨 𝐯𝑨 − 𝒄𝑨 ( )
𝒄
𝒄𝑨
𝑱∗𝑨 = 𝑵𝑨 − (𝒄 𝐯 + 𝒄𝑩 𝐯𝑩 )
𝒄 𝑨 𝑨
𝑱∗𝑨 = 𝑵𝑨 − 𝒙𝑨 (𝑵𝑨 + 𝑵𝑩 ) … … … … … . (2)

Substituting Eq.(2) into Eq.(1) to get:

𝑵𝑨 − 𝒙𝑨 (𝑵𝑨 + 𝑵𝑩 ) = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

𝒄𝟐
𝟏𝟑. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝒋𝑨 = − 𝑴𝑨 𝑴𝑩 𝑫𝑨𝑩 𝞩 𝒙𝑨
𝝆
𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … … . (1)

𝑴
𝑱∗𝑨 = 𝒋 … … … … … . . (2)
𝑴𝑨 𝑴𝑩 𝑨

Substituting Eq.(2) into Eq.(1) to get:

𝑴 𝒄
𝒋 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 → 𝒋𝑨 = − 𝑴 𝑴 𝑫 𝞩 𝒙𝑨
𝑴𝑨 𝑴𝑩 𝑨 𝑴 𝑨 𝑩 𝑨𝑩

𝒄 𝒄𝟐
→ 𝒋𝑨 = − 𝑴 𝑴 𝑫 𝞩 𝒙𝑨 = − 𝑴𝑨 𝑴𝑩 𝑫𝑨𝑩 𝞩 𝒙𝑨
(𝝆⁄𝒄) 𝑨 𝑩 𝑨𝑩 𝝆

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝝆𝟐
𝟏𝟒. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝑱∗𝑨 = −( ) 𝑫𝑨𝑩 𝞩 𝒘𝑨
𝒄 𝑴𝑨 𝑴𝑩

𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … … . (1)

𝑴𝟐
𝞩 𝒙𝑨 = 𝞩 𝒘𝑨 … … … … … . (2)
𝑴𝑨 𝑴𝑩

Substituting Eq.(2) into Eq.(1) to get:

𝑴𝟐 (𝝆⁄𝒄)𝟐
𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒘𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒘𝑨
𝑴𝑨 𝑴𝑩 𝑴𝑨 𝑴𝑩

𝝆𝟐
𝑱∗𝑨 = − ( ) 𝑫𝑨𝑩 𝞩 𝒘𝑨
𝒄 𝑴𝑨 𝑴𝑩

𝒄 𝑫𝑨𝑩
𝟏𝟓. 𝑷𝒓𝒐𝒗𝒆 𝒕𝒉𝒂𝒕; 𝒄 (𝐯𝑨 − 𝐯𝑩 ) = − 𝞩 𝒙𝑨
𝒙𝑨 𝒙𝑩
𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨 … … … … … . (1)

𝑱∗𝑨 = 𝒄𝑨 (𝐯𝑨 − 𝐯 ∗ )

𝑱∗𝑨 = 𝒙𝑨 𝒄 (𝐯𝑨 − 𝐯 ∗ ) … … … … … . (2)

Substituting Eq.(2) into Eq.(1) to get:

𝒙𝑨 𝒄 (𝐯𝑨 − 𝐯 ∗ ) = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

𝑫𝑨𝑩
𝐯𝑨 − 𝐯 ∗ = − 𝞩 𝒙𝑨 … … … … … … (3)
𝒙𝑨

𝑱∗𝑨 + 𝑱∗𝑩 = 𝟎 → 𝑱∗𝑨 = −𝑱∗𝑩 … … … … … … . (4)

𝑱∗𝑩 = 𝒄𝑩 (𝐯𝑩 − 𝐯 ∗ ) … … … … … . . (5)

Substituting Eq.(5) into Eq.(4) to get:

𝑱∗𝑨 = −𝒄𝑩 (𝐯𝑩 − 𝐯 ∗ ) → 𝑱∗𝑨 = 𝒄𝑩 (𝐯 ∗ − 𝐯𝑩 )

𝑱∗𝑨 = 𝒙𝑩 𝒄 (𝐯 ∗ − 𝐯𝑩 ) … … … … … . . . (6)

Substituting Eq.(6) into Eq.(1) to get:

𝒙𝑩 𝒄 (𝐯 ∗ − 𝐯𝑩 ) = −𝒄 𝑫𝑨𝑩 𝞩 𝒙𝑨

𝑫𝑨𝑩
𝐯𝑩 − 𝐯 ∗ = − 𝞩 𝒙𝑨 … … … … … … (7)
𝒙𝑩

Adding Eq.(7) and Eq.(3) to get:

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑫𝑨𝑩 𝑫𝑨𝑩
(𝐯𝑨 − 𝐯 ∗ ) + (𝐯𝑩 − 𝐯 ∗ ) = (− 𝞩 𝒙𝑨 ) + (− 𝞩 𝒙𝑨 )
𝒙𝑨 𝒙𝑩
− 𝑫𝑨𝑩 𝑫𝑨𝑩
𝐯𝑨 − 𝐯𝑩 = 𝞩 𝒙𝑨 − 𝞩 𝒙𝑨
𝒙𝑨 𝒙𝑩

1 1
𝐯𝑨 − 𝐯𝑩 = − 𝑫𝑨𝑩 𝞩 𝒙𝑨 ( + )
𝒙𝑨 𝒙𝑩

𝒙𝑨 + 𝒙𝑩
= − 𝑫𝑨𝑩 𝞩 𝒙𝑨 ( ) 𝐰𝐡𝐞𝐫𝐞: 𝒙𝑨 + 𝒙𝑩 = 𝟏
𝒙𝑨 𝒙𝑩
𝟏
= − 𝑫𝑨𝑩 𝞩 𝒙𝑨 ( )
𝒙𝑨 𝒙𝑩

− 𝑫𝑨𝑩
𝐯𝑨 − 𝐯𝑩 = 𝞩 𝒙𝑨 … … … … … … … (8)
𝒙𝑨 𝒙𝑩

Multiplying Eq.(8) by 𝒄 to get:

−𝒄 𝑫𝑨𝑩
𝒄(𝐯𝑨 − 𝐯𝑩 ) = 𝞩 𝒙𝑨 … … … … … … … (8)
𝒙𝑨 𝒙𝑩

17.9 The Maxwell-Stefan Equations For Multicomponent Diffusion in

Gases at Low Density

For multicomponent diffusion in gases at low density it has been shown that to a very
good approximation:

𝑁 𝑁
𝒙𝑨 𝒙𝑩 𝒄𝑨 𝒄𝑩
𝟏. 𝞩 𝒙𝑨 = − ∑ (𝐯𝑨 − 𝐯𝑩 ) = − ∑ 𝟐 (𝐯 − 𝐯𝑩 ) (17.9 − 1)
𝑫𝑨𝑩 𝒄 𝑫𝑨𝑩 𝑨
𝐵=1 𝐵=1

𝑁
𝟏
𝟐. 𝞩 𝒙𝑨 = − ∑ (𝒙 𝑵 − 𝒙𝑨 𝑵𝑩 ) (17.9 − 1)
𝒄 𝑫𝑨𝑩 𝑩 𝑨
𝐵=1

Where:
𝑨 = 1 ,2 ,3 ,……,𝑁
𝑫𝑨𝑩 = are the binary diffusivities calculated from Eq. 17.3-11 or Eq. 17.3-12

Equations 17.9-1 are referred to as the Maxwell-Stefan equations, since Maxwell suggested
them for binary mixtures on the basis of kinetic theory, and Stefan generalized them to describe
the diffusion in a gas mixture with N species.

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

For the first relation of Eq.(17.9-1):

𝑁
𝒄𝑨 𝒄𝑩
𝞩 𝒙𝑨 = − ∑ (𝐯 − 𝐯𝑩 )
𝒄𝟐 𝑫𝑨𝑩 𝑨
𝐵=1

For binary mixture of A and B:

𝒄𝑨 𝒄𝑩 𝒄𝟐 𝑫𝑨𝑩
𝞩 𝒙𝑨 = − 𝟐 (𝐯 − 𝐯𝑩 ) → (𝐯𝑨 − 𝐯𝑩 ) = − 𝞩 𝒙𝑨
𝒄 𝑫𝑨𝑩 𝑨 𝒄𝑨 𝒄𝑩

𝒅𝒙𝑨
𝒘𝒉𝒆𝒓𝒆: 𝞩 𝒙𝑨 =
𝒅𝒛

𝑫𝑨𝑩 𝒅𝒙𝑨 𝑫𝑨𝑩 𝒅𝒙𝑨

(𝐯𝑨 − 𝐯𝑩 ) = − 𝒄 𝒄 → (𝐯𝑨 − 𝐯𝑩 ) = − × 𝒄𝟐
𝑨 𝑩 𝒅𝒛 𝒙𝑨 𝒙𝑩 𝒅𝒛
𝒄𝟐
𝒅𝒙𝑨
𝒙𝑨 𝒙𝑩 𝒄𝟐 (𝐯𝑨 − 𝐯𝑩 ) = − 𝒄𝟐 𝑫𝑨𝑩
𝒅𝒛

𝒅𝒙𝑨 𝒅𝒙𝑨
𝒄𝑨 𝒄𝑩 (𝐯𝑨 − 𝐯𝑩 ) = − 𝒄𝟐 𝑫𝑨𝑩 → 𝒄𝑨 𝒄𝑩 𝐯𝑨 − 𝒄𝑨 𝒄𝑩 𝐯𝑩 = − 𝒄𝟐 𝑫𝑨𝑩 ÷𝒄
𝒅𝒛 𝒅𝒛

𝒅𝒙𝑨
𝒄𝑨 𝒙𝑩 𝐯𝑨 − 𝒙𝑨 𝒄𝑩 𝐯𝑩 = − 𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒅𝒙𝑨
𝒄𝑨 (𝟏 − 𝒙𝑨 ) 𝐯𝑨 − 𝒙𝑨 𝒄𝑩 𝐯𝑩 = − 𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯𝑨 𝒙𝑨 − 𝒙𝑨 𝒄𝑩 𝐯𝑩 = − 𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − 𝒙𝑨 (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) = − 𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒄𝑨 𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) = − 𝒄 𝑫𝑨𝑩
𝒄 𝒅𝒛

𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − 𝒄𝑨 ( ) = − 𝒄 𝑫𝑨𝑩
𝒄 𝒅𝒛

𝒅𝒙𝑨 𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯 ∗ = − 𝒄 𝑫𝑨𝑩 → 𝑱∗𝑨 = − 𝒄 𝑫𝑨𝑩
𝒅𝒛 𝒅𝒛

𝒅𝒙𝑨 𝑱∗𝑨
→ =−
𝒅𝒛 𝒄 𝑫𝑨𝑩

𝑴 𝑴𝟐
𝒘𝒉𝒆𝒓𝒆: 𝑱∗𝑨 = 𝒋 𝒂𝒏𝒅 𝞩 𝒙𝑨 = 𝞩 𝒘𝑨
𝑴𝑨 𝑴𝑩 𝑨 𝑴𝑨 𝑴𝑩

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Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑴𝟐 𝑴 𝒋𝑨
→ 𝞩 𝒘𝑨 = −
𝑴𝑨 𝑴𝑩 𝑴𝑨 𝑴𝑩 𝒄 𝑫𝑨𝑩

𝒋𝑨
→ 𝑴 𝞩 𝒘𝑨 = −
𝒄 𝑫𝑨𝑩

𝒅𝒘𝑨 𝒅𝒘𝑨
→ 𝒋𝑨 = − 𝒄 𝑴 𝑫𝑨𝑩 → 𝒋𝑨 = − 𝝆 𝑫𝑨𝑩
𝒅𝒛 𝒅𝒛

For the second relation of Eq.(17.9-1):

𝑵
𝟏
𝜵 𝒙𝑨 = − ∑ (𝒙 𝑵 − 𝒙𝑨 𝑵𝑩 )
𝒄 𝑫𝑨𝑩 𝑩 𝑨
𝑩=𝟏

For binary mixture of A and B:

𝟏
𝜵 𝒙𝑨 = − (𝒙 𝑵 − 𝒙𝑨 𝑵𝑩 )
𝒄 𝑫𝑨𝑩 𝑩 𝑨

𝒅𝒙𝑨
𝒙𝑩 𝑵𝑨 − 𝒙𝑨 𝑵𝑩 + 𝒙𝑨 𝑵𝑨 − 𝒙𝑨 𝑵𝑨 = −𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒅𝒙𝑨
(𝒙𝑩 + 𝒙𝑨 ) 𝑵𝑨 − (𝑵𝑨 + 𝑵𝑩 ) 𝒙𝑨 = −𝒄 𝑫𝑨𝑩 , 𝒘𝒉𝒆𝒓𝒆: 𝒙𝑩 + 𝒙𝑨 = 𝟏
𝒅𝒛

𝒅𝒙𝑨
𝑵𝑨 − 𝒙𝑨 (𝑵𝑨 + 𝑵𝑩 ) = −𝒄 𝑫𝑨𝑩
𝒅𝒛

𝒄𝑨 𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − (𝒄𝑨 𝐯𝑨 + 𝒄𝑩 𝐯𝑩 ) = −𝒄 𝑫𝑨𝑩
𝒄 𝒅𝒛

𝒅𝒙𝑨 𝒅𝒙𝑨
𝒄𝑨 𝐯𝑨 − 𝒄𝑨 𝐯 ∗ = −𝒄 𝑫𝑨𝑩 → 𝒄𝑨 ( 𝐯𝑨 − 𝐯 ∗ ) = −𝒄 𝑫𝑨𝑩
𝒅𝒛 𝒅𝒛

𝒅𝒙𝑨 𝒅𝒙𝑨
𝒄𝑨 ( 𝐯𝑨 − 𝐯 ∗ ) = −𝒄 𝑫𝑨𝑩 → 𝑱∗𝑨 = −𝒄 𝑫𝑨𝑩
𝒅𝒛 𝒅𝒛

𝒅𝒙𝑨 𝑱∗𝑨
→ =−
𝒅𝒛 𝒄 𝑫𝑨𝑩

𝑴 𝑴𝟐
𝒘𝒉𝒆𝒓𝒆: 𝑱∗𝑨 = 𝒋 𝒂𝒏𝒅 𝞩 𝒙𝑨 = 𝞩 𝒘𝑨
𝑴𝑨 𝑴𝑩 𝑨 𝑴𝑨 𝑴𝑩

26
Advanced Mass Transfer Master Course Prof. Dr. Ahmed Daham

𝑴𝟐 𝑴 𝒋𝑨
→ 𝞩 𝒘𝑨 = −
𝑴𝑨 𝑴𝑩 𝑴𝑨 𝑴𝑩 𝒄 𝑫𝑨𝑩

𝒋𝑨
→ 𝑴 𝞩 𝒘𝑨 = −
𝒄 𝑫𝑨𝑩

𝒅𝒘𝑨 𝒅𝒘𝑨
→ 𝒋𝑨 = − 𝒄 𝑴 𝑫𝑨𝑩 → 𝒋𝑨 = − 𝝆 𝑫𝑨𝑩
𝒅𝒛 𝒅𝒛

27