1. Introduction to Mass Transfer

2. Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion
Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
Equimolar Counterdiffusion
Combined Diffusion and Convection
Uni-component Diffusion

3. Mass Transfer Mechanisms
1. Convective Mass Transfer
2. Diffusion
Analogous to heat convection and conduction

4. Mass Transfer Mechanisms
3. Convective and Diffusion

5. Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion
Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
Equimolar Counterdiffusion
Combined Diffusion and Convection
Uni-component Diffusion

6. Fick’s Law for Molecular Diffusion
We’ll first consider diffusion of molecules when the bulk fluid is not moving…
𝐽 𝐴𝑧 ∗ =− 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧
𝑑 𝑐 𝐴 =𝑑 𝑐 𝑥 𝐴 =𝑐𝑑 𝑥 𝐴
𝐽 𝐴𝑧 ∗ =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧
For a binary mixture of A and B
J = molar diffusivity
C = total concentration
Xa = mole fraction of A in the mixture
Other driving forces for diffusion: temperature, pressure, electrical potential, etc. (Will not be discussed in 131)

7. Molecular Transport Equations
RECALL:
MOMENTUM
HEAT
MASS

8. Fick’s Law for Molecular Diffusion
Example A mixture of He and N2 gas is collected in a pipe at 298 K and 1 atm total pressure which is constant throughout. At one end of the pipe at point 1 the partial pressure pA1 of He is 0.60 atm and at the other end 0.2 m pA2 = 0.20 atm. Calculate the flux of He at steady state if DAB of the He-N2 mixture is x 10-4 m2/s.
Steady state diffusion
Concentration at the boundaries are constant
Diffusion is limited to molecular motion
Diffusivity is independent from concentration
No temperature gradient
Assume ideal gas, calculate c values from the P given.
Answer: J = 5.63 x 10-6 kg mol of A/s-m2

9. Convective Mass Transfer Coefficient
For fluids in convective flow… 𝑘 𝑐 is very similar to h,
𝑁 𝐴 = 𝑘 𝑐 ( 𝐶 𝐿1 − 𝐶 𝐿𝑖 )
𝑁 𝐴 = 𝑘 𝐺 ( 𝑝 𝐴1 − 𝑝 𝐴𝑖 )
𝑁 𝐴 = 𝑘 𝑦 ( 𝑦 𝐴1 − 𝑦 𝐴𝑖 )
To have a fluid in convective flow usually requires the fluid to be flowing past another immiscible fluid or a solid surface.
Na = convective molar flux, k = MTC with different units depending on the units of the driving force
*MTC is a function of system geometry, fluid properties, and flow velocity.
We will discuss convective mass transfer in more detail LATER…
What factors influence 𝑘 𝑐 ?

10. Outline Mass Transfer Mechanisms 2. Fick’s Law for Molecular Diffusion
Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
Equimolar Counterdiffusion
Combined Diffusion and Convection
Uni-component Diffusion

11. Molecular Diffusion in Gases
Equimolar Counterdiffusion
Flux of one gaseous component is equal to but in the opposite direction of the second gaseous component A B
Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients.
Examples: distillation, alcohol/H2O, HOAc/H2O A B
𝐽 𝐴𝑧 ∗ =− 𝐽 𝐵𝑧 ∗

12. Molecular Diffusion in Gases
Equimolar Counterdiffusion
At constant pressure, 𝑃= 𝑝 𝐴 + 𝑝 𝐵
Then, 𝑐= 𝑐 𝐴 + 𝑐 𝐵
and 𝑑 𝑐 𝐴 =−𝑑 𝑐 𝐵
Fick’s law for B,
𝐽 𝐵𝑧 ∗ =− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 A B
Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients. A B

13. Molecular Diffusion in Gases
Equimolar Counterdiffusion
Substitution of Fick’s law
into the equation
for equimolar counter diffusion, A B
𝐽 𝐴𝑧 ∗ =− 𝐽 𝐵𝑧 ∗
Consider 2 gases A and B in a chamber, initially separated by a partition. At some instant in time, the partition is removed, and A and B diffuse in opposite direction as a result of the concentration gradients. A B
− 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 )

14. Molecular Diffusion in Gases
Equimolar Counterdiffusion
− 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 𝑑 𝑐 𝐵 𝑑𝑧 ) A B
− 𝐷 𝐴𝐵 𝑑 𝑐 𝐴 𝑑𝑧 =−(− 𝐷 𝐵𝐴 (− 𝑑 𝑐 𝐴 𝑑𝑧 ))
For a binary mixture of A and B the diffusivity coefficient DAB for A diffusing in B is the same as DBA for B diffusing into A. A B
𝐷 𝐴𝐵 = 𝐷 𝐵𝐴

15. Molecular Diffusion in Gases
Equimolar Counterdiffusion
For gases, A B
𝑐 𝐴1 = 𝑝 𝐴1 𝑅𝑇 = 𝑛 𝐴 𝑉
𝐽 𝐴𝑧 ∗ =− 𝐷 𝐴𝐵 𝑅𝑇 𝑑 𝑝 𝐴 𝑑𝑧 A B

16. Molecular Diffusion in Gases
Equimolar Counterdiffusion
In terms of mole fraction, A B
𝑐 𝐴 =𝑐 𝑥 𝐴
𝐽 𝐴𝑧 ∗ =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 A B

17. Molecular Diffusion in Gases
Example A large tank filled with a mixture of methane and air is connected to a second tank filled with a different composition of methane and air. Both tanks are at 100 kN/m2 and 0°C. The connection between the tanks is a tube of 2 mm inside diameter and 150 mm long. Calculate the steady state rate of transport of methane through the tube when the concentration of methane is 90 mole percent in one tank and 5 mole percent in the other. Assume that transport between the tanks is by molecular diffusion. The mass diffusivity of methane in air at 0°C and 100 kN/m2 is 1.57 x 10-5 m2/s.

18. Molecular Diffusion in Gases
Diffusion plus Convection
𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀
Multiplying by 𝑐 𝐴 ,
𝑐 𝐴 𝑣 𝐴 = 𝑐 𝐴 𝑣 𝐴𝑑 + 𝑐 𝐴 𝑣 𝑀
𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀
𝑣 𝑀
𝑣 𝐴
Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity
Vad = diffusion velocity of A, measured relative to the moving fluid
Vm = molar average velocity of the whole fluid relative to a stationary point
𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀
𝐽 𝐴 ∗ = 𝑣 𝐴𝑑 𝑐 𝐴

19. Molecular Diffusion in Gases
Diffusion plus Convection
𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀
𝑣 𝑀
Total
convective flux of A
wrt stationary pt
𝑁 𝐴
𝑣 𝐴
Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity
Vad = diffusion velocity of A, measured relative to the moving fluid
𝐽 𝐴 ∗
𝑣 𝐴 = 𝑣 𝐴𝑑 + 𝑣 𝑀
Diffusion flux
wrt moving fluid
𝐽 𝐴 ∗ = 𝑣 𝐴𝑑 𝑐 𝐴
Convective flux
wrt to stationary point
𝑐 𝐴 𝑣 𝑀

20. Molecular Diffusion in Gases
Diffusion plus Convection
𝑁= 𝑐𝑣 𝑀 = 𝑁 𝐴 + 𝑁 𝐵
𝑁 𝐴 = 𝐽 𝐴 ∗ + 𝑐 𝐴 𝑣 𝑀
Replacing 𝐽 𝐴 ∗ and 𝑣 𝑀 ,
𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 + 𝑁 𝐵 𝑐 )
Solving for 𝑣 𝑀 ,
𝑣 𝑀 = 𝑁 𝐴 + 𝑁 𝐵 𝑐
Va= velocity of A relative to a stationary point, sum of diffusion velocity and the average or convective velocity
Vad = diffusion velocity of A, measured relative to the moving fluid

21. Molecular Diffusion in Gases
Diffusion plus Convection
𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 + 𝑁 𝐵 𝑐 )
General equation for diffusion plus convection
For equimolar counterdiffusion, Na = -Nb and the convective term becomes zero.

22. Molecular Diffusion in Gases
Uni-component Diffusion
One component (A)diffuses,
while the other (B) remains stagnant
A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through.
Example: Gas Absorption, evaporation of organic solvent in air
Since B cannot diffuse, 𝑁 𝐵 =0

23. Molecular Diffusion in Gases
Uni-component Diffusion
Since B cannot diffuse, 𝑁 𝐵 =0
A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through.
Example: Gas Absorption, evaporation of organic solvent in air
𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 ( 𝑁 𝐴 +0 𝑐 )

24. Molecular Diffusion in Gases
Uni-component Diffusion
𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑑 𝑥 𝐴 𝑑𝑧 + 𝑐 𝐴 𝑁 𝐴 𝑐
A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through.
Example: Gas Absorption, evaporation of organic solvent in air
𝑁 𝐴 =− 𝑐𝐷 𝐴𝐵 𝑥 𝐵 𝑑 𝑥 𝐴 𝑑𝑧

25. Molecular Diffusion in Gases
Uni-component Diffusion
When P is constant,
𝑁 𝐴 =− 𝐷 𝐴𝐵 𝑃 𝑅𝑇 𝑑 𝑝 𝐴 𝑑𝑧 + 𝑁 𝐴 𝑝 𝐴 𝑃
A special case wherein, one boundary at the end of the diffusion path is impermeable to component B, so it cannot pass through.
Example: Gas Absorption, evaporation of organic solvent in air
𝑁 𝐴 =− 𝐷 𝐴𝐵 𝑃 𝑅𝑇 𝑃 𝐵 𝑑 𝑝 𝐴 𝑑𝑧

26. Molecular Diffusion in Gases
Example Water in the bottom of a narrow metal tune is held a t a constant temperature of 293 K. The total pressure of air (assumed dry) is  105 Pa and the temperature is 293 K.
Water evaporates and diffuses through the air in the tube, and the diffusion path z2-z1 is m long. Calculate the rate of evaporation of water vapor at 293 K and 1 atm pressure. The diffusivity of water in air is x 10-4 m2/s. Assume that the system is isothermal.

27. Long Exam Results LE 1 LE 2 Mean 33.32 36.55 Median 32.00 30.75 Mode
39.00
25.50
Passing Rate
0.00
9.09

28. Quizzes Machine Problems Target Average Scores Student No. Total Q 5/5
Quizzes
Machine Problems
Target Average Scores
Student No.
Total
Q 5/5
Total*
M 15/15
LE3
L 60/60
Final
F 20/20 30
1.50
240 12 82
30.4
16.4 46
2.30 90
27.8
18.0 36
1.80 31 26
1.30 89
29.1
17.8 47
2.35 85 29
17.0 67
33.3
13.4 50
2.50 93
26.9
18.6 23
1.15 92
28.7
18.4
1.55 73
31.9
14.6 77
31.3
15.4 33
1.65
102
20.4 88
29.6
17.6 19
0.95 81
30.9
16.2 74
31.6
14.8 61
3.05 51 35
10.2 8
0.40 99
28.1
19.8 48
2.40 70
32.1
14.0 9
0.45
29.5
104
5.20 45
34.1
9.0 14
0.70
100
27.3
20.0 57
2.85 65
32.5
13.0 21
1.05 91
28.9
18.2