1. No category

Introduction to
Mass Transfer
Outline
1. Mass Transfer Mechanisms
1. Molecular Diffusion
2. Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
1. Equimolar Counterdiffusion
2. Combined Diffusion and Convection
3. Uni-component Diffusion
Mass Transfer Mechanisms
1. Convective Mass Transfer
2. Diffusion
http://www.timedomaincvd.com/CVD_Fundamentals/xprt/xprt_conv_diff.html
Mass Transfer Mechanisms
3. Convective and Diffusion
http://www.timedomaincvd.com/CVD_Fundamentals/xprt/xprt_conv_diff.html
Outline
1. Mass Transfer Mechanisms
1. Molecular Diffusion
2. Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
1. Equimolar Counterdiffusion
2. Combined Diffusion and Convection
3. Uni-component Diffusion
Fick’s Law for Molecular Diffusion
We’ll first consider diffusion of molecules when the
bulk fluid is not moving…
For a binary mixture of A and B
∗
𝐽𝐴𝑧
𝑑𝑐𝐴
= −𝐷𝐴𝐵
𝑑𝑧
𝑑𝑐𝐴 = 𝑑 𝑐𝑥𝐴 = 𝑐𝑑𝑥𝐴
∗
𝐽𝐴𝑧
𝑑𝑥𝐴
= −𝑐𝐷𝐴𝐵
𝑑𝑧
Molecular Transport Equations
RECALL:
driving force
rate of transport =
resistance
d(v x  )
 yx  
dy
MOMENTUM
qy
A
 
d(  c p T)
HEAT
dy
J  DAB
*
Ay
MASS
dcA
dy
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 0.687 x 10-4 m2/s.
Convective Mass Transfer Coefficient
For fluids in convective flow…
𝑁𝐴 = 𝑘𝑐 (𝐶𝐿1 − 𝐶𝐿𝑖 )
𝑁𝐴 = 𝑘𝐺 (𝑝𝐴1 − 𝑝𝐴𝑖 )
𝑁𝐴 = 𝑘𝑦 (𝑦𝐴1 − 𝑦𝐴𝑖 )
𝑘𝑐 is very similar to h,
What factors influence 𝑘𝑐 ?
Outline
1. Mass Transfer Mechanisms
1. Molecular Diffusion
2. Convective Mass Transfer
2. Fick’s Law for Molecular Diffusion
3. Molecular Diffusion in Gases
1. Equimolar Counterdiffusion
2. Combined Diffusion and Convection
3. Uni-component Diffusion
Molecular Diffusion in Gases
Equimolar Counterdiffusion
A
B
A
B
Flux of one gaseous
component is equal to
but in the opposite
direction of the second
gaseous component
∗
𝐽𝐴𝑧
=
∗
−𝐽𝐵𝑧
Molecular Diffusion in Gases
Equimolar Counterdiffusion
At constant pressure,𝑃 = 𝑝𝐴 + 𝑝𝐵
A
B
Then, 𝑐 = 𝑐𝐴 + 𝑐𝐵
and 𝑑𝑐𝐴 = −𝑑𝑐𝐵
Fick’s law for B,
A
B
∗
𝐽𝐵𝑧
𝑑𝑐𝐵
= −𝐷𝐵𝐴
𝑑𝑧
Molecular Diffusion in Gases
Equimolar Counterdiffusion
Substitution of Fick’s law
A
B
into the equation
for equimolar counter diffusion,
∗
∗
𝐽𝐴𝑧
= −𝐽𝐵𝑧
A
B
𝑑𝑐𝐴
𝑑𝑐𝐵
−𝐷𝐴𝐵
= −(−𝐷𝐵𝐴
)
𝑑𝑧
𝑑𝑧
Molecular Diffusion in Gases
Equimolar Counterdiffusion
A
B
𝑑𝑐𝐴
𝑑𝑐𝐵
−𝐷𝐴𝐵
= −(−𝐷𝐵𝐴
)
𝑑𝑧
𝑑𝑧
𝑑𝑐𝐴
𝑑𝑐𝐴
−𝐷𝐴𝐵
= −(−𝐷𝐵𝐴 (−
))
𝑑𝑧
𝑑𝑧
A
B
𝐷𝐴𝐵 = 𝐷𝐵𝐴
Molecular Diffusion in Gases
Equimolar Counterdiffusion
For gases,
A
A
𝑐𝐴1
𝑝𝐴1 𝑛𝐴
=
=
𝑅𝑇
𝑉
∗
𝐽𝐴𝑧
𝐷𝐴𝐵 𝑑𝑝𝐴
=−
𝑅𝑇 𝑑𝑧
B
B
Molecular Diffusion in Gases
Equimolar Counterdiffusion
In terms of mole fraction,
A
B
𝑐𝐴 = 𝑐𝑥𝐴
∗
𝐽𝐴𝑧
A
B
𝑑𝑥𝐴
= −𝑐𝐷𝐴𝐵
𝑑𝑧
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.
Molecular Diffusion in Gases
Diffusion plus Convection
𝑣𝐴 = 𝑣𝐴𝑑 + 𝑣𝑀
𝑣𝑀
𝑣𝐴
𝑣𝐴 = 𝑣𝐴𝑑 + 𝑣𝑀
𝐽𝐴∗ = 𝑣𝐴𝑑 𝑐𝐴
Multiplying by 𝑐𝐴 ,
𝑐𝐴 𝑣𝐴 = 𝑐𝐴 𝑣𝐴𝑑 + 𝑐𝐴 𝑣𝑀
𝑁𝐴 = 𝐽𝐴∗ + 𝑐𝐴 𝑣𝑀
Molecular Diffusion in Gases
Diffusion plus Convection
𝑁𝐴 = 𝐽𝐴∗ + 𝑐𝐴 𝑣𝑀
𝑣𝑀
𝑣𝐴
𝑣𝐴 = 𝑣𝐴𝑑 + 𝑣𝑀
𝐽𝐴∗ = 𝑣𝐴𝑑 𝑐𝐴
𝑁𝐴
𝐽𝐴∗
𝑐𝐴 𝑣𝑀
Total
convective flux of A
wrt stationary pt
Diffusion flux
wrt moving fluid
Convective flux
wrt to stationary point
Molecular Diffusion in Gases
Diffusion plus Convection
𝑁 = 𝑐𝑣𝑀 = 𝑁𝐴 + 𝑁𝐵
Solving for 𝑣𝑀 ,
𝑁𝐴 + 𝑁𝐵
𝑣𝑀 =
𝑐
𝑁𝐴 = 𝐽𝐴∗ + 𝑐𝐴 𝑣𝑀
Replacing 𝐽𝐴∗ and 𝑣𝑀 ,
𝑁𝐴 =
𝑑𝑥𝐴
−𝑐𝐷𝐴𝐵
𝑑𝑧
+
𝑁𝐴 +𝑁𝐵
𝑐𝐴 (
)
𝑐
Molecular Diffusion in Gases
Diffusion plus Convection
𝑁𝐴 =
𝑑𝑥𝐴
−𝑐𝐷𝐴𝐵
𝑑𝑧
+
𝑁𝐴 +𝑁𝐵
𝑐𝐴 (
)
𝑐
Molecular Diffusion in Gases
Uni-component Diffusion
One component (A)diffuses,
while the other (B) remains stagnant
Since B cannot diffuse, 𝑁𝐵 = 0
http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm
Molecular Diffusion in Gases
Uni-component Diffusion
Since B cannot diffuse, 𝑁𝐵 = 0
𝑁𝐴 =
𝑑𝑥𝐴
−𝑐𝐷𝐴𝐵
𝑑𝑧
+
𝑁𝐴 +0
𝑐𝐴 (
)
𝑐
http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm
Molecular Diffusion in Gases
Uni-component Diffusion
𝑑𝑥𝐴
𝑁𝐴
𝑁𝐴 = −𝑐𝐷𝐴𝐵
+ 𝑐𝐴
𝑑𝑧
𝑐
𝑐𝐷𝐴𝐵 𝑑𝑥𝐴
𝑁𝐴 = −
𝑥𝐵 𝑑𝑧
http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm
Molecular Diffusion in Gases
Uni-component Diffusion
When P is constant,
𝐷𝐴𝐵 𝑃 𝑑𝑝𝐴
𝑝𝐴
𝑁𝐴 = −
+ 𝑁𝐴
𝑅𝑇 𝑑𝑧
𝑃
𝐷𝐴𝐵 𝑃 𝑑𝑝𝐴
𝑁𝐴 = −
𝑅𝑇𝑃𝐵 𝑑𝑧
http://sst-web.tees.ac.uk/external/U0000504/Notes/ProcessPrinciples/Diffusion/Default.htm
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
1.01325  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 0.1524m long. Calculate the rate of evaporation
of water vapor at 293 K and 1 atm pressure. The diffusivity of water
in air is 0.250 x 10-4 m2/s. Assume that the system is isothermal.
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
Student No.
2011-18077
2011-57319
2010-04141
2010-01283
2010-31873
2011-07217
2011-03676
2010-36588
2011-18143
2011-18147
2011-09522
2011-30507
2011-09270
2010-53270
2011-14930
2009-21119
2011-21884
2011-19280
2011-26790
2010-21409
2011-01530
2011-30255
Quizzes
Total
Q 5/5
30
46
36
26
47
26
50
23
31
31
33
23
19
36
61
8
48
9
104
14
57
21
1.50
2.30
1.80
1.30
2.35
1.30
2.50
1.15
1.55
1.55
1.65
1.15
0.95
1.80
3.05
0.40
2.40
0.45
5.20
0.70
2.85
1.05
Machine Problems
Total* M 15/15
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
240
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
12
Target Average Scores
LE3
L 60/60 Final F 20/20
82
90
82
89
85
67
93
92
73
77
102
88
81
74
51
99
70
92
45
100
65
91
30.4
27.8
31
29.1
29
33.3
26.9
28.7
31.9
31.3
26
29.6
30.9
31.6
35
28.1
32.1
29.5
34.1
27.3
32.5
28.9
82
90
82
89
85
67
93
92
73
77
102
88
81
74
51
99
70
92
45
100
65
91
16.4
18.0
16.4
17.8
17.0
13.4
18.6
18.4
14.6
15.4
20.4
17.6
16.2
14.8
10.2
19.8
14.0
18.4
9.0
20.0
13.0
18.2