1
00:00:19,490 --> 00:00:25,690
Welcome to the fourth lecture of module one. This lecture will be on steady state molecular
2
00:00:25,690 --> 00:00:31,230
diffusion in fluids under stagnant and laminar
flow conditions.
3
00:00:31,230 --> 00:00:38,230
So, before proceeding to this lecture we will
have some recap on the previous lecture. In
4
00:00:40,109 --> 00:00:47,109
the previous lecture, we have discussed fixed
first law of diffusion, where for one-dimensional
5
00:00:53,379 --> 00:01:00,379
diffusion in a particular direction x, we
can write the molar flux J A, x is equal to
6
00:01:06,890 --> 00:01:13,890
minus D AB dC A dx; dC A is the concentration
gradient and dx is the distance and D AB is
7
00:01:23,420 --> 00:01:30,420
the diffusion coefficient. The negative sign
indicates the drop in concentration towards
8
00:01:30,780 --> 00:01:37,780
the directions of diffusion. And then we have
seen, for an ideal gas mixture the diffusion
9
00:01:38,900 --> 00:01:44,560
coefficient between component A and B, we
call the mutual diffusion coefficients are
10
00:01:44,560 --> 00:01:51,560
equal and we can write D AB is equal to D
BA.
11
00:01:55,130 --> 00:02:02,130
And then we have discussed the unsteady state
diffusion, where rate of diffusion changes
12
00:02:05,180 --> 00:02:12,180
with respect to time. And we have derived
the governing equations containing unsteady
13
00:02:12,610 --> 00:02:19,610
state diffusion, and governing equations are
del C A del t is equal to D AB del C A del
14
00:02:26,969 --> 00:02:33,969
x 2 plus del 2 C A del y 2 plus del 2 C A
del Z 2 and this is known as fixed second
15
00:02:49,310 --> 00:02:56,310
law, and this is frequently applicable in
solids and less frequently or limited situations
16
00:03:21,209 --> 00:03:28,209
in fluid.
17
00:03:29,239 --> 00:03:36,239
So, today we will start with steady state
molecular diffusion in fluids and in this
18
00:03:43,739 --> 00:03:50,739
case there are stagnant conditions and laminar
flow condition. And also, this can occur in
19
00:03:52,999 --> 00:03:59,999
two different geometry, one is constant area,
constant area and variable area. So, today
20
00:04:14,409 --> 00:04:21,409
we will discuss only on constant area and
later we will continue with the variable area.
21
00:04:23,020 --> 00:04:30,020
So, for constant area let us assume
diffusion in x-direction, no chemical reaction,
binary system and under steady state condition.
22
00:05:17,260 --> 00:05:24,260
So, then for diffusion of particular species
A we can write N A is equal to minus C D AB
23
00:05:34,740 --> 00:05:41,740
d Y A dx plus Y A N. Now, if we separate the
variables we can write minus d Y A by N A
24
00:05:59,640 --> 00:06:06,640
minus Y A N is equal to dx divided by C D
AB.
25
00:06:11,590 --> 00:06:18,590
Now, consider mixture of gases, gas mixture,
which is at constant pressure and temperature.
26
00:06:30,260 --> 00:06:37,260
Then, the concentration and diffusion coefficient
D AB, these are constant and independent of
27
00:06:46,740 --> 00:06:53,740
position. So, we can write the boundary condition
as, at x is equal to x 1, the mole fraction
28
00:07:03,640 --> 00:07:10,640
of component 1 will be Y A1; at x is equal
to x 2, the mole function of component 2,
29
00:07:15,660 --> 00:07:22,660
Y A2. So, using these boundary conditions
if we integrate the previous equations, so
30
00:07:25,700 --> 00:07:32,700
we can write Y A1 Y A2 minus d Y A by N A
minus Y A N is equal to integral x 1 to x
31
00:07:48,100 --> 00:07:55,100
2 dx divided by C D A B. So, this is equation
number 2.
32
00:08:03,130 --> 00:08:10,130
Let N A minus Y A N is equal to z. So, then
we
can write minus d Y A N will be dz and hence,
33
00:08:31,280 --> 00:08:38,280
d minus d Y A would be equal to d z by N.
So, if we substitute this over here, with
34
00:08:47,200 --> 00:08:54,200
the change of limit we can write, z 1 to z
2 dz by z 1 by N will be equal to 1 by C D
35
00:09:10,480 --> 00:09:16,560
AB integral x 1 to x 2 dx.
36
00:09:16,560 --> 00:09:23,560
So, now, upon integration write 1 by
N ln z 2 by z 1 is equal to 1 by C D AB x
2 minus x 1. So, if we substitute the limit
37
00:09:49,210 --> 00:09:56,210
z 1, then we can write ln N A minus Y A 2
N divided by N A minus Y A1 N would be equal
38
00:10:12,270 --> 00:10:19,270
to N by C D AB into x 2 minus x 1. So, again,
we can write ln N into N A by N minus Y A2
39
00:10:39,500 --> 00:10:46,500
divided by N N A by N minus Y A1 is equal
to N by C D AB into x 2 minus x 1 and this
40
00:11:06,620 --> 00:11:13,140
will cancel out.
And if we rearrange this one we can write,
41
00:11:13,140 --> 00:11:20,140
1 is equal to C D AB by x 2 minus x 1 and
1 by N ln N A
by N minus Y A 2 divided by N A by N minus
42
00:11:55,220 --> 00:12:02,220
Y A1. So, if we multiplied both sides by N
A, so these equations will become N A and
43
00:12:20,250 --> 00:12:27,250
N A by N this is equation three. So, this
is the final form of the steady state molecular
44
00:12:33,000 --> 00:12:39,220
diffusion of a component A to a constant area.
45
00:12:39,220 --> 00:12:46,220
And then if we use these equations for a particular
case, steady state diffusion through non-diffusing
46
00:12:47,550 --> 00:12:54,550
components, so consider A is diffusing
and B is non-diffusing. In this case since
B is non-diffusing, so flux for B component,
47
00:13:16,330 --> 00:13:23,330
N B will be 0 and the flux for component A
will be constant. Now, if we go back to the
48
00:13:30,430 --> 00:13:37,430
governing equation we can write, N A is equal
to N A by N C D AB by x 2 minus x 1 ln N A
49
00:13:46,610 --> 00:13:53,610
by N minus Y A2 by N A by N minus Y A1.
Now, since N is equal to N A plus N B and
50
00:14:17,709 --> 00:14:24,709
N B is equal to 0, in this case, so N will
be N A. So, this term N A by N will be equal
51
00:14:32,860 --> 00:14:39,860
to N A by N A is equal to 1. So, now, this
equation will become N A is equal to C D AB
52
00:14:53,360 --> 00:15:00,360
by x 2 minus x 1 ln. So, this term becomes
1, this becomes 1 and this becomes 1. So,
53
00:15:11,850 --> 00:15:18,850
this will be 1 minus Y A 2 by 1 minus Y A
1. So, this is equation number 4.
54
00:15:26,649 --> 00:15:33,649
Now, consider the system is ideal gas. So,
in this case we can write, the concentration
55
00:15:39,380 --> 00:15:46,380
is equal to total pressure P t divided by
RT; mole fractions Y A, we can write, partial
56
00:15:52,170 --> 00:15:59,170
pressure of component A by total pressure.
So, in terms of partial pressure we can write
57
00:16:00,420 --> 00:16:07,420
the earlier equations, N A is equal to D AB
P t by RT x 2 minus x 1 into ln P t minus
58
00:16:27,899 --> 00:16:34,899
p A2, partial pressure of component A active,
divided by total pressure minus partial pressure
59
00:16:41,519 --> 00:16:48,519
of component A at location 1. So, p A1 and
p A2 are the partial pressure at location
60
00:16:56,850 --> 00:17:01,940
2 and 1.
And for binary gas mixtures we know, that
61
00:17:01,940 --> 00:17:08,940
p A plus p B, at any location will be total
pressure, so that we can write P t minus p
62
00:17:24,309 --> 00:17:31,309
A2 will p B2. Similarly, P t minus p A1 is
p B1 and p A1 minus p A2 will be p B2 minus
63
00:17:53,309 --> 00:18:00,309
p B1. From this we can write, p A1 minus p
A2, partial pressure at location 1 and 2 for
64
00:18:04,929 --> 00:18:11,929
component A divided by partial pressure of
component B at location 2 minus partial pressure
65
00:18:15,549 --> 00:18:22,549
of, component, component B at location 1 will
be equal to 1.
66
00:18:24,340 --> 00:18:31,340
So, this if we substitute in this relation
equation 5, so now if we substitute this P
67
00:18:36,980 --> 00:18:43,980
t minus p A2 by p B2 and P t minus p A1 by
p B1 and then if we multiply it, this equations
68
00:18:49,450 --> 00:18:56,450
by this ratio, which is 1, we will essentially
have N A is equal to D AB P t by RT x 2 minus
69
00:19:16,740 --> 00:19:23,740
x 1 into p A1 minus p A2 by p B2 minus p B1
ln p B2 by p B1. So, this is equation 6.1.
70
00:19:44,900 --> 00:19:51,900
Now, if we rearrange this equation D AB P
t divided by R T x 2 minus x 1 into p A1 minus
71
00:20:09,250 --> 00:20:16,250
p A2 divided by p B2 minus p B1 divided by
ln p B2 by p B1. So, this term, essentially,
72
00:20:29,720 --> 00:20:36,720
we called the partial pressure difference,
so we can write p BLM. Then, we can write
73
00:20:40,340 --> 00:20:47,340
d flux N A will be D AB P t, total pressure,
divided by RT x 2 minus x 1 p BLM into the
74
00:21:01,520 --> 00:21:08,520
partial pressure difference p A1 minus p A2.
So, this is equation 6.2.
75
00:21:12,559 --> 00:21:19,559
Now, consider carbon dioxide is diffusing
through non-diffusing air under steady state
76
00:21:20,429 --> 00:21:27,429
conditions at a total pressure of 1 atmosphere
and temperature 300 Kelvin. The partial pressure
77
00:21:28,169 --> 00:21:35,169
of CO 2 is 20 kilo Pascal at one point and
5 kilo Pascal at other point. The distance
78
00:21:37,620 --> 00:21:44,620
is given between the two points is 5 centimeter
and we have to calculate the flux of CO 2.
79
00:21:45,679 --> 00:21:52,530
The diffusivity at a particular condition
is given CO 2, 2 into 10 to the power minus
80
00:21:52,530 --> 00:21:59,530
5 meter square per second.
Let us assume ideal gas
81
00:22:14,610 --> 00:22:21,610
and component, B, air, air, component air
is considered as B. So, we can write the flux
82
00:22:28,840 --> 00:22:35,840
equations of component A to non-diffusing
B. We can write flux of CO 2 is equal to D
83
00:22:41,600 --> 00:22:48,600
CO 2 air divided by RT x 2 minus x 1 P t by
P BLM into p CO 2 at 1 minus p CO 2 at 2.
84
00:23:11,290 --> 00:23:18,290
The data, which is given is diffusion coefficient,
total pressure, R and T are known, distance
85
00:23:22,080 --> 00:23:27,220
between the two points are known, partial
pressure were given, we have to calculate
86
00:23:27,220 --> 00:23:29,120
P BLM.
87
00:23:29,120 --> 00:23:36,120
So, let us consider D CO 2 air, which is 2
into 10 to the power minus 5 meter square
88
00:23:46,960 --> 00:23:53,960
per second; P t, total pressure is given 1
atmosphere, which is 101.3 kilo Pascal, which
89
00:24:02,100 --> 00:24:09,100
is 1.013 10 to the power 5 Pascal; t is given
300 Kelvin; x 2 minus x 1, the distance between
90
00:24:20,330 --> 00:24:27,330
the two points is 5 centimeter, which is 0.05
meter; partial pressure of CO 2 at 0.2 is
91
00:24:37,240 --> 00:24:44,240
20 kilopascal is equal to 20000 Pascal; partial
pressure of CO 2 at 0.2 is given 5 kilopascal,
92
00:24:52,700 --> 00:24:59,700
which is 5000 Pascal and R is known to us,
8314 in SI unit Pascal meter cube per K mole
93
00:25:10,160 --> 00:25:17,160
Kelvin.
And then we have to calculate the P BM, p
94
00:25:18,530 --> 00:25:25,530
B, 1 is P t minus partial pressure of CO 2
at 0.1, which is equal to 101.3 minus 20 kilopascal,
95
00:25:35,390 --> 00:25:42,390
which is equal to 81.3 kilopascal; p B, 2,
P t minus p CO 2 at point 2, which is 101.3
96
00:25:55,450 --> 00:26:02,450
minus 5 kilo Pascal, which is 96.3 kilo Pascal.
So, putting this value we can calculate P
97
00:26:10,390 --> 00:26:17,390
BLM, which is p B, 2 minus p B, 1 by ln p
B2 by P B1. So, putting the values, 96.3 minus
98
00:26:30,360 --> 00:26:37,360
81.3 divided by ln 96.3 by 81.3 kilo Pascal
is 88.59 kilo Pascal, so which is equal to
99
00:26:51,080 --> 00:26:53,630
88590 Pascal.
100
00:26:53,630 --> 00:27:00,630
So, now, if we substitute in the flux equation,
N CO 2 we can write, this is diffusion coefficient,
101
00:27:04,870 --> 00:27:11,870
2 into 10 to the power minus 5 meter square
per second. And then the total pressure, 1.013
102
00:27:13,630 --> 00:27:20,630
into 10 to the power 5 Pascal divided by R,
8314 Pascal meter cube per K mole Kelvin into
103
00:27:34,610 --> 00:27:41,610
300 Kelvin. The distance 0.005 meter into
partial pressure difference is of component
104
00:27:48,600 --> 00:27:55,600
B, 88590 Pascal into 20000 partial pressure
of component A at point one minus 5000 Pascal.
105
00:28:06,750 --> 00:28:13,750
So, this will cancel out and this will cancel
out. So, if we calculate, the flux will be
106
00:28:23,640 --> 00:28:30,640
2.75 into 10 to the power minus 6 K mole per
meter square second.
107
00:28:37,059 --> 00:28:44,059
Now, consider the other case where steady
state equimolar counter diffusion is occurring.
108
00:28:45,100 --> 00:28:52,100
If there is an equimolar counter current diffusion
for our governing equations, as we know, N
109
00:28:55,669 --> 00:29:02,669
A is equal to minus C D AB dY A dx plus Y
A N. This is equation number 1, as we have
110
00:29:15,909 --> 00:29:21,700
discussed earlier.
And for equimolar counter current diffusion,
111
00:29:21,700 --> 00:29:28,700
N A is equal to minus N B N is equal to constant.
So, N, which is equal to N A plus N B, would
112
00:29:44,030 --> 00:29:51,030
be equal to 0. Since this is 0, so we can
write N A is equal to minus C D AB dY A dx.
113
00:30:06,320 --> 00:30:13,320
Assume ideal gas, if we assume ideal gas,
let C is equal to total pressure by RT.
114
00:30:23,100 --> 00:30:30,100
So, this equations will be N A, is equal to
minus D AB P t by RT dY A dx. So, this is
115
00:30:54,570 --> 00:31:01,570
equation number 7. Now, if we use the boundary
conditions, at x is equal to x 1, Y A will
116
00:31:12,539 --> 00:31:19,539
be Y A1; at x is equal to x 2, Y A will be
Y A 2. So, using this boundary condition we
117
00:31:25,320 --> 00:31:32,320
can write, flux will be equal to D AB P t
by RT x 2 minus x 1 into Y A1 minus Y A2.
118
00:31:48,090 --> 00:31:55,090
And hence, we can write, D AB by RT x 2 minus
x 1. If it multiplied by the total pressure,
119
00:32:00,940 --> 00:32:07,940
this will be partial pressure p A1 minus p
A2 of component A. So, this is equation 8.
120
00:32:18,039 --> 00:32:25,039
Now, let us have an example where the equimolar
counter current diffusion is occurring. Consider
121
00:32:26,090 --> 00:32:33,090
the similar gas as we have considered in the
example 1, CO 2 is diffusing at steady state
122
00:32:35,000 --> 00:32:42,000
through a straight tube of 0.5 meter long.
The distance is given with an inside diameter
123
00:32:43,470 --> 00:32:50,470
0.05 meter, which contains the nitrogen gas
at 300 Kelvin and 1 atmosphere pressure.
124
00:32:57,669 --> 00:33:03,320
The partial pressure is given at two points,
at one point 15 kilopascal and other point
125
00:33:03,320 --> 00:33:10,320
is 5 kilopascal and the diffusion coefficient,
d CO 2 nitrogen is given 4 into 10 to the
126
00:33:14,419 --> 00:33:20,289
power of minus 5 meter square per second.
And we have to calculate molar flow rate of
127
00:33:20,289 --> 00:33:27,289
component A, that is, CO 2 and the molar flow
rate of component B, that is N 2.
128
00:33:28,400 --> 00:33:35,400
Let us assume ideal gas
and then we know the flux equation N CO 2
will be equal to D CO 2 into nitrate to the
129
00:33:52,080 --> 00:33:59,080
nitrogen, RT x 2 minus x 1 into partial pressure
of CO 2 at point 1 minus partial pressure
130
00:34:05,710 --> 00:34:12,710
of CO 2 at point 2. So, the data, which are
given, diffusion coefficient of CO 2 nitrogen,
131
00:34:18,849 --> 00:34:25,849
4 into 10 to the power minus 5 meter square
per second pressure is given; P t, which is
132
00:34:36,300 --> 00:34:43,300
1 atmosphere, 101.3 kilopascal, which is equal
to 1013 into 10 to the power 5 Pascal and
133
00:34:51,710 --> 00:34:58,710
temperature is given, which is 300 Kelvin.
And the distance, x 2 minus x 1, between two
134
00:35:10,280 --> 00:35:17,280
points is 0.5 meter. The partial pressure
of CO 2 at point 1 is 15 kilo Pascal, which
135
00:35:26,540 --> 00:35:33,540
is 15000 Pascal and
the partial pressure of CO 2 at point 2 is
5 kilo Pascal, which is 5000 Pascal and R
136
00:35:47,589 --> 00:35:54,589
is given, R is known to us, 8314 Pascal meter
cube K mole Kelvin.
137
00:36:02,000 --> 00:36:09,000
Now, if we substitute in this equation, the
flux N CO 2 will be equal to diffusivity,
138
00:36:12,940 --> 00:36:19,940
4 into 10 to the power minus 5 meter square
per second and RT, 8314 Pascal meter cube
139
00:36:30,920 --> 00:36:37,920
per K mole Kelvin into 300 Kelvin into 0.5
meter multiplied by the partial pressure difference,
140
00:36:47,890 --> 00:36:54,890
15000 minus 5000 Pascal. So, if we calculate,
this will be equal to 3.21 into 10 to the
141
00:37:09,500 --> 00:37:16,500
power minus 7 K mole per meter square second.
So, this is the flux of CO 2.
142
00:37:21,490 --> 00:37:28,490
We have to calculate the molar flow rate of
CO 2. So, to calculate the molar flow rate,
143
00:37:28,609 --> 00:37:35,609
the molar flow rate is equal to flux of CO
2 multiplied by the cross-sectional area of
144
00:37:45,910 --> 00:37:52,910
the tube which is A. Internal diameter is
given, say, D i is 0.05 meter and then we
145
00:38:03,150 --> 00:38:10,150
can calculate the cross-sectional area, A
is equal to pi D i square by 4 is equal to
146
00:38:15,480 --> 00:38:22,480
pi by 4 0.05 square meter square, which is
equal to 1.96 into 10 to the power minus 3
147
00:38:35,380 --> 00:38:42,380
meter square. So, the molar flow rate, CO
2, is equal to, flux we have obtained, N,
148
00:38:52,810 --> 00:38:59,810
earlier which is 3.21 into 10 to the power
minus 7 K mole per meter square second.
149
00:39:01,680 --> 00:39:08,680
So, 3.21 into 10 to the power minus 7 K mole
per meter square second multiplied by the
150
00:39:13,750 --> 00:39:20,750
cross-sectional area, 1.96 10 to the power
minus 3 meter square. So, if we multiply,
151
00:39:23,270 --> 00:39:30,270
this will cancel out and then it will be 6.29
10 to the power minus 10 K mole per second.
152
00:39:39,070 --> 00:39:46,070
Similarly, for component B, that is, nitrogen
we can calculate, N, nitrogen is equal to
153
00:39:52,619 --> 00:39:59,619
D N 2 CO 2 divided by RT x 2 minus x 1. Partial
pressure of N 2 at point 1 minus partial pressure
154
00:40:12,260 --> 00:40:19,260
of N 2 at point 2, now p N 2 at point 1 will
be P t minus p CO 2 point 1, which is 101.3
155
00:40:32,680 --> 00:40:39,680
minus 15 kilo Pascal, which is equal to 86.3
kilo Pascal and 86300 Pascal; p N 2 at point
156
00:40:56,000 --> 00:41:03,000
2 is P t minus p CO 2 at point 2 is 101.3
minus 5 kilo Pascal is equal to 96.3 kilo
157
00:41:15,540 --> 00:41:22,540
Pascal, which is equal to 96300 Pascal.
We know that for equimolar counter current
158
00:41:26,460 --> 00:41:33,460
diffusion, D CO 2 N 2 is equal to D N 2 CO
2, so we can use the same diffusion coefficient
159
00:41:39,109 --> 00:41:46,109
and calculate the flux N N 2. If we incorporate
this data, so it will be minus 3.21 into 10
160
00:41:50,890 --> 00:41:57,890
to the power minus 7 K mole per meter square
second.
161
00:42:01,849 --> 00:42:08,849
Similarly, the molar flow rate of N 2 we can
calculate, minus 3.21 into 10 to the power
162
00:42:19,619 --> 00:42:26,619
minus 7 K mole per meter square second into
1.96 into 10 to the power minus 3 meter square.
163
00:42:31,460 --> 00:42:38,460
So, it will be essentially same value, but
with a negative sign, which indicates, that
164
00:42:40,880 --> 00:42:47,880
the flux is in opposite direction compared
to the carbon dioxide, so K mole per second.
165
00:42:54,520 --> 00:43:01,520
So, another type of diffusion occurs where
the component are not equimolecular in nature.
166
00:43:05,220 --> 00:43:12,220
So, there are many situations, they are non-equimolar
counter diffusion. Let us consider the following
167
00:43:12,220 --> 00:43:19,220
reaction, twice A plus B for being twice C,
where 1 mole of A diffuses towards, towards
168
00:43:37,839 --> 00:43:44,839
B, whereas 2 moles of B diffuses back or in
opposite direction. So, in this case, the
169
00:44:07,579 --> 00:44:14,579
N A will be minus N B by 2. And we can derive
the governing equations considering our basic
170
00:44:21,609 --> 00:44:28,609
equations. So, this is end of discussion of
lecture 2 and in the next lecture we will
171
00:44:31,099 --> 00:44:37,579
consider the steady state diffusion through
non-uniform geometries.
172
00:44:37,579 --> 00:44:38,540
Thank you.