WEBVTT
Kind: captions
Language: en
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welcome to the section of this course so we
have in the last class we have already seen
00:00:25.980 --> 00:00:30.830
how the um system performance in osmotic
pressure controlling filtration can be
00:00:30.830 --> 00:00:35.930
can be obtain in case of alter filtration
system and then we have looked into the revers
00:00:35.930 --> 00:00:40.870
osmosis nanofiltration system by replacing
the definition of real intention by the solute
00:00:40.870 --> 00:00:47.079
mass flux to the . membrane in here to connect
the membrane surface constriction and the
00:00:47.079 --> 00:00:53.510
hermit constriction of the solute so we are
then looking in to the various you know . of
00:00:53.510 --> 00:00:58.570
the solution diffusion model and how to include
those in our calculations inward to get a
00:00:58.570 --> 00:01:04.610
system performance so next . model the modification
we be looking into the kedem katechalsky model
00:01:04.610 --> 00:01:12.819
for the r o n f system . for reverse osmosis
and nanofiltration system in this case the
00:01:12.819 --> 00:01:20.569
defuse the the osmotic equation module of
the . law is modified as l p delta p minus
00:01:20.569 --> 00:01:26.320
sigma . delta pi where sigma is known as the
reflection coefficient
00:01:26.320 --> 00:01:38.649
.
00:01:38.649 --> 00:01:50.549
so one can get the expression of solute
ha solvent flux as l p delta p minus a sigma
00:01:50.549 --> 00:01:57.170
c m minus c p so we are assuming that pi is
a linear function of constriction a
00:01:57.170 --> 00:02:04.009
a times c a times c in order get a simplification
. in our calculation so sigma is a reflection
00:02:04.009 --> 00:02:17.100
coefficient and then will be using the film
theory model . film theory model as c m is
00:02:17.100 --> 00:02:31.030
equal to c p plus c naught minus c p exponential
v w by k and then we have the solution diffusion
00:02:31.030 --> 00:02:49.569
model . that is v w times c p is equal
to b times c m minus c p so we have three
00:02:49.569 --> 00:02:54.700
equations three are knows v w c m c p we have
three independent equations involving these
00:02:54.700 --> 00:02:59.620
three parameters again three equations three
unknown system this is highly solvable in
00:02:59.620 --> 00:03:05.890
by using . a trial and errors method this
set of equations can be solved . go obtain
00:03:05.890 --> 00:03:11.879
the permit constriction permit flux and ambient
surface constriction
00:03:11.879 --> 00:03:15.930
and . the it has to be just noted that if
we are talking about a completely detentive
00:03:15.930 --> 00:03:22.209
membrane then reflection coefficient sigma
is equal to one we will be looking into the
00:03:22.209 --> 00:03:34.519
next variant of hm . of solution defection
model this is known as the modified . solution
00:03:34.519 --> 00:03:48.099
diffusion model
for r o n f and it is used sometimes for u
00:03:48.099 --> 00:03:54.849
f also so what is the modification modifications
here is we are we are invoking a term because
00:03:54.849 --> 00:04:00.279
if we are go for a more open membrane for
example reverse osmosis membrane is is assume
00:04:00.279 --> 00:04:06.069
. to be almost an impermeably imperiousness
membrane and its its such a dense membrane
00:04:06.069 --> 00:04:11.549
in case of nanofiltration lower cutoff membranes
the pro size is increased compared to the
00:04:11.549 --> 00:04:17.890
. reverse osmosis membranes so there are appearance
of small post into the mem well defined small
00:04:17.890 --> 00:04:20.180
post inside the membrane matrix
00:04:20.180 --> 00:04:25.900
now if you go for the higher cutoff of nanofiltration
membrane for example six hundred eight hundred
00:04:25.900 --> 00:04:32.650
nine hundred . cutoff then the pro size of
the membranes will be more and the defined
00:04:32.650 --> 00:04:39.080
force will be appearing more in number in
the membrane matrix so therefore if the defined
00:04:39.080 --> 00:04:45.180
force will appear then in . addition to the
diffusive flux of solute through the membrane
00:04:45.180 --> 00:04:51.190
the convective flux through the membrane will
be also predominant so later in the . literal
00:04:51.190 --> 00:04:56.800
monoculate cutoff of nanofiltration membrane
both convective and diffusive flux will be
00:04:56.800 --> 00:05:02.800
equally important and there will be correction
factor or . modification factors will be appearing
00:05:02.800 --> 00:05:08.710
in the solution diffusion model in order to
incorporate the convective flux to the membrane
00:05:08.710 --> 00:05:17.350
so therefore the modified equation will be
the . modified solution defection model can
00:05:17.350 --> 00:05:34.139
be written as v w c p is equal to beta c m
minus c p plus one minus sigma v w c average
00:05:34.139 --> 00:05:44.199
ok so this is the additional term that will
be taking care of the convective flux . through
00:05:44.199 --> 00:05:50.509
the membrane and what is the average c average
have so sigma is the reflection coefficient
00:05:50.509 --> 00:05:56.110
v w is the permit flux c average is the average
solute constriction in the membrane matrix
00:05:56.110 --> 00:06:02.722
and c average is basically a logmein constriction
difference between the . c m and c p it will
00:06:02.722 --> 00:06:11.830
be c m minus c p divided by l n c m divided
by c p this is a more realistic model for
00:06:11.830 --> 00:06:17.870
the solute flux to the membrane surface in
case of open nanofiltration membrane or ultrafiltration
00:06:17.870 --> 00:06:23.009
membrane so in case of ultrafiltration membrane
the pro sizes even larger and so therefore
00:06:23.009 --> 00:06:28.900
the so convective flux will be sometimes predominant
compared to the defensive flux
00:06:28.900 --> 00:06:34.900
so in that case the one has to take care of
the convective flux as well so therefore
00:06:34.900 --> 00:06:41.111
. we can modified the solution diffusion model
in this fashion and if you really um look
00:06:41.111 --> 00:06:46.410
into the you know calculations you have the
film theory by . in the solvent flux to
00:06:46.410 --> 00:06:56.759
the membrane matrix by the . law so that will
be nothing but v w l p delta p minus sigma
00:06:56.759 --> 00:07:06.130
delta pi and we have the film theory model
. for the flow through by the flow through
00:07:06.130 --> 00:07:14.430
the flow chancel outside the membrane v w
is equal to k l n c m minus c p divided by
00:07:14.430 --> 00:07:21.430
c naught minus c p and delta pi can be expressed
in terms of c m and c p if you . construct
00:07:21.430 --> 00:07:28.199
b is pi is a you know pi pi is a linear
function of concentration
00:07:28.199 --> 00:07:35.540
so again three equations and three unknowns
v w c m and c p this three equation three
00:07:35.540 --> 00:07:43.639
unknowns solid systems can be solvable three
equations and three unknown and they can be
00:07:43.639 --> 00:07:52.349
solvable and one can get a system perdition
v w c p and c m ok so this are the various
00:07:52.349 --> 00:07:58.290
variants of solution defection model in order
to get the system performance in a in a very
00:07:58.290 --> 00:08:03.570
predictive mode only some of the parameters
. like l p real intention b which is basically
00:08:03.570 --> 00:08:07.940
the solution diffusion . construct for the
solution diffusion model this has to be this
00:08:07.940 --> 00:08:13.659
have to be estimated separately by by conducting
separate set of experiments and once these
00:08:13.659 --> 00:08:19.819
are done then if you know the operating conditions
if you know the transport co efficients if
00:08:19.819 --> 00:08:26.979
you know the geometry and hm the the solution
properties then one can have a system performance
00:08:26.979 --> 00:08:33.409
in a predictive mode next will be looking
into the so that's all about the one dimensional
00:08:33.409 --> 00:08:38.289
film theory model for modeling the transport
you know concentration polarized in the mass
00:08:38.289 --> 00:08:46.350
transfer boundary layer and combining it with
the . the transport processes those are occurring
00:08:46.350 --> 00:08:52.540
within the membrane surface membrane matrix
combination of these will be really . going
00:08:52.540 --> 00:08:55.910
in to the predict the performance of the system
completely
00:08:55.910 --> 00:09:00.740
next will be looking into the short coming
of one dimensional . modeling or the film
00:09:00.740 --> 00:09:05.929
theory model and then will be adding completion
to the system by considering a two dimensional
00:09:05.929 --> 00:09:11.770
model on before that we are to just look into
the short comings or limitations of the one
00:09:11.770 --> 00:09:39.939
dimensional model . so the the first major
limitation of the one dimensional model is
00:09:39.939 --> 00:09:57.959
the constant thickness of .
thickness of . mass transfer boundary layer
00:09:57.959 --> 00:10:05.000
so as you have describe earlier . is that
if there is a membrane surface there will
00:10:05.000 --> 00:10:12.459
be a development of mass transfer boundary
layer . the membrane surface ok and in the
00:10:12.459 --> 00:10:20.260
in case of film theory we are assuming a constant
thickness of the membrane surface now it let
00:10:20.260 --> 00:10:25.820
us let us look into two two cases separately
it it should not super pose on each of them
00:10:25.820 --> 00:10:33.100
so this is a constant thickness of film or
solute over the membrane surface this is case
00:10:33.100 --> 00:10:42.650
case a and in the second case there is a developing
mass transfer boundary layer over the membrane
00:10:42.650 --> 00:10:50.920
surface so as we have seen earlier that the
thickness of mass transfer boundary layer
00:10:50.920 --> 00:10:56.709
will be offering a resistance against the
solvent flux and against the solvent flow
00:10:56.709 --> 00:11:02.829
if the thickness is more the less self . permit
flux we are going to get if the thickness
00:11:02.829 --> 00:11:08.150
of mass transfer boundary layer is less then
we are going to get a higher permit flux or
00:11:08.150 --> 00:11:13.180
the solvent flux across the membrane so in
case of constant thickness of mass transfer
00:11:13.180 --> 00:11:21.260
boundary layer the permit flux we going to
get as same we are going to get same permit
00:11:21.260 --> 00:11:25.490
flux through the membrane as you go along
the length of the module
00:11:25.490 --> 00:11:32.630
but in case of developing masters and so actually
this is a very realistic case and very unrealistic
00:11:32.630 --> 00:11:36.260
case when were you are considering the masters
thickness of the mass transfer boundary layer
00:11:36.260 --> 00:11:43.840
to be constant why because let us look into
the you know entrance length of any boundary
00:11:43.840 --> 00:11:51.170
layer for entrance length of a hydrodynamic
boundary layer is point zero five times reynolds
00:11:51.170 --> 00:12:03.630
number ok so this is a case of hydrodynamic
boundary layer or velocity boundary layer
00:12:03.630 --> 00:12:09.030
. . now if you considered the equipment dynamics
in the ordered of millimeter and reynolds
00:12:09.030 --> 00:12:14.890
number is in the order of two thousands so
there in laminar the . l e so if it is the
00:12:14.890 --> 00:12:23.699
order of you know millimeter . and reynolds
number is in the order of let set two thousand
00:12:23.699 --> 00:12:31.370
then entrance length will be order of you
know few centimeter so one to five centimeter
00:12:31.370 --> 00:12:37.709
that means if your if your construing a chancel
a module which will be having a length of
00:12:37.709 --> 00:12:43.579
one meter initial apart from initial five
centimeter or four centimeter the full chancel
00:12:43.579 --> 00:12:47.880
will be having the develop fully developed
hydrodynamic boundary layer or velocity boundary
00:12:47.880 --> 00:12:54.549
layer or let us look into the entrance length
for the mass transfer boundary layer for mass
00:12:54.549 --> 00:13:01.339
transfer boundary layer the entrance length
will be l e by d e is equal to point . zero
00:13:01.339 --> 00:13:04.770
five reynolds times spid ok
00:13:04.770 --> 00:13:10.150
so what is reynolds number reynolds number
we said say laminar two thousand ok and d
00:13:10.150 --> 00:13:14.329
inequalities is one millimeter in the order
of one millimeter let us look into the spid
00:13:14.329 --> 00:13:21.620
number spid number is mu by rho d now in case
of membrane filtration we are we will be selecting
00:13:21.620 --> 00:13:26.189
solutes which will be higher pro size i mean
i know larger size compare to the pro size
00:13:26.189 --> 00:13:30.919
of the membrane so that there will be separated
out of the membrane by by the physical separation
00:13:30.919 --> 00:13:37.130
processes now let us if you even if you taken
a viscosity of water so . minus three and
00:13:37.130 --> 00:13:42.169
density of water is ten cube and the diffusivity
typical diffusivity is in the order of ten
00:13:42.169 --> 00:13:46.790
to the minus eleven to ten to the minus twelve
meter square per second if it is ten to the
00:13:46.790 --> 00:13:53.040
minus eleven then the spid number will
be order of ten to the power five if if diffusivity
00:13:53.040 --> 00:13:58.140
is in the order of ten to the power minus
twelve then the spid number will be in the
00:13:58.140 --> 00:14:02.780
order of ten to the six so therefore . in
case of membrane separation we are talking
00:14:02.780 --> 00:14:07.709
about the solutes which will be having very
high spid number if the spid number is very
00:14:07.709 --> 00:14:12.950
high and then if you really calculated the
entrance length the entrance length will be
00:14:12.950 --> 00:14:18.059
in the order of d e will be one millimeter
means ten to the minus three its yes spid
00:14:18.059 --> 00:14:23.330
number will be ten to the power five or ten
to the power six into reynolds number so it
00:14:23.330 --> 00:14:25.010
will be ten to the power three
00:14:25.010 --> 00:14:35.110
so this will be in the order of ten to the
power five meter so entrance length for the
00:14:35.110 --> 00:14:40.900
mass transfer boundary layer in a membrane
chancel will be very very high so it is not
00:14:40.900 --> 00:14:47.020
that if you talk of if you consider if
you are considering two meter five meter . even
00:14:47.020 --> 00:14:50.950
even ten meter chancel length or the tube
length the maters of boundary layer will be
00:14:50.950 --> 00:14:56.209
still developing now let us look into the
interpretation of developing boundary layer
00:14:56.209 --> 00:15:01.430
if the boundary layer is developing then at
the begging of the chancel . the thickness
00:15:01.430 --> 00:15:06.309
of the boundary layer is less so therefore
it offers less resistance against the solving
00:15:06.309 --> 00:15:12.300
flow so the solvent flux will be maximum ok
as the thickness increases it offers more
00:15:12.300 --> 00:15:17.310
resistance and solvent flux is lower so it
will be less so like that it will it will
00:15:17.310 --> 00:15:21.990
be less so as you go down across the chancel
length and later on when it will be growth
00:15:21.990 --> 00:15:26.900
will be almost sluggish then you will be getting
the constant value almost constant value of
00:15:26.900 --> 00:15:33.079
permit flux so at the form down stream of
the chancel were growth of the mass transfer
00:15:33.079 --> 00:15:38.650
boundary layer will be almost sluggish that
means its thickness is almost equal then you
00:15:38.650 --> 00:15:42.950
are going to get a constant thickness of the
film over the membrane surface where you are
00:15:42.950 --> 00:15:47.710
getting the constant permit flux but the majority
of the chancel length will be night five percent
00:15:47.710 --> 00:15:52.220
of the chancel will be having a developing
mass transfer boundary layer where the permit
00:15:52.220 --> 00:15:53.559
flux is very very high
00:15:53.559 --> 00:15:59.319
so now if you assumed a constant thickness
of mass transfer boundary layer is formed
00:15:59.319 --> 00:16:05.049
over the membrane surface . you will be assuming
a constant permit flux over the membrane surface
00:16:05.049 --> 00:16:12.059
so the excess flux due to the developing mass
transfer boundary layer is already world looked
00:16:12.059 --> 00:16:18.530
so therefore if ones goes for a an assumption
of film theory which assumed a constant thickness
00:16:18.530 --> 00:16:24.810
of film one has to under predict or under
consider the permit flux so what are what
00:16:24.810 --> 00:16:29.980
are permit flux one will be getting that will
be always less than whatever in the actually
00:16:29.980 --> 00:16:35.510
scenario so therefore it is essential to go
for a two dimensional model where mass transfer
00:16:35.510 --> 00:16:41.819
boundary layer is a function of both x and
y and one will be getting an actual realistic
00:16:41.819 --> 00:16:46.490
picture for a developing mass transfer boundary
layer so there is the first assumption so
00:16:46.490 --> 00:16:50.419
what is the limitation of the one dimensional
model the limitation of one dimensional model
00:16:50.419 --> 00:16:56.250
is that if you want adopts the one dimensional
model then the one will be under predict the
00:16:56.250 --> 00:16:57.740
permit the flux
00:16:57.740 --> 00:17:04.610
what will be the second . assumption the second
assumption is masters for co efficient that
00:17:04.610 --> 00:17:22.809
we have used in one dimensional model . does
not include . the effects of suction .
00:17:22.809 --> 00:17:28.360
ok typically the schedule number relations
are obtain from the hidden master were . for
00:17:28.360 --> 00:17:33.980
example this hidden master were . and the
heat transfer nusselt number relations are
00:17:33.980 --> 00:17:40.430
basically for the . convict there are all
for impervious convict now in case of membrane
00:17:40.430 --> 00:17:44.880
separation system will be having the wall
or we having the convicts which will be having
00:17:44.880 --> 00:17:50.220
the . walls so therefore the effects of the
. will be will be really dominant in some
00:17:50.220 --> 00:17:54.780
cases following the example in case of reverse
osmosis the wall . will be not not that much
00:17:54.780 --> 00:17:59.950
the effect may not be very important for reverse
osmosis and nanofiltration for ultrafiltration
00:17:59.950 --> 00:18:06.240
. and microfiltration the . sufficiently large
and one will be the the mass transfer boundary
00:18:06.240 --> 00:18:11.950
the masters co efficient will be a function
of you know wall section in that case the
00:18:11.950 --> 00:18:17.700
effects of all positive will be we can not
be neglected so this two were the limitations
00:18:17.700 --> 00:18:22.390
of the . one dimensional film theory model
for modeling mass transfer boundary layer
00:18:22.390 --> 00:18:25.330
outside the membrane surface inside the flow
chancel
00:18:25.330 --> 00:18:35.750
now in order to . will be considering a two
dimensional . flow situation flow visualization
00:18:35.750 --> 00:18:50.070
. or modeling two dimensional flow modeling
in the . mass transfer boundary layers so
00:18:50.070 --> 00:18:56.560
we assumed the boundary layer which will be
the practically the . actual case is developing
00:18:56.560 --> 00:19:06.390
what the membrane system what is . the membrane
chancel this . y direction this x direction
00:19:06.390 --> 00:19:13.880
so this the half height of chancel so y is
equal to h here y is equal to zero here x
00:19:13.880 --> 00:19:20.100
is equal to l here at any point of time delta
is the thickness mass transfer boundary layer
00:19:20.100 --> 00:19:37.940
and delta will be a function a growing function
of x now let us consider a study state .
00:19:37.940 --> 00:19:44.270
laminar flow so let us let us right down the
mass transfer . you know solute balance equation
00:19:44.270 --> 00:19:51.910
within the mass transfer for boundary layer
and see what we get if we write down a solute
00:19:51.910 --> 00:20:07.640
balance equation .
within mass transfer boundary layer let see
00:20:07.640 --> 00:20:18.270
what we get at the study u del c del x plus
v del c and del y is equal to d del square
00:20:18.270 --> 00:20:26.620
c del x square plus d del square c del y square
ok
00:20:26.620 --> 00:20:32.780
so first term represent convention in the
retraction or in second term reflects the
00:20:32.780 --> 00:20:38.770
convective flow in the y direction the first
term on the right side is the diffusive term
00:20:38.770 --> 00:20:44.690
diffusive flux in the x direction and this
is diffusive flux in the y direction but we
00:20:44.690 --> 00:20:51.890
already know that convection u is in the order
of centimeters and v is in the order of ten
00:20:51.890 --> 00:20:58.310
to the power minus six . in the order centimeter
one centimeter per seconds v is the order
00:20:58.310 --> 00:21:04.290
of ten to the power minus five to ten to the
power minus six . centimeter per second so
00:21:04.290 --> 00:21:09.900
therefore the convective flux in the x
direction will be much much dominate compare
00:21:09.900 --> 00:21:16.200
to the diffusive flux in the x direction so
we can neglect this term compare to the convective
00:21:16.200 --> 00:21:26.120
term x direction so u del c del x plus v del
c del y will be is equal to d del square c
00:21:26.120 --> 00:21:33.760
and del y square so now let us look into the
velocity profiles so what is u u is equal
00:21:33.760 --> 00:21:47.770
to for a for a laminar flow in rectangular
chancel three u zero by two one minus h minus
00:21:47.770 --> 00:21:54.110
y divided by h square so will be having a
typical the probability velocity profile as
00:21:54.110 --> 00:22:02.400
an as a laminar flow within the rectangular
chancel and h is the half height . and u naught
00:22:02.400 --> 00:22:09.700
is the processional average velocity so now
as we have already seen that in our case
00:22:09.700 --> 00:22:16.850
our case is a developed hydrodynamic boundary
layer because entrance length required for
00:22:16.850 --> 00:22:21.140
hydrodynamic boundary layer we completely
developed is few centimeter compare to the
00:22:21.140 --> 00:22:25.220
chancel chancel length of in orders of meter
00:22:25.220 --> 00:22:30.800
similarly but on the other hand the mass transfer
boundary layer will be really developing through
00:22:30.800 --> 00:22:36.530
out the whole chancel length so therefore
it is a case of fully developed master . hydrodynamic
00:22:36.530 --> 00:22:42.929
boundary layer and a developing mass transfer
boundary layer so it is case of fully developed
00:22:42.929 --> 00:22:59.480
.
hydrodynamic boundary layer and . developing
00:22:59.480 --> 00:23:12.980
. mass transfer boundary layer so now let
us again go back to the geometry of the system
00:23:12.980 --> 00:23:17.930
if you really go back to the geometry of the
system this is the chancel we are considering
00:23:17.930 --> 00:23:22.710
and this is the middle line of the chancel
y is equal to h this is y is equal to zero
00:23:22.710 --> 00:23:32.970
this is y is equal to two h and membrane is
placed . at the bottom and these are completely
00:23:32.970 --> 00:23:39.000
developed thermo you know hydrodynamic
boundary layer and the mass transfer boundary
00:23:39.000 --> 00:23:47.110
layer is still developing ok so beyond the
so . concentration within the mass transfer
00:23:47.110 --> 00:23:53.030
boundary layer is changing as a function is
the function of both x and y beyond the mass
00:23:53.030 --> 00:23:58.050
transfer boundary layer concentration is sme
it is equal to the bulk concentration within
00:23:58.050 --> 00:24:01.020
the boundary layer it is a function of x and
y .
00:24:01.020 --> 00:24:07.380
so therefore let us so . about the governing
equation whatever we have written of the solute
00:24:07.380 --> 00:24:20.240
mass balance equation that is valid for solute
balance equation that is u del c del x plus
00:24:20.240 --> 00:24:33.010
v del c del y is equal to d del square c del
y square is valid for . y is lying between
00:24:33.010 --> 00:24:39.460
zero and thickness of mass transfer boundary
layer now if that is the case and the velocity
00:24:39.460 --> 00:24:44.520
profile you have . take taken the complete
velocity profile is a parabolic velocity profile
00:24:44.520 --> 00:24:51.390
is three by two u naught one minus h minus
y divided by h square of that parabolic one
00:24:51.390 --> 00:24:56.790
we are talking . about the this complete velocity
profile but if you look into the mass transfer
00:24:56.790 --> 00:25:01.760
boundary . mass transfer boundary layer which
is the dominate of . validity of this equation
00:25:01.760 --> 00:25:07.059
we are only considering only this part of
the complete velocity profile and this part
00:25:07.059 --> 00:25:13.760
is a linear one so that means the mass transfer
boundary layer is really very very small compare
00:25:13.760 --> 00:25:21.020
to the actual dimension of the chancel so
if you look into that . the thickness of the
00:25:21.020 --> 00:25:26.820
mass transfer boundary layer can be can be
roughly assumed you know approximately by
00:25:26.820 --> 00:25:33.860
this equation delta by d e will be inversely
proportional to the spid number and delta
00:25:33.860 --> 00:25:40.690
if you if you talk about . if you put at this
spid numbers around ten to the power five
00:25:40.690 --> 00:25:47.700
this will be delta by d e will be less than
point point one or point zero one ok that
00:25:47.700 --> 00:25:56.680
means in in in with respect to a actual dimension
geometry dimension of the chancel the thickness
00:25:56.680 --> 00:26:01.130
mass transfer boundary layer is only one percent
of the whole chancel length .
00:26:01.130 --> 00:26:08.309
so therefore he can really neglect that means
y by h is very very small and we can neglect
00:26:08.309 --> 00:26:13.620
y square and h square and higher order terms
if you really do that we can expended this
00:26:13.620 --> 00:26:20.410
quadratic term you can find out the after
simplification this will becomes a linear
00:26:20.410 --> 00:26:26.470
one you can neglect y square by h square term
by opening up by opening up this square is
00:26:26.470 --> 00:26:32.790
one one will be canceling out and y square
by h square can be neglected and it will boil
00:26:32.790 --> 00:26:38.100
down to a linear velocity profile and that
will be corresponding to this figure that
00:26:38.100 --> 00:26:43.850
the part that that lies the part of the velocity
profile that lies within the mass transfer
00:26:43.850 --> 00:26:49.150
boundary layer is really a linear part of
the total parabolic profile of the velocity
00:26:49.150 --> 00:26:55.300
profile so this linear part within the mass
transfer boundary layer is replaced by this
00:26:55.300 --> 00:27:02.140
similarly we can now let us look into the
what is . . replaced by v
00:27:02.140 --> 00:27:14.210
v is the y component velocity . in mass transfer
boundary layer and we assumed that . the . we
00:27:14.210 --> 00:27:21.900
have already . founded that delta is much
much less than effective diameter or chancel
00:27:21.900 --> 00:27:29.510
half height so therefore the thickness of
mass transfer boundary layer is probably less
00:27:29.510 --> 00:27:35.590
than one percent of the chancel dimension
and it is really very very small so in order
00:27:35.590 --> 00:27:42.150
to have a continuity we can assumed that within
this small chancel the y dimensional so this
00:27:42.150 --> 00:27:48.580
is basically v w which will be essentially
coming to the negative y direction so we can
00:27:48.580 --> 00:27:56.930
simply assumed v is equal to minus v w which
will be a basically function of x so why this
00:27:56.930 --> 00:28:01.220
profile is can be can be a velocity profile
in the y component . can be assumed because
00:28:01.220 --> 00:28:07.310
the thickness of mass transfer boundary layer
is much much small compare to h so we are
00:28:07.310 --> 00:28:12.770
assumed even if there is a variation of v
as a function of x and y within the within
00:28:12.770 --> 00:28:16.960
mass transfer boundary layer but since
the thickness of the mass transfer boundary
00:28:16.960 --> 00:28:22.570
is very very really very small then it can
be assumed that at the wall that the velocity
00:28:22.570 --> 00:28:28.170
we know that it is minus v w so we can assume
if it is really small so every where it is
00:28:28.170 --> 00:28:30.059
minus v w in the chancel
00:28:30.059 --> 00:28:37.480
so therefore the solute balance equation is
now looks like three u zero y by h del c del
00:28:37.480 --> 00:28:49.420
x minus v w del c del y is equal to d del
square c del y square so i stop in this class
00:28:49.420 --> 00:28:54.490
in the next class what will be doing will
be sitting up the boundary conditions of this
00:28:54.490 --> 00:28:59.400
to solve this this equation and we will be
looking into the complete solution of this
00:28:59.400 --> 00:29:04.850
equation . along which is boundary condition
and how the system performance can be calculated
00:29:04.850 --> 00:29:10.320
in this case of two dimensional analysis and
that will be giving a very very realistic
00:29:10.320 --> 00:29:14.540
solution of an . in actual what is happening
the physical phenomena that were those are
00:29:14.540 --> 00:29:16.270
happening in an actual membrane chancel
00:29:16.270 --> 00:29:17.580
thank you very much