WEBVTT
Kind: captions
Language: en
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welcome to the session we will be looking
into the you will as we have talked in
00:00:23.900 --> 00:00:30.530
last class we have finished the . as the modeling
aspects of osmotic pressure control filtration
00:00:30.530 --> 00:00:36.050
in a cross flow system now we will be talking
about the importance of unstirred batch system
00:00:36.050 --> 00:00:44.089
and do things . and then will be looking
into the modeling of unstirred batch filtration
00:00:44.089 --> 00:00:53.989
system membrane filtration system . why the
unstirred batch system is important that has
00:00:53.989 --> 00:01:00.680
to be understood first because if if if we
are talking dealing with a . with a . real
00:01:00.680 --> 00:01:06.960
time with a solution having a solute with
with unknown properties then it is very easy
00:01:06.960 --> 00:01:14.800
to solve the system using a predictive
mode but suppose we are talking about an industrial
00:01:14.800 --> 00:01:20.229
effluent . or fruit juice or unknown or real
life application where there will be not a
00:01:20.229 --> 00:01:26.090
single component there are many . components
present into the system and it is very difficult
00:01:26.090 --> 00:01:30.980
to identify each and every component number
one number two is that even if you identify
00:01:30.980 --> 00:01:36.100
every . majority of the components then it
will be quite difficult to you know estimate
00:01:36.100 --> 00:01:40.290
their concentration even if you estimate the
concentration now its is it will be very difficult
00:01:40.290 --> 00:01:45.829
to estimate their physical property of each
and individual material component for example
00:01:45.829 --> 00:01:49.900
their diffusivity osmotic pressure the of
the solution things like that
00:01:49.900 --> 00:01:57.430
so therefore what is generally done one
has take an effective diffusivity and effective
00:01:57.430 --> 00:02:02.670
osmotic pressure with a variation of concentration
. concentration may be in terms of total solids
00:02:02.670 --> 00:02:08.500
or total dissolved solid solids based on the
systems and and you will be having an effective
00:02:08.500 --> 00:02:13.340
osmotic pressure coefficient a one c plus
a two c square plus a three c cube this a
00:02:13.340 --> 00:02:19.299
one a two a three may be the you know
the effective coefficient so osmotic pressure
00:02:19.299 --> 00:02:25.329
so lets write down what are the various parameters
those will be appearing a one c plus a two
00:02:25.329 --> 00:02:30.640
c square plus a three c cube that will be
the osmotic pressure of the solution but we
00:02:30.640 --> 00:02:35.940
do not know the values of a one a two and
a three and even if you you can know if
00:02:35.940 --> 00:02:41.660
you have an accurate osmometer but if you
if you do not know the value of effective
00:02:41.660 --> 00:02:47.850
diffusivity effective viscosity and viscosity
can be measured of course what the diffusivity
00:02:47.850 --> 00:02:53.160
value cannot be known so it will be so in
in a typical system you will be having four
00:02:53.160 --> 00:02:59.220
unknowns the osmotic coefficients a one a
two a three and diffusivity effective diffusivity
00:02:59.220 --> 00:03:04.699
of the system . so how how this system is
then model then what we what is generally
00:03:04.699 --> 00:03:11.829
done we conduct large number of experiments
you know . in the order of thirty forty lot
00:03:11.829 --> 00:03:17.120
number of experiments and vary the operating
conditions and see the permeate flux and permeate
00:03:17.120 --> 00:03:22.199
concentration and may be in terms of
total solids or may be in terms of total dissolve
00:03:22.199 --> 00:03:23.199
solids
00:03:23.199 --> 00:03:29.480
once this is done then the . then we we
take we go through the model but we we gave
00:03:29.480 --> 00:03:34.970
we have a have a profile optimization or the
optimization with respect to the experimental
00:03:34.970 --> 00:03:40.800
values so we have a guess values we do a guess
values of a one a two a three or d then we
00:03:40.800 --> 00:03:45.840
go to the model solve the model get the calculated
values of permeate flux and permeate concentration
00:03:45.840 --> 00:03:52.250
for different operating conditions and then
we come we compare those values with the experimental
00:03:52.250 --> 00:04:02.500
data that means we we minimize the sum c p
calculated minus . c p experimental divided
00:04:02.500 --> 00:04:15.900
by c p experimental square of that plus v
w calculated minus v w experimental by v w
00:04:15.900 --> 00:04:21.840
experimental square of that we minimize the
sum and compare and optimize the whole system
00:04:21.840 --> 00:04:27.790
and compare with the experimental data and
the module calculated data then if this
00:04:27.790 --> 00:04:33.840
[vocalized noise] sum is is not less
is not less than a tolerance value then again
00:04:33.840 --> 00:04:40.310
we have to redo a you know reguess this
set of parameters which will be you know directed
00:04:40.310 --> 00:04:46.750
by a proper algorithm or minimization algorithm
or optimization algorithm so during . so we
00:04:46.750 --> 00:04:52.680
will be doing an optimization method optimization
for evaluating the system parameters so then
00:04:52.680 --> 00:05:01.360
a one a two a three and d of the system will
be predicted will be will be estimated . once
00:05:01.360 --> 00:05:06.990
this will be estimated then for an unknown
operating condition we can run the our model
00:05:06.990 --> 00:05:12.320
equations in a predictive mode in order to
able to predict the system performance
00:05:12.320 --> 00:05:17.290
so what is the drawback of the system if you
really do a cross flow system then every cross
00:05:17.290 --> 00:05:21.950
flow system will be under a particular set
of operating condition will be giving you
00:05:21.950 --> 00:05:27.990
one value of permeate concentration and permeate
flux but you will be you require probably
00:05:27.990 --> 00:05:32.530
huge number of experiments so therefore you
have to conduct the cross flow experiment
00:05:32.530 --> 00:05:37.570
large number of times may be forty times fifty
times in order to generate you know forty
00:05:37.570 --> 00:05:42.700
fifty values of permeate concentration and
permeate flux and so you can optimize
00:05:42.700 --> 00:05:49.490
the four parameter a one to a three and d
effective now that will be . you know incurring
00:05:49.490 --> 00:05:56.390
then you will be incurring more cost man power
energy so and material to to conduct those
00:05:56.390 --> 00:06:02.720
experiments on other hand if you can conduct
a batch cell experiment . the the permeate
00:06:02.720 --> 00:06:07.830
in a in a typical batch cell the . because
of the growth of the mass transfer boundary
00:06:07.830 --> 00:06:14.020
layer one will be getting the permeate
flux and permeate concentration profile of
00:06:14.020 --> 00:06:21.490
that as a function of . as a function of time
so we will be getting n number of data so
00:06:21.490 --> 00:06:29.770
these will be these all be the parameters
. and usual function of time the permeate
00:06:29.770 --> 00:06:35.520
flux will keep on decreasing as a function
of time this will be the permeate flux p w
00:06:35.520 --> 00:06:40.889
the non dimensional permeate flux and this
how the permeate concentration one dimensional
00:06:40.889 --> 00:06:44.260
permeate concentration keep on growing as
a function of time
00:06:44.260 --> 00:06:50.530
so what you can do so at any fixed time intervals
you can measure the permeate concentration
00:06:50.530 --> 00:06:57.340
and permeate flux so you can keep on measuring
them you conduct an experiment for five hours
00:06:57.340 --> 00:07:02.310
one is single experiment for five hours and
then . you can measure n number of values
00:07:02.310 --> 00:07:07.200
of permeate concentration and you know
pear or permeate concentration permeate flux
00:07:07.200 --> 00:07:11.900
at different time points so therefore if from
a single experiment one can generated data
00:07:11.900 --> 00:07:16.830
fifty sixty data if you take a run of lets
say two hours or three hours and if you
00:07:16.830 --> 00:07:21.820
if you take you know data accurately within
every ten minutes or five minutes so depending
00:07:21.820 --> 00:07:27.710
on the duration of the experiment one can
collect large number of data and those data
00:07:27.710 --> 00:07:33.450
can be minimized by doing an optimization
so that is know as profile optimization so
00:07:33.450 --> 00:07:38.310
you will you have the you known the module
value of the permeate flux profile and module
00:07:38.310 --> 00:07:44.430
value of calculated value of permeate concentration
profile you measure the . experimental values
00:07:44.430 --> 00:07:51.830
and plot the experimental values like this
.
00:07:51.830 --> 00:07:56.930
and from a single experiment single batch
cell experiment one can get a large number
00:07:56.930 --> 00:08:02.490
of data thats why the modeling of the batch
cell . all we know filtrations set up is very
00:08:02.490 --> 00:08:07.830
very important and will be looking into how
these modeling can be done so for an actual
00:08:07.830 --> 00:08:13.680
real system the reger of experiment can be
reduced or [vocalized noise] in order to estimate
00:08:13.680 --> 00:08:18.220
the parameters quite easily from the bath
cell experiment
00:08:18.220 --> 00:08:23.300
so let us look into the modeling of batch
cell experiment what is a batch cell so its
00:08:23.300 --> 00:08:29.670
basically a pressure vessel having three parts
there will be a top flange there will be a
00:08:29.670 --> 00:08:34.250
body or shell there will be a bottom flange
and this in the bottom flange it will be a
00:08:34.250 --> 00:08:39.020
circular disc so here will be placing the
membrane in the top flange and all these will
00:08:39.020 --> 00:08:44.480
be connect you know connected to
the um filtration . you know body of
00:08:44.480 --> 00:08:50.119
the cell by using nuts and bolts and then
it will pressurized by a nitrogen cylinder
00:08:50.119 --> 00:08:58.990
. but before that you fill up the cell with
lets say five hundred m l or seven hundred
00:08:58.990 --> 00:09:04.400
m l of solution . so what will be let us write
down so there will be a growth of concentration
00:09:04.400 --> 00:09:11.010
boundary layer over the membrane surface as
we have seen earlier but in this case there
00:09:11.010 --> 00:09:16.180
is a difference the difference is that in
the case of the cross flow filtration the
00:09:16.180 --> 00:09:21.089
growth of the mass transfer boundary layer
will be restricted or arrested by the cross
00:09:21.089 --> 00:09:25.930
flow of the retented flow but in this case
so there will be a steady state reached in
00:09:25.930 --> 00:09:32.070
case of the cross flow filtration but in the
case of unstirred batch cell there is no external
00:09:32.070 --> 00:09:38.890
agent to curb or arrest growth of the boundary
layer hence it will be growing undisturbed
00:09:38.890 --> 00:09:43.810
and the permeate flux will keep on decreasing
as a function of time and permeate concentration
00:09:43.810 --> 00:09:49.370
will keep on increasing as function of time
so there will be no attainment of steady state
00:09:49.370 --> 00:09:53.800
that is the unlike the cross flow filtration
cell there is the typical characteristic of
00:09:53.800 --> 00:10:09.400
a batch cell . . . no attainment of the steady
state
00:10:09.400 --> 00:10:20.990
so thus what we will do next we will write
down the solute mass balance . within . mass
00:10:20.990 --> 00:10:34.320
transfer boundary layer so if you write it
del c del t minus plus v del c del y
00:10:34.320 --> 00:10:40.029
is equal to d del square c by del y square
since mass transfer boundary layer is small
00:10:40.029 --> 00:10:47.271
we can assume v is equal to minus v w so that
is coming out . and in this case v w will
00:10:47.271 --> 00:10:57.640
be function of time so del c del t will be
minus v w del c del y is equal to d del square
00:10:57.640 --> 00:11:02.861
c by del y square so now let us set the . bounder
set up the boundary conditions t is equal
00:11:02.861 --> 00:11:10.060
to zero c is equal to c naught at y is equal
to infinity as argued earlier c is equal to
00:11:10.060 --> 00:11:17.930
c naught at y is equal to zero we have
. all the . steady state boundary condition
00:11:17.930 --> 00:11:24.589
all the mass fluxes towards membrane surface
will be equal to zero v w c m minus c p plus
00:11:24.589 --> 00:11:34.580
d del c del y will be equal to zero so this
will set up the governing equation and the
00:11:34.580 --> 00:11:40.310
three boundary conditions next what we will
do we we make the system non dimensional from
00:11:40.310 --> 00:11:47.350
very beginning so let us do that so we define
these non dimensional parameters as c star
00:11:47.350 --> 00:11:55.850
is equal to c by c naught y star is equal
to y by r r is radius of the batch cell
00:11:55.850 --> 00:12:02.300
so now let us look into governing equations
so this becomes del c star del t . minus v
00:12:02.300 --> 00:12:15.420
w del c star r del y star and d del square
c star divided by r square del y star square
00:12:15.420 --> 00:12:23.980
ok now multiplied by r square by d so its
becomes r square by d del c star del t minus
00:12:23.980 --> 00:12:36.860
v w r over d del c star del y star is equal
to del square c star . del y star square so
00:12:36.860 --> 00:12:44.080
the right hand side is completely non dimensional
the second term in left hand side the this
00:12:44.080 --> 00:12:50.460
is non dimensional so this is again non dimensional
permeate flux this case here so here c star
00:12:50.460 --> 00:12:55.750
is non dimensional t is dimensional that means
t d by r square has to be non dimensional
00:12:55.750 --> 00:13:06.020
that means we . we define non dimensional
time as t d . divided by r square . so this
00:13:06.020 --> 00:13:18.170
becomes del c star del tau minus p w del c
star del y star is equal to del square c star
00:13:18.170 --> 00:13:29.490
del y star square . so this is the governing
equation . where p w is equal to v w r over
00:13:29.490 --> 00:13:37.300
d ok and now we will be looking we will be
making the boundary to be non dimensional
00:13:37.300 --> 00:13:46.600
as well so at tau is equal to zero c star
is equal to one at y star is equal to zero
00:13:46.600 --> 00:14:07.149
del c star del y star plus p w c m star r
r equal to zero . at y star equals to infinity
00:14:07.149 --> 00:14:08.399
c star is equal to one
00:14:08.399 --> 00:14:14.519
so that sets up the governing equation and
the boundary condition so next again will
00:14:14.519 --> 00:14:19.460
be looking will be taking request to the similarity
solution because this is a parabolic partial
00:14:19.460 --> 00:14:27.720
differential equation . and one of the boundary
is residing at infinity so we can have a similarity
00:14:27.720 --> 00:14:32.700
solution so let us have a similarity solution
and before that we have to find out what is
00:14:32.700 --> 00:14:37.670
the similarity parameter so again will be
doing the same order of magnitude by evaluating
00:14:37.670 --> 00:14:42.870
this concentration profile at edge of the
boundary layer so this will if we really do
00:14:42.870 --> 00:14:52.279
that del c star . will be delta c star and
delta tau will be tau minus zero the second
00:14:52.279 --> 00:15:02.209
term will be gone so that will be delta c
star divided by delta square . so delta square
00:15:02.209 --> 00:15:09.180
will be function will be will be will be
varying as root of tau in this case
00:15:09.180 --> 00:15:16.180
so now we will be getting a similarity parameter
similarity parameter is y star by del star
00:15:16.180 --> 00:15:23.010
it will be y star by root of tau so that
will be the similarity parameter will be working
00:15:23.010 --> 00:15:29.089
with in this particular problem next as you
have done earlier will be . you know replacing
00:15:29.089 --> 00:15:34.230
all the partial derivative in the governing
equation in terms of the similarity parameter
00:15:34.230 --> 00:15:39.459
and let us first find out the similarly the
partial derivative of the those will be
00:15:39.459 --> 00:15:44.060
appearing in the governing equation in terms
of the similarity parameter once we get those
00:15:44.060 --> 00:15:48.120
expressions we are going to substitute those
expressions in governing equation
00:15:48.120 --> 00:16:01.000
so del c star . del t del tau will be nothing
but d c star d eta . del eta del tau . and
00:16:01.000 --> 00:16:14.270
these becomes minus one by two y star tau
root tau d c star d eta and y star by root
00:16:14.270 --> 00:16:22.730
tau is nothing but eta this is nothing but
eta by two dc star d eta then del c star del
00:16:22.730 --> 00:16:36.720
y star . becomes dc star d eta del eta del
y star and this become one over root over
00:16:36.720 --> 00:16:50.709
tau . d c star d eta and del square c star
del y star square is . del del y star del
00:16:50.709 --> 00:17:02.040
c star del y star and this becomes one over
tau . d square c star d eta square . so now
00:17:02.040 --> 00:17:06.930
we are going to substitute this partial derivatives
in the governing equation and see what we
00:17:06.930 --> 00:17:16.929
get so what we get is that d square c star
d eta square is equal to minus eta by two
00:17:16.929 --> 00:17:33.049
plus v w root over tau dc star . d eta and
as we have already seen that seen that . delta
00:17:33.049 --> 00:17:44.980
v w is inversely proportional to the delta
so this will be the p w right . p w is inversely
00:17:44.980 --> 00:17:52.600
proportional to delta star and delta star
is directly proportional to root of a tau
00:17:52.600 --> 00:17:59.720
therefore p w root over tau is constant and
this constant let us a . say this is a
00:17:59.720 --> 00:18:09.749
if that is the case then the ordinary differential
equation of second order now becomes minus
00:18:09.749 --> 00:18:20.169
eta by two plus a d c star d eta and now you
change the boundary conditions in terms of
00:18:20.169 --> 00:18:30.139
similarity parameter eta eta equal to zero
d c star d eta plus a c m star r r equal to
00:18:30.139 --> 00:18:41.340
zero and at eta equal to infinity c star equals
one so we can will be able solve this equation
00:18:41.340 --> 00:18:49.679
so dc star so solution is as we have see earlier
d c star d eta is equal to k one exponential
00:18:49.679 --> 00:18:56.769
minus eta square by four minus a eta an then
one more integration we will be giving you
00:18:56.769 --> 00:19:07.919
c star is function of eta k one . exponential
minus eta square by four minus a eta d eta
00:19:07.919 --> 00:19:15.880
plus k two and then using the boundary conditions
this two boundary condition one can evaluate
00:19:15.880 --> 00:19:25.700
the integration constant k one and k two as
k one plus a r r k two is equal to zero
00:19:25.700 --> 00:19:35.840
and k one i one plus k two is equal to one
so where i one is basically the definite interval
00:19:35.840 --> 00:19:48.950
zero to infinity p exponential minus eta square
by four minus a eta d eta ok and k one becomes
00:19:48.950 --> 00:20:02.549
minus a r r one minus a r r i one and k two
is nothing but c m star is equals to . one
00:20:02.549 --> 00:20:08.809
minus a r r i one and we have the osmotic
pressure model
00:20:08.809 --> 00:20:17.380
so um darcys law so the porous medium porous
membrane . p w as the function of time will
00:20:17.380 --> 00:20:24.609
be nothing but b into one minus delta pi by
delta p you have already seen that delta pi
00:20:24.609 --> 00:20:32.429
can be expressed as a function of c m star
so now the the modeling model equations complete
00:20:32.429 --> 00:20:40.129
let us look into the algorithm . of calculation
the algorithm calculation is very simple first
00:20:40.129 --> 00:20:51.830
at a particular at a particular time point
tau . we do following one guess . c m star
00:20:51.830 --> 00:21:05.780
. evaluate p w from equation one .
once you get p m w from equation one then
00:21:05.780 --> 00:21:19.929
we can get the expression of a from the from
expression of a a a can be evaluated .
00:21:19.929 --> 00:21:40.239
from the expression of p w ok once that is
done then you can evaluate . i one from equation
00:21:40.239 --> 00:22:04.950
two . once that is done one can evaluate c
m star from equation three . ok so therefore
00:22:04.950 --> 00:22:21.220
the equation expression of p w is nothing
but a is equal to p w root over tau . so
00:22:21.220 --> 00:22:25.059
a is equal to p w root over tau so at that
particular tau we put the value of tau we
00:22:25.059 --> 00:22:29.019
have already evaluated p w so p w multiplied
by root by tau will be giving you the value
00:22:29.019 --> 00:22:36.510
of a substitute a in integral i one and evaluate
definite integral using . rule and get the
00:22:36.510 --> 00:22:40.639
value of i one from the equation two and from
equation three you will be getting the value
00:22:40.639 --> 00:22:48.929
of c m star and check whether this calculated
c m star now c m star calculated minus c m
00:22:48.929 --> 00:22:55.859
star guess the less than epsilon naught
if they are less than epsilon go for next
00:22:55.859 --> 00:23:01.909
time step t plus delta t if not then you have
to re guess the value of c m star. and repeat
00:23:01.909 --> 00:23:03.389
the iterative calculation
00:23:03.389 --> 00:23:07.490
so in the process what you will be getting
is that if the process will be ultimately
00:23:07.490 --> 00:23:17.570
getting c m star as a function of . time c
p star as a function of time and p w as a
00:23:17.570 --> 00:23:25.269
function of time so ultimately you will be
getting p w as the function of time and c
00:23:25.269 --> 00:23:32.789
p star as a function of time . and you will
be having measure measured value of p w at
00:23:32.789 --> 00:23:38.979
different time points and measured value of
c p star defined time points then you can
00:23:38.979 --> 00:23:45.429
do the optimizations and can estimate the
system parameters so thats how the . you
00:23:45.429 --> 00:23:52.000
the batch cell filtration can be done so this
completes our whole you know formulation for
00:23:52.000 --> 00:23:58.129
the modeling or content for the modeling of
. of of the osmotic pressure filtration both
00:23:58.129 --> 00:24:03.099
in the cross flow cell and the batch cell
. from one dimensional model we have first
00:24:03.099 --> 00:24:08.059
started with one dimensional model we went
ead with the two dimensional model and by
00:24:08.059 --> 00:24:12.599
by by observing several short comings of the
one dimensional model and we have seen that
00:24:12.599 --> 00:24:16.559
one dimensional model will really will give
the under prediction of permeate flux and
00:24:16.559 --> 00:24:20.809
therefore we went ahead with two dimensional
model completely solve the two dimensional
00:24:20.809 --> 00:24:26.820
model for the cross flow . filtration system
as well as the batch system filtration system
00:24:26.820 --> 00:24:32.639
also we have iterated that discussed about
the importance of batch cell why it is so
00:24:32.639 --> 00:24:37.879
important for a for evaluating the system
parameter or the process parameters whenever
00:24:37.879 --> 00:24:42.580
you will be working with a real industrial
field or real fruit juice sample
00:24:42.580 --> 00:25:01.690
next type of . you know modeling will be looking
into the gel layer and control modeling . . under
00:25:01.690 --> 00:25:07.370
two circumstances a gel are very highly viscous
solution is prepared on the membrane is is
00:25:07.370 --> 00:25:14.779
formed on the membrane surface there are two
ways first way is first way is that the
00:25:14.779 --> 00:25:27.909
filtration may be osmotic pressure control
initially .
00:25:27.909 --> 00:25:33.019
and then during the osmotic pressure control
filtration membrane surface solute concentration
00:25:33.019 --> 00:25:39.509
of the membrane surface keep on increasing
as . as you move down stream of the channel
00:25:39.509 --> 00:25:45.760
and therefore at a particular location it
may happen that it exceeds the solubility
00:25:45.760 --> 00:25:59.609
limit . exceeds or approaches . the solubility
limit in that case once it it it it it does
00:25:59.609 --> 00:26:17.979
that then . a very viscous . solution is
formed . near membrane surface . and if the
00:26:17.979 --> 00:26:24.729
solubility limit is exceeded at that particular
temperature there may precipitation of the
00:26:24.729 --> 00:26:30.369
solute as solute phase or there may be another
way is a natural gel forming agency there
00:26:30.369 --> 00:26:48.429
are some natural gel forming agents .
for example polyvinyl alcohol pectin . etcetera
00:26:48.429 --> 00:26:53.999
these solutes they form a gel over the membrane
surface from the very beginning
00:26:53.999 --> 00:27:00.149
so you do not need to give time of them for
formation of gel they already form they from
00:27:00.149 --> 00:27:05.119
the very beginning of the filtration . the
form a solute gel type of . they form a highly
00:27:05.119 --> 00:27:12.129
viscous solution so highly viscous layer
near the membrane surface and this gel keeps
00:27:12.129 --> 00:27:17.269
on going in time with with time of operation
or duration of operation in the in the in
00:27:17.269 --> 00:27:23.820
the y direction over the membrane surface
so there are two typical ways how the gel
00:27:23.820 --> 00:27:31.009
formed and in the next class we will be looking
into the various types of you know . gel polarization
00:27:31.009 --> 00:27:36.909
model and how the various parameters will
be estimated in a typical gel . polarization
00:27:36.909 --> 00:27:43.159
model which will be basically wll be taking
few from the classical filtration theory and
00:27:43.159 --> 00:27:49.799
and and combine it and will bring those you
know concepts in the gel layer formation and
00:27:49.799 --> 00:27:52.700
then will looking into how the gel layer modeling
can be done
00:27:52.700 --> 00:27:53.659
thank you very much .