WEBVTT
Kind: captions
Language: en
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welcome to the last session of this course
and we have we were looking into the chemical
00:00:24.980 --> 00:00:30.630
process enhanced separation in a batch
dialysis process we have written down . we
00:00:30.630 --> 00:00:35.720
have taken a example of aniline separation
from the feed using sulfuric acid in the dialysate
00:00:35.720 --> 00:00:41.550
side and we have seen that how the sulfuric
aniline reacts with sulfuric acid you know
00:00:41.550 --> 00:00:46.960
forming the anilinium on maintaining the zero
concentration of aniline in the dialysate
00:00:46.960 --> 00:00:51.670
side so there where there where we can maintaining
the maximum concentration gradient or driving
00:00:51.670 --> 00:00:57.350
force so that we can have the maximum separation
possible so we have written down the governing
00:00:57.350 --> 00:01:01.260
equation of the solute one in the feed side
we have written down the . governing equation
00:01:01.260 --> 00:01:06.020
of the solute one in the dialysate side and
now we look into the initial condition of
00:01:06.020 --> 00:01:16.810
this problem the initial conditions are . at
time t is equal to zero we have c one f is
00:01:16.810 --> 00:01:23.270
equal to c one f naught and c one d is equal
to zero because there is no solute initially
00:01:23.270 --> 00:01:29.180
and c three d equal to zero because there
is anilinium ion also formed initially
00:01:29.180 --> 00:01:35.000
now c three d then be obtained by the equilibrium
relationship . because it is equilibrium govern
00:01:35.000 --> 00:01:43.060
reaction so it is k equilibrium is equal to
c three d divided by c one d times c two d
00:01:43.060 --> 00:01:50.250
where c two d is constant because this sulfuric
acid two is a sulfuric acid and this is constant
00:01:50.250 --> 00:02:02.390
because it is in excess . two is in excess
ok now . we do . we take the laplace transform
00:02:02.390 --> 00:02:06.810
and of both equation the equation like before
and and see what we get so from the first
00:02:06.810 --> 00:02:15.890
equation we will be get in k by v f c i d
bar is transform variable c i f bar is equal
00:02:15.890 --> 00:02:29.299
s c c one f bar minus c one f naught and from
the second equation we get k by v d c one
00:02:29.299 --> 00:02:41.430
f bar minus c one d bar is equal to . s c
one d bar minus c one f naught plus s three
00:02:41.430 --> 00:02:48.829
d bar minus c three d zero which will be basically
zero at time t equal to zero so from these
00:02:48.829 --> 00:03:00.240
two equations what is resulted is that
we can write down k by v f c one d bar is
00:03:00.240 --> 00:03:15.750
equal to k by v f plus s one plus v d over
k one plus c two d c two d is a constant times
00:03:15.750 --> 00:03:25.919
s c one d bar minus c one f naught ok and
by simplifying we can get c one d expression
00:03:25.919 --> 00:03:37.729
of c one d bar is equal to c one f zero divided
by a s plus b s square where a is equal to
00:03:37.729 --> 00:03:51.489
v d by v f one plus c two d and b is equal
to v d by k one plus c two d and we can take
00:03:51.489 --> 00:04:05.700
a inverse laplace and can obtain c . one d
as a function of time so . these a these straight
00:04:05.700 --> 00:04:12.370
forward you you can really do that i can get
the profile of concentration of the of aniline
00:04:12.370 --> 00:04:18.229
in the dialysate side and how it values other
function of time so this is the modeling of
00:04:18.229 --> 00:04:24.570
a batch cell which is or the dialysis is separation
of dialysis enhance by an external chemical
00:04:24.570 --> 00:04:25.570
reagent
00:04:25.570 --> 00:04:30.380
so this must also be known to a student that
how we can maximize the you know driving
00:04:30.380 --> 00:04:36.569
force during a dialysis process and how to
model the situation next will be going a detailed
00:04:36.569 --> 00:04:41.990
two dimensional analysis in actual dialysis
operation or a continuous study stat system
00:04:41.990 --> 00:05:13.930
. so will be doing a detailed two dimensional
analysis of dialysis process . .
00:05:13.930 --> 00:05:24.699
so lets say this is the feed side ok the feed
is flowing as a laminar flow laminar parabolic
00:05:24.699 --> 00:05:31.249
profile so these feed going in to the system
in the feed side and will be will be setting
00:05:31.249 --> 00:05:37.960
up your our x axis in the middle of the
channel these y in these direction so these
00:05:37.960 --> 00:05:49.789
y equal to h this is y equal to zero and um
we have placed a membrane here dialysis
00:05:49.789 --> 00:05:58.370
membrane and this is the dialysate side so
there is a feed side this is a dialysate side
00:05:58.370 --> 00:06:06.199
and these these a dialysate side as well . both
side and it will be having a i know concentration
00:06:06.199 --> 00:06:18.960
profile something like this in the middle
so now so its basically the feed is like this
00:06:18.960 --> 00:06:22.370
is a is a parabolic . velocity profile
00:06:22.370 --> 00:06:36.860
so the if you consider a newtonian fluid laminar
and one dimensional flow the velocity profile
00:06:36.860 --> 00:06:43.509
to as we have already we already know its
its a parabolic velocity profile one minus
00:06:43.509 --> 00:06:57.539
y square over h square so y is the mid plane
of the channel .
00:06:57.539 --> 00:07:04.270
can we invert this is fixed here this is y
equal to . zero ok y is the mid plane of the
00:07:04.270 --> 00:07:17.830
channel and h is the half height .
00:07:17.830 --> 00:07:29.379
next we write down the q q is equal to two
volumetric flow rate u zero w h ok is the
00:07:29.379 --> 00:07:47.699
volumetric flow rate . ok so . we are assuming
that the channel is quite white compared to
00:07:47.699 --> 00:07:54.650
the general height ok w is much much greater
than two h so there is assumption now at the
00:07:54.650 --> 00:08:06.469
steady state we write down the solute balance
with in the channel . . solute balance . in
00:08:06.469 --> 00:08:15.319
the channel that means the channel is arranging
from minus h to h is the half height minus
00:08:15.319 --> 00:08:23.159
h two plus h and we write down the governing
equation u del c del x is equal to d del square
00:08:23.159 --> 00:08:25.089
c del y square
00:08:25.089 --> 00:08:31.410
so now these an ideal dialysis process
where the concentration gradient the only
00:08:31.410 --> 00:08:35.789
driving force that is appearing in the system
is the concentration of gradient and unlike
00:08:35.789 --> 00:08:40.880
the earlier cases is the pressure driven membrane
process is we did not have any pressure gradient
00:08:40.880 --> 00:08:45.080
pressure operating into the system so therefore
that there is no convective plux in the y
00:08:45.080 --> 00:08:51.700
direction will be having the solute plux in
the x direction only so at x . so and there
00:08:51.700 --> 00:08:58.580
is no v del c del y term is present in
this equation so at x equal to zero we have
00:08:58.580 --> 00:09:05.010
c is equal to c naught . and at symmetry in
the mid plane at y is equal to zero we have
00:09:05.010 --> 00:09:16.890
del c del y will be equal to zero and at y
is equal to h we have molar flux minus d del
00:09:16.890 --> 00:09:34.120
c del y should be equal to p times c here
p is the membrane permeability . membrane
00:09:34.120 --> 00:09:44.040
permeability towards the solute so therefore
we we are assuming that the . dialysate
00:09:44.040 --> 00:09:56.260
concentration is extremely dilute c d is very
dilute . and can be considered to be zero
00:09:56.260 --> 00:09:59.440
at any any any any any point of time
00:09:59.440 --> 00:10:07.400
so these . system is a typical eigen value
problem and let us make it non dimensional
00:10:07.400 --> 00:10:17.980
and see how it . how the solution can be obtained
so c star we defined as c by c naught y star
00:10:17.980 --> 00:10:31.360
we defined as . y over h x star we defined
as x d divided by u zero h square and p star
00:10:31.360 --> 00:10:36.910
is membrane biot number they called it so
this will be ph over d and this is also known
00:10:36.910 --> 00:10:49.910
as the membrane biot number .
so let us look into the the governing equation
00:10:49.910 --> 00:10:58.200
and the boundary conditions in the non dimensional
version so three by two one minus y star square
00:10:58.200 --> 00:11:11.790
del c star . del x star is equal to del square
c star del y star plus square these a governing
00:11:11.790 --> 00:11:19.250
equation the boundary conditions become
at x star equal to zero my c star is equal
00:11:19.250 --> 00:11:32.120
to one at y star is equal to zero del c star
del y star equal to zero and at y star is
00:11:32.120 --> 00:11:43.860
equal to one we have del c star del y star
plus p star c star equal to zero now these
00:11:43.860 --> 00:11:53.740
equation is . we can able to solve this equation
by using separation of variable . and all
00:11:53.740 --> 00:11:59.660
of you must be knowing the separation of variable
and analytical solution can be obtain the
00:11:59.660 --> 00:12:10.390
analytical solution . will be in the form
of c star as a function of x star y star will
00:12:10.390 --> 00:12:17.940
be summation it will be represented in the
summation series m is equal to one to infinity
00:12:17.940 --> 00:12:33.930
a m exponential minus two by three lambda
m square x star and summation of a m n y star
00:12:33.930 --> 00:12:38.040
where the n trans from zero to infinity
00:12:38.040 --> 00:12:45.890
so the eigen value of the systems so these
these an these standard eigen value problem
00:12:45.890 --> 00:12:50.529
and the eigen value so this a standard eigen
value problem was sturm louirille problem
00:12:50.529 --> 00:13:06.570
.
00:13:06.570 --> 00:13:13.450
so eigen value of the systems are lambda m
lambda m are the eigen values of the system
00:13:13.450 --> 00:13:19.910
and then we will see how the solution will
involve so once if that is the case then
00:13:19.910 --> 00:13:33.980
rate of removal of the solute can be written
as
00:13:33.980 --> 00:13:40.080
lets say this is the mass . per unit time
that will be in kg per second lets say is
00:13:40.080 --> 00:13:47.800
an kg per second we removed will be two u
zero h times w w is the width channel width
00:13:47.800 --> 00:14:01.640
multiplied the c zero minus c c m what is
c c m c c m is the . cup mixing concentration
00:14:01.640 --> 00:14:15.280
across the channel so its some kind of average
concentration
00:14:15.280 --> 00:14:27.370
of solute in the channel ok so c naught
is the feed concentration h is the half height
00:14:27.370 --> 00:14:32.350
two is two h is the total height usually is
the velocity w is the width so we can now
00:14:32.350 --> 00:14:37.190
now next will be finding out what is the cup
mixing concentration cup mixing concentration
00:14:37.190 --> 00:14:42.910
is given as . by these expression it is some
kind of cross section average concentration
00:14:42.910 --> 00:14:57.230
zero to h three by two u naught one minus
y square by h square d y and there will be
00:14:57.230 --> 00:15:10.250
zero to h three . by two u naught one minus
y square by h square c is the function of
00:15:10.250 --> 00:15:16.410
x and y d y so these expression can be integrate
quite easily there is absolute no problem
00:15:16.410 --> 00:15:20.530
in that we have already found out what is
the expression of c as a function of x and
00:15:20.530 --> 00:15:27.590
y that can be . that can be integrate across
the y so the . will be intact so i will i
00:15:27.590 --> 00:15:31.170
will be getting ultimately c c m as a function
of concentration
00:15:31.170 --> 00:15:36.090
so the how the cup mixing concentration of
the average concentration in the feed side
00:15:36.090 --> 00:15:41.180
will be varying along the x that expression
can be obtain by these so if we looking to
00:15:41.180 --> 00:15:47.630
the . you know expression of that one dimensional
version of that c c m star is equal to nothing
00:15:47.630 --> 00:15:54.970
but c c m divided by c naught and this will
be after integration will be getting this
00:15:54.970 --> 00:16:06.140
expression summation of m is equal to one
to infinity . a m exponential minus two by
00:16:06.140 --> 00:16:19.720
three lambda m square x star summation of
n equal to zero to infinity a n m divided
00:16:19.720 --> 00:16:31.070
by n plus one into n plus three at the eigen
values or the . are roots of the polonium
00:16:31.070 --> 00:16:51.320
polynomial lambda m are roots of polynomial
of the form zero equal to p star minus two
00:16:51.320 --> 00:17:03.650
by three plus five by twelve p star lambda
m square plus . one by twenty plus one by
00:17:03.650 --> 00:17:13.910
forty five p star lambda m to the power four
so this will be the polynomial . that the
00:17:13.910 --> 00:17:17.471
roots or the eigen values so what is the a
zero m
00:17:17.471 --> 00:17:27.999
a zero m is equal to one a one is equal to
zero so basically the even terms will survive
00:17:27.999 --> 00:17:36.049
of the series minus lambda m square by two
and a three m is equal to zero likewise a
00:17:36.049 --> 00:17:44.649
four m will be equal to lambda m square two
plus lambda m square divided by twenty four
00:17:44.649 --> 00:17:57.629
a five m is equal to zero likewise it will
go and capital a m will be given as . summation
00:17:57.629 --> 00:18:10.470
of n equal to zero to infinity . a n m divided
by n plus one into n plus three and there
00:18:10.470 --> 00:18:18.830
will be double summation in the denominator
n equal to zero to infinity p is equal to
00:18:18.830 --> 00:18:32.539
zero to n and this will be a p m a n minus
p times m divided by n plus one into n plus
00:18:32.539 --> 00:18:40.659
three ok so now let us do a certain simplification
simplification is that as we can see that
00:18:40.659 --> 00:18:48.620
the solution as a x dependence in form
of exponential minus lambda square x star
00:18:48.620 --> 00:18:53.980
so lambda m will be . will be large value
and x square will be large so therefore exponential
00:18:53.980 --> 00:18:59.480
decay will be faster so if you that if you
if you take care of that and we we considered
00:18:59.480 --> 00:19:05.149
the since the . exponential decay is faster
we can take the first term as the first eigen
00:19:05.149 --> 00:19:14.570
value
decay is fast and we can consider the first
00:19:14.570 --> 00:19:26.590
eigen value that will be lambda one square
is equal to p star divided by two by three
00:19:26.590 --> 00:19:44.539
plus five by twelve p star and considered
the first term of series . if we do that then
00:19:44.539 --> 00:19:51.710
a typical nth plot log c c m star verses p
star x star and will be getting a performance
00:19:51.710 --> 00:19:53.760
curve .
00:19:53.760 --> 00:19:59.460
we compute taking the first eigen value and
the performance curve will be looking something
00:19:59.460 --> 00:20:11.129
like this . .
will be having log c c m star here and p star
00:20:11.129 --> 00:20:21.690
x star will be nothing but p x by u zero h
and the curves will be like this for different
00:20:21.690 --> 00:20:28.410
value of p star these for p star is equal
to one these for p star is equal to point
00:20:28.410 --> 00:20:34.559
one these for p star equal to zero that means
there is no permission of the solute to the
00:20:34.559 --> 00:20:42.269
. to the membrane it is impermeable so now
if you if you talk about the you know if you
00:20:42.269 --> 00:20:46.320
if you like to design what is the length of
the . you know dialyzer dialyzer length for
00:20:46.320 --> 00:20:52.019
a particular removable of the solute then
we can from the from that particular removable
00:20:52.019 --> 00:20:56.950
of solute we can estimate the value of c c
m star we can directly go of the value of
00:20:56.950 --> 00:21:02.780
p star if p star is point one we can directly
come here . and get the value of x star we
00:21:02.780 --> 00:21:07.840
know this coordinate system so p is known
to us usually known to us channel height is
00:21:07.840 --> 00:21:13.499
known to us so you can get the value of x
so likewise for different value of for different
00:21:13.499 --> 00:21:18.840
type of membrane where the p star will be
known we can get the curve for a . lets say
00:21:18.840 --> 00:21:22.309
ninety nine percent separation ninety eight
percent separation of the solute from the
00:21:22.309 --> 00:21:26.860
feed side so you can get the length of the
of the dialyzer if the length of the dialyzer
00:21:26.860 --> 00:21:32.070
is known then from here one can find out what
will be the each of the height of the dialyser
00:21:32.070 --> 00:21:38.499
so if the p star is not known so we can . then
you have to interpolate between the two
00:21:38.499 --> 00:21:43.309
known values can find out the design of the
dialyser or can find out the length of the
00:21:43.309 --> 00:21:49.139
dialyzer or the height of the dialyzer if
you one of them is fixed then other can be
00:21:49.139 --> 00:21:55.980
evaluated so thus in a continuous system for
a two dimensional model which will be a basically
00:21:55.980 --> 00:22:02.990
and a very . realistic model can be obtained
. can be can be can be utilized to design
00:22:02.990 --> 00:22:07.809
a dialysis next what will be looking will
be looking into a very simplified version
00:22:07.809 --> 00:22:15.159
of this of this model . so so that the analytical
solution will be will be in front of us .
00:22:15.159 --> 00:22:27.980
so let us look into a simplistic approach
.
00:22:27.980 --> 00:22:40.179
first we considered the feed velocity is
large . is large and will be having a plug
00:22:40.179 --> 00:22:57.379
flow dialyaste is dilute and third it is a
study state operation so if it is a plug flow
00:22:57.379 --> 00:23:06.159
or will be having a uniform velocity profile
.
00:23:06.159 --> 00:23:11.590
u is equal to u naught the our own analysis
become simplified and will be having u zero
00:23:11.590 --> 00:23:19.240
del c del x is equal to d del square c del
x square ok and take the non dimensional version
00:23:19.240 --> 00:23:27.980
s star equal to c by c naught y star is equal
to y by h and x star is equal to x by l so
00:23:27.980 --> 00:23:37.269
if that is the case it becomes a del c del
star del x star is equal to del square c star
00:23:37.269 --> 00:23:45.190
del y star square so now these these equation
is are easy to solve and there will be a . analytical
00:23:45.190 --> 00:23:52.629
. solution is . is obtained here so a is equal
to u zero h square by d l this can be written
00:23:52.629 --> 00:24:00.559
as one by four reynolds schmidt d by l as
we have done earlier the non dimensional version
00:24:00.559 --> 00:24:08.610
of the . governing equation the boundary conditions
will be we can write directly that at y stat
00:24:08.610 --> 00:24:18.400
is equal to zero my del c star del y star
will be equal to zero and y star is equal
00:24:18.400 --> 00:24:30.659
to one we have del c star del y star plus
p m c star is equal to zero so what is p m
00:24:30.659 --> 00:24:38.390
star p m star nothing but p h over d d is
the diffusivity of solute through the membrane
00:24:38.390 --> 00:24:45.400
matrix and at x star equal to zero we have
c star equal to one
00:24:45.400 --> 00:24:54.960
now you can use a separation of variable type
of solution
00:24:54.960 --> 00:25:02.749
and we can have . we can assume the concentration
can be a function of sol function of .
00:25:02.749 --> 00:25:09.879
you know x and sol function of y and if you
really do that and put into the governing
00:25:09.879 --> 00:25:21.460
equation these will become a by x dx dx star
is equal to one over y d square y d y star
00:25:21.460 --> 00:25:25.710
square the left hand side is the function
of x alone the right hand side is a function
00:25:25.710 --> 00:25:31.279
of y alone they will be equal and they will
be equal to some constant lets say this constant
00:25:31.279 --> 00:25:40.539
is is . minus lambda n square so the solution
is x is equal c one exponential minus lambda
00:25:40.539 --> 00:25:56.179
n square x star by a and the y varying part
is d square y d y star square plus lambda
00:25:56.179 --> 00:26:02.809
n square y will be equal to zero and . will
be . having the boundary condition that at
00:26:02.809 --> 00:26:12.639
y star equal to zero d y d y star equal to
zero and y star is equal to one will be having
00:26:12.639 --> 00:26:23.210
d y d y star plus p m y is equal to zero so
this will be having an solution y is equal
00:26:23.210 --> 00:26:32.789
to some c you know c n cos lambda n y star
and the eigen values will be nothing but the
00:26:32.789 --> 00:26:46.139
. roots of the transcendental equation of
lambda n tan lambda n . is equal to p m
00:26:46.139 --> 00:26:50.119
so from this transcendental equation roots
of this equation will be the eigen values
00:26:50.119 --> 00:26:58.149
and one can get the final solution as x star
y star is summation of all these solutions
00:26:58.149 --> 00:27:10.309
so this is becomes c n . cos sin lambda n
y star exponential minus lambda n square x
00:27:10.309 --> 00:27:17.759
star by a ok so from by using a boundary condition
initial condition that at x star equal to
00:27:17.759 --> 00:27:25.480
zero c star is equal to one we can we can
we can get the value of c n by using the
00:27:25.480 --> 00:27:31.679
orthogonal properties of the sin and cos sin
function so if you really do that the c n
00:27:31.679 --> 00:27:42.879
will become . zero to one cos lambda n y star
d y star divided by cos square lambda n y
00:27:42.879 --> 00:27:52.230
star . d y star and one can really do a
calculation of the and evaluate the integrals
00:27:52.230 --> 00:28:03.320
and these becomes that c n becomes two sin
lambda n divided by . lambda n p m square
00:28:03.320 --> 00:28:14.570
plus lambda square divided p m square plus
p m plus lambda m square and ultimately
00:28:14.570 --> 00:28:21.269
we can get the expression of c star as function
of x star and y star is equal to summation
00:28:21.269 --> 00:28:32.710
of c n cos sin lambda n y star exponential
minus lambda n square x star divided by a
00:28:32.710 --> 00:28:40.190
n equal to one to infinity and then we can
we can we can evaluate the cup mixing concentration
00:28:40.190 --> 00:28:44.560
and by definition as we have done earlier
the cup mixing concentration in this case
00:28:44.560 --> 00:28:52.659
will be we carry out the integration and lets
lets write down the expression first v x y
00:28:52.659 --> 00:29:01.309
so . this uniform here u zero c x y d y from
zero to one . and here it will be zero to
00:29:01.309 --> 00:29:07.580
one u zero d y . so we will be getting the
expression of cup mixing concentration as
00:29:07.580 --> 00:29:25.249
c n exponential minus lambda n square x star
by times sin lambda n divided by lambda
00:29:25.249 --> 00:29:40.200
n that will be expression of c c m star and
mass removed is rate of removal of pollutant
00:29:40.200 --> 00:29:49.759
will be u zero h times w c naught minus c
c m then this can be you know after multiplying
00:29:49.759 --> 00:29:54.659
the ca naught if you can put this expression
one can get the rate of removal of the pollute
00:29:54.659 --> 00:29:56.529
through the dialysis process
00:29:56.529 --> 00:30:01.679
so this is how the dialysis process can be
model either by . one dimensional model or
00:30:01.679 --> 00:30:07.489
detailed two dimensional model we have discussed
and i hope now now you have really come down
00:30:07.489 --> 00:30:12.440
to the end of our course so let us summarize
whatever we have learned in this course in
00:30:12.440 --> 00:30:17.650
these course we have learned the basic principles
and the various membrane based fundamentals
00:30:17.650 --> 00:30:23.140
of various moment based processes their you
know . underline transport mechanism then
00:30:23.140 --> 00:30:28.460
we have categorized various processes in terms
of reverse osmosis nano filtration micro filtration
00:30:28.460 --> 00:30:33.600
ultra filtration and then membrane modules
we have discussed membrane modules as well
00:30:33.600 --> 00:30:39.320
as the dialysis process we have looked into
the various the important terms and definitions
00:30:39.320 --> 00:30:44.350
those were . those were required tremendously
for modeling the systems we have looked into
00:30:44.350 --> 00:30:50.909
the how from you know irreversible thermodynamics
the transport loss in case of reverse osmosis
00:30:50.909 --> 00:30:56.279
through the plus membrane can be obtained
then we have looked into the fouling irreversible
00:30:56.279 --> 00:31:01.450
fouling and as well as the concentration polarization
then we . have looked into the one dimensional
00:31:01.450 --> 00:31:07.450
analysis of concentration profile concentration
polarization and summarize the short coming
00:31:07.450 --> 00:31:12.970
of one dimensional model and looked into the
osmotic . detailed two dimensional model for
00:31:12.970 --> 00:31:15.289
the osmotic pressure control filtration
00:31:15.289 --> 00:31:19.979
we looked into details of the osmotic pressure
control filtration for dilute solution for
00:31:19.979 --> 00:31:24.639
concentrate solution for developing for one
dimensional model two dimensional model things
00:31:24.639 --> 00:31:30.119
like that and then we moved into the gel layer
polar polarization filtration system and we
00:31:30.119 --> 00:31:34.759
have looked into the detailed modeling of
gel air controlling filtration after that
00:31:34.759 --> 00:31:41.230
then went forward to the membrane modules
and how to covered how to do the modeling
00:31:41.230 --> 00:31:46.799
and design how to format the design equations
to model the membrane module and how to obtain
00:31:46.799 --> 00:31:52.029
the solutions as well and again in this case
as we have gone into step by step complicated
00:31:52.029 --> 00:31:58.059
step initially first the very first instance
we have we we consider the a very simplistic
00:31:58.059 --> 00:32:02.989
case for constant permeate flux then we have
. second the complexity to a second level
00:32:02.989 --> 00:32:07.629
where the permeate flux is proportional to
the delta p or the pressure gradient and then
00:32:07.629 --> 00:32:12.330
we got the solution then went into the next
step there is a most complicated step where
00:32:12.330 --> 00:32:17.019
the it is proportional to delta p minus delta
pi where the osmotic pressure is not negligible
00:32:17.019 --> 00:32:22.399
at all . solve that case then we have gone
into the dialysis process and looked into
00:32:22.399 --> 00:32:29.570
the you know design equation for the basically
batch dialysis continuous dialysis then . dialysis
00:32:29.570 --> 00:32:34.220
batch dialysis with enhanced chemical reaction
enhance separation and then finally we have
00:32:34.220 --> 00:32:38.039
looked into the two dimensional model of a
continuous dialysis process
00:32:38.039 --> 00:32:43.809
i hope will be getting a . you have got a
fair idea how to model and design the membrane
00:32:43.809 --> 00:32:48.619
based separation pressure systems starting
from the pressure driven processes up to the
00:32:48.619 --> 00:32:55.739
dialysis operation and i hope that this course
will be of eminence help for you to your future
00:32:55.739 --> 00:33:00.139
research of your profile of future career
in research or or in industry .
00:33:00.139 --> 00:33:00.429
thank you very much .