WEBVTT
Kind: captions
Language: en
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welcome to the session of our class so we
we are looking into the module design of a
00:00:23.560 --> 00:00:28.670
membrane module design of an actual case where
osmotic pressure is quite significant and
00:00:28.670 --> 00:00:33.770
we are looking into the spiraling module that
is a flow through a channel and as we have
00:00:33.770 --> 00:00:41.750
seen we have derived the governing equation
of . governing by equation of axial
00:00:41.750 --> 00:00:49.370
pressure drop governing equation of velocity
in the retentive as . channel and governing
00:00:49.370 --> 00:00:53.739
equation of concentration we have seen that
these are basically already differential equation
00:00:53.739 --> 00:00:59.800
for the axial pressure drop will be a second
order o d e the for the for the velocity will
00:00:59.800 --> 00:01:05.059
be a first order . o d e and the concentration
it is a first order o d e so three ordinary
00:01:05.059 --> 00:01:10.000
differential equations are to be hooked up
but there are membrane surface concentration
00:01:10.000 --> 00:01:14.540
that will be coming all the time in the picture
so this membrane surface concentration has
00:01:14.540 --> 00:01:19.560
to be related to the bulk concentration and
as we have seen that this is related to to
00:01:19.560 --> 00:01:24.440
the definition of mass transfer coefficient
which will be result in you an result in an
00:01:24.440 --> 00:01:30.730
algebraic equation so ultimately you will
be getting up landing up into a set of ordinary
00:01:30.730 --> 00:01:35.350
differential equation coupled with algebraic
equation it's a d a e system differential
00:01:35.350 --> 00:01:40.070
algebraic equation system so let us just continue
with that and finish of this problem
00:01:40.070 --> 00:01:43.310
so we have already seen that mass transfer
coefficient the the algebraic equation that
00:01:43.310 --> 00:01:50.150
will be connecting the membrane surface concentration
and bulk concentration through this equation
00:01:50.150 --> 00:01:56.230
but a e is also a function of x so this is
the relationship between k and x and let us
00:01:56.230 --> 00:02:04.310
look into what is I I is the definite integral
zero to infinity exponential minus . eta cube
00:02:04.310 --> 00:02:12.819
by three minus point four two lambda theta
d eta and lambda is the length of rich permeate
00:02:12.819 --> 00:02:21.610
flux d e is equivalent diameter l u d square
etcetera . u u is the velocity and d is the
00:02:21.610 --> 00:02:28.450
solute diffusivity to the power one one
third and v w bar is nothing but one over
00:02:28.450 --> 00:02:38.730
l zero to l v w d x . so d is the length of
it's permeate flux so this has become so will
00:02:38.730 --> 00:02:43.620
be having three ordinary differential equation
and one algebraic equation as to be solved
00:02:43.620 --> 00:02:48.700
so this is very very complicated situation
now because the integration I also invoke
00:02:48.700 --> 00:02:53.170
the length of it's permeate flux so what is
the algorithm so we we will be having three
00:02:53.170 --> 00:02:57.580
let us let us just summarize what we are having
this can also be solved analytically like
00:02:57.580 --> 00:03:02.650
the earlier cases so will be having three
ordinary differential . equations and these
00:03:02.650 --> 00:03:08.730
equations are basically governing equations
for delta p governing equation of u and governing
00:03:08.730 --> 00:03:15.120
equation of c concentration bulk eutectic
concentration and one algebraic equation the
00:03:15.120 --> 00:03:32.140
algebraic equation is the relationship
between . c m and c via definition of mass
00:03:32.140 --> 00:03:37.430
transfer coefficient k which will be essentially
a function of x and that will be that will
00:03:37.430 --> 00:03:43.319
be a function of I and I will be basically
a function of v w bar or the length average
00:03:43.319 --> 00:03:44.790
permeate flux
00:03:44.790 --> 00:03:53.180
so what we will be doing so the the algorithm
goes like this . the algorithm will be like
00:03:53.180 --> 00:04:03.380
this we assume v w bar a length average . permeate
flux once the length length average permeate
00:04:03.380 --> 00:04:16.400
flux is assumed then I can be estimated .
once I can be definite integral I can be estimated
00:04:16.400 --> 00:04:22.720
by using a trapezoidal rule than I will I
can get k as a function of x once I get the
00:04:22.720 --> 00:04:34.180
k as a function of x then I can go to the
relation . between c m and c and estimate
00:04:34.180 --> 00:04:40.760
the value of c m and then you will be hooking
hooking up this equation with the . with the
00:04:40.760 --> 00:04:51.280
d a e system so this four will be basically
the algebraic equation . and three o d es
00:04:51.280 --> 00:05:02.780
lets say by r k four algorithm so three o
d es . and one algebraic equation this will
00:05:02.780 --> 00:05:08.730
be formulating a d a e system the differential
algebraic equation system an this has to be
00:05:08.730 --> 00:05:13.710
solved simultaneously and then we will be
in the in the process what will be getting
00:05:13.710 --> 00:05:18.520
the output output will be will be getting
k as a function of will be getting v w as
00:05:18.520 --> 00:05:27.040
a function of x delta p as a function of x
c as a function of x u as a function of x
00:05:27.040 --> 00:05:28.940
and c p as a function of x
00:05:28.940 --> 00:05:34.470
after that what we will be doing we will be
finding out the average length length average
00:05:34.470 --> 00:05:41.430
permeate flux as one by l zero to l v w d
x and we will see whether these length average
00:05:41.430 --> 00:05:46.990
permeate flux is coming close to the . value
or not if this is not coming close then you
00:05:46.990 --> 00:05:51.340
have to curb you have to assume another value
of length average permeate flux redo this
00:05:51.340 --> 00:05:57.910
calculation so this will be very complicated
and it will be basically this at every
00:05:57.910 --> 00:06:03.940
step of r k four this algebraic equation has
to be . evaluated by using newton raphson
00:06:03.940 --> 00:06:10.120
method so . four nested in newton . with newton
raphson method at every step of r k four and
00:06:10.120 --> 00:06:15.170
then the whole loop will be um hm with there
is an outer look which will be carrying at
00:06:15.170 --> 00:06:23.350
a particular value of x so this will be continued
till it goes for the so uh . it goes goes
00:06:23.350 --> 00:06:29.400
for the full length of the channel at at a
particular x so these whole calculations has
00:06:29.400 --> 00:06:34.650
to be calculated as the particular value
of x and then it has to be you have to go
00:06:34.650 --> 00:06:39.940
forward forward x to plus delta x it will
be converged and likewise we have to do the
00:06:39.940 --> 00:06:44.080
calculations at every x location we have to
calculate the permeate flux then you have
00:06:44.080 --> 00:06:48.710
to do a length average permeate flux using
newton raphson and then you will check whether
00:06:48.710 --> 00:06:53.090
this calculated average permeate flux will
be um hm coming close to the case value or
00:06:53.090 --> 00:06:57.860
not if not then it is you have to alterate
if yes then you will be getting a converged
00:06:57.860 --> 00:06:58.860
solution
00:06:58.860 --> 00:07:04.860
so likewise the . in an axial mo[dule]- module
system the whole calculation to be done or
00:07:04.860 --> 00:07:09.700
a module can be designed so at the end of
the design what you will be getting you will
00:07:09.700 --> 00:07:15.120
be getting a length average permeate flux
length average permeate concentration . pressure
00:07:15.120 --> 00:07:19.540
drop across the module um hm so once you know
the . pressure drop across the module one
00:07:19.540 --> 00:07:24.820
can select a pump which will be suitable to
pump the fluid to overcome these . pressure
00:07:24.820 --> 00:07:31.210
drops across the module so that completes
the module um hm you know design in
00:07:31.210 --> 00:07:37.650
a linear actual scale and similar thing can
be translated to a tubular channel or a tubular
00:07:37.650 --> 00:07:40.440
module as well
00:07:40.440 --> 00:07:45.850
so next we will be going to the going to our
next topic that is the you know dialysis
00:07:45.850 --> 00:07:51.570
operation and dialysis becomes very very important
in most of the membrane separation process
00:07:51.570 --> 00:07:57.820
but the there is a basic difference of dialysis
between the pressure driven . processes and
00:07:57.820 --> 00:08:15.550
this the major driving force in dialysis is
concentration . gradient . concentration gradient
00:08:15.550 --> 00:08:23.639
so we will be having a feed chamber ok there
will be two chambers those will be separate
00:08:23.639 --> 00:08:31.160
by a by the membrane will be having a feed
side or feed chamber and then will be having
00:08:31.160 --> 00:08:43.990
a dialysate side
dialysate chamber so in a continuous system
00:08:43.990 --> 00:08:58.990
.
00:08:58.990 --> 00:09:11.140
feed is going . into the feed side feed in
and feed is going out . and from here it's
00:09:11.140 --> 00:09:19.230
a counter current you have seen counter current
is always effective dialysate in and here
00:09:19.230 --> 00:09:30.980
you will be having the dialysate out . so
the the the . solute which will be having
00:09:30.980 --> 00:09:36.350
a very small size and molecular weight they
will be having you know zero concentration
00:09:36.350 --> 00:09:40.660
here and they will be going from feed size
to the dialysate side we are maintaining a
00:09:40.660 --> 00:09:45.550
dialysate flow rate so therefore whenever
the the targeted solutes of smaller molecular
00:09:45.550 --> 00:09:49.330
weight or size they will be coming permeating
through the membrane coming to the dialysate
00:09:49.330 --> 00:09:54.130
side immediately they will be washed away
so the concentration maintained in the dialysate
00:09:54.130 --> 00:09:58.350
side will be will be will be almost equal
to zero so therefore we are maintaining the
00:09:58.350 --> 00:10:03.460
maximum concentration gradient of solute . targeted
solute from the feed side to the dialysate
00:10:03.460 --> 00:10:04.460
side
00:10:04.460 --> 00:10:10.100
so therefore one can expect a very high mass
transfer across the membrane module now let
00:10:10.100 --> 00:10:16.980
us see what are the different you know
. the transfer resistances those are involved
00:10:16.980 --> 00:10:33.980
in dialysate process so diffusion diffusion
flux . diffusion is the main mechanism
00:10:33.980 --> 00:10:39.990
is the . because of the concentration gradient
so let us say n I is the mass flux of ith
00:10:39.990 --> 00:11:05.160
species across the membrane . so d will be
nothing but d I m d c I m . d x ok so . so
00:11:05.160 --> 00:11:20.110
d is nothing but d I m delta c I m by l d
I m is the diffusivity .
00:11:20.110 --> 00:11:36.279
of the ith solute in membrane matrix
and delta c I m is nothing but the concentration
00:11:36.279 --> 00:11:41.220
difference in the feed side and the of ith
species from the in the feed side and the
00:11:41.220 --> 00:11:53.930
dialysate side so these will be n I be nothing
but d I m over l c I m in the feed side minus
00:11:53.930 --> 00:12:01.970
c I m in the dialysate side so det[ermination]-
determination of . d I m is very very important
00:12:01.970 --> 00:12:08.459
so if we can determine the the concentration
the diffusivity of solute in the membrane
00:12:08.459 --> 00:12:13.520
matrix then by by measuring the concentration
in the feed side in the dialysate side we
00:12:13.520 --> 00:12:19.920
can really look into the we can can . really
find what is the total . molar flux or the
00:12:19.920 --> 00:12:21.390
mass flux across the matrix
00:12:21.390 --> 00:12:26.390
now let us look into the different type of
resistances those will be occurring appearing
00:12:26.390 --> 00:12:35.930
in the dialysate process so there will
be apparently three resistances one can come
00:12:35.930 --> 00:12:42.250
across during dialysate process so one will
be the feed side resistance mass transfer
00:12:42.250 --> 00:12:47.070
coefficient you know mass transfer of
boundary layer in the feed side and there
00:12:47.070 --> 00:12:52.830
will be a master from the dialysate side and
there will be the membrane resistance there
00:12:52.830 --> 00:12:58.620
are three resistances you will be encountering
one is r f in the feed side and the I is r
00:12:58.620 --> 00:13:14.730
d in the dialysate side and is the membrane
. so r f is liquid film resistance
00:13:14.730 --> 00:13:31.959
in feed side r d is liquid film resistance
.
00:13:31.959 --> 00:13:44.070
in dialysate side and d I and this will
be he diffusivity in the membrane so these
00:13:44.070 --> 00:13:50.390
are membrane resistance we must write down
the mass flux of I th species which will be
00:13:50.390 --> 00:14:04.360
nothing but c I f bar minus c I f divided
by r f this is across . film in feed side
00:14:04.360 --> 00:14:16.070
this bar . representing the concentration
in the bulk these will be equal to d I m divided
00:14:16.070 --> 00:14:25.519
by l c I f minus c I d c I f is the ith
species concentration in the feed in the dialysate
00:14:25.519 --> 00:14:31.620
side d I m is the diffusivity of ith species
in the membrane matrix and l is the membrane
00:14:31.620 --> 00:14:50.649
thickness and there will be dialysate resistance
c I d minus c I d bar divided . by r d
00:14:50.649 --> 00:14:57.170
so now we can we can add this all up and see
what we get so we will be getting an overall
00:14:57.170 --> 00:15:05.970
resistance so from the three equations . we
will be getting c I f bar minus c I f is equal
00:15:05.970 --> 00:15:22.899
to n I r f c I f minus c I d is equal to n
I l over d I m and then c I d minus c I d
00:15:22.899 --> 00:15:30.210
bar will be giving you n I r d if you add
all them up then these will be canceling out
00:15:30.210 --> 00:15:40.740
and you will be getting c I f bar minus c
I d bar is equal to n I all the resistances
00:15:40.740 --> 00:15:52.139
are added in series .
so we can get an overall resistance n I is
00:15:52.139 --> 00:16:02.269
equal to c I f bar minus c I d bar and divided
by r overall . so overall resistance can be
00:16:02.269 --> 00:16:12.639
written as r f plus l over d I m plus r d
and you can you can write the overall mass
00:16:12.639 --> 00:16:22.320
transfer coefficient so k overall is equal
to one over k in the feed side plus l over
00:16:22.320 --> 00:16:30.899
d I m plus one over k in the dialysate side
this ks are basically the mass transfer coefficient
00:16:30.899 --> 00:16:40.940
.
00:16:40.940 --> 00:16:46.839
so one can get an expression of overall mass
transfer coefficient as one over one over
00:16:46.839 --> 00:17:02.670
k f plus . l over d I m plus one over k
d this will be the expression of . overall
00:17:02.670 --> 00:17:09.589
mass transfer coefficient so once we get the
overall . mass transfer coefficient we will
00:17:09.589 --> 00:17:15.919
be able to get the net mass flux across the
dialysate module from the feed side to the
00:17:15.919 --> 00:17:27.509
permeate side or the dialysate side so net
mass flux .
00:17:27.509 --> 00:17:48.190
across the membrane from feed to dialysate
side so that will be giving you a system in
00:17:48.190 --> 00:18:00.580
a for a counter current flow .
00:18:00.580 --> 00:18:14.210
so c in . c out in the feed c in in dialysate
c out in dialysate typically in the dialysate
00:18:14.210 --> 00:18:18.509
we will be giving the pure solvent so typically
c in in the dialysate side will be zero it
00:18:18.509 --> 00:18:25.399
will be some final variant c out so this will
be exactly like a counter current heat exchanger
00:18:25.399 --> 00:18:35.710
the analogy is counter current . heat exchanger
and in our heat transfer course we have already
00:18:35.710 --> 00:18:40.850
found out that how the overall in the heat
transfer heat transfer that is taking place
00:18:40.850 --> 00:18:46.580
in a counter current heat exchanger now in
the case of dialysis it is instead of a counter
00:18:46.580 --> 00:18:52.610
current heat exchanger it is a counter current
mass exchanger and we can write down in in[stead]-
00:18:52.610 --> 00:18:58.840
instead of overall heat transfer coefficient
we can write the overall mass transfer coefficient
00:18:58.840 --> 00:19:04.399
delta t log min temperature difference . can
be replaced by delta c l f t d
00:19:04.399 --> 00:19:12.340
so we can write down the mass flux mass
flow rate kg per unit time kg per hour will
00:19:12.340 --> 00:19:25.789
be nothing but k zero a delta c l m t d what
is delta c l m t d k zero a delta c out
00:19:25.789 --> 00:19:40.789
delta c in minus delta c out divided by l
n delta c out divide by delta c in divided
00:19:40.789 --> 00:19:50.820
by delta c out ok so this will be the expression
of mass . mass mass exchange or mass transfer
00:19:50.820 --> 00:19:57.489
across the channel and k zero is the overall
mass transfer coefficient and we have already
00:19:57.489 --> 00:20:04.499
ob[tained]- we have already seen the . exp[ression]-
. the expression of overall mass transfer
00:20:04.499 --> 00:20:12.169
coefficient it will give one over k f plus
l over d I m plus one over k in the dialysate
00:20:12.169 --> 00:20:13.169
side
00:20:13.169 --> 00:20:17.780
so therefore we know the mass transfer coefficient
in the feed side we know the mass transfer
00:20:17.780 --> 00:20:22.730
coefficient in the dialysate side and if we
can determine what is the diffusivity of the
00:20:22.730 --> 00:20:27.799
. of the solute we are taking about in the
membrane phase we can calculate the resistance
00:20:27.799 --> 00:20:32.549
occurred by the membrane l by d m l is the
membrane thickness this can be determined
00:20:32.549 --> 00:20:38.309
if we put a cross section under the scanning
electro microscopic we can estimate the thickness
00:20:38.309 --> 00:20:43.950
of the membrane but d I m may not be possible
to know directly so there are if we if we
00:20:43.950 --> 00:20:49.389
if somehow we can estimate the value of diffusivity
of ith solute within the membrane matrix we
00:20:49.389 --> 00:20:54.330
can evaluate the overall mass transfer coefficient
once we can evaluate the overall mass transfer
00:20:54.330 --> 00:20:59.679
coefficient we can find out the concentration
steam all the steams coming out of the system
00:20:59.679 --> 00:21:04.779
going into the system . once we get that we
will be able to estimate the the net delta
00:21:04.779 --> 00:21:10.169
c l t d term and once if we if you know the
value of d I m we and if you know the membrane
00:21:10.169 --> 00:21:14.529
area we can find out what will be the mass
that will be transferred through the dialysis
00:21:14.529 --> 00:21:20.269
or unit time if the concentration is fixed
mass exchange is fixed we can evaluate what
00:21:20.269 --> 00:21:25.519
is the area of the dialysis membrane that
is required for designing such purposes so
00:21:25.519 --> 00:21:31.009
once we know we can you can determine d
I m these equation can be utilized for the
00:21:31.009 --> 00:21:35.879
design purposes for the given design these
equation can be utilized for the symbolization
00:21:35.879 --> 00:21:37.359
purposes as well
00:21:37.359 --> 00:21:42.729
now in the next what we are going to do we
are going to do a how to evaluate the or
00:21:42.729 --> 00:21:49.710
the method of procedure how to evaluate the
diffusivity of ith species in in the membrane
00:21:49.710 --> 00:21:56.059
matrix for that exactly like the case of permeability
and real retention you conduct an independent
00:21:56.059 --> 00:22:03.830
set of experiment in a small batch dialysis
process or a small batch . dialysis cell so
00:22:03.830 --> 00:22:21.759
lets look into the batch dialysis analysis
.
00:22:21.759 --> 00:22:27.029
in a batch dialysis analysis there will be
no continuous flow the cell is exactly the
00:22:27.029 --> 00:22:35.640
same like before the two chambers have to
be separated by a semI permeable barrier that
00:22:35.640 --> 00:22:41.649
is the volume of the feed chamber is v f volume
of the dialysate chamber is d you put a dialysate
00:22:41.649 --> 00:22:46.909
solute solute solution in the dialysate chamber
in most of the cases it is pure water and
00:22:46.909 --> 00:22:51.399
we will put a solution in the feed chamber
and there will be mass exchange from the feed
00:22:51.399 --> 00:22:53.749
side to the dialysate side
00:22:53.749 --> 00:23:16.049
now v f and v d are . . volumes of feed and
dialysate chambers and the solute will be
00:23:16.049 --> 00:23:24.659
permeating from the feed chamber to the dialysate
side now we we should have stirring here as
00:23:24.659 --> 00:23:31.809
well as here so a very good stirring will
be will be preventing formation of any film
00:23:31.809 --> 00:23:49.809
over the membrane surface so continuous and
efficient stirring is required that will lead
00:23:49.809 --> 00:24:08.679
to for prevention of formation . .
of any film resisting the solute movement
00:24:08.679 --> 00:24:29.309
any film over membrane surface ok both in
feed side and in dialysate side
00:24:29.309 --> 00:24:34.946
so if that is done we will we we do not have
r f and we do not have r f and we do not have
00:24:34.946 --> 00:24:42.580
r d in this case so there is a stirring should
be provided so what what resistance is existing
00:24:42.580 --> 00:24:52.739
is the membrane resistance d I m over l so
membrane resistance should be there resistance
00:24:52.739 --> 00:24:59.929
. due to membrane on it
00:24:59.929 --> 00:25:11.489
so next we what we we are . going to do we
are writing the solute mass balance
00:25:11.489 --> 00:25:28.879
in feed chamber . and in dialysate chamber
if you do that the accumulation should be
00:25:28.879 --> 00:25:36.799
the material that is going out of the system
so accumulation will be d d t of c I d and
00:25:36.799 --> 00:25:54.899
v d is equal to a m d I m divided by l c I
f minus c I d so d is the solute balance in
00:25:54.899 --> 00:26:04.570
the dialysate side
d is the accumulation of solute in the dialysate
00:26:04.570 --> 00:26:09.850
chamber and that is the material this accumulation
is because of the material that is the solute
00:26:09.850 --> 00:26:15.109
that is coming in from the feed side similarly
you will be writing the solute balance in
00:26:15.109 --> 00:26:20.869
the feed side so that will be equal to the
amount of solute that will be going out of
00:26:20.869 --> 00:26:27.980
the feed side so it will be a depletion in
the in the case of feed chamber so d d t should
00:26:27.980 --> 00:26:43.129
be equal to c I f v f is equal to minus a
m d I m over l the amount remain same c I
00:26:43.129 --> 00:26:54.679
f minus c I d this is a feed side solute balance
so dialysate side will be enriched by the
00:26:54.679 --> 00:27:00.919
solute transport and the feed side will be
stripped by the solute transport
00:27:00.919 --> 00:27:06.469
so what will . be the bound[ary]- initial
conditions at for both of them at t is equal
00:27:06.469 --> 00:27:16.669
to zero we have c I f is equal to c I f naught
and t is equal to zero c I d equal to zero
00:27:16.669 --> 00:27:22.580
so it was a pure dialysate initially there
is no solute present there so c I d was zero
00:27:22.580 --> 00:27:29.960
and in the feed side there is a feed concentration
corresponding to c I f naught now these two
00:27:29.960 --> 00:27:34.739
equations have to be solved simultaneously
these two ordinary differential equations
00:27:34.739 --> 00:27:38.739
initial way initial volume problem so they
are since their initial conditions are specified
00:27:38.739 --> 00:27:43.450
these are basically initial value problem
these two equations have to be solved simultaneously
00:27:43.450 --> 00:27:48.639
in order to get the concentration profile
in the dialysate side so we take a sample
00:27:48.639 --> 00:27:54.429
from the dialysate side and monitor the concentration
of the solute in the dialysate side and then
00:27:54.429 --> 00:28:01.110
we will be comparing with the if the theoretical
value of the experimental profile of the concentration
00:28:01.110 --> 00:28:06.080
in the . dialysate side should be compared
with the predicted or the theoretical values
00:28:06.080 --> 00:28:12.169
and from that comparison the value of d I
m will be evaluated so I will stop in this
00:28:12.169 --> 00:28:17.610
class in the next class I will we will show
how the value of d I m will be it will be
00:28:17.610 --> 00:28:20.299
coming out from this analysis
00:28:20.299 --> 00:28:20.709
thank you very much .