WEBVTT
Kind: captions
Language: en
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welcome to this section now will be as we
have discuss in this in the last class that
00:00:24.130 --> 00:00:29.270
another typical filtration physical
phenomena occurring in membrane filtration
00:00:29.270 --> 00:00:34.450
is gel formation over the membrane surface
now in the last class we have seen how the
00:00:34.450 --> 00:00:39.160
gel is formed over the membrane surface there
are two ways one the the filtration may be
00:00:39.160 --> 00:00:45.780
. osmotic pressure control initially later
on it can get transition into the gel layer
00:00:45.780 --> 00:00:50.110
control in phenomena in the second case there
may be solutes which will be gel forming from
00:00:50.110 --> 00:00:56.629
the very beginning of the process like polyphenol
alcohol um . and which are known as the
00:00:56.629 --> 00:01:01.760
wellknown gel forming agency so in the in
these class we will be looking into the various
00:01:01.760 --> 00:01:06.900
modelling approaches of the gel layer controlling
filtration the first one will be looking into
00:01:06.900 --> 00:01:16.299
the . steady state . model and this will be
same as one dimensional film theory model
00:01:16.299 --> 00:01:24.320
.
that we have considered in the . in case of
00:01:24.320 --> 00:01:26.210
osmotic pressure control filtration
00:01:26.210 --> 00:01:34.430
we assume a constant film of thickness delta
constant film of solutes of thickness delta
00:01:34.430 --> 00:01:40.580
is formed over the . membrane surface and
it will be having a constant . constant solute
00:01:40.580 --> 00:01:51.750
concentration within the gel so cg is known
as the gel layer concentration . and at the
00:01:51.750 --> 00:01:58.280
steady state there will be a . there will
there will be a constant growth of mass transfer
00:01:58.280 --> 00:02:03.830
bounder layer over this . so these will be
these will be actually the delta the mass
00:02:03.830 --> 00:02:12.709
transfer bounder layer and these will be the
gel thickness so this is the thickness of
00:02:12.709 --> 00:02:24.170
mass transfer bounder layer is a . gel thickness
and we assume that the solute will be suffering
00:02:24.170 --> 00:02:32.520
a concentration gradient or polarization from
the bulk at c naught to gel solution interface
00:02:32.520 --> 00:02:37.250
there is gel concentration and then the concentration
of gel remains same within the gel layer then
00:02:37.250 --> 00:02:44.440
permit you will be getting the permeate flux
now at the . at the steady state we can
00:02:44.440 --> 00:02:49.140
write down the solute balance within these
mass transfers boundary layer if you do a
00:02:49.140 --> 00:03:00.530
if you write a solute balance . in mass transfers
boundary layer . then will be getting v w
00:03:00.530 --> 00:03:08.400
. c plus d c d y will be equal to zero
00:03:08.400 --> 00:03:15.430
there is no solute concentration in the permeate
stream in this case because in most of the
00:03:15.430 --> 00:03:21.819
cases the gel forming material are larger
in size they will be having bigger . sizes
00:03:21.819 --> 00:03:26.680
and the larger molecular weight and typically
they will be return by the membrane surface
00:03:26.680 --> 00:03:31.880
and the permit concentration is zero in most
of the cases so if you really carry out this
00:03:31.880 --> 00:03:38.470
you know integral from zero to delta with
the concen[tration]- with the variation of
00:03:38.470 --> 00:03:47.739
concentration from cg to c naught then we
will be getting . v w is equal to k l . n
00:03:47.739 --> 00:03:56.220
c g by c naught . so this is the steady state
equation of the permeate flux in the gel polarization
00:03:56.220 --> 00:04:01.410
case and mass transfers of coefficient can
be estimated depending . on the domain
00:04:01.410 --> 00:04:10.380
of the flow domain as well as the . as
well as the . whether its it will be a laminar
00:04:10.380 --> 00:04:15.170
flow and turbulent flow and or whether you
will be working with a tubular module or or
00:04:15.170 --> 00:04:20.009
whether you with a working with a flat sheet
geometry or in a hollow fibre module
00:04:20.009 --> 00:04:24.759
so depending on the module i realise the flow
domain mass transfers coefficient will be
00:04:24.759 --> 00:04:30.380
evaluated whether its a start cell . whether
in a start cell unless than reynolds from
00:04:30.380 --> 00:04:33.349
the less than thirty two thousand about thirty
two thousand all those correlations we have
00:04:33.349 --> 00:04:37.410
already of correlations of mass transfers
coefficient we have already discuss earlier
00:04:37.410 --> 00:04:42.010
those are coming from the heat and mass transfers
analogy will you valid and depending on the
00:04:42.010 --> 00:04:46.230
flow domain and the flow geometry one will
be estimating the mass transfers coefficient
00:04:46.230 --> 00:04:50.470
once the mass transfers coefficient will be
estimated the gel layer concentration if the
00:04:50.470 --> 00:04:55.240
gel layer concentration is known to you then
one can estimate the permeate flux at the
00:04:55.240 --> 00:05:02.380
steady state now let us look into the . . so
what is the typical parameter in this case
00:05:02.380 --> 00:05:07.400
the typical parameter in this case is the
gel layer concentration how to evaluate the
00:05:07.400 --> 00:05:08.870
gel layer concentration
00:05:08.870 --> 00:05:14.610
so it is very easy to evaluate the gel layer
concentration the governing equation for calculating
00:05:14.610 --> 00:05:20.600
or estimating the gel layer concentration
c g is this steady state equation so we conduct
00:05:20.600 --> 00:05:28.180
the experiment at different concentration
of the feed solute . and then measure the
00:05:28.180 --> 00:05:34.490
permeate flux at a particular k k means at
particular turbulence so we plot them v w
00:05:34.490 --> 00:05:41.010
versus feed concentration in a lock skill
this will be a lock skill l n c naught and
00:05:41.010 --> 00:05:48.350
then will be seeing that permeate flux will
be decreasing as the feed concentration increases
00:05:48.350 --> 00:05:56.199
so it will be a straight line and we . we
extra pull it this line and were it cuts the
00:05:56.199 --> 00:06:02.430
gel layer . in y equal to a where you cuts
x axis . that concentration is the gel layer
00:06:02.430 --> 00:06:07.550
concentration so when c naught is equal to
c g l n one will be equal to zero so you will
00:06:07.550 --> 00:06:11.960
be getting the g o value of flux layer but
in the in the in an actual experiment you
00:06:11.960 --> 00:06:17.500
will be always getting some value of finite
value of flux so therefore these line will
00:06:17.500 --> 00:06:20.410
be extra polluted and we will be getting the
gel layer concentration
00:06:20.410 --> 00:06:27.320
if you operate if you conduct the same set
of experiment with another value of turbulence
00:06:27.320 --> 00:06:34.710
or the mass transfers coefficient then you
are um you will be getting another so but
00:06:34.710 --> 00:06:41.650
if you extra pollute that will be boil down
into the same value um of c g so that is how
00:06:41.650 --> 00:06:47.110
the gel layer concentration can be can be
obtained experimentally and this parameter
00:06:47.110 --> 00:06:51.259
can be estimated so once you you estimate
the parameter the gel layer concentration
00:06:51.259 --> 00:06:56.389
can be put you know the unknown value of . for
a for a known value of c naught then you could
00:06:56.389 --> 00:07:00.759
be getting the value of permeate flux knowing
the estimating the mass transfers coefficient
00:07:00.759 --> 00:07:14.400
. but remember in case of gel layer control
filtration . control in case c g must be very
00:07:14.400 --> 00:07:20.479
very high and sometimes c g is around may
be as high as one hundred fifty times to two
00:07:20.479 --> 00:07:26.389
hundred times of c naught that means in a
gel layer control filtration the concentration
00:07:26.389 --> 00:07:31.280
suffers maximum from the in the within the
mass transfer boundary layer from c naught
00:07:31.280 --> 00:07:36.550
to c g and sometimes c g can be as high as
one hundred fifty times to two hundred times
00:07:36.550 --> 00:07:38.669
of feed bulk concentration
00:07:38.669 --> 00:07:45.930
therefore variation of of the transport coefficients
will be extremely high in case of gel layer
00:07:45.930 --> 00:07:52.100
control filtration compare to the osmotic
pressure control filtration . as we have discussed
00:07:52.100 --> 00:07:57.300
there are three major parameters which will
be appearing in your system . the transport
00:07:57.300 --> 00:08:02.729
coefficients one is a density viscosity and
diffusibility . density is the weakest function
00:08:02.729 --> 00:08:08.139
of concentration diffusibility is a little
bit stronger viscosity is the strongest function
00:08:08.139 --> 00:08:14.600
of concentration and typically viscosity varies
exponentially concentration so therefore correction
00:08:14.600 --> 00:08:20.100
by sieder tate correction is very very important
in case of the gel layer control filtration
00:08:20.100 --> 00:08:27.130
and one can get a correction factors something
like this one point eight five renyold schmidt
00:08:27.130 --> 00:08:35.329
d e by l for the laminar flow in a rectangular
channel schmidt at bulk divided by schmidt
00:08:35.329 --> 00:08:44.060
at the wall so raise to the power point two
seven what is . sometimes these schmidt number
00:08:44.060 --> 00:08:53.120
. ratio can be approximated as mu at bulk
divide by mu at wall because the if if variation
00:08:53.120 --> 00:08:58.770
of diffusibility and density are minimal then
it will be measure mainly the variation of
00:08:58.770 --> 00:09:06.600
viscosity . and mu at bulk means mu is evaluated
at bulk concentration that is c bulk or c
00:09:06.600 --> 00:09:12.660
naught as we have discussed c naught is the
bulk concentrate in these case and mu at wall
00:09:12.660 --> 00:09:18.890
will be nothing but mu evaluated at gel layer
concentration
00:09:18.890 --> 00:09:25.480
then as we have discussed earlier that since
this value is viscosity is increasing with
00:09:25.480 --> 00:09:30.720
concentration and c gel is always greater
than c naught the denominator will be denominator
00:09:30.720 --> 00:09:36.029
will be always greater than numerator and
we will be and will be basically under . you
00:09:36.029 --> 00:09:41.510
will basically the value of mass transfers
coefficient will be decreased because of the
00:09:41.510 --> 00:09:46.770
variation of the properties in the mass transfers
bounder layer hence one has to invoke the
00:09:46.770 --> 00:09:51.010
seider tate correction factor in order to
estimate mass transfers coefficient in case
00:09:51.010 --> 00:09:56.160
of gel layer control filtration because here
the viscous the concentration variation is
00:09:56.160 --> 00:10:01.750
a maximum therefore leading to the maximum
variation . of the thermo physical or you
00:10:01.750 --> 00:10:07.260
know transport coefficient or transport properties
so this is a must in case of a this correction
00:10:07.260 --> 00:10:11.570
factor is a must in case of gel layer control
filtration compared to the osmotic pressure
00:10:11.570 --> 00:10:12.990
control filtration
00:10:12.990 --> 00:10:33.690
now next will be looking into a most likely
gel polarized filtration cased .
00:10:33.690 --> 00:10:44.570
filtration so what is that a will be having
a membrane . there is a formation of gel over
00:10:44.570 --> 00:10:51.700
it and over their there will be a formation
of mass transfers boundary layer because of
00:10:51.700 --> 00:10:58.890
the external flow so c is equal to c naught
here c is equal to c g here and c g remains
00:10:58.890 --> 00:11:07.150
same in the gel layer . and if you put your
axis at y equal to zero these will be y is
00:11:07.150 --> 00:11:12.220
equal to delta but delta is a thickness of
the mass transfers boundary layer and then
00:11:12.220 --> 00:11:19.290
these will be y is equal to delta plus l where
l is the thickness of gel layer at any point
00:11:19.290 --> 00:11:26.160
of time and then you will be getting a permeate
flux as time progresses these gel layer grows
00:11:26.160 --> 00:11:31.020
in grows in the white direction and your
thickness which will be essentially function
00:11:31.020 --> 00:11:37.850
of time so therefore this thickness l it increases
with time so if you look in to the value so
00:11:37.850 --> 00:11:43.089
you know variation the gel layer thickness
grows with time and then it will be reaching
00:11:43.089 --> 00:11:49.140
at steady state probably because of the you
know existence of external flow and as you
00:11:49.140 --> 00:11:54.110
have seen that there will be two resistance
there will be three resistance acting in random
00:11:54.110 --> 00:11:58.700
in this case one is the mass transfers boundary
layer resistance mass transfers resistance
00:11:58.700 --> 00:12:04.630
. another is the gel layer resistance another
is the membrane resistance
00:12:04.630 --> 00:12:09.920
so these three resistance acting in tandem
here and since the gel layer resistance is
00:12:09.920 --> 00:12:15.190
increasing as a function of time here and
these membrane resistance will be remaining
00:12:15.190 --> 00:12:19.670
constant and mass transfers boundary layer
will be also will remaining constant and then
00:12:19.670 --> 00:12:25.310
you will be getting a constant decrease in
permeate flux so this is how the permeate
00:12:25.310 --> 00:12:30.519
flux varying with time this is how the gel
grows in time when gel layer thickness becomes
00:12:30.519 --> 00:12:36.700
constants steady state permeate flux also
becomes steady state so we are talking about
00:12:36.700 --> 00:12:42.680
a system or lets say moving boundary because
l is growing as a function of time the value
00:12:42.680 --> 00:12:47.630
of l is increasing so its the boundary is
moving we we are talking about a moving boundary
00:12:47.630 --> 00:12:54.340
system so now we write down the mass balance
equation between a within the mass transfers
00:12:54.340 --> 00:13:04.890
transfer boundary layer so solute balance
equation we write .13:00) so these becomes
00:13:04.890 --> 00:13:15.250
rho g d l d t this is the accumulation term
is equal to v w c minus d d c d y and the
00:13:15.250 --> 00:13:21.600
boundary condition it should satisfy is that
at y equal to zero c is equal to c naught
00:13:21.600 --> 00:13:27.089
and evaluate at y equal to delta c is equal
to c g
00:13:27.089 --> 00:13:31.860
so within the this two boundaries we evaluate
this and if you look into this type of this
00:13:31.860 --> 00:13:39.180
equation now here l is a soul function of
time soul function of t so l is a soul function
00:13:39.180 --> 00:13:47.140
of time and c is a function of y alone
so because c is the concentration that is
00:13:47.140 --> 00:13:51.000
solute concentration occurring in the mass
transfers boundary layer and mass transfers
00:13:51.000 --> 00:13:56.930
boundary layer is entirely depending on the
hydrodynamics of the system so i can take
00:13:56.930 --> 00:14:03.490
the i can do this integration over y . keeping
left hand side is constant so this can be
00:14:03.490 --> 00:14:08.329
treated as constant therefore if you if you
can identify this equation these a these a
00:14:08.329 --> 00:14:14.240
first order ordinary differential equation
but its a not homogenous its a non homogenous
00:14:14.240 --> 00:14:27.839
homogenous first order ordinary differential
equation and the non homogenous term is basically
00:14:27.839 --> 00:14:33.300
these term rho g d l d t rho g will be treated
as constant during the integration over y
00:14:33.300 --> 00:14:39.920
so if you really you know carry out this integration
then ultimately you will be getting rho g
00:14:39.920 --> 00:14:53.050
d l d t is equal to v w c g minus c naught
exponential v w by k one minus exponential
00:14:53.050 --> 00:15:03.790
v w by k if you look into the . steady state
so . . steady state solution at the steady
00:15:03.790 --> 00:15:12.820
state . d l d t will be zero and will be getting
back you steady state solution v w equal to
00:15:12.820 --> 00:15:21.730
k l n c g by c naught so these will be the
governing equation of the transient . you
00:15:21.730 --> 00:15:29.050
know transient . value of the gel layer
thickness as well as the permeate flux now
00:15:29.050 --> 00:15:35.730
if you look into the expression of permeate
flux as a resist from the phenomenological
00:15:35.730 --> 00:15:41.620
point of view this will be nothing but the
driving force divided by the resistance so
00:15:41.620 --> 00:15:49.490
there will be two resistances r m plus r g
the gel layer resistance and membrane resistance
00:15:49.490 --> 00:15:55.339
and the mass transfers resistance is already
taken care of in the in the mass transfers
00:15:55.339 --> 00:15:56.380
coefficient
00:15:56.380 --> 00:16:10.470
so what is r g r g is known as the gel layer
resistance and what is gel resistance this
00:16:10.470 --> 00:16:17.430
is be nothing but alpha into one minus epsilon
g rho g times l what is alpha alpha is called
00:16:17.430 --> 00:16:25.579
as the known as the specific . gel layer resistance
its a gel characteristic is known as the specific
00:16:25.579 --> 00:16:45.949
gel layer resistance . it has a unit metre
per k g epsilon is the gel layer . porosity
00:16:45.949 --> 00:16:56.410
rho g is gel layer density .
typically this will be . this gel layer density
00:16:56.410 --> 00:17:01.470
slightly higher than the water density typical
around one zero one five eleven hundred one
00:17:01.470 --> 00:17:06.329
zero five zero k g per metre to like that
and l is the gel layer thickness which will
00:17:06.329 --> 00:17:10.970
be essentially a function of time so therefore
since l is a function of time rest are all
00:17:10.970 --> 00:17:16.760
gel characteristics v w this r g will be function
of time since l is an increasing function
00:17:16.760 --> 00:17:21.850
of time r g will be also increasing function
of time leading to decreasing in flux as a
00:17:21.850 --> 00:17:25.139
function of operation
00:17:25.139 --> 00:17:38.600
so what is alpha alpha is specific gel layer
resistance .
00:17:38.600 --> 00:17:44.139
and these will be obtained form the kozeny
carman equation as one hundred eighty one
00:17:44.139 --> 00:17:54.640
minus epsilon g divided by epsilon g cube
rho g d p square so these comes from kozeny
00:17:54.640 --> 00:18:13.559
carman equation .
classical filtration theory . so rho g is
00:18:13.559 --> 00:18:24.960
the gel . gel resist gel layer density epsilon
is the gel porosity d p is the diameter . of
00:18:24.960 --> 00:18:35.289
gel forming particles .
sometimes we are talking about the filtration
00:18:35.289 --> 00:18:39.859
of the polyvinyl alcohol the . the polymers
sometimes we are talking about the filtration
00:18:39.859 --> 00:18:45.519
of the fruit juice involve in pectin but this
polysaccharides or polymers they will be basically
00:18:45.519 --> 00:18:51.080
. they will not be forming the heart spheres
having a definite particle diameter d p so
00:18:51.080 --> 00:18:55.389
they will be forming a very viscous network
over the membrane surface forming which will
00:18:55.389 --> 00:18:58.799
be . which will be calling we will be calling
as gel
00:18:58.799 --> 00:19:05.240
now d p means . it will be the gel layer . that
means it will be forming a gel layer which
00:19:05.240 --> 00:19:10.690
will be equivalent to the resistance offering
by particles of averaged diameter d p the
00:19:10.690 --> 00:19:15.039
it will be interpreted that way the kozeny
carman specific gel layer resistance will
00:19:15.039 --> 00:19:21.249
be interpreted like that although you may
not be having exactly diol defined particles
00:19:21.249 --> 00:19:27.059
of diameter d p it will be the resistance
that will be offered by the polysaccharides
00:19:27.059 --> 00:19:32.530
of polymers which will be a natural gel forming
agent compared which will be so basically
00:19:32.530 --> 00:19:36.950
they will be offering a gel layer resistance
which will be equivalent to a particle having
00:19:36.950 --> 00:19:42.979
a diameter . equivalent diameter d p so that
will be the interpretation of the specific
00:19:42.979 --> 00:19:47.019
gel layer resistance now if you look into
the governing equation we have the governing
00:19:47.019 --> 00:19:56.519
equation of gel layer . thickness rho g d
l d t is equal to v w c g minus c naught exponential
00:19:56.519 --> 00:20:08.109
v w by k divided by . one minus exponential
v w by k and d l . and v w is equal to delta
00:20:08.109 --> 00:20:21.720
p by mu r m plus r g r g is alpha one minus
epsilon g rho g times l where l is a function
00:20:21.720 --> 00:20:27.320
of time and the governing equation of l is
here and these equation as to be solved at
00:20:27.320 --> 00:20:34.249
l is equal to . at t is equal to zero l is
equal to zero that means at the starting there
00:20:34.249 --> 00:20:35.710
is no gel layer thickness
00:20:35.710 --> 00:20:41.210
so this is the ordinary . differential equation
but these cannot be solved . you know these
00:20:41.210 --> 00:20:45.590
not not not linear this cannot be because
v w will be essenti[al]- is appearing in the
00:20:45.590 --> 00:20:51.260
right hand side there are three . places and
v w will be essentially a function of l so
00:20:51.260 --> 00:20:55.989
you will be getting an ordinary differential
equation d l d t with a initial condition
00:20:55.989 --> 00:21:06.960
t equal to zero l equal to zero so one can
take recurs to the runse kutta four . method
00:21:06.960 --> 00:21:13.570
. to solve these equation as a function
of time so after solution we will be getting
00:21:13.570 --> 00:21:19.169
l as a function of time once you . get a
get l is a function of time insert here so
00:21:19.169 --> 00:21:25.349
you will be getting the permeate flux profile
as a function of time so this is how the gel
00:21:25.349 --> 00:21:31.710
layer control um filtration will be will
will be model in an actual scenario but if
00:21:31.710 --> 00:21:37.130
you . if you have must have understood that
there are several parameters those are appearing
00:21:37.130 --> 00:21:42.210
in the gel layer . control in filtration next
what will be looking into this how to estimate
00:21:42.210 --> 00:21:48.470
this parameters once this parameters are estimated
then only i will be able to solve the governing
00:21:48.470 --> 00:21:51.090
equation for the gel layer control in filtration
00:21:51.090 --> 00:21:57.779
now let us look in to the how various parameters
are estimated now first will let us list down
00:21:57.779 --> 00:22:07.059
the what are the parameters those are appearing
in the gel layer model one is the gel concentration
00:22:07.059 --> 00:22:13.729
and i have already seen how the gel concentration
is estimated and specific gel layer resistance
00:22:13.729 --> 00:22:22.009
alpha epsilon g d p epsilon g gel porosity
d p is a gel forming particle and rho g is
00:22:22.009 --> 00:22:28.740
the gel layer density so lets find out how
the estimation estimation of gel layer concentration
00:22:28.740 --> 00:22:34.499
i have already done now let us look into the
how to estimate the specific gel layer resistance
00:22:34.499 --> 00:22:55.320
. alpha so for that we have to conduct experiment
conduct a separate . experiment in a batch
00:22:55.320 --> 00:23:10.690
cell .
00:23:10.690 --> 00:23:15.700
batch cell and we derive we develop a theory
for you know permeate flux profile how to
00:23:15.700 --> 00:23:21.139
obtain . in a batch cell and then we will
see how the . you know . specific gel layer
00:23:21.139 --> 00:23:26.429
resistance can be estimated quite easily by
conducting a separate set of experiment in
00:23:26.429 --> 00:23:28.289
the batch cell ok
00:23:28.289 --> 00:23:34.349
so in a batch cell if you write the expression
of v w permeate flux it will be nothing but
00:23:34.349 --> 00:23:48.039
one over a d v d t where a is the membrane
area . or filtration area v is the cumulative
00:23:48.039 --> 00:24:06.700
. cumulative permeate volume so we can write
one over a d v d t is equal to delta p divided
00:24:06.700 --> 00:24:15.269
by mu r m plus r g so this is also known as
the . series model . so two resistance there
00:24:15.269 --> 00:24:26.690
are acting in series so one by a d v d t will
be is equal to delta p by mu i take r m outside
00:24:26.690 --> 00:24:41.049
so these becomes non dimensional r g and what
is v w naught this pure water flux . that
00:24:41.049 --> 00:24:51.059
is nothing but delta p by mu r m ok so because
if there is no solution only pure water
00:24:51.059 --> 00:24:55.179
there is no osmotic pressure there is no gel
layer resistance nothing so we will be getting
00:24:55.179 --> 00:25:01.099
delta p by mu r m or l p delta p so this
is a pure water of flux .25:00) so we can
00:25:01.099 --> 00:25:17.269
write the . d v d t one over a d v d t
is equal to v w naught divided by one plus
00:25:17.269 --> 00:25:31.519
r g by r m so next we write down a solute
mass balance .
00:25:31.519 --> 00:25:41.749
solute mass balance in gel layer . if you
really do that this becomes l a one minus
00:25:41.749 --> 00:25:48.980
epsilon g rho g terms c naught terms v v is
the cumulative volume of the filtrate so at
00:25:48.980 --> 00:25:52.299
a particular time this much volume has been
filtered out
00:25:52.299 --> 00:25:57.019
so that will be carrying c naught concentration
of the solute so total mass that will be deposited
00:25:57.019 --> 00:26:02.700
as the gel is c naught terms v so in terms
of the . gel . gel properties it will be gel
00:26:02.700 --> 00:26:05.909
layer thickness multiplied by l cross section
that will be a gel layer volume multiplied
00:26:05.909 --> 00:26:10.659
by one minus epsilon g there will be a actual
material that is there in the gel layer multiplied
00:26:10.659 --> 00:26:21.039
by gel layer density is the amount of gel
that is presented so we can write r g . r
00:26:21.039 --> 00:26:28.159
g is basically as we have written earlier
r g is alpha one minus epsilon g rho g terms
00:26:28.159 --> 00:26:36.559
l so by combining this two we can get r g
is equal to alpha c naught v by a so once
00:26:36.559 --> 00:26:43.840
we get that we can write the governing equation
of cumulative volume as one over a d v d t
00:26:43.840 --> 00:26:58.820
is equal to v w naught divided by one plus
alpha c naught by a r m terms v ok so we
00:26:58.820 --> 00:27:05.929
we take it on the other side . and do the
integration so it will be one plus alpha c
00:27:05.929 --> 00:27:17.619
naught divided by a r m terms v d v is equal
to v w naught a d t where you integrate from
00:27:17.619 --> 00:27:22.999
zero to t and these will be integrate on left
hand side from zero to v
00:27:22.999 --> 00:27:28.669
so let us write down what do we get after
the integration after the integration we will
00:27:28.669 --> 00:27:41.360
be getting v plus alpha c naught divided by
two a r m terms v square a v w naught d t
00:27:41.360 --> 00:27:47.109
and then we do a you know rearrangement to
make it linear form of equation t by v is
00:27:47.109 --> 00:28:02.059
equal to one over a v w naught plus alpha
c naught divided by two a square v w naught
00:28:02.059 --> 00:28:10.159
. r m terms v then what we do we measure the
cumulative volume we conduct a experiment
00:28:10.159 --> 00:28:16.219
batch cell experiment and measure cumulative
volume as a function of time and then plot
00:28:16.219 --> 00:28:26.799
t by v . as a function of v so y equal to
m x plus c so we will be getting value these
00:28:26.799 --> 00:28:37.529
for a particular delta p ok so from the intersect
will be getting one over a v w naught that
00:28:37.529 --> 00:28:49.049
we can verify and from the slope we can get
alpha c naught divided by two a square v w
00:28:49.049 --> 00:28:55.289
naught r m so if you remember that c naught
is known to us there is a fit concentration
00:28:55.289 --> 00:29:00.869
membrane resistance is known to us v w naught
is known to us that is delta p by mu r m . so
00:29:00.869 --> 00:29:05.659
that is also known to us area is the membrane
resistance membrane area so membrane filtration
00:29:05.659 --> 00:29:10.119
areas as well known to us so we know everything
in this case slope we have determine so will
00:29:10.119 --> 00:29:11.960
be getting the value of alpha
00:29:11.960 --> 00:29:20.119
so alpha is determined from the slope now
we conduct the experiments for different pressure
00:29:20.119 --> 00:29:25.549
drop and then carry out the . and estimate
the value of alpha then put a correlation
00:29:25.549 --> 00:29:33.639
of alpha is equal to alpha naught delta p
to the power n so if n is equal to zero then
00:29:33.639 --> 00:29:40.130
the that cake or gel is not compressibility
incompressible n is equal to zero incompressible
00:29:40.130 --> 00:29:53.559
cake or gel ok otherwise the gel is compressible
and typically the value of n is less than
00:29:53.559 --> 00:29:59.419
one generally ok it is compressible so it
gets compressed its epsilon g or the porosity
00:29:59.419 --> 00:30:03.279
decreases as we increase the . trans membrane
pressure drop
00:30:03.279 --> 00:30:08.899
so this is how the parameter specific gel
layer resistance is estimated as in case of
00:30:08.899 --> 00:30:14.089
gel forming gel gel controlling filtration
in the next class we will looking into the
00:30:14.089 --> 00:30:20.669
estimation of gel layer porosity gel layer
. particles size equivalent particle diameter
00:30:20.669 --> 00:30:26.529
for a typical you know polymeric or polysaccharide
gel and gel density then that will completely
00:30:26.529 --> 00:30:31.960
with then that that will completely wind up
the filtration of gel controlling membrane
00:30:31.960 --> 00:30:32.960
separation system
00:30:32.960 --> 00:30:34.839
thank you very much .