WEBVTT
Kind: captions
Language: en
00:00:18.570 --> 00:00:24.050
good morning everyone so will be start as
we have seen in the last class that we
00:00:24.050 --> 00:00:30.560
were look we were looking into the modeling
of membrane modules and . we have
00:00:30.560 --> 00:00:37.600
looked into the very simple version of the
module modeling we have constituted the
00:00:37.600 --> 00:00:41.730
constant permeate flux that is coming out
from the wall or from the membrane surface
00:00:41.730 --> 00:00:47.390
initially we have we have considered the
flow through a rectangular geometry which
00:00:47.390 --> 00:00:52.120
will be nothing but simulating the spiral
wound module and similarly we have looked
00:00:52.120 --> 00:00:57.140
into the tubular module and although we
have not derived the detailed equations from
00:00:57.140 --> 00:01:00.850
the tubular module but this could have been
done in the exactly the same way we have done
00:01:00.850 --> 00:01:06.740
. from the overall material balance and solute
balance and the and the pressure drop calculations
00:01:06.740 --> 00:01:13.369
in a in in rectangular channel so modeling
of membrane module indicates that we have
00:01:13.369 --> 00:01:18.210
we have to calculate the axial pressure drop
that will be developing inside the membrane
00:01:18.210 --> 00:01:23.740
module and this axial pressure drop calculation
is very very important in almost in almost
00:01:23.740 --> 00:01:27.780
all chemical engineering applications whether
it will be a . whether it will be distillation
00:01:27.780 --> 00:01:32.310
column or it is absorption column whether
it is absorption column because once you know
00:01:32.310 --> 00:01:35.090
the pressure drop or or flow through a pipe
00:01:35.090 --> 00:01:43.170
so you . whenever you you know that pressure
drop then you will be able to know the rating
00:01:43.170 --> 00:01:50.320
of the pump by which the fluid has to be transported
so that the design of the pump or the . in
00:01:50.320 --> 00:01:56.420
case of the in case of the liquid or design
of the compressor in case of gas flow can
00:01:56.420 --> 00:02:02.380
be can be calculated easily if you know
the pressure drop . in across the equipment
00:02:02.380 --> 00:02:07.329
so the selection of the equipment now driving
agent for example pump and compressor is very
00:02:07.329 --> 00:02:12.640
very essential in you to know the knowledge
to to know the knowledge of the pressure drop
00:02:12.640 --> 00:02:17.560
in the process stream so that is why the pressure
drop calculations are always very important
00:02:17.560 --> 00:02:23.360
in any chemical engineering applications and
so in our application as well in the membrane
00:02:23.360 --> 00:02:28.220
module so now we will be doing step by step
we will be adding step by step complication
00:02:28.220 --> 00:02:33.000
to the whole theoretical modeling as we have
done earlier next analysis we will be doing
00:02:33.000 --> 00:02:37.770
that the permeate flux is not constant it
will be proportional to the trans membrane
00:02:37.770 --> 00:02:40.310
pressure drop neglecting the osmotic pressure
00:02:40.310 --> 00:02:44.940
so you will be in the next analysis we will
be doing the i have negligible osmotic pressure
00:02:44.940 --> 00:02:49.790
case where permeate flux is proportional to
the trans membrane pressure drop in the third
00:02:49.790 --> 00:02:54.560
analysis and final analysis we will be considering
the axis number three where osmotic pressure
00:02:54.560 --> 00:03:00.970
is not at all negligible and we will be doing
doing a real and . modeling of the membrane
00:03:00.970 --> 00:03:10.530
module . so in these case . case two will
be considering the permeate flux is proportional
00:03:10.530 --> 00:03:23.729
to trans membrane pressure drop we we have
negligible osmotic pressure of the solution
00:03:23.729 --> 00:03:35.840
so first we will talking about the spiral
wound module .
00:03:35.840 --> 00:03:40.860
so as we have done earlier that the governing
equation of trans membrane pressure drop will
00:03:40.860 --> 00:03:50.900
be remaining same that we will be assuming
that the velocity profile . remain is undisturbed
00:03:50.900 --> 00:04:08.350
.
by . small permeation . in the wall . wall
00:04:08.350 --> 00:04:13.540
so the the pressure drop profile the governing
equation for trans membrane pressure drop
00:04:13.540 --> 00:04:22.440
or axial pressure drop . for delta p remains
same and this becomes we have seen in the
00:04:22.440 --> 00:04:31.440
last class that d square del p d x square
is equal to three mu h cube times v w but
00:04:31.440 --> 00:04:38.250
in this case v w will be nothing but l p del
p ok where l p is the membrane permeability
00:04:38.250 --> 00:04:47.090
and the boundary conditions that we have already
one was the pressure drop in the . at the
00:04:47.090 --> 00:04:56.540
inlet at x is equal to zero these delta p
is known to us and also the flow rate is known
00:04:56.540 --> 00:05:02.140
to us and the dilute condition and that will
be . at equal to zero
00:05:02.140 --> 00:05:18.070
d delta p d x is equal to minus three mu divided
by two h cube w q in so since both the boundary
00:05:18.070 --> 00:05:22.430
conditions specify on the same boundary and
located at z equal to zero this is a typical
00:05:22.430 --> 00:05:28.110
quasi boundary condition and we have already
seen earlier that equalization of these
00:05:28.110 --> 00:05:34.680
two boundary conditions in in the previous
class so now this is a simple first
00:05:34.680 --> 00:05:39.550
standard in second ordinary differential equations
two boundary conditions are specified so this
00:05:39.550 --> 00:05:44.290
problem can be solved quite easily and i am
just writing the solution directly . a couple
00:05:44.290 --> 00:05:51.500
of steps in between so kindly do the derivation
by yourself so delta p x divided by delta
00:05:51.500 --> 00:06:09.900
p in is equal to cos hyperbolic lambda x minus
three by two . nu q i h cube this q in w lambda
00:06:09.900 --> 00:06:26.360
delta p in sin hyperbole lambda x where w
is nothing but the channel width ok and the
00:06:26.360 --> 00:06:35.270
parameter lambda is nothing but root over
three mu l p divide by h square
00:06:35.270 --> 00:06:43.030
so so this is the expression of pressure drop
at any distance x in the module or in
00:06:43.030 --> 00:06:53.120
the channel now the pressure axial pressure
drop .
00:06:53.120 --> 00:07:17.741
over the full length module length . l becomes
delta p i minus so it becomes delta p in minus
00:07:17.741 --> 00:07:33.590
delta p x is equal to delta p i one minus
cos hyperbolic lambda x plus three by two
00:07:33.590 --> 00:07:44.280
mu q in divide by h cube w lambda sin hyperbolic
lambda x and for point x is equal to l that
00:07:44.280 --> 00:07:53.580
will give you the axial pressure drop in the
across the channel so delta p in minus
00:07:53.580 --> 00:08:08.090
delta p l is equal to delta p in p in . one
minus cos hyperbolic lambda l plus three by
00:08:08.090 --> 00:08:21.169
two mu q in h cube w lambda sin hyperbolic
lambda l now one can get a an expression of
00:08:21.169 --> 00:08:39.769
fractional recovery of feed a fractional recovery
of feed . over the entire model becomes f
00:08:39.769 --> 00:08:45.050
is equal to the total flow rate in the permeate
stream divide by total flow rate going into
00:08:45.050 --> 00:08:57.040
the system q in so this will be nothing but
two w v w d x over the entire length l divide
00:08:57.040 --> 00:09:09.310
by q in so this will be nothing but two w
. l p zero two l delta p x d x and divided
00:09:09.310 --> 00:09:15.310
by q in and we know the profile of delta p
x so this can be inserted here and this can
00:09:15.310 --> 00:09:25.010
be integrated it out and the final expression
becomes two w l p delta p in divided by . lambda
00:09:25.010 --> 00:09:42.580
q in sin hyperbolic lambda l minus three mu
q in divide by two h cube w lambda delta p
00:09:42.580 --> 00:09:48.810
in cos hyperbolic lambda l minus one
00:09:48.810 --> 00:09:57.290
so we will be getting an expression of fractional
feed recovery so we can also know we can calculate
00:09:57.290 --> 00:10:07.080
the profile of permeate flux because . v
w is equal to nothing but l p del p as a function
00:10:07.080 --> 00:10:13.920
of x so we can we can put the expression of
delta p x and can get the permeate flux as
00:10:13.920 --> 00:10:22.450
a function of x if you really do that the
expression of permeate flux becomes . l p
00:10:22.450 --> 00:10:37.070
delta p in . cos hyperbolic lambda x minus
three mu q in divide by two h cube w lambda
00:10:37.070 --> 00:10:48.740
del p in sin hyperbolic lambda x now that
profile of v can be obtained the profile of
00:10:48.740 --> 00:10:54.230
cross the retended velocity in the channel
that can also be obtained from this expression
00:10:54.230 --> 00:11:00.530
so if you remember that the governing equation
of q from the elemental mass balance that
00:11:00.530 --> 00:11:06.900
we have . already done in the last case in
the previous class so elemental mass
00:11:06.900 --> 00:11:14.210
balance is so basically mass balance over
an element in the in the channel will give
00:11:14.210 --> 00:11:25.740
you minus d q d x is equal to two w v w and
in this case we have to put q is equal to
00:11:25.740 --> 00:11:40.890
two x h times u where u is the velocity u
is the axial velocity in the channel .
00:11:40.890 --> 00:11:45.430
in channel h is the height and two x is
the length
00:11:45.430 --> 00:11:53.899
so that will be the cross sectional area
so you are a going to get d q d x ok so
00:11:53.899 --> 00:12:00.529
now will be getting and v w in this case will
be l p del p so d u you just substitute
00:12:00.529 --> 00:12:09.820
it over here . so we will be getting d u d
x is equal to minus v w by h and you know
00:12:09.820 --> 00:12:17.029
the expression of v w v w is nothing but minus
l p del p divided by h but del p will be a
00:12:17.029 --> 00:12:22.600
function of x and we have already calculated
how profile of pressure drop varies along
00:12:22.600 --> 00:12:27.339
the x so we will be substituting that over
here and then we can we can integrate at integrate
00:12:27.339 --> 00:12:37.570
over u so if you really do that so it will
be nothing but minus l p by h del p in
00:12:37.570 --> 00:12:44.160
by h and then we integrate over the you put
the expression of delta p and integrate over
00:12:44.160 --> 00:12:52.290
zero to x and there will be from u in to u
at in x location will be giving you these
00:12:52.290 --> 00:13:06.310
expression three mu q in divided by two h
cube w lambda . delta p in sin hyperbolic
00:13:06.310 --> 00:13:16.790
lambda x minus cos hyperbolic lambda x d x
and if we can if you if you really carry out
00:13:16.790 --> 00:13:23.480
the integration the final expression will
be nothing but u x divided by u in will be
00:13:23.480 --> 00:13:39.070
equal to one minus l p delta p in divided
by h lambda u in sin hyperbolic lambda x minus
00:13:39.070 --> 00:13:53.450
three mu q in divided by two h cube w lambda
delta p in cos h hyperbolic lambda x minus
00:13:53.450 --> 00:13:54.740
one
00:13:54.740 --> 00:13:58.910
so this will be complicated equation and the
. if you if you really evaluate this this
00:13:58.910 --> 00:14:04.519
will be a positive quantity so . one
minus some positive terms so it will be decreasing
00:14:04.519 --> 00:14:09.560
so as you go along the length the velocity
of the in the feed channel will decrease
00:14:09.560 --> 00:14:14.029
why the velocity will there in the feed channel
will decrease simply because we are extracting
00:14:14.029 --> 00:14:19.050
some amount of material or the in the in the
boundary of both the walls so its velocity
00:14:19.050 --> 00:14:24.490
has to go down and this is the expression
of velocity in the flow channel as a function
00:14:24.490 --> 00:14:31.269
of . module length and then we can really
calculate the concentration in the channel
00:14:31.269 --> 00:14:49.190
as well so assuming . a completely retentive
membrane . retentive membrane . that means
00:14:49.190 --> 00:15:04.080
permeate concentration is zero so . will be
having d of d d x of u . c times two w h will
00:15:04.080 --> 00:15:11.910
be equal to zero so d is basically the solute
mass balance . solute mass balance over the
00:15:11.910 --> 00:15:13.649
differential length of the channel
00:15:13.649 --> 00:15:22.760
so therefore u at inlet c at inlet should
be is equal to u at any x location c at any
00:15:22.760 --> 00:15:31.470
x location so therefore one can get an expression
of c at x c in is equal to u in divide by
00:15:31.470 --> 00:15:40.380
u x and it will be inverse of that so will
be getting the profile of that one minus l
00:15:40.380 --> 00:15:57.450
p del p in divided by h lambda u i sin hyperbolic
lambda x minus three nu q in divided by two
00:15:57.450 --> 00:16:12.260
h cube w . lambda delta p in cos hyperbolic
lambda x minus one so therefore the concentration
00:16:12.260 --> 00:16:15.940
in the feed channel will be increasing as
a function of x because the denominator will
00:16:15.940 --> 00:16:20.680
be decreasing as the as the length of the
channel so inverse of that so therefore the
00:16:20.680 --> 00:16:24.690
concentration will be increasing in the in
the in the feed channel and at the outlet
00:16:24.690 --> 00:16:34.329
if the concentration can be obtained by putting
by putting x is equal to . l we can get concentration
00:16:34.329 --> 00:16:41.380
at the outlet of the module and we have already
got the expression of . of pressure drop
00:16:41.380 --> 00:16:45.990
and from that we can get the rating of the
pump through whatever is required what
00:16:45.990 --> 00:16:51.399
what is or kilo whatever pump is required
to pump the fluid to overcome this pressure
00:16:51.399 --> 00:16:52.860
drop in the module
00:16:52.860 --> 00:16:58.510
so that gives the rough idea about the design
of the module whenever we are talking about
00:16:58.510 --> 00:17:03.209
a rectangular channel where in the permeate
flux . is proportional to the trans membrane
00:17:03.209 --> 00:17:08.370
pressure drop so next will be carrying
out an analysis for the we will be pulling
00:17:08.370 --> 00:17:13.170
out the results in the tubular module we will
doing the exactly the same thing and the finally
00:17:13.170 --> 00:17:16.850
we will be i will be writing that the governing
equations and the final solution of the tubular
00:17:16.850 --> 00:17:23.619
module so you just do the exact the derivations
following the all the in the steps exactly
00:17:23.619 --> 00:17:28.659
the same way in the rectangular channel it
will be getting the results in the tubular
00:17:28.659 --> 00:17:41.700
module so in case of tubular module the velocity
is given as r square the the across section
00:17:41.700 --> 00:17:49.080
average velocity is given as u is equal to
r square by eight mu minus d delta p d x and
00:17:49.080 --> 00:17:55.480
we will be giving the govern[ing]- getting
the governing equation of d delta p d x is
00:17:55.480 --> 00:18:05.659
equal to eight . mu by r square times u . and
if you do a differential mass balance um
00:18:05.659 --> 00:18:24.940
mass balance . over a differential element
. element will give us d u d x is equal to
00:18:24.940 --> 00:18:35.960
minus two v w by r you can put minus two l
p del p divided by r and similarly one will
00:18:35.960 --> 00:18:43.239
be getting an expression of you just you can
differentiate it once again and get d u d
00:18:43.239 --> 00:18:44.239
x
00:18:44.239 --> 00:18:47.799
substitute it over here combining these two
equations one will be getting the governing
00:18:47.799 --> 00:18:57.009
equation of d square delta p d x square is
equal to sixteen mu l p over . r cube times
00:18:57.009 --> 00:19:01.750
delta p and the solution of this equation
will be straight forward . and the solution
00:19:01.750 --> 00:19:16.109
will be delta p i cos hyperbolic m x minus
beta sin hyperbolic m x and m becomes a parameter
00:19:16.109 --> 00:19:22.850
which is nothing but sixteen mu l p over r
cube r is the radius of the tube and beta
00:19:22.850 --> 00:19:31.419
is the parameter eight mu u in divided by
m r square so one can calculate the axial
00:19:31.419 --> 00:19:37.369
pressure drop as a function of channel length
a tube length the axial pressure drop becomes
00:19:37.369 --> 00:19:50.830
. at any point any any location x delta p
in minus delta p x is equal to delta p in
00:19:50.830 --> 00:20:02.659
one minus cos hyperbolic mx plus beta sin
hyperbolic mx . and one can get the axial
00:20:02.659 --> 00:20:17.940
pressure drop across the module . module can
be obtained by putting . x is equal to l in
00:20:17.940 --> 00:20:20.309
the above expression ok
00:20:20.309 --> 00:20:26.950
and similarly the other parameters can be
calculated u over u i u in is equal to one
00:20:26.950 --> 00:20:42.470
minus two l p divided by m into r . delta
p in sin hyperbolic m x plus beta one minus
00:20:42.470 --> 00:20:53.159
cos hyperbolic m x and you can get an expression
of feed concentration retended concentrations
00:20:53.159 --> 00:21:00.960
is a function of c in is equal to u in by
u x it will be just you know reverse . of
00:21:00.960 --> 00:21:07.740
this expression so this will be u in and the
denominator will be putting up this expression
00:21:07.740 --> 00:21:13.279
over here and one can get so the axial pressure
drop where one will be getting the the idea
00:21:13.279 --> 00:21:20.450
is that the axial pressure drop will be vary
ing as a function of x and whenever will be
00:21:20.450 --> 00:21:25.730
calculating the permeate flux v w l p delta
p minus delta pi so delta p is no longer a
00:21:25.730 --> 00:21:31.169
constant at every x location the delta p will
be varying as a function of x and that will
00:21:31.169 --> 00:21:35.360
be taken care of and one will be taking the
axial pressure drop across the module the
00:21:35.360 --> 00:21:39.660
permeate flux profile across the module the
velocity profile across the module the concentration
00:21:39.660 --> 00:21:41.559
profile in the retended stream
00:21:41.559 --> 00:21:47.750
so once that is done now let us look into
a more and the most realistic case of the
00:21:47.750 --> 00:21:53.840
module when we are not neglecting the osmotic
pressure difference so we will be looking
00:21:53.840 --> 00:22:02.690
for case number three now while we are talking
about a newtoninan fluid newtonian . steady
00:22:02.690 --> 00:22:23.979
state . laminar and spiral wound module ok
so the governing equation of trans membrane
00:22:23.979 --> 00:22:35.470
pressure drop remains the same so it becomes
d delta p d x is equal to minus three mu q
00:22:35.470 --> 00:22:48.090
divided by two h cube w and by putting q is
equal to two w h u this can be expressed in
00:22:48.090 --> 00:22:58.639
terms of velocity so d delta p dx becomes
now minus three mu u divided by h square ok
00:22:58.639 --> 00:23:05.999
the permeate flux u w is now . no longer proportional
to delta p now you have l del p minus del
00:23:05.999 --> 00:23:19.120
pi and what is in terms of when you
express express phi is equal to b one c plus
00:23:19.120 --> 00:23:27.979
b two c square plus b three c cube and we
we we replace so delta pi is nothing but pi
00:23:27.979 --> 00:23:34.609
m minus pi p so pi evaluated and membrane
surface concentrations c m and phi p is phi
00:23:34.609 --> 00:23:37.659
evaluated and membrane at permeate concentration
c p
00:23:37.659 --> 00:23:43.739
so then we will replacing r r in we will
be replacing c p in terms of r r and c m so
00:23:43.739 --> 00:23:48.639
by defining r r is equal to one minus c p
by c m and we will be getting an expression
00:23:48.639 --> 00:24:00.199
of v w in terms of c m only l p del p minus
b one c m r r plus minus b two c m square
00:24:00.199 --> 00:24:16.960
. one minus one minus r r square minus b three
c m cube one minus one minus r r cube so this
00:24:16.960 --> 00:24:22.549
is the expression of permeate flux now if
there and now we will doing a solute mass
00:24:22.549 --> 00:24:28.880
balance in the differential area if we do
that will be will be will be and the
00:24:28.880 --> 00:24:33.320
and if and if you doing overall material balance
then we will be doing a solute material . solute
00:24:33.320 --> 00:24:43.779
balance so doing an overall material balance
in the channel in the elemental area . exactly
00:24:43.779 --> 00:24:51.999
like the previous case we will be getting
d u d x is equal to minus v w by h so this
00:24:51.999 --> 00:24:58.269
will be the governing equation of u in the
channel and v w is nothing but l p led p minus
00:24:58.269 --> 00:25:14.149
led pi so what is the then you do a solute
. mass balance . in elemental area in this
00:25:14.149 --> 00:25:23.820
case these becomes d u c d x becomes minus
v w c p divided by h h is the upper w channel
00:25:23.820 --> 00:25:30.049
c p is the concentrated . permeate concentration
and you just open it up the left hand side
00:25:30.049 --> 00:25:40.279
it becomes u d c d x plus c d u d x is equal
to minus v w c p over h
00:25:40.279 --> 00:25:45.309
then we can replace d u d x from this equation
where and ultimately you will be getting the
00:25:45.309 --> 00:25:50.619
governing equation of d c d c u governing
equation of concentration so these becomes
00:25:50.619 --> 00:25:59.590
v w by h c minus c p so this will be by
the . governing equation of they will be
00:25:59.590 --> 00:26:04.700
governing equation of trans memb[rane] . trans
membrane pressure drop or axial pressure drop
00:26:04.700 --> 00:26:09.590
this will be the governing equation for permeate
flux these will be the governing equation
00:26:09.590 --> 00:26:14.899
for velocity these will be the governing equation
for concentration in the feed channel so
00:26:14.899 --> 00:26:20.599
i will be having three ordinary differential
equations for the accounting trans membrane
00:26:20.599 --> 00:26:25.820
pressure drop velocity cos . velocity in the
retended channel as well as the concentration
00:26:25.820 --> 00:26:36.479
in the retended channel so three ordinary
differential equations for delta p u and c
00:26:36.479 --> 00:26:44.309
in channel or module and they will be having
the conditions we know the boundary
00:26:44.309 --> 00:26:50.049
conditions on delta p at x is equal to zero
so at x equal to zero we have two boundary
00:26:50.049 --> 00:26:55.919
conditions on delta p delta p is equal to
delta p in and another one . what that um
00:26:55.919 --> 00:27:01.450
related that q in d delta p d x that will
be known to us at x equal to zero u is . equal
00:27:01.450 --> 00:27:05.979
to u in and at x equal to zero c is equal
to c in
00:27:05.979 --> 00:27:11.559
so these will be coupled but c n is still
not known so for that what we are going to
00:27:11.559 --> 00:27:18.690
do next is we do a we relate the bulk concentration
with the membrane surface concentration so
00:27:18.690 --> 00:27:25.990
we have to relate c m to c so this is bulk
concentration this is the membrane concen[tration]-
00:27:25.990 --> 00:27:35.729
membrane surface concentration . and c is
the bulk concentration how these are related
00:27:35.729 --> 00:27:43.879
these are related through the mass transfer
coefficient so for every x in the module for
00:27:43.879 --> 00:27:57.429
every x location . in the module are we have
this relation holds good k c m minus c is
00:27:57.429 --> 00:28:03.379
equal to minus d del c del y at y equal to
zero . ok so this is the definition of mass
00:28:03.379 --> 00:28:09.669
transfer coefficient that is that is valid
at every location and at steady state we have
00:28:09.669 --> 00:28:26.969
seen earlier also sum of all fluxes towards
. membrane equal to zero so if we do that
00:28:26.969 --> 00:28:38.969
this becomes v w c m minus c p is equal to
minus d del c del y at y equal to zero therefore
00:28:38.969 --> 00:28:48.349
we can equate this two and we will be getting
k c m minus c is equal to v w c m minus c
00:28:48.349 --> 00:28:54.260
p this can be replaced v w c m minus c p can
be replaced in terms of real retention this
00:28:54.260 --> 00:28:59.249
becomes v w c m r r ok
00:28:59.249 --> 00:29:05.330
now we will be getting these . algebraic equation
to be solved at every step so then we will
00:29:05.330 --> 00:29:09.279
be what we will be doing we will be putting
the expression of permeate flux here as l
00:29:09.279 --> 00:29:14.109
p to delta p minus delta pi and we will be
expressing delta pi in terms of c n and lets
00:29:14.109 --> 00:29:28.220
see what we get so by doing that one will
get k c m minus c v c m r r and v w will
00:29:28.220 --> 00:29:41.279
be l p del p minus delta pi . k c m minus
c is equal to c m r r l p and . this will
00:29:41.279 --> 00:29:57.429
be delta p minus a one c m plus a two c m
square plus a three c m cube or k c m minus
00:29:57.429 --> 00:30:13.279
c divided by c m r r . l p is equal to delta
p minus a one c m plus a two c m square plus
00:30:13.279 --> 00:30:22.739
a three c m cube and the . the constance a
one or related to original osmotic coefficient
00:30:22.739 --> 00:30:31.710
and real intention so a two is equal to b
two one minus one minus r r square of that
00:30:31.710 --> 00:30:44.109
then a three is equal to b three one minus
one minus r r cube of that so now we have
00:30:44.109 --> 00:30:48.440
to find out what is the expression of these
algebraic . expression has to be solved also
00:30:48.440 --> 00:30:57.859
k is the function of x as we have seen earlier
that k of x is equal to one over i u d square
00:30:57.859 --> 00:31:03.830
by h x to the power one upon three if you
. remember the mass transfer coefficient analysis
00:31:03.830 --> 00:31:08.529
for the flow through a rectangular channel
so what i will be doing now i will be stopping
00:31:08.529 --> 00:31:13.359
here in this class in the next class i will
be looking of mass transfer coefficient of
00:31:13.359 --> 00:31:18.440
function of x and writing the complete solution
complete expression of the algebraic equation
00:31:18.440 --> 00:31:24.129
that will be that has been solved at every
step of three governing equations of delta
00:31:24.129 --> 00:31:31.379
p u and c and then we will be seeing how the
um module design can be done in a very complicated
00:31:31.379 --> 00:31:32.889
and realistic case
00:31:32.889 --> 00:31:33.429
thank you very much .