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So, now let us make material balance over time on del t.
So, now we have W dot, which is moles per second, at what rate naphthalene evaporates
over del t, and this we equate with rate of change in the mass. So, the volume is 4 by
3 pi r cube. So, this would be the volume multiplied by the density divide by the molecule.
So, essentially we are saying that over this del t time, this naphthalene ball evaporates,
and the size decreases by del r. So, we have del r and over this, over this time, the mass
decreases or the radius decreases from r to r minus del r. So, which means if we take
a limit of del t tends to 0, we can write del W dot as d by dt of 4 by 3 pi r cube cube
rho over M, which can be written as 4 by 3 pi r square rho over m dr by dt.
So, we have rate of change of radius of this naphthalene ball equated with respect to with
rate of change of this W dot, at what rate the naphthalene diffuses over the length r
to infinity. And already we have obtained this expression for W dot diffusion flux moles
per second or kg per seconds in our previous lecture, previous slide here. So, if we just
substitute W dot, we have ln P over P minus P 0 4 pi P, P is a total pressure, D AB binary
diffusivity - diffusion coefficient of naphthalene in air - over R T equals 4 by 3 pi r square
rho over M dr by dt. So, we can simplify r with this 4, 4 pi, pi will get cancelled with
this, and now we can integrate. So, entire expressions which is x actually W dot, we
are assuming that it remains constant over the integral.
So, dr by dt if we integrate from r 1 to r 2, we will have total time for decrease in
the radius from r 1 to r 2 as t ln P over P minus P 0 P D AB M over R T rho equals r
1 is square minus r 2 is square by 2. So, this is on integrations. Now, all we have
to do is, substitute the correct values of r 1 and r 2 and find out the total time, which
is the question asked. So, the volume V is 4 by 3 pi r 1 cube equal
to 0; so, r 1 is 3V over 4 pi 1 over 3, and the volume we can write in terms of 4 by rho
into 1 over 4 pi; so, rho is a density here, and if we recall this t equal to 0, to start
with the weight is 4 kg, of the naphthalene’s weight.
So, we have r 1 here, which if we substitute, we should get 9. 426 10 to power minus 3 meter;
similarly at r 2, when the mass becomes 3 kg here, if we just do normal substitutions,
we can get r 2 as 5 . 94 10 to power minus 3 meter. And then, that if we substitute in
our expression, which we have obtain on integrations we have, t ln, t is a total pressure 1.013
over 1.013 minus the vapour pressure or sublimation pressure is 0.8054 over 760 equals 1.013 10
to power 5 total pressure diffusion coefficient is given 6.9210 power minus 6, molecular weight
is 128, r is 8314, temperature 38 Kelvin, density is 1140; this equals r 1 square minus
r 2 square and r 1 is 9.426 10 to power minus 3 square minus 5.9410 to power minus 3 square
over 2. And then, if we calculate t will equal approximately
8000 seconds, which is nothing but 222 hours. So, it takes a long time for the mass of naphthalene
ball to get down from 4 kg to 3 kg. So, the example, which we have chosen here is nothing
but diffusion in a spherical coordinates, and you should be able to compare this example
to the previous example which we had, where we had this Cartesian coordinate system.
So, two examples are similar: one is in Cartesian coordinate, and one is in spherical coordinates;
the difference is that under study state, in case of coordinates - Cartesian coordinates
kg per second or kg per second per meter square of diffusion plane remain constant. In case
of spherical coordinate, this only the kg per second, which is conserved, kg per second
per meter square or moles per second per meter square, it changes with the radius.
So, these are the three different examples we have taken, for very basic examples to
calculate diffusion flux any over a diffusion length del Z given operating conditions like
pressure, temperature and the concentration gradient.
Now, what we do next? We take up a new topic - Mass Transfer Coefficients; we may be familiar
with this Mass transfer coefficients; now, before we get into this, we must ask this
question - what is this coefficient? Is it a fundamental variable? Is it a physical variable
defined here or it something engineering parameters? The answer is that mass transfer coefficient
is not a fundamental variable, fundamental quantity, it is a engineering parameters;
it has been introduce or it has been define to make our computation simpler. Now, if we
can go back to all those three examples, we said that we had one example, where we had
counter diffusion or where we had a stagnant, you know, medium B etcetera; we could calculate,
how much the diffusion flux over this diffusion length set, the problem comes when the flow
becomes complex. So, it is a hydrodynamics; we said in the
beginning class, you know, you should have some knowledge of hydrodynamics. So, it is
a hydrodynamics, the fluid flow and if the fluid flow is complex, it may be difficult,
by difficult to solve for the velocity profile, and then predict is diffusion flux; in those
cases and mass transfer coefficients become an engineering parameters, which is we have
to estimate from some given operating conditions then, we can we can apply this to calculate
this diffusion flux. So, let us take few examples, where if we
are trying to say that, we should make use of mass transfer coefficients or we may not
be able to make use of that general expression, which we obtain for diffusion flux, in terms
of diffusion length etcetera. So for example, so the question we are trying
to answer here is that, why there is a necessity for defined for defining this mass transfer
coefficient? or what is this? Motivation; so, if we recall, if we have a tube, and if
the Reynolds number is less than 2100, we have well defined velocity profile, which
is parabolics; V equal to we can write 2 V bar, V bar is velocity 1 minus r square over
r square. So, this would be the radius of; now, if we
have the velocity profile like this, in a tubular, and if we are asked to calculate
diffusion flux N A, it is not be a difficult thing, in the sense that we can substitute
this velocity profile in our Navier Stokes equations, one can solve for the concentration
profile and then, we can calculate diffusion flux etcetera. We know the bulk transport,
it would be V into C. So, once we know the diffusion flux, once we know the bulk transport,
one can predict the diffusion flux. So, the problem is not here; this is a straightforward
case, where we have a tubular flow and we have well defined velocity profile, things
cloud be more complicated, if the Reynolds number goes very high; say turbulent, if we
have turbulent, then we call from hydrodynamics, you will have a tubular flow and the velocity
profile will start becoming, will become flatter. So, here we have velocity profile, where in
the core, the velocity is almost really constant at some V velocity or V bar, but near the
wall, we have a gradient - a velocity gradient; now as it is, in the case of turbulence or
turbulent flow, and the Reynolds number is much larger or larger than 2100, there is
un predictions, there is a un certainty in predicting this velocity; as a consequence,
we cannot apply exactly the Navier stokes equations, and we can solve for this diffusion
flux; in those cases, we will like to say that yes, if we know the concentration gradient
or we know the concentration difference from C wall to say C bulk; one can one should be
able to write that diffusion flux equals some constant K into this concentration difference.
So, this is what we are trying to say the here that this K is nothing but mass transfer
coefficients. In principle if hydrodynamic is well defined,
if we have the velocity profile like in laminar flow or viscous flow, there is no necessity
for mass transfer coefficients or defining this mass transfer coefficients; it is very
important that we understand the motivation, mass transfer coefficient is an engineering
parameter. It is an engineering parameter, it is not a fundamental variable and this
type of parameter is required, when we have uncertainty in hydrodynamics, in the prediction
of velocity profiles, there this mass transfer coefficients become become quite.
We will take, for example, once again suppose, we have this sphere and this naphthalene ball,
it evaporates similar to the previous case, where we have this velocity profile like this;
so the fluid approaches from infinity to this V infinity, we have this radius of this ball,
which is R and we have been asked the same questions - what is the diffusion flux here?
N r; now, if the velocity profile is known to us; V r we can one can predict this diffusion
flux N A over a length say between r r 1 or between r 2.
If the velocity profile is well known defined, one can calculate this N A starting from navier
stokes equation, spherical coordinates etcetera; the problems comes, when we have very high
reynolds number; and also very large say, but say 100 Reynolds number based on the particle
size if it is very large say 100, then there is uncertainty in the velocity profile; and
if you recall from hydrodynamics, you will have cases of boundary layer separation etcetera
or the eddies, which is form at the of this rear end.
In which case, it is a to predict diffusion flux, in terms of your bulk transport VCA
and diffusion flux, it becomes quite, it involves lot of error; in the in these cases is possible
to define a mass transfer coefficient quantity and to say that, we have the flux now, which
we can write as diffusion fluxed N A as mass transfer coefficient into some concentration
difference, where del C is concentrations difference between C S at the surface and
this C infinity. So, we are trying to make a case here, what
is the motivation of mass transfer coefficients ? Why we have to define this quantity and
where we can get away without defining this mass transfer coefficients, we saw two examples,
where we have very straightforward case, where we have a tube and certain pool of the liquid
evaporates or we have a naphthalene ball, which evaporates; in those cases we said that,
let us assume that N B is 0, you know bulk transport is 0, and then we can calculate
this diffusion flux.
So, if suppose we have a packed bed, you know certain more complex situations, like we have
packed bed, in which different spherical particles or stack stack like this, and the fluid goes
like this. So, if we look at one sphere or one spherical
particles, there is a quite complex velocity profile surrounded around this, to predict
velocity profile exactly, one requires you know, things likes software engineering to
like CFD - Computational Fluid Dynamics - then maybe you to some extent, we can predict the
exact nature of this process; otherwise it is a very difficult to predict in the fluid
flow inside the pack. In such cases, again the concept of mass transfer
coefficients is quite important here. So, we will like to go back to again the general
expression, which we obtain for diffusion flux and see, how the same expression can
also be rearranged or can also be modified to introduce these engineering parameters,
K mass transfer coefficients. So, let us re-visit this expressions for,
now we assume a study state, no chemical reactions. So, we have N A equals N A over N A plus N
B we wrote D AB C over Z ln NA over N A plus N B minus x A2 we can write in terms of mole
fractions at plane 2 over N A N A plus N B minus x A1.
So, this Z is nothing but diffusion flux - diffusion length - and x A1 and x A2 is your mole fractions
across the plane 1 and 2; now, if you take the again the simpler case of case 1, where
we say that B is a stagnant, stagnant B, we can write N B equal to 0 with N B equal to
0, N A over N A plus B is 1.
And then, N A equals or simplified to D AB C Z ln 1 minus x A2 over 1 minus x A1.We can
also write like D AB C Z we can divide and multiply by x A1 minus x A2 over x A1 minus
x A2 ln 1 minus x A2 over 1 minus x A1 all we are trying to do the similar expressions
we have we have rearrange written like this, and then further we can write D AB C over
Z del X. So, now we are introducing this mole fraction
difference between plane 1 and 2, which is nothing but x A1 and x A2 and here what we
have, we have ln x B2 mole fraction of B at plane 2 1 minus x A2 and B is a stagnant here,
x A1 and x A1 minus x A x A 2 can also be written sorry this should be x B1.
So, we have x B1 and then, x A1 minus x A2 can also be written as x B2 minus x B1. So,
subtract 1 minus this, which we now we have an expression for mole flux, in terms of D
AB C Z into ln if you recall, if you try to recognize this term here, we have x B2 minus
x B1, a quantity like this, we call it x BM. So, logarithmic average which is nothing but
x B1 minus x B2 over ln x B1 x B2. We just like, we have an arithmetic average
or geometrical average or harmonic average, we have been able to obtain an expressions
for diffusion flux of A, when B is a stagnant in terms of known quantities here, and concentration
difference we have x BM. Now, here comes the concept of this mass transfer
coefficients, the entire quantity on the left in the parenthesis, we can call it as K x
into del X. So, what we have obtained here? We are trying to say that diffusion flux,
it is a proportional to del X and the proportional constant, which we got as K x or we are writing
as K x, we are calling it as a Mass Transfer Coefficient.
So, again let us go back, what we said in the beginning? What is the motivation of introducing
this mass transfer coefficients? What was the objective - we said that in those cases,
where the hydrodynamics are complex, it is difficult to solve the velocity profile or
there is uncertainty in solving the velocity profile of fluid flow; just like a flow in
a packed bed very complex situation there or we have high Reynolds number, when we have
a turbulent flow not viscous, not laminar then, there is a uncertainty in predicting
the velocity profile. In those cases, it is not possible to solve
analytically, an expression for diffusion flux which we have obtain so far, we took
two examples of case 1 and case 2. So, all we are trying to say here that, we are trying
to introduce the concept of mass transfer coefficient, is the weight is defined, diffusion
flux N A equals to quantity K X, which we are calling at a mass transfer coefficients
into del X; del X is nothing but your concentration difference, is your mole fraction across the
plane 1 and 2.
we can also write N A equals K X some different prime, we can give into del C, where C is
now the concentration difference C A1 minus C A2.
So, there are different ways of writing this diffusion flux; in case of the case 1, when
the when the other medium, other component B was stationaries, we can write N A like
K X or some proportional to concentration difference or mole fractions; let us take
case 2, when we said that we have equimolar diffusion. Let us see, if we can write in
a similar expression or we can develop similar expressions for diffusion flux. So, case two
we studied in the last class earlier classes, we had N A equals minus N A which is equimolar
counter diffusion. So, when we have N A equals N B, then N A
equals we can go back to the same expressions, which we have started earlier; we can substitute
N A equal to minus N B to write N A D AB Z C x A1 minus x A2, which is even simpler than
the previous case, and again we have the same situations we are obtaining expressions for
N A, we are seeing that is proportional to mole fractions across the plane 1 and 2, and
here the proportional constant K equals D AB C over Z.
So two cases which we studied here, very simple very fundamental; they form the basis for
all our calculations or all the examples, which we had in the previous classes.
Now, we are obtained to say that we have same expressions, we have arrange in a different
way, and we are getting the similar expressions that diffusion flux is a proportional to concentration
across plane 1 and 2 or mole fractions across 1 and 2 or if we have gas instead of liquid,
we can say that diffusion flux is a proportional to difference between the partial pressure
at plane 1 and plane 2. In these cases, of course, there is no necessity,
the example which we took to define mass transfer coefficients, but as we said earlier that
if we have very complex hydrodynamic situations where the velocity profile cannot be solved
with accuracies mass transfer coefficient becomes an engineering tool, we can define
N A diffusion flux equals a quantity K M mass transfer coefficient multiplied by concentration
difference - the potential difference between the mole fraction, difference between the
concentration difference between the partial pressure of 1 and 2, and all complexities
of hydrodynamics will be now embedded in mass transfer coefficients.
So that means, we can conclude that flux moles per second per meter square can be written
as K M into del C or can be written as del X or it can be written as K into del P; of
course, the values of K will be different, different in all cases.
Generally we write K x, we write k g, if it is a gas here and instead of if we using concentrations,
we can write as k c, all of this is a mass transfer coefficients k m mass transfer coefficients.
So all it means, if we have gas system, we can also write N A as k y into y 1 minus y
2; here y 1, y 2 are the mole fractions or we can write as k g P 1 minus P 2 where P
1 and P 2, these are the partial pressure of 1 at the plane 1 and 2.
Let us write down here, the In principle, in principle we do not require K for mass
transfer coefficient k m in laminar flow, why because laminar flow is well defined is
a flow, we have seen in case of tube, we have analytical Navier Stokes solutions V equal
to V max 1 minus r square over r square. So those cases, where the velocity profiles
are well known; there is no need to to define K m to calculate a molar molar diffusion flux,
but we do not require K M in laminar flow, because molecular diffusion prevails prevails
in such flows. In fact, this should like to recall, the hydrodynamics
- the concept of friction factor and track coefficients, if you are asked to calculate
pressure drop in a tube, you can use fanning equations, equations which contains friction
factor, we can also use Eigen Poisson’s equations, if the Reynolds number is less
than 2100. But if the Reynolds number is larger, much
larger than 2100, then is use friction factor fanning equations, which contains friction
factor, then we rely on experimental data or Moody’s plot, which is obtain from some
experimental data to predict this to calculate or estimate this friction factor.
So, the idea is that use of fanning equations, we can use fanning equations very well in
case of laminar as well in case of turbulent; in case of laminar flow, if we use friction
factor is 16 over Reynolds number, which can be obtain if we are equate this fanning equation
with Eigen Poisson’s equations or we did Eigen Poisson’s equation, we had the velocity
profile, we said that the pressure drop equate with the shear stress multiplied by the surface
area etcetera. One can calculate and expression for pressure,
then you define friction factor in terms of pressure drop and the kinetic energy per unit
volume etcetera; show that friction factor 16 by reynolds number, but if the flow is
turbulent then, we have to rely on certain experimental data, because in turbulent there
is uncertainty uncertainty in predicting velocity profile.
Similarly, C D track coefficients, we said C D equals to D 4 by Reynolds number, if Reynolds
number is less than 1 or if we have a viscous form, but how do we get C D equal to 24 by
Reynolds number, we start from the velocity profile, we take a spherical geometries, solve
Navier Stokes equation, the spherical coordinates, we obtain the velocity profile we solved the
for the gradient then, we calculate shear stress and then, we obtain the force 6 by
mu v r which we call it stokes law. If we take this force and defined track coefficients
then one can show that C D equal to 24 by Reynolds number is nothing but a calculated,
its analytical expression for that coefficients, what happens? If the Reynolds number is very
large in that case C D 1 has to rely on certain plots or experimental data.
So, the motivation of defining mass transfer coefficient here is also the same, if the
flow is well described like in laminar flow in viscous flow, there is no need to define
mass transfer coefficient; we can use expressions for N A in terms of diffusion length Z and
we can do the calculations, but if there is a complexities like in a packed bed tower
or turbulent flow in a tube, otherwise or the flow past is simple sphere, but at high
Reynolds number. So, wherever there is a complex situations,
velocity profile is not accurately known; mass transfer coefficient quantity is defined
to predict this diffusion flux in terms of mass transfer coefficient and concentration
difference, but the partial pressure across that diffusion flame and diffusion planes.
How do we get mass transfer coefficients? One has to rely extensively on experimental
data and in those cases, we will see that we defined a quantity like Sherwood number,
which is k m l over diffusion coefficients this l is characteristics length which means
different geometry, different situations l will be different, d is diffusion coefficients,
which is meter square per second, l is meter and k m is meter per second mass transfer
coefficients. If we define as N A moles per second per meter
square equals k m into del C, concentration different moles per cubic meter. So, when
literature you will see that, there are several cases, which people have a studied and there
they have estimated mass transfer coefficient in term in terms of certain one-dimensionless
groups like Reynolds number, Schmidt number to from, which we obtain this mass transfer
coefficient game and then, we can predict diffusion coefficients.
So, you are required to visit a table in which is table 3.1, you will see that table given
in 3.3, where we have several such cases 1, 2 listed; we will just note down few of them
1, 1 is very common, if you have flow past is sphere.
So, if you have a flow past sphere or you have flow in in a tube or flow in a packed
column. So, there are several such examples given in this table, where Sherwood number
in all these cases, Sherwood number has been given in terms of mostly, in terms of Reynolds
number and Schmidt. So, S C is Schmidt number, which is nu over d. So, kinematic viscosity
nu beta square per second and we have this binary mass transfer coefficients beta square
per second dimensionless numbers. So, you are required to visit this table 3.3, and
see how we can defined this mass transfer coefficients ; the idea is that all these
correlations, which have been developed in terms of Reynolds number and Schmidt number.
They are extensively dependent on experimental data. So, when you take up a problem or in
example, you must do a literature survey you must go to visit such table like 3.3 or various
chemical engineering handbook to use an appropriate correlations, from which you can calculate
this mass transfer coefficients.
Then you can predict there is you know diffusion flux we just for an example, suppose we have
what is given you have singular sphere and we have a flow like this right. So, it is
a similar example that we have a sugar ball or salt or naphthalene ball and we are we
have been asked to predict this mass diffusion flux N.
So, if we have some fluid flow here, there is a Reynolds number based on the particle
size, which we can define asV infinity flow velocity at a very far a stream is v infinity,
d p diametric of the particle, rho of the fluid and mu a rho f and mu f these are the
physical properties of the fluid. We can define Reynolds number; similarly, we have Schmidt
number nu f over D AB binary diffusion coefficients of A in B. So, in these cases correlation
is given as Sherwood number equals 2 plus 0.347 Reynolds number, Schmidt number, it
is 0.0 .5, 0.62 more important to note here is that the author has given the range of
Reynolds number into Schmidt number to the power 0.5 this quantity. This expression holds
good with the values is between 1.8 to 600000, and Schmidt number between 0.6 to 3200.
Now, this is an empirical correlations empirical correlations all it means, this has been obtained
based on certain experimental conditions, where we meet these two restrictions and then,
once we know Sherwood number, we can calculate K m, once we have K m, we can predict diffusion
flux N A as K m into del C. The idea here is that Reynolds number and a Schmidt number
certain conditions is not possible to defined this velocity profile or to calculate the
velocity profile and to obtain the expressions for N A, if fluid is a stagnant you can make
out the this term will drop and then, we have Sherwood number equal to 2.
Ah Sherwood number equal to 2 when the fluid is stagnant, one can obtain analytically it
is an exercise does not it is a size, you can try Sherwood number to obtain Sherwood
number equal to 2, if the fluid is stagnant. So, we are trying to give the same message
again and again at definition for mass transfer coefficient, its purely arbitrarily it is
an engineering parameter it is not a fundamental variable it has been defined to make our calculations
simpler in those cases where there is a complexities of defining or solving this velocity profile.
So, similarly we can take one more example like, we have flow inside a circular pipe.
So, if the flow is inside this pipe, the Reynolds number is very large or say Reynolds number
is 4000 and greater than the velocity profile will not be parabolic, and we have some flat
velocity profile; in this case if you are ask to calculate, diffusion flux from the
wall let us say, we have certain species was diffuses from the wall or to the wall; in
this case N A can also be written as mass transfer coefficient into concentration difference
which means; we can write k m as C wall minus c.
So, we have C W wall and we have very far from the wall C. So, this is the advantage
of defining this mass transfer coefficients, complexities of this complexities of the velocity
profile, Reynolds number is embedded now in mass transfer coefficients. And how do we
get came? So, there is a correlations given in terms of we get Sherwood number, which
would be 0.0023 Reynolds number, the power 0.8 Schmidt number the power 1 by 3 and then
the author has given a range that this equation is valid between 4000 and 60000, look at the
range here is the turbulent flow, it is not like the previous case or the simpler case.
When we have the flow is laminar, and we have the velocity parabolic velocity profile like
V equals 2 V bar 1 minus r square over R square. So, in case of turbulent flow, where there
is uncertainty uncertainty predict in this velocity profile, you can come out with this
correlation author has come out with this correlation based on the experimental data
with the range with the restriction the Reynolds number is between 4000 and 60000 Schmidt number
is between 0.6 to 3000 So, once again we have given this example
that would why where there is a necessity of defining this mass transfer coefficients;
similarly, we can take one more example, which is more engineering flow in a packed bed.
So, if you recall in the first first two lectures, we talked of this packed columns, we have
the absorption column and we have sulphur dioxide which air and we want to remove by
absorptions. So, generally in these cases, you have a column pack with certain packings
to improve contact between the gas and the liquid.
So, if you have the packing like this, where we have certain packings, start now could
be spherical could be cylindrical etcetera. So, quite complex velocity profile, the liquid
goes like this and gas flows through this, cross this between the pores - micro pores
of cylinders and this air this gap here, the flow is quite complex, in which case again
the necessity of mass transfer coefficient becomes indispensible.
And this case, also we can write mass diffusion flux N A as K m into del C concentration difference
between the surface and bulk. So, infinity away from this packing etcetera. So, in this
case also Sherwood number, now in this case diffusion flux or sorry I am mass transfer
coefficient is estimated from factor called J D, which is related to 2.06 over epsilon
Reynolds number minus 5.75 equal to friction factor by 2, but this J D is also equals Sherwood
number over Reynolds number into Schmidt number 1 over 3.
Again, we have the similar story here to calculate mass transfer coefficients, we required Sherwood
number, which is K m characteristics length over diffusivity; in this case l would be
equal to the diameter of this particle of the packing etcetera or equivalent diameter
of this packing. So, from here, we can calculate the mass transfer coefficients and can predict
the diffusion flux; in this case also there is a restriction the Reynolds number must
be between 90 to 4000. So, in this Reynolds number is we can defined
with respect to the diameter of the particles and whether Reynolds the Reynolds number is
90 to 4000; all it means, must larger than 1 and we have a turbulent flowing.
So, the velocity profile is quite complex and they hence, there is a necessity for defining
mass transfer coefficient Schmidt number is given as 0 .6.
So, essentially, again we have taken three examples to show that mass transfer coefficient
is an engineering parameters, it has to be estimated, it has to be correlated from the
correlations available in the literature, if the situation which you are trying to model
or which you want to study for that situation, if the mass transfer coefficient is not given
and one has to obtain experimentally. In those situations where the Reynolds number
is very small flow is well defined viscous laminar, one can you know invoke this Navier
Stokes equation calculate velocity profile or the velocity superimpose on species balance
and can solved for the concentration gradients, once we know the concentration, we can solve
for diffusion flux.
So, what we do here now we take up three well known theories. So, we are writing here that
all this Sherwood number in generals can be written as a function of Reynolds number,
hence with in general of course, there could be more complexities here.
But the question is that in all these cases A a and b this this numbers have to be determined
experimentally, there has to be some experiment in this to a calculate this this different
coefficients here there are well theory, well known theories like film theory surface renewal
theory and boundary layer theories these three theories are quite popular to which tells
us that this mass transfer coefficient; in some cases will be proportional to diffusion
coefficient, in some cases in this will be a proportional to square root of diffusion
coefficient etcetera. So, these theories are also widely used to predict or to obtain such
correlations between Sherwood number, Reynolds number and Schmidt number.
So, we will take up these theories 1 by 1, we will not get into mathematical details,
but just to get an idea that how does this Sherwood number is dependent on this diffusion
coefficients. So, there are well three theories there are several theories the 1 the 3, which
are very common one is film theory the second is surface renewal or we will see there is
some difference between surface renewal theory and penetration theory.
And third is boundary layer theory, you may have done some extent boundary layer theory
in your previous class etcetera, on fluid mechanics or hydro dynamics, we will just
go through very briefly through this theories to see that, how does mass transfer coefficients
is depends on what is K m depends on diffusion coefficients.
So, the first very popular theory is film theory and this film theory is applicable,
in general if you assume that, we have a steady state concentration profile and we have we
have stagnant film. So, essentially what this theory says that, we have say two streams
of suppose, we have gas here and we have liquid across this we assume that the film here the
concentration profile, in this liquid phase is linear that is one it is a steady state
profile does not change with time. So, we have say C g here and we have concentration
gradient just at the interface and we can use C i. So, we have C i here and we have
this C bulk or concentration at if of very far distance we have this diffusion length
del Z. So, the fluid fluid is a stagnant film is a stagnant here, and we have a steady state
concentration profile; in this case, one can show that this mass the film theory predicts
that mass transfer coefficient is approximately equal to D AB over del.
Now, again we are not going into details, but theory suggets suggests that mass transfer
coefficient is proportional to diffusion coefficients and one over del. So, which means larger the
thickness film thickness smaller is the mass transfer coefficients
Similarly, we have penetration theory, this was originally proposed by higbie and this
penetration theory it says that steady state profile is not possible profile is not possible.
So, this is in contrary to what we have in case the diffusion flux.
So, let us see what try to understand what is the penetration theory says, which is a
different from the film theory here also we have suppose, we have this gas here and we
have the liquid here. So, as what we have in case of penetration theories, there is
something called eddies; if you recall from here hydro dynamics at high reynolds number,
we have this hypothetical pockets of liquid, which moves from the bulk to the surface.
This eddy is stays here for sometime, in which this gas molecule the diffuses with insulate.
So, there is certain contact time of this eddies at this surface. So, this is your liquid
gas surface at which, there is a contact time of theta when the eddies is stays, before
it gets back to the bulk. So, in other words is a very short contact
short contact between the two phases and we have a mass transfer from the gas to the liquid,
in which case, it is not possible for this steady state profile to establish or or profile
to reach a steady state, in that case penetration theory becomes quite popular and more realistic,
it has been shown that in such cases and a steady state species balance can be written
del C over del theta, if A is component, which is diffusing from gas to liquid phase equals
D AB del square C A over del Z square. So, z would be this direction, one can start
from here. So, t equal to 0, may be the concentration here was C A 0 and at certain time this interface
is exposed to C i then we can write this expression I can solve to obtain concentration profile
of C A in this liquid, and once we have concentration file in the liquid, we can calculate the concentration
gradient and one can calculate this diffusion flux.
In this case, it has been shown that diffusion flux N A will also be equal to k L C A i minus
C A 0 this what we will expect proportional to mass transfer coefficients and concentration
difference between the interface and the bulk concentrations except this K L, now equals
to square root of D AB over pi theta. So, this is a difference between the film
theory here now k L is proportional to the square root of D AB, unlike the film theory
is predicts mass transfer coefficients is linearly proportional to diffusion coefficients.
So, one we had film theory, when we assume that the profile is a steady state film is
a stagnant and it is linear concentration gradient is linear between the two planes,
unlike in case of penetration theory, which says that short there is a very short contact
between the gas and liquid; in other words there is no time for the concentration profile
to reach a steady state. In that case, we write another steady state balance, one can
solve for concentration gradient and can calculate the diffusion flux show that mass transfer
coefficient is a square root proportional to square root diffusion coefficients.
Similarly, we have one more theory, which is a boundary layer theories you may done
in case of hydrodynamics.
So, we will like to conclude with the third theory, which is also popular boundary layer
theory; in this case also we have what we have is a flat plate and it is approach approach
by velocity fluid with U 0 and C A 0 concentrations, the surface concentration is C Ai, you can
think of certain solutes is dissolved in this liquid or this fluid.
Then, there is a boundary layer developed by this del M, we can have similar to this
boundary momentum, boundary layer there is a concentration boundary layer del C, and
if you are ask to calculate the diffusion flux N A, we can also write here N A equals
mass transfer coefficients into C Ai minus C Ao.
And in this case, it has been shown this one can also calculate this mass transfer coefficients
in terms of Sherwood number etcetera which is k m into the length of the plate L over
diffusion coefficients D AB. So, these are the three theories, which are quite popular
to estimate mass transfer coefficients; of course, we have to rely on the extensive experimental
data to predict this mass transfer coefficients.