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Welcome to the second lecture of module 2
which is on mass transfer coefficients. So,
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before going to this lecture, let us have
recap on our previous lecture.
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In the previous lecture, we have discussed
the concept of mass transfer coefficients,
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where we have said that, the mass transfer
coefficient is important for the convective
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mass transfer; and these mass transfer coefficients
depends on the different systems - whether
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it is gas phase, or it is liquid phase, or
whether the diffusion is occurring through
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non diffusing b, or the equimolar counter
current diffusion. So, for each case, we have
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discussed the mass transfer coefficient and
the relations among them. And finally, we
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have calculated the mass transfer coefficient
for different systems and the typical values;
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typical values of mass transfer coefficients,
coefficients for gas phase is about, K c is
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about 10 to the power minus 2 meter per second,
and in case of liquid phase, K l is approximately
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10 to the power minus 5 meter per second.
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In this lecture, we will consider on two topics;
one is dimensionless groups and the correlations
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for convective mass transfer coefficients.
The dimensionless groups is generally important
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for the simplicity to represent the mass transfer
coefficient and other variables, or physiochemical
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properties of the system. So, like in heat
transfer, heat flux can be correlated with
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the heat transfer coefficient and the temperature
gradient. This heat transfer coefficient,
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which is h, can be related with the Nusselt
number Nu, which is Nusselt number. The other
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important dimensionless term in heat transfer
is the Reynolds number
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and Prandtl number. For experimentally obtained
data, under post convection in heat transfer,
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the Nusselt number can be related as a function
of Renolds number and Prandtl number. The
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very important and useful correlations in
case of heat transfer like this, is known
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as Dittus-Boelter equation.
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Similarly, like heat transfer, two important
dimensionless groups in case of mass transfer,
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is the Sherwood number, Sherwood number, and
the other one is Schmidt number. So, like
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in heat transfer, in case of Nusselt number,
we define convective heat flux divided by
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heat flux for conduction
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through a stagnant medium of thickness l for
same delta t, which is equal to h delta t
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divided by K by l into delta t, which is equal
to h l by K; K is the thermal conductivity.
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Similarly, for mass transfer, the Sherwood
number can be defined as convective mass,
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or molar flux, divided by mass, or molar flux
for molecular diffusion through a stagnant
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medium, medium
of thickness l, under same driving force.
So, in case of diffusion of, diffusion of
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a gas phase species through non diffusing
B, convective flux, we can write K G into
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delta P A; P A is the partial pressure difference.
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Now, the mass flux
due to molecular diffusion, is equal to, we
have learned in the last class, D AB P t divided
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by R T l P B L M delta P A. So, if we substitute
in the Sherwood number definition, we will
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get, this K G into delta P A, which is convective
flux, divided by D AB P t by R T l P B L M
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into delta P A, is equal to K G P B L M R
T l divided by D AB P t. It is the total pressure.
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We can write K C l P B L M divided by D AB
P t. If we consider, the transport occurs
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through a liquid phase, and at low concentration,
for liquid phase, at low concentration, x
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L B M approximately equal to 1 and convective
mass flux will be K L, concentration gradient,
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and the diffusive flux is equal to D AB by
l delta C A. So, the Sherwood number, in this
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case, is equal to K L delta C A divided by
D AB by l into delta C A, which is equal to
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K L, small l divided by D AB. The l is the
characteristic length; for sphere, this d
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is the diameter, is the characteristic length;
for cylinder, d dia is the characteristic
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length; for flat plate, distance from the
leading edge, say z, is the characteristic
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length.
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Let us consider another important dimensionless
group in case of mass transfer, which is Schmidt
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number. We know the dimensionless group in
heat transfer is Prandtl number Pr, which
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is defined as the momentum diffusivity, momentum
diffusivity
divided by thermal diffusivity, which we can
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write, mu by rho divided by K, thermal conductivity
by rho C P, which is mu C P by K. The analogous
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number in case of mass transfer is Schmidt
number, which is equal to the momentum diffusivity
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divided by the molecular diffusivity; so,
which is equal to mu by rho by D AB, which
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is equal to mu by rho D AB, which we can write,
nu by D AB. So, this dimensionless number,
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Sherwood number and Schmidt number, how the
magnitude of these two dimensionless numbers
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varies for different systems?
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Let us consider a sphere of 2 centimeter dia,
where the gas phase mass transfer is occurring,
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which is flowing first through this sphere;
and the partial pressure
of the solute is low; that is, P B L M, log
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means special gradient, by total pressure,
will be approximately 1. The Sherwood number
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in this case, will be equal to K C d P B L
M by D A B P B. As we know, the mass transfer
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coefficient is in the order of 10 to the power
of minus 2 meter per second and the diameter
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is given 2 centimeter, which is 2 into 10
to the power minus 2 meter, and the typical
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values for the diffusivity is 10 to the power
minus 5 meter square per second in case of
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gas, so, from which we can get the typical
Sherwood number is approximately 20.
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But Schmidt number, for this case, is equal
to nu by D A B and nu, which is mu by rho
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will be in the order of 10 to the power minus
5 meter square per second and this value the
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diffusivity 10 to the power minus 5 meter
square per second. From this we can obtain
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Schmidt number is approximately equal to 1.
For common gases, gases, these values, the
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Prandtl number is equal to Schmidt number
and this is equal to 1.0. For the liquid phase,
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and for similar geometry, Sherwood number
is equal to K L d by D A B, which is equal
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to 10 to the power minus 2, into 2 into 10
to the power minus 2 divided by 10 to the
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power minus 9 meter square per second, which
is the diffusion coefficient in case of the
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liquid phase. So, from here, the Sherwood
number is approximately 200. And Schmidt number,
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in this case, is nu by D A B, and in this
case, the nu is to 10 to the power minus 6
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meter square per second, divided by the diffusivity
value is 10 to the power minus 9 meter square
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per second, which leads to Schmidt number
is approximately equal to 1000. So, for common
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liquids, except liquid metals, the Prandtl
number is in the range of greater than 10,
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and less than 100. For the same case, the
Schmidt number is greater than 400 and less
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than 10 to the power 4.
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Now, let us consider another dimensionless
number, which is important in case of mass
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transfer. A similar number in heat transfer
is also exists. For heat transfer, the Stanton
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number S t, Stanton number. The Stanton number
for heat transfer, we define the convective
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heat flux
divided by heat flux
due to bulk flow, which is equal to h delta
T divided by C P rho v into delta T; which
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we can write, h l by K divided by v l rho
by mu into C P mu by K, which is equal to
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Nusselt number divided by Reynolds into Prandtl.
The analogous number for mass transfer is
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Stanton number for mass transfer, we can define
the convective mass transfer, mass flux, divided
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by flux due to bulk flow of the material;
so, which we can write K L delta C divided
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by v into delta C, which we can write K L
l by D A B divided by v l rho by mu into mu
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by rho D A B; which is, we can write in terms
of Sherwood number divided by Reynolds into
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Schmidt.
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The another important dimensionless group
in mass transfer, similar to heat transfer,
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is the Peclet number. This Peclet number,
in case of heat transfer, we can define, is
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the heat flux
due to bulk flow divided by flux due to conduction,
due to conduction across the thickness, thickness
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l; so, which is C P rho v delta T divided
by K by l into delta T, which we can divide
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into two groups, two dimensionless group,
v l rho by mu into C P mu by K, which is Reynolds
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number into Prandtl number. The analogous
number for mass transfer is Peclet number,
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is flux due to bulk flow of the medium
divided by diffusive flux across the thickness,
thickness l, which we can write velocity into
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delta C divided by D A B by l into delta C;
which we can write also in two different dimensionless
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term, v l rho by mu into mu by rho D A B,
which is equal to the Reynolds number into
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Schmidt number. So, these are the 4 important
dimensionless number, or important, in case
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of mass transfer, there are two more dimensionless
terms are available in case of mass transfer.
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Now, the correlations for convective mass
transfer coefficients; so, what are the objectives
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for this? It is to explain the concept and
the importance of dimensional analysis for
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the experimental data and to obtain a useful
correlation. And, how to use the Buckingham
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method to determine dimensionless groups involved
for a particular systems. Correlations are
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required for the mass transfer in turbulent
flow, where the mass transfer coefficient
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calculations is not easy from the theoretical
considerations.
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So, we will discuss the Buckingham method
to obtain the correlations. So, in this method,
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first, we have to consider the certain fundamental
dimensions. What are those? Length, which
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we represent by l. Length, is one fundamental
dimensions and which is symbolized as l; and
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like area, we can write l square; volume,
we can write l cube. Another dimension is
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time. It is symbolized as t and velocity,
we can represent as l length per time, and
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acceleration, we can write, length per time
square.
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Another fundamental dimensions is mass, which
is symbolized as m. Like density, if we consider
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density, then, it is mass per volume; this
is m by l cube. So, we have to have these
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fundamental dimensions initially, and then,
in the Buckingham method we have to identify
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the variables significant to a particular
problem. And then, we have to determine the
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number of dimensionless groups. And this number
of dimensionless groups may be obtained by
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Buckingham pi theorem. What is that? If i
d is the number of dimensionless group for
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a particular system and there are n number
of variables for that particular problem,
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and r is the rank of dimensional matrix, then
we can write, id is equal to n minus r, the
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number of dimensionless groups is equal to
number of variables minus rank of the dimensional
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matrix. So, now, we will talk about what is
dimensional matrix and how to determine the
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number of dimensionless groups.
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So, let us consider a simple example for convective
mass transfer into a dilute stream, in a circular
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tube. So, the first step is to identify the
variables pertaining to that particular problem.
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For this problem, these are the variables
- tube diameter, which is symbolized as d
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and the unit is m and dimension is L; similarly,
fluid density, fluid viscosity, fluid velocity,
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mass diffusivity, mass transfer coefficients.
So, these are the variables which incorporate
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the system geometry, which is diameter. And
then, fluid properties – density, viscosity
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and properties and the other primary quantities,
that is, mass transfer coefficients.
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Now, if we take the fundamental dimensions
M, L and t, and make another table, where
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the M will represent for all the exponent
of the fundamental dimensions, which is in
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case of k C, we have l and t, l 1 and t minus
1. So, l is 1 and t is minus 1 and M is 0,
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in case of k C. Similarly, we will obtain
this table for velocity, density, viscosity,
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diffusivity and diameter. So, M, we can represent
all these variables in terms of fundamental
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dimensions. Then, the exponent of these dimensions
will form a matrix which is known as the dimensional
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matrix. So, this is our dimensional matrix
and we have to find the rank of this matrix.
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The rank of this matrix, we can find out using
Matlab or some other program; for this case,
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the r is equal to rank of A, which is 3.
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So, we can obtain id, the number of dimensionless
group for a particular system will be, the
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number of variables n is, here is, if we can
go back, we can see, there are 6 number of
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variables, we have considered for a particular
system. So, n is 6 and rank of the matrix
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r is 3. So, id is n minus r. So, it will be
6 minus 3, is equal to 3. So, there are 3
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dimensional groups for this particular system.
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Let us symbolize this dimensionless group
as pi 1, pi 2 and pi 3. So, step four, the
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three dimensionless group we represent pi
1, pi 2 and pi 3. Now, from the system, we
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have to choose a core group of r variables
in each pi groups. So, how to choose these
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core groups? The one way to choose these core
groups, is to exclude the effect of a particular
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variables which we want to isolate. Say, in
this problem, we want to isolate the mass
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transfer coefficient k C. And also, let us
arbitrarily exclude other variables velocity
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and mu for the particular system; velocity
and viscosity of the fluid, we can exclude,
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as the variables which will not be included
in the core group. So, the core group now
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consists of diffusivity, diameter and the
density of the fluid.
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Now, we can write pi 1 is equal to D A B to
the power a, rho to the power b and d to the
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power c, that is core group and we will include
the other excluded variables, that is, K c;
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and, pi 2 is equal to D A B to the power d,
rho to the power e and d to the power f and
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v ; and, pi 3 is D A B to the power g, rho
to the power h and d to the power i and mu.
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These are the 3 groups we have identified.
Now, from the dimensional form, we can write,
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M 0, L 0, t 0 is equal to 1, is equal to L
square; for pi 1, we can write L square t
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minus 1 to the power a, M L to the power minus
3 to the power b and L to the power c L t
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minus 1. So, from this, if we solve these
equations, we will have for L, we will have
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0 is equal to twice a minus 3 b plus c plus
1; and 0 for t, 0 is equal to minus a minus
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1 and for M 0 is equal to b. So, if we solve
these three, we will have a is equal to minus
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1; b is equal to 0, and c is equal to 1. So,
with this, we can write, pi 1, if we have
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obtained power of these variables, and if
we put this, pi 1 will be K c d divided by
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D A B, and which is known as Sherwood number;
and it is analogous to heat transfer of the
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dimensionless group which is Nusselt number.
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For pi 2, we will do the similar analysis,
and we will obtain pi 2 is v d divided by
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D A B, which is equal to Peclet number for
mass transfer we have discussed earlier. And,
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for pi 3 also, similar analysis will give
you, mu by rho D A B, so, which is Schmidt
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number. Now, if we divide pi 2 by pi 3, so,
it will give you, v d by D A B into rho D
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A B by mu, so, which is v rho d by mu, so,
which is known as Reynolds number. So, the
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dimensional analysis will give us pi 1 is
a function of pi 2 and pi 3; that means, Sherwood
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number, we can write as function of Reynolds
and Schmidt number, which is in the form of
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phi function of, is phi into R e to the power
alpha, and Schmidt number to the power beta;
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alpha, phi, alpha, beta are the dimensionless
constants. So, this is analogous equations
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in heat transfer; Nusselt number is equal
to function of Reynolds number and Prandtl
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number.
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Similar correlations, we can get for different
systems. Let us consider few examples of correlations
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like this system, Range and correlation. Laminar
flow through a circular tube - if Reynolds
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number less than 2100, then, Sherwood number
is 1.62; Reynolds Schmidt d by l to the power
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1 by three. Turbulent flow through a tube,
turbulent flow through a tube - if Reynolds
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number greater than equal to 4000 and less
than equal to 60000, and Schmidt number greater
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than 0.6 and less than 3000. So, we have Sherwood
number is equal to 0.023; Reynolds number
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to the power 0.83 and Schmidt number to the
power 0.33; similarly, liquid flow through
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a, liquid flow through packed bed. So, here,
Reynolds greater than equal to 3 and less
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than 10000; the Sherwood number is equal to
2 plus 1.1, Reynolds to the power 0.6 and
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00:45:07,400 --> 00:45:10,420
Schmidt to the power 0.33.
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00:45:10,420 --> 00:45:17,420
Now, let us consider one example. A sphere
of naphthalene of diameter 20 millimeter is
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00:45:19,270 --> 00:45:26,270
suspended in a flowing air at 45 degree Centigrade.
The velocity of air is 1 meter per second.
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00:45:26,619 --> 00:45:32,510
The diffusivity of naphthalene in air at 45
degree Centigrade is given. Given that the
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00:45:32,510 --> 00:45:39,510
density and viscosity of the air and sublimation
pressure of naphthalene as 1 kilopascal and
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00:45:40,490 --> 00:45:47,490
using this correlation, calculate the mass
transfer coefficient and flux of mass transfer.
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00:45:48,050 --> 00:45:55,050
So, given the data, we can calculate, Reynolds
number will be d v rho by mu, which is equal
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00:46:01,240 --> 00:46:08,240
to 0.02 into 1 into 1.2, divided by 1.9 into
10 to the power minus 5, which is 1260. And,
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00:46:15,630 --> 00:46:22,630
Schmidt number also we can calculate, mu by
rho D A B, which is equal to 1.9 into 10 to
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00:46:24,369 --> 00:46:31,369
the power minus 5, divided by 1.2 into 6.9
into 10 to the power minus 6; so, it will
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00:46:33,910 --> 00:46:40,910
be 2.29. And, Sherwood number, as the equations
given, we can calculate 2 plus 0.55, Reynolds
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00:46:42,860 --> 00:46:49,860
to the power 0.53 and Schmidt to the power
0.33. Substituting the data, it will be 33.9.
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00:46:54,220 --> 00:47:01,220
Now, we can write the diffusion, diffusion
of A through non diffusing B. So, we can calculate
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00:47:12,170 --> 00:47:19,170
Sherwood number is equal to K G P B L M R
T l divided by D A B P t and at 45 degree
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00:47:28,520 --> 00:47:35,520
Centigrade, the vapour pressure is small.
So, P t by P B L M is approximately 1; R is
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00:47:40,859 --> 00:47:47,859
equal to 8.3066 meter cube kilopascal per
K mol Kelvin and t is 318 Kelvin. So, we can
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00:47:57,600 --> 00:48:04,600
write, Sherwood number is given, 33.9 is equal
to K G into 8.3066 into 318 divided by 6.9
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00:48:10,730 --> 00:48:17,730
into 10 to the power minus 6. So, from this
we can calculate K G, which is 8.855 into
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00:48:21,980 --> 00:48:28,980
10 to the power minus 8 K mol per meter square
second kilopascal.
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00:48:29,600 --> 00:48:36,600
Flux, we can calculate, N A is equal to K
G P A1 minus P A2. So, P A1 is given; is 1
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00:48:47,020 --> 00:48:54,020
kilopascal and P A2 can be considered approximately
equal to 0. So, it will be 8.855 into 10 to
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00:48:59,690 --> 00:49:06,690
the power minus 8, 1 minus 0. So, it will
be 8.855 into 10 to the power minus 8 K mol
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00:49:10,410 --> 00:49:17,410
per meter square second. So, this is end of
this lecture.