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Welcome to the seventh lecture of module one
which is on diffusion mass transfer. So, before
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going to this lecture, let us have recap on
the previous lecture what we have discussed.
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In our previous lecture, we have considered
the diffusion coefficient measurements and
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predictions. First, we have considered the
gas phase diffusion coefficient measurements.
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As we have discussed, that there are several
methods available to measure the gas-phase
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diffusivity, but we have considered only two
simple methods to determine experimentally
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the gas-phase diffusivity; one is twin bulb
method, and the second one we have considered
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is Stefan tube method. So, for both the cases
we have obtained the governing equations to
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calculate the diffusivity from the experiments,
and we have explained both the methods with
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relevant examples. So, in this lecture we
will continue with our discussion of gas-phase
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diffusivity prediction by different correlations
available in the literature.
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In this lecture we will consider gas-phase
diffusivity prediction, and then we will consider
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liquid-phase diffusivity measurements, and
liquid-phase diffusivity prediction. Let us
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consider the first case gas-phase diffusivity
predictions.
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Although there are several experimental data
available for different gaseous components,
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there are many components, which are required
experimental data or shortage of experimental
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data. In those cases, instead of experimental
data we can use the empirical correlations
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by which we can determine the diffusivity
of the gas-phase components.
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A very simple, and reasonably accurate empirical
equations which was suggested by Fuller, Schettler
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and Giddings in 1966 for binary, binary gas-phase
diffusivity, gas-phase diffusivity up to moderate
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pressure, say up about 20 atmosphere.
The correlations they proposed is diffusivity
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of component A to diffusivity in B is equal
to 10 to the power minus 7 T to the power
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1.75 divided by P t into summation over V
A to the power one-third plus summation over
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V B to the power one-third whole square multiplied
by square root of 1 by M A plus 1 by M B and
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the unit in meter square per second, where
T is equal to temperature
in Kelvin; M A, M B, these are the molecular
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weights of A and B; P t, total pressure in
atmosphere; V A and V B are the atomic diffusion
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volume in meter cube. So, the diffusivity
obtained from this relation is in meter square
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per second.
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Now, let us take an example to calculate the
diffusivity from this correlation. For example,
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for mixtures of carbon monoxide and hydrogen,
predict the diffusivity using Fuller et al.
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1966 formula for the following conditions.
The first one is at 1 atmosphere pressure
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and 100 degree centigrade temperature; the
second problem is at 2 atmosphere pressure
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and 100 degree centigrade temperature and
the third one, at 1 atmosphere pressure, 200
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degree centigrade temperature.
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So, let us consider the first one at 1 atmosphere
presser and 100 degree centigrade temperature.
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We know that diffusivity D AB will be equal
to 10 to the power minus 7 T to the power
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1.75 divided by P t summation over V A to
the power one-third plus summation over V
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B to the power one-third whole square and
into root over bar 1 by M A plus 1 by M B.
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So, the atomic values, values, can be obtained
from text book.
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Let us consider A is equal to carbon monoxide
and B is equal to hydrogen. Now, summation
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of V A would be 1 into carbon plus 1 into
oxygen, so this will be equal to 1. The molecular
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volume of carbon is 16.5 plus for oxygen it
is 5.48, so it will be equal to 21.98. And
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molecular weight of carbon monoxide M A would
be 1 into 12 plus 1 into 16, would be 28 kg
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per kmol. Similarly, for molecular volume
for V B would be 2 into hydrogen, 2 into molecular
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volume 2.31, it is 4.62 and M B is 2 into
1 is equal to 2 kg per kmol.
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The temperature is given as 100 degree centigrade,
is equal to 100 plus 273, this much Kelvin,
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so 373 Kelvin. And P t is
given 1 atmosphere, so if we substitute, D
AB would be equal to 10 to the power minus
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7, 373 temperature to the power 1.75 divided
by 1 pressure into molecular volume of A is
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21.98 to the power one-third plus, for hydrogen
4.62 to the power one-third whole square into
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root over 1 by 28 plus 1 by 2 meter square
per second. So, this will give you 1.159 into
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10 to the power minus 4 meter square per second.
Now, let us consider the second problem, that
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is, at 2 atmosphere and 100 degree centigrade
temperature. So, for this case, since we have
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already determined for the same system the
diffusivity at a particular temperature and
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pressure, there is no need to calculate the
diffusion coefficient for the same system
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at different conditions from scratch. So,
we can use the diffusion co-efficient determined
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in the earlier problem and calculate with
respect to the temperature and pressure corrections.
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So, keeping all the properties same for the
components in problem two, at constant temperature,
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which is temperature is 100 degree centigrade
remains same, the diffusivity D AB from the
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Fullers equation is inversely proportional
with the total pressure. So, we can write,
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at that temperature it is inversely proportional
with pressure, so we can write D AB1 or D
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AB2 by D AB1 will be P t 1 by P t 2.
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So, in this problem we can write D AB2 will
be equal to P t 1 by P t 2 into D AB1. So,
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P t 1 is 1 atmosphere, P t 2 is 2 atmosphere,
and D AB is we have obtained 1.156 into 10
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to the power minus 4 meter square per second.
So, we can calculate D AB2 is equal to 1 by
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2 into 1.156 into 10 to the power minus 4
meter square per second, which is equal to
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0.58 into 10 to the power minus 4 meter square
per second.
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So, for the third problem, similarly, since
the temperature T 2 is given as 200 degree
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centigrade temperature and P t 2 is 1 atmosphere,
which is remain same for the first problem,
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so we can write D AB is directly proportional
to the T to the power of 1.75. So, we can
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write D AB2 by D AB1 will be equal to T 1,
T 2 by T 1 to the power 1.75.
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T 2 is given as 200 plus 273 is equal to 473
K and D AB1 is 1.159 into 10 to the power
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minus 4 meter square per second and T 1 is
100 degree centigrade, which is 373 Kelvin.
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So, D AB2 would be 473 by 373 to the power
1.75 into 1.159 into 10 to the power minus
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4 meter square per second, which will be about
1.761 into 10 to the power minus 4 meter square
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per second. So, this is different conditions
for the same systems.
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There is another important useful correlation,
which is Chapman-Enskog equation. In this
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case, the equation is derived based on the
kinetic theory of gasses and the diffusion
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coefficient calculated from this equation
is strongly dependent on binary interaction
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parameters of the mixture pair.
So the equation, which is given D AB is equal
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to 1.858 into 10 to the power minus 7 T to
the power 1.5 divided by P t sigma AB square
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omega D into square root of 1 by M A plus
1 by M B, this is in meter square per second,
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where T is the temperature in Kelvin; M A,
M B are the molecular weights of A and B;
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P t is the total pressure, total pressure,
in atmosphere; sigma AB the characteristic
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length, characteristic length, of binary mixture
in angstrom, this we can calculate using sigma
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A plus sigma B divided by 2, and omega D is
the collision integral, integral, which is
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a function of K T by epsilon AB. This epsilon
AB can be calculated from root over epsilon
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A into epsilon B.
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Now, let us take an example, very simple example
of mixture of ammonia and hydrogen for this.
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Predict the diffusivity using this formula
at 1 atmosphere pressure and 100 degree centigrade
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temperature.
Now, let us consider A is ammonia, so M A
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is equal to NH 3 is equal to 17 and M B is
hydrogen is 2, T is given, 100 degree centigrade,
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so which is equal to 100 plus 273 So, 373
Kelvin and P t is the total pressure, which
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is 1 atmosphere.
Now, the Lennard-Jones potential parameters
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we can calculate for ammonia from the literature,
we can, from the text book we can get 2.900
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angstrom, so that we can calculate the epsilon
A by K is 558.3 and for hydrogen sigma B is
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2.827 angstroms and epsilon B by K is 59.7.
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So, using this value for this ammonia-hydrogen
pair, for NH 3, which is A and H 2, which
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is B, this pair, we can calculate sigma AB,
which is sigma A plus sigma B by 2 is equal
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to 2.900 plus 2.827 by 2 is equal to 2.8635
angstrom. So, epsilon AB by K we can obtain,
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epsilon A by K, epsilon B by K to the power
half. So, it will be 558.3 into 59.7 to the
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power half is equal to 182.6. So, with this
value K T by epsilon AB will be 373 by 182.6
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will give you 2.04. With these values from
the collision integral
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we can calculate omega D, we can obtain 1.075.
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Now, if we substitute all these values in
the equation D AB, will be 1.858 into 10 to
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the power minus 7, T is 100 degree centigrade,
is 373 to the power 1.5 divided by 1 into
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2.8635 square into 1.075 multiplied by root
over bar 1 by 17 plus 1 by 2. So, this will
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give 1.134 into 10 to the power minus 4 meter
square per second.
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Let us consider another technique for liquid
phase diffusion coefficient measurements.
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There are many methods available to calculate
the liquid phase diffusivity; one of the simplest
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methods is known as the diaphragm cell method.
In this case, a simple cell is divided into
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two parts with a diaphragm, this diaphragm
porous in nature and in two compartments,
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compartment 1 and comportment 2. We can take
two different concentrations of the liquids
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and allow them to diffuse between the compartments.
After certain period of time take the sample
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from different compartments and analyze, and
then from the mole balance we can calculate
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the diffusivity.
The assumption is, that one very dilute solution
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is placed in compartment 1 and relatively
little lower concentrations are placed in
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compartment 2, and molecular diffusions take
place through the diaphragm. Let the volume
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of the compartment 1 is D A1 and the concentration
in comportment 1 is C A1 and the volume of
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compartment 2 is D 2 and the concentration
is C A2. Both the comportments are well mixed.
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Let the area of diaphragm
is A, and porosity is epsilon, so effective
area for diffusion
for diffusion is A epsilon. The diffusion
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path is not uniform, so to account the non-uniformity
of the diffusion path we can use a term tortuosity
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factor tau, tortuosity, tortuosity factor
tau. Since the solute concentration is very
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low, solute concentration is very low, so
the bulk flow term can be neglected and taken
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as 0. So, in that case we can write flux equation
N A will be minus D AB dC A divided by dh,
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which is equal to C A1 minus C A2 divided
by h 2 minus h 1 into tau, which is equal
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to D AB C A1 minus C A2 by x tau, where x
is equal to h 2 minus h 1, the thickness of
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the diaphragm.
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So, we can write flux equations, flux equations
for both sides of the diaphragm. For compartment
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one it is v 1 d CA 1 dt would be equal to
epsilon N A and for compartment 2 will be
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v 2 d C A2 dt is equal to A epsilon N A. So,
combining these two equations we can get,
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d C A1 minus C A2 dt would be equal to A epsilon
D AB by x d tau C A1 minus C A2 into 1 by
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v 1 plus 1 by v 2.
So, if the initial concentrations are
between the compartments C A1, 0 and C A2,
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0, then with the following conditions at t
is equal to 0, C A1 will be C A1, 0; and C
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A2 will be C A2, 0; t is equal t F, C A1 will
be C A1, F and C A2 will be C A2, F.
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Using this condition we can integrate the
flux equations and it will give D AB will
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be x d tau by A epsilon t F 1 by v 1 plus
1 by
v 2 inverse ln C A1, 0 minus C A2, 0 divided
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by C A1, F minus C A2, F, which we can write
1 by alpha t F ln C A1, 0 minus C A2, 0 divided
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by C A1, F minus C A2, F, where alpha is the
cell constant, constant, which is equal to
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A epsilon by x d tau 1 by v 1 plus 1 by v
2. So, this can be determined, alpha can be
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obtained using a solute of known diffusion
coefficient.
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Now, let us consider a simple example. To
measure the diffusivity of acetone in water
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at 20 degree centigrade a diaphragm cell is
used. Initially, compartment one of volume
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50 centimetre cube is filled with 0.5 molar
acetones in water and the compartment 2 of
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volume 55 centimetre cube is filled with water.
The molar concentration of acetone dropped
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to 0.4 molar in compartment 1 after 40 hours.
The cell constant is given as 0.3 per centimetre
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square. Calculate the diffusivity of acetone.
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So, the data, which are given is, volume of
compartment 1 is 50 centimetre cube and for
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compartment 2 is 55 centimetre cube. t F,
total time, which is 40 hours is 40 into 60
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into 60 seconds, so it will be 14400 seconds.
Alpha cell constant is given, 0.3 centimetre
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minus 2; at t is equal to 0, C A1, 0 is equal
to 0.5 and C A2, 0, because of pure water
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it is 0. Concentration of acetone is 0 at
t is equal to t F, C A1, F is 0.4, dropped
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in compartment 1 and C A2, F we have to calculate
from the material balance.
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So, if you do the material balance, v 1 C
A1, 0 plus v 2, plus v 2 into C A2, 0 will
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be v 1 C A1, F plus v 2 C A2, F. So, 50 into
0.5 plus 55 into 0 will be equal to 50 into
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104 plus 55 into C A2, F, so C A2, F is equal
to 0.091.
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So, then diffusivity D AB we can use, 1 by
alpha t F ln concentration of C A1, 0 minus
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C A2, 0 by C A1, F minus C A2, F. Now, putting
the value 1 by 0.3 into 144000 seconds ln
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0.5 minus 0 divided by 0.4 minus 0.091, so
this will give 1.11 into 10 to the power minus
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5 centimetre square per second, so which is
equal to 1.11 into 10 to the power of minus
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9 meter square per second.
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There is some empirical correlations available
to determine the liquid phase diffusivity
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and one of the simplest correlations, which
is known as Wilke-Chang equation and it gives
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D AB is equal to 1.173 into 10 to the power
minus 16 phi M B to the power 0.5 into T divided
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by mu B v A to the power 0.6 in meter square
per second.
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Here, D AB is the diffusivity of solute A
in B, solvent B; phi is the association factor
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M B is the molecular weight of solvent B,
solvent B, and T is the temperature in Kelvin;
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mu B is the solution viscosity in kg per meter
second; v A is the molar volume at normal
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boiling point in meter cube per kmol.
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So, let us consider simple example to predict
the diffusivity of acetone in water at 20
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degree centigrade using Wilke-Chang equation.
The association factor for water is given
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2.26, the molar volume of acetone is given
0.074 meter cube per kmol and viscosity of
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water at 20 degree centigrade is 1.002 into
10 to the power minus 3 kg per meter second.
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So, using this values we can calculate, D
AB will be 1.173 into 10 to the power of minus
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16 into phi M B to the power 0.5 into T divided
by mu B v A
to the power 0.6. The temperature is 20 degree
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centigrade, which is 20 plus 273 Kelvin, so
293 Kelvin. So, if you substitute, D AB would
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be 1.173 into 10 to the power of minus 16
into association factor phi is given 2.26
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00:43:48,039 --> 00:43:55,039
and M B is 18 for water. So, substituting
2.26 into 18 to the power 0.5 into 293 divided
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by viscosity is given, 1.002 into 10 to the
power minus 3 multiplied by the volume, molar
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volume is given 0.074 to the power 0.6. This
will give 1.04 into 10 to the power minus
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00:44:26,450 --> 00:44:30,700
9 meter square per second.
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There is another correlation, which is Stokes-Einstein
equation, which also says that the liquid
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phase diffusivity for different components
varies linearly with absolute temperature,
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and inversely proportional to the viscosity
of the medium. So, we can write D AB mu by
184
00:44:56,750 --> 00:45:12,300
T as constant. So, this is end of lecture
7 and in the next lecture we will continue
185
00:45:12,329 --> 00:45:16,480
diffusivity for the multi-component mixtures.
Thank you.