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In the previous lecture, we defined two parameters, numbers of transfer unit and height of a transfer
unit to design or to find or to determine the height of a column. So, we said that number
of transfer units should be seen as the degree of difficulty in separations. In other words,
if you want to purify a gas, gas is a stream say from 10 percent to 30 percent, 40 percent
then the entire absorption column can be divided into several number of units, and each units
we achieve a certain degree of rectifications or separations. And then we came out with
the mathematical expression for number of transfer units as some integral, which we
simplify to say that this equals y 1 minus y 2, that is total overall composition change
over divided by the average driving force for mass transfer.
So, in other words; if you have a large number of transfer units all it reflects that we
have a very small mass transfer driving force for a given separation desire. At the same
time, if you say that this absorption column is equivalent to very fewer number of transfer
units, we are trying to say here that the average driving force for the mass transfer
is quite significantly higher there. Similarly, we defined height of transfer units, if you
have one absorption column, divide into 5 number of transfer units; each transfer unit
has a height, how much height we required to achieve desired degree of separations and
height of a transfer units, we also said that mathematically, it was shown to be equal to
or define as the ratios of the flow rate; gas flow rates over mass transfer coefficients
or volumetric mass transfer coefficients Kg a.
All it means that if you have large mass transfer coefficients, because of say large Reynolds
number then we have then the height of a transfer unit is the small. We require only a small
height to achieve separations or to improve this purity. Similarly, if you have a larger
interfacial area between the two phases - gas and liquid, then we have a smaller mass transfer
height STG, and the product of two numbers of transfer units and height of transfer units
gives us the total height of the column. Now, if you go back to that lecture, we obtained
this expressions based on the gaseous phase. We did the species balance or we wrote down
the operating condition; operating line for gaseous phase. In principle it is possible
to come out with the similar expressions for the liquid phase concentrations. So, that
means number of transfer units; height of a transfer unit can be written for the gaseous
phase, it can also be written for the liquid phase and can also write it for overall mass
transfer coefficient or bulk quantities. In this lecture, we will write down this number
of transfer units, the height of transfer units in the different ways, giving us the
same results and then we will take a few examples.
So, recall in the previous lecture, we had N T G as dy over y minus y i from y 2 to y
1 and this is for assuming dilute solutions simplified case we have dilute solutions and
we had H TG, G prime over K y a. So, the total height of the column Z is a product of N T
G multiplied by H T G. So, G signifies here of the gaseous phase and T for the total height
of the columns. So, Z N T G into H T G and we are saying that we can also write as N
T L into H T L. So, in other words this N T L can be written as or can be shown to be
equivalent leave this as an exercise just make an analogy between the gas phase and
the liquid phase one can also obtain this as x 1 to x 2 dx over x i minus x, so see
the difference y minus y i. Now, we have x i minus x i and that means
we have neglected 1 minus x 1 over 1 minus x 2 as 1 for very dilute solutions. Similarly,
this H T L this quantity here can also be written as L prime over K x a, so notice this
K y a that means, we had a quantity here 1 minus x logarithmic average of interfacial
concentration with the mole fractions, so which has been taken as 1, if you assume it
is a dilute solutions. So, in both ways whether you work on the gas phase or on the liquid
phase, we should be able to obtain the same results for the total height of these columns.
We will one can also define this N T G and N T S G based on bulk concentration, that
is y star or capital Y star or x or capital X star. So, all of this based on this bulk
quantities, one can also obtain number of transfer unit or height of transfer unit.
So, if you recall that we have the interface and then the concentration drops heights,
so there is some bulk phase then drops here, then in the liquid phases drops then we have
bulk phase for the liquid and we have this gas phase. So, we have y, we have x, y i and
x i. So, for given y, if you recall we can have x star and for given x, we can have y
star, that means this one can also obtain as N t o G, number of transfer unit as y 2,
y 1, dy over y minus y star, so notice here now we are using N t o G, o means overall.
So, that means we have a avoided working on y i interfacial concentrations and instead
of this, we work on x star and y star, so x star would be equilibrium concentrations
as if the entire bulk phase concentrations is brought to this interface in equilibrium
with this bulk phase concentration. Similarly, we have y star for a given x. So, we can also
define N t o G instead of N T G working on y star instead of y i or if you want to work
on solute-free concentrations capital Y to capital Y 1 then this becomes dy, y minus
y star, which means analogous to this we will have H t o G, which can be written as G prime
capital K y a, so this is overall mass transfer coefficient based on the gas phase. So, even
these two should give us the same results Z equal to N t o G multiplied by H t o G,
the difference comes here o, o through y star or y star and capital K y as overall mass
transfer coefficients, so which means one can also define as N t o L multiplied by H
t o L working on small mole fraction x star or capital X star and then we can have overall
mass transfer coefficient based on this liquid phase.
So, several ways of there is writing here for determining this height of a absorption
columns Z equal to N T G multiplied by H t G equals N t L multiplied by H t L, that will
also equal to N t o G multiplied by H t o G, that will also be equal to N t o L multiplied
by H t o L. The idea is the same, which we did in couple of lectures earlier when we
discussing this interface. Generally, we want to avoid working on this interfacial concentrations
y i and x i, which cannot be measured, so instead of working on y and i x i, we have
y star and x star or we have instead of working on individual mass transfer coefficients k
y and k x, one can work on capital K y overall mass transfer coefficient based on the gas
phase or overall mass transfer coefficient based on this liquid phase.
Either ways depending upon type of problems or the quantities, which are known or which
we can evaluate, one can obtain this height of this column. The main idea is we must understand
that is an approach, it is an approach in which case height of a column or column has
been divided into several such small transfer units. Larger the mass transfer larger the
driving force, we have a small number of units and larger the mass transfer coefficient base
depending upon the Reynolds number height of a transfer units also smaller, so that
is the most fundamentals and basics you must understand the meaning of this meaning of
this and definition the mathematical expression, which we obtained for N T G and S T G etcetera.
Just to recap here, if you recall again we had k y capital A mass transfer coefficient
as individual gas fill mass transfer coefficient plus m slope of the equilibrium curve k x
a and then we can write H t o G, so this is definition we wanted to point out here that
H t o G equals G prime k y a; G prime over k y a, which is H t G over G prime plus m
and k x can also be written as k x a, L prime H T L. So, H t o G one can also write as H
T G overall mass overall height of transfer units based on the gas phase equals individual
H T G based on the gas phase plus m G prime over L prime, so essentially you are multiplied
by G prime both sides of expression m G prime, L prime H T L. So, this is another way of
you know writing this H t o G in terms of H T G and H T L.
Similarly, one can also show that H T O L, height of a transfer unit based on overall
this liquid phase will equal H T L individual 1 over m L prime over G prime H T G, so this
is a different ways of writing this. We take some special cases here that as given in the
text dilute solutions of course we are discussing here that N t o G; we are writing as y 2 y
1, so N t o G overall mass overall transfer units. So, this will equal dy over y minus
y star. So we do not have any other terms, because we are assuming it’s a dilute solution.
Now, if equilibrium curve is linear, sometime it happens that over the concentration range
equilibrium curve is linear, that means we can write as y star m x plus c.
We have the equilibrium curve, which is linear or may be linear over certain range. In this
case, we can write down this equation of this curve as y star equal to m x plus c then,
if you want draw this operating curve, which is y equals L by G x minus x 2 plus y 2.
So, this is our operating curve and this is equilibrium curve one can substitute one from
the another as y minus y star equals L by G, x minus x 2 plus y 2 minus m x minus c.
So, we are trying to evaluate this integral for N t o G at dy over d minus y star. We
are trying to see, if we can obtain an analytical expression. So, in this case when, we have
the dilute solutions plus the equilibrium curve is linear then we can write y minus
y star as this, you should be able to recognize that this also form of q x plus c, everything
is a constant here x 2, m, y 2, L by G ratios given L by G, so we have y minus y star as
q x plus R. Now, one can go back and integrate expression for N t o G to obtain L by G x
2 2 x 1 dx and we have q x plus r. So, actually instead of q x plus C, lets us
to use q x plus r for all the constant y 2 minus c L by G x 2 that is r here constant
instead of c, so we have L by G, x 2 to x1, dx by q x plus r. One can again integrate
this to bring L over a q L n, y minus y star at 1 over x minus excuse me as y minus y start
at 2, so here we have again L by G, so L by G q, this q comes of the integrations and
then we substitute the limits and go back and use my previous expressions for operating
curves to obtain y minus y star at 1 and y minus y star at 2 and this will also be equal
to then y 1 minus y 2 L by G q, again you go back and substitute the previous equation
from the operating lines then we have y minus y star at 1, it is a simple mathematics as
in exercise you just try this y minus y star at 1 and 2 over l n, y minus t star at 1 over
y minus y star at 2. we have a simple expressions for a very special
case, when the equilibrium curve is linear and we have the dilute solutions y 1 minus
y 2 over y minus y star logarithmic average, so that will logarithmic average between 1
and 2. So, this is expressions who one can obtained analytically for number of transfer
units. Height of a transfer units is H t o G equals G prime over capital K y a, so this
is your overall mass transfer coefficients. So, we have expression for N t o G and H t
o G in this special case Z product of the 2, this will become G prime K y a into N t
o G, y 1 minus y 2 over y minus y star m between 1 and 2. Sometimes the same equation is also
rearranged to obtain this G 1 prime y 1 minus y 2.
So, how much amount of sulphur dioxide the impurity is removed, this also equals capital
K y a, Z y minus y star m. So, this also very frequently encounters in this type of expressions
for a very, very special case, where we have this equilibrium curve being linear and we
have this dilute solutions. There is another very special case that, if we follow Henry’s
law. If we have Henry’s law and linear that means we have y equal to m x, so c being zero,
so if we have the equilibrium curve like. This also very you know; when the intercept
is very small or at very low concentration, we have very small mole fractions in the gas
in the gas phase then we can write the equilibrium curve as y equal to m x.
In that case, N t o G can be shown to be obtain as a big expressions y 1 minus m x 2, so m
is a slope of this equilibrium curve, y 2 minus m, though we have x 2 and here we have
excuse me this should be m x 1, 1 minus 1 over A plus 1 over A over 1 minus 1 over A
and A should be familiar at A is L by m g absorption constant. This is not the first
time you are using this N t o G, if you recall in our previous first two to three lectures
when we talked about the cascades in series or the devices in series, we said that one
can obtain the expression analytically in that case also we had this similar expressions
for number of devices or number of a stages in series required for certain rectifications.
So it is a familiar expression and there also we defined this quantity A.
Similarly, for the liquid phase N t o G, if we write N T O L, one can show you should
obtain L n, x 2, y 1 over m, x 1 minus y 1 over m, 1 minus A plus A, so we correct here
this is x 2 minus y 2 over m, 1 minus A plus A over 1 minus A and both of the expressions
in the literature or the textbook you will find there is a graphical representations.
So, you can use this analytical expression or you can use the graphs directory. So, this
is what we have as a theory for today’s lecture or last two lectures, if you combine
to obtain the height of a column given the operating conditions. So, very simple case
given a gas flow rates, we know what is the composition of this gas at inlet to do this
we have been given a solvent. We know the composition of the solvent and
we have been asked to find out the height 1, diameter 2 and the liquid flow rates. So,
if you recall already we have addressed how to obtain the liquid flow rates, liquid flow
rates comes from the minimum solvent requirement and then we take 1.5 times as in operating
conditions. So, once we know the gas flow rate, we know the liquid flow rates, we know
the two compositions one can obtained all y 1, y 2, x 1 and x 2, but we should also
go back and check the hydrodynamics. In the sense that, since we have a counter current
flow of liquid and gas they should not be loading and the flooding.
So, we have supposed to go back and use the classical that Eckerd plot or the Pickford
plot for packing or the loading and flooding in the packing’s, that will give us what
should be there permitted gas flow rates per second per meter square cross-sectional area
of this column, so from that we get G prime, G and G primes gives us the diameter of the
column and from there we can get gives us the cross-sectional area of the column and
from that we get the diameter of the column. So, the height of the column is now the only
quantity left. We know G, we know L, we know all boundaries y 1, y 2, x 1, x 2, and we
know the diameter of the column, so that we know the velocity. Once we know the velocity,
we can obtain mass transfer coefficients. If you recall that given the velocity, we
know the Reynolds number, based on the Reynolds number, we can obtain Sherwood number and
the Sherwood number will give us what is the mass transfer coefficient in the gas phase
or in the liquid phase, equilibrium we have m, so we can find out also the overall mass
transfer coefficients. So, we take this example to determine what is the height of this column
based on the number of transfer units and the height of the transfer units.
Let us to take this simple example to do the calculations. We have absorption column and
we start with the similar approach or the similar type of problems, where this flow
rate of the sulphur dioxide or total flue gases is given to us 1500 kg per hour, this
contains 10 mole percent of sulphur dioxide, this should immediately we should write y
1 equal to 0.1, capital Y 1 equal to 0.1 divided by 0.9 to obtain 0.111. Since this 10 mole
percent sulphur dioxide in air, we should be able to obtain molecular weight as 0.1
into 64 plus 0.9 into 28.8 for air to obtain the molecular weight or average molecular
weight of this flue gas 30. 32.3. So, that we can obtain this G in kilomoles 1500 divided
by 32.3 molecular weight equals 46.41 kilomole per hour and once we know this G, we can obtain
G solid free basis solvent-free basis as G 1 minus y 1.
This is our one inlet here to obtain 41.77 kilomoles per hour, so we know G s, we know
Y 1 and now we have been told that 97 percent of the sulphur dioxide is removed. So, 97
percent removal, once we know 97 percent removal, we have this, this end is 2, we obtain capital
Y 2 as 0.111 into 0.03, so we are removing 97 percent 3 percent is left we have 0.00333.
So, we know y 2, we know G s, G s will remain the same. Now, we have been given a solvent,
which is water; pure water, so we have x 2 equals to capital X 2 equal to 0. So, there
is no sulphur dioxide in this pure solvents and we have the exit concentrations where
we have this L S, L S will remain the same x 1 becomes unknown quantities, so L S and
x 1 are unknown quantities. We have the packing’s, it is a very standard
type problems, which we addressed earlier also P T equal to given 1 atmosphere temperature
is given 30 degree centigrade and it is packed with 25 millimeter of ceramic raschig rings.
So, as soon as we see this 25 millimeter ceramic raschig rings, we can go to that table and
we can get the different information we wanted for this packing factor, which we require
to do this hydrodynamic calculations. This is also given to us that mass transfer coefficient
K x A is 1.25 kilomole per cubic meter per second per del X, so notice this K x individual
mass transfer coefficient based on the gas film is K x A 1.25 kilometer moles per cubic
meter per second per del X. So, check it is consistent with the definitions
of mass transfer coefficients units and K Y is given as 0.75, this for the liquid phase,
this for the gas phase, kilomole per cubic meter per second per del Y. So, this is given
to us and we have been asked very typical questions what is number 1, calculate L mean
minimum amount of solvent require, calculate L, which is 1.5 times the minimum solvent
require our operating condition is at 70 percent of flooding velocity, based on this determine
the diameter of the column and the last but not the list which we are going to decide
and is going to calculate here determine the height of this column.
First three already we have answered in our previous lecture. So, we will go very quickly
here and then we have also been given this table to calculate; to draw this equilibrium
curve, so this is thermodynamic data mole fraction 000.562, 0.79, 1.403, we have 2.23,
2.8, we have 6.19, 19.65, we have 68.5 then we have 27.9 we have 104. So, this x in 10
to power 4, all it means, I leave this as exercise we calculate capital X and capital
Y, as x over 1 minus x, y as y over 1 minus Y and based on this, we should be able to
draw this curve equilibrium curve x and y and if you it draw carefully, we will see
that there is some convex down here, so there is some possibility of pinch occurring somewhere
here, so you have to be very, very careful when you draw this curve here.
So, let us to mark the points here, we have been given 1 and 2, these was all column at
location 1 and 2. We knew x 2, pure solvent and we knew y 2, so we can immediately mark
this point as 0 for the solvents and we had this y 2 given here 0.00333. Y 1 we know this
also given to us, y 1 as 0.111, so be careful when we plot not necessarily we have to plot
from 0 to 1 for larger resolutions we should work only between the two limits point 0.00333
and 0.111. Now, if we realize there is a pinch here, we can draw this curve as a tangent
so wherever it makes a tangent from there, we can get either X 1 or we can take the slope
as L s by G s mean, so whichever way you think is comfortable you find the slope and then
take 1.5 times, the slope would be 1.5 times the minimum or you can write down the expression
for this operating curve and then you can obtained L s by G s operating curve, so either
way we should be able to work. So, if you write down the equations for operating
curves we have G s, Y 1 minus Y 2 equals L s, X 1 minus X 2. So, we know these quantities
41.77, Y 1 we know 0.11, Y 2 we know 0.00333, L S mean we have to determine based on X 1,
so this X 1 based on this tangent where it makes tangent will turn out to be 0.00272,
so if we look at very carefully this is point 0.00272 and X 2 is of course 0. So, based
on this we are can obtain L S mean as 1655 kilo mole per hour, that means operating curve
L S is 1.5 times L S mean, to make it 2069 kilomole per hour. So, idea of doing this
exercise is either you write down the expressions or directly from here, if you know the slope
take 1.5 times a slope, obtain this curve and you can get X 1 operating, which will
we can show that this quantity will be point 0.0002174.
So, either again you can go back and use this expressions knowing this putting the new values
of L S to obtain this X 1, as points 0.0002174. So, the idea is that either use the graph
to obtain this X 1 or you can go back and put the values of L S to obtain this X 1 quantity,
so you should be familiar, we did in the last example earlier case also that either way
graphical way or the equation will give the same results. So, these are first part of
your question given L mean, we have determine L mean minimum amount of solvent require based
on this driving force being 0, wherever it makes a tangent based on that we got obtain
one 1.5 operating curves, we got the new values of X 1, so we have marked our all over boundaries.
The next is we have to go to hydrodynamics and see there is no flooding. There is no
loading or flooding here and it is given that we have to operate at 70 percent of the operating
curves about the flooding conditions, from that information we should be able to what
is a cross-sectional area of the column required to avoid this flooding. So, that will give
us G prime kg per second per meter square cross-sectional area and from that we will
get the cross-sectional area and the diameter of the column.
Let us quickly do this exercise here. Now, if you recall in that plot, we have to suppose
to get L S by G s. So, first get L at the bottom of the column, why this also we have
discuss earlier, that the flow rate of the liquid is a maximum at the bottom, so the
calculations to be more conservative site, we should do at the bottom of the column.
So, at the bottom of the column, we have two 2069 kilomoles and then we have 18 water flow
rate kg per hour, so this was the solvent 2069, so 2069 plus 18 kg per hour, plus we
have sulphur dioxide in this, so how much is amount of sulphur dioxide, well we know
the G s 4.0, 41.477 that is our G s, multiplied by 0.97, because 97 percent is removed, so
whatever is removed it gets into this water face.
So, this is a amount of sulphur dioxide, which has gone into this water face into 64 molecular
weight of sulphur dioxide, so this will be 37529.83 kg per hour of liquid flow rate,
maximum flow rate which we have the liquids. So, now we can go back and draw this again
very quickly. We should recall this curve for loading and flooding where you had certain
quantities on the X, certain quantities in the Y, we have supposed to mark this curve
here, go to this flooding that will give you flooding conditions, you operate on 70 percent
of this pressure drop. So, when you work on the 70 percent of this quantity, you will
get G prime from here. So, all of this I leave it as an exercise, already we have done this
in the previous lecture. This will give us G prime 1921 kg per meter
square per hour, so look at this now we are getting this kg per meter square per hour.
We know this G total gas flow rate as 1500 kg per hour. So, we know G we know G prime
here, this will give us cross-sectional area, as G over G prime and this will give us 0.781
meter square from this, we get the diameter of the column as 1 meter. So far minimum amount
of solvent, operating liquid flow rates, operating conditions, the boundaries and we got this
diameter of this column to avoid this loading and flooding 70 percent of this business;70
percent of this flooding condition and now we start this very quickly. We do this height
of the columns based on the number of transfer units and the height of a transfer unit.
Again, we start from here Z t is N t G multiplied by H t G, this will also be equal to N t L
multiplied by H t L, one can also work on overall transfer units are N t o G then we
choose as S t o G, one can also work on N t o L multiplied by H t o L. So, they are
all four different ways of writing this, height of a column or number of transfer units or
height of transfer units. It is up to us, we have careful, you read the problem correctly,
what are the quantities which are given to us, because you have to calculate you have
to do these integrations for N t G and you have to calculate the height of a transfer
unit based on this mass transfer coefficient. In principle, each any of the four equations
will work and should give you the almost approximately the same result with in this numerical calculations
or numerical air. So, here if we recall the individual mass transfer coefficient for the
gas phase and the liquid phase is given to us, so we should make use of and we should
work on this N t G and H t g. There is no one can still work on N t o G and H t o G,
but the compression will be slightly more cumbersome.
First is the very simplest H t G, which we know G prime K y a. This G prime again be
careful, this is kg mole, it is not kg of loading and flooding and graph, so this is
kg mole hour per meter square, this is the place where generally we make a mistake here.
This G prime varies, so the question again comes what G prime you should use in this
calculations, because it is not G s, so as the gas goes through this passes with the
packed bed column the flow rate of the gas this kg mole per hour per meter square, it
changes as there is mass transfer from the gas phase to the liquid phase. So, as a very
good design or optimum design one should work on the average at G 1 and G 2 at the 1 and
2 inlet and the outlet of this column. So, we know G 1 prime, this is 1 this is 2,
this quantity we know we have calculated earlier 1921 divide by the molecular weight to give
us kilomole per hour per meter square. So, this is what we calculated from G prime from
flooding conditions. So, this was the 70 percent of the flooding limit from where we got 1921
kg, so we divide by the molecular weight to make it kilomole per hour per meter square
and G 2 prime is nothing, but what we have G s over 1 minus y 2, so this is solvent-flow
rate, which remains constant 1 and 2, so we have G s over 1 minus 1.2 into 1 over 0.7812
give you the same quantities 53.66 kilomole per hour per meter square.
So, what is this 0.781, this is the cross-sectional area we got, so this was kilomole per hour
divided by per meter square, because we know this quantity again, this is the cross-sectional
area to make it; to give us kilo hour per meter square and kilo per hour per meter square
at 1 and 2. From this you put simple calculations, this will give us 56. 44, as the average mass
flow rate based on this m t cross of superficial or the cross-sectional area of this column,
so we know S t G here, so we know this is your G prime. So, we know the G prime all
we have to do is to substitute our expressions, which is G prime over mass transfer coefficients.
So, remember now you have to go back and see what mass transfer coefficient was given to
us, which is 270. So, how do we get this 270, it is given 0.075
individual film mass transfer coefficients multiplied by 3600, because this is per hour
and what is given the problem is per seconds, so we have multiplied this to obtain 270,
this will give us 0.21 meter, so height of a transfer unit is 0.21, so that simplest
quantity we have obtain. If you want to work on H t o G then remember we have to and if
you want to work on this quantity, then we have to do some more calculations here, so
we have to see what the problems given, so that our calculation is simpler here, so we
obtain H t G now we can obtain N t G. So, now we have to calculate number of transfer
units N t G based on this gas phase individual gas phase.
So again, here if you read the problems it is given that sulphur dioxide at the inlet
is 1 percent, so generally we treat 1 percent as a dilute of very small concentrations,
in which case we can make this approximation and we can neglect 1 minus y logarithmic average.
So, we write down this N t G expressions simplified form of N t G as dy over y minus y i, y 2
to y 1 so this what we have suppose to evaluate then we can obtain the height of the column
Z here. So, this integral is now between y and y i, so this is the place where we requires
numerical integrations, so you have to pay attention here how do we calculate this quantity,
the best would be to do this graphically. So, we have this equilibrium curve y and x,
so now we are working on a small x and small y. This is the original curve, equilibrium
curve which we can for example we can plot like this. Now, y minus y i that means we
have to obtain operating curve; we have to draw the operating curve.
So, what was the operating curve, operating curve is G s, capital Y 1 minus capital 1
or small y 1 over 1 minus y 1 minus y 2 over 1 minus y 2 equals L s, x 1, 1 minus x 1 minus
x 2 over 1 minus x 2. So, in general we can write, replace y 1, y 2 or x 1 x 2 with y
and x, so if you do this you will see that equilibrium curve. Now it will not to be a
straight line, because we are working on the small y and small x. So, this would be the
equilibrium curve and this is the operating curve. Note neither this is linear or nor
this is linear. So, this is small x and this is y, we have
plotted on this and we have to mark we have to integrate from y 2 to y 1, recall we had
this y 2 as 0.0033, so after 97 percent removal we had 0.003 and here we have 10 percent of
this inlet sulphur dioxide, so y 1 is given as 0.1, so we know the limit y 2 and y 1 dy,
y minus y i. So, all we have to do is, we have to take several small units del y, so
this is numerical integrations, we have to work on del y. So, if you start from here
how do we get y i, well you have to take a slope of minus K x over K y. So, this was
very first few lectures said that from the based on the driving force we can obtain this
y i here, so this is the interfacial concentrations. Given the bulk phase concentration what is
the interfacial concentrations, recall this schematics here y to y i and x i to x, so
x i and y i and we have y here, so we have; we can take the slope here to obtain this
y i. Similarly, you take some other points here, so you take several data points to obtain
all these slopes in principal should be parallel, because we assumed that mass transfer coefficient
do not change along this length, so take as many as points here between 1 and 2 preferentially
you should take at least 10 number of points and then you make a table of given y what
is y i, y minus y i and of course you have del y, you make a write down 1, 2, 3 add all
of these or perform this calculations to obtain N t G, number of transfer unit and you should
be able to obtain this N t G as around 21.5. So, there it was a problem in which we have
addressed that we obtained number of transfer unit as 21.5, height of a transfer unit already
we have calculated, so height of a transfer units and number of a transfer units which
is 0.21 meter, so the total height of the column is 0.21 multiplied by 21.5 meter, so
try to understand this that entire column is approximately divided. So, we are saying
that entire absorption column is divided into 22 units 21.5 approximately as a conservative
design, height of this column has been divided in 22 small, small units and each units we
are having a certain degree of rectifications. So, total rectification we wanted to starting
from 1 percent 2.003 percents, so from 0.1 to 0.003 percent we achieved this in the small
22 of such units, each units we have this fraction of total rectifications and we required
certain height to achieve this, so 22 units multiplied by height of 1 unit will you give
us the height of the column. So, this example we have shown one minimum amount of solvent
require, two operating curves operating line 1.5, so that we can mark the boundaries then
we obtain the diameter of the column or the cross-sectional area of the column to avoid
the flooding and now we know the height of the columns and height of the columns we have
obtain or we have calculated based on this number of transfer units and height of a transfer
units approach. One can also since liquid mass transfer coefficient
is also given to us, you must have noticed that in the beginning we said that both K
y a and K x a are given. So, you are most welcome to work on the liquid phase number
of transfer units or liquid phase over height of transfer units to obtain very close results.
One can also work on overall mass transfer coefficients, since we know the slope of the
equilibrium curve m, one can also in principle can work on overall mass transfer coefficient
based on the liquid phase or on the gas phase, each of this should give the same results.
So, as an exercise you can try each of these four ways of calculating number of transfer
units. So this finishes our chapter on absorptions may be we will take couple of more examples
in the next lecture to address some of the issues in the absorption columns. Thank you.