In this screen cast we are going to perform sample material balance calculations on a
simple one stage batch still.
In our still we have 50 moles of an ethanol water mixture.
And at the beginning of our distillation process the head temperature is recorded, or the vapor
temperature is recorded, and that is determined to be 84 C. The still is going to be run until
the head temperature is 89 C and what we are trying to determine are both the total amount
of distillate collected and also the composition of distillate and waste products.
In these calculations, capital W is going to represent total number of moles of waste
in the pot still or the liquid that is in the pot still at any time.
xw is the instantaneous composition of that liquid in the still.
And then we have D being the amount of distillate that is produced and also xd is the composition
of the distillate.
And xd for the case of these calculations is an instantaneous composition.
In a batch still operation the compositions of both the distillate product and the composition
in the pot still at any given time is going to be varying.
So what we need is some relationship that we can use for the material balances in that
system over time.
And that relationship is the Rayleigh equation shown here on the right.
The Rayleigh equation simply relates the amount of moles of liquid in the still at the beginning
time point to the amount of moles in the still at the end of the batch still operation and
relates that to the integral of this term here which relates the instantaneous composition
of the distillate and the instantaneous composition of what is in the still, that liquid phase,
itself.
We integrate this term on the right hand side with respect to the composition of the liquid
inside the still at any given time and we integrate over the range from the final composition
to the initial composition.
Now this is a general form of the Rayleigh equation relating the liquid composition to
the distillate composition.
Here we have a special case where we actually have the distillate actually being in equilibrium
with the pot still at any given time because this whole process is driven by vapor liquid
equilibrium.
So in this special case we can replace the composition in the distillate with a composition
of the vapor phase when we are dealing with vapor liquid equilibrium.
We can also replace the composition of the liquid in the pot still at any given time
with an instantaneous composition associated with a liquid phase.
And then these x and y compositions are related through vapor liquid equilibrium.
We changed our limits of equilibrium to correspond to taking the integral with respect to the
liquid composition and now we are integrating from the final liquid composition to the original
liquid composition, much as we had before for the original Rayleigh equation.
Now we have an expression we can solve for the final number of moles in the still when
we are done running the still.
our plan for solving this equation is going to be to get the xy data or the equilibrium
data for this particular ethanol water system.
Do the numerical integration for the right hand side of this system and then use this
to solve for that final amount of moles in the still.
So as a first step we went to Perry's handbook and pulled out xy data or equilibrium data
for this ethanol water system and we have plotted it here.
You will also note that I have the temperature data associated with this VLE data.
This will allow us to determine the limits of integration for the liquid composition
when we start the system and the still head system is 84 C, we can see that the composition
of the liquid is approximately 17%.
Then as we run that still, the temperature is going to increase until we finally get
to a point where we have the temperature of the still is approximately 89 degrees.
And at that point the composition of the liquid phase is only going to be 7% ethanol.
Now for our calculations using the Rayleigh equation we know we are doing the integration
with respect to that liquid composition.
so the important numbers we need to pay attention to now are the 17% composition of the liquid
phase when we start that pot still operation.
And the 7% composition of the liquid phase at the end of the system.
So now that we have the VLE data to do the numerical integration of the right hand side,
all that is left for us to do is actually execute that integration.
There are several options to chose from with varying degrees of accuracy associated with
each.
A popular choice is included, the Simpson's rule, which fits the data that you have to
a curve and solves for the area under the curve and then adds and sums those areas up.
The trapezoidal rule fits your data as a series of trapezoids and sums up those trapezoids
to get the integration.
And finally we have the right hand rule which fits your data curve to a series of rectangles
and sums up those rectangles to determine the area under the curve.
Well the right hand rule has the highest amount of error associated with approximating that
integral.
We are going to use it here just for the sake of simplicity.
So our general approach is to find a value for dx divided by y-x and then sum up those
values to determine the overall integral value.
When we return to our spreadsheet you can see that I have added a column here for dx
and that column has values associated with the difference in the x compositions.
So on this particular case we are taking the difference between this x composition and
the one that proceeded it.
Now to calculate the full term of dx/y-x I have taken our calculated values for dx in
this particular instant and divided that by y which is the vapor phase composition, here
it is 0.17 minus the liquid phase composition and here it is 0.019.
And now to approximate the integral we just need to add up the sections of this dx/y-x
that lie between our starting point and our ending point.
And when we sum those up we get an approximated value for that integral.
Now when we return to the Rayleigh equation we can substitute in for that integral our
value for the sum of 0.27 and also for the starting amount of material in the still we
can insert 50 mol from our original problem statement.
Now it is a simple matter of solving our equation to determine the final number of moles in
the pot still when we are finished with our operation is going to be 38 moles.
To determine the total number of moles in the distillate that have generated we simply
have to do a material balance on the whole process where we take the total number of
moles that entered the system and the pot still, being the initial amount of added to
the still at the beginning, and that is going to be what ended up in the still at the end
of the distillation operation and also what exited the still in the form of distillate.
When we substitute in the values that we know for what was in the still as liquid at the
beginning of the batch still operation and at the end of the batch still operation we
can determine that 12 moles left the system in the form of distillate.
To get the composition of the distillate that exited the top of the still all we have to
do is total ethanol balance of what started in the system, what ended in that pot still
and what exited the pot still in terms of distillate.
When we plug in the values we previously determined in terms of what initially started in the
still, the composition that started from the still, and we determine that just from our
Txy diagram, what remains in the still based on our Rayleigh Equation, the final composition
of what remains in the still based on the Txy diagram and then the amount of distillate
that left based on our material balances, all that is left to find is the composition
of the distillate on average.
Run those numbers and we determine that we have roughly 46% ethanol in that product.
If we want to check to see if this composition makes any sense according to what we have
in the system all we have to do is look at our Txy diagram and see what the composition
should have been from the start of the distillate and to the end of that batch still operation.
When we started we could see the vapor in the system was approximately 51% ethanol.
Then when we ran that system all the way to the end we could see the composition was approximately
39% ethanol.
So our composition of 46% ethanol lies between those two bounds and is a reasonable number
to have as an answer.
Now as far as the last part of the problem of determining the composition of the waste,
for that all we have to do is see what the composition of the liquid was at the end of
distillation operation and there we could have read that directly off the Txy diagram
and seen the composition at the end was only approximately 7% ethanol.