In this screencast we will work through a material balance problem that involves a
mixing point and a recycle stream. We will use a degree of freedom analysis to determine
where to start and determine our strategy to solve. Since we have multiple units, and
we may or may not be able to actually solve this problem. So we will start there, but
before we get going, let me introduce you to the concept of leaching. A separation process
that removes a component that is a solid into a liquid phase. A classic example of this
is making coffee, but lets look at an industrial example, for leaching used with coffee beans.
Caffeine is removed to create a decaffeinated version of the coffee bean. Now I have drawn
out a schematic of the processes with some information that may have been collect. Say
from lab data. Coffee beans enter a mixing tank along with a decaffeinated solvent, which
I have written here as DCS. The beans are then removed to a dryer, and this dryer evaporates
some of the solvent off and that solvent comes back into the mixing tank. Also out of the
dryer is our treated coffee beans. That now has less caffeine then they did before, but
also some solvent residue that was not collect out of the dryer and sent back to the mixing
tank. So also out of the mixing tank is the dirty solvent that now contains the extracted
caffeine. So this is places into the separation unit, which two streams emerge. One stream
is heavy in the solvent. That stream is sent back to our incoming freed stream, and back
to our mixing tank, and the other stream is a high purity caffeine product. So here for
the sake for scalier purposes we want to know how much DCS is needed for 100 kg of coffee
beans. We are also interesting in using some kind of quality control. To know if our separation
unit process is working properly. So we need to know the ideal composition of the stream
that is being recycled back to the feed, and then we can compare this to some measurement
that we may take. So we know the following facts about this process. First that each
100 kg of coffee beans contain 1.5 kg of caffeine. Two is that our solvent the DCS removes 90
percent of the caffeine in the coffee beans. Third for each 100 kg of coffee beans 20 kg
of our solvent leaves with the coffee beans into the dryer, and 90 percent of that solvent
is recovered and passed back into the mixing tank. Lastly, out solvent entering the mixing
tank is 95 percent solvent, and that entering the settling unit is 88 percent solvent, and
out waste solution is 5 percent solvent. The remaining amount of those streams are caffeine.
So again we are trying to determine the amount of solvent we need for 100 kg of coffee beans
and our composition of the recycle stream that is leaving our separation unit. A good
place to start with any material balances to fill all our unknowns and unknown variables
in our streams. So lets start with our basis of 100 kg of coffee beans that enter our process,
and what do we know about the coffee beans. We are told that 1.5 percent of the coffee
beans is caffeine. So I could write down the composition under the stream, but I decided
to do it in terms of our masses. So 1.5 kilograms of our caffeine, and 98.5 kg of our just beans.
We are told that the process removes 90 percent of the caffeine, which could be true for some
decaffeinated products.This means that 0.15 kg, which is the 10 percent remaining leaves
with the beans, So here we have 9.5 kg of beans, because nothing is happening to them
in the process entering and exiting as a whole unit. We have some unknown amount of caffeine
that is leaving with the beans, but we are given information about that, and we know
it is 10 percent of what entered. So our unknown caffeine in this stream is 0.15 kg, and that
is something we can set equal to m1. Now in part 3 we are told that 20 kg of our solvent
will be leaving the mixing tank and entering the dryer. 10 percent of the solvent that
entered the dryer is not recovered. So that means 2 kg leave with our decaffeinated coffee
beans. So we have completely defined out entering coffee bean stream and our exiting coffee
bean stream, and we can use other information to fill in appropriate compositions and unknown
mass flow rates as we need to. So out of our dryer is the other 90 percent of DCS, that
gives us 18 kg going back into the mixing tank. We don't know how much solvent is coming
into the process. That is what we are looking for. So I will write that here, and m2 as
a variable that we don't know. We also don't know the mass flow of the stream coming into
the tank. So I will label that as m3. We do have an idea of the composition of m3. which
we are told is 95 percent solvent. So I have written the mass fractions for the solvent
and the caffeine that is entering the mixing tank. We also know the composition leaving
the mixing tank and entering the extraction unit. As 88 percent solvent. That leaving
the extraction unit. Was given to us as 5 percent solvent, but we don't know the composition
of our recycles stream. Nor do we know the mass flow rate. So I have labeled this as
an unknown m6, and our composition will say is some mass fraction x for our solvent, and
therefore 1 minus x is going to be the remaining balance in that stream, which is caffeine.
So now we have labeled all our known and unknowns for this processes using some of the information
given to us in the problem statement. So we need to figure out an appropriate place to
start. Lets use degree of freedom analysis to determine if we have enough information
to solve it and how we would go about doing this. Lets first do a degree of freedom analysis
on the overall process. Draw a box around process that would give us an idea of what
comes in and what comes out, and anything that is within the process we really don't
care about. So here is looking at this system we start to write out how many unknowns we
have. Well m2 is an unknown, we also forgot to label 2 of our streams. The one leaving
the mixing tank and the one leaving the extraction unit. So I have labeled those as m4 and m5,
and m5 is one of our unknowns. Now are there any other unknowns? Not that we see. We have
all the information necessary for our coffee bean stream both entering and exiting. We
have the composition of our m5 and we know the composition of m2. It is 100 percent of
our solvent. So we can write 2 species balances. We can write one for or DCS and another one
for caffeine. So you might ask why cant we write one for our beans. Well we have already
used that information at the entrance and the exit and nothing changes. So there really
is no bean balance to write at this point. So doing a degree of freedom analysis. We
see that there is 2 unknowns and 2 overall balances that we can write. So our degree
of freedom is 0, and we can solve for n2 and n5. So where do we go from here. We can do
a degree of freedom analysis around our mixing tank. So again we write out our unknowns.
We know the information about what is coming in from our coffee bean stream. As well as
what is coming in from our dryer, and what is leaving to our dryer. SO we have 2 unknowns.
Again our species balances are caffeine and DCS. So our degree of freedom analysis at
this point is equal to 0. So we can solve for m3 and m4. So at this point we have both
m6 and x as our only variables that we don't know. We have a choice to do a balance around
the mixing point or around the extraction unit. Either one will help you solve for x6
and m6. So let me do it around the extraction unit. We have 2 unknowns m6 and x, and again
our species balances are caffeine and DCS. So 2 unknowns and 2 balances gives us a degree
of freedom of 0. We can solve for m6 and x, and all of our unknowns at this point are
solved. What if we did not start with the mixing tank or the overall balance? Lets start
with the mixing point. If we started at our mixing point. We would have m3, m6, and x
as well as m2 as our unknowns, and for our species balances we can do caffeine and DCS.
This gives us a degree of freedom of 2. So we couldn't solve for that. Even if we would
have solve for the overall balance, again that will give use a degree of freedom of
1, and we still couldn't solve for it. So just doing a quick degree of freedom analysis
gives us an idea of whether that is the place we should start, and whether or not this problem
is solvable at all. So lets recall our plan for this problem. First we solve for overall
caffeine balance. This will give us m5. Then we are going to solve or overall DCS balance,
this will give use m2. So that will at least give us an idea for the first part of the
problem on how much solvent we would need for 100 kg of coffee beans. The next step
would be to solve both species balance simultaneously around the mixing tank. In this case gives
us m3 and m4. Then we can solve for m6 using an overall balance either around the mixing
point or extraction unit. Lets say around the mixing point. That gives us m6, and our
last step of the process will be to solve for DCS or caffeine species balance around
the same point, and that should give us our composition or mass fraction x. So now we
have our plan. I will quickly run through the math. So you have something to refer too.
So our overall caffeine balance we have 1.5 kg of caffeine entering. It must equal to
what is exiting, and that is 0.15 kg in the coffee bean stream. Plus 95 percent of our
product stream m5. So this solves for m5, and we get 1.42 kg for that stream. Our second
step is to solve for m2 using an overall solvent balance. We know what is coming in m2 has
to be what is leaving. We have 2 kg leaving in our coffee bean stream and have 5 percent
of our product stream. We know m5 so I will plug that in. Therefore we can solve for m2
like we said and that gives us 2.07 kg. So we have m2 and m5. Then our next step was
to solve for m3 and m4 simultaneously using both species balances around the mixing tank.
So lets first write our caffeine balance around the mixing tank. We know that we have 1.5
kg entering with also or m3 stream, which is 5 percent in caffeine. This must be equal
to what is leaving the mixing tank. So this is 0.12 times m4 plus 0.15 kg that leaves
with our coffee beans. So then again we have 1 equation and 2 unknowns. We need that second
balance. So we will do it for the solvent. We have 95 percent of stream 3. That equals
88 percent of stream 4 plus the 2 kg that leaves with the coffee beans. Now all do we
do have 20 kg entering the dryer we have 18 coming back in. Not even bothering with those
2 streams. I am just looking at the stream that leaves. So now we have our 2 equations.
So we can solve for m3 and m4. I get m3 is equal to 20.4 kg and m4 is equal to 19.75
kg. So I have filled those 2 stream in. Our last two steps were to solve for m6 and x.
So we can do an overall balance around the mixing point. We know that 2.07 kg plus our
recycle stream m6 must be equal to the 20.4 kg that are entering our mixing tank. This
m6 is 18.33 kg. We can do the same for species around the mixing point. So I will choose
our solvent DCS. Since we are looking for x. We know that 2.07 kg of DCS plus some composition
x times m6, which we know as 18.33 now. Is going to be equal to our 95 percent of our
steam entering our mixing tank. This gives us a composition of roughly 94.4 percent or
a mass fraction of 0.944. So now everything is solved. So we would need 2.07 kg of our
solvent per 100 kg of coffee beans to keep the conditions as stated from a material balance
perspective. In our recycle stream is about 94.4 percent our solvent. So some caffeine
is lost back into the mixing tank. Therefore maybe more efficiency separation unit or train
can improve this and affect the overall requirements. Hopefully this gives you a good example on
how to approach a material balance problem with multiple units.