In this screencast we are going to work on a fairy complex material balance problem that
involves multiple units and a recycle stream. Now I am going to focus on an idea of a solution
and what happens when you have a slurry in a mixture of solids and solutions together.
So let me read the problem statement as I am reading this try to sketch a picture, and
I will set up our diagram and we will work through these material balances to solve this
problem. So we have a salt potassium dichromate is to be recovered from a 25 weight percent
aqueous solution. Now the solution is joined by a recycle stream and fed to a crystallizer/centrifuge
where enough water is removed so that out going solution is 85 weight percent water.
Exiting the crystallizer are the crystals with 10 percent solution. So that is our slurry
and the remaining solution which is the recycle stream. Now the filter cake, contains 85 weight
percent crystals and the rest solution. This filter cake is fed to a dryer where it is
contracted with dry air. This causes the remaining water to evaporated. Thus leaving pure potassium
dichromate crystals. So we are given some information. The air leaves with a 0.08 mole
fraction that is water. We are trying to produce a 1000 kg/hr of potassium dichromate. So given
our basis of 1000 kg/hr of potassium dichromate crystals exiting the dryer we want to determine,
one the water evaporated in the crystallizer centrifuge, 2 the mass flow rate of the recycle
stream, and lastly the moles of air that flow through the dryer. So let me take this information
and start sketching our picture. So first we know we have a feed that enters our crystallizer
centrifuge combination. There is a mixture that leaves the crystallizer/centrifuge that
enters a dryer. Then leaving this dryer is our final products. So we know this is a 1000
kg/hr of our potassium dichromate, which I will call KD. So there is a couple of other
streams that we still need do fill in. One we have a recycle stream. So we know that
there is a stream that comes out of the centrifuge and crystallizer that is the solution that
is fed back into the main feed line to the crystallizer and centrifuge, and we have our
recycle stream there. We also know that water is being evaporated that allows the crystals
to precipitate out and get our solids. So we have a water line and into the dryer we
have air that is removing the water from the crystals. So we will have a stream that is
coming out of the dryer that has both air and water. So now I need to label all of all
of our streams with known information from the prompt statement as well as our unknown
variables. I will do this in a slightly neater fashion. So we are told that our main feed
line is 25 percent potassium dichromate. We are told that out of the centrifuge/ crystallizer
we get a solution that is 15 weight percent potassium dichromate the rest water. We are
also told that filter cake that enters the dryer is 85 percent solids. So that means
that 85 weight percent of this stream is our solid crystals, and the other 15 weight percent
is our solution, which is saturated with potassium dichromate. So we have potassium dichromate
that has not hasn't come out of the solution yet. Now we are told a basis of 1000 kg/hr.
We have a mole fraction of water in our exit stream of our dryer, and it is 0.08. we know
that the recycle stream is going to have the same composition and the solution that is
leaving the crystallizer/centrifuge, and the only other information that we don't have
worked out on our chart. Is that this stream is 90 percent of the solution. So 10 percent
of the solution goes with the filter cake, and the other 90 percent is our recycle stream.
So know we will label all of our unknowns. We don't know the mass flow rate of the fed.
We don't know the mass flow of our recycle stream, which is something we are looking
for. Thus we don't know the mass flow rate or the composition entering our crystallizer
and centrifuge. We are looking for the amount of water that is being evaporated from our
crystallizer and centrifuge. We don't know the mass flow rate that is entering our dryer
nor do we know the molar flow rates of water into the dryer, which we are looking for or
the total molar flow rate out. So the three pieces of information that we need to find
to solve this problem are all are the mass flow rate of our recycle stream, the mass
flow rate of water that we are evaporating off and the molar flow rate of air that passes
through the dryer. So. where do we start? Typically the best way to start is doing a
degree of freedom analysis, and I will start by doing a degree of freedom analysis on the
overall process. Now by doing this we have the following unknowns. Mass flow rate of
our feed m1, the exit flow rate of our water from our crystallizer and centrifuge. The
molar flow rate into our dryer and the molar flow rate the exit of the dryer. Now the composition
of the air in is just pure air, and the composition of n3 is just pure water. So we have 4 unknowns.
In terms of relationships that we know. We can write 3 material balances. Since we have
3 different species' potassium dichromate, water and air. Now this gives us a degree
of freedom of 1, which means we cannot solve for these unknowns. However under more careful
examination we can see that n1 and n2 are just water and air, and m3 is just water.
So we have one stream that has our potassium dichromate. So we can do a potassium dichromate
balance to solve for m1. So lets do that. We know that 25 weight percent of our entering
feed stream m1 must be equal to our production rate. This means that m1 is 4000 kg/hr. So
unfortunately we have removed one of or unknowns, that doesn't mean that we can write 3 species
balances to find our other 3 unknowns, because as we start n1, n2, and n3 only have water,
and air . So we cant do a balances for the potassium dichromate. So we need to move else
where to solve. Now using that same logic as before. We can do the same balance around
the dryer. Since only one stream in and one stream out has the potassium dichromate. So
lets look around the dryer and do a potassium dichromate balance around the dryer. So we
know that 85 weight percent of stream 4 that enters the dryer is our potassium dichromate
solids. However we still have potassium dichromate in the solution. So 15 weight percent of that
solution, which is 15 weight percent of the main mass flow rate is our potassium dichromate.
Now these 2 values has to equal to all of the potassium dichromate coming out of the
system, which is 1000 kg/hr. So now we can solve for m4. When I solve for m4 I get 1146
kg/ hr. So at this point we can go through a degree of freedom analysis on the mixing
point, on the crystallizer/centrifuge, again on the dryer or on the overall balance. Sometimes
it is nice just to step back and look at each of these units and figure out what we are
missing and if we look at our crystallizer and centrifuge, the only variable that we
don't know is m3, and what I mean by that is we can take out this recycle stream and
do it around this area, and we know what is coming into this system, and we know what
is leaving. So the only variable of unknown is the water that is being evaporated, which
is something we are looking for. So we don't even need to worry about m2 or m5. So lets
do an overall balance around the crystallizer/centrifuge recycle stream area and solve for m3. So we
have 4000 kg/hr coming into this area. This has to be equal to what leaves, which is a
1146 kg/hr plus m3. Now we can just solve for m3. We get 2854 kg/hr. So we have solve
for one part of the original problem statement, how much water is being evaporated that is
being evaporated from the crystallizer and centrifuge. So we still need to figure out
m5 the mass flow rate of the recycle stream as well as n1 the molar flow rate of our air
coming in to the dryer. So lets take a look at our recycle stream. Recall that the problem
statement that 90 percent of the solution is recycled, but we have the idea about what
isn't recycled and we know the mass flow rate that is associated with that. So we should
be able to figure out the amount of solution that is recycled. So lets set this up So 15
weight percent of our m4 stream, which we solve for 1146 kg/hr. Now that is equal to
10 percent of the total solution. So one way we can write this is multiply this by the
ratio of the 2 streams. We know that 90 percent is recycled, and 10 percent isn't. So using
that ratio we can solve for the mass flow rate of m5, and when we do that. We get that
m5 is equal to 1547 kg/hr. Now for those of you that did not like the ratio system. We
can write it out as such, again 0.15 times m4 so that is our solution. That is equal
to 10 percent of the total solution. So 10 percent times the total solution, which is
0.15,m4, plus now m5. SO given the information we have for m4. You can plug in that value
and solve for m5, and you will get the same thing. It may also be good to practice to
simplify this equaution to see how we got the ratio in the first place. So now the question
is how do we solve for n2, but we dont really need it. All that is left to solve for in
this problem is the molar flow rate of air entering the dryer. So to get that we have
information about the water that is removed. We have a relationship between the water that
comes in and the water that leaves. So we need to convert the water that comes onto
the dryer into a molar flow rate. So we can solve for our air. So lets first figure out
how much water we have coming into the dryer. So we know that 85 weight percent of the solution
is water. So 0.85 times our solution, which is 15 weight percent of m4. So that's the
water that enters, which means that this is going to be the water that leaves, which we
now is 0.08 times n2. Now the problems is we have a mass flow rate in the left side,
and a molar flow rate on the right. So we need to divide the left side by the molecular
weight of water, which is 18 times 10 to the minus kg. So then again we can plug in m4.
Solve for n2, and I get that the molar flow rate of n2 is equal to 1.01 times 10 to the
5 moles per hour. So now we have n2. We can easily solve for n1. The final part of this
problem. So now for air we know that 92 mole percent of n2 is equal to n1. Thus n1 the
molar flow rate of air that enters our system is 9.3 time 10 to the 4th moles per hour.
So now we have solve for the 3 molar flow rates we were looking for. Hopefully this
helped you get a good idea on how to approach a multiple unit material balance, and how
we can use specific information about the streams to help us solve for unknowns.