In this screencast we will work through a material balance problem that involves a mixing point and a recycle stream. We'll use a degree of freedom analysis to determine where to start and set up our strategy to solve since we have multiple units and we may or may not be able to actually solve this problem. So we'll start there, but before we get going let me introduce you to the concept of leaching, a separation process that removes a component that's in a solid into a liquid phase. A classic example of this is making coffee. So let's look at an industrial example of leaching used with coffee beans where caffeine is removed to create a decaffeinated version of the coffee bean.
Now I've drawn out a schematic of the process with some information that may have been collected say from lab data. Coffee beans enter a mixing tank along with a decaffeinating 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 the stream of our treated coffee beans that now has less caffeine than they did before, but also some solvent residue that wasn't collected 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 placed into a separation unit in which two streams merge. One stream is heavy in the solvent and that stream is sent back to our incoming feed stream and then back to our mixing tank. And the other stream is a high purity caffeine product.
So here for the sake of scale up purposes we want to know how much DCS is needed per 100 kg of coffee beans. We are also interested in using some kind of quality control to know that our separation unit 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 against some measurements that we may take.
So we know the following facts about this process. First, that each 100 kilograms of coffee beans contains 1.5 kilograms of caffeine. Two is that our solvent, the DCS, removes 90% of the caffeine in the coffee beans.
Third, for each 100 kilograms of coffee beans, 20 kilograms of our solvent leaves with the coffee beans into the dryer and 90% of that solvent is recovered through the drying and passed back into the mixing tank. Lastly, our solvent entering the mixing tank is 95% solvent, and that entering the settling unit is 88% solvent. Our waste solution is 5% solvent, the remaining amount of those streams being our 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 recycled stream that is leaving our separation unit.
A good place to start with any material balance is to fill in all our known and unknown variables on our stream. So let's start with our basis of 100 kg of coffee beans that enter our process. What do we know about the coffee beans? We are told that 1.5% of the coffee beans is caffeine.
So I could write the composition under the stream. I have decided to do it in terms of our masses. So 1.5 kg of caffeine and 98.5 kg of just beans. We are told that the process removes 90% of the caffeine which could be true for some decaffeinated products. This means that 0.15 kg which is the 10% remaining leaves with the beans.
So here I have our 98.5 kg of beans since nothing is happening to them in the process. They are 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 that as 10% of what entered. So our unknown caffeine in the bean 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 leaves the mixing tank and enters the dryer. 10% of the solvent that enters the dryer is not recovered, so that means 2 kg leaves with our decaffeinated coffee beans. So we have completely defined our entering coffee bean stream and our exiting coffee bean stream. We could 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% of the DCS.
That gives us 18 kg of solvent going back to 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 as M2 as a variable that we don't know. We also don't know the mass flow of the stream coming into the mixing tank. So I will label that as M3.
We do have an idea of the composition of M3. We are told that it is 95% 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% solvent. That leaving the extraction unit was given to us as 5% solvent, but we don't know the composition that's in our recycle stream. Nor do we know the mass flow rate. So I have labeled this as an unknown M6 and our composition we 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 knowns and unknowns for the process using some of the information given to us in the problem statement. So we need to figure out an appropriate place to start. Let's use degree of freedom analysis to determine if we have enough information to solve this and how we would go about doing this. Let's first do a degree of freedom analysis.
on the overall process. Draw a box around a process that would give us an idea of what comes in and what comes out and anything that is within the process we don't really care about. So here is looking at this system and we start to write out how many unknowns we have.
Well M2 is an unknown. We also forgot to label two 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 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's 100% of our solvent. So we could write two species balances.
We could write one for our DCS and the other one for caffeine. Now you might be asking why we can't write one for our beans. Well we've already used that information at the entrance and the exit and nothing changes.
So there is really no bean balance to write at this point. So doing a degree of freedom analysis we see that there are two unknowns and two overall balances that we could write. So our degree of freedom is 0 and we could solve for M2 and M5.
So where do we go from here? We could 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 two 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 M6 and X. So let me do it around the extraction unit. We have two unknowns.
M6 and X and again our species balances are caffeine and DCS. So two unknowns and two balances gives us a degree of freedom of zero. We can solve for M6 and X and all our unknowns at that point are solved. What if we didn't start with the mixing tank or the overall balance?
Let's 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 could do caffeine and DCS. This gives us a degree of freedom of 2. So we couldn't solve for it.
Now even if we had N2 from doing the overall balance, again that would give us 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 let's recall our plan for this problem. First we solve overall caffeine balance. This will give us...
Then we are going to solve our overall DCS balance. This will give us M2. So that will at least give us an idea of the first part of this problem of how much solvent we would need for our 100 kg of coffee beans.
The next step would be to solve both species balances simultaneously around the mixing tank. This gives us M3 and M4. Then we can solve for M6 using an overall balance either around the mixing point or the extraction unit. So I will say around the mixing point. That gives us M6.
And then our last step of the process will be to solve a DCS or caffeine species balance around the same point. This should give us our composition or mass fraction X. So now we have our plan. I will quickly run through the math so that you have something to refer to. So our overall caffeine balance we have 1.5 kg of caffeine entering.
That must equal what is exiting. 0.15 kilograms in the coffee bean stream. This is plus 95% of our product or byproduct stream M5.
So this solves for M5 and we get 1.42 kilograms for that stream. Our second step was 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 kilograms leaving in our coffee bean stream and we have 5% of our byproduct stream. Now we know m5 so I will plug that in and therefore we can solve for m2 like we said and that gives us 2.07 kg. So we have m2 and m5. Our next step was to to solve for m3 and m4 simultaneously using both species balances around the mixing tank. So let's first write our caffeine balance around the mixing tank.
We know that we have 1.5 kilograms entering with also our m3 stream which is 5% caffeine. This must be equal to what is leaving the mixing tank. So this is 0.12 times m4 plus the 0.15 kilograms that leaves with our coffee beans.
So again we have one equation, two unknowns, we need that second balance so we will do it for the solvent. We have 95% of stream 3, that equals 88% of stream 4 plus the 2 kg that leaves with the coffee beans. Now although we do have 20 kg entering the dryer we have 18 coming back in. Not even bothering with those two streams, I am just looking at the stream that leaves. So now we have our two equations, 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 filled those two streams in.
Our last two steps were to solve for M6 and X. So we could 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. Thus M6 is equal to 18.75 kg.
We can do the same for species balance 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% of our stream entering our mixing tank. This gives us a composition of roughly 94. 94.4% or a mass fraction of.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 and our recycle stream is about 94.4% of our solvent. So some caffeine is lost back into the mixing tank and therefore maybe a more efficient separation unit or train could improve upon this and maybe affect the overall requirements. Hopefully this gives you a good example of approaching a material balance problem with multiple units.