# Introduction to Chemical Engineering Processes/How to Analyze a Recycle System

## Differences between Recycle and non-Recycle systems

The biggest difference between recycle and non-recycle systems is that the extra splitting and recombination points must be taken into account, and the properties of the streams change from before to after these points. To see what is meant by this, consider any arbitrary process in which a change occurs between two streams:

Feed -> Process -> Outlet


If we wish to implement a recycle system on this process, we often will do something like this:

The "extra" stream between the splitting and recombination point must be taken into account, but the way to do this is not to do a mass balance on the process, since the recycle stream itself does not go into the process, only the recombined stream does.

Instead, we take it into account by performing a mass balance on the recombination point and one on the splitting point.

### Assumptions at the Splitting Point

The recombination point is relatively unpredictable because the composition of the stream leaving depends on both the composition of the feed and the composition of the recycle stream. However, the spliitng point is special because when a stream is split, it generally is split into two streams with equal composition. This is a piece of information that counts towards "additional information" when performing a degree of freedom analysis.

As an additional specification, it is common to know the ratio of splitting, i.e. how much of the exit stream from the process will be put into the outlet and how much will be recycled. This also counts as "additional information".

### Assumptions at the Recombination Point

The recombination point is generally not specified like the splitting point, and also the recycle stream and feed stream are very likely to have different compositions. The important thing to remember is that you can generally use the properties of the stream coming from the splitting point for the stream entering the recombination point, unless it goes through another process in between (which is entirely possible).

## Degree of Freedom Analysis of Recycle Systems

Degree of freedom analyses are similar for recycle systems to those for other systems, but with a couple important points that the engineer must keep in mind:

1. The recombination point and the splitting point must be counted in the degree of freedom analysis as "processes", since they can have unknowns that aren't counted anywhere else.
2. When doing the degree of freedom analysis on the splitting point, you should not label the concentrations as the same but leave them as separate unknowns until after you complete the DOF analysis in order to avoid confusion, since labeling the concentrations as identical "uses up" one of your pieces of information and then you can't count it.

As an example, let's do a degree of freedom analysis on the hypothetical system above, assuming that all streams have two components.

• Recombination Point: 6 variables (3 concentrations and 3 total flow rates) - 2 mass balances = 4 DOF
• Process: Assuming it's not a reactor and there's only 2 streams, there's 4 variables and 2 mass balances = 2 DOF
• Splitting Point: 6 variables - 2 mass balances - 1 knowing compositions are the same - 1 splitting ratio = 2 DOF

So the total is 4 + 2 + 2 - 6 (in-between variables) = 2 DOF. Therefore, if the feed is specified then this entire system can be solved! Of course the results will be different if the process has more than 2 streams, if the splitting is 3-way, if there are more than two components, and so on.

## Suggested Solving Method

The solving method for recycle systems is similar to those of other systems we have seen so far but as you've likely noticed, they are increasingly complicated. Therefore, the importance of making a plan becomes of the utmost importance. The way to make a plan is generally as follows:

1. Draw a completely labeled flow chart for the process.
2. Do a DOF analysis to make sure the problem is solvable.
3. If it is solvable, a lot of the time, the best place to start with a recycle system is with a set of overall system balances, sometimes in combination with balances on processes on the border. The reason for this is that the overall system balance cuts out the recycle stream entirely, since the recycle stream does not enter or leave the system as a whole but merely travels between two processes, like any other intermediate stream. Often, the composition of the recycle stream is unknown, so this simplifies the calculations a good deal.
4. Find a set of independent equations that will yield values for a certain set of unknowns (this is often most difficult the first time; sometimes, one of the unit operations in the system will have 0 DOF so start with that one. Otherwise it'll take some searching.)
5. Considering those variables as known, do a new DOF balance until something has 0 DOF. Calculate the variables on that process.
6. Repeat until all processes are specified completely.

## Example problem: Improving a Separation Process

It has been stated that recycle can help to This example helps to show that this is true and also show some limitations of the use of recycle on real processes.

Consider the following proposed system without recycle.

Example:

A mixture of 50% A and 50% B enters a separation process that is capable of splitting the two components into two streams: one containing 60% of the entering A and half the B, and one with 40% of the A and half the B (all by mass):

If 100 kg/hr of feed containing 50% A by mass enters the separator, what are the concentrations of A in the exit streams?

A degree of freedom analysis on this process:

4 unknowns ($\dot{m}_2,x_{A2},\dot{m}_3,\mbox{ and } x_{A3}$), 2 mass balances, and 2 pieces of information (knowing that 40% of A and half of B leaves in stream 3 is not independent from knowing that 60% of A and half of B leaves in stream 2) = 0 DOF.

Methods of previous chapters can be used to determine that $\dot{m}_2 = 55 \frac{kg}{hr}, x_{A2} = 0.545, \dot{m}_3=45 \frac{kg}{hr}$ and $x_{A3} = 0.444$. This is good practice for the interested reader.

If we want to obtain a greater separation than this, one thing that we can do is use a recycle system, in which a portion of one of the streams is siphoned off and remixed with the feed stream in order for it to be re-separated. The choice of which stream should be re-siphoned depends on the desired properties of the exit streams. The effects of each choice will now be assessed.

### Implementing Recycle on the Separation Process

Example:

Suppose that in the previous example, a recycle system is set up in which half of stream 3 is siphoned off and recombined with the feed (which is still the same composition as before). Recalculate the concentrations of A in streams 2 and 3. Is the separation more or less effective than that without recycle? Can you see a major limitation of this method? How might this be overcome?

This is a rather involved problem, and must be taken one step at a time. The analyses of the cases for recycling each stream are similar, so the first case will be considered in detail and the second will be left for the reader.

#### Step 1: Draw a Flowchart

You must be careful when drawing the flowchart because the separator separates 60% of all the A that enters it into stream 2, not 60% of the fresh feed stream.

Note: there is a mistake in the flow scheme. m6 and xA6 before the process is actually m4 and xA4

#### Step 2: Do a Degree of Freedom Analysis

Recall that you must include the recombination and splitting points in your analysis.

• Recombination point: 4 unknowns - 2 mass balances = 2 degrees of freedom
• Separator: 6 unknowns (nothing is specified) - 2 independent pieces of information - 2 mass balances = 2 DOF
• Splitting point: 6 unknowns (again, nothing is specified) - 2 mass balances - 1 assumption that concentration remains constant - 1 splitting ratio = 2 DOF
• Total = 2 + 2 + 2 - 6 = 0. Thus the problem is completely specified.

#### Step 3: Devise a Plan and Carry it Out

First, look at the entire system, since none of the original processes individually had 0 DOF.

• Overall mass balance on A: $0.5* 100\frac{kg}{h} = \dot{m}_2*x_{A2} + \dot{m}_6*x_{A6}$
• Overall mass balance on B: $50\frac{kg}{h} = \dot{m}_2*(1-x_{A2}) + \dot{m}_6*(1-x_{A6})$

We have 4 unknowns and 2 equations at this point. This is where the problem solving requires some ingenuity. First, lets see what happens when we combine this information with the splitting ratio and constant concentration at the splitter:

• Splitting Ratio: ${m}_6 = \frac{\dot{m}_3}{2}$
• Constant concentration: $x_{A6} = x_{A3}$

Plugging these into the overall balances we have:

• On A: $50 = \dot{m}_2*x_{A2} + \frac{\dot{m}_3}{2}*x_{A3}$
• Total: $50 = \dot{m}_2*(1-x_{A2}) + \frac{\dot{m}_3}{2}*(1-x_{A3})$

Again we have more equations than unknowns but we know how to relate everything in these two equations to the inlet concentrations in the separator. This is due to the conversions we are given:

• 60% of entering A goes into stream 2 means $\dot{m}_2*x_{A2} = 0.6*x_{A4}*\dot{m}_4$
• 40% of entering A goes into stream 3 means $\dot{m}_3*x_{A3} = 0.4*x_{A4}*\dot{m}_4$
• 50% of entering B goes into stream 2 means $\dot{m}_2*(1-x_{A2}) = 0.5*(1-x_{A4})*\dot{m}_4$
• 50% of entering B goes into stream 3 means $\dot{m}_3*x_{A3} = 0.5*(1-x_{A4})*\dot{m}_4$

Spend some time trying to figure out where these equations come from, it's all definition of mass fraction and translating words into algebraic equations.

Plugging in all of these into the existing balances, we finally obtain 2 equations in 2 unknowns:

 On A: $50 = 0.6 \dot{m}_4*x_{A4} + \frac{0.4}{2} \dot{m}_4*x_{A4}$ On B: $50 = 0.5 \dot{m}_4*(1-x_{A4})+ \frac{0.5}{2} \dot{m}_4*(1-x_{A4})$

Solving these equations gives:

 $\dot{m}_4 = 129.17\frac{kg}{h}, x_{A4} = 0.484$
 Note: Notice that two things happened as expected: the concentration of the stream entering the evaporator went down (because the feed is mixing with a more dilute recycle stream), and the total flowrate went up (again due to contribution from the recycle stream). This is always a good rough check to see if your answer makes sense, for example if the flowrate was lower than the feed rate you'd know something went wrong

Once these values are known, you can choose to do a balance either on the separator or on the recombination point, since both now have 0 degrees of freedom. We choose the separator because that leads directly to what we're looking for.

The mass balances on the separator can be solved using the same method as that without a recycle system, the results are:

 $\dot{m}_2 = 70.83 \frac{kg}{hr}, x_{A2} = 0.530, \dot{m}_3 = 58.33 \frac{kg}{hr}, x_{A3} = 0.429$

Now since we know the flowrate of stream 3 and the splitting ratio we can find the rate of stream 6:

 $\dot{m}_6 = \frac{\dot{m}_3}{2} = 29.165\frac{kg}{hr}, x_{A6}=x_{A3}= 0.429$
 Note: You should check to make sure that m2 and m6 add up to the total feed rate, otherwise you made a mistake.

Now we can assess how effective the recycle is. The concentration of A in the liquid stream was reduced, by a small margin of 0.015 mole fraction. However, this extra reduction came at a pair of costs: the flow rate of dilute stream was significantly reduced: from 45 to 29.165 kg/hr! This limitation is important to keep in mind and also explains why we bother trying to make very efficient separation processes.