# PFR vs. CSTR reactor size for various order reactions

1. Sep 4, 2014

### Maylis

Hello,

I have been working on a very interesting problem out of Fogler's Chemical Reaction Engineering. I have completed the problem (parts (a) through (c) which is what i'll do because I've been working on this problem for a while and I'm too tired to do part (d) right now), but I want to share my results and ask for some help in the interpretation of the results.

The equations used were the design equations for PFR and CSTR, which are

$$V_{CSTR} = \frac {F_{out} - F_{in}}{r_{A}}$$

$$\frac {dF_{A}}{dV_{PFR}} = r_{A}$$

where $V$ is the volume of the reactor, $F$ is the molar flow rate of the species, and $r_{A}$ is the reaction rate of the species.

I went beyond the scope of the problem and did some work in Matlab to write some code to plot reactor size vs. input stream flow rate, for fun and to gain a deeper understanding for this as well as keep up with my matlab skills! For the sake of context, I will post the problem as well as share my m-file and plots.

One preliminary detail I want to ask about is the following. Why is it that when you have the reaction rate for species A $r_{A} = -kC_{A}$, $C_{A}$ is the concentration of A in the exit stream, and not some sort of instantaneous concentration?

So now for the meat of the interpretation.

So to begin with the zeroth order reaction, why is it that they require the same volume for a given input feed? As I worked and refined through this problem, it became very apparent to me that the reaction order makes a great deal of difference.

For the first and second order reactions, why is it that the CSTR requires such a larger volume to have the same conversion than the PFR? The difference between zeroth and first order is extremely obvious, but even for the second order reaction, it appears that the relative difference in size requirement decreases more than for the first order. My prediction for a third order reaction would be that the line for the PFR would become even flatter compared to the CSTR. What is the physical reason for this?

My observation is that as the order of the reaction increases, the relative difference in reactor volume between the PFR and CSTR increases. It seems that the PFR begins to flatten out, and the PFR continues to slope up. I haven't done the test, but I would suppose at some $n^{th}$ order reaction, the line for the PFR would eventually have zero slope.

Why does the relative difference between the two seem to decrease as the order of the reaction increases?

You can see that the actual value of the reactor size increases by orders of magnitude, from $10^3$ to $10^4$ to $10^5$ liters for zeroth, first, and second order reactions, respectively. Why does the order of a reaction affect the actual size of the reactor necessary to convert a input the same amount, so much as an increase of a magnitude of order? I am looking for some sort of physical reasoning.

Lastly, to follow up with this problem I was wondering if there are any other relations I should look at and plot to gain physical insight into sizing of a chemical reactor?

Thank you

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2. Sep 4, 2014

### Staff: Mentor

I've been very busy with family matters today, but I'll try to address your questions soon.

Chet

3. Sep 4, 2014

### Maylis

I had the wrong titles for my graphs!

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4. Sep 5, 2014

### Staff: Mentor

This is true only for a Continuous Stirred Tank Reactor (CSTR), not for a Plug Flow Reactor (PFR).

CSTR: In an ideal CSTR, the mixing is so intense that there are no concentration variations within the reactor. Each parcel of fluid within the tank has the same number of molecules of the reactant; and, therefore, the stream coming out of the reactor has the same concentration of reactant as within the reactor.

Mixing is a process that involves a combination of convection and diffusion. The mixing convection creates very thin striations of fluid adjacent to one another. Even if these striations have quite different concentrations, because the striations are very thin, the process of diffusion rapidly removes these concentration differences. In the limit of infinite mixing rate, the concentration in the CSTR is perfectly uniform.

If you examine any parcel of liquid within an ideal CSTR, it contains molecules of reactant having all different ages (relative to when the molecules entered the tank). It contains molecules that entered the tank just a short time ago, and molecules that entered the tank a long time ago. This is what the high degree of mixing accomplishes. However, the total number of reactant molecules per unit volume (of all ages) in every parcel will be exactly the same.

This is why, in an ideal CSTR, there is only one reactant concentration value (i.e., uniform concentration) in the tank, and why it is also the concentration of reactant in the exit stream.

PFR: In an ideal PFR, the situation is reversed. There is absolutely no mixing taking place, and the concentration within the PFR varies with axial position along the pipe. Furthermore, we neglect the fact that there is a no-slip boundary condition at the wall, and assume that the velocity profile in the pipe is perfectly flat. We also neglect axial diffusion resulting from axial concentration gradients. Under these circumstances, at steady state, the differential mass balance equation for the PRF becomes:
$$v\frac{dC}{dx}=r$$
where v is the axial velocity and x is axial position in the pipe. If we multiply numerator and denominator of the rhs of this equation by the cross sectional area of the pipe, we obtain:
$$F\frac{dC}{dV}=r$$
where F is the volumetric flow rate in the pipe, and V is the differential pipe volume between x and x + dx. These equations can also be written in another form:
$$\frac{dC}{dt}=r$$
where t =x/v is the cumulative residence time from the inlet to the reactor x = 0 to some arbitrary location x. In this form, you can recognize the equation as the same as for a batch reactor. If you situated yourself on a little parcel of fluid as it passes through a PFR, your parcel would essentially be a little batch reactor, and you would measure the same concentration at time t as in a batch reactor at time t. This is what happens in analyzing a PFR reactor when you adopt a so called Lagrangian (material) frame of reference.

If the reaction rate doesn't depend on the concentration of the reactant (zero order reaction), the reaction rate at every location in a PFR will be exactly the same as the reaction rate at every location in a CSTR. So the only thing that matters is the average amount of time that the fluid elements spend in the reactor. This is equal to the volume of the reactor divided by the volumetric flow rate.

Gotta go now. I'll get back to your other questions later.

Chet

5. Sep 9, 2014

### Staff: Mentor

I see where you sent me a thanks for this post. Do you still want me to try to field your other questions, or are you OK with all this now?

Chet

6. Sep 9, 2014

### Maylis

Oh, well I still was wondering about the other parts too thank you

I was thanking you for what you've already written

Last edited: Sep 9, 2014
7. Sep 10, 2014

### Staff: Mentor

Let's look at a first order reaction, and let's assume that the exit concentration from the CSTR is the same as the exit concentration from the PFR, for the same feed rate. In the CSTR, the concentration of reactant throughout the reactor is Cfinal, and this is the concentration in the exit stream. In the PFR, the concentration of reactant throughout the reactor is higher than the exit concentration Cfinal (it is decreasing with axial location along the reactor). Since the rate of reaction is kC, the average rate of reaction in the PFR reactor is higher than the average rate of reaction in the CSTR. So the volume of the PFR can be less.

If the reaction is second order, the reaction rate goes as the square of the concentration, rather than the concentration to the first power. So, for a second order reaction, the ratio of the average rate of reaction in a CSTR to the average rate of reaction in the PFR is even higher than for a first order reaction.

Chet

8. Sep 11, 2014

### Maylis

But why does the PFR volume not increase so significantly for an increasing molar flow rate, as compared to the CSTR?

9. Sep 11, 2014

### Staff: Mentor

Who says? Are we talking about absolute increase or relative increase? Far a first order reaction, the percentage increases in volume should be the same.

Chet

10. Sep 11, 2014

### Maylis

Look at my graph!

11. Sep 11, 2014

### Staff: Mentor

Well, consider this:

PFR: $C_f=C_0e^{-\frac{kV}{F}}$

CSTR: $C_f=\frac{C_0}{1+\frac{kV}{F}}$

Note that V/F occurs in combination. So, if you double F, you have to double V. How do the above equations compare with your results in the graph? Your graph shows that, in both cases, V
is proportional to F. So a 10% increase in F will be accompanied by a 10% increase in V in each case.

Chet