Optimizing Reactor Sizing for Gas-Phase Irreversible Reaction

[gfd43tg]

Homework Statement

A 400-liter CSTR and a 100-liter PFR are available to process 1.0 liter of feed per second. The feed contains 41% A, 41% B, and 18% inerts. The irreversible gas-phase reaction

$$A + B → C$$

is to be carried out at 10 atm and 227 °C. The rate of reaction in mol/liter min is given below as a function of conversion

$$\begin{pmatrix} -r_{A} & 0.2 & 0.0167 & 0.00488 & 0.00286 & 0.00204 \\ \hline X & 0.0 & 0.1 & 0.4 & 0.7 & 0.9 \end{pmatrix}$$

(a) What is the maximum conversion that can be achieved with these two reactors connected in series (Ans. $X_{C,P}$ = 0.445, $X_{P,C}$ = 0.515.)

(b) What would be the overall conversion if two 400-liter CSTRs were connected in series for the same feed and operating conditions? (Ans: X = 0.595)

(c) What would be the overall conversion if two 400-liter CSTRs were connected in parallel with half of the feed going to each reactor? (Ans: X = 0.52)

(d) What is the volume of a single tubular reactor necessary to achieve 60% conversion if the molar feed rate is 2 mol A/min? (Ans: V = 180 liters.)

(e) If the total pressure were reduced by a factor of 10, would the conversion increase, decrease, or remain the same?

(f) Give a critique of the answers to this problem.

Homework Equations

The Attempt at a Solution

Right now I am working on part (a), and have some troubles. First off, is it fair to say that if I use a basis of one second, then the initial concentration of A, which I am using as my limiting reactant, is 0.41 mol/1 L, for 0.41 M?

In my attempt, I start off with a CSTR connected to a PFR to see the overall conversion, which I will then do a PFR connected to a CSTR and compare the overall conversion.

My material balance on the CSTR
$$\frac {dN_{A}}{dt} = F_{A0} - F_{A1} + r_{A1}V_{1}$$

Knowing that $F_{A1} = F_{A0} - F_{A0}X_{1}$, and assuming steady state, the mole balance simplifies to

$$0 = F_{A0} - F_{A0} + F_{A0}X_{1} + r_{A1}V_{1}$$

$$\frac {-F_{A0}X_{1}}{V_{1}} = r_{A1}$$

$$\frac {-0.41X_{1}}{400} = r_{A1}$$

Here I run into a problem. I don't know what the conversion of the CSTR is, hence I don't know what the reaction rate is, so I have two unknowns and only one equation.

I put the data into matlab, and here is my code

rA = [-0.2 -0.0167 -0.00488 -0.00286 -0.00204];
X = [0 0.1 0.4 0.7 0.9];
FA0 = 0.41;
plot(X,-FA0./(rA/60))
xlabel('X')
ylabel('-FA0/rA')
title('FA0/-rA vs. Conversion')

This reaction appears to be zero order, based on the nearly straight, which implies a zero order reaction. This may be a clue, but I'm not sure.

 

[Chestermiller]

The reaction rate is a function of conversion r = r(X), so you have one equation in one unknown for the conversion. The equation is non-linear, and you have to interpolate in your table to solve for the conversion in the CSTR. This non-linear equation can be solved in a variety of ways, like secant method or half-interval technique.

chet

 

[gfd43tg]

I did a polynomial fit to the data given (finding $r_{A} = f(X)$ is proving not to be so trivial),

polyfit(X,rA/60,2)

ans =

   -0.0081    0.0097   -0.0024

Hence my $r_{A} = -0.0081X^2 + 0.0097X - 0.0024$

Then I equate it $\frac {-0.41X_{1}}{400} = -.0081X_{1}^{2} + 0.0097X_{1} - 0.0024$ and find the roots,

 roots([-.0081 0.010725 -0.0024])

ans =

    1.0389
    0.2852

So I take it that $X_{1} = 0.285$, well that solves the conversion of the CSTR, but I still need to find the conversion of the PFR, $X_{2}$.

Is it that case that $r_{A2}$ would not be the same function as $r_{A1} = f(X)$, or will they be the same function of $X$?

I then try to find the conversion of the PFR using the integral form to find the volume of a PFR

$$V = F_{A0}\int_{X_{in}}^{X_{out}} \frac {dX}{-r_{A}}$$

keep in mind that when I used matlab, I plotted $X$ against $r_{A}$, and not against $-r_{A}$, which is why I omit the negative sign in the integral when I do it below.

$$0.41 \int_{0.285}^{X_{2}} \frac {dX}{-.0081X^2 + 0.0097X - .0024} = 100$$

Then I run with guess and check in matlab

 f = @(x) 1./(-.0081*x.^2 + .0097*x - .0024)

f = 

    @(x)1./(-.0081*x.^2+.0097*x-.0024)

0.41*quad(f,.285,.445)

ans =

 -107.3562

0.41*quad(f,.285,.446)

ans =

  351.9439

But this is obviously behaving wildly

 

[Chestermiller]

It looks like you have the right idea. How does your polynomial fit to the data look if you compare it vs the data. My guess is that, when you make the plot, you will not be happy with what you see, since the function r(X) is monotonic, and your fit is not going to be monotonic. A polynomial fit is going to turn out to be not such a good choice. Try plotting log r vs X and see how it looks. Then fit a polynomial to logr (or lnr) vs X. I think you will be much happier with the fit. Then re-solve the CSTR case and see what you get. In the case of the PFR, might I suggest doing the integration numerically, rather than trying to get an analytic solution. Using an automatic integrator (if you have access to one) might things easier, especially one that will automatically stop the integration when you satisfy the final condition on the total volume.

Chet

 

[gfd43tg]

Okay, I will try that. The polynomial fit is not that great. By the way, when I am using the function quad in MATLAB, that is a numeric integrator, I guess it has an input function. I can just use trapz to integrate the data only though.

This problem is really making me dig deep into using matlab, got to figure out how to do curve fitting beyond a polynomial fit, awesome!

EDIT: Wow, after spending all this time with matlab, I realized that Excel can do this a million times more easily, and I got a logarithmic curve fit,

$r_{A} = 0.0001 \ln{X} + 1*10^{-5}$

I solve and find $X_{1} = 0.1656$.

I can see why you say to numerically integrate. When I try to integrate

$$0.41 \int_{0.1656}^{X_{2}} \frac {dX}{0.0001 \ln{X} + 1*10^{-5}}$$,

and

fhan = @(x) 1./(0.0001*log(x) + 1e-5); 
0.41*quad(fhan,0.1656,0.4)

ans =

 -841.2465

I get extremely erroneous results.

The plot

x = linspace(0,1,1000);
plot(x,fhan(x))

looks completely different from

plot(X,fhan(X))

This is so weird, when I plot $r_{A}$ vs. $X$, it has a logarithmic fit, but when I do $1/r_{A}$ vs. $X$, it has a linear fit.

 

[gfd43tg]

This has become quite irritating. I believe I found a better fit than a logarithm, and this is my equation

$$r_{A} = \frac{1}{-31412X - 122.76}$$

When I fit this, it actually fits pretty nicely. I uploaded my plots in the PDF as well as code.

This is how I calculated the conversion through the CSTR, $X_{1}$.

fZero = 

    @(x)(1./(-31412*x-122.76))+0.001025*x

EDU>> fzero(fZero,0.5)

ans =

    0.1743

However, when I try to find the area

fThis

fThis = 

    @(x)-31412*x-122.76

EDU>> FA0*abs(quad(fThis,0.1743,0.445))

ans =

   1.0932e+03

I am off by a factor of 10 for the volume, so I can't explain what is going on!

 

[Chestermiller]

This has become quite irritating. I believe I found a better fit than a logarithm, and this is my equation

$$r_{A} = \frac{1}{-31412X - 122.76}$$

When I fit this, it actually fits pretty nicely. I uploaded my plots in the PDF as well as code.

This is how I calculated the conversion through the CSTR, $X_{1}$.

fZero = 

    @(x)(1./(-31412*x-122.76))+0.001025*x

EDU>> fzero(fZero,0.5)

ans =

    0.1743

However, when I try to find the area

fThis

fThis = 

    @(x)-31412*x-122.76

EDU>> FA0*abs(quad(fThis,0.1743,0.445))

ans =

   1.0932e+03

I am off by a factor of 10 for the volume, so I can't explain what is going on!

None of the functionalities you showed look like they fit the data. Why don't you just do the problem with linear interpolation between the data points and see what you get? It's got to be better than what you have been getting so far. This will at least give you a start. You can then try to develop a better functionality later.

Chet

 

[Chestermiller]

Try this:

For 0≤x≤0.1, $r=-0.2e^{-24.829X}$

For 0.1≤X≤0.4, $r=-0.02517e^{-4.101X}$

For 0.4≤X≤0.7, $r=-0.0.00995e^{-1.781X}$

For 0.7≤X≤0.9, $r=-0.00933e^{-1.6895X}$

Chet

 

[gfd43tg]

I don't understand how linearly interpolating between the points will help me solve for the conversion. Don't I need to simultaneously know $r_{A}$?

 

[Chestermiller]

I don't understand how linearly interpolating between the points will help me solve for the conversion. Don't I need to simultaneously know $r_{A}$?

Interpolating r vs X between the data points is conceptually the same as fitting your polynomial or your other functions to the r vs X data. At least the interpolation is monotonic and passes through all the data points. I prefer the interpolation I provided in my most recent post, involving logarithmic/exponential interpolation. Each segment is even curved in the correct way.

For the CSTR, you are just solving 2 equations in two unknowns (X1 and r(X1)), correct?

I ran the calculation for a CSTR, and got a conversion of about 0.2, indicating that the solution lies in the second interpolation interval.

chet