Chapter 2: Conversion and Reactor Sizing

Numerical Evaulation of an Integral

Consider the liquid phase reaction

which is to take place in a PFR. The following data was obtained in a batch reactor.



X

0

0.4

0.8

-r_A(mol/dm^3.s) 0.01 0.008
0.002

If the molar feed of A to the PFR is 2 mol/s, what PFR volume is necessary to achieve 80% conversion under identical conditions as those under which the batch data was obtained?

(1) What is the PFR design equation and how should the data be plotted?

$\dfrac{dF_A}{dt} = r_AV$

$V=\int_{0}^{X}\dfrac{F_{A0}dX}{-r_A}$

$V = \int_{0}^{X}\dfrac{F_{A0}dX}{r_A}$

(2) How can you evaluate the design equation?

$-r_A$ as a function of X

$\dfrac{1}{-r_A}$ as a function of X

$\dfrac{X}{-r_A}$ as a function of X

(3) What is the Simpson's three-point formula for numbered integration?

Full Solution

Hint 1
F_A0 = 2 mol/s, fed to a plug flow reactor

(a) PFR $\therefore V = F_{A0}\int_{0}^{X}\dfrac{1}{-r_A}dX = \int_{0}^{X}\dfrac{F_{A0}}{-r_A}dX$

(b) Thus, one needs (1/-r_A) as a function of X

Back to Problem

Part (c)

You can either take the area under the curve and multiply by F_A0 or you can use one of the formulas in Appendix A.4.
$V=F_{A_0}\int_{0}^{X}\dfrac{1}{-r_A}dX=\dfrac{\Delta X}{3}\Big{[}\dfrac{1}{-r_A(X=0)} + \dfrac{4}{-r_A(X=0.4)} + \dfrac{1}{-r_A(X=0.8)}\Big{]}$
Back to Problem

Solution

F_A0 = 2 mol/s, fed to a plug flow reactor

Thus, one needs (1/-r_A) as a function of X

You can either take the area under the curve and multiply by F_A0 or you can use one of the formulas in Appendix A.4.
For Simpson's three point formula we have:

To reach 80% conversion, your PFR must be 293.3 dm³.

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Chapter 2: Conversion and Reactor Sizing

Numerical Evaulation of an Integral