For the reaction of SO2 oxidation (see Fogler or Levenspiel), from the stoichiometric table one can derive the following rate expression, -1/rA vs. X, with k=200 L/mol/s. Please verify that it is correct (see p. 100 in Fogler's textbook).

((1 )/(-(Subscript[r, A]))) = ((0.5 (((1 - 0.14  X))^2))/((1 - X) (0.54 - 0.5  X)))

Let's now construct this graph, which is by far among the most important ones for easily visualizing the strategies available in chemical reactor analysis and design.

recipRate = 
        0.5*(1 - 
                  0.14  X)^2/((1 - X)*(0.54 - 
                    0.5  X));
Plot[recipRate, {X, 0, 0.95}, 
        GridLines -> Automatic];

Note the large increase at high conversions, which suggests that an MFR would be much more disadvantageous than a PFR for very high conversions. This is an important point and it is often illustrated in CRE courses. Let's prepare the relevant graph.

recipRate2 = 
      1/(200*(0.1^2)*(1 - xA)*(2 - 
                xA))
Plot[recipRate2, {xA, 0, 0.9},  
        GridLines -> Automatic];

You can now analyze the relative sizes of PFR vs. MFR as a function of conversion. For MFR at 99.9% conversion:

recipRate2 /. xA -> 0.999

So the MFR reactor volume is (in liters):

%*0.999*0.1*400

For PFR at X=0.999:

NIntegrate[recipRate2, {xA, 0, 0.999}]

So the PFR volume is:

%*0.1*400

I'll let you explore the solution to this important problem for various xA values... For example, you can verify that, as you go from 50 to 99.9% conversion, the ratio of MFR-to-PFR volume increases from ca. 3 to ca. 160 (!!!).

(lrr3@psu . edu , 01/15/08)