with
and
Calculate the conversion predicted by the maximum mixedness model for a second order reaction in a reactor with the following RTD
E(t) = 0 for 0 < t 10s
E(t) = 0.01 (t–10) for 10 < t < 20
E(t) = 0.01 (30–t) for 20 < t < 30
E(t) = 0 for t > 30
The rate law iswithand
ODE Solver Solution
We will use the polymath solution rule than the graphical solution
The maximum time at which E(t) is not zero is 30 seconds. Therefore
Let z = 30 – l or l = 30 – z
The RTD function can be expressed as
However, remember E(t) = E(l)
Numerical Solution
|
l |
0 |
10 |
12 |
14 |
16 |
18 |
20 |
22 |
24 |
26 |
28 |
29 |
29.5 |
30 |
|
E(l) |
0 |
0 |
0.02 |
0.04 |
0.06 |
0.08 |
0.1 |
0.08 |
0.06 |
0.04 |
0.02 |
0.01 |
0.005 |
0 |
|
F(l) |
0 |
0 |
0.02 |
0.079 |
0.18 |
0.32 |
0.5 |
0.68 |
0.82 |
0.92 |
0.98 |
0.99 |
0.9986 |
0 |
|
EF=E(l) |
0 |
0 |
0.025 |
0.044 |
0.073 |
0.118 |
0.2 |
0.25 |
0.33 |
0.5 |
1.0 |
2.1 |
4.1 |
0 |
|
1–F(l) |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Calculate backwards starting at l = 30
l = 30 X = 0
l = 28 X = 0 + (–2) [–0.12 (1–0)2 + (0) (1.0)] = 0.24
l = 26 X = 0.24 + (–2) [–0.12 (1–0.24)2 + (0.24) (0.5)] = 0.14
l = 24 X = 0.14 + (–2) [–0.12 (1–0.14)2 + (0.14) (6.33)] = 0.22
l = 22 X = 0.22 + (–2) [–0.12 (1–0.22)2 + (0.22) (0.25)] = 0.26
l = 20 X = 0.26 + (–2) [–0.12 (1–0.26)2 + (0.26) (0.2)] = 0.287
l = 18 X = 0.341
l = 16 X = 0.395
l = 14 X = 0.45
l = 12 X = 0.45 + (–2) [–0.12 (1–0.45)2 + (0.45) (0.25)] = 0.5
l = 10 X = 0.5 + (–2) [–0.12 (1–0.5)2 + (0.5) (0)] = 0.56
l = 8 X = 0.56 + (–2) [–0.12 (1–0.56)2 + (0.56) (0)] = 0.666
l = 6 X = 0.643
l = 4 X = 0.678
l = 2 X = 0.7
l = 0 X = 0.72
One notes a slight difference between the conversion predicted by the software solution (X = 0.69) and that predicted by the backward integration using the very approximate Euler method. We would have achieved closer agreement if we had chosen a small step size, especially at the beginning.