Parametric Estimation - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 314

X 16
i ¼1

P 16
i ¼1

C Ci 1 C Ci

16 16

16

C Ci ¼ 249:8577 C Ci ¼ 59:03 ¼ 2:13826

16 1

i ¼ 1 i ¼ 1

r

1 0:01109
2:13826

R ¼ ¼ 0:997403

The value of R is very close to unity and thus the regression is reliable. Visually from the
graphical representation in Fig. 7.13 , the data are scattered around the regression model
predication (line) and there is no obvious bias in the data. Therefore, the regression is a
success.
However, from a chemical kinetics point of view, the power-law index of 1.373 is not
pleasing. One may suggest choosing a more acceptable number, say 1.5 for n . If this is
done, we can go back to the Excel worksheet and set n to 1.5. Now when we use the Solver
to minimize the variance of the data for C C around the regression model, we can only vary k f
as we wish to fix the value of n . In this, the final solution is shown in Fig. 7.14 . One can
observe that visually the solution is good: no bias in data as they scattered around the regres-
sion model or the line. The variance is slightly higher than the previous solution. Neverthe-
less, the solution is still acceptable. In this case, R

0.997368.
Still, we are not satisfied with the regression analysis. The reaction orders of 1.5 for C and
0.75 for A and B do not seem to agree with our instinct on the order of this type of reactions.
What happens if the orders of A and B are 1? If these orders can be satisfied, the reaction may
be explained as elementary. Even if the reaction is not elementary, these orders will be more
convincing to a chemical engineer than the previous solutions. Let us set n

¼

¼ 2 and repeat the

regression analysis. Figure 7.15 shows the solution for this case.
One can observe from Fig. 7.15 that the solution is visually acceptable: the data points
are scattered around the regression model and there are no obvious bias in the data. The
variance of the experimental data around the regression model is unfortunately higher
than the previous two solutions. However, the difference is not significant and this solution
makes more sense physically. Therefore, this last solution is our solution to the problem.
The correlation coefficient in this case is given by R ¼ 0.996588. The kinetic parameters

10 3 (mol/L) 1 min 1 , n 10 4

are given by k f ¼ 1.32 ¼ 2, and k b ¼ 2.653 (mol/

L) 1 min 1 .
This section shows that physical acceptance should overwrite the small detailed regres-
sion analysis criteria. While everything else being acceptable, the smaller the variance is
better. However, if some parameters are to be adjusted to satisfy physical constrictions, slight
increase in the variance should not be alarmed.
Comparing the solution obtained here: k f ¼ 10 3 (mol/L) 1 min 1 , n

1.32 ¼ 2 and

10 4 (mol/L) 1 min 1 with those obtained from the traditional integral and
differential methods, sections Regression Models and Classification of Regression Models ,
we notice a slight difference in k f . As the error being minimized is the experimental data
compared with the model prediction in this general approach, we can trust the solution
obtained here. In terms of complexity in solution, the general approach requires more

k b ¼ 2.653