

|
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Problem 14-13-f |
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Calculated values of DEQ variables
|
|
Variable |
Initial
value |
Minimal
value |
Maximal
value |
Final
value |
|
1 |
a |
0.93 |
0.93 |
0.93 |
0.93 |
|
2 |
C1 |
0.02 |
7.465E-08
|
0.02 |
7.465E-08
|
|
3 |
C2 |
0 |
0 |
0.0080946
|
1.075E-07
|
|
4 |
E |
0 |
0 |
0.0809462
|
1.075E-06
|
|
5 |
Et |
0 |
0 |
0.443293
|
0.0001075
|
|
6 |
Et2 |
0 |
0 |
5.594346
|
0.0107546
|
|
7 |
k1 |
0.1 |
0.1 |
0.1 |
0.1 |
|
8 |
k2CA0 |
0.1 |
0.1 |
0.1 |
0.1 |
|
9 |
t |
0 |
0 |
100. |
100. |
|
10 |
tau |
10. |
10. |
10. |
10. |
|
11 |
v |
10. |
10. |
10. |
10. |
|
12 |
X1 |
0 |
0 |
0.9999546
|
0.9999546
|
|
13 |
X1E |
0 |
0 |
0.0309204
|
1.075E-06
|
|
14 |
X2 |
0 |
0 |
0.9090909
|
0.9090909
|
|
15 |
X2E |
0 |
0 |
0.0255554
|
9.777E-07
|
Differential equations
|
1 |
d(C2)/d(t)
= 2*(a+1)/tau*(C1-C2) |
|
2 |
d(C1)/d(t)
= 2*a*C2/tau-2*(a+1)/tau*C1 |
Explicit equations
|
1 |
a = 0.93
|
|
2 |
tau = 10 |
|
3 |
v = 10 |
|
4 |
E = v*C2
|
|
5 |
Et = E*t
|
|
6 |
Et2 =
E*t^2 |
|
7 |
k2CA0 =
0.1 |
|
8 |
k1 = 0.1
|
|
9 |
X1 =
1-exp(-k1*t) |
|
10 |
X2 =
k2CA0*t/(1+k2CA0*t) |
|
11 |
X1E =
X1*E |
|
12 |
X2E =
X2*E |
General
|
Total
number of equations |
14
|
|
Number
of differential equations |
2
|
|
Number
of explicit equations |
12
|
|
Elapsed
time |
0.000
sec |
|
Solution
method |
RKF_45
|
|
Step
size guess. h |
0.000001
|
|
Truncation
error tolerance. eps |
0.000001
|
Data file:
h:\coursework\che-7250\hw solutions-4th ed\14-13-f.pol