| Name: |
Teneq1a - Chemical Equilibrium
Problem - R = 10 |
| Source: |
Hiebrt, K. L., ACM Trans.
Math. Software 8, 1(1982) |
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| Reference/s |
Bullard, L. G. and Biegler,
L. T., Computers Chem. Engng. 15(4), 239 (1991). |
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Shacham, M., pp. 891-923 in
A. W. Westerberg and H. H. Chien (Eds), |
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Proc. of FOCAPD 2, CACHE Publications,
Ann Arbor, Michigan, 1984. |
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| Model: |
10 implicit equations, indep.
variables x1 to x10 |
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Lower difficulty level |
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Constraints: xi are nonnegative
for I=1,2,…10 |
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Discontinuities: Undefined
for negative values of x1,x2,x3,x4 |
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Initial estimates: 1. (1,
1, 10, 1, 1, 1, 0, 0, 0, 0); 2.(2, 2, 10, 1, 1, 2, 0, 0, 0, 0); |
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| Solved by |
Shacham, M., POLYMATH 5.1,
build 19, April 9, 2001 |
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| Model Eqs. |
Chemical Equilibrium Problem
- R = 10 |POLVER05_1 |
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EXCEL
FILE |
f(x1) = x1+x4-3 # |
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TEXT
FILE |
f(x2) = 2*x1+x2+x4+x7+x8+x9+2*x10-R
# |
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POLYMATH
FILE |
f(x3) = 2*x2+2*x5+x6+x7-8
# |
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f(x4) = 2*x3+x5-4*R # |
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f(x5) = x1*x5-0.193*x2*x4
# |
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f(x6) = x6*sqrt(x2)-0.002597*sqrt(x2*x4*xs)
# |
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f(x7) = x7*sqrt(x4)-0.003448*sqrt(x1*x4*xs)
# |
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f(x8) = x8*x4-1.799e-5*x2*xs
# |
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f(x9) = x9*x4-0.0002155*x1*sqrt(x3*xs)
# |
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f(x10) = x10*x4^2-3.846e-5*x4^2*xs
# |
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R = 10 # |
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xs = x1+x2+x3+x4+x5+x6+x7+x8+x9+x10
# |
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x1(0)=1 |
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x2(0)=1 |
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x3(0)=10 |
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x4(0)=1 |
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x5(0)=1 |
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x6(0)=1 |
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x7(0)= |
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x8(0)=0 |
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x9(0)=0 |
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x10(0)=0 |
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| Variable/function values |
Variable |
Value |
f(x) |
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x1 |
1 |
-1.000 |
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x2 |
1 |
-6 |
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x3 |
10 |
-3.000 |
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x4 |
1 |
-19.000 |
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x5 |
1 |
0.807 |
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x6 |
1 |
0.989941862 |
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x7 |
0 |
-0.01335405 |
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x8 |
0 |
-0.00026985 |
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x9 |
0 |
-0.00263933 |
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x10 |
0 |
-0.0005769 |
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R |
10 |
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xs |
15 |
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| Solution |
Variable |
Value |
f(x) |
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x1 |
2.88010599840556 |
0 |
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x2 |
3.95067493980017 |
0 |
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x3 |
19.9841296101664 |
-8.882E-16 |
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x4 |
0.11989400159444 |
-7.105E-15 |
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x5 |
0.03174077966720 |
-2.776E-17 |
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x6 |
0.00468458194156 |
-1.735E-18 |
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x7 |
0.03048397912369 |
1.735E-18 |
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x8 |
0.01608812124188 |
-6.505E-19 |
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x9 |
0.12055939838135 |
3.469E-18 |
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x10 |
0.00104378152368 |
0 |
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R |
10 |
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xs |
27.1394051918459 |
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