| Name: |
Twoeq3 - Equilibrium conversion
in a chemical reactor |
| Source: |
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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| Reference/s |
Shacham, M., Computers
Chem. Engng. 9(2), 103 (1985). |
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Bullard, L. G. and Biegler,
L. T., Computers Chem. Engng. 15(4), 239 (1985). |
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| Model: |
2 implicit equations, indep.
variables x and T |
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Higher difficulty level |
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Constraints: 0<=x<=1 |
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Discontinuities: Undefined
for x>=1and for T<=0 |
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Initial estimates: 1. x=0.5,
T= 1700; 2. x=0, T= 1600; 3. x=0, T= 1650; |
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4. x=0.9, T= 1600; 5. x=0.9,
T= 1700 |
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| Solved by |
Shacham, M., POLYMATH 5.1,
build 19, April 2, 2001 |
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| Model Eqs. |
Equilibrium conversion in
a chemical reactor |POLVER05_3 |
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EXCEL
FILE |
f(x) = k*sqrt(1-x)*((0.91-0.5*x)/(9.1-0.5*x)-x^2/((1-x)^2*Kp))
# |
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TEXT
FILE |
f(T) = T*(1.84*x+77.3)-43260*x-105128
# |
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POLYMATH
FILE |
k = exp(-149750/T+92.5) # |
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Kp = exp(42300/T-24.2+0.17*ln(T))
# |
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x(0)=0.5 |
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T(0)=1700 |
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| Variable/function values |
Variable |
Value |
f(x) |
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x |
0.5 |
-3.97 |
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T |
1700 |
6216 |
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k |
82.41 |
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Kp |
7.01 |
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| Solution |
Variable |
Value |
f(x) |
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x |
0.5333728995523 |
-9.3123E-16 |
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T |
1637.70322946500 |
0.0000E+00 |
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k |
2.88916283060 |
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Kp |
17.94002000367 |
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| Additional Information |
Most methods won't converge
from initial guess 2 |
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because of singular points |