Difference between revisions of "Oil Shale Pyrolysis"
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| − | == | + | == Solutions == |
| − | + | * [[:Category:Gekko | GEKKO Python code]] at [[Oil shale pyrolysis (GEKKO)]] | |
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== References == | == References == | ||
Latest revision as of 03:02, 15 March 2019
| Oil Shale Pyrolysis | |
|---|---|
| State dimension: | 4 |
| Differential states: | 4 |
| Continuous control functions: | 1 |
| Discrete control functions: | 0 |
| Path constraints: | 4 |
| Interior point equalities: | 0 |
The following problem is an example from the global optimal control literature and was introduced in [Wen1977]The entry doesn't exist yet.. The process starts with kerogen and is decomposed into pyrolytic bitumen, oil and gas, and residual carbon. The objective is to maximize the fraction of pyrolytic bitumen. There are 5 reactions including:
Each reaction is governed by a rate described by:
Mathematical formulation
![\begin{array}{lll}
\displaystyle \max_{T} & \displaystyle &x_2(t_N) \\[1.5ex]
\mbox{s.t.} & \displaystyle \dot{x}_1 &= -k_1x_1-(k_3+k_4+k_5)x_1x_2\\
& \displaystyle \dot{x}_2 &= k_1x_1-k_2x_2 + k_3x_1x_2\\
& \displaystyle \dot{x}_3 &= k_2x_2 + k_4x_1x_2\\
& \displaystyle \dot{x}_4 &= k_5x_1x_2\\
& \displaystyle k_i &= a_i e^{-\frac{b_i}{RT}},\quad \forall i\in \{1,\dots,5\} \\ [1.5ex]
& \displaystyle t &\in \left[t_0,t_N\right] \\
& \displaystyle T(t) &\in \left[698.15K,748.15K\right]\\
& \displaystyle x(t_0) &= (1,0,0,0)^T\\
\end{array}](https://mintoc.de/images/math/2/b/0/2b07ecd81011ae22420bbdafe9eaa73e.png)
Parameters
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[698.15,748.15] |
Solutions
References
| [Wen1977] | The entry doesn't exist yet. |
