Difference between revisions of "Oil Shale Pyrolysis"
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{{Dimensions | {{Dimensions | ||
− | |nd = | + | |nd = 4 |
− | |nx = | + | |nx = 4 |
|nu = 1 | |nu = 1 | ||
|nw = 0 | |nw = 0 | ||
− | |nre = | + | |nc = 4 |
+ | |nre = 0 | ||
}} | }} | ||
− | The following problem is an example from the global optimal control literature and was introduced in <bib id="Wen1977" />. | + | The following problem is an example from the global optimal control literature and was introduced in <bib id="Wen1977" />. 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: |
+ | <math>A_1 \xrightarrow{k_1} A_2</math> | ||
+ | |||
+ | <math>A_2 \xrightarrow{k_2} A_3</math> | ||
+ | |||
+ | <math>A_1 + A_2 \xrightarrow{k_3} A_2 + A_2</math> | ||
+ | |||
+ | <math>A_1 + A_2 \xrightarrow{k_4} A_3 + A_2</math> | ||
+ | |||
+ | <math>A_1 + A_2 \xrightarrow{k_5} A_4 + A_2</math> | ||
+ | |||
+ | Each reaction is governed by a rate described by: | ||
+ | |||
+ | <math>k_i = k_{i0} \exp{\left(-E_i/RT\right)}, (i=1,2,3,4,5)</math> | ||
== Mathematical formulation == | == Mathematical formulation == | ||
Line 14: | Line 28: | ||
<math> | <math> | ||
\begin{array}{lll} | \begin{array}{lll} | ||
− | \displaystyle \ | + | \displaystyle \max_{T} & \displaystyle &x_2(t_N) \\[1.5ex] |
− | \mbox{s.t.} & \displaystyle \dot{x} | + | \mbox{s.t.} & \displaystyle \dot{x}_1 &= -k_1x_1-(k_3+k_4+k_5)x_1x_2\\ |
− | & \displaystyle \dot{x} | + | & \displaystyle \dot{x}_2 &= k_1x_1-k_2x_2 + k_3x_1x_2\\ |
− | & \displaystyle k_i &= a_i e^{- | + | & \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 &\in \left[t_0,t_N\right] \\ | ||
− | & \displaystyle | + | & \displaystyle T(t) &\in \left[698.15K,748.15K\right]\\ |
− | & \displaystyle x(t_0) &= (1,0)^T\\ | + | & \displaystyle x(t_0) &= (1,0,0,0)^T\\ |
\end{array} | \end{array} | ||
</math> | </math> | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== Parameters == | == Parameters == | ||
− | |||
− | |||
{| class="wikitable" | {| class="wikitable" | ||
Line 40: | Line 48: | ||
|Initial value (<math>t_0</math>) | |Initial value (<math>t_0</math>) | ||
|- | |- | ||
− | |<math> | + | |<math>x_1(t_0)</math> |
|<math>1</math> | |<math>1</math> | ||
|- | |- | ||
− | |<math> | + | |<math>x_2(t_0)</math> |
+ | |<math>0 </math> | ||
+ | |- | ||
+ | |<math>x_3(t_0)</math> | ||
+ | |<math>0 </math> | ||
+ | |- | ||
+ | |<math>x_4(t_0)</math> | ||
|<math>0 </math> | |<math>0 </math> | ||
|} | |} | ||
Line 54: | Line 68: | ||
|- | |- | ||
|<math>a_1</math> | |<math>a_1</math> | ||
− | |<math> | + | |<math>20.3</math> |
|- | |- | ||
|<math>a_2</math> | |<math>a_2</math> | ||
− | |<math> | + | |<math>37.4</math> |
|- | |- | ||
|<math>a_3</math> | |<math>a_3</math> | ||
− | |<math> | + | |<math>33.8</math> |
|- | |- | ||
|<math>a_4</math> | |<math>a_4</math> | ||
− | |<math> | + | |<math>28.2</math> |
|- | |- | ||
|<math>a_5</math> | |<math>a_5</math> | ||
− | |<math> | + | |<math>31.0</math> |
|- | |- | ||
|<math>b_1</math> | |<math>b_1</math> | ||
− | |<math> | + | |<math>\exp{(8.86)}</math> |
|- | |- | ||
|<math>b_2</math> | |<math>b_2</math> | ||
− | |<math> | + | |<math>\exp{(24.25)}</math> |
|- | |- | ||
|<math>b_3</math> | |<math>b_3</math> | ||
− | |<math> | + | |<math>\exp{(23.67)}</math> |
|- | |- | ||
|<math>b_4</math> | |<math>b_4</math> | ||
− | |<math> | + | |<math>\exp{(18.75)}</math> |
|- | |- | ||
|<math>b_5</math> | |<math>b_5</math> | ||
− | |<math> | + | |<math>\exp{(20.7)}</math> |
|} | |} | ||
Line 90: | Line 104: | ||
|Interval | |Interval | ||
|- | |- | ||
− | |<math> | + | |<math>T(t)</math> |
− | |[ | + | |[698.15,748.15] |
|} | |} | ||
− | + | == Solutions == | |
− | + | ||
− | == | + | |
− | + | ||
− | + | ||
− | + | ||
− | + | ||
− | + | * [[:Category:Gekko | GEKKO Python code]] at [[Oil shale pyrolysis (GEKKO)]] | |
== 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
Parameters
Symbol | Initial value () |
Symbol | Value |
Symbol | Interval |
[698.15,748.15] |
Solutions
References
[Wen1977] | The entry doesn't exist yet. |