Oil Shale Pyrolysis: Difference between revisions

Oil Shale Pyrolysis
State dimension:	1
Differential states:	2
Continuous control functions:	1
Discrete control functions:	0
Path constraints:	4
Interior point equalities:	2

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Revision as of 01:49, 15 March 2019

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:

$A_{1} \overset{k_{1}}{\to} A_{2}$

$A_{2} \overset{k_{2}}{\to} A_{3}$

$A_{1} + A_{2} \overset{k_{3}}{\to} A_{2} + A_{2}$

$A_{1} + A_{2} \overset{k_{4}}{\to} A_{3} + A_{2}$

$A_{1} + A_{2} \overset{k_{5}}{\to} A_{4} + A_{2}$

Each reaction is governed by a rate described by:

$k_{i} = k_{i 0} \exp - E_{i} / R T, (i = 1, 2, 3, 4, 5)$

Mathematical formulation

$\begin{array}{lll} \min_{T} & - x_{1} (t_{N}) \\ s.t. & {\dot{x}}_{1} & = - k_{1} x_{1} - (k_{3} + k_{4} + k_{5}) x_{1} x_{2} \\ {\dot{x}}_{2} & = k_{1} x_{1} - k_{2} x_{2} + k_{3} x_{1} x_{2} \\ {\dot{x}}_{3} & = k_{2} x_{2} + k_{4} x_{1} x_{2} \\ {\dot{x}}_{4} & = k_{5} x_{1} x_{2} \\ k_{i} & = a_{i} e^{- \frac{b_{i}}{R T}}, \forall i \in {1, \dots, 5} \\ [1.5 e x] & t & \in [t_{0}, t_{N}] \\ T (t) & \in [698.15 K, 748.15 K] \\ x (t_{0}) & = (1, 0, 0, 0)^{T} \end{array}$

Parameters

State variables
Symbol	Initial value ( $t_{0}$ )
$x_{1} (t_{0})$	$1$
$x_{2} (t_{0})$	$0$
$x_{3} (t_{0})$	$0$
$x_{4} (t_{0})$	$0$

Parameters
Symbol	Value
$a_{1}$	$20.3$
$a_{2}$	$37.4$
$a_{3}$	$33.8$
$a_{4}$	$28.2$
$a_{5}$	$31.0$
$b_{1}$	$\exp 8.86$
$b_{2}$	$\exp 24.25$
$b_{3}$	$\exp 23.67$
$b_{4}$	$\exp 18.75$
$b_{5}$	$\exp 20.7$

Control variable
Symbol	Interval
$T (t)$	[698.15,748.15]

Reference solution

Coming soon.

Source Code

Model descriptions are not yet available.

References

[Wen1977]

The entry doesn't exist yet.

@@ Line 8: / Line 8: @@
 }}
-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{-E_i/RT}, (i=1,2,3,4,5)</math>
 == Mathematical formulation ==
@@ Line 15: / Line 28: @@
 <math>
 \begin{array}{lll}
-  \displaystyle \min_{u} &  \displaystyle &-x_1(t_N)^2  \\[1.5ex]
+  \displaystyle \min_{T} &  \displaystyle &-x_1(t_N)  \\[1.5ex]
-  \mbox{s.t.} &  \displaystyle \dot{x}_0 &= -k_0x_0-(k_2+k_3+k_4)x_0x_1\\
+  \mbox{s.t.} &  \displaystyle \dot{x}_1 &= -k_1x_1-(k_3+k_4+k_5)x_1x_2\\
-  &  \displaystyle \dot{x}_1 &= k_0x_0-k_1x_1 + k_2x_0x_1\\
+  &  \displaystyle \dot{x}_2 &= k_1x_1-k_2x_2 + k_3x_1x_2\\
-  &  \displaystyle k_i &= a_i e^{-u\frac{b_i}{R}},\quad \forall i\in \{1,\dots,5\} \\ [1.5ex]
+ &  \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 u(t) &\in \left[698.15/748.15,1\right]\\
+  &  \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}
 </math>
-where this is the normalized form with
-<math> u(t)= \frac{1}{u_{temp}} </math>, with
-<math> u_{temp} \in \left[698.15,748.15\right] </math>
 == Parameters ==
 {| class="wikitable"
@@ Line 41: / Line 48: @@
 |Initial value (<math>t_0</math>)
 |-
-|<math>x_0(t)</math>
+|<math>x_1(t_0)</math>
 |<math>1</math>
 |-
-|<math>x_1(t)</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>
 |}
@@ Line 55: / Line 68: @@
 |-
 |<math>a_1</math>
-|<math>8.86</math>
+|<math>20.3</math>
 |-
 |<math>a_2</math>
-|<math>24.25</math>
+|<math>37.4</math>
 |-
 |<math>a_3</math>
-|<math>23.67</math>
+|<math>33.8</math>
 |-
 |<math>a_4</math>
-|<math>18.75</math>
+|<math>28.2</math>
 |-
 |<math>a_5</math>
-|<math>20.7</math>
+|<math>31.0</math>
 |-
 |<math>b_1</math>
-|<math>20.3</math>
+|<math>\exp{8.86}</math>
 |-
 |<math>b_2</math>
-|<math>37.4</math>
+|<math>\exp{24.25}</math>
 |-
 |<math>b_3</math>
-|<math>33.8</math>
+|<math>\exp{23.67}</math>
 |-
 |<math>b_4</math>
-|<math>28.2</math>
+|<math>\exp{18.75}</math>
 |-
 |<math>b_5</math>
-|<math>31.0</math>
+|<math>\exp{20.7}</math>
 |}
@@ Line 91: / Line 104: @@
 |Interval
 |-
-|<math>u(t)</math>
+|<math>T(t)</math>
-|[698.15748.15,1]
+|[698.15,748.15]
 |}
-'''Measurement grid'''
 == Reference solution ==