Oil Shale Pyrolysis: Difference between revisions

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

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Latest revision as of 13:53, 16 October 2025

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

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

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]

Solutions

GEKKO Python code at Oil shale pyrolysis (GEKKO)

References

[Wen1977]

The entry doesn't exist yet.

@@ Line 27: / Line 27: @@
 <math>
-\begin{array}{lll}
+\begin{array}{llll}
-  \displaystyle \max_{T} &  \displaystyle &x_2(t_N)  \\[1.5ex]
+  \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\\
+  \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}_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}_3 &=& k_2x_2 + k_4x_1x_2\\
-  &  \displaystyle \dot{x}_4 &= k_5x_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 k_i &=& a_i \exp\left(-\frac{b_i}{RT}\right),\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 T(t) &\in \left[698.15K,748.15K\right]\\
+  &  \displaystyle T(t) &\in& \left[698.15K,748.15K\right]\\
-  &  \displaystyle x(t_0) &= (1,0,0,0)^T\\
+  &  \displaystyle x(t_0) &=& (1,0,0,0)^T\\
 \end{array}
 </math>