Oil Shale Pyrolysis

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

The following problem is an example from the global optimal control literature and was introduced in [Wen1977]The entry doesn't exist yet..

Mathematical formulation

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

where this is the normalized form with

$u (t) = \frac{1}{u_{t e m p}}$ , with

$u_{t e m p} \in [698.15, 748.15]$

Parameters

State variables
Symbol	Initial value ( $t_{0}$ )
$x_{0} (t)$	$1$
$x_{1} (t)$	$0$

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

Control variable
Symbol	Interval
$u (t)$	[698.15/748.15,1]

Measurement grid

Reference solution

Coming soon.

References

[Wen1977]

The entry doesn't exist yet.