Oil Shale Pyrolysis

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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:

A_1 \xrightarrow{k_1} A_2

A_2 \xrightarrow{k_2} A_3

A_1 + A_2 \xrightarrow{k_3} A_2 + A_2

A_1 + A_2 \xrightarrow{k_4} A_3 + A_2

A_1 + A_2 \xrightarrow{k_5} A_4 + A_2

Each reaction is governed by a rate described by:

k_i = k_{i0} \exp{\left(-E_i/RT\right)}, (i=1,2,3,4,5)

Mathematical formulation

 \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\\


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
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]



[Wen1977]The entry doesn't exist yet.