Oil Shale Pyrolysis - Revision history

RobertLampel: /* Mathematical formulation */

2025-10-16T13:53:13Z

Mathematical formulation

← Older revision		Revision as of 13:53, 16 October 2025
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	& \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]\\

SebastianSager: /* Mathematical formulation */

2025-07-24T13:18:33Z

Mathematical formulation

← Older revision		Revision as of 13:18, 24 July 2025
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	<math>		<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 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 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}		\end{array}
	</math>		</math>

JohnHedengren at 02:02, 15 March 2019

2019-03-15T02:02:13Z

← Older revision		Revision as of 02:02, 15 March 2019
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	\|}		\|}

	== ~~Reference solution~~ ==		== Solutions ==

	~~Coming soon.~~		* [[:Category:Gekko \| GEKKO Python code]] at [[Oil shale pyrolysis (GEKKO)]]

	~~== Source Code ==~~

	~~Model descriptions are not yet available.~~

	== References ==		== References ==

JohnHedengren at 01:54, 15 March 2019

2019-03-15T01:54:31Z

← Older revision		Revision as of 01:54, 15 March 2019
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	{{Dimensions		{{Dimensions
	\|nd = 1		\|nd = 4
	\|nx = 2		\|nx = 4
	\|nu = 1		\|nu = 1
	\|nw = 0		\|nw = 0
	\|nc = 4		\|nc = 4
	\|nre = 2		\|nre = 0
	}}		}}

JohnHedengren: /* Mathematical formulation */

2019-03-15T01:52:52Z

Mathematical formulation

← Older revision		Revision as of 01:52, 15 March 2019
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	<math>		<math>
	\begin{array}{lll}		\begin{array}{lll}
	\displaystyle \~~min_~~{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\\

JohnHedengren: /* Mathematical formulation */

2019-03-15T01:51:49Z

Mathematical formulation

← Older revision		Revision as of 01:51, 15 March 2019
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	<math>		<math>
	\begin{array}{lll}		\begin{array}{lll}
	\displaystyle \min_{T} & \displaystyle &-~~x_1~~(t_N) \\[1.5ex]		\displaystyle \min_{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\\

JohnHedengren at 01:51, 15 March 2019

2019-03-15T01:51:20Z

← Older revision		Revision as of 01:51, 15 March 2019
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	Each reaction is governed by a rate described by:		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>		<math>k_i = k_{i0} \exp{\left(-E_i/RT\right)}, (i=1,2,3,4,5)</math>

	== Mathematical formulation ==		== Mathematical formulation ==
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	\|-		\|-
	\|<math>b_1</math>		\|<math>b_1</math>
	\|<math>\exp{8.86}</math>		\|<math>\exp{(8.86)}</math>
	\|-		\|-
	\|<math>b_2</math>		\|<math>b_2</math>
	\|<math>\exp{24.25}</math>		\|<math>\exp{(24.25)}</math>
	\|-		\|-
	\|<math>b_3</math>		\|<math>b_3</math>
	\|<math>\exp{23.67}</math>		\|<math>\exp{(23.67)}</math>
	\|-		\|-
	\|<math>b_4</math>		\|<math>b_4</math>
	\|<math>\exp{18.75}</math>		\|<math>\exp{(18.75)}</math>
	\|-		\|-
	\|<math>b_5</math>		\|<math>b_5</math>
	\|<math>\exp{20.7}</math>		\|<math>\exp{(20.7)}</math>
	\|}		\|}

JohnHedengren at 01:49, 15 March 2019

2019-03-15T01:49:16Z

← Older revision		Revision as of 01:49, 15 March 2019
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	}}		}}

	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 ==		== Mathematical formulation ==
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	<math>		<math>
	\begin{array}{lll}		\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 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}		\end{array}
	</math>		</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 ==		== Parameters ==



	{\| class="wikitable"		{\| class="wikitable"
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	\|Initial value (<math>t_0</math>)		\|Initial value (<math>t_0</math>)
	\|-		\|-
	\|<math>~~x_0~~(t)</math>		\|<math>x_1(t_0)</math>
	\|<math>1</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>		\|<math>0 </math>
	\|}		\|}
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	\|-		\|-
	\|<math>a_1</math>		\|<math>a_1</math>
	\|<math>8.86</math>		\|<math>20.3</math>
	\|-		\|-
	\|<math>a_2</math>		\|<math>a_2</math>
	\|<math>24.25</math>		\|<math>37.4</math>
	\|-		\|-
	\|<math>a_3</math>		\|<math>a_3</math>
	\|<math>23.67</math>		\|<math>33.8</math>
	\|-		\|-
	\|<math>a_4</math>		\|<math>a_4</math>
	\|<math>18.75</math>		\|<math>28.2</math>
	\|-		\|-
	\|<math>a_5</math>		\|<math>a_5</math>
	\|<math>20.7</math>		\|<math>31.0</math>
	\|-		\|-
	\|<math>b_1</math>		\|<math>b_1</math>
	\|<math>20.3</math>		\|<math>\exp{8.86}</math>
	\|-		\|-
	\|<math>b_2</math>		\|<math>b_2</math>
	\|<math>37.4</math>		\|<math>\exp{24.25}</math>
	\|-		\|-
	\|<math>b_3</math>		\|<math>b_3</math>
	\|<math>33.8</math>		\|<math>\exp{23.67}</math>
	\|-		\|-
	\|<math>b_4</math>		\|<math>b_4</math>
	\|<math>28.2</math>		\|<math>\exp{18.75}</math>
	\|-		\|-
	\|<math>b_5</math>		\|<math>b_5</math>
	\|<math>31.0</math>		\|<math>\exp{20.7}</math>
	\|}		\|}

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	\|Interval		\|Interval
	\|-		\|-
	\|<math>u(t)</math>		\|<math>T(t)</math>
	\|[698.~~15748~~.15,1]		\|[698.15,748.15]
	\|}		\|}

	~~'''Measurement grid'''~~

	== Reference solution ==		== Reference solution ==

FelixMueller at 08:33, 27 July 2016

2016-07-27T08:33:15Z

← Older revision		Revision as of 08:33, 27 July 2016
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	\|nu = 1		\|nu = 1
	\|nw = 0		\|nw = 0
			\|nc = 4
	\|nre = 2		\|nre = 2
	}}		}}

FelixMueller: /* Parameters */

2016-02-22T16:07:49Z

Parameters

← Older revision		Revision as of 16:07, 22 February 2016
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	\|-		\|-
	\|<math>u(t)</math>		\|<math>u(t)</math>
	\|[~~\frac{~~698.~~15}{748~~.15},1]		\|[698.15748.15,1]
	\|}		\|}