TomSym Refinery Optimization - Revision history

Mbot at 13:34, 8 November 2011

2011-11-08T13:34:45Z

← Older revision		Revision as of 08:34, 8 November 2011
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	<pre>		<pre>
	Problem type appears to be: lp		Problem type appears to be: lp
	Time for symbolic processing: 0.~~23983~~ seconds		Time for symbolic processing: 0.23556 seconds
	Starting numeric solver		Starting numeric solver
	===== * * * =================================================================== * * *		===== * * * =================================================================== * * *

Mbot at 21:25, 2 November 2011

2011-11-02T21:25:43Z

← Older revision		Revision as of 16:25, 2 November 2011
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	After the distillation, crude 1 gives 3% butane, 15% naphtha, 40% gasoil, and 15% residue. Crude 2 results in 5% butane, 20% naphtha, 35% gasoil, and 10% residue. The reforming of the naphtha gives 15% butane and 85% of reformate (reformed naphtha). The catalytic cracking of the residue results in 40% of cracked naphtha and 35% of cracked gasoil (note that these percentages do not add up to 100% because the process also produces 15% of gas, 5% coke and another type of residue that are not taken into consideration in our example). The petrol is produced with three ingredients: reformed naphtha (reformate), butane, and cracked naphtha. The diesel oil is obtained by blending sweetened gasoil, cracked gasoil, and cracked naphtha. The heating oil may contain gasoil and cracked naphtha without any restrictions on their proportions.		After the distillation, crude 1 gives 3% butane, 15% naphtha, 40% gasoil, and 15% residue. Crude 2 results in 5% butane, 20% naphtha, 35% gasoil, and 10% residue. The reforming of the naphtha gives 15% butane and 85% of reformate (reformed naphtha). The catalytic cracking of the residue results in 40% of cracked naphtha and 35% of cracked gasoil (note that these percentages do not add up to 100% because the process also produces 15% of gas, 5% coke and another type of residue that are not taken into consideration in our example). The petrol is produced with three ingredients: reformed naphtha (reformate), butane, and cracked naphtha. The diesel oil is obtained by blending sweetened gasoil, cracked gasoil, and cracked naphtha. The heating oil may contain gasoil and cracked naphtha without any restrictions on their proportions.

			<pre>
	Crudes		Crudes
	\|		\|
Line 31:		Line 32:
	\| To Petrol, Heating Oil and Diesel Blending		\| To Petrol, Heating Oil and Diesel Blending
	V		V
			</pre>

			<pre>
	To Heating Oil and Diesel Blending		To Heating Oil and Diesel Blending
			</pre>

	Simplified representation of a refinery		Simplified representation of a refinery
Line 40:		Line 44:
	Characteristics of intermediate and final products		Characteristics of intermediate and final products

			<pre>
	+--------------+------+---------+-------+-------+-------+------+------+		+--------------+------+---------+-------+-------+-------+------+------+
	\|Characteristic\|Butane\|Reformate\|Cracked\|Cracked\|Desulf \|Petrol\|Diesel\|		\|Characteristic\|Butane\|Reformate\|Cracked\|Cracked\|Desulf \|Petrol\|Diesel\|
Line 49:		Line 54:
	\|Sulfur (in %) \| - \| - \| 0.12 \| 0.76 \| 0.03 \| - \|<=0.05\|		\|Sulfur (in %) \| - \| - \| 0.12 \| 0.76 \| 0.03 \| - \|<=0.05\|
	+--------------+------+---------+-------+-------+-------+------+------+		+--------------+------+---------+-------+-------+-------+------+------+
			</pre>

	In the next month the refinery needs to produce 20,000 tonnes of butane, 40,000 tonnes of petrol, 30,000 tonnes of diesel oil, and 42,000 tonnes of heating oil. 250,000 tonnes of crude 1 and 500,000 tonnes of crude 2 are available. The monthly capacity of the reformer are 30,000 tonnes, for the desulfurization 40,000 tonnes and for the cracking 50,000 tonnes. The cost of processing is based on the use of fuel and catalysts for the different operations: the costs of distillation, reforming, desulfurization, and cracking are $ 2.10, $ 4.18, $ 2.04 and $ 0.60 per tonne respectively.		In the next month the refinery needs to produce 20,000 tonnes of butane, 40,000 tonnes of petrol, 30,000 tonnes of diesel oil, and 42,000 tonnes of heating oil. 250,000 tonnes of crude 1 and 500,000 tonnes of crude 2 are available. The monthly capacity of the reformer are 30,000 tonnes, for the desulfurization 40,000 tonnes and for the cracking 50,000 tonnes. The cost of processing is based on the use of fuel and catalysts for the different operations: the costs of distillation, reforming, desulfurization, and cracking are $ 2.10, $ 4.18, $ 2.04 and $ 0.60 per tonne respectively.
Line 54:		Line 60:
	==Variables==		==Variables==

			<pre>
	demand Demand of each product		demand Demand of each product
	supply Supply of crudes		supply Supply of crudes
Line 70:		Line 77:
	dieselspec_L Lower bounds (contents in diesel)		dieselspec_L Lower bounds (contents in diesel)
	dieselspec_U Upper bounds (contents in diesel)		dieselspec_U Upper bounds (contents in diesel)
			</pre>

	==Results==		==Results==

			<pre>
	4 Final products butane, petrol, diesel, heating oil		4 Final products butane, petrol, diesel, heating oil
	3 Intermediates to Petrol petbutane, reformerate, petcrknaphtha		3 Intermediates to Petrol petbutane, reformerate, petcrknaphtha
Line 81:		Line 90:
	2 Intermediates from Cracking crknaphtha, crkgasoil		2 Intermediates from Cracking crknaphtha, crkgasoil
	2 Crudes crude1, crude2		2 Crudes crude1, crude2
			</pre>

	==Reference==		==Reference==

Mbot: Created page with "{{Part Of Manual|title=the TomSym Manual|link=TomSym Manual}} ==Problem description== A refinery produces butane, petrol, diesel oil, and heating oil from two crudes..."

2011-11-02T21:02:47Z

Created page with "{{Part Of Manual|title=the TomSym Manual|link=TomSym Manual}} ==Problem description== A refinery produces butane, petrol, diesel oil, and heating oil from two crudes..."

New page

{{Part Of Manual|title=the TomSym Manual|link=[[TomSym|TomSym Manual]]}}

==Problem description==

A refinery produces butane, petrol, diesel oil, and heating oil from two crudes. Four types of operations are necessary to obtain these products: separation, conversion, upgrading, and blending.

The separation phase consists of distilling the raw product into, among others, butane, naphtha, gasoil, and a residue. The residue subsequently undergoes a conversion phase (catalytic cracking) to obtain lighter products. The different products that come out of the distillation are purified (desulfurization or sweetening) or upgraded by a reforming operation that augments their octane value. Finally, to obtain the products that will be sold, the refinery blends several of the intermediate products in order to fulfill the prescribed characteristics of the commercial products. The following drawing gives a simplified overview on the production processes in this refinery.

After the distillation, crude 1 gives 3% butane, 15% naphtha, 40% gasoil, and 15% residue. Crude 2 results in 5% butane, 20% naphtha, 35% gasoil, and 10% residue. The reforming of the naphtha gives 15% butane and 85% of reformate (reformed naphtha). The catalytic cracking of the residue results in 40% of cracked naphtha and 35% of cracked gasoil (note that these percentages do not add up to 100% because the process also produces 15% of gas, 5% coke and another type of residue that are not taken into consideration in our example). The petrol is produced with three ingredients: reformed naphtha (reformate), butane, and cracked naphtha. The diesel oil is obtained by blending sweetened gasoil, cracked gasoil, and cracked naphtha. The heating oil may contain gasoil and cracked naphtha without any restrictions on their proportions.

Crudes
|
V
Distillation -----------------------+
| |
+-------+--------+--------------+ |
| | | |
Gasoil Residue Naphta |
| | | |
V V V V
Desulf. Cat. crack Reform --> Butane
| | | | |
Sw.Gasoil C.Naphta C.Gasoil Reformate |
| | | | V
| | | | To Petrol Blending and Butane
| | | V
| | | To Petrol Blending
| | V
| | To Heating Oil and Diesel Blending
| V
| To Petrol, Heating Oil and Diesel Blending
V

To Heating Oil and Diesel Blending

Simplified representation of a refinery

Certain conditions on the quality of the petrol and diesel oil are imposed by law. There are three important characteristics for petrol: the octane value, vapor pressure and volatility. The octane value is a measure of the anti-knock power in the motor. The vapor pressure is a measure of the risk of explosion during storage, especially with hot weather. The volatility is a measure for how easy the motor is started during cold weather. Finally, the maximum sulfur content of the diesel oil is imposed by antipollution specifications. The following table summarizes the required characteristics of the final products and the composition of the intermediate ones. Fields are left empty if no particular limit applies. We work with the assumption that all these characteristics blend linearly by weight (in reality, this is only true for the sulfur contents).

Characteristics of intermediate and final products

+--------------+------+---------+-------+-------+-------+------+------+
|Characteristic|Butane|Reformate|Cracked|Cracked|Desulf |Petrol|Diesel|
| | | |naphtha|gasoil |gasoil | | oil |
+--------------+------+---------+-------+-------+-------+------+------+
|Octane value | 120 | 100 | 74 | - | - |>=94 | - |
|Vapor pressure| 60 | 2.6 | 4.1 | - | - |<=12.7| - |
|Volatility | 105 | 3 | 12 | - | - |>=17 | - |
|Sulfur (in %) | - | - | 0.12 | 0.76 | 0.03 | - |<=0.05|
+--------------+------+---------+-------+-------+-------+------+------+

In the next month the refinery needs to produce 20,000 tonnes of butane, 40,000 tonnes of petrol, 30,000 tonnes of diesel oil, and 42,000 tonnes of heating oil. 250,000 tonnes of crude 1 and 500,000 tonnes of crude 2 are available. The monthly capacity of the reformer are 30,000 tonnes, for the desulfurization 40,000 tonnes and for the cracking 50,000 tonnes. The cost of processing is based on the use of fuel and catalysts for the different operations: the costs of distillation, reforming, desulfurization, and cracking are $ 2.10, $ 4.18, $ 2.04 and $ 0.60 per tonne respectively.

==Variables==

demand Demand of each product
supply Supply of crudes
capacity Capacity of reformer, desulfer and cracker.
costs Cost for each step.
supplycomp Contents of crudes.
numvbls Number of variables
octane Octane in intermediates
vappres Vapor Pressure in intermediates
volatility Volatility of intermediates
sulfur Sulfur in intermediates
reformer Output of components from reformer
cracker Output of components from cracker
petrolspec_L Lower bounds (contents in petrol)
petrolspec_U Upper bounds (contents in petrol)
dieselspec_L Lower bounds (contents in diesel)
dieselspec_U Upper bounds (contents in diesel)

==Results==

4 Final products butane, petrol, diesel, heating oil
3 Intermediates to Petrol petbutane, reformerate, petcrknaphtha
3 Intermediates to Diesel dslgasoil, dslcrknaphtha, dslcrkgasoil
3 Intermediates to Heating Oil hogasoil, hocrknaphtha, hocrkgasoil
4 Intermediates from Distilling distbutane, naphtha, residue, gasoil
2 Intermediates from Reforming refbutane, reformerate
2 Intermediates from Cracking crknaphtha, crkgasoil
2 Crudes crude1, crude2

==Reference==

Applications of optimization... Gueret, Prins, Seveaux

<source lang="matlab">
% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com
% Copyright (c) 2005-2009 by Tomlab Optimization Inc., $Release: 7.2.0$
% Written Oct 7, 2005. Last modified Apr 12, 2009.
</source>

==Problem setup==

<source lang="matlab">
demand = [20;40;30;42]*1000;
supply = [250;500]*1000;
capacity = [30;40;50]*1000;
idistcosts = [0 4.18 .6 2.04];
crudecost = [2.1 2.1];
supplycomp = [.03 .15 .15 .40; .05 .2 .1 .35];

octane = [120 100 74];
vappres = [60 2.6 4.1];
volatility = [105 3 12];
sulfur = [.03 .12 .76];

reformer = [.15; .85];
cracker = [.40; .35];

% PRODUCTS
% butane, petrol, diesel, heating (FINAL PRODUCTS, 4) 4
% petbutane, reformerate, petcrknaphtha (IPETROL, 3) 7
% dslgasoil, dslcrknaphtha, dslcrkgasoil (IDIESEL, 3) 10
% hogasoil, hocrknaphtha, hocrkgasoil (IHO, 3) 13
% distbutane, naphtha, residue, gasoil (IDIST, 4) 17
% refbutane, reformate (IREF, 2) 19
% crknaphtha, crkgasoil (ICRACK, 2) 21
% crude1, crude2 (CRUDES, 2) 23

toms butane petrol diesel heating
toms petbutane reformerate petcrknaphtha
toms dslgasoil dslcrknaphtha dslcrkgasoil
toms hogasoil hocrknaphtha hocrkgasoil
toms distbutane naphtha residu gasoil
toms refbutane reformate
toms crknaphtha crkgasoil
toms crude1 crude2

idist = [distbutane; naphtha; residu; gasoil];
crudes = [crude1; crude2];

objective = idistcosts*idist + crudecost*crudes;

% Production constraint
con1 = cell(length(idist),1);
for i=1:length(idist)
con1{i} = {idist(i) <= supplycomp(:,i)'*crudes};
end

% Reformer constraint
iref = [refbutane; reformate];
con2 = cell(length(iref),1);
for i=1:length(iref)
con2{i} = {iref(i) <= reformer(i)*naphtha};
end

% Cracker constraint
icrack = [crknaphtha; crkgasoil];
con3 = cell(length(icrack),1);
for i=1:length(icrack)
con3{i} = {icrack(i) <= cracker(i)*residu};
end

% Product equality constraints
con4 = [];
con4{1} = {crknaphtha >= petcrknaphtha + hocrknaphtha + dslcrknaphtha};
con4{2} = {crkgasoil >= hocrkgasoil + dslcrkgasoil};
con4{3} = {gasoil >= hogasoil + dslgasoil};

% Final produts equalities
con5 = [];
con5{1} = {butane + petbutane == distbutane + refbutane};
ipetrol = [petbutane; reformerate; petcrknaphtha];
con5{2} = {petrol == sum(ipetrol)};
idiesel = [dslgasoil; dslcrknaphtha; dslcrkgasoil];
con5{3} = {diesel == sum(idiesel)};
iho = [hogasoil; hocrknaphtha; hocrkgasoil];
con5{4} = {heating == sum(iho)};

% Octane constraint
con6 = [];
con6{1} = {octane*ipetrol >= 94*petrol};
con6{2} = {vappres*ipetrol <= 12.7*petrol};
con6{3} = {volatility*ipetrol >= 17*petrol};
con6{4} = {sulfur*idiesel <= 0.05*diesel};

% Reformerate are the same
con7 = {reformerate == reformate};

bnds1 = {[naphtha; residu; gasoil] <= capacity};
bnds2 = {crudes <= supply};
bnds3 = {[butane; petrol; diesel; heating] >= demand};

bnds4 = [butane; petrol; diesel; heating;...
petbutane; reformerate; petcrknaphtha;...
dslgasoil; dslcrknaphtha; dslcrkgasoil;...
hogasoil; hocrknaphtha; hocrkgasoil;...
distbutane; naphtha; residu; gasoil;...
refbutane; reformate;...
crknaphtha; crkgasoil;...
crude1; crude2] >= 0;

constraints = {con1, con2, con3, con4, con5, con6, con7};
bounds = {bnds1, bnds2, bnds3, bnds4};

options = struct;
options.solver = 'cplex';
options.name = 'Refinery';
sol = ezsolve(objective,{bounds, constraints});

PriLev = 1;
if PriLev > 0
disp('4 Final products')
disp([' butane - ' num2str(sol.butane)])
disp([' petrol - ' num2str(sol.petrol)])
disp([' diesel - ' num2str(sol.diesel)])
disp([' heating oil - ' num2str(sol.heating)])
disp('3 Intermediates to Petrol')
disp([' petbutane - ' num2str(sol.petbutane)])
disp([' reformerate - ' num2str(sol.reformerate)])
disp([' petcrknaphtha - ' num2str(sol.petcrknaphtha)])
disp('3 Intermediates to Diesel')
disp([' dslgasoil - ' num2str(sol.dslgasoil)])
disp([' dslcrknaphtha - ' num2str(sol.dslcrknaphtha)])
disp([' dslcrkgasoil - ' num2str(sol.dslcrkgasoil)])
disp('3 Intermediates to Heating Oil')
disp([' hogasoil - ' num2str(sol.hogasoil)])
disp([' hocrknaphtha - ' num2str(sol.hocrknaphtha)])
disp([' hocrkgasoil - ' num2str(sol.hocrkgasoil)])
disp('4 Intermediates from Distilling')
disp([' distbutane - ' num2str(sol.distbutane)])
disp([' naphtha - ' num2str(sol.naphtha)])
disp([' residue - ' num2str(sol.residu)])
disp([' gasoil - ' num2str(sol.gasoil)])
disp('2 Intermediates from Reforming')
disp([' refbutane - ' num2str(sol.refbutane)])
disp([' reformerate - ' num2str(sol.reformerate)])
disp('2 Intermediates from Cracking')
disp([' crknaphtha - ' num2str(sol.crknaphtha)])
disp([' crkgasoil - ' num2str(sol.crkgasoil)])
disp('2 Crudes')
disp([' crude1 - ' num2str(sol.crude1)])
disp([' crude2 - ' num2str(sol.crude2)])
end

% MODIFICATION LOG
%
% 051007 med Created
% 060104 med Updated and corrected
% 060110 per Added documentation
% 060125 per Moved disp to end
% 090411 med Converted to tomSym
</source>

<pre>
Problem type appears to be: lp
Time for symbolic processing: 0.23983 seconds
Starting numeric solver
===== * * * =================================================================== * * *
TOMLAB - TOMLAB Development license 999007. Valid to 2011-12-31
=====================================================================================
Problem: --- 1: f_k 1175400.000000000000000000
f(x_0) 0.000000000000000000

Solver: CPLEX. EXIT=0. INFORM=1.
CPLEX Dual Simplex LP solver
Optimal solution found

FuncEv 9 Iter 9
Elapsed time: 0.002000 sec.
4 Final products
butane - 20000
petrol - 40000
diesel - 30000
heating oil - 42000
3 Intermediates to Petrol
petbutane - 6500
reformerate - 25500
petcrknaphtha - 8000
3 Intermediates to Diesel
dslgasoil - 29178.0822
dslcrknaphtha - 0
dslcrkgasoil - 821.9178
3 Intermediates to Heating Oil
hogasoil - 20821.9178
hocrknaphtha - 8000
hocrkgasoil - 13178.0822
4 Intermediates from Distilling
distbutane - 22000
naphtha - 30000
residue - 40000
gasoil - 50000
2 Intermediates from Reforming
refbutane - 4500
reformerate - 25500
2 Intermediates from Cracking
crknaphtha - 16000
crkgasoil - 14000
2 Crudes
crude1 - 0
crude2 - 440000

</pre>