# Refinery Optimization

## Contents

## 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

% 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.

## Problem setup

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 residue gasoil toms refbutane reformate toms crknaphtha crkgasoil toms crude1 crude2 idist = [distbutane; naphtha; residue; 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)*residue}; 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; residue; 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; residue; 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.residue)]) 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

Problem type appears to be: lp Starting numeric solver ===== * * * =================================================================== * * * TOMLAB - Tomlab Optimization Inc. Development license 999001. Valid to 2010-02-05 ===================================================================================== 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 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