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Cutlip and Shacham: Problem Solving in Chemical and Biochemical Engineering
Chapter 5
Problem Solving with MATLAB
Cheng-Liang Chen
PSELABORATORYDepartment of Chemical Engineering
National TAIWAN University
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where P = pressure in atm
V = molar volume in liters/g-mol
T = temperature in K
R = gas constant (R= 0.08206 (atm-liter/g-mol-K))T e = critical temperature in K
P e = critical pressure in atm
The compressibility factor is given by
z = P VRT
Equation (1) can be written, after considerable algebra, in terms of thecompressibility factor as a cubic equation (see Seader and Henley)
f(z) =z
3
z2
qz r= 0 (5)where r = A2B, q = B2 + B A2
A2 = 0.42747
Pr
T5/2r
B = 0.0866
PrTr
in whichPr is the reduced pressure (P/Pc) andTr is the reduced temperature
(T /Tc). Equation (5) can be solved analytically for three roots. Some of these
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roots are complex. Considering only the real roots, the sequence of calculationsinvolves the steps
C =f
33
+g
22
f = 3q 1
3 g =
27r 9q 227
IfC >0, there is one real solution for z given by
z = D+ E+ 1/3
D = (g/2 +
C)1/3 E = (g/2
C)1/3
IfC
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the maximal zk is selected as the true compressibility factor. (Note: let z = 0 forC
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%filename P5_01A_CCL
R = 0.08206;%Gas constant (L-atm/g-mol-K)
Tc = 647.4; %Critical temperature (K)
Pc = 218.3; %Critical pressure (atm)
a = 0.42747 * R ^ 2 * Tc ^ (5 / 2) / Pc;%Eq. (4-2), RK equation constb = 0.08664 * R * Tc / Pc; %Eq. (4-3),RK equation consta
Pr = 10 %1.2;%0.1; %Reduced pressure (dimensionless)
Tr = 3 %1; %Reduced temperature (dimensionless)
Asqr = 0.42747 * Pr / (Tr ^ 2.5);%Eq. (4-8)
B = 0.08664 * Pr / Tr; %Eq. (4-9)
r = Asqr * B; %Eq. (4-6)q = B ^ 2 + B - Asqr; %Eq. (4-7)
g = (-27 * r - (9 * q) - 2) / 27;%Eq. (4-12)
f = (-3 * q - 1) / 3; %Eq. (4-11)
C = (f / 3) ^ 3 + (g / 2) ^ 2; %Eq. (4-10)
if (C > 0)E1 = 0 - (g / 2) - sqrt(C); %Eq. (4-15)
D = (0 - (g / 2) + sqrt(C)) ^ (1 / 3);%Eq. (4-14)
E = sign(E1) * abs(E1) ^ (1 / 3);%Eq. (4-15)
z = D + E + 1 / 3;%Eq. (4-13), Compressibility factor (dimensionl
else
z = 0 ;
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end
P = Pr * Pc; %Pressure (atm)
T = Tr * Tc; %Temperature (K)
V = z * R * T / P %Molar volume (L/g-mol)
Pr =
10
Tr =
3
V =
0.0837
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%filename P5_01B_CCL
clear, clc, format compact, format short g
Tc = 647.4; %Critical temperature (K)
Pc = 218.3; %Critical pressure (atm)
Tr_set=[1 1.2 1.5 2 3];Pr_set(1) = 0.1;
Pr_set(2) = 0.2;
i = 2 ;
while Pr_set(i)
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end
end
end
disp( Compressibility Factor Versus Tr and Pr);
disp( Tabular Results);disp( );
disp(Pr\Tr 1.0 1.2 1.5 2 3 );
Res=[Pr_set z];
disp(Res);
subplot(1,2,1)
plot(Pr_set,z(:,1),-,Pr_set,z(:,2),+,Pr_set,z(:,3),*,...Pr_set,z(:,4),x,Pr_set,z(:,5),o,LineWidth,2);
set(gca,FontSize,14,Linewidth,2)
legend(Tr=1,Tr=1.2,Tr=1.5,Tr=2,Tr=3);
title(\bf Compressibility Factor Versus Tr and Pr,FontSize,12)
xlabel(\bf Reduced Pressure (Pr),FontSize,14);
ylabel(\bf Compressibility Factor (z),FontSize,14);
disp( Pause; Please press any key to continue ... )
pause
P_set=Pr_set.*Pc;
disp( Molar Volume Versus Tr and P);
disp( Tabular Results);
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disp( );
disp(Pr\Tr 1.0 1.2 1.5 2 3 );
Res=[P_set V];
disp(Res);
subplot(1,2,2)loglog(P_set,V(:,1),-,P_set,V(:,2),+,P_set,V(:,3),*,...
P_set,V(:,4),x,P_set,V(:,5),o,LineWidth,2);
legend(Tr=1,Tr=1.2,Tr=1.5,Tr=2,Tr=3);
set(gca,FontSize,14,Linewidth,2)
title(\bf Molar Volume Versus Tr and P,FontSize,12)
xlabel(\bf Pressure (atm),FontSize,14);ylabel(\bf Molar Volume (L/g-mol),FontSize,14);
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function [z, V] = RKfun(Tr,Pr,Tc,Pc)
R = 0.08206; %Gas constant (L-atm/g-m
a = 0.42747 * R ^ 2 * Tc ^ (5 / 2) / Pc;%Eq. (4-2), RK equation const
b = 0.08664 * R * Tc / Pc; %Eq. (4-3),RK equation constant
Asqr = 0.42747 * Pr / (Tr ^ 2.5); %Eq. (4-8)B = 0.08664 * Pr / Tr; %Eq. (4-9)
r = Asqr * B; %Eq. (4-6)
q = B ^ 2 + B - Asqr; %Eq. (4-7)
g = (-27 * r - (9 * q) - 2) / 27; %Eq. (4-12)
f = (-3 * q - 1) / 3; %Eq. (4-11)
C = (f / 3) ^ 3 + (g / 2) ^ 2; %Eq. (4-10)if (C > 0)
E1 = 0 - (g / 2) - sqrt(C); %Eq. (4-15)
D = (0 - (g / 2) + sqrt(C)) ^ (1 / 3); %Eq. (4-14)
E = sign(E1) * abs(E1) ^ (1 / 3); %Eq. (4-15)
z = D + E + 1 / 3; %Eq. (4-13), Compressibility factor (dimenselse
z = 0 ;
end
P = Pr * Pc; %Pressure (atm)
T = Tr * Tc; %Temperature (K)
V = z * R * T / P; %Molar volume (L/g-mol)
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Calculation of Flow Rate In A Pipeline
Concepts Utilized: Application of the general mechanical energy balance for
incompressible fluids, and calculation of flow rate in a pipeline for various pipediameters and lengths.
Numerical Methods: Solution of a single nonlinear algebraic equation andalternative solution using the successive substitution method.
Problem Statement:The following figure shows a pipeline that delivers water at a constanttemperature T = 60oF from point 1 where the pressure is P1= 150 psig and theelevation is z1= 0 ft to point 2 where the pressure is atmospheric and theelevation is z2= 300 ft.
The density and viscosity of the water can be calculated from the following
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equations.
= 62.122 + 0.0122T 1.54 104T2 + 2.65 107T3 2.24 1010T4
ln = 11.0318 + 1057.51
T+ 214.624
where T is in oF, is in lbm/ft3 , and is in lbm/fts.
(a) Calculate the flow rate q (in gal/min) for a pipeline with effective length of
L= 1000 ft and made of nominal 8-inch diameter schedule 40commercial steelpipe. (Solution: v= 11.61 ft/s, gpm =1811 gal/min)
(b) Calculate the flow velocities in ft/s and flow rates in gal/min for pipelines at60oF with effective lengths ofL= 500, 1000, . . . 10, 000 ft and made of nominal
4-, 5-, 6- and 8-inch schedule 40 commercial steel pipe. Use the successivesubstitution method for solving the equations for the various cases and presentthe results in tabular form. Prepare plots of flow velocity v versusD andL, andflow rate qversus D andL.
(c) Repeat part (a) at temperatures T = 40, 60, and 100oF and display the results
in a table showing temperature, density, viscosity, and flow rate.
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Solution:
The general mechanical energy balance on an incompressible liquid applied to thiscase yields
1
2v2 + gz+
gcP
+ 2
fFLv2
D = 0 (4
20)
where v is the flow velocity in ft/s, g is the acceleration of gravity given byg= 32.174 ft/s2, z =z2 z1 is the difference in elevation (ft), gc is aconversion factor (in English units gc= 32.174ft-lbm/lbfs2), P =P2P1 is thedifference in pressure lbm/ft
2), fF is the Fanning friction factor, L is the length of
the pipe (ft) andD is the inside diameter of the pipe (ft). The use of thesuccessive substitution method requires Equation (4-20) to be solved for v as
v=
gz+
gcP
12 2fFLv
2
D
The equation for calculation of the Fanning friction factor depends on theReynolds number, Re= vD/, where is the viscosity in lbm/ft-s. For laminarflow (Re
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function P5_2C_CCL
clear, clc, format short g, format compact
D_list=[4.026/12 5.047/12 6.065/12 7.981/12]; % Inside diameter of pi
T = 60; %Temperature (deg. F)
for i = 1:4D = D_list(i);
j=0;
for L=500:500:10000
j = j+1;
L_list(j)=L; % Effective length of pipe (ft)
[v(j,i),fval]=fzero(@NLEfun,[1 20],[],D,L,T);if abs(fval)>1e-10
disp([No Conv. for L = num2str(L) and D = num2str(D)]);
end
q(j,i) = v(j,i) * pi * D ^ 2 / 4* 7.481 * 60; %Flow rate (gpm)
end
end
disp( Flow Velocity (ft/s) versus Pipe Length and Diameter);
disp( Tabular Results);
disp();
disp( L\D D=4" D=5" D=6" D=8");
Res=[L_list v];
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disp(Res);
subplot(1,2,1)
plot(L_list,v(:,1),-,L_list,v(:,2),+,L_list,v(:,3),*,...
L_list,v(:,4),x,LineWidth,2);
set(gca,FontSize,14,Linewidth,2)legend( D=4", D=5", D=6", D=8");
title(\bf Flow Velocity,FontSize,12)
xlabel(\bf Pipe Length (ft),FontSize,14);
ylabel(\bf Velocity (ft/s),FontSize,14);
disp( Pause; Please press any key to continue ... )
pausedisp( Flow Rate (gpm) versus Pipe Length and Diameter);
disp( Tabular Results); disp();
disp( L\D D=4" D=5" D=6" D=8");
Res=[L_list q(:,1) q(:,2) q(:,3) q(:,4)];
disp(Res);
subplot(1,2,2)
plot(L_list,q(:,1),-,L_list,q(:,2),+,L_list,q(:,3),*,...
L_list,q(:,4),x,Linewidth,2);
set(gca,FontSize,14,Linewidth,2)
legend( D=4", D=5", D=6", D=8");
title(\bf Flow rate,FontSize,14)
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xlabel(\bf Pipe Length (ft),FontSize,14);
ylabel(\bf Flow rate (gpm),FontSize,14);
function fv = NLEfun(v,D,L,T)epsilon = 0.00015;%Surface rougness of the pipe (ft)
rho = 62.122+T*(0.0122+T*(-0.000154+T*(0.000000265-...
(T*0.000000000224)))); %Fluid density (lb/cu. ft
deltaz = 300; %Elevation difference (ft)
deltaP = -150; %Pressure difference (psi)
vis = exp(-11.0318 + 1057.51 / (T + 214.624)); %Fluid viscosity (lbm/pi = 3.1416; %The constant pi
eoD = epsilon / D; %Pipe roughness to diameter ratio (dimensionless)
Re = D * v * rho / vis; %Reynolds number (dimesionless)
if (Re < 2100) %Fanning friction factor (dimensionless)
fF = 16 / Re;
else
fF=1/(16*log10(eoD/3.7-(5.02*log10(eoD/3.7+14.5/Re)/Re))^2);
end
fv=v-sqrt((32.174*deltaz+deltaP*144*32.174/rho)...
/(0.5-(2*fF*L/D))); %velocity (ft/s)
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Flow Velocity (ft/s) versus Pipe Length and Diameter
Tabular Results
L\D D=4" D=5" D=6" D=8"
500 10.773 12.516 14.15 17.035
1000 7.4207 8.6048 9.7032 11.6131500 5.9721 6.9243 7.8051 9.3295
2000 5.1188 5.9361 6.6912 7.9953
2500 4.5409 5.2674 5.9382 7.0953
3000 4.1168 4.7769 5.3861 6.4362
3500 3.7888 4.3975 4.9592 5.927
4000 3.5255 4.093 4.6166 5.51854500 3.3082 3.8416 4.3338 5.1815
5000 3.1249 3.6297 4.0953 4.8973
5500 2.9677 3.4478 3.8907 4.6535
6000 2.8309 3.2896 3.7128 4.4415
6500 2.7106 3.1504 3.5561 4.2548
7000 2.6036 3.0266 3.4169 4.0889
7500 2.5077 2.9156 3.292 3.9402
8000 2.4211 2.8154 3.1793 3.8059
8500 2.3424 2.7244 3.0769 3.6838
9000 2.2706 2.6412 2.9832 3.5723
9500 2.2046 2.5648 2.8972 3.4698
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9000 90.099 164.71 268.65 557.05
9500 87.48 159.94 260.91 541.07
10000 85.064 155.54 253.76 526.33
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Adiabatic Operation of A Tabular Reactor for
Cracking of Acetone
Concepts Utilized: Calculation of the conversion and temperature profile in anadiabatic tubular reactor. Demonstration of the effect of pressure and heatcapacity change on the conversion in the reactor.
Numerical Methods:
Solution of simultaneous ordinary differential equations.
Problem Statement:
The irreversible, vapor-phase cracking of acetone (A) to ketene (B) and methane(C) that is given by the reaction
CH3COCH3 CH2CO+ CH4is carried out adiabatically in a tubular reactor. The reaction is first order withrespect to acetone and the specific reaction rate can be expressed by
ln(k) = 34.3434222
T (4-26)
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wherek is in s1 andT is in K. The acetone feed flow rate to the reactor is 8000kg/hr, the inlet temperature is T= 1150 K and the reactor operates at theconstant pressure ofP= 162 kPa (1.6 atm). The volume of the reactor is 4 m3.
The material balance equations for the plug-flow reactor are given by
dFAdV
= rA (4-27)
dFBdV
= rA (4-28)dFC
dV =
rA
(4-29)
where FA, FB, andFCare flow rates of acetone, ketene, and methane in g-mol/s,respectively and rA is the reaction rate ofA in g-mol/m
3s. The reaction is firstorder with respect to acetone, thus
rA= kCAwhere CA is the concentration of acetone in g-mol/m
3. For a gas phase reactor,using the appropriate units of the gas constant, the concentration of the acetonein g-mol/m3 is obtained by
CA=
1000yAP
8.31T
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reaction and the molar heat capacities.
H = 80770 + 6.8(T 298) 0.00575(T2 2982) 1.27 106(T3 2983)CpA = 26.60 + 0.183T
45.86
106T2
CpB = 20.04 + .0945T 30.95 106T2CpC = 13.39 + 0.077T 18.71 106T2
Galculate the final conversion and the final temperature ofP = 1.6, 1.8, . . . , 5.0atm, for acetone feed rates ofFA0= 10, 20, 30, 35, and38.3 mol/s where nitrogenis fed to maintain the total feed rate38.3 mol/s in all cases. Present the results intabular form and prepare plots of final conversion versus P andFA0 and finaltemperature versus P andFA0.
function P5_03C_CCL
clear, clc, format short g, format compactFA0set = [10 20 30 35 38.3]; %Feed rate of acetone in kg-mol/s
P_set(1)=1.6;
i=1;
while P_set(i)
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end
n_P = size(P_set);
for i=1:5
FA0=FA0set(i);
for j=1:n_PP=P_set(j)*101.325; % Pressure in kPa
y0=[FA0; 0; 0; 1035; 0];
[V,y] = ode45(@ODEfun,[0 4],y0,[],FA0,P);
Xfin(j,i)=y(end,5);
Tfin(j,i)=y(end,4);
endend
% -------------------------------------------------------------------
disp( Final Conversion versus FA0 and Pressure);
disp( Tabular Results);
disp();
disp( Pressure FA0=10 FA0=20 FA0=30 FA0=35 FA0=38.3 );
Res=[P_set Xfin(:,1) Xfin(:,2) Xfin(:,3) Xfin(:,4) Xfin(:,5)];
disp(Res);
subplot(1,2,1)
plot(P_set,Xfin(:,1),-,P_set,Xfin(:,2),+,P_set,Xfin(:,3),*,...
P_set,Xfin(:,4),x,P_set,Xfin(:,5),o,LineWidth,2);
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set(gca,FontSize,14,Linewidth,2)
legend(FA0=10,FA0=20,FA0=30,FA0=35,FA0=38.3);
title(\bf Final Conversion versus FA0 and Pressure,FontSize,12)
xlabel(\bf Pressure (atm),FontSize,14);
ylabel(\bf Final Conversion,FontSize,14);disp( Pause; Please press any key to continue ... )
pause
disp( Final Temperature versus FA0 and Pressure);
disp( Tabular Results);
disp();
disp( Pressure FA0=10 FA0=20 FA0=30 FA0=35 FA0=38.3 );Res=[P_set Tfin(:,1) Tfin(:,2) Tfin(:,3) Tfin(:,4) Tfin(:,5)];
disp(Res);
subplot(1,2,2)
plot(P_set,Tfin(:,1),-,P_set,Tfin(:,2),+,P_set,Tfin(:,3),*,...
P_set,Tfin(:,4),x,P_set,Tfin(:,5),o,Linewidth,2);
set(gca,FontSize,14,Linewidth,2)
legend(FA0=10,FA0=20,FA0=30,FA0=35,FA0=38.3);
title(\bf Final Temperature versus FA0 and Pressure,FontSize,12)
xlabel(\bf Pressure (atm),FontSize,14);
ylabel(\bf Temperature (K),FontSize,14);
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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function dYfuncvecdV = ODEfun(V,Yfuncvec,FA0,P);
FA = Yfuncvec(1);
FB = Yfuncvec(2);
FC = Yfuncvec(3);
T = Yfuncvec(4);XA = Yfuncvec(5);
k = 8.2E14 * exp(-34222 / T); %Reaction rate constant in s-1
FN = 38.3 - FA0; %Feed rate of nitrogene in kg-mol/s
yA = FA / (FA + FB + FC + FN); %Mole fraction of acetone
CA = yA * P * 1000 / (8.31 * T); %Concentration of acetone in k-mol/m
yB = FB / (FA + FB + FC + FN); %Mole fraction of keteneyC = FC / (FA + FB + FC + FN); %Mole fraction of methane
rA = -k * CA; %Reaction rate in kg-mole/m3-s
deltaH = 80770 + 6.8 * (T - 298) - .00575 * (T ^ 2 - 298 ^ 2)...
- 1.27e-6 * (T ^ 3 - 298 ^ 3);
CpA = 26.6 + .183 * T - 45.86e-6 * T ^ 2; %Heat capacity of acetone
CpB = 20.04 + 0.0945 * T - 30.95e-6 * T ^ 2; %Heat capacity of ketene
CpC = 13.39 + 0.077 * T - 18.71e-6 * T ^ 2; %Heat capacity of methane
CpN = 6.25 + 8.78e-3 * T - 2.1e-8 * T ^ 2; %Heat capacity of nitrogen
dFAdV = rA; %Differential mass balance on acetone
dFBdV = -rA; %Differential mass balance on ketene
dFCdV = -rA; %Differential mass balance on methane
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dTdV = (-deltaH)*(-rA)/(FA*CpA+FB*CpB+FC*CpC+FN*CpN);
%Differential enthalpy balance
dXAdV = -rA / FA0; %Conversion of acetone
dYfuncvecdV = [dFAdV; dFBdV; dFCdV; dTdV; dXAdV];
Final Conversion versus FA0 and Pressure
Tabular Results
Pressure FA0=10 FA0=20 FA0=30 FA0=35
1.6 0.31358 0.27759 0.26394 0.25964
1.8 0.32043 0.28346 0.26946 0.26505
2 0.32651 0.28867 0.27436 0.269862.2 0.33197 0.29335 0.27877 0.27419
2.4 0.33693 0.2976 0.28277 0.27811
2.6 0.34147 0.30149 0.28643 0.2817
2.8 0.34565 0.30508 0.2898 0.28501
3 0.34952 0.3084 0.29293 0.288083.2 0.35313 0.31149 0.29584 0.29094
3.4 0.3565 0.31438 0.29856 0.29361
3.6 0.35967 0.3171 0.30112 0.29612
3.8 0.36266 0.31966 0.30353 0.29849
4 0.36548 0.32208 0.30581 0.30073
4.2 0.36815 0.32438 0.30797 0.30285
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0.30752
Pause; Please press any key to continue ...
Final Temperature versus FA0 and Pressure
Tabular Results
Pressure FA0=10 FA0=20 FA0=30 FA0=35Columns 1 through 5
1.6 911.84 908.14 907.47 907.47
1.8 909.02 905.32 904.66 904.67
2 906.5 902.81 902.16 902.18
2.2 904.24 900.55 899.91 899.93
2.4 902.18 898.49 897.87 897.892.6 900.3 896.61 895.99 896.02
2.8 898.56 894.87 894.26 894.29
3 896.94 893.26 892.66 892.69
3.2 895.44 891.76 891.16 891.2
3.4 894.03 890.35 889.76 889.8
3.6 892.7 889.03 888.44 888.48
3.8 891.45 887.78 887.19 887.24
4 890.27 886.59 886.02 886.07
4.2 889.15 885.47 884.9 884.95
4.4 888.08 884.4 883.84 883.89
4.6 887.07 883.39 882.82 882.88
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Column 6
907.53
904.74
902.25900.01
897.97
896.1
894.38
892.78
891.29889.89
888.58
887.34
886.17
885.05883.99
882.98
882.02
881.1
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Correlation of The Physical Properties of Ethane
Concepts Utilized:
Correlations for heat capacity, vapor pressure, and liquid viscosity for an ideal gas.
Numerical Methods:
Polynomial, multiple linear, and nonlinear regression of data with linearization andtransformation functions.
Problem Statement:
Tables F-l through F-4 of Appendix F present values for different propertiesof ethane (ideal gas heat capacity, vapor pressure, and liquid viscosity) as functionof temperature. Various regression models will be fitted to the properties ofAppendix using MATLAB.
(a) Construct a MATLAB function which solves the linear regression problemXb = y, where X is the matrix of the independent variable values, y is thevector of dependent variable values, and b is the vector of the linear regressionmodel parameters. The input parameters of the function are X, y, and alogical variable which indicates whether there is a free parameter. The returnedparameters are: and the respective confidence intervals, the calculated values
of the dependent variable ycalc, the linear correlation coefficient R2
, and the
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variance. Test the function by fitting the Wagner equation to vapor pressuredata of ethane from TableF-3 of Appendix F.
(b) Fit 3rd- and 5th-degree polynomials to the heat capacity of ethane for for the
data given in Tables F-l and F-2 of Appendix F by using the multiple linearregression function developed in (a). Compare the quality of the representationof the various data sets with the polynomials of different degrees.
(c) Fit the Antoine equation to liquid viscosity of ethane given in Table F-4 ofAppendix F.
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% filename P5_04A_CCL
clear, clc, format short g, format compact
prob_title = ([Vapor Pressure Correlation for Ethane]);
ind_var_name=[\bf Functions of Reduced Temp.];
dep_var_name=[\bf Logarithm of Reduced Pressure];fname=input(Please enter the data file name > ); % VPfile.txt
xyData=load(fname);
X=xyData(:,2:end);
y=xyData(:,1);
[m,n]=size(X);
freeparm=input(Input 1 if there is a free par., otherwise input 0>)[Beta,ConfInt, ycal,Var, R2]=MlinReg(X,y,freeparm);
disp([ Results, prob_title]);
Res=[];
if freeparm==0, nparm = n-1; else nparm = n; end
for i=0:nparm
if freeparm==0; ii=i+1; else ii=i; end
Res=[Res; ii Beta(i+1) ConfInt(i+1)];
end
disp( Parameter No. Beta Conf_int);
disp(Res);
disp([ Variance , num2str(Var)]);
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disp([ Correlation Coefficient , num2str(R2)]);
disp( Pause; Please press any key to continue ... )
pause
subplot(1,2,1)
plot(y,y-ycal,*,Linewidth,2) % residual plotset(gca,FontSize,14,Linewidth,2)
title([\bf Residual, prob_title],FontSize,12)
xlabel([dep_var_name \bf (Measured)],FontSize,14)
ylabel(\bf Residual,FontSize,14)
disp( Pause; Please press any key to continue ... )
pausesubplot(1,2,2)
plot(X,ycal, r-,X,y,bo,Linewidth,2)
title([\bf Cal/Exp Data prob_title],FontSize,12)
set(gca,FontSize,14,Linewidth,2)
xlabel([ind_var_name],FontSize,14)
ylabel([dep_var_name],FontSize,14)
% VPfile.txt : data file provided elsewhere
Please enter the data file name > VPfile.txt
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Input 1 if there is a free parameter, otherwise input 0> 0
Results,Vapor Pressure Correlation for Ethane
Parameter No. Beta Conf_int
1 -6.4585 0.09508
2 1.2895 0.21514
3 -1.6712 0.26773
4 -1.2599 0.29417
Variance 9.3486e-005
Correlation Coefficient 1
Pause; Please press any key to continue ...
Pause; Please press any key to continue ...
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%
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% filename P5_04B_CCL
clear, clc, format short g, format compact
prob_title = ([Heat Capacity of Ethane]);
ind_var_name=[\bf Normalized Temp];
dep_var_name=[\bf Heat Capacity J/kmol*K ];fname=input(Please enter the data file name > ); % CPfile.txt
xyData=load(fname);
X=xyData(:,2:end);
y=xyData(:,1);
[m,n]=size(X);
freeparm=input(Input 1 if there is a free par., otherwise input 0>)[Beta, ConfInt,ycal, Var, R2]=MlinReg(X,y,freeparm);
disp([ Results, prob_title]);
Res=[];
if freeparm==0, nparm = n-1; else nparm = n; end
for i=0:nparm
if freeparm, ii=i+1; else ii=i; end
Res=[Res; ii Beta(i+1) ConfInt(i+1)];
end
disp( Parameter No. Beta Conf_int);
disp(Res);
disp([ Variance , num2str(Var)]);
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Chen CL 41
C l Ch i l E ilib i
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Complex Chemical Equilibrium
by Gibbs Energy Minimization
Concepts Utilized:Formulation of the chemical equilibrium problem as a Gibbs energy minimizationproblem with atomic balance constraints. The use of Lagrange multipliers tointroduce the constraints into the objective function. Conversion of theminimization problem into a system of nonlinear algebraic equations.
Numerical Methods:
Solution of a system of nonlinear algebraic equations.
Problem Statement:
Ethane reacts with steam to form hydrogen over a cracking catalyst at a
temperature ofT= 1000 K and pressure ofP = 1 atm. The feed contains 4moles ofH2O per mole ofCH4. Balzisher et al. suggest that only the compoundsshown in Table 4-10 are present in the equilibrium mixture (assuming that nocarbon is deposited). The Gibbs energies of formation of the various compoundsat the temperature of the reaction (1000K) are also given in Table 4-10. Theequilibrium composition of the effluent mixture is to be calculated using these
data.
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Table 4-10: Compounds Present in Effluentof Steam Cracking Reactor
Gibbs Energy Feed Effluent
No. Comp. kcal/gm-mol gm-mol Ini. Est.1 CH4 4.61 0.001
2 C2H4 28.249 0.001
3 C2H2 40.604 0.001
4 CO2
94.61 0.993
5 CO 47.942 1.6 O2 0. 0.0001
7 H2 0. 5.992
8 H2O 46.03 4 1.
9 C2H6 26.13 1 0.001
Formulate the problem as a constrained minimization problem. Introduce theconstraints into the objective function using Lagrange multipliers and differentiatethis function to obtain a system of nonlinear algebraic equations.
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Solution:
The objective function to be minimized is the total Gibbs energy given by
min
ni
G
RT
= i=1
cni Goi
RT
+ ln ni
ni (4 49)
where ni is the number of moles of component i, c is the total number ofcompounds, R is the gas constant, and Go is the Gibbs energy of pure componenti at temperature T. The minimization of Equation (4-49) must be carried outsubject to atomic balance constraints
Oxygen Balance 0 =g1 = 2n4+ n5+ 2n6+ n7 4 (4 50)Hydrogen Balance 0 =g2 = 4n1+ 4n2+ 2n3+ 2n7+ 2n8+ 6n9 14 (4 51)
Carbon Balance 0 =g3 = n1+ 2n2+ 2n3+ n4+ n5+ 2n9 2 (4 52)
The identification of the various components is given in Table 4-10. These threeconstraints can be introduced into the objective functions using Lagrangemultipliers: 1, 2,3. The extended objective function is
minni,j
F = i=1
cni GoiRT
+ ln ni
ni+
3
j=1
jgj (4 53)
Chen CL 44
Th diti f i i f this f ti t ti l i t is th t ll th
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The condition for minimum of this function at a particular point is that all thepartial derivatives ofF with respect to ni andj vanish at this point. The partialderivative ofF with respect to n1 for example, is
F
n1 = Go
1RT + ln
n1ni + 4
2+ 3 (4 54)
The other partial derivatives with respect to ni can be obtained similarly. If it isexpected that the amount of a particular compound at equilibrium is very close tozero, it is preferable to rewrite the equation in a form that does not require
calculation of the logarithm of a very small number. Rearranging Equation (4-54),for example, yields
n1
niexp
Go1RT
+ 42+ 3
= 0
The partial derivatives ofFwith respect to 1, 2, and 3 areg1, g2, andg3,respectively.
Chen CL 45
function P5 05A1 CCL
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function P5_05A1_CCL
clear, clc, format short g, format compact
xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001 ...
0.001 0.001 0.001 0.0001]; % initial guess vector
disp(Variable values at the initial estimate);
fguess = MNLEfun(xguess);
disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([xint2str(i) num2str(xguess(i)) num2str(fguess(i))]);
end
options = optimset(Diagnostics,[off],TolFun,[1e-9],...TolX,[1e-9]);
xsolv = fsolve(@MNLEfun,xguess,options);
disp(Variable values at the solution);
fsolv = MNLEfun(xsolv);
disp( Variable Value Function Value)
for i=1:size(xguess,2);disp([xint2str(i) num2str(xsolv(i)) num2str(fsolv(i))])
end
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function fx = MNLEfun(x)
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function fx = MNLEfun(x)
lamda1 = x(1);
lamda2 = x(2);
lamda3 = x(3);
H2 = x(4);
H2O = x(5);
CO = x(6);
CO2 = x(7);
CH4 = x(8);
C2H6 = x(9);
C2H4 = x(10);C2H2 = x(11);
O2 = x(12);
R = 1.9872;
sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;
fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4; %Oxygen balance
fx(2,1) = 4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo
fx(4,1) = log(H2 / sum) + 2 * lamda2;
fx(5,1) = -46.03 / R + log(H2O / sum) + lamda1 + 2 * lamda2;
fx(6,1) = -47.942 / R + log(CO / sum) + lamda1 + lamda3;
fx(7,1) = -94.61 / R + log(CO2 / sum) + 2 * lamda1 + lamda3;
Chen CL 47
fx(8 1) = 4 61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;
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fx(8,1) = 4.61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;
fx(9,1) = 26.13 / R + log(C2H6 / sum) + 6 * lamda2 + 2 * lamda3;
fx(10,1) = 28.249 / R + log(C2H4 / sum) + 4 * lamda2 + 2 * lamda3;
fx(11,1) = C2H2-exp(-(40.604/R+2*lamda2+2*lamda3))*sum;
fx(12,1) = O2 - exp(-2 * lamda1) * sum;
Variable values at the initial estimate
Variable Value Function Value
x1 10 -0.0138
x2 10 0
x3 10 0x4 5.992 19.5944
x5 1 4.6407
x6 1 -6.3214
x7 0.993 -19.8127
x8 0.001 43.2161
x9 0.001 84.0454
x10 0.001 65.1117
x11 0.001 0.001
x12 0.0001 9.9981e-005
Optimization terminated: no further progress can be made.
Trust-region radius less than 2*eps.
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Problem may be ill-conditioned or Jacobian may be inaccurate
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Problem may be ill conditioned or Jacobian may be inaccurate.
Try using exact Jacobian or check Jacobian for errors.
Variable values at the solution
Variable Value Function Value
x1 10+7.44066e-015i -0.013798+4.9091e-
x2 10-3.7436e-008i -0.015082-0.0002223
x3 10-1.4705e-008i -0.0054569-7.4103e-
x4 5.992+8.1803e-010i 19.5947+4.04813e
x5 1+8.1803e-010i 4.6411+4.0488e-006i
x6 1+8.182e-010i -6.3211+4.109e-006i
x7 0.993+8.182e-010i -19.8124+4.10898ex8 0.00029165+6.9845e-010i 41.9842+6.3
x9 -0.00037496-3.7053e-005i 83.0696-3.
x10 -5.5221e-009-1.0635e-009i 53.0235-
x11 0.0010003+8.1796e-010i 0.0010003+8
x12 0.00010027+8.1823e-010i 0.00010025
Chen CL 49
function P5 05A2 CCL
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function P5_05A2_CCL
clear, clc, format short g, format compact
xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...
0.001 0.001 0.001 0.0001]; % initial guess vector
disp(Variable values at the initial estimate);
fguess = MNLEfun(xguess);
disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([xint2str(i) num2str(xguess(i)) num2str(fguess(i))]);
end
options = optimset(Diagnostics,off,TolFun,1e-9,...TolX,1e-16,NonlEqnAlgorithm,gn);
xsolv = fsolve(@MNLEfun,xguess,options);
disp(Variable values at the solution);
fsolv=MNLEfun(real(xsolv));
disp( Variable Value Function Value)
for i=1:size(xguess,2);disp([x int2str(i) num2str(real(xsolv(i))) num2str(fsolv(i))
end
Chen CL 50
function fx = MNLEfun(x)
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function fx = MNLEfun(x)
lamda1 = x(1);
lamda2 = x(2);
lamda3 = x(3);
H2 = x(4);
H2O = x(5);
CO = x(6);
CO2 = x(7);
CH4 = x(8);
C2H6 = x(9);
C2H4 = x(10);C2H2 = x(11);
O2 = x(12);
R = 1.9872;
sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;
fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4;
fx(2,1) =4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo
fx(4,1) = log(H2 / sum) + 2 * lamda2;
fx(5,1) = -46.03 / R + log(H2O / sum) + lamda1 + 2 * lamda2;
fx(6,1) = -47.942 / R + log(CO / sum) + lamda1 + lamda3;
fx(7,1) = -94.61 / R + log(CO2 / sum) + 2 * lamda1 + lamda3;
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Chen CL 52
Variable Value Function Value
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Variable Value Function Value
x1 24.4197 0
x2 0.25306 1.7764e-015
x3 1.5598 0
x4 5.3452 -1.1102e-016
x5 1.5216 2.1094e-015
x6 1.3885 2.2204e-016
x7 0.54492 -3.3307e-015
x8 0.066564 0
x9 1.6707e-007 1.3323e-015
x10 9.5412e-008 1.3323e-015x11 3.157e-010 1.4387e-020
x12 7.0058e-021 1.5466e-021
Chen CL 53
function P5 05B CCL
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function P5_05B_CCL
clear, clc, format short g, format compact
xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...
0.001 0.001 0.001 0.0001]; % initial guess vector
disp(Variable values at the initial estimate);
fguess = MNLEfun(xguess);
disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([x int2str(i) num2str(xguess(i)) num2str(fguess(i))])
end
pote=[0 0 0 2 2 2 2 2 2 2 2 2];tol=1e-9;
maxit=100;
derfun=0;
print=0;
[xsolv,y,dy,info]=conles(@MNLEfun,xguess,pote,[],...
tol,maxit,derfun,print);disp(Variable values at the solution);
fsolv = MNLEfun(xsolv);
disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([x int2str(i) num2str(xsolv(i)) num2str(fsolv(i))])
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Chen CL 55
function fx = MNLEfun(x)
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function fx MNLEfun(x)
lamda1 = x(1);
lamda2 = x(2);
lamda3 = x(3);
H2 = x(4);
H2O = x(5);
CO = x(6);
CO2 = x(7);
CH4 = x(8);
C2H6 = x(9);
C2H4 = x(10);C2H2 = x(11);
O2 = x(12);
R = 1.9872;
sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;
fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4;
fx(2,1) = 4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo
fx(4,1) = log(H2 / sum) + 2 * lamda2;
fx(5,1) = -46.03 / R + log(H2O / sum) + lamda1 + 2 * lamda2;
fx(6,1) = -47.942 / R + log(CO / sum) + lamda1 + lamda3;
fx(7,1) = -94.61 / R + log(CO2 / sum) + 2 * lamda1 + lamda3;
Chen CL 56
fx(8,1) = 4.61 / R + log(CH4 / sum) + 4 * lamda2 + lamda3;
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fx(8,1) 4.61 / R log(CH4 / sum) 4 lamda2 lamda3;
fx(9,1) = 26.13 / R + log(C2H6 / sum) + 6 * lamda2 + 2 * lamda3;
fx(10,1) = 28.249 / R + log(C2H4 / sum) + 4 * lamda2 + 2 * lamda3;
fx(11,1) = C2H2-exp(-(40.604/R+2*lamda2+2*lamda3))*sum;
fx(12,1) = O2 - exp(-2 * lamda1) * sum;
Variable values at the initial estimate
Variable Value Function Value
x1 10 -0.0138
x2 10 0
x3 10 0x4 5.992 19.5944
x5 1 4.6407
x6 1 -6.3214
x7 0.993 -19.8127
x8 0.001 43.2161
x9 0.001 84.0454
x10 0.001 65.1117
x11 0.001 0.001
x12 0.0001 9.9981e-005
Variable values at the solution
Variable Value Function Value
Chen CL 57
x1 24.4197 0
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x2 0.25306 -1.7764e-015
x3 1.5598 0
x4 5.3452 -1.1102e-016
x5 1.5216 -1.6653e-015
x6 1.3885 -1.5543e-015
x7 0.54492 5.5511e-015
x8 0.066564 -4.4409e-016
x9 1.6707e-007 -1.6964e-013
x10 9.5412e-008 -2.589e-013
x11 3.157e-010 9.3058e-025x12 5.4592e-021 -3.9873e-035
Chen CL 58
function P5_05C_CCL
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clear, clc, format short g, format compact
xguess = [10. 10. 10. 5.992 1. 1. 0.993 0.001...
0.001 0.001 0.001 0.0001]; % initial guess vector
disp(Variable values at the initial estimate);
fguess = MNLEfun(xguess);
disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([x int2str(i) num2str(xguess(i)) num2str(fguess(i))]);
end
pote=[0 0 0 2 2 2 2 2 2 2 2 2];tol=1e-9; maxit=100;
derfun=0; print=0;
[xsolv,y,dy,info] = conles(@MNLEfun,xguess,pote,[],...
tol,maxit,derfun,print);
disp(Variable values at the solution);
fsolv = MNLEfun(xsolv);disp( Variable Value Function Value)
for i=1:size(xguess,2);
disp([x int2str(i) num2str(xsolv(i)) num2str(fsolv(i))]);
end
H2 =xsolv(4);
Chen CL 59
H2O =xsolv(5);
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;
CO =xsolv(6);
CO2 =xsolv(7);
CH4 =xsolv(8);
C2H6=xsolv(9);
C2H4=xsolv(10);
C2H2=xsolv(11);
O2 =xsolv(12);
R = 1.9872;
sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;
G_O2 = O2 * log(abs(O2 / sum));G_H2 = H2 * log(H2 / sum);
G_H2O = H2O * (-46.03 / R + log(H2O / sum));
G_CO = CO * (-47.942 / R + log(CO / sum));
G_CO2 = CO2 * (-94.61 / R + log(CO2 / sum));
G_CH4 = CH4 * (4.61 / R + log(abs(CH4 / sum)));
G_C2H6 = C2H6 * (26.13 / R + log(abs(C2H6 / sum)));G_C2H4 = C2H4 * (28.249 / R + log(abs(C2H4 / sum)));
G_C2H2 = C2H2 * (40.604 / R + log(abs(C2H2 / sum)));
ObjFun=G_H2+G_H2O+G_CO+G_O2+G_CO2...
+G_CH4+G_C2H6+G_C2H4+G_C2H2
Chen CL 60
function fx = MNLEfun(x)
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lamda1 = x(1);
lamda2 = x(2);
lamda3 = x(3);
H2 = x(4);
H2O = x(5);
CO = x(6);
CO2 = x(7);
CH4 = x(8);
C2H6 = x(9);
C2H4 = x(10);C2H2 = x(11);
O2 = x(12);
R = 1.9872;
sum = H2 + O2 + H2O + CO + CO2 + CH4 + C2H6 + C2H4 + C2H2;
fx(1,1) = 2 * CO2 + CO + 2 * O2 + H2O - 4; %Oxygen balance
fx(2,1) = 4*CH4+4*C2H4+2*C2H2+2*H2+2*H2O+6*C2H6-14; %Hydrogen balancefx(3,1) = CH4 + 2 * C2H4 + 2 * C2H2 + CO2 + CO + 2 * C2H6 - 2; %Carbo
fx(4,1) = log(H2 / sum) + 2 * lamda2;
fx(5,1) = -46.03 / R + log(H2O / sum) + lamda1 + 2 * lamda2;
fx(6,1) = -47.942 / R + log(CO / sum) + lamda1 + lamda3;
fx(7,1) = -94.61 / R + log(CO2 / sum) + 2 * lamda1 + lamda3;
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Chen CL 62
x1 24.4197 0
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x2 0.25306 -1.7764e-015
x3 1.5598 0
x4 5.3452 -1.1102e-016
x5 1.5216 -1.6653e-015
x6 1.3885 -1.5543e-015
x7 0.54492 5.5511e-015
x8 0.066564 -4.4409e-016
x9 1.6707e-007 -1.6964e-013
x10 9.5412e-008 -2.589e-013
x11 3.157e-010 9.3058e-025x12 5.4592e-021 -3.9873e-035
ObjFun =
-104.34
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Thank You for Your Attention
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