How to find an unknown in an integral equation

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Narasimha Varma Hemadri 2021-1-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/716488-how-to-find-an-unknown-in-an-integral-equation

评论： Alan Stevens 2021-1-15

采纳的回答： Alan Stevens

Hi, I am experiencing errors in calculating the following. Kindly help me solve these equations.

a = 0.0859;

b = -0.1650;

c = 0.0344;

lambda = 2.8e-9;

alphaA = 0.52;

T = 303;

F = 96487;

R = 8.314;

Ioref = 0.0013;

K = 7.5e-4;

Icell = 0.02;

A = 1000;

Cmeoh = (a .* (z.^2)) + (b.*z) + c;

fun = @(z,nA) Cmeoh./(Cmeoh + (lambda.*exp((alphaA.*nA.*F)./(R.*T))));

eqn = Icell == A.*Ioref.*K.*exp((2.*alphaA.*nA.*F)./(R.*T)).*(integral(fun,0.015,0.0173));

"nA" is the unknown and it has to be found from the above equation.

Thank you for your time and help.

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Answer 1

Alan Stevens 2021-1-14

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https://ww2.mathworks.cn/matlabcentral/answers/716488-how-to-find-an-unknown-in-an-integral-equation#answer_597788

在 MATLAB Online 中打开

Here's one way:

nA0 = 1; % Initial guess
nA = fzero(@fn, nA0);
disp(nA)
 
function Z = fn(nA)
    alphaA = 0.52;
    T = 303;
    F = 96487;                  
    R = 8.314; 
    Ioref = 0.0013;
    K = 7.5e-4;
    Icell = 0.02;
    A = 1000;
    a = 0.0859;
    b = -0.1650;
    c = 0.0344;
    lambda = 2.8e-9;            
    Cmeoh = @(z) a*z.^2 + b*z + c;
    fun = @(z) Cmeoh(z)./(Cmeoh(z) + (lambda.*exp((alphaA.*nA.*F)./(R.*T))));
    I = integral(fun,0.015,0.0173);
    Z = A.*Ioref.*K.*exp((2.*alphaA.*nA.*F)./(R.*T)).*I - Icell;
end

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Narasimha Varma Hemadri 2021-1-14

New Text Document.txt

Thanks a lot. It worked !

Now I want to do the same calcualtion for different values of T, Icell, Cb. Kindly help me out.

T has 5 values, Icell has 10 values and Cb has 5. So, you will see I have used 3 nested for loops.

I have to find "nA" for all these different inputs. I am attaching the text file which contains the code.

Thank you for your time !

Alan Stevens 2021-1-15

在 MATLAB Online 中打开

More like this, with the loops around the fzero function:

nA = zeros(5,10,5);
nA0 = 1; % Initial guess
for t = 1:1:5
    for i = 1:1:10
        for j = 1:1:5
nA(t,i,j) = fzero(@(nA)fn(nA,t,i,j), nA0);
        end
    end
end
disp(nA)
 
function Z = fn(nA,t,i,j)
    alphaA = 0.52;
    T = [303,313,323,333,343];      % Kelvin
    Cb = [0.05,0.1,0.2,0.3,0.5];    %Concentration of Methanol
    delA = 0.0023;                  % cm
    delB = 0.015; 
    n = 6;
    F = 96487;                  
    R = 8.314; 
    K = 7.5e-4;
    A = 1000;
    Dm = 4.9e-6.*exp(-2436.*((1/333)-(1./T)));
    K2 = 0.8;
    K1 = 0.8;
    Db = 8.7e-6;            % cm^2/s
    delM = 0.018;           % cm
    Da = 2.8e-5.*exp(-2436.*((1/353)-(1./T)));
    xmeoh = Cb./(Cb+55.55);
    Emeoh = 2.5.*xmeoh;
    Icell = linspace(0.02,2,10); %0.02:0.02:0.2;  *******Only 10 values for i             
    Ioref = 9.425e-3.*exp((35570/R).*((1/353)-(1./T)));
    lambda = 2.8e-9;            % mol/cm^3
                      
C1A = (delA.*Dm(t).*K2.*((Db.*Cb(j))-(Icell(i).*delB)./(12*F)) + delM.*Da(t).*((Db.*Cb(j))-(1+6.*Emeoh(j)).*(Icell(i).*delB)./(6*F)))...
    ./((Db.*K1.*(delA.*Dm(t).*K2+delM.*Da(t)))+(delB.*Da(t).*Dm(t).*K2));
C2A = (delM.*((Da(t).*Db.*Cb(j)) - (delA.*Db.*K1.*(1+12.*Emeoh(j)).*(Icell(i))./(2*n*F)) - (delB.*Da(t).*(1+6.*Emeoh(j)).*(Icell(i))./(6*F))))...
    ./((Db.*K1.*(delA.*Dm(t).*K2+delM.*Da(t)))+(delB.*Da(t).*Dm(t).*K2));            
a = Icell(i)./(12.*F.*delA.*Da(t));
b = (((C2A - C1A)./delA) - ((Icell(i)./(12.*F.*delA.*Da(t))).*((2.*delB)+delA)));
c = C1A - (((C2A-C1A)./delA).*delB) + (Icell(i)./(12.*F.*delA.*Da(t)).*delB.*(delB+delA));
Cmeoh = @(z) a.*(z.^2) + b.*z + c;
f1 = @(z) Cmeoh(z)./(Cmeoh(z) + (lambda.*exp((alphaA.*nA.*F)./(R.*T(t)))));
I = integral(f1,0.015,0.0173);
Z = A.*Ioref(t).*K.*exp((2.*alphaA.*nA.*F)./(R.*T(t))).*I - Icell(i); 
end