Boling point problem with bisection method

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Arctgx
Arctgx 2020-4-16
评论： GeotechnicalEngr 2020-4-23
Hi everybody! im trying to solve a boiling problem with bisection method. but there is something wrong.
Question is:
To solve it, i wrote these codes: but there is an error (too many input arguments)
function bisection(a, b, eps)
iter=0;
global k A B C
k=[0.05 0.15 0.5 0.3];
A=[15.8333 15.8366 15.8737 15.9426];
B=[2477.07 2697.55 2911.32 3120.29];
C=[-39.94 -48.78 -56.51 -63.63];
xm = (a+b)/2;
x=[a b xm];
f = fonk(x);
for i=1:4
exp(A-B/(x+C))*k*133.32-1e5;
end
while abs(a-b)>eps
if f(1)*f(3) <=0
b = xm;
elseif f(2)*f(3)<=0
a = xm;
else
disp('You have a ill conditioned function or multiple root');
end
iter=iter+1;
end
fprintf('Bulunan kök değeri (sürtünme katsayısı) = %8.5f \n', (a+b)/2);
fprintf('%i. iterasyonda sonuca ulaşıldı. \n', iter);
end
function [f] = fonk(x)
global k A B C
f=exp(A-B/(x+C))*k*133.32-1e5;
end
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Ameer Hamza
Ameer Hamza 2020-4-20
Which line gives this error? What are the values of a, b, and eps when you call this function? vector A, B, and C are 1x4, but x is 1x3 so this expression
exp(A-B/(x+C))*k*133.32-1e5;
will also give error.
Arctgx
Arctgx 2020-4-20
darova, i used "k" for x. and i used "x" for T. i want to find mixture's boiling point.
Ameer, what is the solution?

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采纳的回答

Ameer Hamza
Ameer Hamza 2020-4-20
编辑：Ameer Hamza 2020-4-20
Try this code. Check the differences with your code to see the mistakes
x = bisection(300, 400, 0.001);
function xm = bisection(a, b, eps)
iter = 0;
global k A B C
k=[0.05 0.15 0.5 0.3];
A=[15.8333 15.8366 15.8737 15.9426];
B=[2477.07 2697.55 2911.32 3120.29];
C=[-39.94 -48.78 -56.51 -63.63];
while abs(a-b)>eps
xm = (a+b)/2;
if fonk(a)*fonk(xm) <=0
b = xm;
elseif fonk(xm)*fonk(b)<=0
a = xm;
else
disp('You have a ill conditioned function or multiple root');
end
iter=iter+1;
end
fprintf('Bulunan kök değeri (sürtünme katsayısı) = %8.5f \n', (a+b)/2);
fprintf('%i. iterasyonda sonuca ulaşıldı. \n', iter);
end
function [f] = fonk(x)
global k A B C
f=sum(exp(A-B./(x+C)).*k*133.32)-1e5;
end
  9 个评论
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Ameer Hamza
Ameer Hamza 2020-4-23
try this
x = bisection(300, 400, 0.001);
function xm = bisection(a, b, eps)
iter=0;
a=330;
b=398.75;
eps=1e-5;
iter=0;
global k A B C
k=[0.05 0.15 0.50 0.30];
A=[15.8333 15.8366 15.8737 15.9426];
B=[2477.07 2697.55 2911.32 3120.29];
C=[-39.94 -48.78 -56.51 -63.63];
while abs(a-b)>eps
xm = (a+b)/2;
x=[a b xm];
f(1) = fonk(x(1));
f(2) = fonk(x(2));
f(3) = fonk(x(3));
if f(1)*f(3) <=0
b = xm;
elseif f(2)*f(3)<=0
a = xm;
else
disp('You have a ill conditioned function or multiple root');
end
iter=iter+1;
end
fprintf('Root value) = %8.5f \n', (a+b)/2);
end
function [f] = fonk(x)
global k A B C
f=sum(exp(A-B./(x+C)).*k*133.32)-1e5;
end
GeotechnicalEngr
GeotechnicalEngr 2020-4-23
Code is working now. Thank you so much for your help!

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