Boling point problem with bisection method

1 vue (au cours des 30 derniers jours)
Arctgx
Arctgx le 16 Avr 2020
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
  5 commentaires
Ameer Hamza
Ameer Hamza le 20 Avr 2020
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 le 20 Avr 2020
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?

Connectez-vous pour commenter.

Réponse acceptée

Ameer Hamza
Ameer Hamza le 20 Avr 2020
Modifié(e) : Ameer Hamza le 20 Avr 2020
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 commentaires
Ameer Hamza
Ameer Hamza le 23 Avr 2020
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 le 23 Avr 2020
Code is working now. Thank you so much for your help!

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Mathematics dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by