Problems with the application of Newton's method

function [] = newton_raphson(func, diff, x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here

x = x0;
maxiter = 200;
tol = 10^(-5);
eps = 0.4;
c_s = 5.67*10^(-8);
alpha_k = 4;
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;

func = @(x) eps * c_s * x^4 + (alpha_k + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k * T_l);
diff = @(x) 4 * eps * c_s * x^3 + (alpha_k + 1/(s1/lamda1+s2/lamda2));
newton_raphson(func, diff, 200)

for i = 1:maxiter
    
    if
        diff(x(i)) < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    
    x(i+1) = x(i) - func(x(i))/diff(x(i));
    abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
    
    if
        abs(x(i+1) - x(ix)) < tol
        fprintf('The Root has converged at x = %.10f\n', x(i+1));
        
    else
        fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));
        
        
    end
   
end
end

Can someone help me please. Unfortunately, I'm not quite fit in Matlab and have recently started working with functions. I don't know what's wrong with this code. Unfortunately I don't get a result. It had to come out with an X value of around 290.

Thanks a lot

Answer 1

[Torsten]

function main
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 20;
maxiter = 200;
tol = 10^(-5);
eps = 0.4;
c_s = 5.67*10^(-8);
alpha_k = 4;
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
func = @(x) eps * c_s * x^4 + (alpha_k + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k * T_l);
diff = @(x) 4 * eps * c_s * x^3 + (alpha_k + 1/(s1/lamda1+s2/lamda2));
xsol = newton_raphson(func, diff, x0)  
end

function xsol = newton_raphson(func, diff, x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);

for i = 1:maxiter
    
    if  diff(x(i)) < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    
    x(i+1) = x(i) - func(x(i))/diff(x(i));
    abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
    
    if abs(x(i+1) - x(i)) < tol
        fprintf('The Root has converged at x = %.10f\n', x(i+1));      
    else
        fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));      
    end
    
end
xsol = x(end);
end

[Aryo Aryanapour]

Wow, thank you very much. It works. I will study it to understand.

[Aryo Aryanapour]

clear 
clc
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 280;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;

func = @(x) eps * c_s * x^4 + (alpha_k + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k * T_l);
diff = @(x) 4 * eps * c_s * x^3 + (alpha_k + 1/(s1/lamda1+s2/lamda2));
xsol = newton_raphson(func, diff, x0)  


function xsol = newton_raphson(func, diff, x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
T_l = 10;


for i = 1:maxiter
    
    if  diff(x(i)) < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    
    x(i+1) = x(i) - func(x(i))/diff(x(i));
    abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
    
    Gr = (g*L.^3*betha*(x(i+1)-T_l))/v;
    Ra = Gr*Pr;
    alpha_k = ((0.825+(0.387*Ra^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;

    if abs(x(i+1) - x(i)) < tol
        fprintf('The Root has converged at x = %.10f\n', x(i+1));      
    else
        fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));      
    end
    
end
xsol = x(end);
alpha_k
end

Hey Torsten, Can you help me again? With the same code, alpha_k is no longer constant, but changes with temperature. I have to calculate Gr as a function of the temperature and then re-calculate alpha_k with Gr. At the end, the x value or xsol must be calculated using the alpha_k. Unfortunately the alpha_k cannot be calculated for me. Can you see why Thanks so much

[Torsten]

function main
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 80;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;

Ra = @(x)Pr*(g*L^3*betha*(x-T_l))/v;
alpha_k = @(x)((0.825+(0.387*Ra(x)^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;
func = @(x) eps * c_s * x^4 + (alpha_k(x) + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k(x) * T_l);
xsol = newton_raphson(func,x0)  
end
function xsol = newton_raphson(func,x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);

for i = 1:maxiter
    
    diff_xi = (func(x(i)+1e-6)-func(x(i)))*1e6;
    if  diff_xi < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    x(i+1) = x(i) - func(x(i))/diff_xi;
    abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
    
    if abs(x(i+1) - x(i)) < tol
        fprintf('The Root has converged at x = %.10f\n', x(i+1));      
    else
        fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));      
    end
    
end
xsol = x(end);
end

If you are willing to spend the effort to get an analytical expression for the derivative of "func" with respect to x, you can do so. But I don't have the time.

And it works with a finite-difference approximation for the derivative, as you can see.

[Aryo Aryanapour]

Dear Torsten

I'm very sorry for the late reply. I'm pretty busy with some projects. Your code is working fine again. Thank you very much and happy new year.

[Aryo Aryanapour]

function main
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x0 = 80;
eps = 0.4;
c_s = 5.67*10^(-8);
s1 = 0.250;
s2 = 0.015;
lamda1 = 0.35;
lamda2 = 22.7;
Tw_1 = 1200;
T_l = 10;
Tw_1 = 1200+273.15;
T_l = 10+273.15;
lambda_L = 25.12;
betha = 3.543;
v = 144.0;
L = 7;
g = 9.81;
Pr = 0.7095;
Ra = @(x)Pr*(g*L^3*betha*(x-T_l))/v;
alpha_k = @(x)((0.825+(0.387*Ra(x)^1/6)/(1+(0.492/Pr)^9/16)^8/27)^2)*lambda_L/L;
func = @(x) eps * c_s * x^4 + (alpha_k(x) + 1/(s1/lamda1+s2/lamda2)) * x - ((1/(s1/lamda1+s2/lamda2)) * Tw_1 + alpha_k(x) * T_l);
xsol = newton_raphson(func,x0)  
end
function xsol = newton_raphson(func,x0)
%UNTITLED Summary of this function goes here
%   Detailed explanation goes here
x(1) = x0;
maxiter = 200;
tol = 10^(-5);
for i = 1:maxiter
    
    diff_xi = (func(x(i)+1e-6)-func(x(i)))*1e6;
    if  diff_xi < tol
        fprintf('Pitfall hast occured a better initial guess\n');
        return;
        
    end
    
    x(i+1) = x(i) - func(x(i))/diff_xi;
    abs_error(i+1) = abs((x(i+1)-x(i))/x(i+1))*100;
    
    if abs(x(i+1) - x(i)) < tol
        fprintf('The Root has converged at x = %.10f\n', x(i+1));      
    else
        fprintf('Iteration no: %d,current guess x = %.10f, error = %.5f', i, x(i+1), abs_error(i+1));      
    end
    
end
xsol = x(end);
end

Hello Torsten

sorry if i have to annoy you again. can you help me again I'm calculating the heat flow in watts here, but matlab won't give me the value. I also need the value for Alpha end printed out so I know how big alpha is. Can you help me to determine these two values. Oh god I need a good read on Matlab, especially functions. There are many youtube videos. But somehow they couldn't help me. I thank you in advance.

by the way this is not new code. This is the one from the previous comment. I just re-pasted it here.

[Torsten]

After the line

xsol = newton_raphson(func,x0)

add

eps*c_s*xsol^4 + alpha_k(xsol)*(xsol-T_l)
1/(s1/lamda1+s2/lamda2)*(xsol-Tw_1)
alpha_k(xsol)

to see the values you are looking for in the command window.

[Aryo Aryanapour]

Thank you Torsten It works. But could you explain to me how you come up with alpha-ges and heat flow. Because my formula for calculating alpha-ges and heat flow is:

alpha_ges = alpha_k + (eps *c_s*(xsol^4+T_l^4)/(xsol-T_l))  % Total heat transfer coefficient
Q_ges = (1/((s1/lamda1+s2/lamda2)+1/alpha_ges))*A*(Tw_1-T_l) % heat flow

and it works for the code where the alpha value is constant 4.0.

Your code seems to me to be some sort of combination of my heat flow code. can you explain it to me please Isn't alpha_k(xsoll) = 65.2597 very high?

Thanks so much

[Torsten]

Sorry, but I'm no engineer. I can't help you in this respect.

I thought the question was how to obtain deduced quantities from the result xsol in general.

[Aryo Aryanapour]

Thorsten

dont be sorry

You are the best. You have been able to help me a lot more than I thought. Thank you again for always replying so quickly.