How to solve this equation with matlab

clear all;
clc;
data8={'1','10','20','23','12','11'};
kk{1}='Internal pipe roughness (mm)';
kk{2}='Hydraulic diameter of pipe (m)';
kk{3}='Reynolds number ';
kk{4}='Pipe line length (m)';
kk{5}='Average pipe line velosity (m/s)';
kk{6}='Acceleration due to gravity (m/s^2)';
kkk=inputdlg(kk,'Head loss',1,data8);
e11=str2num(kkk{1});
d12=str2num(kkk{2});
re1=str2num(kkk{3});
l2=str2num(kkk{4});
v11=str2num(kkk{5});
g12=str2num(kkk{6});

eqn=(1/sqrt(f9)==(-2*log((e11/3.7*d12)+(2.51/re1*sqrt(f9)))));
solve('eqn','sqrt(f9)');
hf=(f12*l2*(v11^2))/(2*g12*d12);

fprintf('Friction factor in pipe = %f \n  Head loss = %f   (m) \n-----------------------------------------------\n',f1,hf);

Undefined function or variable 'f9'.

Error in Untitled (line 150)

eqn=(1/sqrt(f9)==(-2*log((e11/3.7*d12)+(2.51/re1*sqrt(f9)))));

[Walter Roberson]

e11=str2num(kkk{1});
d12=str2num(kkk{1});
re1=str2num(kkk{1});
l2=str2num(kkk{1});
v11=str2num(kkk{1});
g12=str2num(kkk{1});

are you sure that you want to use kkk{1} as the source for all of those str2num() ? That assigns the same value to each of the variables.

[hadi mohammadian]

yes . you're right . I changed them . thanks alot .

in matlab they're correct but they were wrong in here .

Answer 1

[Walter Roberson]

syms f9 f9sqrt
eqn=(1/sqrt(f9)==(-2*log((e11/3.7*d12)+(2.51/re1*sqrt(f9)))));
F9sqrt = solve(subs(eqn, f9, f9sqrt^2), f9sqrt)

[hadi mohammadian]

thanks . but no.

the eror still there .

[hadi mohammadian]

in direct words , I gave all values to program (with inputdlg so the user able to change numbers) and I want to program solve the equation and give me the value of (f9) , so I can use it in other equations . that's all .lol

[Walter Roberson]

e11 = 1;
d12 = 10;
re1 = 20;
i2 = 23;
v11 = 12;
g12 = 11;
syms f9sqrt
eqn=(1/f9sqrt==(-2*log((e11/3.7*d12)+(2.51/re1*f9sqrt))));
sol1 = vpasolve(eqn, f9sqrt, -15)
sol2 = vpasolve(eqn, f9sqrt, -1)

The results are -13.261186218170519738583506098119 and -0.51544996325405629560226567032353 . Those are the square roots of the values, so f9 itself will be complex valued

Please re-check whether the expression should be (2.51/re1)*sqrt(f9) like you have coded, or if it should instead be 2.51/(re1*f9sqrt) . If it is the second of those, there is a closed form solution involving WrightOmega that leads to sqrt(f9) about -0.5517 (also complex valued.)

[hadi mohammadian]

thanks man . my issue solved .

[Walter Roberson]

What was the solution ?

Answer 2

[Alan Stevens]

Strange way of entering data! However, one way to find the friction factor is a simple fixed point iteration method, such as

e11 = 1;
d12 = 10;
re1 = 20;
l2 = 23;
v11 = 12;
g12 = 11;

f9 = 1; % initial guess
tol = 10^-6;
err = 1;
% fixed point iteration
while err>tol
      f9old = f9;
      f9 = (-2*log10(e11/(3.7*d12)+2.51/(re1*sqrt(f9))))^-2;
      err = abs(f9-f9old);
end
disp(f9)

Note that I've used log10 rather than log (the latter is log to the base e).

[hadi mohammadian]

thanks man .

but It's not I'm looking for .

and the log10 don't path the eror .