How to solve this equation with matlab

3 visualizaciones (últimos 30 días)
hadi mohammadian
hadi mohammadian el 30 de Dic. de 2020
Comentada: Walter Roberson el 30 de Dic. de 2020
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)))));
  2 comentarios
Walter Roberson
Walter Roberson el 30 de Dic. de 2020
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
hadi mohammadian el 30 de Dic. de 2020
Editada: hadi mohammadian el 30 de Dic. de 2020
yes . you're right . I changed them . thanks alot .
in matlab they're correct but they were wrong in here .

Iniciar sesión para comentar.

Respuesta aceptada

Walter Roberson
Walter Roberson el 30 de Dic. de 2020
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)
  5 comentarios
hadi mohammadian
hadi mohammadian el 30 de Dic. de 2020
thanks man . my issue solved .
Walter Roberson
Walter Roberson el 30 de Dic. de 2020
What was the solution ?

Iniciar sesión para comentar.

Más respuestas (1)

Alan Stevens
Alan Stevens el 30 de Dic. de 2020
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).
  1 comentario
hadi mohammadian
hadi mohammadian el 30 de Dic. de 2020
thanks man .
but It's not I'm looking for .
and the log10 don't path the eror .

Iniciar sesión para comentar.

Categorías

Más información sobre Symbolic Math Toolbox en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by