Colebrook equation can be used for calculation of friction factor in the pipes.The implicit form of the equation can be solved for (Re>4000) using Newton-Raphson Method.Following code can be used to solve the equation iteratively.
"D=.1; %diameter
Q=6;%flow
d=1;% density
u=.001;%viscocity
Vel=Q/(.25*pi*D^2); % velocity
Re=D*Vel*d/u % Reynolds number
f=64/Re; % initial guess
e=.0021;%roughness
ee=e/D
er=1;
n=1;
while er>1e-6
F =( 1/sqrt(f)+2*(log10( (e/(3.7*D))+(2.51/(Re*sqrt(f))))));
H=((e/(3.7*D))+(2.51/(Re*sqrt(f))));
G= -.5*f^-1.5+(.8686*( -1.255/(Re*f^1.5)))/H;
h=-F/G;
fn=f+h;
er=fn-f;
n=n+1;
f=fn;
end
n
er
fn
The next post will be on calculating exit velocity in a circular pipe.
"D=.1; %diameter
Q=6;%flow
d=1;% density
u=.001;%viscocity
Vel=Q/(.25*pi*D^2); % velocity
Re=D*Vel*d/u % Reynolds number
f=64/Re; % initial guess
e=.0021;%roughness
ee=e/D
er=1;
n=1;
while er>1e-6
F =( 1/sqrt(f)+2*(log10( (e/(3.7*D))+(2.51/(Re*sqrt(f))))));
H=((e/(3.7*D))+(2.51/(Re*sqrt(f))));
G= -.5*f^-1.5+(.8686*( -1.255/(Re*f^1.5)))/H;
h=-F/G;
fn=f+h;
er=fn-f;
n=n+1;
f=fn;
end
n
er
fn
The next post will be on calculating exit velocity in a circular pipe.
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