colebrook iterative equation in a for loop

DoM = [0.7620    0.8128    0.8636    0.9144    0.9652    1.0160    1.0668    1.1176    1.1684    1.2192]
Re=   [319130    299180    281590    265940    251940    239350    227950    217590    208130    199460]
Ks=0.000045
Er=10.^-8
Fo=0.01;
for I=0:1:10^8;
    Fn=(1./(-4.*log10(Ks./(3.71.*DoM))+(1.26./(Re.*sqrt(Fo))))).^2
    E=abs((Fn-Fo)/Fn)
    
    if  E<=Er
        Fn=Fn ;break;
        
    elseif E>Er
        Fn=Fo;
        
    end
end

i have an iterative equation where i have to set an initial value for Fo=0.01 to ahieve the Left hand side (Fn)

then calculate the absolute error (E)

after that compare the absolute error (E) with the specified Erorr (Er)

if the absolute error (E) is greater than the specified Error (Er)

replace Fo with the new Fn achived earlier

and calculate for a new Fn

until the absolute error is less than the specified error (Er)

then show the final value of Fn

Answer 1

[KALYAN ACHARJYA]

This way you can more easily implement the same code, please check the login of the code.

DoM=[0.7620    0.8128    0.8636    0.9144    0.9652    1.0160    1.0668    1.1176    1.1684    1.2192];
Re=[319130    299180    281590    265940    251940    239350    227950    217590    208130    199460];
Ks=0.000045;
Er=10.^-8;
Fo=0.01;
E=10.^-20; %initialize any value less than Er
while E<Er
  % Check the internal while loop carefully
  Fn=(1./(-4.*log10(Ks./(3.71.*DoM))+(1.26./(Re.*sqrt(Fo))))).^2;
  E=abs((Fn-Fo)/Fn); % may be change required, it run for once only then E>Er, while loop break
  Fo=Fn;    
end
disp(Fn);