How do I complete my code to plot the Moody Chart?

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function [f] = frictionFactor(Re, ed) %Re = Reynolds Number, ed = eps/d, relative roughness
colebrook = @(f) 1/sqrt(f)+2*log10((ed/3.7)+(2.51)/(Re*sqrt(f)));
if Re > 4000 %turbulent
f = fzero(colebrook, [0.008, 0.1]);
elseif Re < 2000 %laminar
f = 64/Re;
else %transitional
f = (((Re-2000)/(4000-2000))*(0.1-0.008))+0.008;
end
array = linspace(0.000001, 0.05, 21);
for i = array
colebrook(i)
end
end
  2 Commenti
Walter Roberson
Walter Roberson il 1 Mar 2018
It is not clear to me why you calculate f and then ignore it when you for i = array ?
I somehow suspect that the values in array are intended to represent different Re values that you want to evaluate colebrook with after figuring out what f value you want to use ?
Seanice Thompson
Seanice Thompson il 1 Mar 2018
Modificato: Seanice Thompson il 1 Mar 2018
I was told to create an array of roughnesses and plug them into the colebrook equation. Then, plot the moody. I'm just not sure how to go about that. The for loop portion is what I need help with.

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Risposte (2)

Walter Roberson
Walter Roberson il 1 Mar 2018
function [f, rough] = frictionFactor(Re, ed) %Re = Reynolds Number, ed = eps/d, relative roughness
colebrook_fed = @(f, ed) 1/sqrt(f)+2*log10((ed/3.7)+(2.51)/(Re*sqrt(f)));
if Re > 4000 %turbulent
f = fzero(@(f) colebrook_fed(f, ed), [0.008, 0.1]);
elseif Re < 2000 %laminar
f = 64/Re;
else %transitional
f = (((Re-2000)/(4000-2000))*(0.1-0.008))+0.008;
end
array = linspace(0.000001, 0.05, 21);
narray = length(array);
rough = zeros(1, narray);
for K = 1 : narray
i = array(K);
rough(K) = colebrook_fed(f, i);
end

Nikolaj Maack Bielefeld
Nikolaj Maack Bielefeld il 28 Mar 2020
Modificato: Nikolaj Maack Bielefeld il 28 Mar 2020
You could also have used the symbolic math implicit plot command:
close all; clear; clc
% symbolic math: y = f, x = Re
syms x y
% relative roughness
relrough = [0 2e-7 1e-6 5e-6 10e-6 50e-6 100e-6 200e-6 400e-6 600e-6 ...
800e-6 1e-3 2e-3 4e-3 6e-3 8e-3 10e-3 15e-3 20e-3 30e-3 40e-3 50e-3];
% Colebrook equation
eqn = 1/sqrt(y) == -2*log10(relrough/3.7+2.51/(x*sqrt(y)));
% implicit plot with x- and y-limits
fimplicit(eqn,[2000 10e8 0.006 0.1])
set(gca, 'XScale', 'log') % logarithmic x-axis
set(gca, 'YScale', 'log') % logarithmic y-axis
title('Moody Chart')
xlabel('Reynolds Number')
ylabel('Friction Factor')
  2 Commenti
Walter Roberson
Walter Roberson il 28 Mar 2020
Because of the three different ranges of values, your lower bound on x for the fimplicit should be 4000. You would need to also hold on and plot the other two parts (you could plot both together if you used piecewise)

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