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

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

댓글 수: 2

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 2018년 3월 1일
편집: Seanice Thompson 2018년 3월 1일
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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답변 (2개)

Walter Roberson
Walter Roberson 2018년 3월 1일
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 2020년 3월 28일
편집: Nikolaj Maack Bielefeld 2020년 3월 28일
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')

댓글 수: 2

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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질문:

2018년 3월 1일

댓글:

2020년 3월 29일

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