Question

Fluid Mechanics Friction Problem: Write one MATLAB m-file that solves the Type I and II problems presented in class based on the file posted for the Type III problem (use Colebrook to estimate f). Type I: Solve hL for v=0.74x10-5ft^2/s, D=3 in, L=1000 ft, e=0.006 in, and Re=80000. f=0.0258 from Moody Chart. Type II: Solve Q for v=10^-6 m^2/s, D=0.2 m, L=500 m, e=0.046 mm, and hL=30m. Use “rough” Colebrook to generate an estimate for f.

Answer 1

Type III: matlab function to calculate friction factor by colbrook equation:

function F = colebrook(Re,K)
% F = COLEBROOK(Re,K) computation of the
%     Darcy-Weisbach friction factor F according to the Colebrook equation:
%                             -                       -
%      1                     |    K        2.51        |
% --------- = -2 * Log_10 | ----- + ------------- |
%   sqrt(F)                  |   3.7     R * sqrt(F)   |
%                             -                       -
% INPUT:
%   R : Reynolds' number (should be >= 2300).
%   K : Equivalent sand roughness height divided by the hydraulic
%       diameter (default K=0).
%
% OUTPUT:
%   F : Friction factor.
%
% FORMAT:
%   R, K and F are either scalars or compatible arrays.
%
% EXAMPLE: F = colebrook([3e3,7e5,1e100],0.01)
%
% Edit the m-file for more details.

% Check for errors.
if any(Re(:)<2300) == 1,
   warning('The Colebrook equation is valid for Reynolds'' numbers >= 2300.');    
end,
if nargin == 1 || isempty(K) == 1,    
   K = 0;
end,
if any(K(:)<0) == 1,
   warning('The relative sand roughness must be non-negative.');
end,

% Initialization.
X1 = K .* Re * 0.123968186335417556;              % X1 <- K * R * log(10) / 18.574.
X2 = log(Re) - 0.779397488455682028;              % X2 <- log( R * log(10) / 5.02 );                 

% Initial guess.                                            
F = X2 - 0.2;   

% First iteration.
E = ( log(X1+F) - 0.2 ) ./ ( 1 + X1 + F );
F = F - (1+X1+F+0.5*E) .* E .*(X1+F) ./ (1+X1+F+E.*(1+E/3));

% Second iteration (remove the next two lines for moderate accuracy).
E = ( log(X1+F) + F - X2 ) ./ ( 1 + X1 + F );
F = F - (1+X1+F+0.5*E) .* E .*(X1+F) ./ (1+X1+F+E.*(1+E/3));

% Finalized solution.
F = 1.151292546497022842 ./ F;                   % F <- 0.5 * log(10) / F;
F = F .* F;                                      % F <- Friction factor.

Save this function in your matlab directory first:

type 1: matlab code: with output

clc;
clear;
%given values converted into SI units
Re=80000;
vis=0.74*10^(-5)*(.3048)^2; %viscosity
D=3*0.0254;       %diameter of pipe
L=1000*0.3048;    %length of pipe
e=0.006*0.0254;   %surface roughness
g=9.81;        %acceleration due to gravity
v=Re*vis/D;    %velicity

%output
f=colebrook(Re,e); %calculate f using colebrook equation function
hL= (f*L*v^(2))/(2*g*D);    %head loss
fprintf('head loss in pipe is %.4f m \n',hL);

output:

Type II: matlab code with output

clc;
clear;
%given values converted into SI units
vis=10^(-6); %viscosity
D=0.2;       %diameter of pipe
L=500;    %length of pipe
e=0.046*10^(-3);   %surface roughness
g=9.81;        %acceleration due to gravity
hL=30;       %head loss

%output
%using colebrook equation and inserting reynold equation and head loss equation,
%derive a formula for velocity

v= -sqrt(hL*2*g*D/L)*(1.8)*log(((e/D)/3.7)+2.51*vis/(D*sqrt(2*hL*g*D/L))); %velocity
A=(pi*D^2)/4;   %area
Q=A*v; %flowrate

fprintf('flow rate in pipe is %.4f m^2/s \n',Q);

OUTPUT:

this equation is used for velocity