# type3 - Newtons method is x = x0 F0(dF/dx_0 Here F(x = 0...

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% DOL 10/13/09 % Octave (Matlab) script to solve Type III pipe flow proplem. % Find Diameter, given head loss. % Use a system demand curve. clc; clear; %---------- Given these: mu = 1.985E-5; % kg/m*s rho = 1.145; % kg/m3 L = 150; % m eps = 0.0; % m Q = 0.35; % m3/s hLwant = 20; % m, desired head loss %--------- g = 9.80665; % m/s2 D = linspace(0.01,1, 1000); D = D'; v = Q./(pi/4*D.^2); Re = rho/mu*D.*v; %--------- Compute f % Initial guess from Haaland equation; x = sqrt(f) x = 1.0./(-1.8*log10(6.9./Re + (eps./D/3.7).^1.11)); % x is sqrt(f) % Solve Colbrook equation with Newton's method, 5 iterations (plenty)
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Unformatted text preview: % Newtons method is x = x0 - F0/(dF/dx)_0 % Here, F(x) = 0; with F(x) = 1/x+2*log10(eps/3.7/D + 2.51/Re/x) % This is just the Colbrook equation rearranged to equal zero for i=1:5 x = x + (1./x + 2*log10(eps./D/3.7 + 2.51./Re./x)) ./ . .. (1./x./x + 2*2.51./Re./x./x./log10(eps./D/3.7+2.51./Re./x)); end % Recover f f = x.^2; %---------- Compute head loss hL = f*L./D .*v.^2/2/g; %---------- Plot the results semilogy(D,hL); xlabel('D (m)'); ylabel('hL (m)'); %---------- interpolate D for desired head loss Dwant = interp1(hL,D,hLwant)...
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## This note was uploaded on 03/11/2012 for the course CHE 374 taught by Professor Davidlignell during the Fall '12 term at Brigham Young University, Hawaii.

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