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@MURPHYENGINEERING
Last active April 17, 2021 02:52
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HW 7 Problem 2(a)
clear; clc; cla; clf; close all; format long;
global Ccb Chp Chl Cre zB;
% Physical constants
g = 9.81; % m / s^2
rho = 1000; % kg / m^2
nu = 0.9e-6; % m^2 / s
% Given properties
D = 100e-3; % m
L = 50; % m
zB = 12; % m
epsilon = 0.3e-3; % m
P_pump = 4.5e3; % W
% Derived properties
A = pi/4 * D^2; % m^2
% Precomputed constants
Ccb = epsilon / D / 3.72; % Colebrook constant term
Chp = P_pump / (g*rho*A); % Pump head constant factor
Chl = L / (D * 2*g); % Head loss constant factor
Cre = D / nu / 2.51; % Reynolds number as used in Colebrook
% Initial point
f0 = 0.0038; % ASU ID: xxxxxxxx 38
U0 = 1; % m/s
x = num2cell(fsolve(@pipeSystem, [f0, U0]));
[f,U] = x{:}
function F = pipeSystem(x)
global Ccb Chp Chl Cre zB;
f = x(1);
U = x(2);
Re = Cre * U;
% Colebrook equation
sqrt_f_i = 1 / sqrt(f);
F(1) = sqrt_f_i + 2*log10(Ccb + sqrt_f_i/Re);
% Pump head, head loss, and gravitational potential
F(2) = (Chp / U) - (f * Chl * U^2) - zB;
end
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