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+//Example 8_4

+clc;clear;funcprot(0);

+// Given values

+P=1;// atm

+T=35;// degree celsius

+L=150;// m

+h_L=20;// m

+v=0.35;// m^3/s

+g=9.81;// m/s^2

+//Properties

+rho=1.145;// kg/m^3

+mu=1.895*10^-5;// kg/m.s

+nu=1.655*10^-5;// m^2/s

+

+//Calculation

+// V=y(1); Re=y(2); f=y(3);D=y(4)

+function[X] = Diameter(y) 

+    X(1)=(v/(%pi*(y(4)^2)/4))-y(1);

+    X(2)=((y(1)*y(4))/(nu))-y(2);

+    X(3)=(-2.0*log10(2.51/(y(2)*sqrt(y(3)))))-(1/sqrt(y(3)));

+    X(4)=(y(3)*(L/(y(4))*((y(1)^2)/(2*g))))-h_L;

+endfunction

+y=[1 100000 0.01 0.1];

+z=fsolve(y,Diameter);

+V=z(1);// m/s

+Re=z(2);// Reynolds number

+f=z(3);

+D=z(4);// m

+printf('The minimum diameter of the duct,D=%0.3f m\n',D);

+//The diameter can also be determined directly from the third Swamee–Jain formula to be

+y=0;

+D=0.66*(((y^1.25*((L*v^2)/(g*h_L))^4.75))+(nu*v^9.4*(L/(g*h_L))^5.2))^0.04;

+printf('The diameter can also be determined directly from the third Swamee–Jain formula to be D=%0.3f m\n',D);