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+// Example 9_3
+clc;funcprot(0);
+// Given data
+dp=100;// The pressure drop in psi
+rho=1*10^3;// The density of water in kg/m^3
+g=9.807;// The acceleration due to gravity in m/s^2
+Q=2000;// The flow rate of water in gal/min
+D=4;// The next pipe size in inch
+L=100;// Length in m
+nu=1*10^-6;// m^2/s
+
+// Calculation
+deltah=(dp*6.895*10^3)/(rho*g);// m
+printf("\nh_in-h_out=%2.2f m",deltah);
+D=D*2.54*10^-2;// m
+Q=Q*(3.782*10^-3)/60;// m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=5*10^-5;// physical height
+function[X]=frictionfactor(y)
+    X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+deltah_f=K_f*((V^2)/(2*g));// The head loss in m
+printf("\nD=%0.4f m \nQ=%1.3e m^3/s \nV=%2.2f m/s \nRe_D=%1.3e \nf=%1.3e \nK_f=%2.2f \nh_in-h_out=%3.1f m",D,Q,V,Re_D,f,K_f,deltah_f)
+printf("\nThe head loss of 205.9 m is greater than the allowable los s of 70.31 m.");
+
+// If we try the next size pipe, D = 6 in,
+D=6;// inch
+D=D*2.54*10^-2;// m
+Q=2000;// The flow rate of water in gal/min
+Q=Q*(3.782*10^-3)/60;// m^3/s
+V=(4*Q)/(%pi*D^2);// m/s
+Re_D=(V*D)/nu;// Reynolds number
+epsilon=5*10^-5;// physical height
+function[X]=frictionfactor(y)
+    X(1)=-(2.0*log10(((epsilon/D)/3.7)+(2.51/(Re_D*sqrt(y(1))))))-(1/sqrt(y(1)));
+endfunction
+// Guessing a value of f=1*10^-2;
+y=[1*10^-2];
+f=fsolve(y,frictionfactor);
+K_f=f*((L)/D);// The head loss coefficient
+deltah_f=K_f*((V^2)/(2*g));// The head loss in m
+printf("\nD=%0.4f m \nQ=%1.3e m^3/s \nV=%1.3f m/s \nRe_D=%1.3e \nf=%1.3e \nK_f=%2.2f \nh_in-h_out=%2.2f m",D,Q,V,Re_D,f,K_f,deltah_f)
+printf("\nThis is smaller than the allowable head loss so that a 6 in diameter pipe is acceptable.")