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+// Example 9_6
+clc;funcprot(0);
+// Given data
+L=1.5;// The length in km
+D=6;// Diameter in inch
+h=80;// m
+// Assume 
+deltah_l=20;// m
+g=9.807;// The acceleration due to gravity in m/s^2
+nu=1*10^-6;// m/s^2
+epsilon=5*10^-5;// roughness in m
+
+// Calculation
+D=D*2.54*10^-2;// m
+sqrtoffintoRe_D=((2*g*deltah_l*D^3)/(((nu)^2)*L*10^3))^(1/2);
+Re_D=-2*sqrtoffintoRe_D*log10(((epsilon/D)/3.7)+(2.51/(sqrtoffintoRe_D)));// Reynolds number
+Q=(%pi*D*nu*Re_D)/4;// The volume flow rate in m^3/s
+Q_20=(Q*60)/(3.782*10^-3)//  The volume flow rate in gal/min
+deltah=150*(1-(Q_20/1000)^2);// m
+dh_20=deltah-(h+deltah_l);// m
+deltah_l=40;// m
+sqrtoffintoRe_D=((2*g*deltah_l*D^3)/(((nu)^2)*L*10^3))^(1/2);
+Re_D=-2*sqrtoffintoRe_D*log10(((epsilon/D)/3.7)+(2.51/(sqrtoffintoRe_D)));// Reynolds number
+Q=(%pi*D*nu*Re_D)/4;// The volume flow rate in m^3/s
+Q_40=(Q*60)/(3.782*10^-3)//  The volume flow rate in gal/min
+deltah=150*(1-(Q_40/1000)^2);// m
+dh_40=deltah-(h+deltah_l);// m
+Q=((((dh_20)/(dh_20-dh_40))*(Q_40-Q_20))+Q_20);// GPM
+deltah=150*(1-(Q/1000)^2);// m
+deltah_l=deltah-h;// m
+printf("\nThe flow rate through the system,Q=%3.1f GPM \ndeltah=%3.1f m \ndeltah_l=%2.2f m",Q,deltah,deltah_l);
+printf("\nContinuing this process to improve our estimate of Q and Ah we finally arrive at:Q=527.7(GPM);deltah=108.3 m")
+