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+
+
+//example 5.12
+//calculate radius of zero drawdown
+//coefficient of permeability
+//drawdown in well
+//specific capacity
+//maximum rate at which water can be pumped
+clc;
+//given
+d=0.6;            //diameter of well;
+rw=d/2;
+H=40;             //depth of water in well before pumping
+Q=2000;           //discharge from well
+s1=4;             //drawdown in well
+B1=10;            //distance between well
+s2=2;
+B2=20;
+//Part (a)
+h1=H-s1;
+h2=H-s2;
+t=(H^2-h2^2)/(H^2-h1^2);
+R=(B2/(B1^t))^(1/(1-t));
+R=round(R*100)/100;
+mprintf(" radius of zero drawdown=%f m",R);
+//Part (b)
+r=10;
+k=Q*log10(R/r)*60*24/(1.36*(H^2-h1^2)*1000);
+k=round(k*100)/100;
+mprintf("\ncoefficient of permeability=%f m/day.",k);
+
+//part (c)
+Ho=(H^2-(Q*log10(R/rw)*24*60/(1000*1.36*k)))^0.5;
+D=H-Ho;
+D=round(D*100)/100;
+mprintf("\ndrawdown in well=%f m.",D);
+
+//part (d)
+C=Q/(1000*R);
+//for R=1 m;Q=Sc
+//hence on putting the values in discharge equation  we get
+//Sc*log10(61.2*Sc)=0.3223.
+//on solving this by trial and error method we get Sc=0.266 m^2/min.
+mprintf("\nSpecific capacity=0.266 cubic metre/minutes/metre.");
+
+//part (e)
+//this is obtained when Q=H
+//hence from equation of discharge,we get
+//Q*log10(69.2*Q)=6.528.
+//solving it by trial and error method we get Q=2.85 m^3/min.
+mprintf("\nmaximum rate at which water can be pumped=2.85 cubic metre/min");
+