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+

+

+//example 16.12

+//calculate average boundary shear stress;

+//percentage of earth work is saved in lined section

+clc;funcprot(0);

+//given

+s=1.5;        //side slope

+Q=15;         //discharge

+S=1/4000;     //bed slope

+Nl=0.014;      //manning n for lined channel

+Nu=0.028;      //manning n for ulined channel 

+fb=0.75;         //free board

+

+//considering the perimeter of trapezoidal section

+//taking minimum perimeter for given area

+//i.e dP/dD=0

+//we get

+//A=2.1D^2; R=D/2; and P=4.2D

+

+//for linrd channel

+//Q=AR^(2/3)*S^0.5

+//substituting above values we get

+D=(10.0396)^(3/8);

+B=0.6*D;

+R=D/2;

+tau=9.81*R*S*1000;

+tau=round(tau*1000)/1000;

+mprintf("for lined canal:");

+mprintf("\naverage boundary shear stress=%f N/square m.",tau);

+Dc=D+fb;            //total depth of cutting

+A1=(B+1.5*Dc)*Dc;

+

+//for unlined channel

+//Q=AR^(2/3)*S^0.5

+//substituting above values we get

+D=3.08;

+B=0.6*D;

+R=D/2;

+tau=9.81*R*S*1000;

+tau=round(tau*100)/100;

+mprintf("\n\nfor unlined canal:");

+mprintf("\naverage boundary shear stress=%f N/square m.",tau);

+Dc=D+fb;            //total depth of cutting

+A2=(B+1.5*Dc)*Dc;

+per=(A2-A1)*100/A2;   

+per=round(per*100)/100;

+mprintf("\n\npercent saving of earth=%f percent.",per);