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+

+

+//example 14.11

+//find channel section and discharge

+clc;funcprot(0);

+//given

+S=1/5000;            //bed slope

+s=1/2;               //side slope

+f=0.9;                 //laecy silt factor

+

+Q=(f^(5/3)/(3340*S))^6;

+R=f^3/(4980*S)^2;

+P=4.75*Q^0.5;

+A=P*R;

+//using the value of A and P in equations we get,

+//equation in D as

+y=poly([-6.961,9.41,-1.73],'x','c');

+D=roots(y);

+//we get D=4.5561754 and 0.8831309.

+//taking

+D=0.8831309;

+B=9.41-D*2.23;

+B=round(B*100)*100;

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

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

+mprintf("Width of channel section=%f m.",B);

+mprintf("\nDepth of channel section=%f m.",D);

+mprintf("\n Discharge=%f cumecs.",Q);