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+
+
+//example 16.14
+//calculate to what extent discharge can be increased without changing bed slope
+clc;funcprot(0);
+//given
+B=5;         //bed width
+D=2;         //bed depth
+S=1/1600;    //bed slope
+n=0.015;     //manning n
+
+A=B+2*D;     //area of lining
+//let B1 and D1 be new  width and depth of bed
+//for getting maximum discharge we diffrentiate Q and equating it to zero
+//Q=S^0.5*B1D1^5/3/n
+//we get
+D1=45/16;
+B1=9-2*D1;
+Q1=S^0.5*B1*D1^5/3/n;
+D1=round(D1*10000)/10000;
+mprintf("new width of bed=%f m.",B1);
+mprintf("\nnew depth of bed=%f m.",D1);
+mprintf("\n maximum discharge=%f cumec.",Q1);
+R=D;
+V=R^(2/3)*S^0.5/n;
+F=V/(9.81*D)^0.5;       //froud number
+R=D1;
+V=R^(2/3)*S^0.5/n;
+F=V/(9.81*D1)^0.5;       //froud number
+mprintf("\nFroud number is less than 1 in both case.\nHence,flow doesnot change from sub-critical to super critical.");