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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Finding of discharge through triangular channel
+//Given
+y=4;
+theta=60;
+b=theta/2;
+n=0.0182;
+sb=1/1000;
+T=2*tan(b)*y;z=tan(b); disp(z);
+A=0.5*T*y;
+P=2*sqrt(y^2+(y*tan(b))^2);
+R=A/P;disp(A);disp(P);disp(R);
+//To Find
+q=A*(1/n)*(R)^2/3*(sb)^1/2;
+disp("discharge through triangular channel ="+string(q)+" m^3/sec");