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+//Example 13_3
+clc;clear;funcprot(0);
+// Given values
+b=4;// Bottom width in m 
+V=51;// Flow rate in ft^3/s
+// Properties
+n=0.014;//The Manning coefficient 
+// Calculation
+//The cross-sectional area, perimeter, and hydraulic radius of the channel are A_c=4y;p=4+2y;R_h=A_c/p=(4y)/(4+y);
+S_0=2/1000;
+
+//Using the Manning equation, the flow rate through the channel can be expressed as Vdot=(a/n)*A_c*R_h^(2/3)*S_0^(1/2)
+// y=y(1)
+function[X]=flowdepth(y);
+    X(1)=real(((1.486/n)*(4*y(1))*((4*y(1))/(4+(2*y(1))))^(2/3)*(S_0)^(1/2))-V);
+endfunction
+y=[1];
+z=fsolve(y,flowdepth);
+printf('If S_0=2/1000=0.002.The flow depth is determined to be y=%0.1f ft\n',z(1));
+
+// If the bottom drop were just 1 ft per 1000 ft length, the bottom slope would be
+S_0=0.001;
+// y=y(2)
+function[X]=flowdepth(z);
+    X(1)=real(((1.486/0.014)*(4*z(1))*((4*z(1))/(4+(2*z(1))))^(2/3)*(0.001)^(1/2))-51);
+endfunction
+y=[1];
+y=fsolve(z,flowdepth);
+printf('If the bottom slope would be S_0=.001, and the flow depth would be y=%0.1f ft\n',y(1));