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+//Example 13_4
+clc;clear;
+// Given values
+S_0=0.003;// Bottom slope
+n_1=0.030;
+n_2=0.050;
+
+// Calculation 
+s=sqrt(3^2+3^2);
+//Then the flow area, perimeter, and hydraulic radius for each subsection and the entire channel become
+// Subsection 1:
+A_c1=21;// m^2
+p_1=10.486; // m
+R_h1=A_c1/p_1;// m
+// Subsection 2:
+A_c2=16;// m^2
+p_2=10;// m
+R_h2=A_c2/p_2;// m
+// Entire channel
+A_c=A_c1+A_c2;// m^2 
+p=p_1+p_2;// m
+R_h=A_c/p;// m
+//Using the Manning equation for each subsection,
+a=1;//m^(1/3)/s
+v_1=(a/n_1)*(A_c1*R_h1^(2/3))*(S_0)^(1/2);// m^3/s
+v_2=(a/n_2)*(A_c2*R_h2^(2/3))*(S_0)^(1/2);// m^3/s
+v=v_1+v_2;// m^3/s
+printf('The total flow rate through the channel,V=%0.0f m^3/s\n',v);
+n_eff=(a*A_c*R_h^(2/3)*S_0^(1/2))/v;
+printf('The effective Manning coefficient for the entire channel ,n_eff=%0.3f \n',n_eff);