Explanation:
assume maximum permissible is 4 ft/s and n=0.025 we have the formula:
[tex]V=\frac{1.49}{n} R_{h} ^2^/^3S_{0} ^1^/^2[/tex]
[tex]R_{h} =(Vn/1.49S_{0} ^1^/^2)^3^/^2[/tex]
where
[tex]V= 0.0005\\S_{0} =1000cfs[/tex]
we get
[tex]R_{h} =((0.0005*0.025/1.49*1000^1^/^2))^3^/^2=1.36 ft[/tex]
[tex]A=Q/V=1000cfs/4.00ft/s=250ft^2[/tex]
assume side slopes will be 1 vertical to 2 horizontal
[tex]P=A/R_{h} =250ft^2/1.36 ft=183ft[/tex]
also
[tex]P=b+2y\sqrt{1+4} =b+2y\sqrt{5} =183ft\\\\A=by+2y^2=250ft^2[/tex]
Solve the above two equations for the bottom width b and depth y.
