Answer:
(a) 1.125 ft, Section factor = 22.78
(b) 42.75 ft
Explanation:
Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where
A = Cross sectional area of flow and
P = Perimeter h
Since the cross section is a circle then at depth 4 of 4.5 the perimeter
[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex] where r = 2.25 and θ = 102.1 °
perimeter = 10.1 ft and the area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] = 11.39 ft²
Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]
Section factor is given by for critical flow = Z = A×√D
= 11.39 ft² × √(4 ft) = 22.78
for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78
(b) The alternate depth of flow is given by
for a given flow rate, we have from chart for flow in circular pipes
Alternative depth = 0.9×45 = 42.75 ft
