Question 9.4: By using the regime approach, determine the cross section of......
By using the regime approach, determine the cross section of an alluvial channel for a design flow of 8 m³ /s. The sediment carried by water is 0.4 mm sand.
Given:
Q = 8 m³ /s;
d = 0.4 mm.
Determine:
B_{o} = ?
y = ?
\quad\quad\quad\quad P=4.75\sqrt{8}=13.44 m\\\quad\quad\quad\quad f_{s}=1.76(0.4)^{\frac{1}{2}}=1.11\\\quad\quad\quad\quad R=0.47(\frac{8}{1.11})^{0.333}=0.907 m
Let the side slopes be 1H to 2 V. Then,
\quad\quad\quad\quad A=(B_{o}+0.5y)y=PR=13.44\times0.907=12.18 m\\\quad\quad\quad\quad P=B_{o}+2\sqrt{1+(0.5)^{2}}y\\\quad\quad\quad\quad\quad=B_{o}+2.24y=13.44 m
Substituting this expression for B_{o} into equation for A, we obtain
\quad\quad\quad\quad1.74y² – 13.44y + 12.18 = 0
Solution of this equation yields y = 1.05 m. Let us use a freeboard of 0.6 m. Then, the depth of the section is 1.05 + 0.6 = 1.65 m and the width, B_{o} = 13.44 – 2.24 × 1.05 = 11.1 m.
\quad\quadNow,
\quad\quad\quad\quad S=3\times 10^{-4}\times(1.11)^{1.67}(8)^{0.167}\\\quad\quad\quad\quad\quad=5.05\times 10^{-4}