Question

What will be the regime scour depth for a rectangular channel having 30 m width carrying 430 cumecs water if the average grain size of bed material is 0.14 mm.

• A. 2.03 m
• B. 5.92 m
• C. 7.12 m
• D. 9.16 m

Answer 1

The correct option is D 9.16 m

Given,

$d_{mm}=0.14,Q=430cumecs,B=30m$

Silt factor,
$f=1.76\sqrt{d_{mm}}$

$=1.76\sqrt{0.14}=0.658$

$\text{Regime scour depth = 1.35}{\left(\frac{q^2}{f}\right)}^{1/3}$

$=1.35{\left[\frac{(430/30{)}^2}{0.658}\right]}^{1/3}$

$=9.16m$