Question

A rectangular channel 6.0 m wide carries a discharge of 16.0$m^3/s$ under uniform flow condition with normal depth of 1.60 m. Manning's 'n' is 0.015.

The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition is

• A. 3.0 m
• B. 3.8 m
• C. 4.1 m
• D. 4.5 m

Answer 1

The correct option is C 4.1 m

If the channel is contracted, the depth of flow at this location decreases. Thus width of the channel can be decreased till the specific energy becomes minimum. If the channel is contracted further, the depth of flow starts decreasing.

$\therefore y+\frac{Q^2}{2g{A}^2}=E_{min}$

$\Rightarrow 1.6+\frac{(16{)}^2}{2\times 9.81\times (9.6{)}^2}=\frac{3}{2}{y}_c$

$\Rightarrow 1.74=\frac{3}{2}\times {\left(\frac{Q^2/b^2}{g}\right)}^{1/3}$

$\Rightarrow 1.74=\frac{3}{2}\times {\left[\frac{(1.6{)}^2}{9.81}\right]}^{1/3}\times \frac{1}{(b{)}^{2/3}}$

$\Rightarrow b^{2/3}=2.557$

$\Rightarrow b=4.1m$