Question

A triangular open channel has a vertex angle of ${90}^o$ and carries flow at a critical depth of 0.30 m. The discharge in the channel is

• A. $0.15m^3/s$
• B. $0.2m^3/s$
• C. $0.08m^3/s$
• D. $0.11m^3/s$

Answer 1

The correct option is D $0.11m^3/s$

The side slope of triangular open channel is given as z horizontal to 1 vertical.
When vertex angle is ${90}^o,z=1$



$Now,y_c={\left(\frac{2Q^2}{g{z}^2}\right)}^{1/5}$ for a triangular channel

$\Rightarrow Q^2=\frac{g{z}^2(y_c{)}^5}{2}$

$\Rightarrow Q^2=\frac{9.81\times (1{)}^2\times (0.30{)}^5}{2}$

$\Rightarrow Q=0.11m^3/s$