Question

A triangular open channel has a vortex angle of 90° and carries flow at a critical depth of 0.3 m. The discharge in the channel is
1. 0.41 m³ / s
2. 0.11 m³ / s
3. 0.21 m³ / s
4. 0.31 m³ / s

Answer 1

Correct Answer - Option 2 : 0.11 m³ / s

Concept:

The relation between discharge and critical depth of flow for triangular channel is given as:

\({y_c} = \;{\left( {\frac{{2{Q^2}}}{{g{m^2}}}} \right)^{\frac{1}{5}}}\)

Where,

M is side slope of channel

Q is the discharge

Y_c is the critical depth of flow

Calculation:

Given

y_c = 0.3 m; Side slope when vortex angle is 90⁰, m = 1

Now

\(0.3 = {\left( {\frac{{2{Q^2}}}{{9.81\; \times \;{1^2}}}} \right)^{\frac{1}{5}}}\)

⇒ Q = 0.11 m³/s

Note:

The above same relation in case of rectangular channel is given as:

\({y_c} = {\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}}\)

Where, q is the discharge per unit width.