Question

An open channel of symmetric right-angled triangular cross-section is conveying a discharge \( Q \). If \( g \) is the acceleration due to gravity, what is the critical depth for this channel?

• A. \( \left( \frac{Q^2}{g} \right)^{\frac{1}{3}} \)
• B. \( \left( \frac{Q^2}{g} \right)^{\frac{1}{5}} \)
• C. \( \left( \frac{Q^2}{g} \right)^{\frac{1}{4}} \)
• D. \( \left( \frac{Q^2}{g} \right)^{\frac{1}{2}} \)

Hint:

For open channels, the critical depth can be calculated using the discharge and gravitational acceleration, where the exponent depends on the type of channel geometry.

Answer 1

The Correct Option is A

The critical depth (\( y_c \)) for an open channel with a triangular cross-section is calculated using the formula: \[ y_c = \left( \frac{Q^2}{g} \right)^{\frac{1}{3}} \] In this formula, \( Q \) represents the discharge, and \( g \) denotes the acceleration due to gravity. This equation originates from the principles of specific energy in open channel flow.
Final Answer: \[ \boxed{\left( \frac{Q^2}{g} \right)^{\frac{1}{3}}} \]