Question

The critical flow condition in a channel is given by \underline{\hspace{1.5cm}.} [Note: $\alpha$ – kinetic energy correction factor; $Q$ – discharge; $A_c$ – cross-sectional area of flow at critical flow condition; $T_c$ – top width of flow at critical flow condition; $g$ – acceleration due to gravity]

• A. $\dfrac{\alpha Q^2}{g}=\dfrac{A_c^3}{T_c}$
• B. $\dfrac{\alpha Q}{g}=\dfrac{A_c^3}{T_c^2}$
• C. $\dfrac{\alpha Q^2}{g}=\dfrac{A_c^3}{T_c^2}$
• D. $\dfrac{\alpha Q}{g}=\dfrac{A_c^3}{T_c}$

Hint:

At critical flow, always use $Fr=1$. Replace velocity by $Q/A$ and depth by $A/T$. This gives the relation $\dfrac{\alpha Q^2}{g}=\dfrac{A^3}{T}$ for critical state.

Answer 1

The Correct Option is A

Step 1: Recall the condition for critical flow.
Critical flow in open channels occurs when the Froude number $Fr=1$.
\[ Fr=\frac{V}{\sqrt{gD}} \] where $V=\dfrac{Q}{A}$ is the mean velocity and $D=\dfrac{A}{T}$ is the hydraulic depth.

Step 2: Apply the critical flow condition.
At critical flow, \[ \frac{Q}{A_c}\cdot \frac{1}{\sqrt{g\,(A_c/T_c)}}=1. \]

Step 3: Rearrange.
\[ \frac{Q^2}{g} = \frac{A_c^3}{T_c}. \]

Step 4: Include the kinetic energy correction factor $\alpha$.
Considering $\alpha$, \[ \frac{\alpha Q^2}{g}=\frac{A_c^3}{T_c}. \] \[ \boxed{\dfrac{\alpha Q^2}{g}=\dfrac{A_c^3}{T_c}} \]