Question

The discharge through a trapezoidal channel is maximum when

• A. $b = 2d \tan \theta/2$
• B. $d = 2b \tan \theta/2$
• C. $d = 2b$
• D. $b = 2d$

Hint:

For the most efficient channel design:
- Trapezoidal: Sloping side = Half of top width. This simplifies to \(b = 2d\tan(\theta/2)\).
- Rectangular: Bed width is twice the depth of flow (\(b = 2d\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a given cross-sectional area and bed slope, a channel section that maximizes discharge is referred to as the "most economical" or "most efficient" section.
According to Manning's or Chezy's formula, discharge is maximized when the wetted perimeter is minimized, which maximizes the hydraulic radius.
Key Formula or Approach:
For a trapezoidal channel with a bottom width \(b\), flow depth \(d\), and side slopes making an angle \(\theta\) with the horizontal:
The condition for the most economical section requires the length of the sloping side to be equal to half of the top width:
\[ d\sqrt{1 + n^2} = \frac{b + 2nd}{2} \] Where \(n = \cot \theta\) represents the horizontal-to-vertical side slope ratio.

Step 2: Detailed Explanation:
Let us simplify this condition using trigonometric relationships:
\[ 2d\sqrt{1 + \cot^2 \theta} = b + 2d\cot \theta \] Since \(\sqrt{1 + \cot^2 \theta} = \csc \theta\):
\[ 2d\csc \theta = b + 2d\cot \theta \] Rearranging the equation to solve for the bed width \(b\):
\[ b = 2d\csc \theta - 2d\cot \theta \] \[ b = 2d(\csc \theta - \cot \theta) \] Now, substitute the trigonometric definitions:
\[ b = 2d\left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right) \] \[ b = 2d\left( \frac{1 - \cos \theta}{\sin \theta} \right) \] Using half-angle trigonometric identities:
\[ 1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) \] \[ \sin \theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \] Substituting these into the equation:
\[ b = 2d \left[ \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} \right] \] \[ b = 2d\tan\left(\frac{\theta}{2}\right) \] This mathematical relationship represents the optimal bottom width required to maximize discharge in a trapezoidal channel.

Step 3: Final Answer:
The discharge is maximum when \(b = 2d\tan\left(\frac{\theta}{2}\right)\).