If ‘m’ is hydraulic mean depth and ‘d’ is a depth of flow, then the discharge through rectangular channel is maximum when
- Am = d/3
- Bm = d/2Correct
- Cm = 2d
- Dm = 3d/2
Solution
Concept
Most economical rectangular channel hydraulic section
Formula
NONE
Step-by-step solution
1. For a given cross-sectional area, a channel section is hydraulically most efficient when its wetted perimeter is minimized. 2. For a rectangular channel of width b and depth d, the most efficient section occurs when width b = 2d. 3. Hydraulic mean depth m = Area / Wetted Perimeter = (b × d) / (b + 2d) = (2d × d) / (2d + 2d) = 2d² / 4d = d/2. Answer: B
Explanation
A channel section provides maximum discharge for a given cross-sectional area when its wetted perimeter is minimized. For a rectangular open channel, this condition of maximum efficiency occurs when the bed width equals twice the depth of flow. Under these optimal dimensions, the hydraulic mean depth simplifies directly to half the flow depth.
Why the other options are wrong
A) m = d/3 does not satisfy the condition for the most economical or maximum discharge rectangular channel. C) m = 2d yields physically impossible channel dimensions for maximum discharge. D) m = 3d/2 does not meet the condition that hydraulic radius must equal half the depth of flow.
Shortcut
Maximum discharge for a rectangular channel occurs when hydraulic depth m = d/2.
Common mistake
Confusing the relation between hydraulic mean radius and depth (m = d/2) with the channel width to depth ratio (b = 2d).
Correct answer: B) m = d/2