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Hydraulics, open channels and fluid power
Most efficient rectangular channel width
Bed width of the open rectangular section that has the smallest wetted perimeter for a given flow area A.
- Ref
- HYD-25
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- bopt [m]
For a given flow area A, the wetted perimeter of an open rectangular channel is P = b + 2A/b. Setting dP/db = 0 gives b = √(2A), so b = 2y at the optimum: a width twice the depth. This gives the least perimeter, so the greatest hydraulic radius, and the largest flow for the slope and roughness. Construction cost, freeboard and land constraints often override this.
Bed width of the open rectangular section that has the smallest wetted perimeter for a given flow area A. At the optimum the width is twice the depth.
Calculator
- Optimum depth, b/22 m
- Wetted perimeter at the optimum, 2√(2A)8 m
- Hydraulic radius at the optimum, y/21 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 8 m² section is most efficient at 4 m wide and 2 m deep.
- At this shape the hydraulic radius is half the depth, 1 m, the largest possible for this area.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| A Flow area | 0.01 to 1,000 m² | 0.1414 | 44.72 | 44.58 | +10% | +4.88 % |
Sweep of Flow area
| Flow area (m²) | Optimum bed width (m) |
|---|---|
| 0.01 | 0.1414 |
| 0.04217 | 0.2904 |
| 0.1778 | 0.5964 |
| 0.7499 | 1.225 |
| 3.162 | 2.515 |
| 13.34 | 5.164 |
| 56.23 | 10.61 |
| 237.1 | 21.78 |
| 1,000 | 44.72 |
Sweep one variable
Variables and how each one contributes
Optimum bed width
What the equation gives you. Everything else feeds it.
Flow area
The flow area, in m². Raises the optimum width with the square root of the area. Raising it by 10 % (to 8.8) moves the result by +4.88 %, from 4 to 4.195 m.
Worked example with the values above
- Double the area: 2A = 16 m².
- Take the square root.
- b = 4 m, with a depth of 2 m.
Assumptions and limits
- Prismatic channels with a steady, nearly uniform flow, in SI units.
- A free surface at atmospheric pressure and a roughness that is the same all round the wetted perimeter.
Common mistakes
- Applying the result to a trapezoid or circle, which have different optimum shapes.
- Using the result as the final design without checking the velocity.
- Using flow in place of the flow area.
Statistics and uncertainty
This is an exact geometric result for a rectangle, so the only uncertainty is the flow area. It ignores freeboard, construction cost and side slopes, which often matter more than the small gain in hydraulic efficiency.
Where it is used
- Designing canals, drains, culverts and flumes.
- Estimating flood depth and checking bed shear for scour.
Maths you need
- Wetted perimeter: the length of channel boundary in contact with the water
- Manning’s equation: V = R^(2/3) S^(1/2) / n
- Hydraulic radius: area divided by wetted perimeter