---
title: "Circular culvert diameter and slope"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:circular-culvert-diameter-and-slope"
source: https://www.bctengineer.com/q/semester_4/hydraulics/circular-culvert-diameter-and-slope
site: BCT Engineering
---

# Circular culvert diameter and slope

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2074 Bhadra, 7 marks
- Video: [Normal Depth for a Partially-Full Circular Pipe (Part 1)](https://www.youtube.com/watch?v=gyGixsNdLGo) — HydroMohsen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/circular-culvert-diameter-and-slope
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

A circular culvert has a capacity of 0.5 m³/s when flowing full. Velocity should not be less than 0.7 m/s if the depth is one-fourth the diameter. Assuming uniform flow, determine diameter and slope taking manning's n = 0.012.

## Short answer

Velocity condition at y=d/4 gives full-flow velocity 0.999 m/s. For Q=0.5 m³/s, diameter d ≈ 0.80 m. The required slope S₀ ≈ 0.00123 (about 1 in 811).

## Answer

#### Given data
$Q_{full}=0.5$ m³/s, $n=0.012$. At depth $y=d/4$ the velocity must be at least $0.7$ m/s (so the limiting case is $V=0.7$ m/s at $y=d/4$).

#### Geometry at $y=d/4$
$$
\cos\frac\theta2=1-\frac{2y}{d}=0.5\;\Rightarrow\;\theta=120^\circ=2.0944\ \text{rad}
$$
$$
R'=\frac d4\left(1-\frac{\sin\theta}{\theta}\right)=\frac d4\left(1-\frac{0.8660}{2.0944}\right)=0.14663\,d
$$
Full flow: $R_f=\dfrac d4=0.25\,d$.

#### Relating velocities (same $n$ and $S_0$)
From Manning, $V\propto R^{2/3}$:
$$
\frac{V_{full}}{V_{d/4}}=\left(\frac{R_f}{R'}\right)^{2/3}=\left(\frac{0.25}{0.14663}\right)^{2/3}=1.4272
$$
$$
V_{full}=1.4272\times0.7=0.9990\ \text{m/s}
$$

#### Diameter
$$
Q=V_{full}\frac{\pi d^2}{4}\;\Rightarrow\;d=\sqrt{\frac{4\times0.5}{\pi\times0.9990}}=0.798\ \text{m}\approx0.80\ \text{m}
$$

#### Slope
$$
V_{full}=\frac1n\left(\frac d4\right)^{2/3}S_0^{1/2}\;\Rightarrow\;S_0=\left[\frac{nV_{full}}{(d/4)^{2/3}}\right]^2=\left[\frac{0.012\times0.9990}{(0.1996)^{2/3}}\right]^2=0.001232
$$
i.e. about 1 in 811.

**Answer: diameter $d\approx0.80$ m (provide 0.80 m) and bed slope $S_0\approx0.00123$ (1 in 811).**
Check with $d=0.80$ m: the slope needed for $Q=0.5$ m³/s is $0.00122$ (1 in 821), and then $V$ at $y=d/4$ is $0.697$ m/s, about 0.7 m/s.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The velocity should not be less than 0.7 m/s at depth d/4. Which case is used to size the culvert?
   A. V = 0.7 m/s at y = d/4 (the limiting case)
   B. V = 0.7 m/s when flowing full
   C. V = 0.7 m/s at y = d/2
   D. Maximum velocity at y = 0.81 d
   Answer: A. The minimum allowed velocity is just met at y = d/4, so V = 0.7 m/s there fixes the design. (Key point: Limiting case)

2. How are the full-flow and d/4-depth velocities related?
   A. Equal, since n and S0 are the same
   B. V_full/V_(d/4) = R_f/R'
   C. V_full/V_(d/4) = (R_f/R')^(2/3) from Manning's equation
   D. V_full/V_(d/4) = (R'/R_f)^(2/3)
   Answer: C. With the same n and S0, Manning gives V proportional to R^(2/3), so the ratio is (0.25/0.1466)^(2/3) = 1.427. (Key point: Relating partial and full velocity)

3. At y = d/4 (theta = 120 degrees) the hydraulic radius compared with the full-pipe value d/4 is:
   A. Equal to d/4
   B. R' = 0.1466 d, which is smaller than d/4
   C. R' = 0.3 d, which is larger than d/4
   D. R' = d/2
   Answer: B. R' = (d/4)(1 - sin(theta)/theta) = 0.1466 d, less than the full value 0.25 d. (Key point: Hydraulic radii)

4. After V_full = 1.427 x 0.7 = 0.999 m/s, how are d and S0 found?
   A. d from Q = V pi d^2/4 (0.80 m), then S0 from Manning with R = d/4
   B. S0 first from Chezy, then d from continuity of d/4
   C. d = 0.5 m from Q = 0.5
   D. d from V = sqrt(g d)
   Answer: A. d = sqrt(4Q/(pi V)) = 0.798 m, then S0 = [n V/(d/4)^(2/3)]^2 = 0.00123. (Key point: Diameter and slope)

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