---
title: "Depth for maximum velocity in circular channel"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:most-economical-circular-max-velocity-depth"
source: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-circular-max-velocity-depth
site: BCT Engineering
---

# Depth for maximum velocity in circular channel

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2081 Chaitra, 6+2 marks; 2068 Bhadra, 5+2 marks (asked 2 times)
- Video: [CONDITION FOR MAXIMUM VELOCITY IN MOST ECONOMICAL CIRCULAR SECTION. OPEN CHANNEL FLOW.](https://www.youtube.com/watch?v=fcHYQ1bdcYU) — ROYAL EXPERT ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-circular-max-velocity-depth
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Find an expression for the theoretical depth for maximum velocity in a closed circular channel in terms of the diameter d. Compare the discharge at maximum velocity with that when the channel is running full, assuming that the Chezy's coefficient is unaltered, and the pressure remains atmospheric.

## Short answer

For a circular channel flowing partly full, theoretical depth for maximum velocity is y≈0.81d, found by maximizing hydraulic radius (tanθ=θ, θ≈257.5°). At this depth, velocity is 1.103 times full-flow velocity and discharge is 0.960 times full-pipe discharge.

## Answer

#### Geometry of a partly full circular channel
Let $\theta$ be the angle (radians) subtended at the centre by the wetted arc, $d$ the diameter and $y$ the depth.
$$
A=\frac{d^2}{8}(\theta-\sin\theta),\qquad P=\frac{d\,\theta}{2},\qquad R=\frac{A}{P}=\frac d4\left(1-\frac{\sin\theta}{\theta}\right),\qquad y=\frac d2\left(1-\cos\frac\theta2\right)
$$

```text
    ___
  /  |  \      y = depth
 |   |th |     th = angle at centre
  \  | _/      subtended by wetted arc
    ---
```

#### Depth for maximum velocity
Chezy: $V=C\sqrt{RS_0}$. With $C$ and $S_0$ constant, $V$ is maximum when $R$ is maximum:
$$
\frac{dR}{d\theta}=\frac d4\left[-\frac{\theta\cos\theta-\sin\theta}{\theta^2}\right]=0\;\Rightarrow\;\theta\cos\theta=\sin\theta\;\Rightarrow\;\tan\theta=\theta
$$
Solving by trial, $\theta=4.4934$ rad $=257.45^\circ$ (the root between $\pi$ and $2\pi$).
$$
y=\frac d2\left(1-\cos\frac{\theta}{2}\right)=\frac d2\left(1-\cos 128.73^\circ\right)=0.8128\,d
$$
**So the depth for maximum velocity is $y\approx0.81\,d$.**

#### Discharge comparison with full flow
At maximum velocity: $A=\dfrac{d^2}{8}(\theta-\sin\theta)=0.6837\,d^2$ and $R=0.3043\,d$.

Running full: $A_f=\dfrac{\pi}{4}d^2=0.7854\,d^2$, $R_f=\dfrac d4=0.25\,d$.

$$
\frac{V_{max}}{V_f}=\sqrt{\frac{R}{R_f}}=\sqrt{\frac{0.3043}{0.25}}=1.1033
$$
$$
\frac{Q_{V_{max}}}{Q_f}=\frac{A\sqrt R}{A_f\sqrt{R_f}}=\frac{0.6837\sqrt{0.3043}}{0.7854\sqrt{0.25}}=0.9604
$$
**Answer: depth for maximum velocity $=0.81\,d$; velocity is 1.103 times the full-flow velocity, and the discharge is 0.960 times (about 96.0% of) the full-pipe discharge.** So the maximum velocity occurs at a discharge slightly less than the full-flow value.

## Check yourself

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

1. In a partly full circular channel with C and S0 constant, velocity is maximum when:
   A. Area is maximum
   B. Hydraulic radius R is maximum
   C. Wetted perimeter is maximum
   D. Top width is maximum
   Answer: B. V = C sqrt(R S0), so V is maximum where R is maximum. (Key point: Condition for max velocity)

2. Setting dR/d(theta) = 0 leads to which condition?
   A. sin(theta) = 0
   B. cos(theta) = 0
   C. tan(theta) = theta
   D. theta = pi
   Answer: C. R = (d/4)(1 - sin(theta)/theta); differentiating gives theta cos(theta) = sin(theta), i.e. tan(theta) = theta. (Key point: Equation for theta)

3. What depth gives maximum velocity?
   A. 0.50 d
   B. 0.95 d
   C. 0.81 d
   D. 1.00 d
   Answer: C. theta = 4.4934 rad gives y = (d/2)(1 - cos(128.73 deg)) = 0.8128 d. (Key point: Depth)

4. How does the discharge at maximum velocity compare with the full-pipe discharge?
   A. About 96% of full discharge
   B. About 110% of full discharge
   C. Exactly equal
   D. About 50%
   Answer: A. Velocity is 1.103 times the full velocity but area is smaller, giving Q/Qf = 0.960. (Key point: Discharge comparison)

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