---
title: "94% depth for maximum discharge in circular channel"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:circular-channel-94-percent-depth-max-discharge"
source: https://www.bctengineer.com/q/semester_4/hydraulics/circular-channel-94-percent-depth-max-discharge
site: BCT Engineering
---

# 94% depth for maximum discharge in circular channel

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2072 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/circular-channel-94-percent-depth-max-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Using Manning's equation, show that the depth of flow is equal to 94% of the diameter for the partially filled most economical circular channel considering maximum discharge.

## Short answer

For maximum discharge in a circular channel, differentiate Manning's equation with respect to the angle θ, leading to 5θ(1-cosθ)=2(θ-sinθ). Solving gives θ=302.4°, and depth y = 0.5d(1-cos(θ/2)) ≈ 0.94d, i.e., 94% of the diameter.

## Answer

#### Condition for maximum discharge
Manning: $Q=\dfrac1nAR^{2/3}S^{1/2}=\dfrac{S^{1/2}}{n}\dfrac{A^{5/3}}{P^{2/3}}$. For constant $d$, $n$ and $S$, $Q$ is maximum when $A^{5/3}P^{-2/3}$ is maximum:
$$
\frac{dQ}{d\theta}=0\;\Rightarrow\;5P\frac{dA}{d\theta}=2A\frac{dP}{d\theta}
$$

#### Geometry
Let $\theta$ be the angle (rad) subtended at the centre by the wetted arc:
$$
A=\frac{d^2}{8}(\theta-\sin\theta),\quad P=\frac{d\theta}{2},\quad \frac{dA}{d\theta}=\frac{d^2}{8}(1-\cos\theta),\quad\frac{dP}{d\theta}=\frac d2
$$

```text
    ___
  /  |  \     y = depth
 |   |th |    d = diameter
  \  |__/
    ---
```

#### Solution
$$
5\left(\frac{d\theta}{2}\right)\frac{d^2}{8}(1-\cos\theta)=2\,\frac{d^2}{8}(\theta-\sin\theta)\frac d2
$$
$$
5\theta(1-\cos\theta)=2(\theta-\sin\theta)
$$
$$
3\theta-5\theta\cos\theta+2\sin\theta=0
$$
Solving by trial (or Newton's method): $\theta=5.2781$ rad $=302.41^\circ$.

| $\theta$ (deg) | $3\theta-5\theta\cos\theta+2\sin\theta$ |
|---|---|
| 290 | 4.649 |
| 302.4 | 0.000 |
| 310 | -2.690 |

#### Depth of flow
$$
y=\frac d2\left(1-\cos\frac\theta2\right)=\frac d2\left(1-\cos151.21^\circ\right)=0.9382\,d
$$
**So the depth for maximum discharge is $y\approx0.94\,d$ (94% of the diameter).**

## Check yourself

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

1. For fixed d, n and S, Manning's discharge in a circular channel is maximum when which quantity is maximum?
   A. A^(2/3)/P^(5/3)
   B. A x P
   C. A^(5/3)/P^(2/3)
   D. A/P^2
   Answer: C. Q = (S^(1/2)/n) A^(5/3) P^(-2/3), so the maximising condition is on A^(5/3)/P^(2/3). (Key point: Condition for maximum Q)

2. Setting dQ/d(theta) = 0 gives:
   A. 5P dA/dtheta = 2A dP/dtheta
   B. 2P dA/dtheta = 5A dP/dtheta
   C. P dA/dtheta = A dP/dtheta
   D. dA/dtheta = 0
   Answer: A. d/dtheta of A^(5/3)P^(-2/3) = 0 gives 5P A' = 2A P'. (Key point: Differentiated condition)

3. Substituting the circular segment geometry gives 3 theta - 5 theta cos(theta) + 2 sin(theta) = 0. Its root is about:
   A. theta = 180 degrees
   B. theta = 302.4 degrees (5.278 rad)
   C. theta = 240 degrees
   D. theta = 360 degrees
   Answer: B. Trial or Newton's method gives theta = 302.4 degrees. (Key point: Solving for the angle)

4. With theta = 302.4 degrees, the depth y = (d/2)(1 - cos(theta/2)) equals:
   A. 0.81 d
   B. 0.50 d
   C. 0.99 d
   D. 0.94 d
   Answer: D. cos(151.2 degrees) = -0.876, so y = 0.938 d, about 94% of the diameter. (Key point: Depth ratio)

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