---
title: "Depth for maximum discharge, triangular channel"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:triangular-channel-depth-for-max-discharge"
source: https://www.bctengineer.com/q/semester_4/hydraulics/triangular-channel-depth-for-max-discharge
site: BCT Engineering
---

# Depth for maximum discharge, triangular channel

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2072 Magh, 5 marks
- Video: [Most economical triangular channel section](https://www.youtube.com/watch?v=2D24RzHOVs0) — Civil Engineer FBH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/triangular-channel-depth-for-max-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

In a partially full channel having a triangular section as shown in figure, the rate of discharge $Q = KAR^{2/3}$, in which K = a constant; A = flow area and R = hydraulic radius. Determine the depth at which the discharge is maximum. [Figure: triangular channel with both sides at 60° to the horizontal (equilateral section) and flow depth h]

## Short answer

For a closed equilateral triangle with apex up, maximize Q=KA^(5/3)P^(-2/3) to get 8u²-u-5=0, where u=h/H. Solving yields u≈0.856. Thus maximum discharge occurs at depth h ≈ 0.856H ≈ 0.741a, with a the side length.

## Answer

**Reading of the figure:** the section is a closed equilateral triangle conduit (both sides at $60^\circ$ to the horizontal) with the base horizontal at the bottom and apex at the top, partly filled to depth $h$. (For an open V-shaped section the discharge would increase continuously with depth, so a maximum exists only for the closed apex-up section.) Let the side of the triangle be $a$ and its height $H=\dfrac{\sqrt3}{2}a$.

```text
        /\
       /  \         apex up
      / ~~ \  h     water of depth h
     /______\
        a
```

#### Geometry for depth $h$
Width of water surface at height $h$: $T=a\left(1-\dfrac hH\right)$.
$$
A=ah-\frac{a}{2H}h^2=a\left(h-\frac{h^2}{2H}\right)
$$
Wetted perimeter: base + the two sloping sides below the surface ($h/\sin60^\circ=2h/\sqrt3$ each):
$$
P=a+\frac{4h}{\sqrt3}
$$
Let $u=h/H$. Then $\dfrac{4h}{\sqrt3}=\dfrac{4Hu}{\sqrt3}=2au$:
$$
A=aH\left(u-\frac{u^2}{2}\right),\qquad P=a(1+2u)
$$

#### Maximum discharge
$Q=KAR^{2/3}=KA^{5/3}P^{-2/3}$. For maximum $Q$: $\dfrac{dQ}{dh}=0$:
$$
5P\frac{dA}{dh}=2A\frac{dP}{dh}
$$
$$
\frac{dA}{dh}=a(1-u),\qquad \frac{dP}{dh}=\frac{2a}{H}
$$
$$
5\,a(1+2u)\,a(1-u)=2\,aH\left(u-\frac{u^2}{2}\right)\frac{2a}{H}
$$
$$
5(1+u-2u^2)=4u-2u^2\;\Rightarrow\;8u^2-u-5=0
$$
$$
u=\frac{1+\sqrt{1+160}}{16}=\frac{1+12.689}{16}=0.8555
$$
$$
h=uH=0.8555\,H=0.7409\,a
$$
**Answer: the discharge is maximum when the depth is about 0.856 of the height of the triangle ($h\approx0.856H\approx0.741a$, where $a$ is the side).**

## Check yourself

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

1. Why does a maximum discharge exist for this section?
   A. Because the open V-section has a maximum at its brim
   B. Because the section is a closed equilateral triangle with apex up, so the width narrows near the top and perimeter keeps growing with little extra area
   C. Because discharge always peaks at half depth
   D. Because Q = K A R^(2/3) has no dependence on depth
   Answer: B. For an open V the discharge increases continuously; only the closed apex-up triangle has a maximum. (Key point: Which section the figure shows)

2. For depth h with u = h/H, the area and wetted perimeter are:
   A. A = aH u^2, P = a(1 + u)
   B. A = aH(u - u^2/2), P = a(1 + 2u)
   C. A = aH u, P = a(1 + 4u)
   D. A = aH(u + u^2/2), P = a(1 - 2u)
   Answer: B. Surface width falls as a(1-u), so A = aH(u - u^2/2); the base plus two slanting sides give P = a(1+2u). (Key point: Geometry in terms of u = h/H)

3. Applying 5P dA/dh = 2A dP/dh leads to which equation in u?
   A. u^2 - 5u + 1 = 0
   B. 5u^2 - u - 8 = 0
   C. 8u^2 - u - 5 = 0
   D. 2u^2 - 5u + 1 = 0
   Answer: C. 5(1 + u - 2u^2) = 4u - 2u^2 simplifies to 8u^2 - u - 5 = 0. (Key point: Maximum condition)

4. The positive root gives the depth for maximum discharge as:
   A. h = 0.50 H
   B. h = 0.856 H (about 0.741 a)
   C. h = H (full)
   D. h = 0.667 H
   Answer: B. u = (1 + sqrt(161))/16 = 0.8555, so h is about 0.856 of the triangle height. (Key point: Result)

## Related in Hydraulics

- [Economical triangular section hydraulic radius](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Amost-economical-triangular-section-hydraulic-radius) — Hydraulics, chapter 6
- [94% depth for maximum discharge in circular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acircular-channel-94-percent-depth-max-discharge) — Hydraulics, chapter 6
- [Hydraulic jump in 90° triangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ajump-90-degree-triangular-channel-flow-rate) — Hydraulics, chapter 9
- [Variation of hydraulic radius with depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahydraulic-radius-variation-wide-deep-channel) — Hydraulics, chapter 5, asked 3×
- [Jump relation in horizontal triangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ajump-triangular-channel-froude-relation) — Hydraulics, chapter 9
- [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6

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