---
title: "Flow depth in triangular channel"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:triangular-channel-depth-1-in-2000"
source: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-channel-depth-1-in-2000
site: BCT Engineering
---

# Flow depth in triangular channel

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2082 Chaitra, 6 marks
- Video: [Numerical solution 2078 chaitra || open channel flow || triangular channel || fluid || hydropower](https://www.youtube.com/watch?v=o780EbX3rRQ) — House Of Learners
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-channel-depth-1-in-2000
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Find the depth of flow in a triangular channel having a longitudinal slope of 1 in 2000 and side slopes of 1.5H:1V. The channel carries a discharge of 2.5 m³/s. Take Manning's roughness coefficient n = 0.025.

## Short answer

Using Manning's equation for uniform flow in a triangular channel (m=1.5, S=1/2000, n=0.025), the normal depth is calculated as y ≈ 1.57 m. The geometry gives A=1.5y², R=0.416y, and solving Q=(1/n)AR^(2/3)S^(1/2) yields y=1.57 m.

## Answer

Uniform flow depth is found with Manning's equation $Q = \dfrac{1}{n}AR^{2/3}S^{1/2}$.

#### Geometry (side slope $m = 1.5$ H : 1 V)
$$
\begin{aligned}
A &= my^2 = 1.5y^2 \\
P &= 2y\sqrt{1+m^2} = 2y\sqrt{3.25} = 3.606y \\
R &= \frac{A}{P} = \frac{1.5y^2}{3.606y} = 0.416y
\end{aligned}
$$

#### Manning's equation
$$
\begin{aligned}
2.5 &= \frac{1}{0.025}(1.5y^2)(0.416y)^{2/3}\left(\frac{1}{2000}\right)^{1/2} \\
2.5 &= 40 \times 1.5 \times 0.5573 \times 0.02236 \times y^{8/3} \\
2.5 &= 0.7477\, y^{8/3} \\
y^{8/3} &= 3.344 \;\Rightarrow\; y = 3.344^{3/8} = 1.572\ \text{m}
\end{aligned}
$$

#### Check
$A = 1.5(1.572)^2 = 3.709$ m², $P = 5.670$ m, $R = 0.654$ m, $V = 2.5/3.709 = 0.674$ m/s, and $\frac{1}{0.025}(3.709)(0.654)^{2/3}(0.02236) = 2.50$ m³/s. Correct.

**Answer:** Depth of flow $y \approx 1.57$ m.

## Check yourself

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

1. Which equation is used to find the uniform flow depth?
   A. Manning's equation Q = (1/n) A R^(2/3) S^(1/2)
   B. Joukowsky's equation
   C. Darcy-Weisbach with f
   D. Specific energy equation
   Answer: A. Uniform flow depth comes from Manning's equation with A and R as functions of y. (Key point: Equation used)

2. For side slope m = 1.5 the area and hydraulic radius are:
   A. A = 1.5 y^2, R = 0.416 y
   B. A = 3 y^2, R = 0.416 y
   C. A = 1.5 y^2, R = 0.75 y
   D. A = 1.5 y, R = y/2
   Answer: A. A = m y^2 = 1.5 y^2, P = 2y sqrt(1+m^2) = 3.606 y, so R = 0.416 y. (Key point: Geometry)

3. Combining the terms in Manning's equation leaves y to what power?
   A. y^2
   B. y^(8/3)
   C. y^(5/3)
   D. y^(1/2)
   Answer: B. A R^(2/3) is proportional to y^2 x y^(2/3) = y^(8/3). (Key point: Solving)

4. The depth of flow is:
   A. 0.57 m
   B. 1.57 m
   C. 2.57 m
   D. 3.34 m
   Answer: B. y^(8/3) = 3.344, so y = 3.344^(3/8) = 1.572 m. (Key point: Result)

## Close variants set in other exams

- [Normal depth in triangular channel (1:1)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-0-0004) — Hydro Power, chapter 1, asked 2×

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