---
title: "Normal depth in triangular channel (1:1)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:triangular-channel-normal-depth-0-0004"
source: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-channel-normal-depth-0-0004
site: BCT Engineering
---

# Normal depth in triangular channel (1:1)

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

- Asked in: 2074 Bhadra, 8 marks
- 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-normal-depth-0-0004
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Find the normal depth of flow in a triangular channel having longitudinal slope of 0.0004 and side slopes of 1:1 when it carries 1 m³/sec, manning roughness coefficient (n) = 0.014.

## Short answer

For a triangular channel with 1:1 side slopes (m=1), longitudinal slope 0.0004, discharge 1 m³/s, and Manning's n=0.014, the normal depth is calculated using Manning's equation as y ≈ 1.13 m. The geometry gives A=y² and R=0.3536y.

## Answer

Normal depth is the uniform flow depth from Manning's equation.

#### Geometry (side slope 1:1, $m = 1$)
$$
\begin{aligned}
A &= my^2 = y^2 \\
P &= 2y\sqrt{1+1^2} = 2\sqrt{2}\,y = 2.828y \\
R &= \frac{y^2}{2.828y} = 0.3536y
\end{aligned}
$$

#### Manning's equation
$$
\begin{aligned}
Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\
1 &= \frac{1}{0.014}\,y^2\,(0.3536y)^{2/3}(0.0004)^{1/2} \\
1 &= 71.43 \times 0.5 \times 0.02\, y^{8/3} = 0.7143\, y^{8/3} \\
y^{8/3} &= 1.400 \;\Rightarrow\; y = 1.400^{3/8} = 1.134\ \text{m}
\end{aligned}
$$
(Here $0.3536^{2/3} = 0.5$.)

#### Check
$A = 1.134^2 = 1.287$ m², $P = 3.209$ m, $R = 0.401$ m, $V = 1/1.287 = 0.777$ m/s, $Q = 71.43 \times 1.287 \times 0.401^{2/3} \times 0.02 = 1.00$ m³/s.

**Answer:** Normal depth $y_n \approx 1.13$ m.

## Check yourself

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

1. For a 1:1 triangular channel, A and R in terms of y are:
   A. A = y^2, R = 0.3536 y
   B. A = 2y^2, R = 0.3536 y
   C. A = y^2, R = y/2
   D. A = y^2/2, R = 0.5 y
   Answer: A. A = y^2, P = 2.828 y, so R = 0.3536 y. (Key point: Geometry)

2. Substituting into Manning's equation reduces it to which form?
   A. 1 = 0.7143 y^(8/3)
   B. 1 = 0.7143 y^2
   C. 1 = 71.43 y
   D. 1 = 0.02 y^(5/3)
   Answer: A. 71.43 x 0.5 x 0.02 = 0.7143, with 0.3536^(2/3) = 0.5, and A R^(2/3) proportional to y^(8/3). (Key point: Manning setup)

3. How is y obtained from y^(8/3) = 1.400?
   A. y = 1.4^(8/3)
   B. y = 1.4^(3/8)
   C. y = 1.4^(1/2)
   D. y = 1.4 x 3/8
   Answer: B. Raise both sides to 3/8: y = 1.4^(0.375) = 1.134 m. (Key point: Solving)

4. How can the answer be verified?
   A. Recompute Q from A, R and S with y = 1.134 and confirm Q = 1 m^3/s
   B. Check that V = 1 m/s
   C. Check that R = y
   D. Check that P = A
   Answer: A. Substituting y back should reproduce the given discharge of 1 m^3/s. (Key point: Check)

## Close variants set in other exams

- [Flow depth in triangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-depth-1-in-2000) — Hydro Power, chapter 1, asked 2×

## Related in Hydro Power

- [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2×
- [Efficient trapezoidal section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-trap) — Hydro Power, chapter 1, asked 2×

## Related in other subjects

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- [Bed slope for critical normal depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abed-slope-for-critical-flow-at-normal-depth) — Hydraulics, chapter 7
- [Equivalent roughness of compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-equivalent-roughness-horton-einstein) — Hydraulics, chapter 6