---
title: "Hydraulic exponent for triangular section 16/3"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:hydraulic-exponent-triangular-16-3"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-exponent-triangular-16-3
site: BCT Engineering
---

# Hydraulic exponent for triangular section 16/3

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2073 Magh, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-exponent-triangular-16-3
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Define hydraulic exponent. Show that the value of hydraulic exponent for triangular section is equal to 16/3.

## Short answer

For a triangular section with side slope z, area A ∝ y^2 and wetted perimeter P ∝ y, so hydraulic radius R ∝ y. Conveyance K ∝ y^2 * y^(2/3) = y^(8/3), therefore K^2 ∝ y^(16/3) and N=16/3.

## Answer

#### Hydraulic exponent
The **second hydraulic exponent $N$** (for uniform flow) is defined by $K^2\propto y^N$, where the conveyance is
$$
K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}
$$
(The first exponent $M$ is defined by $A^3/T\propto y^M$ for critical flow.)

#### General expression
$$
\frac{y}{K}\frac{dK}{dy}=\frac53\frac yA\frac{dA}{dy}-\frac23\frac yP\frac{dP}{dy}=\frac N2
$$

#### Triangular section (side slope $z$)
$$
A=zy^2,\qquad \frac{dA}{dy}=2zy,\qquad \frac yA\frac{dA}{dy}=\frac{y\cdot2zy}{zy^2}=2
$$
$$
P=2y\sqrt{1+z^2},\qquad \frac{dP}{dy}=2\sqrt{1+z^2},\qquad \frac yP\frac{dP}{dy}=1
$$
$$
\frac N2=\frac53(2)-\frac23(1)=\frac{10}{3}-\frac23=\frac83
$$
$$
\boxed{N=\frac{16}{3}}
$$
Directly: $R=\dfrac{zy}{2\sqrt{1+z^2}}\propto y$, so $K\propto y^2\cdot y^{2/3}=y^{8/3}$ and $K^2\propto y^{16/3}$ ✓. This is exact for a triangle for all depths because $A\propto y^2$ and $P\propto y$.

## Check yourself

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

1. For a triangular section with side slope z, how do A and P vary with depth?
   A. A proportional to y and P proportional to y^2
   B. A proportional to y^2 and P proportional to y
   C. Both proportional to y
   D. Both proportional to y^2
   Answer: B. A = z y^2 and P = 2y sqrt(1+z^2). (Key point: Triangular geometry)

2. For the triangle, (y/A)dA/dy = 2 and (y/P)dP/dy = 1. Then N/2 equals:
   A. (5/3)(2) + (2/3)(1) = 4
   B. (5/3)(1) - (2/3)(2) = 1/3
   C. (5/3)(2) - (2/3)(1) = 8/3
   D. (2)(1) = 2
   Answer: C. N/2 = 10/3 - 2/3 = 8/3, so N = 16/3. (Key point: Using the log-derivative)

3. Why is N = 16/3 exact for a triangle at every depth?
   A. Because K is independent of y
   B. Because the top width is constant
   C. Because the section is wide
   D. Because R is proportional to y, so K is proportional to y^2 x y^(2/3) = y^(8/3) and K^2 to y^(16/3)
   Answer: D. A is proportional to y^2 and R to y exactly, so K goes as y^(8/3) and K^2 as y^(16/3). (Key point: Direct check)

## Close variants set in other exams

- [Hydraulic exponent for rectangular section 10/3](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahydraulic-exponent-rectangular-10-3) — Hydraulics, chapter 6, asked 2×

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