---
title: "Hydraulic exponent for rectangular section 10/3"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:hydraulic-exponent-rectangular-10-3"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-exponent-rectangular-10-3
site: BCT Engineering
---

# Hydraulic exponent for rectangular section 10/3

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2073 Bhadra, 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-rectangular-10-3
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Define hydraulic exponent. Show that the value of hydraulic exponent for rectangular section is equal to 10/3.

## Short answer

Hydraulic exponent N is defined by K^2 proportional to y^N. For a wide rectangular channel A=by, P≈b, hence R≈y and conveyance K=(b/n)y^(5/3). Thus K^2 ∝ y^(10/3), giving N=10/3.

## Answer

#### Hydraulic exponent
In a prismatic channel, a **hydraulic exponent** is the power of the depth to which a section factor is proportional. The **second hydraulic exponent $N$** (for uniform flow) is defined by
$$
K^2\propto y^{N},\qquad K=\frac1nAR^{2/3}=\frac{A^{5/3}}{nP^{2/3}}
$$
where $K$ is the conveyance. (The first exponent $M$ is defined by $A^3/T\propto y^M$ for critical flow.)

#### General expression
Take logarithms and differentiate with respect to $y$:
$$
\ln K=\frac53\ln A-\frac23\ln P-\ln n
$$
$$
\frac{y}{K}\frac{dK}{dy}=\frac53\frac yA\frac{dA}{dy}-\frac23\frac yP\frac{dP}{dy}=\frac N2
$$
(since $K\propto y^{N/2}$, $\dfrac yK\dfrac{dK}{dy}=\dfrac N2$). So
$$
N=\frac{10}{3}\frac{y}{A}\frac{dA}{dy}-\frac43\frac yP\frac{dP}{dy}
$$

#### Rectangular section (wide channel, $b\gg y$)
$A=by$, $\dfrac{dA}{dy}=b$, so $\dfrac yA\dfrac{dA}{dy}=\dfrac{y\,b}{by}=1$.

$P=b+2y\approx b$, $\dfrac{dP}{dy}=2$, so $\dfrac yP\dfrac{dP}{dy}=\dfrac{2y}{b+2y}\to0$.
$$
N=\frac{10}{3}(1)-\frac43(0)=\boxed{\frac{10}{3}}
$$
Directly: for a wide rectangle $R\approx y$, so $K=\dfrac{b}{n}y\cdot y^{2/3}=\dfrac bn y^{5/3}$, and $K^2\propto y^{10/3}$ ✓.

## Check yourself

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

1. The (second) hydraulic exponent N for uniform flow is defined through:
   A. Z^2 proportional to y^N
   B. K^2 proportional to y^N, where K is the conveyance
   C. A^2 proportional to y^N
   D. Q proportional to y^N
   Answer: B. N is defined by K^2 proportional to y^N with K = (1/n) A R^(2/3). (Key point: Definition of hydraulic exponent)

2. Differentiating ln K = (5/3) ln A - (2/3) ln P - ln n with respect to y gives N from:
   A. N = (10/3)(y/A)dA/dy - (4/3)(y/P)dP/dy
   B. N = (5/3)(y/A)dA/dy + (2/3)(y/P)dP/dy
   C. N = (5/3)(y/A)dA/dy - (2/3)(y/P)dP/dy
   D. N = (10/3)(y/A)dA/dy + (4/3)(y/P)dP/dy
   Answer: A. Since (y/K)dK/dy = N/2, doubling the right side gives N = (10/3)(y/A)dA/dy - (4/3)(y/P)dP/dy. (Key point: General expression)

3. For a wide rectangular channel (b >> y), why does N become 10/3?
   A. Because (y/A)dA/dy = 1 and (y/P)dP/dy tends to 0
   B. Because P is proportional to y
   C. Because A is proportional to y^2
   D. Because R is constant
   Answer: A. A = by gives (y/A)dA/dy = 1, and P = b + 2y is nearly constant so its term vanishes: N = 10/3. (Key point: Wide rectangle)

## Close variants set in other exams

- [Hydraulic exponent for triangular section 16/3](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahydraulic-exponent-triangular-16-3) — Hydraulics, chapter 6, asked 2×

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