---
title: "Width equals 8/9 of minimum specific energy"
subject: "Hydraulics"
chapter: 7
key: "semester_4/hydraulics:critical-flow-min-wetted-perimeter-width-8-9"
source: https://www.bctengineer.com/q/semester_4/hydraulics/critical-flow-min-wetted-perimeter-width-8-9
site: BCT Engineering
---

# Width equals 8/9 of minimum specific energy

Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow

- Asked in: 2078 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/critical-flow-min-wetted-perimeter-width-8-9
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7

## Question

A rectangular channel section to have critical flow and at the same time the wetted perimeter is to be minimum. Show that for these two conditions simultaneously, the width of the channel must be equal to 8/9 times minimum specific energy.

## Short answer

Minimizing wetted perimeter P = b + 2yc under critical flow yields b = (4/3)yc. Using yc = (2/3)Emin gives b = (8/9)Emin.

## Answer

#### Given conditions (rectangular channel, width $b$, discharge $Q$)
1. Flow is critical.
2. The wetted perimeter is minimum (for the given discharge).

#### Critical flow in a rectangular channel
$$
q=\frac Qb,\quad y_c=\sqrt[3]{\frac{q^2}{g}}=\left(\frac{Q^2}{gb^2}\right)^{1/3},\qquad E_{min}=\frac32y_c
$$

#### Minimum wetted perimeter
$$
P=b+2y_c=b+2\left(\frac{Q^2}{g}\right)^{1/3}b^{-2/3}
$$
Differentiate with respect to $b$ ($Q$ constant) and equate to zero:
$$
\frac{dP}{db}=1-\frac43\left(\frac{Q^2}{g}\right)^{1/3}b^{-5/3}=0
$$
$$
b^{5/3}=\frac43\left(\frac{Q^2}{g}\right)^{1/3}
$$
Since $\left(\dfrac{Q^2}{g}\right)^{1/3}=y_c\,b^{2/3}$ (from $y_c^3=Q^2/(gb^2)$):
$$
b^{5/3}=\frac43y_cb^{2/3}\;\Rightarrow\;b=\frac43y_c
$$
(The second derivative is positive, so this is a minimum.)

#### Relation with $E_{min}$
From $E_{min}=\tfrac32y_c$, $y_c=\tfrac23E_{min}$:
$$
b=\frac43\times\frac23E_{min}=\frac89E_{min}
$$
$$
\boxed{b=\frac89E_{min}}
$$
Hence proved.

## Check yourself

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

1. For critical flow in a rectangular channel the wetted perimeter written in terms of b is P = b + 2yc where yc is
   A. (Q^2/(g b^2))^(1/3)
   B. Q/(b sqrt(g))
   C. (Q/b)^2/g
   D. (2/3) b
   Answer: A. yc = (q^2/g)^(1/3) with q = Q/b, so P depends on b through yc. (Key point: Wetted perimeter expression)

2. Setting dP/db = 0 with Q constant gives the relation
   A. b = yc
   B. b = 2yc
   C. b = 4yc/3
   D. b = 3yc/2
   Answer: C. Differentiation gives b^(5/3) = (4/3)(Q^2/g)^(1/3), and using (Q^2/g)^(1/3) = yc b^(2/3) gives b = 4yc/3. (Key point: Minimisation)

3. Using Emin = 1.5yc, b = 4yc/3 becomes
   A. b = 2 Emin
   B. b = 4 Emin/9
   C. b = 9 Emin/8
   D. b = 8 Emin/9
   Answer: D. yc = 2Emin/3, so b = (4/3)(2/3)Emin = 8Emin/9. (Key point: Link to Emin)

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