---
title: "Half-hexagon channel dimensions and Manning n"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:half-hexagon-channel-chezy-manning-n"
source: https://www.bctengineer.com/q/semester_4/hydraulics/half-hexagon-channel-chezy-manning-n
site: BCT Engineering
---

# Half-hexagon channel dimensions and Manning n

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2079 Chaitra, 8 marks
- Video: [Most economical trapezoid channel section //open channel flow //](https://www.youtube.com/watch?v=9PJhN517foA) — Civil Engineer FBH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/half-hexagon-channel-chezy-manning-n
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

An open channel of most economical section having the form of a half hexagon with horizontal bottom is required to give a maximum discharge of 20.2 m³/s of water. The slope of the channel bottom is 1 in 2500. Take Chezy's $C = 60\,m^{1/2}/s$. Determine the dimension of the channel section and Manning's roughness coefficient.

## Short answer

For half-hexagon (z=1/√3), solving Q=20.2 m³/s with C=60 and slope 1/2500 gives y=2.85 m, b=3.29 m. Manning's n from C = R¹/⁶/n is 0.0177.

## Answer

#### Properties of the half-hexagon section
The most economical trapezoid with the best side slope is half of a regular hexagon: side slope $60^\circ$ ($z=1/\sqrt3$), bed width $b$ = each sloping side $=\dfrac{2y}{\sqrt3}$, $R=\dfrac y2$.
$$
A=(b+zy)y=\left(\frac{2y}{\sqrt3}+\frac{y}{\sqrt3}\right)y=\sqrt3\,y^2,\qquad P=3b=\frac{6y}{\sqrt3}=2\sqrt3\,y
$$

```text
   b
  ____
 /60  \     b = side = 2y/sqrt(3)
/______\
```

#### Given data
$Q=20.2$ m³/s, $S_0=1/2500$, $C=60$ m$^{1/2}$/s.

#### Depth from Chezy's equation
$$
Q=CA\sqrt{RS_0}=60\,(\sqrt3\,y^2)\sqrt{\frac y2\times\frac{1}{2500}}
$$
$$
y^{5/2}=\frac{20.2}{60\times\sqrt3\times\sqrt{0.5/2500}}=13.7444\;\Rightarrow\;y=2.853\ \text{m}
$$
$$
b=\frac{2y}{\sqrt3}=3.294\ \text{m},\qquad \text{top width}=b+2zy=2b=6.588\ \text{m}
$$
$$
A=\sqrt3y^2=14.095\ \text{m}^2,\qquad R=\frac y2=1.426\ \text{m}
$$
Check: $Q=60\times14.095\times\sqrt{1.426/2500}=20.20$ m³/s ✓.

#### Manning's roughness coefficient
$$
C=\frac{R^{1/6}}{n}\;\Rightarrow\;n=\frac{R^{1/6}}{C}=\frac{1.426^{1/6}}{60}=\frac{1.0610}{60}=0.0177
$$
**Answer: depth $y=2.85$ m, bed width = each sloping side $=3.29$ m, top width $=6.59$ m; Manning's $n=0.0177$.**

## Check yourself

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

1. In the half-hexagon section, how are the bed width and sloping sides related?
   A. Bed width is twice each sloping side
   B. Bed width is half of each sloping side
   C. Bed width equals depth
   D. Bed width equals each sloping side, 2y/sqrt(3), with sides at 60 degrees
   Answer: D. A regular hexagon has equal sides; the half-hexagon has bed = each slant side = 2y/sqrt(3). (Key point: Half-hexagon geometry)

2. For this section, the area and hydraulic radius in terms of depth y are:
   A. A = sqrt(3) y^2 and R = y/2
   B. A = 2y^2 and R = y/2
   C. A = sqrt(3) y^2 and R = y
   D. A = y^2/sqrt(3) and R = y/4
   Answer: A. A = (b + zy)y = sqrt(3) y^2 and P = 2 sqrt(3) y, so R = y/2. (Key point: Area and hydraulic radius)

3. How is the depth found from Q = 20.2 m3/s?
   A. From Manning's equation with n = 0.0177 assumed
   B. By substituting A and R into Q = C A sqrt(R S0), giving y^(5/2) = constant and y = 2.85 m
   C. From critical depth equations
   D. From Q = V x top width
   Answer: B. Q = 60 (sqrt(3) y^2) sqrt(y/2 x 1/2500) reduces to y^(5/2) = 13.74, so y = 2.853 m. (Key point: Finding depth)

4. How is Manning's n obtained once R = 1.426 m is known?
   A. n = C R^(1/6)
   B. n = R^(1/6)/C = 1.061/60 = 0.0177
   C. n = C/R^(1/6)
   D. n = R^(2/3)/C
   Answer: B. Since C = R^(1/6)/n, n = R^(1/6)/C = 0.0177. (Key point: Manning's n from C)

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