# Depth and vertex angle of triangular channel for minimum perimeter

Date of Exam

**Problem**

A triangular shaped channel is to be designed to carry 700 L/s on a slope of 0.0001. Determine what vertex angle and depth of water over the vertex will be necessary to give a section with minimum perimeter, assuming the channel is made of timber, *n* = 0.012. Use Manning’s formula.

A. θ = 45°, h = 1.425 m |
C. θ = 45°, h = 2.125 m |

B. θ = 90°, h = 2.215 m |
D. θ = 90°, h = 1.215 m |

**Answer Key**

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[ D ]

**Solution**

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For minimum perimeter, the vertex angle must be 90°

$A = \frac{1}{2}(2h)(h) = h^2$

$A = \frac{1}{2}(2h)(h) = h^2$

$P = 2\sqrt{2}h$

$R = \dfrac{A}{P} = \dfrac{h^2}{2\sqrt{2}h} = \dfrac{\sqrt{2}}{4}h$

$Q = A \dfrac{1}{n}R^{2/3} S^{1/2}$

$0.7 = h^2 \dfrac{1}{0.012}(\frac{1}{4}\sqrt{2}h)^{2/3}(0.0001^{1/2})$

$h = 1.215 ~ \text{m}$ ← *answer*

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