Question

A triangular channel (n = 0.011) with side slope z = 2 will be used to...

A triangular channel (n = 0.011) with side slope z = 2 will be used to convey 100 gpm down a hill with slope 0.001. What is the normal depth? Iterate by hand (guess and check) and make sure to show your work.

Homework Answers

Answer #1

[Note: As all the required data are given, iterations are not required. Instead, it can be solved diretly]

Given,

n=0.011

Z=2

Q = 100 GPM

= 100 * 0.002228

= 0.2228 ft3/s

S = 0.001

Yn = ?

For the triangular channel,

Cross-sectional area (A) = Z * Yn2

= 2 * Yn2

Perimeter (P) = 2 *Yn * (1+Z2)0.5

= 2 * Yn * (1+22)0.5

= 4.472 * Yn

Using the Mannings equation we can write,

Q = A * (1/n) * (A/P)(2/3) * S0.5

0.2228 = 2 * Yn2 * (1/0.011) * ((2 * Yn2) / (4.472 * Yn ))(2/3) * 0.0010.5

0.2228 = 3.362452 * Yn(2+2/3)

  Yn(8/3)= (0.2228/3.362452)

  Yn = (0.2228/3.362452)(3/8)

  Yn = 0.3614 ft (Answer)

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