Question

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.

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

Y_n = ?

For the triangular channel,

Cross-sectional area (A) = Z * Y_n²

= 2 * Y_n²

Perimeter (P) = 2 *Y_n * (1+Z²)^0.5

= 2 * Yn * (1+2²)^0.5

= 4.472 * Y_n

Using the Mannings equation we can write,

Q = A * (1/n) * (A/P)^(2/3) * S^0.5

0.2228 = 2 * Y_n² * (1/0.011) * ((2 * Y_n²) / (4.472 * Y_n ))^(2/3) * 0.001^0.5

0.2228 = 3.362452 * Y_n^(2+2/3)

  Y_n^(8/3)= (0.2228/3.362452)

  Y_n = (0.2228/3.362452)^(3/8)

  Y_n = 0.3614 ft (Answer)