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 ft^{3}/s

S = 0.001

Y_{n} = ?

For the triangular channel,

Cross-sectional area (A) = Z * Y_{n}^{2}

= 2 * Y_{n}^{2}

Perimeter (P) = 2 *Y_{n} *
(1+Z^{2})^{0.5}

= 2 * Yn * (1+2^{2})^{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_{n}^{2} * (1/0.011) * ((2 *
Y_{n}^{2}) / (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)**

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