Question

A triangular irrigation lined canal carries a discharge of 25 m^3/s at bed slope = 1/6000 . If the side slopes of the canal are 1:1 and mannings coefficient is 0.018, the central depth of flow is equal to

Answer 1

Answer:

4675365336_673747477

Answer 2

Given,

A triangular irrigation lined canal carries a discharge of 25 m³/s at bed slope = 1/6000

The side slopes of the canals are in a ratio 1:1

The Manning's coefficient = 0.018

To Find,

The central depth of flow =?
Solution,

From the formula of Manning's equation we have

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

Where Q = 25 m³/s, n = Manning's coefficient = 0.018, S = 1 / 16000

$5 = \frac{1}{0.018} *(y/2\sqrt{2} )^{\frac{2}{3} } * (1/6000)^{1/2}*(y)^2\\25 = \frac{1}{0.018} *(y )^{\frac{8}{3} } * (1/154.92)\\(y)^{\frac{8}{3} } = 69.72\\y = 4.91$

Hence, the central depth of flow is equal to 4.91m.