Question

A channel designed by lacey's theory has a mean velocity of 1 metre per second the silt factor is unity the hydraulic mean radius will be

Answer 1

Answer: 2.5 cm

Explanation:

Lacey's silt factor s = 5 v^2 / 2R

By substitution we get 2.5 m as answer

Answer 2

Answer:

The correct answer is 2.75m

Explanation:

From the above question,

A channel designed by Lacey's theory has a mean velocity of 1.1 m/s. The silt factor is 1.1. Then hydraulic mean radius will be

V is the mean velocity of the fluid

R is the hydraulic mean radius of the channel

Sf is the silt factor, and C is a coefficient that depends on the shape of the channel.

Hydraulic mean radius = R

                                   R = $\frac{5}{2} . \frac{V^{2} }{f}$

Here,

            V = mean velocity

            V = 1.1 m/s

            f = silt factor

            f = 1.1

            R = $\frac{5}{2}$ x $\frac{1.1^{2} }{1.1}$

            R = 2.75 m

Hence, the hydraulic mean radius of the channel is 2.75 m, this is what given in the problem.

Thus this result confirms that the assumptions of Lacey's theory were consistent with the actual conditions of the channel.

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