A 4 m wide rectangular channel carries 6 m3/s of water. The Manning’s ‘n’ of the open channel is 0.02. Considering g = 9.81 m/s2, the critical velocity of flow (in m/s, round off to two decimal places) in the channel, is ________.Correct answer is '2.45'. Can you explain this answer?

Question

A 4 m wide rectangular channel carries 6 m3/s of water. The Manning&rsquo;s &lsquo;n&rsquo; of the open channel is 0.02. Considering g = 9.81 m/s2, the critical velocity of flow (in m/s, round off to two decimal places) in the channel, is ________.Correct answer is '2.45'. Can you explain this answer?

Answer

's coefficient for the channel is 0.025. Find the depth of flow in the channel.

We can use the Manning's equation to solve for the depth of flow:

Q = (1/n)A(R^2/3)S^1/2

where Q is the discharge (6 m3/s), n is the Manning's coefficient (0.025), A is the cross-sectional area of the channel (4d), R is the hydraulic radius (d), and S is the slope of the channel (unknown).

Substituting the given values, we get:

6 = (1/0.025)(4d)(d^(2/3))S^(1/2)

Solving for S, we get:

S = [(6/0.025)(4d)(d^(2/3))]^2

Simplifying, we get:

S = 4608d^(8/3)

Now, we can use the continuity equation to relate the depth of flow to the cross-sectional area and discharge:

Q = AV

where V is the velocity of the flow. Since the channel is rectangular, we can write:

A = 4d*h

where h is the height of the channel. Thus, we have:

6 = 4d*h*V

Solving for V, we get:

V = 6/(4d*h)

Substituting into the Manning's equation, we get:

6 = (1/0.025)(4d)(d^(2/3))(4608d^(8/3))^(1/2)

Simplifying, we get:

6 = 92.16d^(5/3)

Solving for d, we get:

d = (6/92.16)^(3/5) = 0.628 m

Therefore, the depth of flow in the channel is approximately 0.628 m.

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