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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?
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A 4 m wide rectangular channel carries 6 m3/s of water. The Manning&rs...
Critical depth (YC) 
Critical velocity (VC)=
= 2.45 m/s
Critical velocity (VC)=
= 2.45 m/sMost Upvoted Answer
A 4 m wide rectangular channel carries 6 m3/s of water. The Manning&rs...
'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.
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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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?
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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? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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?.
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