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Water flows in a rectangular, concrete, open channel that is 12m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate (n=0.013)?
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Water flows in a rectangular, concrete, open channel that is 12m wide ...
Solution:
Given data:
Width of the channel (b) = 12m
Depth of the channel (y) = 2.5m
Slope of the channel (S) = 0.0028
Manning’s coefficient (n) = 0.013
Step 1: Calculate the hydraulic radius (R)
- Hydraulic radius (R) is defined as the ratio of the cross-sectional area of the flow (A) to the wetted perimeter (P).
- R = A/P
- For a rectangular channel, A = by and P = b + 2y
- R = (by)/(b + 2y)
R = (12 × 2.5)/(12 + 2 × 2.5) = 1.923 m
Step 2: Calculate the mean velocity (V)
- The mean velocity (V) is given by the Chezy’s formula.
- V = (C/ n) × R^(2/3) × S^(1/2)
- C is the Chezy’s coefficient, which is a constant for a given channel and depends on the roughness of the channel.
- For concrete channels, the value of C is usually taken as 55.
V = (55/0.013) × (1.923)^(2/3) × (0.0028)^(1/2) = 1.394 m/s
Step 3: Calculate the flow rate (Q)
- The flow rate (Q) is given by the equation Q = AV, where A is the cross-sectional area of the flow.
- For a rectangular channel, A = by
Q = (12 × 2.5) × 1.394 = 41.85 m³/s
Therefore, the water velocity is 1.394 m/s and the flow rate is 41.85 m³/s.
Given data:
Width of the channel (b) = 12m
Depth of the channel (y) = 2.5m
Slope of the channel (S) = 0.0028
Manning’s coefficient (n) = 0.013
Step 1: Calculate the hydraulic radius (R)
- Hydraulic radius (R) is defined as the ratio of the cross-sectional area of the flow (A) to the wetted perimeter (P).
- R = A/P
- For a rectangular channel, A = by and P = b + 2y
- R = (by)/(b + 2y)
R = (12 × 2.5)/(12 + 2 × 2.5) = 1.923 m
Step 2: Calculate the mean velocity (V)
- The mean velocity (V) is given by the Chezy’s formula.
- V = (C/ n) × R^(2/3) × S^(1/2)
- C is the Chezy’s coefficient, which is a constant for a given channel and depends on the roughness of the channel.
- For concrete channels, the value of C is usually taken as 55.
V = (55/0.013) × (1.923)^(2/3) × (0.0028)^(1/2) = 1.394 m/s
Step 3: Calculate the flow rate (Q)
- The flow rate (Q) is given by the equation Q = AV, where A is the cross-sectional area of the flow.
- For a rectangular channel, A = by
Q = (12 × 2.5) × 1.394 = 41.85 m³/s
Therefore, the water velocity is 1.394 m/s and the flow rate is 41.85 m³/s.
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Water flows in a rectangular, concrete, open channel that is 12m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate (n=0.013)?
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Water flows in a rectangular, concrete, open channel that is 12m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate (n=0.013)? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Water flows in a rectangular, concrete, open channel that is 12m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate (n=0.013)? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water flows in a rectangular, concrete, open channel that is 12m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate (n=0.013)?.
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