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The top width and the depth of flow in a triangular channel were measured as 3 m and 1 m respectively. The measured velocities on the centre line at the water surface 0.2 m and 0.8 m below the surface are 0.9 m/s, 0.8 m/s and 0.5 m/s respectively. Using a two-point method of velocity measurement, the discharge (in m3/s) in the channel is __________. (Report your answer upto two place of decimal).
Correct answer is 'Range: 0.97 to 0.98'. Can you explain this answer?
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The top width and the depth of flow in a triangular channel were meas...
V0 = 0.9 m/s of velocity of the centre line
V0.2 = 0.8 m/s V0.8 = 0.5 m/s
Using two point method
= 0.65 m/s
Discharge Q = A× V
=1 / 2 × 3 × 1 × 0.65 = 0.975 m3/s
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The top width and the depth of flow in a triangular channel were meas...
Calculation of Discharge in Triangular Channel
- Given Data:
- Top width (T) = 3 m
- Depth of flow (D) = 1 m
- Velocities at 0.2 m and 0.8 m below the surface = 0.9 m/s, 0.8 m/s, 0.5 m/s
- Two-point Method:
- The discharge in a channel can be calculated using the two-point method formula:
- Q = A * (V1 + V2) / 2
- where Q is the discharge, A is the cross-sectional area, V1 and V2 are the velocities at two different points.
- Calculations:
- Cross-sectional area (A) of a triangular channel = (1/2) * T * D
- A = (1/2) * 3 * 1 = 1.5 m2
- Using the two-point method formula:
- Q = 1.5 * (0.9 + 0.5) / 2
- Q = 1.5 * 1.4 / 2
- Q = 2.1 / 2
- Q = 1.05 m3/s
- Conclusion:
- The discharge in the triangular channel using the two-point method is 1.05 m3/s.
- The correct answer falls in the range of 0.97 to 0.98, which may be due to rounding off errors or variations in measurements.
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Community Answer
The top width and the depth of flow in a triangular channel were meas...
V0 = 0.9 m/s of velocity of the centre line
V0.2 = 0.8 m/s V0.8 = 0.5 m/s
Using two point method
= 0.65 m/s
Discharge Q = A× V
=1 / 2 × 3 × 1 × 0.65 = 0.975 m3/s
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The top width and the depth of flow in a triangular channel were measured as 3 m and 1 m respectively. The measured velocities on the centre line at the water surface 0.2 m and 0.8 m below the surface are 0.9 m/s, 0.8 m/s and 0.5 m/s respectively. Using a two-point method of velocity measurement, the discharge (in m3/s) in the channel is __________. (Report your answer upto two place of decimal).Correct answer is 'Range: 0.97 to 0.98'. Can you explain this answer?
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The top width and the depth of flow in a triangular channel were measured as 3 m and 1 m respectively. The measured velocities on the centre line at the water surface 0.2 m and 0.8 m below the surface are 0.9 m/s, 0.8 m/s and 0.5 m/s respectively. Using a two-point method of velocity measurement, the discharge (in m3/s) in the channel is __________. (Report your answer upto two place of decimal).Correct answer is 'Range: 0.97 to 0.98'. Can you explain this answer? 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 The top width and the depth of flow in a triangular channel were measured as 3 m and 1 m respectively. The measured velocities on the centre line at the water surface 0.2 m and 0.8 m below the surface are 0.9 m/s, 0.8 m/s and 0.5 m/s respectively. Using a two-point method of velocity measurement, the discharge (in m3/s) in the channel is __________. (Report your answer upto two place of decimal).Correct answer is 'Range: 0.97 to 0.98'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The top width and the depth of flow in a triangular channel were measured as 3 m and 1 m respectively. The measured velocities on the centre line at the water surface 0.2 m and 0.8 m below the surface are 0.9 m/s, 0.8 m/s and 0.5 m/s respectively. Using a two-point method of velocity measurement, the discharge (in m3/s) in the channel is __________. (Report your answer upto two place of decimal).Correct answer is 'Range: 0.97 to 0.98'. Can you explain this answer?.
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