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In a triangular channel the top width and depth of flow were 2.0 m and 0.9 m respectively. Velocity measurements on the centre line at 18 cm and 72 cm below water surface indicated velocities of 0.60 m/s and 0.40 m/s respectively. The discharge in the channel (in m3/s) is
  • a)
    0.90
  • b)
    1.80
  • c)
    0.45
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In a triangular channel the top width and depth of flow were 2.0 m and...
Discharge in the channel
= x area of channel Area of channel
Area of channel
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Most Upvoted Answer
In a triangular channel the top width and depth of flow were 2.0 m and...
Given data:
Top width (T) = 2.0 m
Depth of flow (D) = 0.9 m
Velocity at 18 cm below water surface (V1) = 0.60 m/s
Velocity at 72 cm below water surface (V2) = 0.40 m/s

To find: Discharge (Q) in the channel

Formula used:
Q = AV
where,
Q = Discharge
A = Cross-sectional area of flow
V = Velocity of flow

Calculation:
1. Cross-sectional area of flow:
Using the formula for the area of a triangular channel, we get:
A = (1/2)TD
A = (1/2) × 2.0 m × 0.9 m
A = 0.9 m²

2. Velocity at the centre of the channel:
To find the velocity at the centre of the channel, we can use the formula for the average velocity of flow in an open channel:
Vavg = (1/2)(V1+V2)
Vavg = (1/2)(0.60 m/s + 0.40 m/s)
Vavg = 0.50 m/s

3. Discharge in the channel:
Using the formula Q = AV, we get:
Q = 0.9 m² × 0.50 m/s
Q = 0.45 m³/s

Therefore, the discharge in the channel is 0.45 m³/s, which is option (c).
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In a triangular channel the top width and depth of flow were 2.0 m and 0.9 m respectively. Velocity measurements on the centre line at 18 cm and 72 cm below water surface indicated velocities of 0.60 m/s and 0.40 m/s respectively. The discharge in the channel (in m3/s) isa)0.90b)1.80c)0.45d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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In a triangular channel the top width and depth of flow were 2.0 m and 0.9 m respectively. Velocity measurements on the centre line at 18 cm and 72 cm below water surface indicated velocities of 0.60 m/s and 0.40 m/s respectively. The discharge in the channel (in m3/s) isa)0.90b)1.80c)0.45d)None of theseCorrect answer is option 'C'. 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 In a triangular channel the top width and depth of flow were 2.0 m and 0.9 m respectively. Velocity measurements on the centre line at 18 cm and 72 cm below water surface indicated velocities of 0.60 m/s and 0.40 m/s respectively. The discharge in the channel (in m3/s) isa)0.90b)1.80c)0.45d)None of theseCorrect answer is option 'C'. 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 In a triangular channel the top width and depth of flow were 2.0 m and 0.9 m respectively. Velocity measurements on the centre line at 18 cm and 72 cm below water surface indicated velocities of 0.60 m/s and 0.40 m/s respectively. The discharge in the channel (in m3/s) isa)0.90b)1.80c)0.45d)None of theseCorrect answer is option 'C'. Can you explain this answer?.
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