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A channel of bed slope 0.0009 carries a discharge of 30 m3/s when the depth of flow is 1.0 m. What is the distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001?
- a)10 m3/s
- b)15 m3/s
- c)60m3/s
- d)90 m3/s
Correct answer is option 'A'. Can you explain this answer?
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A channel of bed slope 0.0009 carries a discharge of 30 m3/s when the ...
Given data:
Discharge (Q1) = 30 m3/s
Bed slope (S1) = 0.0009
Depth of flow (y1) = 1.0 m
New bed slope (S2) = 0.0001
To find: Distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001
Formula used:
Q = AyV
Where,
Q = Discharge
A = Cross-sectional area of flow
y = Depth of flow
V = Velocity of flow
Analysis:
1) Cross-sectional area (A) of flow can be calculated using the formula:
A = yB
Where B is the width of the channel.
2) Velocity (V1) can be calculated using the formula:
V1 = (1.49/n) * R2/3 * S1/2
Where n is the Manning's roughness coefficient, R is the hydraulic radius, and S1 is the bed slope.
3) Using the above formula, we can calculate the value of R:
R = A/P
Where P is the wetted perimeter.
4) Wetted perimeter (P) can be calculated using the formula:
P = B + 2y(1 + S12)1/2
5) Using the above values, we can calculate the discharge (Q1) as:
Q1 = Ay1V1
6) Now, using the new bed slope (S2), we can calculate the new velocity (V2) using the above formula.
7) Using the new velocity (V2) and depth of flow (y1), we can calculate the new cross-sectional area (A2) using the formula:
A2 = Q1/V2
8) Using the new cross-sectional area (A2) and width of the channel, we can calculate the new distance (x) using the formula:
x = A2/B
Solution:
1) Cross-sectional area (A) of flow:
A = y1B
2) Velocity (V1) can be calculated using the formula:
V1 = (1.49/n) * R2/3 * S1/2
Using n = 0.03 (for a smooth channel), we get:
R = A/P = A/[B + 2y1(1 + S12)1/2]
= (1.0 * B)/[B + 2(1.0)(1 + 0.00092)1/2]
= 0.422 B
V1 = (1.49/0.03) * (0.422)2/3 * (0.0009)1/2
= 1.04 m/s
3) Discharge (Q1):
Q1 = Ay1V1 = (1.0 * B) * (1.04) * 30
= 31.2 B m2/s
4) New velocity (V2):
V2 = (1.49/0.03) * (0.422)2/3 * (0.0001)1/2
= 0.35 m/s
5) New cross-sectional area (A2):
A2
Discharge (Q1) = 30 m3/s
Bed slope (S1) = 0.0009
Depth of flow (y1) = 1.0 m
New bed slope (S2) = 0.0001
To find: Distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001
Formula used:
Q = AyV
Where,
Q = Discharge
A = Cross-sectional area of flow
y = Depth of flow
V = Velocity of flow
Analysis:
1) Cross-sectional area (A) of flow can be calculated using the formula:
A = yB
Where B is the width of the channel.
2) Velocity (V1) can be calculated using the formula:
V1 = (1.49/n) * R2/3 * S1/2
Where n is the Manning's roughness coefficient, R is the hydraulic radius, and S1 is the bed slope.
3) Using the above formula, we can calculate the value of R:
R = A/P
Where P is the wetted perimeter.
4) Wetted perimeter (P) can be calculated using the formula:
P = B + 2y(1 + S12)1/2
5) Using the above values, we can calculate the discharge (Q1) as:
Q1 = Ay1V1
6) Now, using the new bed slope (S2), we can calculate the new velocity (V2) using the above formula.
7) Using the new velocity (V2) and depth of flow (y1), we can calculate the new cross-sectional area (A2) using the formula:
A2 = Q1/V2
8) Using the new cross-sectional area (A2) and width of the channel, we can calculate the new distance (x) using the formula:
x = A2/B
Solution:
1) Cross-sectional area (A) of flow:
A = y1B
2) Velocity (V1) can be calculated using the formula:
V1 = (1.49/n) * R2/3 * S1/2
Using n = 0.03 (for a smooth channel), we get:
R = A/P = A/[B + 2y1(1 + S12)1/2]
= (1.0 * B)/[B + 2(1.0)(1 + 0.00092)1/2]
= 0.422 B
V1 = (1.49/0.03) * (0.422)2/3 * (0.0009)1/2
= 1.04 m/s
3) Discharge (Q1):
Q1 = Ay1V1 = (1.0 * B) * (1.04) * 30
= 31.2 B m2/s
4) New velocity (V2):
V2 = (1.49/0.03) * (0.422)2/3 * (0.0001)1/2
= 0.35 m/s
5) New cross-sectional area (A2):
A2
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Community Answer
A channel of bed slope 0.0009 carries a discharge of 30 m3/s when the ...
Q2/Q1=underroot (S2/S1)
Q2= underroot (0.0001/0.0009)*30
Q2= 10m3/s
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A channel of bed slope 0.0009 carries a discharge of 30 m3/s when the depth of flow is 1.0 m. What is the distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001? a)10 m3/sb)15m3/sc)60m3/sd)90m3/sCorrect answer is option 'A'. Can you explain this answer?
Question Description
A channel of bed slope 0.0009 carries a discharge of 30 m3/s when the depth of flow is 1.0 m. What is the distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001? a)10 m3/sb)15m3/sc)60m3/sd)90m3/sCorrect answer is option 'A'. 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 channel of bed slope 0.0009 carries a discharge of 30 m3/s when the depth of flow is 1.0 m. What is the distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001? a)10 m3/sb)15m3/sc)60m3/sd)90m3/sCorrect answer is option 'A'. 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 channel of bed slope 0.0009 carries a discharge of 30 m3/s when the depth of flow is 1.0 m. What is the distance carried by an exactly similar channel at the same depth of flow if the slope is decreased to 0.0001? a)10 m3/sb)15m3/sc)60m3/sd)90m3/sCorrect answer is option 'A'. Can you explain this answer?.
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