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A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.
The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition is
- a)3.0 m
- b)3.8 m
- c)4.1 m
- d)4.5 m
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s und...
At sub-critical flow, when width of the channel is contracted, there will be decrease in depth of flow at the contracted section and depth of contracted section will decrease till it reaches the depth at contracted section equal to critical depth and flow at contracted section will be critical flow means minimum specific energy.
B2/3 = 2.557
∴ B = 4.088 m ≃ 4.1 m
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A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s und...
To determine the minimum width to which the channel can be contracted without affecting the upstream flow condition, we can use the Manning's equation:
\[Q = \frac{k}{n}A R^{2/3} S^{1/2}\]
Where:
Q = Discharge (m³/s)
k = A constant depending on the units used (1.0 for SI units)
n = Manning's roughness coefficient
A = Cross-sectional area of flow (m²)
R = Hydraulic radius (m)
S = Slope of the channel bottom (m/m)
Let's start by finding the cross-sectional area of flow (A) and hydraulic radius (R) using the given information:
Cross-sectional area (A):
A = width × depth
A = 6.0 m × 1.60 m
A = 9.6 m²
Hydraulic radius (R):
R = A / wetted perimeter
The wetted perimeter is twice the depth because the channel is rectangular.
R = 9.6 m² / (2 × 1.60 m)
R = 3.0 m
Next, we rearrange the Manning's equation to solve for the slope (S):
\[S = \left(\frac{Q \cdot n}{k \cdot A \cdot R^{2/3}}\right)^2\]
Substituting the given values:
S = (16.0 m³/s × 0.015) / (1.0 × 9.6 m² × (3.0 m)^(2/3))
S = 0.240 / (1.0 × 9.6 × 1.817)
S = 0.013 m/m
Now, let's determine the minimum width (W) to which the channel can be contracted without affecting the upstream flow condition. We can use the Manning's equation again, but this time we will solve for the width (W):
\[W = \frac{Q}{n \cdot A \cdot R^{2/3} \cdot S^{1/2}}\]
Substituting the given values:
W = 16.0 m³/s / (0.015 × 9.6 m² × (3.0 m)^(2/3) × (0.013 m/m)^(1/2))
W ≈ 4.1 m
Therefore, the minimum width to which the channel can be contracted without affecting the upstream flow condition is approximately 4.1 meters.
\[Q = \frac{k}{n}A R^{2/3} S^{1/2}\]
Where:
Q = Discharge (m³/s)
k = A constant depending on the units used (1.0 for SI units)
n = Manning's roughness coefficient
A = Cross-sectional area of flow (m²)
R = Hydraulic radius (m)
S = Slope of the channel bottom (m/m)
Let's start by finding the cross-sectional area of flow (A) and hydraulic radius (R) using the given information:
Cross-sectional area (A):
A = width × depth
A = 6.0 m × 1.60 m
A = 9.6 m²
Hydraulic radius (R):
R = A / wetted perimeter
The wetted perimeter is twice the depth because the channel is rectangular.
R = 9.6 m² / (2 × 1.60 m)
R = 3.0 m
Next, we rearrange the Manning's equation to solve for the slope (S):
\[S = \left(\frac{Q \cdot n}{k \cdot A \cdot R^{2/3}}\right)^2\]
Substituting the given values:
S = (16.0 m³/s × 0.015) / (1.0 × 9.6 m² × (3.0 m)^(2/3))
S = 0.240 / (1.0 × 9.6 × 1.817)
S = 0.013 m/m
Now, let's determine the minimum width (W) to which the channel can be contracted without affecting the upstream flow condition. We can use the Manning's equation again, but this time we will solve for the width (W):
\[W = \frac{Q}{n \cdot A \cdot R^{2/3} \cdot S^{1/2}}\]
Substituting the given values:
W = 16.0 m³/s / (0.015 × 9.6 m² × (3.0 m)^(2/3) × (0.013 m/m)^(1/2))
W ≈ 4.1 m
Therefore, the minimum width to which the channel can be contracted without affecting the upstream flow condition is approximately 4.1 meters.
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A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer?
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A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer?.
A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular channel 6.0 m wide carries a discharge of 16.0 m3/s under uniform condition with normal depth of 1.60 m. Manning's n is 0.015.The channel width is to be contracted. The minimum width to which the channel can be contracted without affecting the upstream flow condition isa)3.0 mb)3.8 mc)4.1 md)4.5 mCorrect answer is option 'C'. Can you explain this answer?.
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