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A hydraulically efficient trapezoidal channel section has a uniform flow depth of 2m. The bed width (expressed in m) of the channel is
- a)2.31
- b)4
- c)2
- d)1.154
Correct answer is option 'A'. Can you explain this answer?
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To determine the bed width of a trapezoidal channel section with a uniform flow depth, we can use the concept of hydraulic efficiency. The hydraulic efficiency of a channel section refers to how well it utilizes the available cross-sectional area to convey flow.
Given:
Uniform flow depth (y) = 2m
Let's determine the bed width using the Manning's equation, which relates the flow rate (Q) to the channel geometry and Manning's roughness coefficient (n):
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = flow rate
A = cross-sectional area of flow
R = hydraulic radius
S = slope of the channel bed
Since the channel section is trapezoidal, the cross-sectional area can be expressed as:
A = y * ((b1 + b2) / 2)
Where:
b1 = width of the channel at the water surface
b2 = width of the channel at the bed
The hydraulic radius can be calculated as:
R = A / P
P = wetted perimeter = b1 + b2 + 2 * √(y^2 + (b1 - b2)^2)
Assuming a uniform flow and a trapezoidal channel section, the slope of the channel bed (S) can be approximated as:
S = (b1 - b2) / L
L = length of the channel section
Let's substitute these equations into the Manning's equation:
Q = (1/n) * y * ((b1 + b2) / 2) * (A / P)^(2/3) * ((b1 - b2) / L)^(1/2)
Since the flow is uniform, the flow rate (Q) remains constant. We can simplify the equation by canceling out common terms and rearranging:
(b1 + b2) / (b1 - b2) = (2/3) * (n^2) * (y / S)^(1/2)
Now, substituting the given values:
(b1 + b2) / (b1 - b2) = (2/3) * (n^2) * (2 / S)^(1/2)
To maximize hydraulic efficiency, we need to minimize the ratio (b1 + b2) / (b1 - b2). This occurs when (b1 + b2) is as close to 2 times (b1 - b2) as possible.
In this case, the bed width (b2) is unknown. However, we can assume that the trapezoidal channel section is hydraulically efficient. This implies that the side slopes are typically 1:1 (vertical:horizontal). Therefore, the ratio of (b1 + b2) / (b1 - b2) can be approximated as 2.
So, we have:
2 = (2/3) * (n^2) * (2 / S)^(1/2)
Simplifying the equation:
1 = (2/3) * (n^2) * (2 / S)^(1/2)
We can rearrange this equation to solve for the slope (S):
S = (4/3) * (n^2)
Since the slope (S) is constant for a given channel, we
Given:
Uniform flow depth (y) = 2m
Let's determine the bed width using the Manning's equation, which relates the flow rate (Q) to the channel geometry and Manning's roughness coefficient (n):
Q = (1/n) * A * R^(2/3) * S^(1/2)
Where:
Q = flow rate
A = cross-sectional area of flow
R = hydraulic radius
S = slope of the channel bed
Since the channel section is trapezoidal, the cross-sectional area can be expressed as:
A = y * ((b1 + b2) / 2)
Where:
b1 = width of the channel at the water surface
b2 = width of the channel at the bed
The hydraulic radius can be calculated as:
R = A / P
P = wetted perimeter = b1 + b2 + 2 * √(y^2 + (b1 - b2)^2)
Assuming a uniform flow and a trapezoidal channel section, the slope of the channel bed (S) can be approximated as:
S = (b1 - b2) / L
L = length of the channel section
Let's substitute these equations into the Manning's equation:
Q = (1/n) * y * ((b1 + b2) / 2) * (A / P)^(2/3) * ((b1 - b2) / L)^(1/2)
Since the flow is uniform, the flow rate (Q) remains constant. We can simplify the equation by canceling out common terms and rearranging:
(b1 + b2) / (b1 - b2) = (2/3) * (n^2) * (y / S)^(1/2)
Now, substituting the given values:
(b1 + b2) / (b1 - b2) = (2/3) * (n^2) * (2 / S)^(1/2)
To maximize hydraulic efficiency, we need to minimize the ratio (b1 + b2) / (b1 - b2). This occurs when (b1 + b2) is as close to 2 times (b1 - b2) as possible.
In this case, the bed width (b2) is unknown. However, we can assume that the trapezoidal channel section is hydraulically efficient. This implies that the side slopes are typically 1:1 (vertical:horizontal). Therefore, the ratio of (b1 + b2) / (b1 - b2) can be approximated as 2.
So, we have:
2 = (2/3) * (n^2) * (2 / S)^(1/2)
Simplifying the equation:
1 = (2/3) * (n^2) * (2 / S)^(1/2)
We can rearrange this equation to solve for the slope (S):
S = (4/3) * (n^2)
Since the slope (S) is constant for a given channel, we
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A hydraulically efficient trapezoidal channel section has a uniform flow depth of 2m. The bed width (expressed in m) of the channel isa)2.31b)4c)2d)1.154Correct answer is option 'A'. Can you explain this answer?
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