Answer :
The uniform flux through a trapezoidal concrete canal having a side slope of 2H h ZV and the bottom width of 1.5m if the depth of flow is 2m is [tex]0.0588 (3 + 4H) (2 - zV)\sqrt{(1 + 2H^2)} m^3/s[/tex].
In order to determine the uniform flux through a trapezoidal concrete canal having a side slope of 2H h ZV and a bottom width of 1.5m, the following data needs to be provided:
Depth of flow = 2m
Slope of the channel = 2m/km
Use n = 0.012
First, we need to calculate the area of cross-section of the canal:
Area of trapezium = (b1 + b2) x h / 2, where b1 and b2 are the lengths of the parallel sides of the trapezium and h is the perpendicular distance between the two bases.
b1 = 1.5m
b2 = b1 + 2H = 1.5 + 2Hm
Now, we can find the value of h: [tex]h = zV = 2Hh / \sqrt{(1 + 2H^2)}[/tex]
We know that the depth of flow is 2m, so:
[tex]2 = h + zV2 = (2Hh / \sqrt{(1 + 2H^2))} + zV\\\sqrt{(1 + 2H^2)} = 2Hh / (2 - zV)\\h = (2 - zV)\sqrt{(1 + 2H^2)} / 2H[/tex]
Now, we can substitute the values of b1, b2, and h to find the area of cross-section:
[tex]Area = (1.5 + 1.5 + 4H) \times (2 - zV)\sqrt(1 + 2H^2) / 2H\\Area = (3 + 4H) \times (2 - zV)\sqrt{(1 + 2H^2)} / 2H[/tex]
The discharge (Q) can be calculated using the Chezy formula:
Q = C x A x R^(2/3) x S^(1/2), where C is the Chezy coefficient, R is the hydraulic radius, and S is the slope of the channel.
Since we know that the flow is uniform, we can assume that the velocity of flow is constant throughout the cross-section, which means that the hydraulic radius is equal to the depth of flow (2m).
We are given that n = 0.012.
The value of C can be calculated using the Manning formula:
C = 1 / n * R^(2/3) * S^(1/2)
C = 1 / 0.012 * 2^(2/3) * (2 / 1000)^(1/2)
C = 22.02
Using this value of C, we can now find the discharge:
Q = 22.02 x (3 + 4H) x (2 - zV)√(1 + 2H²) / 2H x 2^(2/3) x (2 / 1000)^(1/2)
Q = 0.0588 (3 + 4H) (2 - zV)√(1 + 2H²) m³/s
Hence, the uniform flux through a trapezoidal concrete canal having a side slope of 2H h ZV and the bottom width of 1.5m if the depth of flow is 2m is [tex]0.0588 (3 + 4H) (2 - zV)\sqrt{(1 + 2H^2)} m^3/s[/tex].
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The uniform flow rate through the trapezoidal concrete canal with the given dimensions and flow depth is approximately 21.69 cubic meters per second.
To determine the uniform flow through a trapezoidal concrete canal, we can use the Manning's equation, which relates the flow rate to the channel geometry, hydraulic radius, and channel slope. The Manning's equation is as follows:
Q = (1/n) * A * R^(2/3) * S^(1/2)
where:
Q is the flow rate,
n is the Manning's roughness coefficient,
A is the cross-sectional area of flow,
R is the hydraulic radius, and
S is the channel slope.
Side slope (z) = 2H:1V
Bottom width (B) = 1.5 m
Depth of flow (d) = 2 m
Channel slope (S) = 2 m/km = 0.002
To calculate the cross-sectional area (A), hydraulic radius (R), and Manning's roughness coefficient (n), we need to determine the geometric properties of the trapezoidal canal.
Cross-sectional area (A):
A = (B + z * d) * d
Substituting the given values:
A = (1.5 m + (2 * H) * 2 m) * 2 m
A = (1.5 m + 4 m) * 2 m
A = 11 m²
Hydraulic radius (R):
R = A / P
where P is the wetted perimeter.
To calculate the wetted perimeter (P), we need to determine the side slope (z) and the bottom width (B).
Wetted perimeter (P) = B + 2 * d * √(1 + z²)
Substituting the given values:
P = 1.5 m + 2 * 2 m * √(1 + (2H)²)
P = 1.5 m + 4 m * √(1 + 4)
P = 1.5 m + 4 m * √(17)
P ≈ 1.5 m + 4 m * 4.12
P ≈ 1.5 m + 16.48 m
P ≈ 17.98 m
Now, we can calculate the hydraulic radius:
R = A / P
R = 11 m² / 17.98 m
R ≈ 0.612 m
Manning's roughness coefficient (n):
Given that n = 0.012
Now, we can calculate the flow rate (Q) using Manning's equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)
Q = (1/0.012) * 11 m² * (0.612 m)^(2/3) * (0.002)^(1/2)
Calculating the values:
Q = 83.33 * 11 * 0.723 * 0.0447
Q ≈ 21.69 m³/s
Therefore, the uniform flow rate through the trapezoidal concrete canal with the given dimensions and flow depth is approximately 21.69 cubic meters per second.
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