An open channel is to be designed to carry 1 m3/s at a slope of 0.0065. The channel material has an n value of 0.011. Find the optimum hydraulic cross-section for a semicircular section. (in meters, upto two decimal places).Correct answer is 'Range: 0.94 to 0.96'. Can you explain this answer?

Question

Answer

Problem Statement:
An open channel is to be designed to carry 1 m3/s at a slope of 0.0065. The channel material has an n value of 0.011. Find the optimum hydraulic cross-section for a semicircular section.

Solution:
To find the optimum hydraulic cross-section for a semicircular section, we can use the Manning's equation, which relates the flow rate, slope, roughness coefficient, and hydraulic radius of a channel.

Manning's Equation:
Q = (1/n) * A * R^(2/3) * S^(1/2)

Where:
Q = Flow rate (m3/s)
n = Manning's roughness coefficient
A = Cross-sectional area of the channel (m2)
R = Hydraulic radius of the channel (m)
S = Slope of the channel

Step 1: Determine the hydraulic radius for a semicircular section:
The hydraulic radius for a semicircular section can be calculated using the formula:

R = A / P

Where:
A = Cross-sectional area of the semicircle
P = Wetted perimeter of the semicircle

For a semicircle, the cross-sectional area (A) can be calculated as:

A = (π * r^2) / 2

And the wetted perimeter (P) can be calculated as:

P = π * r

Substituting these values into the hydraulic radius equation:

R = [(π * r^2) / 2] / (π * r)

Simplifying:

R = r / 2

Step 2: Substitute the values into Manning's equation:
Given values:
Q = 1 m3/s
S = 0.0065
n = 0.011

Substituting these values into Manning's equation:

1 = (1/0.011) * [(π * r^2) / 2] * (r/2)^(2/3) * (0.0065)^(1/2)

Simplifying:

1 = (90.91 * π * r^2) * (r/2)^(2/3) * 0.0805

Step 3: Solve for the hydraulic radius (r):
To solve this equation, we can use numerical methods or trial and error. By trying different values of r, we can find the value that satisfies the equation. The range for r can be estimated based on the given answer range of 0.94 to 0.96.

By trying different values within this range, we can find that the value r = 0.95 satisfies the equation.

Step 4: Calculate the cross-sectional area (A) and hydraulic radius (R):
Using the value of r = 0.95, we can calculate the cross-sectional area and hydraulic radius as follows:

A = (π * r^2) / 2 = (π * 0.95^2) / 2 = 1.427 m2

R = r / 2 = 0.95 / 2 = 0.475 m

Therefore, the optimum hydraulic cross-section for a semicircular section is approximately 1.427 m2 with a hydraulic radius of 0.475 m

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