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a circular culvert with diameter d = 3.00 ft and roughness coefficient n = 0.025 lies on a slope of 0.0012 ft/ft. if the depth of uniform flow is 1.6200 ft, what are:

User Haz
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Answer:

To find the required values, we can use the Manning's equation for open channel flow:

Q = (1/n) * (A * R^(2/3) * S^(1/2))

Where:

Q is the discharge (flow rate)

n is the Manning's roughness coefficient

A is the cross-sectional area of flow

R is the hydraulic radius

S is the slope of the channel

We are given:

Diameter of the culvert (d) = 3.00 ft

Depth of uniform flow (y) = 1.6200 ft

Slope of the channel (S) = 0.0012 ft/ft

Manning's roughness coefficient (n) = 0.025

We can calculate the required values as follows:

Cross-sectional area (A) of flow:

Since the culvert is circular, the cross-sectional area can be calculated using the formula:

A = (π/4) * d^2

A = (π/4) * (3.00 ft)^2

Hydraulic radius (R):

The hydraulic radius can be calculated as:

R = A / P

where P is the wetted perimeter of the flow.

For a circular culvert, the wetted perimeter is equal to the circumference of the circle:

P = π * d

Now, we can substitute the calculated values into the Manning's equation to find the discharge (Q).

Once we have the discharge (Q), we can calculate the velocity (V) using the formula:

V = Q / A

The required values are:

(a) Discharge (Q)

(b) Velocity (V)

Let's calculate these values:

(a) Discharge (Q):

A = (π/4) * (3.00 ft)^2

P = π * 3.00 ft

R = A / P

Q = (1/n) * (A * R^(2/3) * S^(1/2))

(b) Velocity (V):

V = Q / A

By substituting the calculated values into the equations, we can find the answers.

User Jacole
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