Open Channel Flow (mannings equation) rectangular section

[Stacyg]

A rectangular concrete-lined channel is 2m wide with Manning's n 0.012 and a bed slope of 0.0025 m/m. Obtain the depth of flow when the discharge is 9m^3/s.

Q=V.A I have the discharge but how do I calculate the velocity without the area ?? Once I have the velocity I should be able to work it out.

Thanks

 

[dirk_mec1]

You've got two equations with two unknowns: eliminate v. Write the area A as b*h with b the width and h the height of the rectangular cross-section. Now solve for h.

 

[piercas]

for your question. To calculate the depth of flow in an open channel using Manning's equation, we need to use the formula:

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

where V is the velocity, n is Manning's roughness coefficient, R is the hydraulic radius (equal to the cross-sectional area divided by the wetted perimeter), and S is the slope of the channel bed.

In this case, we are given the width of the channel (2m) and the slope (0.0025 m/m), but we need to calculate the hydraulic radius and the cross-sectional area.

To calculate the hydraulic radius, we can use the formula:

R = A/P

where A is the cross-sectional area and P is the wetted perimeter. In a rectangular channel, the wetted perimeter is equal to the width of the channel (2m) plus twice the depth of flow (y). So, P = 2 + 2y

To find the cross-sectional area, we can use the formula:

A = W * y

where W is the width of the channel (2m) and y is the depth of flow.

Now, we can substitute these values into the formula for the hydraulic radius and the velocity equation:

R = (2 * y) / (2 + 2y)

V = (1/0.012) * [(2 * y) / (2 + 2y)]^(2/3) * (0.0025)^(1/2)

We know that the discharge (Q) is equal to the velocity (V) times the cross-sectional area (A), so we can rearrange the equation to solve for y:

Q = V * A

9 = V * (2 * y)

y = 4.5/V

Now, we can substitute this value of y into the velocity equation to solve for V:

9 = V * (2 * (4.5/V))

V = 1 m/s

Finally, we can use this value of V to calculate the depth of flow:

y = 4.5/1

y = 4.5 m

Therefore, the depth of flow in the rectangular channel is 4.5 meters when the discharge is 9m^3/s.