Question

A rectangular channel with a width of 2 m is carrying 15 m3/s. What are the critical depth and the flow velocity

Answer 1

The critical depth of the rectangular channel is approximately 1.790 meters.

The flow velocity in the rectangular channel is 4.190 meters per second.

Step-by-step explanation:

From Open Channel Theory we know that critical depth of the rectangular channel (
`y_(c)`), measured in meters, is calculated by using this equation:

`y_(c) = \sqrt[3]{(\dot V^(2))/(g\cdot b^(2)) }` (Eq. 1)

Where:

`\dot V` - Volume flow rate, measured in cubic meters per second.

`g` - Gravitational acceleration, measured in meters per square second.

`b` - Channel width, measured in meters.

If we know that
`\dot V = 15\,(m^(3))/(s)`,
`g = 9.807\,(m)/(s^(2))` and
`b = 2\,m`, then the critical depth is:

`y_(c) = \sqrt[3]{(\left(15\,(m^(3))/(s) \right)^(2))/(\left(9.807\,(m)/(s^(2)) \right)\cdot (2\,m)^(2)) }`

`y_(c) \approx 1.790\,m`

The critical depth of the rectangular channel is approximately 1.790 meters.

Lastly, the flow velocity (
`v`), measured in meters, is obtained from this formula:

`v = (\dot V)/(b\cdot y_(c))` (Eq. 2)

If we know that
`\dot V = 15\,(m^(3))/(s)`,
`b = 2\,m` and
`y_(c) = 1.790\,m`, then the flow velocity in the rectangular channel is:

`v = (15\,(m^(2))/(s) )/((2\,m)\cdot (1.790\,m))`

`v = 4.190\,(m)/(s)`

The flow velocity in the rectangular channel is 4.190 meters per second.