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

Answer:

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.

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:

$y_{c} = \sqrt[3]{\frac{\dot V^{2}}{g\cdot b^{2}} }$ (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 $\dot V = 15\,\frac{m^{3}}{s}$, $g = 9.807\,\frac{m}{s^{2}}$ and $b = 2\,m$, then the critical depth is:

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

$y_{c} \approx 1.790\,m$

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:

$v = \frac{\dot V}{b\cdot y_{c}}$ (Eq. 2)

If we know that $\dot V = 15\,\frac{m^{3}}{s}$, $b = 2\,m$ and $y_{c} = 1.790\,m$, then the flow velocity in the rectangular channel is:

$v = \frac{15\,\frac{m^{2}}{s} }{(2\,m)\cdot (1.790\,m)}$

$v = 4.190\,\frac{m}{s}$

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