Question

Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stream is 6.8 m wide and 3.2 m deep, and flows at 2.4 m/s. If the river has width 10.4 m and speed 2.8 m/s, what is its depth?

Answer 1

Answer:

The depth of the resulting stream is 3.8 meters.

Explanation:

Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:

$\dot V_{1} + \dot V_{2} = \dot V_{3}$ (1)

All volume flows are measured in cubic meters per second.

Dimensionally speaking, we can determine the depth of the resulting stream ($h_{3}$), in meters, by expanding (1) in this manner:

$w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}$

$h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}$ (2)

$v_{1}, v_{2}$ - Speed of the merging streams, in meters per second.

$h_{1}, h_{2}$ - Depth of the merging streams, in meters.

$w_{1}, w_{2}$ - Width of the merging streams, in meters.

$w_{3}$ - Width of the resulting stream, in meters.

$v_{3}$ - Speed of the resulting stream, in meters per second.

If we know that $w_{1} = 8.3\,m$, $h_{1} = 3.2\,m$, $v_{1} = 2.2\,\frac{m}{s}$, $w_{2} = 6.8\,m$, $h_{2} = 3.2\,m$, $v_{2} = 2.4\,\frac{m}{s}$, $w_{3} = 10.4\,m$ and $v_{3} = 2.8\,\frac{m}{s}$, then the depth of the resulting stream is:

$h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}$

$h_{3} = 3.8\,m$

The depth of the resulting stream is 3.8 meters.