Question

I have a fluid dynamics question.

If I have a constant stream of water moving 7 ft/s that flows directly into a rectangular 8" x 6" opening (at the collection end) that tapers down over a 15" length to a 8" x 1-1/2" outlet, how much will the speed and pressure increase at the outlet.

Thanks a bunch,

Todd

Answer 1

First we can easily find the change of velocity

Continuity of flow gives us:

$7\cdot(8^{\prime\prime}\times 6^{\prime\prime})=v_{2}\cdot(8^{\prime\prime}\times 1.5^{\prime\prime})$

and we find that

$v_{2}=\frac{7\cdot(8^{\prime\prime}\times 6^{\prime\prime})}{(8^{\prime\prime}\times 1.5^{\prime\prime})}=28ft/s$

Now we can find change of pressure

Bernoulli’s principle tells us, that

$v^{2}/2+p/\rho=const$

where $\rho$ is density of water and $p$ is pressure. Hence

$v_{1}^{2}/2+p_{1}/\rho=v_{2}^{2}/2+p_{2}/\rho$

$p_{2}-p_{1}=\rho\cdot(v_{1}^{2}/2-v_{2}^{2}/2)=34141.8672\,Pa$

Change of pressure approximately is 34 kPa.