Emptying a container and Efflux Speed

[fog37]

TL;DR Summary

fluid efflux speed and how quickly a container is emptied

Hello Forum,

The speed of efflux $v_{efflux}$ of a liquid, say water, from an orifice in the lower part of a container depends on the vertical distance between the free surface of the fluid and the lower position of the orifice in the container itself. The faster the speed $v_{efflux}$ the quicker the container will be emptied.

Assuming the same vertical separation between free surface and orifice position, how do the orifice diameter $D$ and the shape of the orifice impact the speed at which the container is emptied?

I believe that both the orifice diameter $D$ and shape should/could have an effect on how quickly the container gets emptied. Different orifice shapes would probably produce different $v_{efflux}$ for the same orifice diameter $D$. In theory, the diameter $D$ should not matter since the continuity equation stating that $D_1 v_1 = D_2 v_2$ but I wonder if in real life a larger diameter $D$ would empty the container faster...

Thanks!

 

[Chestermiller]

If the orifice has a diameterD, how could it be any other shape but round?

 

[fog37]

Sorry, I was not clear. I understand that there are indeed different orifice types. I guess they can have the same cross-sectional area but different effect:

 

[Chestermiller]

Sorry, I was not clear. I understand that there are indeed different orifice types. I guess they can have the same cross-sectional area but different effect:

View attachment 274974

In the case of approximating with Bernoulli’s equation, this is irrelevant.

 

[fog37]

Ok, so, assuming just a circular orifice, the larger the area the smaller $v_{efflux}$ and vice versa but the time it takes to empty the container is about the same.

As far as the different orifice shapes, they produce, I believe, a different spread/divergence of the exiting fluid from the orifice...

 

[The Fez]

To answer your first question, the exiting velocity will be approximately constant at √2gh, but the flow is Q=v*A, so an orifice with twice the area will discharge twice the flow. This will be modified depending on the "vena contracta" of the exiting fluid, which depends on orifice geometry, but in general, the same orifice shape twice as big will have twice the flow.

The exit velocity is indeed affected by the shape of the orifice. Each orifice shape as a Cv value (not to be confused with the Cv Valve coefficient). For instance, the sharp edge orifice above has a Cv of 0.98, so the velocity is reduced by 2%. The short tube, on the other hand, has a Cv of 0.6 to 0.8 depending on if the fluid separates from the walls of the tube, so the exit velocity will be much less.

The effective area of the orifice will be affected by the vena contracta, as I said above. The reduction in area depends on the orifice geometry again, so each orifice geometry has a Cc value as well, which reduces the effective area. Cc for the sharp edged orifice is 0.63, for instance.

Putting it all together, the flow out of the orifice is Q=A*Cv*Cc*√2gh.

 

[fog37]

To answer your first question, the exiting velocity will be approximately constant at √2gh, but the flow is Q=v*A, so an orifice with twice the area will discharge twice the flow. This will be modified depending on the "vena contracta" of the exiting fluid, which depends on orifice geometry, but in general, the same orifice shape twice as big will have twice the flow.

The exit velocity is indeed affected by the shape of the orifice. Each orifice shape as a Cv value (not to be confused with the Cv Valve coefficient). For instance, the sharp edge orifice above has a Cv of 0.98, so the velocity is reduced by 2%. The short tube, on the other hand, has a Cv of 0.6 to 0.8 depending on if the fluid separates from the walls of the tube, so the exit velocity will be much less.

The effective area of the orifice will be affected by the vena contracta, as I said above. The reduction in area depends on the orifice geometry again, so each orifice geometry has a Cc value as well, which reduces the effective area. Cc for the sharp edged orifice is 0.63, for instance.

Putting it all together, the flow out of the orifice is Q=A*Cv*Cc*√2gh.

Thank you Fez and Chestermiller.

Fez, when the fluid exits the container, the pressure at the air/fluid interface is equal to the existing air pressure, i.e. the air pressure is $p_{air}$ and the fluid pressure is $p_{air}=p_{fluid}$.

I believe the pressure inside the fluid is also equal to $p_{fluid}=p_{air}$. So the speed at which the fluid is moving does not affect the fluid pressure $p_{fluid}$. However, if the fluid is brought to a stop, the pressure at that point(s), called stagnation points will be isotropic and $p_{fluid} >p_{air}$.

Is that correct?