Re: Beginner's Question about Bernoulli and Tanks

From: PD (pdraper_at_yahoo.com)
Date: 02/11/05 

Date: 11 Feb 2005 13:53:35 -0800

anebt2@peoplepc.com wrote:
> PD wrote:
>
> > I'm guessing that the problem has a hole in the side of the tank
> > somewhere near the bottom, and they're asking about the speed of
the
> > fluid coming out of the hole.
>
> Yes.
>
> > What you need to consider in this case is the pressure that is
> pressing
> > on the top surface of the water (1 atm if an open tank), and the
> > pressure that is pushing on the water at the hole (1 atm if it's
the
> > atmosphere that's pushing on the water there).
>
> Supposedly, this is what happens. However, I thought that fluid
> pressure under the atmosphere is p= p(0) + (density)(g)(h). IOW,
> pressure under the surface is the sum of the pressure at the surface
> *plus* the weight of the fluid above point p.

It is if you're not at the hole. If you're at the hole, the pressure is
atmospheric, because the NET force on the bit of water flying out the
hole is zero (it's leaving at constant velocity), and the air pushing
in is obviously at atmospheric pressure, so the water behind the flying
 bit must be pushing with atmospheric as well.

Now, I do think it's worthwhile raising your hand in class and asking
in your best furrowed-brow expression:
"Gee, Professor, I understand the pressure in the fluid right at the
hole is atmospheric, just like at the top of the tank. But what about
at the other side of the tank, at the same depth as the hole? Shouldn't
that be at higher pressure because it's under all that water above it?"
Then it will be your turn to see if the professor squirms and gibbers
for a bit.

PD

>
> How, then, do the two pressures in the open tank cancel?
>
> BTW, thank you to all the folks who took the time to help out.

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