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Re: fluid flowing between 2 tanks



Carl E. Mungan wrote:

Consider a tank with a drain pipe in its bottom opening into an
identical tank directly below the first:

| |
|____ ____|
||
||
| || |
|__________|

The drain pipe is below the water level in the lower tank.

Let point 1 be on the free air-water surface of the upper tank and
let point 2 be on the free air-water surface of the lower tank.

How does the speed v1 compare to v2?

Answer 1. Use the equation of continuity. Both tanks have the same
cross-sectional area A. Hence v1 = v2. In Torricelli problems, this
speed is taken to be nearly zero if A is big enough.

Answer 2. Use Bernoulli's equation where P1 = P2 = P_atm. Therefore
v2^2 - v1^2 = 2gh, where h is the difference in heights of water in
the two tanks.

Obviously both answers cannot be right.

Agreed.

I suspect the second is wrong

Yup.

and that there is *necessarily* some kind of turbulent or viscous
loss of speed.

Also true.

More to the point, though, is that answer 2 is
formally invalid. Bernoulli's principle (in its
simplest form) is predicated on the two parcels
of fluid having the same total mechanical energy.
The principle (in its simplest form) is not
applicable to situations (such as this) where
there is significant dissipation. For details, see
http://www.monmouth.com/~jsd/how/htm/airfoils.html#sec-bernoulli

There's no way water can fall from one reservoir
to the other and be at rest when it gets there,
without significant dissipation.

I am worried about typical end-of-chapter problems involving siphons,
Torricelli tanks, etc where clearly one is supposed to ignore such
losses.

You can blissfully ignore dissipation !!or!! you
can invoke Bernoulli's principle. Your choice.

This posting is the position of the writer, not that of Polly Baker,
Timothy Turnstone, or Richard Saunders.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.