Re: Bernoulli

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

Referring to Bernoulli (P+0.5*rho*v^2+rho*g*y=const.)
John Barrer wrote:

> Isn't it clearer to explain that this formula is simply a
> statement that, in the absence of a change in sysem
> energy, the energy density (J per cubic meter) of the
> system is constant?

Is this a new law of nature? I used to think that this is a
statement about conservation of energy but now I am
less certain. Let me ask the same questions in a different
way.

Consider a pipe which is wider on the left (area A1) and
progressively narrower on the right (area A2). The right
side is at a higher elevation than the left side. A force on
the massless piston on the left is F1=P1*A1, its direction
is the same as its constant speed. The force on the massless
piston on the right is F2=P2*A2, its direction is opposite
to the direction of its constant speed. Presumably both
pistons move horizontally without acceleration. But water
gains speeds as it is pushed up.

The net result is an increased KE and an increase of PE
of each "water particle"; JohnD would probably call
this a "parcel" of water. Each F=P*A is a force exerted
on water by a piston. In this context pressure is viewed
as "force per unit area." But, as JohnB reminds us, N/m^2
is also J/m^3. This allows us to view P as "energy density"
of water. note that P decreases as the parcel climbs up
along the pipe. This reminds me of two interpretations of
work: W as F*ds and W as "energy in transit." Another
logico-linguistic trap  or something very important and
significant?

In any case, is it appropriate to say that

P+0.5*rho*v^2+rho*g*y

is the total mechanical energy (of a parcel ow water)? If so
then P is a form of energy. Is "energy of pressure" kinetic
or potential? If it is potential then what kind of work should
it be associated with?

Another option is to say that only that Emech of a parcel is
the sum of the last two terms. Then we should say that there
is a change in total mechanical energy. A net force acting on
a parcel of water can change its kinetic energy but it can not
change its potential energy. Right?

Referring to:

> The first impression is that KE decreases at the
> expense of PE. But this is not so.

JohnD wrote: "Actually it is so.  That's the point."

I wish I could understand why "it is so" in the above example.
Ludwik Kowalski