Re: Bernoulli (horizontal)

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

Ludwik Kowalski wrote:

> Presumably the system is water between the two pistons
> The naive explanation of the observed increase of KE is
> "because the volume of the liquid must be conserved."
> A more sophisticated explanation is the work-energy
> theorem applied to the CM of the system. The net work
> is equal to deltaKE.
>
> Suppose the setup is horizontal; the tube becomes narrower
> but the average PEgrv of "parcels of water" does not change
> along the tube. In that simple case we have
>
> P1+0.5*rho*v1^2=P2+0.5*rho*v2^2.
>
> where P is pressure. Is it OK to say that P is also the energy
> per unit volume (because N/m^2=J/m^3)? We know that P
> decreases along the tube as the cross sectional area becomes
> smaller. The situation looks like an energy transformation;
> process. Bernoulli tells us that P decreases by the same
> amount by which the KE (also per unit volume) increases.
> What is wrong with saying that P is the "mechanical energy
> of pressure," per unit volume?

Here is one more comment on the above.

If P1 and P2  are "local energies per unit volume" then the
total "pressure energy" of the system is the sum of all P'*V,
where P' is the mean pressure and V is the volume of the
system (liquid between the two pistons). Note that the
volume of our system does not change while P is decreasing.
Therefore the amount of "pressure energy" lost by a small
parcel of water (volume V'), when it is pushed by dx, should
be V'*dP. What is dP? It the product of the gradient of P (along
the progressively narrowing pipe) and dx. The "pressure energy
of each parcel  of water decreases by V'*dP while its kinetic
energy increases by the same amount.

No heat is generated and the total mechanical energy is
conserved. What is wrong with inventing a new form of
energy and keep the "general idea" of conservation of E?

The V'*dP reminds me the P'*dV work for a gas. By
integrating P'*dV, for example along an isothermic path,
we get the total work done on the system. Likewise, by
integrating V'*dP, for example along the constant gradient
pipe, we will calculate the total number of joules. Should
it be called work done on the liquid or should it be called
"pressure energy" lost by the liquid when its kinetic energy
is increased by the same amount, and at the same time?

I suppose somebody will say "neither"; it is only "energy
in transfer" through this mechanical engine. How does
the Bernoulli pipe differ from other "simple machines?"

Silly speculations? Hmm. Where am I wrong?
Ludwik Kowalski