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On 11/21/2010 06:22 PM, John Mallinckrodt wrote:
... Bernoulli's eqn is derived under the assumption
that the flow is "steady" (i.e., not time-dependent) and is,
therefore, only applicable to steady flows. But, except in boring
cases, steady flows are only steady in *one* reference frame. Of
course, to have a non-boring steady flow, one does also have to have
"stationary" boundaries like pipes or river beds or mountaintops or
airfoils, which then also determine the preferred reference frame,
but that is not the fundamental requirement.
I wouldn't have said that. ISTM it misses a couple of fundamental
points.
1) First of all, when I derive the Bernoulli equation, it is not
restricted to steady flows.
2) Even if it were restricted, it would not solve all instances
of the riddle that was posed. In particular, consider the case
of two semi-infinite airmasses, one to the left of the XZ plane
and one to the right. They are sliding past each other. This
is a steady flow.
-- In some frame, airmass A is moving in the +X direction while
airmass B is moving in the -X direction.
-- In another frame, airmass A is moving while airmass B is
stationary.
-- In yet another frame, airmass A is stationary while airmass
B is moving.
So, in accordance with Bernoulli's principle, which airmass has
the lower pressure?
As with all such riddles, the properly-stated laws of physics
do not contain any paradoxes. The so-called paradox only
arises if/when you misstate the laws of physics.