Re: [Phys-l] question about Bernoulli

[brian whatcott <betwys1@sbcglobal.net>]

How about this?

Draw a circle of unit radius. The circle depicts the speed and direction 
of any air particle in the middle of a pipe.
Draw two parallel horizontal lines touching the circle, one on each 
side. The two straight lines represent a pipe.

From the center of the circle, draw an arrow perpendicular to the top 
horizontal.
 The perpendicular arrow represents the speed of one particle hitting 
the pipe, contributing some pressure, proportional to the arrow's length.

Now draw a third short horizontal line with a arrow between the other 
two horizontal lines - representing the speed with which all the air is 
now moved.

This 'wind' arrow will move any particle moving towards a wall.  The 
vertically moving particle now moves instead at a forward slant. So a 
particle with the SAME speed as before now has a forward component. The 
outward component of its speed is reduced for this reason....

This argument cries out for a diagram, in the Newtonian manner!

Brian W

.
On 11/17/2010 6:04 PM, William Robertson wrote:
> I should also mention that I'm hoping for something less abstract--a 
> description of what the molecules in the fluid are doing that explains 
> Bernoulli.
>
> Bill
>
>
> William C. Robertson, Ph.D.
>
>
>
> On Nov 17, 2010, at 4:20 PM, brian whatcott wrote:
>
> > Here's a Just-so explanation, offered without much corroboration.
> >
> > The current theory relating pressure and temperature in gases, due to
> > Clausius,
> >  has it that the kinetic energy of molecules impacting container walls
> > determines the mean pressure on those surfaces.
> >
> > Visualize a spherical surface plotting the (probable, mean, rms??)
> > speed direction vectors of the cumulative air molecule, located at its
> > center.
> >
> > If the molecules are all  moved in one direction, the spherical surface
> > is displaced
> >  forward by that displacement rate vector, with the effect that  the
> > mean molecule is
> > now off centered in its sphere, the mean molecule trailing  the center
> > of its speed
> >  vector sphere.
> >
> > The transverse distance from the molecule to the sphere's speed vector
> > surface is now
> >  reduced from a radius, to  some lesser value. It was this transverse
> > set of trajectories
> >  that contributed the usual pressure, and it is this set of slower
> > transverse trajectories
> >  that now provide a reduced side wall pressure.
> >  The pressure being proportional to kinetic energy, this pressure drops
> > with the
> > drop in the square of this transverse speed.
> >
> > This is a fairly concrete vivid image, no doubt. However - no 
> > guarantees.
> >
> >
> > Brian W
> >
> >
> > On 11/17/2010 3:04 PM, William Robertson wrote:
> > > I have for some time strived to explain the Bernoulli effect in terms
> > > of what's happening at the molecular level. We all know the
> > > mathematical explanation that leads to higher velocities being
> > > associated with lower pressures, but I want something that does not
> > > rely on the mathematics. In other words, what are the molecules doing
> > > that leads to the Bernoulli effect? /snip/
> > > Bill
> > >
> > >
> > > William C. Robertson, Ph.D.
> >
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