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Re: [Phys-l] question about Bernoulli



Oops! Thanks.

If the flow is adiabatic, then pressure and density have to change in the same direction - so it's kind of chicken and egg. On the microsopic level, why does the density lower? Mass continuity requires rho*A*v to be constant in equilibrium flow, so small A will require higher v - but why a rho change? - it's not needed with a liquid obviously - so why with gas molecules with lots of space between them?

If the gas molecules speed up, they must have suffered more collisions from the direction of the wider pipe and fewer from the direction of the narrower pipe - so we're back around the circle again.

Bob at PC

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-
bounces@carnot.physics.buffalo.edu] On Behalf Of John Mallinckrodt
Sent: Tuesday, November 23, 2010 6:02 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] question about Bernoulli

On Nov 23, 2010, at 2:31 PM, LaMontagne, Bob wrote:

Certainly the density is lower. But the gas is flowing through much
faster, so more molecules in a given time will pass a given area of
wall and exert more pressure (hit the wall more often) than a
stationary gas of the same reduced density.

I think you need to think that one through a little better. If twice
as many molecules pass a region in a given time, they each spend half
as much time there.

So which effect dominates?

There isn't any competition. Lower density and a barotropic equation
of state (due to the adiabatic assumption) => lower pressure, period.

I think the spirit in which Bill asked the question requires a
conceptual, non-mathematical answer.

You don't like "Less squeezing => less pressure"?

I don't think I can do any better, but maybe someone else will.

John Mallinckrodt
Cal Poly Pomona
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