On 05/28/2003 11:37 AM, Roger Haar wrote:
>
> I am not sure why you do not like the
> water-rushing-in analogy. The inrushing air has a
> linear momentum that must be "absorbed" by the
> membrane. To the zeroth approximation it is
> similar to water hammer. The first order
> approximation needs to consider the
> compressibility of air compared to water.
That's a start in the right direction.
I believe the key variables are
-- compressibility
-- density
-- viscosity.
Since the speed of sound depends on compressibility
and density, one may equivalently express things in
terms of
-- speed of sound
-- density
-- viscosity.
At the next level of detail: Suppose some fluid
is flowing through a pipe, flowing as fast as it
can under the influence of a given pressure-drop.
We then slam shut a valve. How big is the pressure
spike? How does it depend on the choice of fluid?
We start with:
(d/dt) volume = pi delta(p) r^4 / 8 l eta
which is the implicit definition of viscosity,
for Pousseille flow, where
eta = viscosity
delta(p) = pressure drop
r = radius of pipe
l = length of pipe
The average velocity is therefore something like
v = [(d/dt) volume] / pi r^2
= delta(p) r^2 / 8 l eta
The momentum per unit length is
mpl = v rho r^2
= pi delta(p) r^4 rho / 8 l eta
where rho = density (mass per unit volume)
At this point we get to actually think and apply
a little physics. Due to compressibility effects
(or, equivalently, due to speed-of-sound effects)
when we slam the valve not all of the fluid finds
out about it. In fact, after a time (t), only
the fluid within a distance (c t) of the valve
has found out about it.
So, to stop the fluid, we need to supply a certain
momentum per unit time. This will just be the
momentum per unit length (calculated above) times
the speed of sound.
Force = c delta(p) r^4 rho / 8 l eta
Pressure = c delta(p) r^2 / 8 pi l eta
If we keep the same geometry and the same driving
pressure drop, when we change fluids all that
matters is the factor
X = c rho / eta
If you plug in the numbers for air (compared to
water), you find that the viscosity is smaller,
which would tend to increase the "water" hammer
effect, but the speed of sound is smaller and
the density is much smaller, so at the end of
the day the "water" hammer effect for air is
about eighty times smaller than for real water.
============
On 05/28/2003 03:52 PM, Stefan Jeglinski wrote:
> What is a reasonable v?
> Not large, intuition tells me 10cm/sec into a vacuum is a reasonable
> guess.
My intuition differs.
I would expect that catastrophic venting due
to a broken window would lead to air moving
in with velocity on the order of 340 m/s.
That is the RMS characteristic velocity of air
molecules. They started out with that velocity,
and I don't see any physical process that
will slow them down.
You can get the same answer by approximating
ordinary static air in a box as a "standing
wave" composed of 340 m/s rightward flow
superposed with 340 m/s leftward flow. The
flow turns around at the boundaries. This
turnaround is what produces pressure on the
boundaries. If you remove the right-hand
boundary, the rightward flow just exits
stage right, ballistically. Reality is a
little more complicated than this, since
there is a broad distribution of velocities.
Still, the velocities are peaked around the
characteristic velocity.
==============
SJ also questiond my "sound wave" analysis.
I think I muffed a factor of two. Some of
the ideas can be salvaged, but I need to
start over. Here goes:
It may help to pretend there is some small
amount of air in the chamber before the
rupture. The inrushing air is then joined
to the pre-existing air to form an overall
pressure pattern. The initial condition is
a step-function in pressure. We then watch
the evolution of the pressure pattern starting
from this initial condition. This initial
condition corresponds to a very loud sound.
If the chamber has a shape resembling a horn,
it could be very effective at focussing this
sound onto the membrane.