On May 7, 11:01=A0pm, WaltBJ <waltb...@[EMAIL PROTECTED]
> wrote:
> Come now. Are you going to tell me that a rush of air past a
> standing person will exert no force on that person?
Well, as a physics teacher I suspect you understand the difference
between skin friction (due to viscous interaction between the object
and a fluid flow) and dynamic pressure (as in, for example, an
airstream hitting and being deccelerated by the surface of an object).
These are very different sources of "drag", and in macroscopic objects
the second (usually termed "pressure drag") dominates. So no, the
"rush of air past a person" does very little - it goes past. The
portion of the airstream that hits the "front" of the person, and must
be deccelerated by interaction with that surface, is what's doing the
work: the pressure in the front is then higher than the pressure at
the rear, and that's what accelerates the object.
In the situation of a depressurization, there are at least two
differential pressure fields: the ambient field (the fact that you
have low pressure to one side and higher pressure to the other because
one side is/was a pressurized environment), and the dynamic field (the
fact that airflow, or wind, being brought to rest or redirected by the
object is raised to a higher pressure... think Pitot tube). This
thread started out as an attempt to calcualte the forces on an object
under these conditions, and it turns out for a pressurized environment
"blowing out" to a vacuum to be very very small. Perhaps not so for an
airplane... but then since for an airplane the ambient pressure
differential is even less than the vacuum decompression case, there's
some explaining to do. Thus the thread.
> Remember that a considerable section of the Aloha 737... She
> was an obstruction to free flow, therefore the air between her
> and the outside would be at lower pressure than the air in the
> cabin behind her.
Yes. But if you try to do the calculation based on the assumed size of
the hole and the pressure difference, you find that the entire plane
decompresses in a second or so... so fast that while the pressure
differential forces on her body are significant, they do not have time
to accelerate her to any significant speed. Thus the issue. It's not
that the pressure difference doesn't exist - it's that the pressure
differential doesn't exist long enough to produce a significant
terminal velocity.
> I cite the unfortunate B29 side gunners as cases in point where
> a sudden rupture enabled the outrushing cabin air to force them out of
> the aircraft.
That sounds good, but I'd love to see if the numbers work out to
support this assertion.
> I just this moment recalled cases where submarines, after long
submergence=
,
> had developed sufficient internal pressure from compressed air leakage
and=
> supplemental oxygen out of emergency cylinders to catapult an incautious
> person right out of the sub upon opening the first hatch.
Well, let's do the numbers. An 80 kg man weighs 785 N. If I'm standing
upright in the hatch, I'd have a surface area of about 0.09 m^2 (about
30 cm by 30 cm... I measured my posterior). That means to just
counterbalance the gravitational forces, I'd need a pressure
difference of 785/0.09 =3D 8722 Pa, quite reasonable. If I wanted the
unlucky dolphin to accelerate upward at 9.81 m/s^2, I'd need twice
that, or 17,444 Pa (2.53 lbs/in^2). All this assumes *no* airflow
around them - here I'm treating the person like a piston in a
cylinder, which is certainly overgenerous, but roughly accurate.
The difficulty in transfering this case to one of large-area explosive
decompression is one of timing (how long does the pressure difference
last) and distance (it's easy if you are in the breach and the breach
is small... but a very short distance away from the hole, the wind and
pressure forces drop sharply.
--
Brian Davis
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