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Ability to observe electrically charged drops, preferably uniformly charged, would be very helpful to those who teach nuclear physics. Yes, I am thinking about spontaneous fission, described by the liquid drop model. Droplets of electrolyte, when charged, are not charged uniformly.
Ludwik
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On Oct 10, 2010, at 1:11 PM, Paul Doherty wrote:
BC
Check out the new exhibit at the Exploratorium.
At the exhibit there is a 15 cm diameter upward vertical airflow.
Visitors create water drops and drop them into the airflow.
Drops of a certain size, about 5 mm, are at trerminal velocity in
the airstream and float at a constant altitude.
Squirt a couple of drops and when they collide and coalesce the
resulting larger drop explodes in a spray of small droplets.
I'll try to get images and video.
Paul D
On Oct 10, 2010, at 12:13 AM, Bernard Cleyet wrote:
Bob!
(delayed by trip to birthday dinner a day away, not mine.)
Yes! You're correct. I forgot about the buoyant force, which is
negligible in this case, and the g force is ~ to the cube of the
radius, while the drag is only prop. to the square.***
http://en.wikipedia.org/wiki/Stokes'_law
doesn't apply.
hyperphysics has done the work for me:
http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html#c5
for 2 mm and 5 mm drops their terminal speeds are v. ~ 6.4 and 10.1
m/s.
I cheated and found in Eisberg and Lerner (I've seen it elsewhere
also.) the formula for the position (time) for v^2 drag.
The formulae, as I graphed for two and 5 mm D drops, is on my site.
http://www.cleyet.org/images/Drops%20at%20the%20Exploratorium/
Here's the generic:
X = (1/gamma) (Ln[exp(sqrt(gamma*g)*t )+ exp(sqrt(gamma*g)*t)]/2)
gamma = drag factor / m = 0.5* density (air) * cross section(Pi *
R^2) * coefficient (sphere ~0.5) / density * [(4/3) Pi * R^3]
this reduces to ~ 2.25E-4 / r
I pray no error. The equations for 2 mm and 5 mm diameter are
part of the titles of the graphs. They were generated by the
function feature in Kaleidograph. I extracted the data it created
and subtracted the positions. As one can see if dropped ~ 10 cm
apart after v.~ 600 ms the five mm one will catch up. This is
after falling about five feet.
*** what happens when ones mind's eye model is incomplete.
bc no experience w/ dropping, so doesn't know if practical.
On 2010, Oct 08, , at 19:25, LaMontagne, Bob wrote:
Just curious - what do you mean by "The smaller will catch up and
likely coalesce instead of shatter."? My understanding is that
the fall speed of droplets is proportional to the square root of
their radii. Don't larger drops catch up with smaller ones?
Bob at P
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Ludwik
http://csam.montclair.edu/~kowalski/life/intro.html
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