G
George Herold
Guest
On Sunday, April 12, 2020 at 6:03:01 PM UTC-4, Jasen Betts wrote:
picturing a long tube on the vent... it could be coiled
inside the box.
So here's a 'back of the envelope' calculation question.
The diffusion constant of H2O in air (STP) is
about 0.3 cm^2/sec.
https://en.wikipedia.org/wiki/Mass_diffusivity#Example_values
So if I have a 10 cm (long) tube the time it takes for
water to diffuse along it is about 200 seconds.
t = len^2/(2*D). So how then do I calculate the number
of water atoms at steady state that are diffusing
along the tube?
Do I calculate the number of water atoms entering the
front area of the tube, in one second. And then divide
by 200?
George H.
Hmm well not the cable (carrying wires). I wasOn 2020-04-12, George Herold <ggherold@gmail.com> wrote:
On Saturday, April 11, 2020 at 5:38:40 PM UTC-4, Joe Gwinn wrote:
On Tue, 7 Apr 2020 07:55:27 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:
On 2020-04-06 21:35, jlarkin@highlandsniptechnology.com wrote:
On Mon, 6 Apr 2020 17:43:46 -0400, Phil Hobbs
[snip]
I guess the seive would soak up humidity and reduce the pressure
inside, so a tiny flow through the seals would introduce a little more
humidity. Wouldn't that eventually get to zero humidity and zero
pressure differential?
I guess atmospheric changes would still pump the system slightly.
Simon's the seal expert at this point. A sufficiently stiff box, with
enough screws holding the lid on, and hermetic connectors, ought to be
able to stay sealed pretty well. The pressure changes are nontrivial
though--our box is about 3 x 5 inches, so a 7% pressure change amounts
to about 15 pounds over the surface of the lid. They're also fairly
slow, so it doesn't take much of a leak rate to equalize the pressure.
We're using a cable gland rather than a hermetic connector, primarily
for cost reasons. I suspect that enough air will flow inside the cable
to manage the vent job, but we'll probably have to measure that to find out.
A traditional alternative is a long thin tube whose volume is
sufficient to ensure that no inside air gets out or outside air gets
in, despite the +/- 7% variation in ambient air pressure. Invented by
Louis Pasteur in 1859.
.<https://en.wikipedia.org/wiki/Swan_neck_flask
Joe Gwinn
Nice, I was trying to express this idea, but I was more focused on
keeping the length of the tube long to limit diffusion of the H2O.
George H.
With minimally vented or imperfectly sealed enclosure you have the
problem that if it gets splashed with cold water on a hot day it
cools rapdily and the air inside contracts and sucks surface water
in.
Using the cable to vent the enclosure seems like a good scheme,
assuming that air flows freely enough and it terminates in a dry
location.
picturing a long tube on the vent... it could be coiled
inside the box.
So here's a 'back of the envelope' calculation question.
The diffusion constant of H2O in air (STP) is
about 0.3 cm^2/sec.
https://en.wikipedia.org/wiki/Mass_diffusivity#Example_values
So if I have a 10 cm (long) tube the time it takes for
water to diffuse along it is about 200 seconds.
t = len^2/(2*D). So how then do I calculate the number
of water atoms at steady state that are diffusing
along the tube?
Do I calculate the number of water atoms entering the
front area of the tube, in one second. And then divide
by 200?
George H.
--
Jasen.