Evaporation of a drop

Consider a drop of water, small enough that it would cling to an upside
down horizontal surface. All other factors being equal, would that drop
evaporate faster than an identical drop on a horizontal surface, not
upside down.

Just wondering
Pete

Dear P T:

"P T" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ].webtv.net...

Water vapor is lighter than air. If there were no wind, then the
"upside down horizontal surface" drop should evaporate slower.
The air near this drop will become saturated with water, slowing
the evaporation. For the other surface, the water vapor would
"lift away", carrying away water vapor, and bringing in fresh dry
air. As I see it...

David A. Smith

"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@nospam.com> wrote in
message newsGSke.3733$Xh.460@fed1read07...

Evaporation (at temperatures below the boiling point) is endothermic. It
takes heat out of the air (and water). That is how evaporative coolers
work.

The cooler, humidified air is more dense, and would fall.

If I had access to my old lab I would get a 20 microliter syringe, some DI
water, a microscope slide, and a stopwatch and run the experiment.


Tom Davidson
Richmond, VA

N:dlzc D:aol T:com (dlzc) wrote:

You might be right, but OTOH I wonder about the surface of a
"upside-down" drop hanging from the ceiling (wich should be somewhat
more "droply" shaped) vs. the surface of drop on a flat plane.

I don't know if there's a way to calculate this difference, but it
seems there must be one, which would also have an impact on
evaporation.

Interesting question. Where's a lab when you need one? ;-)

Good luck,

A. Friend

On Wed, 25 May 2005 00:09:31 -0400, "tadchem"
<[Only registered users see links. ]> wrote:



You should--you must---hurry! Moist air is lighter than dry air. The
molecule H20 having a molecular mass of 18 while N2 and O2 has a mass
of 32 and 28 in an 80,20 mixture That's why thunderclouds form,
clouds stay in the sky, and weather works. If the effect of
evaporative cooling on density isn't a second order effect, I would be
very suprised. Let us know.
John Bailey
[Only registered users see links. ]

Dear Clemens W:

"Clemens W" <[Only registered users see links. ]> wrote in message
news:1117014855.857365.224510@g43g2000cwa.googlegr oups.com...

I would think that the surface area difference would be quite
small, in the scheme of things. It is necessary for the water to
displace 100 or so times its volume in order to evaporate. Here
in the arrid desert (I beride people for thinking provincially,
and I may have done the same), when you water your lawn, a gentle
breeze springs up.

Sounds like a good lab experiment for school.

David A. Smith

John Bailey wrote:

<snip repost>


I can't - I'm in Virginia (with a desk job) and the lab is in Colorado.
Any interested student with access to a chemistry or biology lab could
run the experiment, though.


The average molecular weight is lower for wet air, but *freshly
humidified* air is cooler (before the temperature has a chance to
equilibrate). The question to be answered empirically is whether the
density of air is more affected by a cooling of a few degrees or by a
reduction of molecular weight due to admixture of dry air (average
molecular weight about 29.0) with a few percent of water (average
molecular weight 18.015). I'm betting it will depend on the initial
temperature.


Living in the High Plains for several years teaches one that the major
factor in the development of thunderclouds is thermal convection. The
localized heating of the moist air at the lower levels is sufficient to
reduce the density below that of the cooler but dryer air at altitude.


It is a first order effect, by the Gas Laws:

P*V = n*R*T

and density d = M*n/V, so temperature and density are inversely related
in the first order.


Tom Davidson
Richmond, VA

Dear tadchem:

"tadchem" <[Only registered users see links. ]> wrote in message
news:1117026772.615834.63890@z14g2000cwz.googlegro ups.com...
....

If I still care this weekend, I might pull out the psychrometric
charts, wet and dry bulb temperatures, etc. and work out an
empirical solution. I think the actual solution will depend on
the definition of "all things being equal".

And for what its worth, don't bet against Mr. Davidson... I
still *think* I'm right.

David A. Smith

N:dlzc D:aol T:com (dlzc) wrote:







The most obvious influences are temperature, pressure, partial pressure
of water vapor in the air, surface tension, and surface area. Control
for temperature of the water at the start and air at the start, control
for partial pressure of water vapor at the start, surface tension should
be about the same for the same water against the same air at the same
temperature and pressure.

Surface area gets affected by gravity, and by how hygrophobic the glass
is. On clean glass the water droplet will tend to spread out and
increase the surface area. Gravity will support that on the glass and
oppose it under the glass.

Then there's the speed of the air movement. In one case you get the
vapor-laden air tending to travel in a straight line away from the
glass, while the dry air moves in to replace it from all sides. In the
other case you get the vapor-laden air tending to travel in all
directions along the glass while the dry air moves in to replace it
straight toward the glass.

I'm not sure which would go faster, but there's the question how small a
difference can you detect?

On 25 May 2005 06:12:52 -0700, "tadchem" <[Only registered users see links. ]>
wrote:

On further reflection, isn't it obvious that if a cool slab is held
horizontally in moist air, droplets will condense on the lower side
first but not the upper? (well it was, when I thought of it) If this
is the case, arguing thermodynamic reversibility, leads to the
conclusion that droplets on the lower side will evaporate first.

Now that I have expressed these thoughts, the situation is just as
puzzling as the OP, if not more so. Hmmm.


John Bailey
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