Metallus
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Liquid evaporation time dependency
Hello,
back in university I studied properties like vapor pressure, heat of evaporation and what not, but when actually presented with a seemingly trivial
problem such as "how much time it takes for a given volume of water to evaporate?" I fail to connect my knowledge.
Let's make a practical example: what is the time needed for the complete evaporation of 10ml of water homogeneously spread on a flat surface of
stainless steel (AISI304) of 2 mm thickness in normal conditions? No need for dead accurate values, approximations and simplifications are very
welcome. I just want to get a rough idea.
P = 1 atm
T = 25 °C
H vap = 2,26 kJ/g
m = 10 g
V = 10 ml
TC steel aisi304 = 16,2 W/mK
Vapor pressure = 0,023 atm
I know that a drop of water from a burette is ca 0,05 cm3. If we approximate a drop of water to a semisphere, then the "height" of this theoretical
drop would be equal to the half radius:
V = 2/3*pi*r3 ==> r3 = 3V/(2pi) ==> r = 2,88 mm ==> h = 1,44 mm
From this we could get the area of this "disk" of water (ignoring angles) which would be A = V/h = 69.4 cm2.
So now the problem is: how much time is required for a disk of water of 1,44 mm thickness to evaporate? Considering we are in a open environment and
therefore the "container" volume is infinite (or very large), water is bound to fully evaporate since it is not in equilibrium with the environment.
But how do I approach this problem from here?
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Sulaiman
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I think that you can assume that the water temperature will be the same as ambient temperature due to low evaporation rate (heat flow) and good
conductivity of steel and water.
(If you do get a final numericalor algorithmical result then you can come back and check the validity of this assumption)
Now you need to also know the partial vapour pressure of water at ambient temperature,
ambient relative (or absolute) humidity,
airflow (or diffusion) rate
and
find an equation to plug the numbers into 
P.S. Your stainless steel is dirty.
(s/s is hydrophilic) 
P.P.S. I bet you could measure it quicker than you could decide what 'fudge factors' to use in any equation that you do manage to find.
i.e. one drop of water on a sensitive weighing machine vs. time
[Edited on 27-8-2018 by Sulaiman]
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fusso
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Quote: Originally posted by Metallus  | Hello,
Let's make a practical example: what is the time needed for the complete evaporation of 10ml of water homogeneously spread on a flat surface of
stainless steel (AISI304) of 2 mm thickness in normal conditions? No need for dead accurate values, approximations and simplifications are very
welcome. I just want to get a rough idea.
P = 1 atm
T = 25 °C
H vap = 2,26 kJ/g
m = 10 g
V = 10 ml
TC steel aisi304 = 16,2 W/mK
Vapor pressure = 0,023 atm
I know that a drop of water from a burette is ca 0,05 cm3. If we approximate a drop of water to a semisphere, then the "height" of this theoretical
drop would be equal to the half radius:
V = 2/3*pi*r3 ==> r3 = 3V/(2pi) ==> r = 2,88 mm ==> h = 1,44 mm
From this we could get the area of this "disk" of water (ignoring angles) which would be A = V/h = 69.4 cm2.
So now the problem is: how much time is required for a disk of water of 1,44 mm thickness to evaporate? Considering we are in a open environment and
therefore the "container" volume is infinite (or very large), water is bound to fully evaporate since it is not in equilibrium with the environment.
But how do I approach this problem from here?
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10=2/3*pi*r3
r should be 2.88cm instead of mm.
And why would you approximate the shape of a large enough puddle to a hemisphere?!
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walruslover69
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you are going to need a fairly complicated differential equation to solve a problem like this. You essentially need to model the diffusion. You will
also need to know the humidity.
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Sulaiman
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I expect that any equations that you find,
that give a sensible answer,
will be based on empirical data,
not theory.
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Fulmen
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I have to agree with the two previous posters. This is actually a very hard problem to solve, I expect the solution will involving at least heat flux,
diffusion and natural convection. That's a juicy cocktail.
We're not banging rocks together here. We know how to put a man back together.
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Metallus
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| Quote: Originally posted by fusso | 10=2/3*pi*r3
r should be 2.88cm instead of mm.
And why would you approximate the shape of a large enough puddle to a hemisphere?! |
r is 0.288 cm (so 2.88 mm). I approximated the single drop of water to a hemisphere to get a rough idea of the height of the puddle that would form
after continuous leakage on the same area. I made a mistake in my previous calculation, as I didn't need to further divide the result by 2.
This problem actually arises from a practical issue. My boiler is leaking ca 10 ml in one day on the same spot (my stainless steel pan); assuming that
the dimensions of the drops are ca 0.05 cm3, then you should be able to roughly estimate the dimensions of the puddle of water that forms
and the time needed for it to completely evaporate.
Further researching this subject, ventilation and air humidity are indeed necessary to make this calculation as some have stated in this thread.
However I don't really know how to quantify these.
Does anyone have a rough idea of "wind" speed in a small apartment with windows closed? And what would the average humidity be if I live in a small
urban area where the weather is rainy 2-3 days per week but it is far from the sea?
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Sulaiman
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Enjoy fudging the 'fudge factors' :)
Try putting some cloth/wick/sponge to absorb the drops
(on the way down or on the s/s)
to distribute the water by capillary action
then evaporate the water from a large surface area.
Mold/fungus/bacteria etc. will try to colonise the cloth
so treat it with something* or wash and/or replace as required.
* I'd try a little colloidal silver - but I have stock
P.S. you can easily make a wet bulb & dry bulb hygrometer.
(use distilled water etc. to prevent buildup of salts in the wick)
do a few hundred measurements
and fit some equation to the data
[Edited on 29-8-2018 by Sulaiman]
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battoussai114
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Assuming you're not under direct sunlight and there's no strong wind, I feel a pseudo-stationary model of molecular diffusion through stagnant gas in
1D (upwards from your SS surface) might give decent approximation of a worst case time for drying up (including convective diffusion and stuff like
air movement would likely mean it drying quicker but also a uglier model).
I might have something on this in between my transport phenomena notes, will check it later.
Oh, btw: you'd also want to assume your diffusion coefficients to be constant throughout the process. The conditions in your room would most likely
not change due to a puny puddle evaporating, so that's the approximation that you should worry the least about.
[Edited on 30-8-2018 by battoussai114]
Batoussai.
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boilingstone
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Solution
I typed up a step by step solution to find the time it would take for a hemispherical water droplet to diffuse into air.
You will need to use the mass continuity equation, as well as the equation for Fick's law of diffusion. This is only for still air, as if there is
convection then it completely changes things. You will need the diffusivity of water into air at the chosen conditions, the vapor pressure of water,
the partial pressure of water already in the bulk phase (far from the droplet), the density of liquid water, the molecular weight of water, and the
dimensions of the water droplet.
Attachment: Diffusion Problem.pdf (199kB)
This file has been downloaded 235 times
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