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Q & A: how much does wind speed up evaporation?

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Most recent answer: 06/01/2015
Q:
Hi, I am trying to investigate the effects of a wind flow on the evaporation of sweat. Lets say on your hand of Area 160cm^2 you sweat 1.6mL/min. And you have standard conditions say, 37degrees C for body and sweat temperature, sweat you assume is water (enthalpy vaporisation the same). How can you relate the air flow to the evaporation of this sweat / determine the flow rate to evaporate the sweat? HELP !
- duncan cunninghame (age 23)
Auckland
A:

That 1.6 mL/min of evaporation can give a cooling of 1.6*2260J/ml *1 min/60s= 55 Watts via the latent heat of evaporation. Your question amounts to asking whether plain diffusion can carry the water vapor away fast enough to reach that maximum cooling rate, or whether some wind is needed too, and if so how much wind is needed.

How fast does the water vapor diffuse? The hand is something like a two-sided circular surface, each side with area 160 cm2. So the problem looks like 1-D diffusion away from that disk up to a distance of around the size of the hand, ~10 cm, beyond which it looks like a 3-D problem. That means that the concentration of water vapor would fall to a fraction of the saturation level on a distance of roughly 10 cm. So the mass flow of the water would be about:

320 cm2 *(saturation vapor concentration)* D/10 cm,

where D is the diffusion coefficient of water in room-temperature air, ~0.3 cm2/s.

So that gives a diffusion flow rate of ~(saturation vapor concentration)* 10cm3/s. I'm a little unsure what vapor concentration to use, since the temperature of the sweat is somewhere between room temperature (say 25°C) and body temperature (37°C) and there's some salt in it too. The value should be close to 2*10-5 gm/cm3. That would give a flow of about 10-2 gm/min. So that wouldn't be nearly enough to get the full cooling from the 1.6 gm/min sweat.

How much wind would be needed to start to help noticeably? The effective flow velocity from diffusion is roughly D/(typical length) or in this case (0.3 cm2/s)/10cm= 0.03 cm/s. A breeze faster than that would help. A breeze about 100 times faster would be needed to keep up with that sweat rate. So that would be very roughly 3 cm/s or 0.1 km/ hour. It sounds like just plain wallking would be fast enough.

We have some old answers that should help out too, e.g.


.

Mike W.


(published on 06/01/2015)

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