Re: evaporation losses... was: [DIYbio] Electronic requirements for redesign of Arduino PCR thermal cycler

There was also this answer, which gives a formula if you know the
energy of your heat source:
http://physics.stackexchange.com/questions/91810/how-do-i-predict-volume-loss-due-to-evaporation-when-boiling-water

On Sat, Feb 28, 2015 at 1:51 AM, Nathan McCorkle <nmz787@gmail.com> wrote:
> On Fri, Feb 27, 2015 at 10:08 PM, Mac Cowell <mac@diybio.org> wrote:
>> The capillary design is not as complicated as you make it sound, Cathal. I
>> think evaporation is marginal in open 10 uL tubes, so they don't need to be
>> closed,
>
>
> hmm, this comment really interested me, since I've previously done
> some experiments where I watched evaporation over a few days using a
> nice mettler lab scale with a serial port on it for data collection.
>
> I tried finding out how to calculate for evaporation losses, and found
> an empirical equation for things like swimming pools.
>
> TLDR; the calculation I did (which could be wrong) shows about 2 to 15
> nanoliters PER HOUR of evaporation for a 100 micron diameter capillary
> with no air movement from 25 to ~65 C (but if you're really
> interested, please try to verify this number on your own... and let us
> know). However, if you increase the diameter from 100 microns to 1mm,
> the evaporation seems to be more like 200nL/H to 1.5uL/H (1500nL/H).
>
>
> This is where I got the equation:
> http://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html
>
> which says:
>
> gs = Θ A (xs - x) / 3600
>
> or
>
> gh = Θ A (xs - x)
>
> where
>
> gs = amount of evaporated water per second (kg/s)
>
> gh = amount of evaporated water per hour (kg/h)
>
> Θ = (25 + 19 v) = evaporation coefficient (kg/m2h)
>
> v = velocity of air above the water surface (m/s)
>
> A = water surface area (m2)
>
> xs = humidity ratio in saturated air at the same temperature as the
> water surface (kg/kg) (kg H2O in kg Dry Air)
>
> x = humidity ratio in the air (kg/kg) (kg H2O in kg Dry Air)
>
>
> What I ended up plugging in for numbers, to get my nanoliter values:
> http://www.wolframalpha.com/input/?i=%2825*%28+pi*%2850+%2F1000000%29%5E2%29*%280.02-0.0098%29%29+kilograms+water+to+microliters
>
>
> used this page to try getting some values to plug in for [xs]
> http://www.engineeringtoolbox.com/humidity-ratio-air-d_686.html
>
> Used this hard-to-figure-out chart to try some other values for a
> higher temperature (variable [x]):
> http://www.engineeringtoolbox.com/psychrometric-chart-mollier-d_27.html
>
>
>
>
>
> Can someone verify or disprove the numbers, or perhaps hone the
> calculation to get rid of my 'mollier diagram; guesstimation issues?



--
-Nathan

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