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

[Nathan McCorkle]

On Fri, Feb 27, 2015 at 10:08 PM, Mac Cowell <m...@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 McCorkle]

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

--
-Nathan

[John Griessen]

On 02/28/2015 03:57 AM, Nathan McCorkle wrote:
> perhaps hone the
>>calculation to get rid of my 'mollier diagram; guesstimation issues?

I did not see temperature in the model, so it seems odd to miss that...

or is it there as a worst case constant somewhere at about 40 deg C?

As evaporation happens from a capillary, the exposed water surface will shrink back into
the capillary and cause evap slow down more because it is in a well where there will be no breeze.

That part seems to become a diffusion calculation that will vary as depth of the "well".

[Nathan McCorkle]

The two terms xs and x are temperature based terms relating temperature to how much the air can absorb.... Something like that.

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