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The moisture-carrying ability of air goes up as the temperature rises. So if I take a volume of air and heat it or cool it (allowing it to expand/contract so the pressure remains constant) how can I calculate the (approximate) relative humidity of the warmer/cooler air that contains the same amount of water vapor?

Say I blow 30°c air at 20% relative humidity over a cool object (but not cool enough to condense water out of the air.) The air is cooled to 20°. What would the relative humidity of the cooled air be if it contained the same amount of water vapor? How do I calculate that?

I am building an Arduino-based control for a warm air dryer for ski boots and bike gear. I will measure the temperature/humidity of the heated air as it comes out of the dryer, and then measure it again as it escapes the gear that I am drying. While the gear is still damp, the waste air should pick up additional moisture, and so its relative humidity should be higher that it would be if I simply cooled it down. When my gear is completely dry, the cooler waste air should contain the same amount of water vapor that it started with, so the relative humidity should be the amount calculated for the measured change in temperature.

I want a "good enough" solution, not one that's mathematically pure. If it gets me answers that are within a percent but involves less computation, that's better than an ideal solution that involves lots of time-consuming calculations. The Arduino microcontroller is fairly slow and cannot do high speed floating point calculations.

(Would this question be better on an engineering forum? I could see arguments either way.)

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  • $\begingroup$ I meant see this: physics.stackexchange.com/questions/337440/… $\endgroup$ – Buck Thorn Dec 15 '18 at 19:15
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    $\begingroup$ You'll need the temperature and RH of the intake air and the temperature and RH of the exiting air. Then it is about what kind of math the microprocessor can do. I can't see why "slow" calculations would be a problem, even a minutes of extra drying shouldn't matter. In fact you'd probably want to add some extra time just to be sure that the items are really dry. $\endgroup$ – MaxW Dec 15 '18 at 19:27
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    $\begingroup$ You'll of course need to know the vapor pressure of water as a function of temperature (which gives you 100% RH). See Wikipedia for several different formulas and their comparison. $\endgroup$ – MaxW Dec 15 '18 at 19:40
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If you want to convert the %RH from one temperature to another you can use one of various parameterizations, for instance the Antoine equation:

range: 273. - 303 K,
Bridgeman and Aldrich, 1964,
https://webbook.nist.gov/cgi/inchi?ID=C7732185&Mask=4#Thermo-Phase

A = 5.40221;
B = -1838.675;
C = -31.737;

log(P) = A + B/(C+T);

Then obtaining the $RH_2$ at $T_2$ from that at $T_1$ is as simple as $RH_2 = RH_1 P_1/P_2$, where $P_1$ and $P_2$ are the saturation vapor pressures of water computed with the equation above, assuming the ambient pressure (and therefore the partial pressure of water) is constant.

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  • $\begingroup$ That looks like what I want. Is that log base 10, base 2, or natural log? $\endgroup$ – Duncan C Dec 15 '18 at 21:49
  • $\begingroup$ Wikipedia uses A - B/(C+T) gives slightly different coefficients, but good over range from 1 to 99 C. $\endgroup$ – MaxW Dec 15 '18 at 22:04
  • $\begingroup$ Cool. That seems to work. At first I didn't realize the temps were in Kelvin, so I was getting partial pressures of infinity (floating point overflow from crazy-big powers of 10). $\endgroup$ – Duncan C Dec 15 '18 at 22:23
  • $\begingroup$ That's natural log (base e). $\endgroup$ – Buck Thorn Dec 16 '18 at 9:08

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