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I would like to find a formula to approximate RH based on known air temperature and dew point. I need the approximation to be valid at least between -20...+40 degrees Celsius. I read the accepted answer on this question, but unfortunately

  • the link for the document where the tabulated values were retrieved from is no longer working
  • using those values yield weird results. I tried it (the RH formula with the constants in the first row of the table) with T = 16.15 and TD (dew point) = -4.45, and it gives 215% RH (could it be $100^{m*...}$ instead of $100 *10^{m*...}$?...)

Any help would be appreciated. Thanks.

UPDATE

Thanks @BarocliniCplusplus for his answer. If anyone reading this needs an implementation in Python to approximate the RH, here is one:

RH = 100*(math.exp((17.625*TD)/(243.04+TD))/math.exp((17.625*T)/(243.04+T)))

where T is the temperature in Celsius, and TD the dew point in Celsius. This approximation is derived from the "Conclusions" section of this article and is suggested to have relative error of 0.384% or less when used between -40°C and 50°C.

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  • $\begingroup$ The update takes a few leaps and would be best as an answer so people can vote upon it's validity. That said, it appears to be right, as you usually can use the same vapor pressure equations with T and Td respectively to get e and es, and then you canceled the multiplier. That said, explanation of how you got there helps others. But I did get it to match other RH calculators, so seems it may be valid. $\endgroup$ Oct 4 at 19:20
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    $\begingroup$ @JeopardyTempest I added back the *100 part, as RH is usually expressed as percentage rather than as fraction. The formula is the same as the one found here. Validity range/error might be different than what you added, according to this other resource found on the same website. $\endgroup$
    – Darius V.
    Oct 16 at 9:58
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    $\begingroup$ As for the derivation of the formula, I suppose it's directly equation (21) in the paper applied to the RH formula (see the accepted answer). $\endgroup$
    – Darius V.
    Oct 16 at 10:14
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The equation for relative humidity is $$RH=100\times \frac{e}{e_s(T)}=100\times \frac{e_s(T_d)}{e_s(T)} \tag{1}$$ Where $T_d$ is the dewpoint temperature and $T$ is the temperature, $e$ is the water vapor pressure, and $e_s$ is the saturation vapor pressure, which is also known as the Clausius Clapeyeron equation. While the previous link has a decent definition and equation, my preferred equation (particularly because it is derivable) is the low temperature approximation:

$$e_s(T)= e_s(273 \mathrm{K})\exp\left[\frac{L_v}{R_v}\left(273.15^{-1}-T^{-2}\right)\right] \tag{2}$$ where $T$ is the temperature or dewpoint temperature in Kelvin, $L_v$ is the latent heat of vaporization,$e_s(273 \mathrm{K})=6.11 hPa$, and $R_v$ is the specific gas constant for water vapor. Note that $(2)$ is for liquid water. You may be able to replace $L_v$ for $L_s$ for sublimation/deposition. Also note, that this is the "pure" form of the relative humidity equation. The presence of solutes (Cloud condensation nuclei) may lower the saturation vapor pressure, increasing the actual relative humidity. I'll let combining $(1)$ and $(2)$ be an exercise left for the reader.

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  • $\begingroup$ Thanks, the first link did the trick. But just out of curiosity, what's with Lv from (2)? (I don't know meteorology/physics well enough, so I can only guess what's happening). Do I have to compute it for any particular temperature if it were to use the formula derived from (1) and (2)? $\endgroup$
    – Darius V.
    Dec 15 '20 at 12:19
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    $\begingroup$ $L_v$ is the latent heat of vaporization (the amount of energy it takes to turn water vapor into liquid water). Technically, [t does vary with temperature] [en.wikipedia.org/wiki/…, but it is assumed to be a small amount. I usually use 2.5E6 J/kg, Obviously, if the temperature is below freezing, then it is more likely ice would form rather than liquid water (in which case, follow the hyperlinks for deposition). $\endgroup$ Dec 15 '20 at 16:39

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