Convert air vapor pressure to relative humidity

Question

Is it possible to convert Air vapor pressure ($N/m^2$) to relative humidity (%) given a known air temperature?

For example, I know that relative humidity can be estimated as:

$$ RH = 100* \frac{e_{a}}{e_{s}}$$

where $e_{a}$ is the air vapor pressure and $e_{s}$ saturated vapor pressure at air temperature, both with units of mb. Specifically, $e_{s}$ can be estimated by:

$$e_{s} = 6.11*\exp\left(\frac{17.27T_{a}}{237.3+T_{a}}\right)$$

where $T_{a}$ is the air temperature in $^{o}C$.

However, in my case the air vapor pressure has units of $N/m^2$?

How can I use this conversion with these values/units? Is it simply the case of converting $e_{a}$ to mb via:

$$e_{a} (mb) = e_{a} (N/m^{2})*0.01$$

and then using the equation?

Answer 1

$$1\:\mathrm{N/m^2} = 1 \:\mathrm{Pa}$$ $$100 \:\mathrm{Pa} = 1 \:\mathrm{hPa} = 1 \:\mathrm{mb}$$

So yes, you do have valid units of pressure. And can easily convert it.

The calculation $ RH = 100* \frac{e_{a}}{e_{s}}$ will work as long as the pressures are in equal units, whether Pa, mb, Gbar, etc, since it's just a ratio. They just need to be the same. So, you can convert your given $e_a$ into mb by multiplying by 100, convert your $e_s$ into Pa by dividing by 100, or just apply the change in the $e_s$ equation to get $e_s$ in $\:\mathrm{N/m^2}$.

We often use your $e_s=6.11∗exp(17.27T_a/(237.3+T_a))$ in meteorology since we work with pressures in mb (or equivalently hPa). But indeed the same equation simply works with the constant adjusted to the proper units as:

$$e_{s} = 0.0611*\exp\left(\frac{17.27T_{a}}{237.3+T_{a}}\right)$$

giving answers in Pascals ( = $\:\mathrm{N/m^2}$).

So long story short: yes, all of your equations and reasoning are good, your units are all pressures, they just don't match form. But you can indeed adjust at any step to account for that. Simplest is multiplying your $e_a$ (in $\mathrm{N/m^2}$) by 100 or dividing your $e_s$ (in $\mathrm{ mbar}$) by 100 to make them match.

Also of note, the first equation you gave is exactly valid. The second equation, as you indicate, is an estimation. The longer version described in this paper would give exact values if you wish, but it looks like you're unlikely to be off by more than 0.1%, so what you're doing is likely fine unless you're need very exacting precision.