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

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

<p>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 $\:\mathrm{N/m^2}$.</p>

<p>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:</p>

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

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

<p>So long story short: <strong>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.</strong></p>

<p>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 <a href="http://www.vaisala.com/Vaisala%20Documents/Application%20notes/Humidity_Conversion_Formulas_B210973EN-F.pdf" rel="nofollow noreferrer">this paper</a> 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.</p>

CC BY-SA 4.0
2022-12-12T04:21:52.313Z
Broken link to paper. Try https://github.com/David-OConnor/humidity/blob/master/Humidity_Conversion_Formulas_B210973EN-F.pdf
Carl
CC BY-SA 3.0
2017-03-09T22:13:17.777Z
2017-03-10T07:52:10.313Z
2017-03-10T07:52:10.313Z
JeopardyTempest
<p>Is it possible to convert Air vapor pressure ($N/m^2$) to relative humidity (%) given a known air temperature? </p>

<p>For example, I know that relative humidity can be estimated as:</p>

<p>$$ RH = 100* \frac{e_{a}}{e_{s}}$$</p>

<p>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:</p>

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

<p>where $T_{a}$ is the air temperature in $^{o}C$. </p>

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

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

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

<p>and then using the equation?</p>

CC BY-SA 3.0
2017-03-09T11:40:00.300Z
Your logic is sound. I would do the same.
Fred
CC BY-SA 3.0
2017-03-09T17:03:07.520Z
If you are concerned about the units, consider 1 Pascal= 1 Newton per square meter. To remind you, Pressure=Force/Area.
BarocliniCplusplus
<meteorology>
Convert air vapor pressure to relative humidity
Emma Tebbs
# Convert air vapor pressure to relative humidity

(Asked by: username_0 on 2017-03-09)

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?

**Tags:** meteorology

> **Comments:**
> **username_1**: Your logic is sound. I would do the same.  
> **username_2**: If you are concerned about the units, consider 1 Pascal= 1 Newton per square meter. To remind you, Pressure=Force/Area.  

---

## 1 answers

### Answer by username_3 on 2017-03-09. Score: 2 | Accepted answer

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

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](http://www.vaisala.com/Vaisala%20Documents/Application%20notes/Humidity_Conversion_Formulas_B210973EN-F.pdf) 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.

> **Comments:**
> **username_4**: Broken link to paper. Try https://github.com/David-OConnor/humidity/blob/master/Humidity_Conversion_Formulas_B210973EN-F.pdf  

---

