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For my data, I have temperature(F), atmospheric pressure and dew point.

I wanted to get a rough estimate of the air density, using all three of these.

In addition, how would I get an even rougher estimate using just temp and dew?

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  • $\begingroup$ Use the ideal gas law for air density, given atmospheric pressure and temperature. If you only have dew point temperature and air temperature, you can't get an estimate of air density because the vapor pressure of water vapor is independent of air pressure. $\endgroup$ – David White Dec 7 '16 at 23:14
  • $\begingroup$ Ok, so I read up on ideal gas law, and I couldn't find any simply formulas in regards to adding dew. $\endgroup$ – Danny W Dec 8 '16 at 0:19
  • $\begingroup$ An ideal gas has a particle density determined by temperature and pressure. Density, though, depends on gas particle WEIGHT, and H2O is a lighter molecule than O2 or N2. $\endgroup$ – Whit3rd Dec 8 '16 at 1:06
  • $\begingroup$ @DannyW, you (or I) may be missing a fine point here. For a "rough" estimate, ignore the amount of water vapor in the air, if you are talking about ambient temperature. If the temperature is not ambient, please specify conditions that are somewhat more specific. $\endgroup$ – David White Dec 8 '16 at 3:27
  • $\begingroup$ What about simply calculating the densities by the universal gas formula, and adding them? $\endgroup$ – peterh says reinstate Monica Dec 8 '16 at 4:08
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Parameters which you have are temperature, atmospheric pressure and dew point. Parameters which are need for calculating the air density are temperature, atmospheric pressure, relative humidity and saturated vapor pressure.

In this case, the relative humidity is necessary to be calculated from the dew point.

The relative humidity can be obtained by the ratio of the saturated water vapor amount $s(t0)$, $s(t)$ at the dew point $t0$ and the temperature $t$. Namely, the relative humidity $Rh$ can be expressed as follows.

$$ Rh = \frac{s(t0)}{s(t)} \times 100 $$

$s(t)$ can be obtained from the equation of state of water vapor.

$$ s(t) = \frac{217 Ps}{t + 273.15} $$

, where the saturated water vapor pressure $Ps$ [Pa] can be obtained from Tetens's formula.

$$ Ps = 611 \times 10^{7.5 t / (t + 237.3)} $$

Here, the relative humidity could be obtained. As a next step, it is calculated the air density.

The air density can be obtained from Jones's formula. Jones's paper is F. E. Jones, “The air density equation and the transfer of the mass unit”, J. Res. Natl. Bur. Stand. 83, 1978, pp. 419-428.

The air density $\rho$ is

$$ \rho = \frac{0.0034848}{t + 273.15} (P - 0.0037960 \cdot Rh \cdot Ps) $$

, where $t$ [Celsius] and $P$ [Pa] are temperature and atmospheric pressure, respectively. The unit of air density $\rho$ is [kg/m$^3$].

The unit of temperature which used here is Celsius. So if you want to use Fahrenheit as a unit of temperature, please convert it. If my explain is difficult to understand, I apologize. Because my English is poor.

If you want to quickly check the above calculation, you can confirm it using following AWK command. The input values for "echo" are atmospheric pressure, temperature and dew point, respectively.

$ echo '1013.25 25 14' | awk '{ps = 611 * 10^(7.5 * $2 /($2 + 237.3))} {ps0 = 611 * 10^(7.5 * $3 /($3 + 237.3))} {st = 217 * ps / ($2 + 273.15)} {st0 = 217 * ps0 / ($3 + 273.15)} {rh = 100 * st0 / st} {ro = ($1 * 10^2 - 0.003796 * rh * ps) * 0.0034848 / ($2 + 273.15)} END{print "\nAir density is " ro " [kg/m^3]";}'

When atmospheric pressure, temperature and dew point are 1013.25 hPa, 25 degrees C and 14 degrees C, the air density is 1.17693 [kg/m^3].

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