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A cloud has relative humidity of 100% or more. Is it correct to say that the mixing ratio of a cloud is undefined? My reasoning is that dry air in a cloud is negligible therefore $w = \frac{x g}{ 0 kg}$

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No, it is not undefined.

A relative humidity of 100% refers to cases in which the atmosphere holds as much water, as the temperature would allow. The term "dry air" denotes the fraction of the air not contaning any water vapor (or explicitly not any water molecules). Dry air is therefore never negligible in clouds.

The saturation vapor pressure over a plane water surface can be calculated by the Magnus formula (which is an empirical approximation to the Clausius-Clapeyron-equation valid in a temperature range from -45°C to 60°C):

$E_{H_2O}(\vartheta)=6.112hPa \cdot exp{\frac{17.62 \cdot \vartheta}{243.15 ^\circ C + \vartheta}} $,

where $\vartheta$ is the temperature in °C.

The water vapor mixing ration $r$ itself is calculated by:

$r=\frac{0.622e}{p-e}$,

where $p$ is the air pressure and $e$ is the water vapor pressure. In case of a cloud you equate $e$ and $E_{H_2O}(\vartheta)$, since the relative humidity is defined as:

$r.H.=\frac{e}{E_{H_2O} (\vartheta)}$

As a sidenote: Oversaturation in clouds may occur frequently but is usually smaller then 1%. For back-of-the-envelope-calculation this should be sufficient.

Please note, that different text books use diffent symbols for $e, E, ...$

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    $\begingroup$ Indeed, put it this way Reqbisha: when the precipitation falls on you, there's still other air there apart from the water/moisture. You can still breathe. 100% saturated doesn't mean the air is 100% water, it just means that it cannot hold any more water vapor in that air (apart from small amounts of supersaturation due to curvature geometry) because the vapor pressure is too high. Even in the most extreme cases, air is still somewhere around 90-95% non-water. $\endgroup$ – JeopardyTempest Sep 3 '17 at 15:00

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