Precipitable Water Vapor calculation

redcapaussie

Hi there,

I am trying to calculate the precipitable water vapor, or column water vapor amount, from a given volume mixing ratio at a particular pressure. What is the procedure to do this?

I understand you can convert volume mixing ratio to mass mixing ratio, by multiplying by the ratio of the molecular mass of water to the molecular mass of dry air. Multiplying that by 1000 should give the mass mixing ration in g/kg (common unit for mass mixing ratio).

But I don't get how to convert this into the precipitable water vapor amount, which is usually expressed as a height per unit area if the water vapour was condensed into liquid.

Any ideas?
thanks

klimatos

If all the water in the Earth’s atmosphere were precipitated onto the Earth’s surface, it is estimated that it would form a layer about 2.5 cm in depth. This is an estimate, not a calculation, and is based on a large number of assumptions. These assumptions may not all be valid for a given situation.

You don’t say whether you are performing a laboratory experiment or dealing with a column of the free atmosphere. If the latter, you should be aware that the temperature, the pressure, and the mixing ratio will all change dramatically with changes in elevation. These changes are not generally predictable for the real atmosphere, but may be estimated for standard atmospheres. You would need samples throughout the entire column to arrive at a mean value.

In the laboratory, I would convert the mixing ratio to vapor pressure and find the number density using the formula: n = p/kt. Here, n is the number density in number of molecules per cubic meter, p is the vapor pressure in Pascals, k is Boltzmann’s Constant, and T is the temperature in Kelvins. Then multiply n by the VSMOW molecular mass in kilograms and you have your total mass per unit volume. Multiply by the volume of the column, and you have your answer.

redcapaussie

Hi there, thanks for your response.

I am trying to do this given a standard atmosphere profile, which gives the height, pressure, temperature, and volume mixing ratio.

So, I'd be trying to calculate this for an portion of an individual layer of atmosphere, then add up all the layers to get the total.

I think the method is to work out how many moles are in each section (using PV=nRT) and then use the molecular weight & density to get an amount. Though I am not sure how to convert volume mixing ratio to vapour pressure.

Xnn

It's a complex problem and I'm not about to solve it.

Conceptually it's fairly straight forward to calculate the amount of water vapor in a given volume of air. However, in the atmosphere temperature generally decreases with rising altitude and the saturation point of water is reached. This is why there are clouds in the sky. Above clouds there is still air and pressure, but not nearly as much water vapor. So, water vapor is not evenly distributed.

Also, temperature profiles vary with weather systems. High pressure systems are unique from low pressure systems. Inversions and non-inversions.

redcapaussie

Xnn, I think you are missing my point.

I am calculating the precipitable water vapour in a column of air of unit cross sectional area. For example, between 10km and 20km, with a known pressure at each altitude (i.e., using some sort of standard profile).

In the layer of interest, the volume mixing ratio is known, but I don't know how to convert this into amount of precipitable water vapour.

Knowns are: pressure (at either ends of the layer), volume (1m2 * layer height), volume mixing ratio

klimatos

P = nkT is a simpler restatement of PV = RT, since R = Nk and n = N/V. It replaces the awkward and cumbersome molar volume with a simple cubic meter. Molar volume varies continuously from top to bottom of a parcel of the atmosphere; whereas a cubic meter is a cubic meter throughout.

The ratio of vapor pressure to dry air pressure is the same as the ratio of vapor volume to dry air volume or vapor molecular abundance to dry air molecular abundance. The actual vapor pressure, of course depends on the temperature at that point in the parcel as well.

The official mean mass of the water molecule is given by the Vienna Standard Mean Ocean Water (VSMOW) protocol. This protocol takes into account the relative abundances of the nine stable water isotopes. This mean mass is 2.99150512E-26 kilograms.

If your concern is with the total mass of water above a unit area of ground surface, I assume that you will abandon the column of uniform cross-sectional area in favor of a parcel with a truncated pyramid configuration. The column concept always gets applause at the meetings of the Flat Earth Society, but denotes sloppy thinking when used by atmospheric scientists. The area above a square meter of surface increases steadily as you go up.

On the other hand, if you are doing this for a purely fictional parcel of air rather than a parcel of the real atmosphere, then you can give it whatever configuration you find convenient. I do wonder, however, why you would expend this much effort on something that would have extremely limited applicability. I find the whole concept of uniform "layers" of the atmosphere to be particularly disturbing.

redcapaussie

Hi klimatos - I am going to PM you.