On Wed, Jun 22, 2011 at 6:51 PM, Mark Iverson <zeropo...@charter.net> wrote:
> ** > Oh well, I'll run the errand tomorrow... > > As a start, go read about the gas laws and partial pressure and how > humidity is calculated from partial pressure... > > In order to understand how Galantini can ESTIMATE the liquid water content > of the steam, you need to think several steps ahead as in chess, or in a > complex mathematical derivation that involves many steps and applying > theorems at each step in order to derive the final desired answer. Its not > a direct measurement as I've said numerous times. > You're not saying anything. > > The behavior and properties of gases are very different from liquids, and > are dictated by mass or mole fraction, not concentrations. > "Gases dissolve, diffuse, and react according to their partial pressures, > and not according to their concentrations in gas mixtures or liquids." > Still nothing. > > If you vaporized so many grams of liquid water into a cubic meter box with > NO other molecules present, you'd end up with a specific temperature and > pressure, and that could also be communicated as a mixing ratio. For > atmospheric science where we ARE dealing with air, then the mixing ratio is > the mass of water (if you condense the water vapor) to the mass of dry air. > However, you do NOT need other molecules in order to measure humidity. > Humidity exists without other molecules. That's true. But if you want to measure humidity with a device calibrated in air, you need to make the measurement in the same conditions the calibration was performed under. But humidity is not what you want if you're interested in the steam wetness. The relative humidity of steam is 100%. If that device gives the mass of water vapor per unit volume, then it will give the density of steam: 0.6 kg/ m^3, or 1000 g of water vapor per kg of steam. We already know that number. > So you're getting hung up on the denominator thinking that there has to be > some entity or volume of some other molecule(s) when in fact, it might as > well say, "cubic meter of empty space". > But if all you know is x g/m^3, you don't know the mass unless you know how many m^3. And you need the mass for your simple algebra. The mass per unit volume doesn't help.
