Water Extraction

[Tarnished]

Does anyone have a formula or information on how much water condenses out of a fixed volume of air when it is cooled?


ie you know the starting condition of the air and the delta T, can you work out how much liquid water results?

Ta

T

[enicalyth]

If you want an exact answer you need a textbook on thermodynamics e.g by Rogers and Mayhew. I’d need to know process details, enthalpy etc.

But if you want a coarse answer start with http://wahiduddin.net/calc/index.htm and browse Richard Shelquist’s excellent site.

Dry air has a molar mass of 0.029 kg. It is denser than water vapour, which has a molar mass of only 0.018 kg. Strange but true, humid air is lighter than dry air.

Richard’s on-line calculators will do the work for you; I’d guess you’ll pick two temperatures, two baro pressures and proceed on these lines:

Let the total atmospheric pressure = P and the local water vapour pressure = p. Then the partial pressure of the dry air component is [P – p].

The weight or mass ratio of the two components, water vapour and dry air is then = (0.018 *p) / [ 0.029 *(P - p ) ]
= 0.62 *[p / (P - p )] expressed in kg water vapour / kg dry air.

Very roughly under ISA conditions at ground level you could say = (0.0000062*p) kg of water vapour per kg dry air.

At 35 deg C the maximum amount of water that the air can hold is 0.038kg/kg dry air.

Cool the air to 25 deg C and it becomes 0.020kg/kg dry air.

Rgds enicalyth

PS - ask uncle old smokey under what conditions the B707, DC8 and Harrier jump-jet are significantly steam-powered!!

[Volume]

If you do understand a little german, you can find some data here (http://www.holzfragen.de/seiten/taupunkt.html), including an online calculator.

see also this former thread (http://www.pprune.org/forums/showthread.php?s=&threadid=166282)