Water Vapor Question

UWSojourner

I've asked this before somewhere, but still don't understand what's up with water vapor.

I've heard what to my mind right now are two contradictory things. First, some say it acts like any other gas in your breathing mix so, if you have water vapor pressure present, it dilutes the other gases. Second, while modeling Buhlmann tissue compartments, the amount of water vapor pressure you subtract from ambient pressure remains constant. In other words, water vapor pressure does not change with pressure changes!? Can anyone clear this up?

Thanks.

Sutty

That is because we are talking about the "saturated vapour pressure" of water see here. Because the gas in the alveoli in the lungs is saturated with water the pressure exerted by the water vapour will be the saturated vapour pressure of water at body temperature (47mmHg).

Neil

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NeilHolden

[quote=UWSojourner]I've asked this before somewhere, but still don't understand what's up with water vapor.

I've heard what to my mind right now are two contradictory things. First, some say it acts like any other gas in your breathing mix so, if you have water vapor pressure present, it dilutes the other gases. Second, while modeling Buhlmann tissue compartments, the amount of water vapor pressure you subtract from ambient pressure remains constant. In other words, water vapor pressure does not change with pressure changes!? Can anyone clear this up? quote]

Never really thought about this before, as it is not something that I have ever been taught or heard of, but have now thought about it for a short time here goes my theory

Water vapour in the loop is in fact not really a gas - although acting like one. for water to turn gaseous it must have an increase in temperature -100 deg c - boiling, steam (not the stuff you see coming out of the Kettle which is water vapour you can't actually see steam).

As water vapour - tiny tiny droplets of water, they are in fact incompressible, at least at the pressures we are concered with. So their volume remains unchanged with the pressure in the loop. If you increase the pressure they tend to gather together as the gas volume between them reduces - another thread about brownian motion etc.....

Subtract the overall volume of the water vapour in the loop (inconsequentially
small) then work out the remaining partial pressures on the remaining volume and pressure. So strictly speaking yes a dilution, however such a small amount it is not usually taken into account in the loop calculation.

Water vapour is not 'mixed' within the gas but held in suspension and transported around with it, and no doubt many pure scientist will now all over my very simplistic explanation .

So as it is incompressible the amount of water vapour volume to subtract (once at working level) will remain approx constant.

I am of course more than willing to be taught otherwise!

Any Help or just more .

Neil

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UWSojourner

Thanks, that explains the part of it acting like a gas. But now assume you move down to 4 ATAs. Your tissues start loading inert gas due to pressure. But when you determine the inspired inert gas pressure in Buhlmann's model the equation works like this:

At the surface (33fsw): Inspired pressure of nitrogen = (33 - 1.6 ) * 0.79.

At 100ft (133fsw): Inspired pressure of nitrogen = (133 - 1.6 ) * 0.79.

This assumes you are breathing air in both cases. The 1.6 is approximate and represents the water vapor pressure that needs to be subtracted out of ambient pressure to get to your inspired inert gas pressure. My question is why does it (the 1.6fsw vapor pressure) remain constant with changes in pressure if it acts like a gas?

NeilHolden

See what I mean about my explanation being simplistic!

I should have also mentioned boinling point of water at 100 deg C is at 1 ATM, increase the pressure increase the boiling point - ho hum..

Back to doing more practical stuff - like diving

Neil

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Sutty

The water vapour is in equilibrium with the water film in the lung, as you compress an alveolus containing gas saturated with water vapour the water vapour will condense to become water (in the film of water), and the pressure of the water vapour will remain the same = saturated vapour pressure of water at body temp.

Neil - water vapour is gaseous, not tiny droplets.

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UWSojourner

Ok. Is the following statement correct then ... saturation vapor pressure is completely indepenent of ambient pressure so that higher ambient pressure environments hold less water vapor (as a % of total volume) than lower ambient pressure environments?

Sutty

Correct! As a side note, if ambient pressure drops to equal the saturated vapour pressure of a liquid (at its temperature), the liquid will boil.

Neil

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UWSojourner

Thanks much for your help.

Scubascooby

Dug out this old thread hoping it might asnwer my question but maybe someone can help.

Deco tables and computers use a known figure for water vapour (wv) in the inspired gas. I am guessing that this originated from OC diving or chamber dives where the inspired gas is very dry. I think this might mean a lower lung water vapour content than a CC diver.

If this is the case, do CC algorithms use a different figure for water vapour content ?

During the deep part of a dive, the lower wv figure would be in your favour but in the longer deco phase at the end of the dive, the lower wv figure would assume a lower inert gas percentage. The loop gas may also be more moist at this phase of the dive.

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