Oxygen solubility and half times

[The Iceni]

Hi Dennis,

You said. "I think you need to stop thinking of the counterlung(or loop or whatever it's called) as a bag. Instead, consider that it is a rigid body that is being pressurized, kinda like a Scuba tank which is being filled. This seems to be a good analogy since the shape of the bag imitates a rigid body as long as the ambient pressure is matched by injected air.

Absolutely correct! Dennis. It is the model I use but the box volume can be altered with associated changes to the pressures inside!

Dalton's Law tells us that the total pressure of oxygen (pp O2) in any cylinder, box or rebreather loop is the sum of;

1) The partial pressure of oxygen in existing mix
2) The pressure of pure oxygen added (P O2)
3) The partial pressure of oxygen in the top up air (pp air) added to working pressure (P+1).

I find it far, far easier to work in partial pressures at all times and use the following basic equation myself for mixing Nitrox. It is easily modified for any Nitrox top-up.

P O2 = ((Mix O2 – 0.209) x (P+1))/ 0.791

My working of your example to 100 bar abs;

P O2 = ((0.32 - 0.209) x 100/0.791

= 14.03% oxygen (and 86% air) correct!

"I believe the rigid body displacement model is at work in your rebreather calculations."

Yes, I agree it is a fixed volume for most of the time but yours is exactly the sort of comment I have received. Can I work through this with partial pressures to 99 barg (100 bar abs)?

1 and 3 are air in your Nitrox example. (Tank starts with air and you add 14.03 bar of Oxygen and then top-up)

P O2 of pure oxygen will be 14.03 bar

pp of oxygen in all air will be the partial pressure of air multiplied by its fO2.

P = 100 bar abs, so pp air = 100 - P O2 = 85.97 bar

pp O2 in original and added air = 85.97 x 0.209 = 17. 97 bar.

Dalton's Law; total pp O2 = 14.03 + 17.97 = 32 bar.

As total pressure is 99 + 1 = 100 bar this gives an fO2 of 32%.

I think you will find that my maths is correct.

More properly the example should be wroked as follows.

Oxygen is added to an "empty" cylinder of air at 1 bar.

S = 14.24 bar (14.03 + 0.209) at 15.23 bar ABS

Top-up air 100 - 15.23 = 84.97 bar

I = 84.97 x 0.209 = 17.76 bar

pp O2 = 17.76 + 14.24 = 32 bar (In 100 bar ABS = 32%)

Although the principle is very simple the maths is a bit complex and perhaps does not read too easily in the "txt" format but do take a look at the attachment with my second post.

I am coming to the conclusion that the differences in predicted and experienced values are due to sensor delays and averaging.

Kind regards,

Paul

 

[devjr]

Thanks for the tutorial, I enjoyed it. Give me time and I'll reply, hopefully. BTW, I edited my message, just wordsmithing.

Dennis

 

[devjr]

Well, I did a back calculation to see how much O2 would have to be added to the counterlung while on the surface. The dive is to 5 atmospheres(A). The diving set is calibrated for a PP/O2 on the bottom of 1.4 atm. It is assumed that no oxygen is consumed on the descent, so I guess we best make it a rapid one(G).

FO2 (mix) - FO2 (air) / FN2 = FO2 (add at surface)

.28-.21/.79 = .09, .09 X 73.5 = 6.6 psi

(73.5 psi is the pressure on the bottom).

This (6.6 psi) is the oxygen pressure that must be added at the surface. Note that this is less than the surface pressure of 14.7 psi. Normally, this procedure would not be practical in our rigid body model. However, since the counterlung actually is flexible it would only be necessary to inflate the bag about half full with oxygen (6.6/14.7) to get the equivalent volume. If you intend to breath, then add a tad more.

Dennis

 

[Dr Deco]

Hello oxygen breathers:

While solubilities can account for some of the oxygen loss, activities such as EVA by astronauts consume about 1,000 ml/min for a 200 pound individual. Have factors such as metabolic usage been factored into the calculations for oxygen loss. This is quite large.

I do not recall seeing this explanation, thus far - - (or did I miss it?)

Dr Deco:doctor:

 

[devjr]

Dr Deco, thanks for confirming the number. Yes, we're aware, those pesky details are still sitting in the wings while waiting for us to work through the gas law. (G)

 

[The Iceni]

Hi all,

We are all talking about theories and while some make a great deal of sense, at least to me, others are being questioned.

May I first say that I did not want to get into considerable detail about what could happen from the point of view of an armchair rebreather diver becuse those at the sharp end are quick to pick up my obvious lack of experience.

Once upon a time in the UK we had a currency in which there were 12 pennies in a shilling and 20 shillings in a pound. This was Sterling or £sd (pounds, shillings and pence). It was a nightmare so we went decimal in 1971. I can sympathise with those GIs who were over here during the war coping with this because, much as I hate all things "European", I find working in psi and feet of sea water a real headache. Metres and bar are much easier!

Dennis, I think your 6.6 psi of oxygen somehow has to be added to the air in the loop which has surface pp O2 of 3 psi, if only because the pp O2 of air at 40 metres is 1.045 and you want a higher pp O2 than this. I must say that I feel the analogy of mixing Nitrox in a cylinder is not a good one to use when looking at a rebreather loop because of all those conversions to and from fO2 and pp O2. For example, the set point for descent is usually 0.7 which means the diver starts with 70% Nitrox AND with a partially filled loop.

Dr Deco I did touch on metabolism. I have it on good authority that the oxygen consumption in UK rebreather divers is between 1 litre and 1.5 litres per minute (at STP) and this will of course be reflected in a reduction in the pp O2 seen in the loop.

If one litre is removed from a rigid 10 litre box at atmospheric pressure (1 bar) the pressure is reduced by 0.1 bar. It would seem that metabolism will reduce any excess oxygen in the loop by this amount i.e 0.1 to 0.15 bar per minute.

When the diver is as near to the steady state as he can get with a bottom set point pp O2 of 1.3 very little (more) oxygen will go into simple solution, so the pp O2 in the loop will be reduced by metabolism alone. A descent from 30 M to 40 M with a system at equilibrium and a set point of 1.3 should produce an oxygen spike to 1.5 bar but I am told that this is never seen in practice, even immediately on reaching 40 metres.

If this is accurately reported, to my mind it confirms that metabolism cannot account for the difference between what is observed and predicted. I thought we were dealing with large quantities of oxygen dissolving in simple solution but now think this to be highly unlikely.

Perhaps on the initial descent where the tissues are in equilibrium at 0.2 bar and the new inspiratory pp O2 exceeds 1.3 bar, a driving pressure of 1.1 bar could be responsible for the dissolution of a whole litre of oxygen. However, once more, I agree with Dr Deco that the amounts will be very small in such a short time scale, so it seems likely that the brain will be exposed to the pp O2 in the loop, whether this is accurately refected by these low sensor readings is a question I am still unsure of as we can never be certain of the actual loop volume in such a dynamic system.

I would also appreciate it if someone could confirm my understanding of the effects of the fO2 of the diluent;

If the loop is kept at a constant volume by the addition of diluent I believe the loop pp O2 will be increased by the surface pp O2 of the diluent for every 10 metres of descent.

From this must be deducted metabolic loss (fairly constant).

Regards,

Paul

 

[devjr]

Dr. Paul, I've shown that there is no theoretical problem down to 40 meters starting with a partial fill of 100% O2, and air as the diluent. Substitution of O2 with Nitrox does not change this. I was prepared to suggest this very thing but started with a worst case approach, and the result did not indicate any initial tweaking of the mix was necessary but it does assume that, at some point early in the dive, the counterlung will be fully inflated by adding diluent. This is my understanding of "constant volume", which might be different from yours. It needn't make any difference, though; should the bag not be fully inflated the relative fractions can be adjusted at the onset, beginning with oxygen. However, the issue has since become a moving target by a narrowing of the depth scenarios, and real world complexities such as O2 consumption and the mechanics of rebreathers.

This is beyond my expertise since I'm not familiar with the parameters of the rebreather's sensors and system logic. I don't even know the actual capacity of the counterlung.

However, given the simple exercises in partial pressures and relative quantities or fractions, and factoring in oxygen depletion, I'm inclined to believe that there is no spike in PP/O2 beyond the system setpoint.

Perhaps somebody else will jump in. TIA

Dennis

 

[devjr]

Doctor Paul, for the past few moments I've been thinking of you sitting at your spacious desk while adding and rejiggering the partial pressures and wondering why the bridge too far.

I then had an "aha" moment which might explain our mutual failure to communicate; but It's not shillings versus quarters, at least not in the sense you implied.

However the rote of adding "bars" may have its advantages---I suspect, and hopefully without presumption , that it also hinders the capture of a fundamental concept.

In this instance, PP addition, and the modified rigid body model, may have led you to believe that adding diluent(air) to pure oxygen causes the PP/O2 to increase. Actually, it doesn't, not on the surface and not as long as the bag has room to expand. Adding diluent at the surface causes the fraction and the PP/O2 within the counterlung to decrease. This is a flexible body model, for the time being.

A counterlung at the surface which is filled or partially filled with O2 has a PP/O2 of 1 bar. The same counterlung which is fully diluted (per my previous straw man calculation) contains 56% O2 and has a PP/O2 of 0.56 bar, in other words, 56% Nitrox. At this point the diver submerges and the counterlung becomes a rigid body which needs addition of diluent to simply maintain its shape. As external pressure is matched by internal pressure, through addition of diluent, the O2 fraction decreases but the PP/O2 increases. The endpoint arrives but at a lower PP than you supposed in your thesis. That is attributed to a lower starting point than assumed. Remember the flexible body. Thus, at 5 atmospheres the mix is 28% Nitrox and the PP/O2 is 1.4 bar.

I might add that I have no idea if my assumptions are even close to how a rebreather actually works. The oxygen consumption still has to be factored but it should argue against the putative O2 spike, all things being equal.

Dennis

 

[The Iceni]

Dennis,

You said "As external pressure is matched by internal pressure, through addition of diluent, the O2 fraction decreases but the PP/O2 increases. The endpoint arrives but at a lower PP than you supposed in your thesis. That is attributed to a lower starting point than assumed. Remember the flexible body. Thus, at 5 atmospheres the mix is 28% Nitrox and the PP/O2 is 1.4 bar."

I think we are violently in agreement!

All I am saying is that if the diluent added to maintain the working volume of the loop (at its nominal 10 litres) has ANY oxygen in it, this oxygen will be added that already in the loop thus enriching the loop contents. As you say this IS NOT by increasing its fO2 but by increasing the absolute amount of oxygen in the loop. In fact, in all likelihood the f02 will be reduced. The additional oxygen is reflected in the pp O2, which after all is all we are interested in; - A cylinder ful of air contains more oxygen than a quarter full cylinder of Nitrox 36 although the fO2 in the latter is greater.

A good example is a loop filled with air at the surface with air as diluent (S = 0.21). No matter how much air is added during the descent the fO2 in the loop will remain constant at 0.21. However, the pp O2 will increase linearly with depth - 0.21 bar per 10 metres - and will be no different from open circuit air with a pp O2 of 1.045 at 40 metres.

With a rebreather, you start at a higher value for "S".

(On the other hand if helium is used as the diluent no oxygen is ever added to the loop so the pp O2 in the loop will not change except by metabolic loss.)

To prevent a dangerous spike at the bottom of the descent with air as diluent I gather the EFFECTIVE surface pp O2 must be manipulated to below 0.5 bar;

Can I take an example of a loop filled to 50% capacity on the surface with 70% Nitrox which is taken down to 40 metres?

P is ambient pressure at depth in bar
S is the starting pp O2, which IN A FULL LOOP = 0.70/2 = 0.35 bar
p is the starting total pressure in FULL loop =1 bar/2 = 0.5 bar
fO2 is oxygen mix of diluent.
D is the pp O2 in the loop after the addition of diluent to working volume.

D = S + ((P - p) x fO2);-

At 5 metres D = 0.35 + ((1.5 - 0.5) x 0.2) = 0.55
At 8 metres D = 0.35 + ((1.8 - 0.5) x 0.2) = 0.61
At 10 metres D = 0.35 + ((2.0 - 0.5) x 0.2) = 0.65
At 14 metres D = 0.35 + ((2.4 - 0.5) x 0.2) = 0.73
At 16 metres D = 0.35 + ((2.6 - 0.5) x 0.2) = 0.77
At 20 metres D = 0.35 + ((3.0 - 0.5) x 0.2) = 0.85
At 25 metres D = 0.35 + ((3.5 - 0.5) x 0.2) = 0.95
At 30 metres D = 0.35 + ((4.0 - 0.5) x 0.2) = 1.05
At 35 metres D = 0.35 + ((4.5 - 0.5) x 0.2) = 1.15
At 40 metres D = 0.35 + ((5.0 - 0.5) x 0.2) = 1.25 bar

A linear increase of 0.2 bar per 10 metres, simply because of the added air.

Once on the bottom "S" becomes 1.3 - the set point - so IN THEORY there are likely to be significant spikes with any further descent subesquently during the dive. This is what I am trying to understand and prevent.

Does this make sense?

Regards,

Paul

 

[devjr]

Houston, we have a problem.

Hi Dr Paul,

Since the concept of effective partial pressure does not apply here, the bottom PP/O2 exclusive of metabolic loss is actually 1.5 bar, and borderline for extended range safey.

D = S + [(P-p) X FO2]

D = .7 + [(5.0-1.0) X 0.2]

D = 1.5 bar

I understand where you are coming from but as long as the loop remains constant volume (working volume) throughout the sequence from surface to bottom we can use the partial pressures reflective of atmospheric conditions, and add the normal increments. In other words, the loop is assumed to remain half full on average from start to end.

I can think of a number of reasons why the mfgr might want to specify a half full loop. One reason could be that the relative(%) depletion of oxygen, and consequent reduction of PP/O2, is greater than in a full loop.

Otherwise, I don't know what else to say other than that I understand why you are asking this question about.

Dennis