Freef

Here's my thought process, starting with Dragers spec for the Dolphin loop being 10.5L:

At a fixed pO2, for any depth the surface volume of O2 in the loop is the same.

Therefore, for a loop pO2 of 0.5, at the surface the loop will need 50% or 5.25L of O2 in it.

At 10m the loop would need 2.625L of O2 for the same pO2-or with the same fO2 you would get a pO2 of 1

Examples:

pO2 of 2.1 at 30m:

fO2 = 2.1 / 4 = 52.5% O2. At 30m the pressure is 4 bar, therefore the O2 content of the loop is 10.5 x 0.525 x 4 = 22.05L

pO2 of 2.1 at 50m:

fO2 = 2.1 / 6 = 35% O2. At 50m the pressure is 6 bar, therefore the O2 content of the loop is 10.5 x 0.35 x 6 = 22.05L

And to find the loop O2 volume at 1.3 bar for the same depths:

pO2 of 1.3 at 30m:

fO2 = 1.3 / 4 = 32.5% O2. At 30m the pressure is 4 bar, therefore the O2 content of the loop is 10.5 x 0.325 x 4 = 13.65L

pO2 of 1.3 at 50m:

fO2 = 1.3 / 6 = 21.7% O2. At 50m the pressure is 6 bar, therefore the O2 content of the loop is 10.5 x 0.217 x 6 = 13.65L

So therefore you would have a constant surface volume of O2 to metabolise.

This is as far as my thinking went, and where I went wrong. I see what you mean about the changing ratio of gasses the deeper you are, so I ran through the figures again-and it gets worse, not better. For example:

If you take the decrease in loop volume into account, at 40m, in one minute intervals you would get the following, if you assume a vO2 of 1 L/min, 100% CO2 capture and a loop volume of 10.5L that can collapse as the gas is used.

time = 0 pO2 = 2.100 vol [O2] = 22.05 vol [inert] = 30.45 loop fO2 = 0.42

time = 1 pO2 = 2.045 vol [O2] = 21.05 vol [inert] = 30.45 loop fO2 = 0.409

time = 2 pO2 = 1.985 vol [O2] = 20.05 vol [inert] = 30.45 loop fO2 = 0.397

time = 3 pO2 = 1.925 vol [O2] = 19.05 vol [inert] = 30.45 loop fO2 = 0.385

time = 4 pO2 = 1.860 vol [O2] = 18.05 vol [inert] = 30.45 loop fO2 = 0.372

time = 5 pO2 = 1.795 vol [O2] = 17.05 vol [inert] = 30.45 loop fO2 = 0.359

time = 6 pO2 = 1.725 vol [O2] = 16.05 vol [inert] = 30.45 loop fO2 = 0.345

time = 7 pO2 = 1.650 vol [O2] = 15.05 vol [inert] = 30.45 loop fO2 = 0.330

time = 8 pO2 = 1.580 vol [O2] = 14.05 vol [inert] = 30.45 loop fO2 = 0.316

time = 9 pO2 = 1.500 vol [O2] = 13.05 vol [inert] = 30.45 loop fO2 = 0.300

time =10 pO2 = 1.420 vol [O2] = 12.05 vol [inert] = 30.45 loop fO2 = 0.284

time =11 pO2 = 1.330 vol [O2] = 11.05 vol [inert] = 30.45 loop fO2 = 0.266


If you take the decrease in loop volume into account, at 60m, in two minute intervals you would get the following, if you assume 100% CO2 capture and a loop volume of 10.5L that can collapse as the gas is used.

time = 0 pO2 = 2.100 vol [O2] = 22.05 vol [inert] = 51.45 loop fO2 = 0.300

time = 2 pO2 = 1.960 vol [O2] = 20.05 vol [inert] = 51.45 loop fO2 = 0.280

time = 4 pO2 = 1.820 vol [O2] = 18.05 vol [inert] = 51.45 loop fO2 = 0.260

time = 6 pO2 = 1.666 vol [O2] = 16.05 vol [inert] = 51.45 loop fO2 = 0.238

time = 8 pO2 = 1.505 vol [O2] = 14.05 vol [inert] = 51.45 loop fO2 = 0.215

time =10 pO2 = 1.330 vol [O2] = 12.05 vol [inert] = 51.45 loop fO2 = 0.190

So I was wrong to assume a fixed volume, and my figures gave a shorter time to metabolise the pO2. Looks like it's you who is the SCR guru!

Appropriate green sent and humble pie in the oven. If you find these figures wrong I'll sulk though!

The real problem with attempting to rely on the pO2 dropping in this way is that the loop will be collapsing by 2.2L at 40m to drop the pO2. I think that this is about the volume of the exhale CL. It will become difficult to breathe as the loop volume decreases, and when the gas is turned back on the ADV will be likely to fire, adding more O2 to the loop.

__________________
David.

Currently owner of two differently sized ankles.