The Iceni
Medical Moderator
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
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