A helium in CCR question

Mike_Eitel

Hi
I'm not an expert in He diving. But since longer time I'm wondering the mixture behavior in a closed loop:

a. Lets make it easy and use f.x. 0/50 trimix = 50 N2 & 50% He.
b. have an perfect CCR, nit loosing any diluent to environment.
c. now diving to f.x 40m.

As I understand HE is going faster in and out of my body than N2. And in my understanding, this must change the mix in my loop. pp of He will degrade and N2 will have a higher percentage than He.

As my body consumes gas from the loop, finally II have to inject the equal mix diluent…
It will rise the He part in the loop.

That way I’m loading my body with more He then in OC with the same mix ???

I’m aware that this example might be academic, but I’m wondering if anybody ever tried to measure this in real diving and had a look into the eventual outcome for decompression calculations.

trakrat

this is something that should show up in the appoc when it is released!
it has a he sensor so will monitor this and log it !
try asking brad or dave as they have there hands on the appoc !

Wol

It would be interesting to get a feel as to what volume of free He or N2 is absorbed into the average human body when saturated. This would give a gauge as to what volume of inert gases we are talking about and whether this is of any consequence to the diver decompressing and holding a perfect loop with no flushing. - Any bidders out there?

jaksel

EDIT: I realized I made a mistake, I will redo the math...

Kermit

A typical human, fully saturated at 1 atm holds about a litre of N2.

nigelh

Woo....
Where did that come from?
I spent ages trying to come up with a number and ended up at about 1.2L but with very little confidence in the factors I'd used. Finding somebody else is within a factor of 10 would be good. This is cool. :D

Certainly if this is the case we aren't going to see much loop change as we gas on and off.
I got similar final volumes for helium but just about three times as fast.

__________________
nigelh

jaap

Sounds about right.

This link is perhaps of interest:
Rubicon Research Repository: Item 123456789/2892

Looks like helium has about 70% the solubility of nitrogen in blood.

Parts of this thread might also be interesting:

http://www.rebreatherworld.com/gener...tml#post284716

/Anders

jaksel

OK, I made an initial post which I instantly realized was completely wrong. Here's a new one.

If you look at the constant in Henry's law, dealing with concentration of inert gases at equilibrium, assuming T=298 K and solutant is water:
K_N2 = 0.01492
K_He = 0.009051
At equilibrium your loop (with volume V_loop) will have some unknown pN2_loop, and unknown pHe_loop. The molecular concentrations of N2 and He in your body (volume V_body) will be K_N2 x pN2_loop and K_He x pHe_loop, respectively.

Since we know that we inject a diluent with equal fractions N2 and He, the total amount (loop + body) of N2 must be equal to total amount of He:

V_loop x pN2_loop + V_body x (K_N2 x pN2_loop) =
V_loop x pHe_loop + V_body x (K_He x pHe_loop)

What you get if you rearrange is:

pHe_loop = pN2_loop x Q

where Q = (V_loop + K_N2 x V_body )/(V_loop + K_He x V_body)

If we assume V_loop = 10 L and V_body = 80 L, we get:

Q = (10 + 0.01492 x 80)/(10 + 0.009051 x 80) = 1.044

So the ratio pN2_loop/pHe_loop will be 1/1.044, or 0.4892/0.5108, i.e. N2 contents will be 48.9% and He contents 51.1% at saturation at any depth.

However this is at equilibrium (i.e. saturation). Since He diffuses more rapidly than N2, the He fraction might initially drop below 50% as He rushes into your body, then rise back up above 50% as N2 slowly diffuses into your body and He is near equilibrium.

Actually you could use the formula for Q from above to get a worst-case estimate if we assume He reaches equilibrium instantly and N2 has not diffused into body at all:

Q = (10 + 0)/(10 + 0.009051 x 80) = 0.9325

so ratio is now 1/0.9325 or 0.5175/0.4825, i.e. 51.75% N2 and 48.25% He. Again, note that this is just a worst-case estimate.

Please correct me if I'm wrong.

jaksel

Just use Henry's law:

pN2_body = K_N2 x pN2_loop = 0.01492 x 0.79 bar = 0.01179 bar

Total amount of N2 in body = V_body x pN2_body = 80 L x 0.01179 bar = 0.94 L x bar, i.e. about 0.94 litres of N2 at 1 atmosphere.

wedivebc

That number is also referred to in the PDIC OW manual. They don't say how they derived it though.