Mixed Gas Theory

If we have more than only one inert gas (nitrogen N2 with air/nitrox), for e.g. two as in the case of trimix / heliox (nitrogen N2 and helium He) we have to expand the Haldane-equation (6a).

Basically the law of Henry will be valid for each inert gasi separately:
(well, well, except the region, what specialists call "Poyinting effect", i.e.: very high pressures and very complex organic molecules ...)

qi = li * pi * Vk

Q = ∑ qi


The sum encloses all inert gasesi and as well all compartmentsk

Variable Definition
qi amount of diluted inert gas i within a compartment with volume Vk
li solubility coefficient of inert gasi valid for this volume Vk
pi partial pressure of inert gasi
Vk volume of compartment k

For the sum of all partial pressures we will have the law of Dalton:

P = ∑ pi = p1 + p2 + p3 + ... (51)

Thus we will have for a stationary state in a compartment the following:
Pt(0) = ∑pt, i(0) = pt, N2(0) + pt, He(0) (52a)

as well for the alveoli:
Palv(0) = ∑palv, i(0) = palv, N2(0) + palv, He(0) (52b)

as well for all times t:
Ptiss(t) = Pt(t) = ∑ ptiss, i(t) = ptiss, N2(t) + ptiss, He(t) (52c)

Thus we expand the Haldane-equation (6a):

Pt(t) = Palv0,N2 + [Pt0,N2 - Palv0,N2] e-kN2t + Palv0,He + [Pt0,He - Palv0,He] e-kHet (53)

We have to check if this expansion will satisfy the basic differential equation (1a). So we expand (1a) with an additional term for a second inert gas:

dPt(t)/dt = k1 * [Palv,N2(t) - Pt,N2(t)] + k2 * [Palv,He(t) - Pt,He(t)] (54)

We determine the proportionality factors k1 and k2, by calculating d Pt(t) / dt from (53) :

dPt(t)/dt = - kN2 * [Pt0,N2 - Palv0,N2] e-kN2t - kHe * [Pt0,He - Palv0,He] e-kHet (53a)

... and with this terms we replace the left hand side of (54):

- kN2 * [Pt0,N2 - Palv0,N2] e-kN2t - kHe * [Pt0,He - Palv0,He] e-kHet =

k1 * [Palv,N2(t) - Pt,N2(t)] + k2 * [Palv,He(t) - Pt,He(t)] =

k1 * Palv,N2(t) - k1 * Pt,N2(t) + k2 * Palv,He(t) - k2 * Pt,He(t)


For the pt, i(t) we exploit once again (52c) resp. (53), put this into (55), right hand side, re-arrange and recieve:

- kN2 * [Pt0,N2 - Palv0,N2] e-kN2t - kHe * [Pt0,He - Palv0,He] e-kHet =

k1 * Palv,N2(t) + k2 * Palv,He(t) - k1 * [Palv0,N2 + [Pt0,N2 - Palv0,N2] e-kN2t - k2 * [Palv0,He + [Pt0,He - Palv0,He] e-kHet


We multiply the terms of the squared brackets [ ... ] of the right hand side and re-arrange according to constant resp. time-dependant terms:

- kN2 * [Pt0,N2 - Palv0,N2] e-kN2t - kHe * [Pt0,He - Palv0,He] e-kHet =

k1 * Palv,N2(t) + k2 * Palv,He(t) - k1 * Palv0,N2 - k1 * (Pt0,N2 - Palv0,N2)e-kN2t - k2 * Palv0,He - k2 * (Pt0,He - Palv0,He)e-kHet =

k1 * [Palv,N2(t) - Palv0,N2] + k2 * [Palv,He(t) - Palv0,He] - k1 * (Pt0,N2 - Palv0,N2)e-kN2t - k2 * (Pt0,He - Palv0,He)e-kHet


We take now the identical boundary conditions for the both inert gases which guided us to the solution of (6a):

Palv,N2(t) = Palv0,N2 for all times t and analogous
Palv,He(t) = Palv0,He
the equation (55a) will be identically satisfied with:
k1 = kN2 as well: k2 = kHe

Thus the approach (53) satisfies the central differential equation (1a), if the second gas is applied as purely additive:
this yields for the stationary part, i.e. the part with no varation in time in the alveolar gas (52b) and as well for the time-dependant part (52c).

Calculation of decompression times with mixed gases

To start with, we check the very simple situation with only one inert gas (air/nitrox) on a definite deco stage (6m),
before we could ascent to the next one (3m):

NB: the figures are only a first, coarse approximation: the identification of the variables is for getting an overview over equation (21):

t = - τ / ln2 * ln[ (Pt(t) - Palv0) / (Pt0 - Palv0) ]

The actual deco time td is the maximal value out of all calculated deco times which have been assessed for the complete group of all the compartments.

For the terms we have the following:

The extensions for mixed gases with (22) are for the a- und b-coefficients, and as well the sets for τ resp. λ (equation set 56)!

a* = a (He + N2) = [( Pt, He * aHe ) + ( Pt, N2 * aN2)] / ( Pt, He + Pt, N2 )
b* = b (He + N2) = [( Pt, He * bHe ) + ( Pt, N2 * bN2)] / ( Pt, He + Pt, N2 )
λ* = λ (He + N2) = [( Pt, He * λHe ) + ( Pt, N2 * λN2)] / ( Pt, He + Pt, N2 ) (56)
and this is for all the k compartments, e.g.: k = 1 - 16! (within the standard ZH-L framework; if you want to compare other things, like the ANALYST software from COCHRAN, k runs from 1 to 20, that this from approx. 1.25 min to 900 mins.)

As well we have for our two inert gases He and N2 (52a, 52b, 52c) and thus:

If you take now for the calculation of the deco times td the equation (21) with all of the above cited extensions for mixed gases,
you generate an equation with two more unknowns,

i.e.: ptiss, N2(t) und ptiss, He(t) for the time t = td
because of: Pt(t) = ptiss, N2(t) + ptiss, He(t), und this for all compartmentsk! As
a*, b* and as well λ* have been assesssed for time t = t0 , but this is only the start of the deco phase!
At first, td is unknown and thus as well the respective compartment saturations ptiss(td) and thus consequently also a*, b*, λ* at time t = td.

©    albi (at, bei:) divetable (dot, Punkt:) de
Stand: 02 / 2010