EAD, Equivalent Air Depth

[alo100]

I brows around here and saw someone talked about the EAD formula, does anybody know the derive of the formula. Actually, I just tried it briefly and it's not yet done. I am sure I've done something wrong, maybe those who are interested or are familiar with it will be able to tell.... this can save me some time

Dalton's law of mole fraction.

P(total) = Pi/mi

Where Pi is the i-th partial pressure.
mi is the mole fraction of gas i among other gases, say, in this case the tank. So, Avogadro's Law said the volume of gas is proportional to the amount of gas in mole and is the same for all gas, indepent of size or mass of the molecules. So, the value 0.79, fraction for N2 in air, for example, can still be used here for the mole ratio. Others like Ideal gas conditions etc.

True for all depth and assume ideal gas conditions.

P(total)@1 is P(total) at depth1 for a certain tank.
Pi@1 partial pressure at depth1 for the same tank
m1: mole ratio 1 e.g. N2 in depth 1, its mole fraction in tank
m2: mole ratio 2 N2 in depth 2...
Unit: depth in meter and pressure in atm (atmospheric pressure)

P(total)@1 = Pi@1/m1 ........ (1)
P(total)@2 = Pi@2/m2 ........ (2)

A typical EAD formula looks like this:
Assume metric scale:
((EAD/10) + 1) / ((D/10) + 1) = FN2/0.79

Assume every 10 meter decend in depth in sea water, the pressure would increase 1 atm, and the "+ 1" is the 1 atm at sea level.

where EAD is the equivalent depth for air in meter. D
is the desired depth in meter and FN2 is the desired
fracton of N2.
Compare to this, try to use (1) and (2) to derive
the same formula

(1)/(2):
P(total)@1 / P(total)@2 = (Pi@1/Pi@2) x (m2/m1)
Let depth 1 = EAD, depth 2 is the desired depth
Let D be the desired depth, and as usual use N2 = 0.79 as in air.
FN2: fraction of N2 in the mix.

((EAD/10) + 1) / ((D/10) + 1)
= (Pi@EAD / Pi@D) x (FN2/0.79)

The 2 formulas are not necessarily equal unless
Pi@EAD = Pi@D ????

Which step did I do wrong?

Thanks!

 

[Charlie99]

The 2 formulas are not necessarily equal unless
Pi@EAD = Pi@D ????

Which step did I do wrong?

NONE!

I'm having a bit of trouble following your naming conventions, but the result of Pi@EAD = Pi@D is indeed the correct result.

The partial pressure of N2 in air at the equivalent air depth is the same as the partial pressure of N2 in the nitrox mix at the desired depth, D. Although you may or may not find that in your nitrox book, that's the definition of Equivalent Air Depth.

You can get to that same result pretty easily by multiplying out the two fractions of the typical EAD equation that you wrote, to get

[(EAD/10) + 1] * 0.79 = [(D/10) + 1] * FN

The equation on each side is simply ppN2.

-----------------------------------------
Dalton's equation, expressed (using your terms) can be written P(total) * mi = P(i)

Have P(i) of air and a nitrox mix be equal, then solve for depth, and you will get the EAD equation written as above, since (EAD / 10) + 1 is P(total) at the EAD, and (D/10)+1 is the P(total) at the desired depth.

 

[alo100]

It's a little bit long for this message, I hope you wouldn't mind. Thanks a lot for explaining this!

[(EAD/10) + 1] * 0.79 = [(D/10) + 1] * FN

Ha! So that's the meaning of the "equivalent". So that's how we define it, to set the partial pressures of a gas in different depth equal, and see the same effect of that gas in body. The boundary condition for the equation correspondingly.

So do we actually use this formula by setting EAD = 0 (as if we are having the environment at sea level), then set D, our desired depth, and find the concentration of N2 we need?

FN2 = 0.79 / [(D/10) + 1 ]
Or this is something to begin with? It seems to me that if there is a max N2 concentration we can take on land and min N2 concentration on land, we can set EAD = 0 (having an env at sea level) and find FN2 initial. Then with this initial FN2 (a constant now) we can change the 0.79 to N2 max to see the Dmax (deepest we can go with this N2 concentration) and N2 min to find Dmin (shallowest we can go with this N2 concentration). The valid range of depth (Dmin, Dmax) is like a sliding window if Dmin is 0 or negative, we can actually use this mix without a extra bottle? Or if the Dmin > 0 (the gas mix cannot be used onland or shallow water), then we need to find a slightly different FN2 to accomodate?

Similar method for finding FO2?
So the valid range for the mix is a min{O2,N2}, the overlap between 2 windows of N2 and O2?

Is it really how you guys do it?

I was going to say, why don't we just use the Dalton's Law of partial pressure:

P(total) = p(O2) + p(N2)
D=0 [(D/10)+1] = 0.21 + 0.79
D=10 2 = 2 x 0.21 + 2 x 0.79
D=20 3 = 3 x 0.21 + 3 x 0.79
....

For a given D, we know the partial pressure at that depth
e.g. for p(N2)@D = [(D/10) + 1] x 0.79
....