Deeeep diving theory

[DA Aquamaster]

People are getting hung up on the tidal volume when it is the tank pressure that is the limiting factor.

Assuming you have 3015 psi (absolute) in the tank at the surface, at 6768 fsw the pressure differential inside and outside the tank will be zero and no gas flow will occur even if you opened the valve with no reg on the tank (well...assuming the tank is inverted and the bouyancy of the air is not an issue).

So with no pressure differential, even if you had a perfect regulator that would deliver gas right down to zero psi, you would not be able to draw a breath out of your AL 80 at a depth deeper than 6768 ft. However in practice, even a balanced high performance reg gets hard to breath at around 50 psi. Assuming you can still suck air at a useable rate down to 50 psi., the max depth where any air flow would occur would then be 6656 fsw.

At that point it's a rate change calculus problem involving tidal volume, tank volume and the ambient pressure change with depth to figure what slightly shallower max depth would leave you at 50 psi over ambientwhen your lungs become full. And I'll leave that to someone else as it's been 20 years since I cracked a calculus book.

And without doing the calculus, I'd estimate that depth would be around 6600 ft.

 

[dandrian]

Ah um.......I was told when I signed on to this board there would be no math...........:ne_nau:

 

[LSDeep]

People are getting hung up on the tidal volume when it is the tank pressure that is the limiting factor.

Assuming you have 3015 psi (absolute) in the tank at the surface, at 6768 fsw the pressure differential inside and outside the tank will be zero and no gas flow will occur even if you opened the valve with no reg on the tank (well...assuming the tank is inverted and the bouyancy of the air is not an issue).

So with no pressure differential, even if you had a perfect regulator that would deliver gas right down to zero psi, you would not be able to draw a breath out of your AL 80 at a depth deeper than 6768 ft. However in practice, even a balanced high performance reg gets hard to breath at around 50 psi. Assuming you can still suck air at a useable rate down to 50 psi., the max depth where any air flow would occur would then be 6656 fsw.

At that point it's a rate change calculus problem involving tidal volume, tank volume and the ambient pressure change with depth to figure what slightly shallower max depth would leave you at 50 psi over ambientwhen your lungs become full. And I'll leave that to someone else as it's been 20 years since I cracked a calculus book.

And without doing the calculus, I'd estimate that depth would be around 6600 ft.

thx, youre even better than me and my rum and coke state! wasnt even considering that!!!!

 

[chrispete]

heh, AJ has the real world problem down, your tank would actually take on water the moment any valve (tank, 1st, 2nd, octo, etc.) was opened until the pressure was equalized, you'd be having a bad air day at that point.

To keep the exercise going, let's assume that you had a tank that had 80cf of air available at the time that you reached the depth that you could take your single breath at.

 

[Mlody11]

2 answers... pick which ever you like best. I think number 2 is better. Both answer are implied that you are using a perfectly filled 80cuft @ 3000psi and dont take a breathe from it going down

1st answer... You are basically asking what is... At what pressure does 80 cuft compress to .5 liters. Lets get the units straight first... 80cuft = 2265 liters. So, at what pressure does 2265L compress to .5L? P1V1=nRT P2V2=nRT nRT are constant, so P1V1 = P2V2. (2256L)(1atm) = (.5L)(X atm) x = 4512 atm.
(4512 -1) * 33 ft = 148,863 ft or 45,373 meters

2nd answer... Since you could not drawn air from your reg. beyond the outside pressure. The limit of taking a breath is 206 ATM. (206 -1) * 33 = 6,765 feet or 2,062 meters before you can no longer breathe from your tank. So now, when you take a breathe of .5L you are in essence increasing the volume of the tank by .5 liters again...
P1V1 = P2V2. (2265 L)(206 ATM) = (2265.5 L)( X atm) X = 205.95 atm, (X -1)* 33ft = aprox. 6763.35 ft
so only about 1.5 feet above max before you cannot breathe in a PERFECT REGULATOR

EDIT to build on the 50PSI post above:
If you want 50PSI to be your max... 50 psi = 3.4 atm. so your working pressure is 206 atm - 3.4 atm = 202.6 atm
Again P1V1 = P2V2 (202.6 atm)(2265L) = (X atm)(2265.5L) x = 202.55 atm (x-1) * 33 feet = 6651.15 feet

 

[Dave in PA]

....I need more coffee......

 

[The Kraken]

Given: There are no factors involved in this problem other than the volume of gas being consumed and pressure effecting the volume of gas as it is delivered from the tank and has reached its maximum volume condition at the depth inquestion.

OK . . . let's say the diver has a SAC rate of .50 - that is 1/2 cubic foot per minute.
Let us say also that the diver has a respiration rate of 10 cycles per minute.

So, 1 breathing cycle would mean that the diver will intake .05 cubic feet of air.

Following along this basis, the breathing air, once delivered from the regulator would have to be compressed by a factor of 1548 (based on an AL80 @ 3000 psi = 77.4 cu ft) = 77.4/.05 (1 tank to one breath)

That means the diver would have to be at 1548 ATA's.

1548 ATA equates to some serious depth!

I come up with 51,051 feet.

the K-rushed

 

[surfsidedav]

Assuming everything is perfect and the tank holds 80 ft3. Then p1v1=p2v2.
1(atm)*80(ft3)=x(atm)*.5(ft3), x=160atm, so 160*33=5280 ft, minus the initial 1 atm, and you end up with 5247 ft.

 

[The Kraken]

Surfsideav,

Think about it. You have to consume 80 cubic feet of air in one breath.

The .5 you are using is the SAC, or the Surface Air Consumption rate per minute.

To determine how much air one takes in one breath, the SAC has to be divided by the respiratory cycles per minute of the diver.

We'll keep it to round numbers.

SAC rate = .50 cubic feet per minute
Respirations per minute = 10
Volume of 1 respiration cycle = .50/10 therefore 1 respiration cycle = .05 cubic feet.
80 cubic feet (volume of tank at surface) / .05 (volume of 1 resp. at surface) = 1600

therefore the pressure (ATA) required to reduce 80 cubic feet to a volume of .05 cubic feet will be 1600 atmospheres absolute.

D=(ATA-1)x33

D=(1600-1)x33
D=1599x33
D= 52,767'

the K

 

[LSDeep]

ROFL - is it still going??? i will get some coffe and wake up, afterwards i will try to figure out what the rum last night had me calculate here