How Long For Freeflow To Empty Tank?

[Big-t-2538]

It doesn't come out of the tank faster but you use if faster because it takes more gas to fill your lungs with each breath.

Not exactly....yes, you use it faster, but your lungs do not require more gas, your lungs require a constant volume to be filled...it is the fact that there is less gas available for you to draw from.

 

[Rick Murchison]

Flow resistance is based on fluid properties and volumetric flow, not mass flow.

Aha! Now I see where some of y'all are getting confused... What adder70 is saying here is (essentially) correct when comparing two different gasses - if you take a tank of helium and a tank of air side-by-side that are the same size with the same pressure and the same valves and open them both, they'll both drain at essentially the same rate, and the mass flow rate for the helium will be about a seventh that of the air mass flow rate, while the volumetric flow rates will be about the same. (the helium will actually drain slightly faster, but this has more to do with moving mass around corners)
That's true.
But it does not apply to the "how fast does the tank empty at depth compared to the surface."
Let's say the rate of drain at 3000 psi is 1100 liters per minute for either tank. In fact, for clarity let's say the rate is 1100 surface liters per minute. If we increase the pressure outside the tanks to 5 ATM, they'll still empty at the same rate, initially (just slightly less than) 1100 surface liters per minute (the same mass flow for each gas as before, with the helium still draining about a seventh the mass per minute as the air), but if you measure the volume of that 1100 "surface liters" at the 5ATM, you'll only have 220.
I do hope this clears it up for any of you who remain confused.
Rick

 

[MikeFerrara]

Not exactly....yes, you use it faster, but your lungs do not require more gas, your lungs require a constant volume to be filled...it is the fact that there is less gas available for you to draw from.

What are you smokin dude?

At 2 ATA it takes twice the number of gas molecules to fill your lungs.

 

[DA Aquamaster]

But it does not apply to the "how fast does the tank empty at depth compared to the surface."
Let's say the rate of drain at 3000 psi is 1100 liters per minute for either tank. In fact, for clarity let's say the rate is 1100 surface liters per minute. If we increase the pressure outside the tanks to 5 ATM, they'll still empty at the same rate, initially (just slightly less than) 1100 surface liters per minute (the same mass flow for each gas as before, with the helium still draining about a seventh the mass per minute as the air), but if you measure the volume of that 1100 "surface liters" at the 5ATM, you'll only have 220

Right, and that would be correct if the flow occurred at the valve without respect to the reg. In that case the supply pressure feeding the valve is not affected by ambient pressure. (although there would be approx 75 psi less difference in pressure drop across the valve at 5 ATM.

But the reg is part of the equation and all valves are not created equal in terms of flow rates. If you involve the reg in the process, then the reg under 5 ATM of ambient pressue will be operating at an IP of 220 psi rather than 145 psi and will be flowing a greater mass of air than at the surface.

For the more engine inclined, this is working essentially the same way that a supercharger increases the mass flow through an engine by increasing the density of the charge.

 

[Big-t-2538]

What are you smokin dude?

At 2 ATA it takes twice the number of gas molecules to fill your lungs.

I'm not having a good science week...

yes...2X the number of molecules at 2 ATA...when you said more gas, I thought you meant more volume.

 

[Rick Murchison]

But the reg is part of the equation and all valves are not created equal in terms of flow rates. If you involve the reg in the process, then the reg under 5 ATM of ambient pressue will be operating at an IP of 220 psi rather than 145 psi and will be flowing a greater mass of air than at the surface.

IP is only relevant when the valves are closed. In a freeflow all valves are wide open and the IP is irrelevant. The whole reg and K valve combo just acts like one great big complicated open valve, and you get the same maximum flow (less that attributable to the increased ambient pressure) that you get on the surface; the same depletion rate for the tank.
The end.
Rick

 

[truva]

No, no, no... it is mass flow rate The density of the air in the tank - and the air flowing through the restriction - remains the same regardless of depth.
SAC rate principles do not apply to this problem!
Rick

Which of the 2 different problems are you saying I am wrong on?

A free flow at the tank has one pressure drop. This case depth doesn't really doesn't matter.

A free flow at the secondary has 2 pressure drops. This case mass or density does matter. The mass flow rate will be higher as long as there are 2 pressure drops. In this case I have assumed that the first stage can more than keep up with the free flowing secondary.


Truva

 

[Big-t-2538]

This case mass or density does matter. The mass flow rate will be higher as long as there are 2 pressure drops. In this case I have assumed that the first stage can more than keep up with the free flowing secondary.

There is one and only one pressure drop in either case...this is what causes flow to happen...that differential is the pressure of the gas in the tank to the ambietn surronding pressure at whatever depth you are at. How does density matter...educate me.

 

[truva]

There is one and only one pressure drop in either case...this is what causes flow to happen...that differential is the pressure of the gas in the tank to the ambietn surronding pressure at whatever depth you are at. How does density matter...educate me.

You have one pressure drop at the first stage and you have another at the leak. As long as the first stage can keep up with the flow flowing secondary or other leak you have these two pressure drops.

Density: If you have a liter of a gas at one atmosphere and you have a liter of the same gas at 5 atmospheres the gas at 5 atmospheres is 4 times the density.

You pretty much say the same thing in one of your earlier posts.

Back to a first stage being able to keep up with a free flowing secondary, keeping up with only means that the set IP pressure set on the regulator remains there. Now this only matters to the regulator right at the diaphragm, that is where the regulator “measures” it, not at the low-pressure ports or at the end of the hoses.

You can test and see if your regulator is keeping up. Take a full tank and you regulator place your IP gage in one line then free flow you other regulator. You will see the IP go down to like 125 PSIG but then it will level out there until the tank pressure reaches a point that just cannot push enough air through that first valve once this happens the pressure will start to drop off. If your first stage cannot keep up with the free flow at all you will see the pressure start dropping right away (slowly) as the pressure in the tank lowers.

Truva

 

[perpet1]

Most K valves deliver between 1000 and 1100 litres per min wide open. Most regs can match or better this.

An AL80 has an internal volume of about 11 litres. When full that's about 2200 litres of air.

Sooooo.... 2200/1100 gives you about 2 minutes at the surface.

[edit: removed nonsense]

R..

So... you are measuring the time it takes to drain a tank as a function of a specified flow rate (lpm) through the K valve "wide open". This is interesting because last I checked flow rate was dependent on Delta P (Pressure Differential) and flow resistance (head loss)

In fact flow rate = Delta P / Head loss (looks a lot like ohms law)

Sooooo.... your assumption is 100% accurate if you had a constant pressure source (like a compressor) but flawed when you look at the fact that the Delta P is going down as the tank is drained.

You could look at it like you have your stated 1000 lpm when the tank is full but I am thinking you will have 0 lpm when you are at 1ata in the tank. In fact all things being consistant the change in flow rate and pressure will be linear.

Now back to the simplification as stated in a few other posts.

Assume head loss is consistant for the purpose of this discussion.

The flow rate will be proportional to Delta P. So the time it takes to drain a tank is proportional to Delta P and since, as stated in a previous post, the pressure difference in between the tank and its surroundings is less at depth------> the tank will drain slower at depth. Although not significantly so, also as stated in a previous post.