How Long For Freeflow To Empty Tank?

[adder70]

OK,


Something about the rigid container making a difference strikes a bell but what confuses me is this. If your tank expells 1100 litres of air a 5 ata then aren't those bubbles going to expand on the way up? Aren't they going to turn into 5500 litres of bubbles? Doesn't that mean that you must have ejected 5500 litres of air at depth under pressure to make this happen?

My head hurts.

R..

It's 1100 litres when it comes to equilibrium with the water and air around you. When it is constricted, it is (likely) still at a significantly higher pressure, something closer to the halfway point in pressure between the tank and the ambient pressure.

By the same token the expelled air is 5500 litres when it reaches equilibrium at the surface, but the pressure at the point of restricted flow is much higher than the ambient surface pressure, so the volumetric flow - which should be relatively similar - will also result in a somewhat similar (or at least closer than 5 to 1) mass flow rate. (BTW, your tank will have the same mass of air at 3000 psi (or say 250 bar) and 70F (or say 25C) regardless of the outside pressure or the pressure through the restriction.)

 

[truva]

How long would it take for a freeflowing regulator to empty an AL 80 at the surface?

How much less would this time be at ,say, 5ATM (132 feet??)

That's the question

For some reason I chose to think it was a secondary regulator when in fact it doesn’t say which.

In which case (a secondary or cut hose) it will empty faster the deeper you are because of the RELATIVE DENSITY INCREASE.

If you just have the tank valve open you probably couldn’t tell much difference, I’d bet on the surface being a bit faster……

Truva

 

[adder70]

To really irritate the mathematicians out there:

Figure in Open Water you're going to start to ascend at 30'/min as soon as the freeflow begins. So you're not talking about a freeflow at 5ATA, you're talking about a freeflow at decreasing ATAs.

I bring it up because I can't do math and I'm jealous of those who can.

It becomes even more "irritating" if you want accuracy anyway. You have to account for the constantly changing pressure in the tank as well. (Once you hit choked flow, lowering the lower pressure will not matter (OK, it will, but not significantly); the flow rate would be limited by the pressure in the tank and the restriction characteristics.)

You would really have to set up a program (probably a spreadsheet would do) to perform a numerical analysis point by point with smaller time deltas until the answer stays essentially the same. Did I mention this would require some testing of the flow at a few points to form a baseline for the calcs? You could probably get a ballpark from one draining with meaurement of pressure every second or two under one set of circumstances and then apply the calcs to predict it for the other. Otherwise you would need detailed info on the size and shape of all restrictions in the flow path and a sophisticated fluid flow program to predict actual characteristics. Much easier to stick on a flowmeter and pressure sensor drain a tank once.

OK, now MY head hurts...

 

[Big-t-2538]

Where are you getting closed system? We are talking about a secondary free flow and would it empty a tank faster, slower or the same at depth. The density of the gas flowing out of the first stage increases with depth.....

Truva

I'm at a loss to see how desity figures into the equation...it makes no sense. Help me see what I'm missing.

Here's how I see it.

On the tank side of the regulator you have a volume of 80 cu ft at 3000psi at ambient pressure.

In order for there to be flow, there has to be a pressure differential between the tank and the ambient pressure on the other side of the reg.

The flow rate is constricted to the resistance the gas in the tank sees as it moves through the regulator. This will not change with depth...I don't see how it can. The only chage you will see is when the tank and ambient pressure are very close (say within 10 psi) then the flow will begin to slow down.

On the other side of the regulator, you have ambient whatever. For simplicity let's say we're at 99' or 4 times the pressure that we see on the surface. Let's say we've attached an 80 cu. ft capacity balloon to the exiting side of the second stange.

Now...flow will happen at the specified rate until the volume of gas and pressure in the tank is equal to the volume and pressure in the balloon. Let's say if we do this on the surface, it takes 4 minutes.

What happens when we open the system to let gas flow into the balloon at depth?
Simple, gas will flow through the reg until equilibrium occurs

How much gas will be in the balloon?
Simple, since we're at 4 times the ambient pressure and volume rating of the tank, there will be 1/4th of that volume in the balloon....or 20 cu. ft.

How long did that take?
Since the flow rate through the regulator would remain constant, and I don't have any reason why it wouldn't, it would be 1/4th the time it takes the system to equalize at ambient pressure b/c now we're only pumping 20 cu ft of gas through to "empty" the tank.

In conclusion...if it takes 4 minutes to free flow a tank to empty on the surface,it will take 1 minute to free flow a tank to empty at 99'.

Any questions?

 

[Big-t-2538]

In which case (a secondary or cut hose) it will empty faster the deeper you are because of the RELATIVE DENSITY INCREASE.

I really see no reason for the relative density of anything to increase....please point me somewhere that is relative to this situation.

If this were true, a SAC rate in cu ft per minute would be meaningless.

 

[truva]

I'm at a loss to see how desity figures into the equation...it makes no sense. Help me see what I'm missing.



Any questions?

You think that gas in your example isn't 4 times as dense at depth?

Truva

 

[adder70]

QUOTE=Big-t-2538 (Responses in bold) I'm at a loss to see how density figures into the equation...it makes no sense. Help me see what I'm missing.

Here's how I see it.

On the tank side of the regulator you have a volume of 80 cu ft at 3000psi at ambient pressure.

In order for there to be flow, there has to be a pressure differential between the tank and the ambient pressure on the other side of the reg.

The flow rate is constricted to the resistance the gas in the tank sees as it moves through the regulator. This will not change with depth...I don't see how it can. The only change you will see is when the tank and ambient pressure are very close (say within 10 psi) then the flow will begin to slow down.

On the other side of the regulator, you have ambient whatever. For simplicity let's say we're at 99' or 4 times the pressure that we see on the surface. Let's say we've attached an 80 cu. ft capacity balloon to the exiting side of the second stange.

Now...flow will happen at the specified rate until the volume of gas and pressure in the tank is equal to the volume and pressure in the balloon. Let's say if we do this on the surface, it takes 4 minutes.

It took a while, but I finally saw the situation here. The volume in each does not matter. You will get equilibrium when the pressures are the same, whether the balloon ends up with 0.001% of the air or 99.999% of the air.

What happens when we open the system to let gas flow into the balloon at depth?
Simple, gas will flow through the reg until equilibrium occurs which again is related only to pressure, not volume

How much gas will be in the balloon?
Simple, since we're at 4 times the ambient pressure and volume rating of the tank, there will be 1/4th of that volume in the balloon....or 20 cu. ft. HERE is the issue. What was the pressure in the tank? Likely it was 3000 psi. That is about 205 atmospheres! So you went from 205 atm to 4 atm. At the surface you went from 205 atm to 1atm. The total volume is 80cubic feet in the second example, but the volume of air that went through the restriction is far closer to the volume of the air at the tank pressure. That volume would be about 0.025 cubic feet. Even at half the pressure the volume would be 0.05 for the depth example and 0.05 for the surface example (You need more decimal places to see the difference in volumetric flow!) The bottom line is that 99.5% of the air leaves the tank in either example, and it passes the restriction at a pressure much higher than the ambient pressure.

How long did that take?
Since the flow rate through the regulator would remain constant, and I don't have any reason why it wouldn't, see above it would be 1/4th the time it takes the system to equalize at ambient pressure b/c now we're only pumping 20 cu ft again, see above - you're pumping about 78.1cf of gas through to "empty" the tank.

In conclusion...if it takes 4 minutes to free flow a tank to empty on the surface,it will take 1 minute to free flow a tank to empty at 99'.

Any questions?

 

[dc4bs]

Excellent point Rick. One thing to keep in mind though.... If that tank is upright at 2000M, it'll empty as fast as it can fill with water... Maybe you math/physics guys can play around with that for awhile...

Cheers,
James

Unless you remove the valve (yes it should be possible as there will be no pressure differential across the threads) then it's gonna take a LONG time to get through the 'bottleneck' of the tiny air channel through the valve itself. Two way traffic on a one lane road is never very efficient.

You could drill a hole in the bottom of the tank to make it go prety quick though. Kinda like makin a small hole in the cranbery sauce can bottom to get it to come out in one clean, can-shaped lump at thanksgiving...

 

[nyresq]

I read through this thread and boy does my head hurt.
I have seen the logic on all side of the different theorys. But I pose this statement for your consideration:
If I inflate my wing on the surface it takes about 10 seconds from empty to 1/2 full. At depth it takes longer to inflate it to equal size. I use a balanced diaphragm so the reg is flowing the same amount of air and its going through the same inflator... but it takes longer at depth.
So why is it that it takes longer to flow enough air to inflate the wing to the same size if the tank will supposedly empty 5x faster?

 

[Rick Murchison]

Would you all agree that putting a regulator, with all its convoluted gas paths and pressure stepdown seats and levers and springs and poppets & other stuff onto a "K" valve cannot increase the flow of gas through the little hole in the "K" valve?
No? You don't agree? Go back to physics 101.
If you do agree, you're right; let's continue...
Now, since the regulator can only slow the flow that is possible through the "K" valve, let's take the regulator and all it's confusing "constant pressures" and such completely out of the problem, and examine what happens when you just open the "K" valve wide open. Gas will flow through the valve from the area of higher pressure to the area of lower pressure based almost completely on only two factors - those are (1) the pressure drop from one side of the valve to the other (this is the same as gauge pressure) and (2) the resistance to the gas flow provided by the valve itself,
*** physics stuff... skip if you like ***
which can be expressed mathmatically as a cross sectional area of a hole (the size of the mathmatical hole is smaller than the actual hole in the valve because the actual hole has length and turbulence and irregularities, but it will *act* like a hole with no length of an easily calculated size based on the valve's flow characteristics) Flow rates will also be effected by temperature and the specific density of the gas involved - He will exhaust faster than O2, for example, but for the purposes of this explanation those effects aren't significant or relevant.
******************************
Now we can't change the size or shape of the hole in the wide open valve, so that restriction to flow is unchanging - the only thing we can change is the pressure difference between the inside of the tank and the outside of the tank.
*** at this point we have to talk a little physics because there is a fundamental misunderstanding in this thread of what happens to the flowing gas under varying ambient pressures... Flow rate is often expressed in liters/minute, because this is a useful measurement above the water... but it lends confusion to this discussion because the flow rate is actually a mass and not a volume flow rate. So (for air) the 1100 liters/minute (approx 39 CFM) cited earlier, for example, is really about 1430 grams/minute, and this flow rate will not increase as a tank is taken to depth. In fact, it will decrease a bit, because the pressure gradient decreases as the ambient pressure rises. And if you wanted to measure the flow rate in CFM or liters per minute at the ambient pressure of, say 5ATA, then that same 1430 grams/minute would yield not 1100 liters/minute, but 220, and the tank would empty at essentially the same rate as it did on the surface.
If this is still not clear, arrange a demonstration to satisfy yourself. You'll find there is no significant difference in the rate of depletion from the tank at any recreational diving depth.
I promise.
Rick