Need help re: air pressure to fluid pressure

[RobA]

Hi, I'm working on a project and need a bit of help, since I don't want to kill myself.

My question is this: If I have a cylindrical tank (think a big scuba tank) half full of liquid, half full of gas (let's just say it's water/air) and start heating up the liquid, I know that pressure will go up as the liquid heats up the entire tank.

But I also know water is (practically) incompressible. So is the pressure change only in the top half of the tank where the air is? Am I correct in assuming that as the pressure in the top (gas filled) part of the tank, the pressure in the bottome half (liquid filled) is more or less the same?

I'm confused because as a diver, I know that when calculating pressure in terms of diving, we need to add 14.7 psi (1 atm) to the calculation, to compensate for the pressure of the air above the water. So at 33 feet, which is the depth that represents 1 atm, we actually have to calculate it as 2 atm with regards to figuring out dive total and avoiding the bends. This suggests that the pressure above a liquid translates into pressure throughout that liquid.

On the other hand, I also know that when scuba tank manufacturers test tanks hydrostatically, they fill the tank up with water first. Because water cannot compress, the tank will hhave way, way, way less kinetic energy @ 3000+ PSI then it would if it was filled with air @ 3000+ PSI. So even if it fails the test and blows, the damage will be much less. To me, this suggests that air pressing down on a liiquid below does NOT translate that pressure throughout that liquid.

The reason I'm wondering is, I'm doing a chemistry experiment where I need a tank that can withstand high pressures/temps, in the range of 1500 psi and 250 celsius. Pressure tanks are insanely expensive, and scuba tanks can easily handle that kind of pressure, so I'm thinking about modifying one to use as a pressure tank. I'll need to drill a few holes near the bottom, and weld on some npt threaded fittings. I need to know if the ball valves/piping etc connected to those fittings need to be rated to the same PSI pressures, or if they (because they are completely submerged inn the liquid) are "shielded" against the high pressures in the air pocket by the liquid.

And I'd like to not maim myself, so any help from someone who understands pressure and fluid would be a big help.

Thanks in advance.

 

[RobA]

I should clarify, it won't be ME welding on the threaded fittings. It's a professional welder with certifications in welding pressure tanks.

 

[Doug Huffman]

Your fittings should be rated at some engineered multiple of the system design pressure. The welds must be NDT and the system hydrostatically tested to some engineering standard. 1500 psi @ 250°C ain't backyard engineering. I would further submerge it in water during use - much as SCUBA tanks are while being filled and for the same reason.

I directed yield pressure testing of large submarine reactor air bottles (banks of tens of cubic feet NOP 3,000 psi) by filling them gas free and monitoring the change in pressure versus change in test volume to be a straight line up to test pressure.

 

[russ_watters]

Incompressible does not mean you can't pressurize it, it just means it doesn't get smaller when you do.

 

[RobA]

Your fittings should be rated at some engineered multiple of the system design pressure. The welds must be NDT and the system hydrostatically tested to some engineering standard. 1500 psi @ 250°C ain't backyard engineering. I would further submerge it in water during use - much as SCUBA tanks are while being filled and for the same reason.

I directed yield pressure testing of large submarine reactor air bottles (banks of tens of cubic feet NOP 3,000 psi) by filling them gas free and monitoring the change in pressure versus change in test volume to be a straight line up to test pressure.

Thanks for the response Doug, I'll def make sure to test the welds first.

 

[RobA]

Incompressible died not mean you can't pressurize it, it just means it doesn't get smaller when you do.

I see. So, is the pressure uniform throughout? For example, if my gauge on the top of the tank is reading 1500 psi, is the pressure at the bottom of the tank also at 1500 PSI?

So if I were to crack open a ball valve, the liquid would shoot out at that much pressure?

 

[jack action]

So if I were to crack open a ball valve, the liquid would shoot out at that much pressure?

Yes. That is the whole concept of water guns:

Air pressurized reservoir
This type was made famous by the Super Soaker brand of water guns, actually first employed by the Cosmic Liquidator.[1][5] In this device, a pump is used to push air into a partially water-filled reservoir. The reservoir is otherwise air-tight, but it has one valve to let the incoming air in from the pump as well as a manually controlled valve operated by the user, commonly activated by pulling on a trigger. As more air is pumped in, the air in the reservoir is compressed, increasing in pressure; the water is also pressurized by the now compressed air. Upon opening the nozzle valve, the pressurized water is pushed out through the nozzle as the air attempts to re-equilibrate with atmospheric pressure. This system allows pumping energy to be stored and used as needed. As well, unlike the methods noted above, this air pressure system allows production of a solid, continuous stream of water.[3]

The limitation of this design is the need for a large number of strokes to pressurize a larger reservoir. As well, poorly sealed reservoirs would render a water gun useless. Moreover, these water guns cannot be refilled unless emptied and depressurized. Opening a pressurized reservoir blaster while there is pressure remaining in the system can result in copious local water spray or even an unexpected launch of the water gun and/or reservoir out of one's hands.

 

[russ_watters]

I'm hesitant to leave this thread open because as you say, what you propose is dangerous and I'm not sure you are taking the necessary precautions and don't understand the science (which is why you are asking). Pressure vessels are expensive precisely because they are pressure vessels and need to be engineered for the risk. They need inspections and certifications. In addition, in-use pressure vessels (and even scuba tanks) are required to have safety relief valves to prevent them from being over-pressurized.

 

[Bystander]

hesitant to leave this thread op

Water? 250 C? 1500 psi? Scuba gear? Uh-uh. Veerrryyyy baaddd ju-ju.

 

[Chestermiller]

Hi, I'm working on a project and need a bit of help, since I don't want to kill myself.

My question is this: If I have a cylindrical tank (think a big scuba tank) half full of liquid, half full of gas (let's just say it's water/air) and start heating up the liquid, I know that pressure will go up as the liquid heats up the entire tank.

Is the tank closed? I'm assuming it is.

But I also know water is (practically) incompressible. So is the pressure change only in the top half of the tank where the air is?

No. The pressure increases in the bottom half also.

Am I correct in assuming that as the pressure in the top (gas filled) part of the tank, the pressure in the bottome half (liquid filled) is more or less the same?

No. Of course not. The pressure in the bottom half of the tank increases by the same amount.

I'm confused because as a diver, I know that when calculating pressure in terms of diving, we need to add 14.7 psi (1 atm) to the calculation, to compensate for the pressure of the air above the water. So at 33 feet, which is the depth that represents 1 atm, we actually have to calculate it as 2 atm with regards to figuring out dive total and avoiding the bends. This suggests that the pressure above a liquid translates into pressure throughout that liquid.

Yes.

On the other hand, I also know that when scuba tank manufacturers test tanks hydrostatically, they fill the tank up with water first. Because water cannot compress, the tank will hhave way, way, way less kinetic energy @ 3000+ PSI then it would if it was filled with air @ 3000+ PSI. So even if it fails the test and blows, the damage will be much less. To me, this suggests that air pressing down on a liiquid below does NOT translate that pressure throughout that liquid.

Wrong. What it means is that the liquid is not capable of expanding to a great extent. So it can't do damage because it can't blow things at high speed.

The reason I'm wondering is, I'm doing a chemistry experiment where I need a tank that can withstand high pressures/temps, in the range of 1500 psi and 250 celsius. Pressure tanks are insanely expensive, and scuba tanks can easily handle that kind of pressure, so I'm thinking about modifying one to use as a pressure tank. I'll need to drill a few holes near the bottom, and weld on some npt threaded fittings. I need to know if the ball valves/piping etc connected to those fittings need to be rated to the same PSI pressures, or if they (because they are completely submerged inn the liquid) are "shielded" against the high pressures in the air pocket by the liquid.

And I'd like to not maim myself, so any help from someone who understands pressure and fluid would be a big help.

Thanks in advance.

You might as well not have the liquid water in there. The air at 1500 psi and 250 Celsius is extremely dangerous, and the water at the bottom will do very little to reduce this danger.

Chet