HP Air Bottle Sizing for 4.19m³ Tank at 10 Bar

  • Thread starter Thread starter Nasir6992
  • Start date Start date
  • Tags Tags
    Air Sizing
AI Thread Summary
To empty a 4.19 cubic meter tank filled with water at 10 Bar, approximately 0.57 cubic meters of air at that pressure is required, factoring in the need to overcome the water pressure at a depth of 5 meters. The air expands significantly due to the pressure difference, with an expansion ratio of 7.33. However, achieving complete emptying may be challenging due to air tunneling, which can lead to inefficient water removal. Engineering solutions may involve fluid simulations to optimize air introduction or accepting residual water. Understanding these dynamics is crucial for effective tank emptying.
Nasir6992
Messages
1
Reaction score
0
Senario-1

I have a tank of Volume = 4.19 Cubic meter which is filled with water. I want to empty the tank by giving a HP air blow at 10 Bar. How much air is required to empty the tank completely.
If someone can give any formula or any reference that can help me.

Senario-2

I have a tank of Volume = 4.19 Cubic meter which is filled with water and the tank is dipped in water at a depth of 5 meter depth. I want to empty the tank by giving a HP air blow at 10 Bar. How much air is required to empty the tank completely.

If someone can give any formula or any reference that can help me.
 
Engineering news on Phys.org
Scenario 1.
Don't understand.
scenario 2
1 bar = 10 meters of water, so you will need 0.5 bar (1.5 bar absolute) to empty the tank. Your air source is 10 bar (11 bar absolute) so your air supply wil expand by 11/1.5 = 7.33. Your water tank is 4.19 cubic meter, so you will need 4.19/7.33 = 0.57 cubic meters of air at 10 bar (11 bar absolute).
 
You simply need the volume of air to fill the tank at the pressure you are displacing the water to.

So for a 4.19 cubic meters of water at 1 bar is displaced by 4.19 cubic meters of air at 1bar. Now if the water is being displaced to a higher height than the bottom of the tank, or at a higher pressure then you need to account for the difference.

Now what I mentioned is the truth from a physics standpoint. From an engineering standpoint you are not going to get the tank empty with the volume of air required to empty it because the air will tunnel out to the drain at the bottom causing the water to start spraying at the end instead of a solid stream. If you want to get an idea of how much air you are going to need to run some sort of fluid simulations taking into account the size and shape of your tank as well as the way you introduce the air to blow the tanks. Alternatively you can just live with a small amount of water in the bottom and blow the tanks slowly.
 
chayced said:
You simply need the volume of air to fill the tank at the pressure you are displacing the water to.QUOTE]
The tank is 5 meters deep so he needs 0.5 bar psig in the tank to get the water out the bottom. His pressurized air tank is pressurized to 10 bar.
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
Thread 'How can I find the cleanout for my building drain?'
I am a long distance truck driver, but I recently completed a plumbing program with Stratford Career Institute. In the chapter of my textbook Repairing DWV Systems, the author says that if there is a clog in the building drain, one can clear out the clog by using a snake augur or maybe some other type of tool into the cleanout for the building drain. The author said that the cleanout for the building drain is usually near the stack. I live in a duplex townhouse. Just out of curiosity, I...
Back
Top