High pressure water exhaust 2

[Hearticalone]

I have a strange question...and from this post, you will be able to tell that I am not a marine engineer (Actually, I am a software engineer).

Anyway, here is the question. I am trying to understand ocean pressure.
Let me present the scenario. Lets say you had a vessel submerged several hundreds of feet where the ocean pressures are enormous. In this situation, the vessel was partly flooded with water (but structurally sound). In order to evacuate this water, would the vessel require a pump that had a force greater than the ambient outside water? Would this vessel require a huge amount of energy to expel the water? What I am trying to understand is the difference between the pressure against the vessel, and the amount of energy used to expel the water from the vessel.
I've been considering a technique for draining the water:

The vessel could contain special drainage compartments, in which the inside water could be leaked. Then these compartments would be sealed, and then pressurized to the outside pressure, after which a pump could be used to pump from this now pressurized compartment to the outside ocean, and the volume of the compartment could be reduced while the water is being evacuated.
This process could be completed until the water has been drained.
Is this feasible. Would it require energy equal to the force of the outside pressure?


Thanx in advance

 

[zeusfaber]

It all depends on the pressure inside the vessel.

If the internal and external pressures are matched, then moving the water out is easy. If not you have to overcome the pressure difference.

Energy is force times distance it acts over (Energy requirement is this, divided by the efficiency of your pumping system). Power is Energy per unit time.

Treat force as pressure difference times the cross sectional area of the pipe you're exhausting through. Multiply this by the exit velocity, and you'll get a power figure (use Pascals, square metres and metres, and you might even get an answer in watts).

The flaw in your cunning plan is in the bit where you reduce the volume of your pressurised bilge tank. The easiest way to do this would be by making it look like a hydraulic ram. Once you open the ram to sea pressure, a (large) force starts to act on the piston. To reduce the volume, you need to move against this force, and this requires energy (force times distance again).

The mechanism you're describing is simply a positive displacement pump.

A.

 

[Hearticalone]

Thanx very much for your quick answer.
I take you are saying that once the pressures are equal, then the energy to exhaust the bilge tank is not a lot. How do you think submarines remove the water in these type of situations? Do the use a ram or some form of suction pump?

 

[zeusfaber]

Pumping water when there's no pressure differential is easy.

If the tank volume remains the same, the moment you start to pump water out, the pressure within it will plummet (water being more or less incompressible), and pumping becomes more difficult (and energy consuming).

If you try to get round this by allowing the tank volume to change as you pump out, then you have to use energy to maintain the pressure inside.

Submarines use multi-stage HP Bilge/Ballast Pumps to get water overboard when properly submerged, usually backed up with an LP air blower to push larger quantities of water out the ballast tanks once on the surface.

A.

 

[waross]

The minimum energy required to rraise the vessel is the same no matter what method is used and it is not excessive.
Example:
A vessel is lying in 500 feet of water. The pressure will be close to 250 psig. You may arrange a method whereby water may exit from the bottom of a fuel tank, water tank or closed hold. You then connect an air line to the tank and apply air pressure. Nothing much happens until the air pressure equals the water pressure. Once the air pressure in the tank or hold exceeds the sea water pressure at that depth the water will be blown out of the hole in the bottom of the tank. When enough water has been blown out(displaced) that the bouyancy created by the air equals the weight of the vessel (often called the displacement) the vessel will begin to rise. At this time, the water pressure almost equals the air pressure. As the vessel rises towards the surface the water pressure decreases but the air pressure doesn't. Well, it does a little but the water left in the tank and the air that will escape when all the water is gone can't escape fast enough to make a substantial reduction in internal pressure. The tanks and general construction of the vessel cannot stand that magnitude of pressure and long before the vessel reaches the surface the tanks will rupture and the vessel will return to the bottom.
A common method of raising smaller vessels is to use canvas bags with lifting straps attached. The lifting straps are attached to the propellor shafts, the anchor chains and any other strong lifting points. The bags are open on one end. Air is pumped into the bags to displace the water. As the vessel rises and the air expands due to the reducing pressure the excess air escapes from the open bottom of the bag. The more experienced raising crews try to avoid raising the stricken vessel directly below the work boats.
respectfully