Raising a Sunken Vessel: Calculating Air & Water Flow

[Painter1]

This is a real world problem. We have a vessel sunk in about 50' of water and need to raise it. We have devised a way to pump the water out and pump air into replace the water. It is important that we keep the differential pressure very close to Zero and running a "snorkle" tube is not an option.

So we need to replace the water with air while keeping the inside pressure the same. How much air do I need to put in?

I am going to pump the air in with a compressor that puts out 115 psi at 170 cfm and pump the water out with a submersible at 20 gals per minute. The air will need to to pumped in in short bursts so we need to know how long to hold the compressor air valve open (putting air in) for every minute the water pump operates.

Your help is most welcome. Thanks

 

[HallsofIvy]

That depends upon how large and how heavy the boat is. you will need to replace enough of the water in the boat with air so that the total weight of boat, water still in it, and air will be less that the same volume of water.

 

[DaveC426913]

You should not do this.

This is extremely dangerous. You are dealing with very large forces. And you are not experienced.

I am an open water diver and this is way over my head. This is technical diving certification stuff. And technical diving is one of the most dangerous professions in the world.

If you found yourself pushed up by that boat a mere 6 feet, you could die from embolism. It has happened to extremely well-seasoned divers.

This is why professionals do this and how non-professionals get killed.

 

[berkeman]

Thread closed for Moderation...

EDIT -- Thread re-opened for a bit. It may be closed, depending on how the safety questions are addressed...

 

[berkeman]

This is a real world problem. We have a vessel sunk in about 50' of water and need to raise it. We have devised a way to pump the water out and pump air into replace the water. It is important that we keep the differential pressure very close to Zero and running a "snorkle" tube is not an option.

So we need to replace the water with air while keeping the inside pressure the same. How much air do I need to put in?

I am going to pump the air in with a compressor that puts out 115 psi at 170 cfm and pump the water out with a submersible at 20 gals per minute. The air will need to to pumped in in short bursts so we need to know how long to hold the compressor air valve open (putting air in) for every minute the water pump operates.

Your help is most welcome. Thanks

Painter, to Dave's points, can you please give us more details about the diving experience and certifications of your crew? Have you done anything like this before? How big is the vessel? Are you going to pump up an air bag inside the vessel, or are you going to rely on some air-tight compartment that can stably lift the vessel? How are you planning on monitoring the operation under water?

 

[russ_watters]

...and why would you need to pump water out of the vessel? It's going to want to leave on its own when you pump air in!

 

[DaveC426913]

...and why would you need to pump water out of the vessel? It's going to want to leave on its own when you pump air in!

I think that was his point about pumping air in.

Man, this is not the place for amateurs!

Lock this thread before he gets himself killed and they come after PF.

 

[russ_watters]

I think that was his point about pumping air in.

No, the OP refers to a water pump in addition to the air compressor.

Man, this is not the place for amateurs!

Lock this thread before he gets himself killed and they come after PF.

We'll monitor it...

 

[K^2]

First of all, 115 PSI is WAY overkill for this. You should need a little over 20PSI to pump air to 50', and you should not use much more than that for safety reasons.

That said, you obviously don't know what you're doing. Hire professionals.

 

[mrspeedybob]

...and why would you need to pump water out of the vessel? It's going to want to leave on its own when you pump air in!

That's what I was wondering.

The closest I've come to raising a boat was watching the myth-busters do it with ping-ping balls so I don't know what I'm talking about, but why does pumping air in not just displace the water, pushing it out through doors and windows and such. It seems like if the air was added slowly then pressures should stay in equilibrium.

I'm hoping someone with relevant experience can educate us.

 

[DaveC426913]

That's what I was wondering.

The closest I've come to raising a boat was watching the myth-busters do it with ping-ping balls so I don't know what I'm talking about, but why does pumping air in not just displace the water, pushing it out through doors and windows and such. It seems like if the air was added slowly then pressures should stay in equilibrium.

I'm hoping someone with relevant experience can educate us.

Frankly, I'm having trouble imagining how even in principle they're going to pump water out except by pumping air into displace it. What will be left? Vacuum?

If you started a water pump and didn't have air flowing in, well, the pump will just cavitate. It makes no sense.

I guess having a water pump in conjunction with an air pump might take some pressure off the air pump...

 

[K^2]

I guess having a water pump in conjunction with an air pump might take some pressure off the air pump...

That'd make some sense, if the air pump wasn't outputting 5x the pressure they need to pump the air down.

Frankly, I'm having trouble imagining how even in principle they're going to pump water out except by pumping air into displace it. What will be left? Vacuum?

If you started a water pump and didn't have air flowing in, well, the pump will just cavitate. It makes no sense.

I doubt the thing is completely water tight. It will just take in more water as you pump it out.

If it was not for the dangers involved, I would very much like to see what happens when these guys try to put their plan into action.

 

[JeffKoch]

Sounds suspiciously like a homework problem rephrased. It's difficult to imagine someone trying to actually do something like this, and be so totally inexperienced that he would need to post this question on a forum.

 

[FireStorm000]

It wouldn't be terribly difficult to derive equations for this; it's just ideal gas law, but I'll echo what everyone else said. Bad idea more than likely. You sound like you've got about as much clue what you're doing as I do, and I'm smart enough not to try this.

I can post up an equation for you IF you address the concerns listed so far.

Also, it might be enlightening to try this in small scale, with a weighted 2L bottle as your boat, a small pump attached to the mouth and a very small compressor (or just a straw to blow into). It's probably safe to play around in small scale, but just make sure that boat is worth your life if you choose to go though with this for real.

 

[jack action]

First, I'm no specialist in this kind of work and I would like to start by seconding FireStorm000's idea of doing a scale down experiment. Especially since we don't know what kind of vessel we're talking about, its size and condition.

All that being said, I don't understand the big deal about this or I'm missing something. Basic physics tells me the water pressure at 50' is around 20 psi ($\rho$gh), so the air pump pressure doesn't need to be much higher. Using a regulator, you can increase the air pressure until the water pressure is achieved. Then by pumping the air inside the vessel with a pressure slightly above the outside water pressure, the water will get out of the vessel by whatever outlet there is. The cfm needed by the pump only depends on the size of this outlet and the rate desired.

Once there is enough air, the vessel begins rising and as it rises, you must adjust your regulator pressure to match the decreasing water pressure. If pressure differential is critical, you might want to control your rising rate by using some kind of physical restraints. Restraints will probably be a good idea also to make sure the vessel does not turn around itself, flipping over and loosing all the air inside, which would mean a rapid sudden drop. That would be the most dangerous part of the lifting process in my point of view. Inflating a balloon inside the vessel might be a good idea if one doesn't want to take the chance of loosing the air.

Again, not an expert opinion, just my two cents, and do a scale down experiment first.

 

[Integral]

... [Once there is enough air, the vessel begins rising and as it rises, you must adjust your regulator pressure to match the decreasing water pressure. If pressure differential is critical, you might want to control your rising rate by using some kind of physical restraints. Restraints will probably be a good idea also to make sure the vessel does not turn around itself, flipping over and loosing all the air inside, which would mean a rapid sudden drop. That would be the most dangerous part of the lifting process in my point of view. Inflating a balloon inside the vessel might be a good idea if one doesn't want to take the chance of loosing the air.

Again, not an expert opinion, just my two cents, and do a scale down experiment first.

Just not the way it will work. The boat will NOT begin to slowly rise it will hang in the mud until it has sufficient boyuancy to break free. Once free of the mud it will shoot up VERY quickly.

Since you have no idea as to the integrity of the hull, you will have no idea as to what its orientitation will be when (or if) it reaches the surface. I am kinda on Dave's side here, anyone underwater and near this wreck as it breaks free is putting themselves at risk.

 

[DaveC426913]

Just not the way it will work. The boat will NOT begin to slowly rise it will hang in the mud until it has sufficient boyuancy to break free. Once free of the mud it will shoot up VERY quickly.

That is an excellent point that I should have thought of.
Bouyancy increases as it rises, meaning every foot the thing rises it becomes more bouyant. (It is going from 9 atmo to 1 , the air will expand by that amount).

It will break the surface like a breaching whale.


All that aside, there is no issue about the trick in principle - a small scale test will tell you nothing you don't already know. The problem is the real life danger.

 

[cjl]

(It is going from 9 atmo to 1 , the air will expand by that amount).

I fully agree about the danger, but at that depth, it's only going from about 2.5 atm (absolute) to 1. That doesn't change the fact that it should only be attempted by professionals though.

 

[DaveC426913]

I fully agree about the danger, but at that depth, it's only going from about 2.5 atm (absolute) to 1. That doesn't change the fact that it should only be attempted by professionals though.

What's a factor of 4 between friends...

 

[sophiecentaur]

To maintain the same amount of lift on its way to the surface, you would need to produce a volume around the highest point on the wreck and make it airtight but with a large hole underneath. (Effectively, an inverted bucket) This volume would need to be such that it contains the same weight of water as the boat's weight - or a little bit more. If you then pump air and fill this void, water will be displaced and the boat will rise up. As it rises, the air inside will expand but escape through the bottom - maintaining the same volume of displaced water and hence, the same amount of lift so the boat won't accelerate upwards uncontrollably. It's, afaik, normal to use lifting bags, which self vent out of the bottom and work in the way described above. Basically, you need a constant volume system and not a constant pressure system.
What do you intend to do with the wreck, once it has reached the surface? There is always the risk that it will roll when it reaches the surface and go right down again. (How big is it?)