Pumping Out Submerged Tank

The depth of 100 meters equates to 328.08 feet of head, or 142 psi (a tad more if it is salt water). You would have to find out the leakage rate into the tank and size your pump or gas pressurizing system accordingly.

A bucket conveyor is a mechanical system that may be able to empy the tank.

Sounds like a homework question to me.

The tank is fixed to the bottom. As the water is pumped from the tank, it is supplied air from the surface to ensure a vacuum is not created. How do you calculate the power the pump needs to completely empty the tank of water because the volume of water is changing inside the tank over time? It is a cylindrical tank with a volume of 119L. The thing I am trying to figure out is whether you would calculate the pump power required given these unique circumstances: the pressure is equal between the tank and the outside 100m pressure at the beginning, but decreases as the pump is pumping the water out. Thanks so much for the ideas and further questions!

100m is the maximum depth for a WW1 submarine. This is well into the realm that if you have to ask how to do this, you're not qualified.

Now I believe that many of the WW1 & WW2 submarines did have backup pumps that were hand operated. But for obvious reasons you have to exceed the external water pressure levels to get the internal water to now be external water. At these pressures though, unless inside pressure is close to external pressure levels then any leak will produce a lot of water inside. Now unless your tank(?) is resting on something at 100m down it will fairly quickly start to sink from the added weight of the water.

If instead you think that you can remove the water to add some buoyancy to this filled tank then some lift bags on the outside will work much better. But there is no energy here to be recovered. If you wish to fill this tank with something heavier than water, then just pouring the heavier fluid in will work. You may violate multiple environmental emission standards because your leaking tank will leak the heavy fluid. If instead you wish to pump something lighter than water (air) in then you will be putting energy into this system to displace the water.

To go deeper that 40m you will need either a qualified technical diver or an unmanned submersible. Clearly you have neither available since you are asking us.

This is a basic pressure problem at hazardous depths. If this is not a homework problem, you can kill someone. Step away from the problem.

Thanks! I am not planning on diving to this depth or trying to build a submarine or putting anyone in water. This is just a working theory on an energy harnessing system with this one snag. The energy is not from the water coming into the tank, that is just a symptom of the system that I have not been able to solve yet. Thanks for the comments. This is a great website. Thanks for allowing me to bounce ideas off you guys.

I thought you were thinking of filling the talk with a pipe and a turbine, but at 119L?? what's the point?

You sound like me! so vague

Mate,

this sounds like you try to retrieve Energy from the incoming stream of Water into the tank and hope this will be more than what you need to to displace the water with air in order to restart the process.

This sounds like a perpetuum mobile and you will like to hear that it is a wonderful idea but it simply does not work.

Can you elaborate more on the Engery harnessing system?

Icke

Perhaps he is thinking of a hydraulic ram pump?

Chris

At 119 litters that is a rather small volume and given the depth the required pressure is well within a simple positive displacement gear or piston type pumps working pressure range.

Also the overall work energy required is well within that of a simple battery based power source.

The pump only needs to be able to pump up to 150 psi to displace the water.

How fast it displaces it will be relative to the volume of air the pump is able to produce at 150 psi ...more horsepower + cubic inches=faster time.

If you are going to put the pump in the tank, you will need another identical tank and just move the air back and forth between the two tanks.

If it <...fills up with water...> then what is in the tank and at what temperature and pressure when the filling operation starts?

Why does the tank fill up with water? Is it deliberate or is it a leak?

If it is a leak then the first obvious thing to do is to increase the pressure of the material that is in the tank so that the water cannot get in. 142psig or above should do it, assuming the tank is built to withstand the internal and external pressure differences. If it isn't then its volume will change, either by design or by catastrophic mechanical failure.

The second obvious thing to do would be to stop the leak and prevent the water getting in.

If filling it is deliberate, then simply close the aperture that was opened to let the water in. What is it - a valve or similar?

All sorts of pumps could then be used to empty it and their individual efficiencies are of little relevance, because if the water didn't keep getting in then once it has been emptied, no energy at all would be required to keep it empty, though the mountings that are keeping it on the bottom will be straining with the bouyancy of it at that time. Are these adequate or will the tank become a boat (been there, done it, secondhand T-shirt now on eBay <other on-line auction sites are available>)?

Bear in mind that any pump that lifts the water to the surface will need to overcome the pressure difference between the tank and the surface. Its power at any instant is the pressure difference multiplied by the volumetric flowrate of the water coming from the tank at that moment. And the pressure difference will be whatever the difference in depth is between the tank level and the surface multiplied by the density of water and by acceleration due to gravity.

My suggestion is to add pressure to the tank to 3 atmospheres. dry it and move in and stop creating stupid problems, the best cure really is old fashion common brain power or what has been suggested buy a bucket.

the simplest way is just to move the tank to dry land.

In post 5 you stated that the tank has a water line and an air line attached to it.

Use the air line to feed 150 PSIG air to the tank, and the water will be pushed out the water pipe.

The tank will be empty and pressurized slightly above water pressure to prevent any more water from entering the tank.

Since it is fixed to the bottom, it will not float.

Energygod

At least someone is on the same wavelength!

Depends on where the waterline is connected.....

Agreed- but the poster spoke of a line to remove water and an air line to break the possible vacuum.

That SHOULD mean a water line from near the bottom and an air line at the top.

Energygod is correct.

I don't see how you could ever generate power with this system. If you place a turbine in the water line, the pumping pressure would have to be increased to overcome the pressure drop of the turbine.

Just in case I guessed what you are trying to do(??):

If you want to send an empty tank to 100m depth, allowing it to fill in at that depth, compressing the air inside to 10 bars, then using that to drive a turbine , then repeat.... Then:

You could do the 1st step of compression at 100m, then have air bags (good enough to hold 10bar+) with the required volume to displace the tank's metal envelop volume and thus float back to the surface, Empty the water via turbine, recover the air from the bags (...?...), and sink the tank again to repeat. BUT, BUT BUT, the energy to force the tank down, while empty, might cancel the recovered energy, and more. it is all a matter of calculating the whole thing properly.

That is if this is near enough to what you are trying to do...(?)

First, pump it out the side or pump it to the top, it still requires a 142 psi pump. How fast? 4.2 cubic ft per ? This converts to HP. Type of pump, type of motor, and power source all help determine the efficiency. and the calculated HP times the efficiency, plus the calculated HP is the needed energy. But, you have the same calculation for the "Turbine" filling the tank. It also has an efficiency, or inefficiency if you prefer, and the combined losses in both flow it in and pump it out combine to create the total losses. This number will never be zero. So the trick is to make the cost of the additional complexity of the machine be less than the cost of the recovered power. This would be truly a first.

If you REALLY want to try to use a 119L tank to make power, try this-

Evacuate the tank as discussed and fill with liquid refrigerant- probably R-410.

Use a refrigerant pump to extract the refrigerant and send it to a thermal source.

Use a solar collector to heat refrigerant high enough to make "decent" pressure.

Use the refrigerant to drive some type of prime mover connected to a generator.

Use the "air" pipe and the tank to condense the refrigerant gas with heat loss to the large body of water (5 C near the location of the tank), allowing recycling of the refrigerant.

The refrigerant pump keeps the process moving.

You will likely make enough power to offset the relatively small pumping energy- everything else is essentially energy driven.

By the way- 11 C liquid R-410 has a pressure of just under 148 PSIG, so the tank will not be contaminated with water if refrigerant is maintained at or above that temperature- which is easy to do with pressure controls.

Original poster "guest"- sign in to CR4 and you can contact me regarding what type of prime mover would work.

The most efficient way to remove the water!

Climb down into the tank.

Put 10 liters of water into a container.
Boil the water.
Add one can of malt extract.
Cook for 20 minutes
Add 9 cups of sugar.
Add some yeast.
Wait a bit.

Drink the beer.

Climb out of the tank.