# Is it plausible to generate electricity through buoyancy?

#### cabcabby

Is it plausible to use the buoyancy of air in a controlled column of water to turn a drive shaft and, by extension, generate electricity through a low RPM/high torque turbine? See attached PDF for possible configuration. Only energy input would be to compress air tanks located below the water column.

Just wondering if there is a glaring reason why this wouldn't work. Also wondering if there are some formulas for figuring out what kind of RPMs can be generated with the given setup.

Any insight into this idea would be appreciated.

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#### brewnog

Gold Member
Yes the system would work, but you won't get any net energy output. It will take more energy to pump and compress your air than you'll get out from your generator.

#### cabcabby

What if the height of the water column were raised exponentially? Would there come a point where there'd be a net energy output?

#### Mech_Engineer

Gold Member
You can only have so much water before the size/height of the power plant becomes insurmountable.

#### russ_watters

Mentor
What if the height of the water column were raised exponentially? Would there come a point where there'd be a net energy output?
No. The work out is always theoretically limited to exactly the work input in a system like the one you describe. But then after you start factoring in effeciency losses, you end up with less than the work input.

Throw some number into it and do some math: calculate the energy required to expel water from a 55 gallon drum 300 feet under water and compare it to the work you get out of it when it rises.

Anyway, buoyancy can be used for power generation if the input work comes from somewhere else - such waves or the tides.

#### russ_watters

Mentor
Well, I was bored, so I did it myself:

f*d: 458*300=137,445 ft-lb
p*v: 62.4lb/ft^3*300ft*7.3ft^3=137,966 ft-lb

Equal, to within a few decimal places. And notice that 62.4*7.3 is equal to 458. The variable you mentioned, depth, is the same in both equations, so the equations always produce the same result.

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#### cabcabby

Russ -- Thanks for taking the time to consider this idea.

You had mentioned in your previous post that there would be energy required to expel the water at 300 ft., and Brewnog mentioned something about energy required to pump the air. Perhaps I need to better explain the system.

The columns of water are raised above ground, perhaps in concrete reinforced pens. Beneath these raised water columns would be a volume of pressurized air, with timed nozzles at the base of the water column.

As the rising vessel nearly reaches the surface, its lid is mechanically opened, releasing the air. At that same instant, the lid of the lowering vessel is mechanically closed, and the air nozzle below it is activated, filling the vessel with air, naturally expelling the water. This vessel begins to rise, and its now emptied twin vessel begins to sink, its lid still open to prevent drag. The elements of the system could also be made of a buoyant-neutral material, such as a hardened plastic, and using a high-salinity liquid would also increase the efficiency of the system.

The only energy input required in this system is the energy required to pressurize the air in the tanks beneath the water columns. All other functions within the system could be handled completely through mechanical means. I was even thinking that the energy required to pressurize the tanks could come from solar energy, creating a truly green system. But of course, the ideal is a closed system.

So I guess the real question becomes how much energy is required to compress X amount of air, and how much energy would be produced by running X through such a system.

The “free” energy in this system comes from air’s natural buoyancy. By utilizing this property in a raised column of water, its full force is being tapped. And since the whole system is above ground, no energy is needed to pump energy to depth.

Maybe this helps make more sense of things.

#### Danger

Gold Member
It explains your idea more clearly, but it still won't work as a closed system. You can't get something for nothing.

#### russ_watters

Mentor
The columns of water are raised above ground, perhaps in concrete reinforced pens.
A compressor and a pump are the same thing - raising that coulmn of water requires energy.

#### cabcabby

In the sense that it would require energy to construct the whole system, yes, raising the water would require energy, as would filling the water column. But, like the construction of anything, that would be a one time energy investment.

Think of the water column as a very large glass of water on stilts.

#### Danger

Gold Member
You still have to put more energy into compressing the air than you can recover from releasing it.

#### cabcabby

I guess that's what I'm interested in understanding. How much energy does it take to compress the air ... 360 cu. ft. of air into a 1.76 cu. ft. tank? ... that's at 3000 psi.

That's enough to cycle the system 25 times (360 cu.ft./14.66 cu.ft.), exerting 6,898,300 ft-lb of total work, according to russ watters earlier calculations. And how much energy does 6,989,300 ft-lb generate when run through a turbine?

#### wxrocks

I am thinking the same thing -- I'd love to invest in this compressor that doesn't generate any heat or noise or anything and have every bit of energy go into compressing the air. *rolling eyes*

And cabby, if you used solar energy -- why wouldn't you just pump that electricity into the grid?? Why put it through extra cycles of compressing air and actuating valves -- reducing the amount of electricity you would get out.

Another thing to think of is the work done on the water. As the bubbles would pass through the system, the friction would generate heat. At some point your water would boil or increase the pressure or evaporate. then you'd have to invest energy into lifting more water up into your system.

I might suggest you look up some basics on thermodynamics. There is this thing called entropy that should explain why there is no perfect system.

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