Water column object insertion and energy recapture

[HOTTUBTHOMPSON25]

TL;DR

This is just a thought experiment-

What is the difference in energy expenditure for inserting a given bouyant object into the bottom of a vertical water column vs energy produced harnessing the object during ascending the water column.

Given optimized variables for object weight to volume ratio and accounting for losses- how much energy would be lost in that procedure? Would it be less loss than just traditionally hoisting the object?

I am just a layman without any physics background, so I am basically looking for some knowledgeable input.

 

[phinds]

Well, one way to think about it is this: suppose you have a hollow sphere with a thin but strong shell and it's filled with air. It floats easily on water. Now you tie to it a weight that is just enough to make it sink to the bottom. Obviously, it cannot raise that weight or it would not have gone to the bottom, BUT ... it very likely would not take very much weight removal for the balloon to be able to make it back to the top. SO ... my conclusion, without doing any math, is that yes, there will definitely be a loss in the system, but not a huge amount.

 

[russ_watters]

This sounds very much like like the common class of perpetual motion machines that try to harness buoyancy. Simplify the scenario and it becomes easy to see why it doesnt work/why the energy is the same:

Consider a cylinder filled with water 10 units high and a float with a diameter equal to the cylinder except for a negligible gap to let water flow around it and a height of one unit. In order to get the float to the bottom of the cylinder you have to lift the entire column of water by one unit. Whether you put the float in at the bottom or push it down from the top, the end result is the same: the float is at the bottom and the column of water is now 11 units high(including the float).

What people get wrong is trying to think of a mechanism that can let the float in at the bottom with less energy. There isn't one due to the result being the same, but people fantasize that maybe if they could just make it complicated enough they can avoid that result. And then what happens is they make the device so complicated they can't correctly analyze it anymore. And that's where they find perpetual motion.

Note: it is against policy to discuss perpetual motion here but I will leave the thread open for a bit to see if it comes to an easy conclusion.

 

[HOTTUBTHOMPSON25]

Sorry, just to clarify the setup, the object would be outside (not in) the water column and would be forcefully inserted at the very bottom of the water column.

So the question is the amount of force needed to insert the object into the column compared to the energy produced by the buoyant object once in the column.

 

[HOTTUBTHOMPSON25]

Sorry, just to clarify the setup, the object would be outside (not in) the water column and would be forcefully inserted at the very bottom of the water column.

So the question is the amount of force needed to insert the object into the column compared to the energy produced by the buoyant object once in the column.

I know energy will be lost, but how much roughly? Less than if the object is hoisted against gravity?

 

[Ibix]

A simple realisation of this is to build an airlock at the bottom of the column. Push the object in, close the air-side door, open the water-side door, push the object out and let it float up. The energy usage then comes in pumping the water out of the lock. That involves pushing water back in to the column against the water pressure, which (as @russ_watters noted) costs at least as much as lifting the original object; how much more depends on how close to neutrally buoyant the object was and how efficient your pumps are.

You can make this work passively by just dumping water from the airlock at the bottom and refilling it from the top as long as there's a water source at the top. In that case, the energy source is whatever lifts the water to the top - in other words, indirect solar usually.

A standard counterweight lift system would be easier, needing a lot less plumbing.

 

[russ_watters]

I know energy will be lost, but how much roughly? Less than if the object is hoisted against gravity?

What do you mean "hoisted"? Aren't we talking about the energy to push the float to the bottom vs what you get back by letting it float back up? In the ideal case they are exactly the same no matter how you get the float to the bottom of the column.

 

[HOTTUBTHOMPSON25]

Sorry bear with my ignorance.

I was just asking of there would be a difference in energy expenditure in lifting an object through air against gravity vs forcefully inserting that same object (buoyant) into the bottom of a column of water and somehow mechanically recapturing some lost energy through bouyancy as it lifts through the water

 

[Ibix]

Sorry bear with my ignorance.

I was just asking of there would be a difference in energy expenditure in lifting an object through air against gravity vs forcefully inserting that same object (buoyant) into the bottom of a column of water and somehow mechanically recapturing some lost energy through bouyancy as it lifts through the water

Ideally, no, but it depends on the density of the object. It's much less efficient if the density of the object is much different from that of water. It's also a lot messier.

 

[russ_watters]

Sorry bear with my ignorance.

I was just asking of there would be a difference in energy expenditure in lifting an object through air against gravity vs forcefully inserting that same object (buoyant) into the bottom of a column of water and somehow mechanically recapturing some lost energy through bouyancy as it lifts through the water

You're saying:

Case 1: Lift an object in the air.

Case 2: Insert an object into water and recover energy when it floats back up.

Is that right? Case 1 costs energy. Case 2 is ideally zero sum: it neither costs nor gains you energy.

But I don't see a relationship between these cases.

 

[Ibix]

Is that right? Case 1 costs energy. Case 2 is ideally zero sum: it neither costs nor gains you energy.

No, OP says case 2 is about inserting the object at the bottom of the water column. That costs energy to lift the water, ideally the same as case 1.

It's theoretically free to insert the thing at the top, sink it, and recover the energy as it floats back up - that's equivalent to a pair of weights joined by a rope over a pulley.

 

[russ_watters]

No, OP says case 2 is about inserting the object at the bottom of the water column.

It's all one (run-on)sentence: inserting it into the bottom and then recovering energy when it floats back up.

OP needs to be clearer...

 

[Nugatory]

I was just asking of there would be a difference in energy expenditure in lifting an object through air against gravity vs forcefully inserting that same object (buoyant) into the bottom of a column of water and somehow mechanically recapturing some lost energy through bouyancy as it lifts through the water

Ideally (no friction, no mechanical inefficiencies anywhere, 100% recapture) inserting the object into the bottom of the column and letting it float to the top while capturing the energy from the buoyant force pushing it up nets out to the exact same amount of energy as lifting the object in air and setting it on top of the water.

Inserting the object at the bottom requires lifting an equivalent volume of water against gravity, as described by @russ_watters in #3 above - basically we are converting the mechanical energy used to push the object in and the water up into potential energy in the form of a higher water column. When we release the object we get that energy back as the water comes back down, less the energy required to lift the object to its final height. So our net energy expenditure is exactly what was required to the lift the object to its final height.

 

[Baluncore]

Any rotary valve system, that is used to insert an object into the base of the column, will waste an equal or greater volume of high pressure water, with each cycle of operation. That water must be returned, or replaced, at the same or a greater cost in energy, as is gained when a buoyant object rises to the surface.

 

[Lnewqban]

Sorry, just to clarify the setup, the object would be outside (not in) the water column and would be forcefully inserted at the very bottom of the water column.

Welcome, @HOTTUBTHOMPSON25 !

The action of inserting the object is equivalent to pumping a volume of water to a higher level in the column.
That displaced volume of water is exactly the same than the volume of the object.

Therefore, the amount of work or energy that is required for inserting any object into the bottom of a vertical water column is equal to amount of energy required to pump that mass of water to that new height.

Please, see:
https://www.engineeringtoolbox.com/mechanical-energy-equation-d_614.html

Even without considering loses due to friction, the amount of work that the object can return while moving up to the surface will always be less than the amount of work required for inserting the object into the bottom of the water column.

The reason is that some volume of water must remain pumped up in order to keep the object neutrally floating.
That volume of water will weigh exactly as much as the weight of the object was when out of the water.

 

[sophiecentaur]

The action of inserting the object is equivalent to pumping a volume of water to a higher level in the column.
That displaced volume of water is exactly the same than the volume of the object.

Yes!! Archimedes sussed out that floating problems all involve displacement. Whatever the net density of the 'object', when it's been 'inserted' in the bottom of the column it will displace its own volume of water. The work involved will be its volume times the density of water times the depth of water in the tank. That's the same as the work needed to 'pull it down' from the surface on a string.If it floats to the top, it will displace its weight of water. As @Lnewqban says,the change in depth of water will depend on the surface area of the tank so to calculate the work. The h will become h+v/A.
If it is dense enough, it will stay on the bottom.