Buoyancy/Gravity Concept - How to insert buoyant weight into water column?

In summary, the conversation discusses a concept for renewable energy using buoyant objects in a water column. The question is how to insert the object with the least amount of energy. The idea is to seal the top and bottom of the water chamber to insert and release the object. However, it is pointed out that this would violate the law of conservation of energy. Other related concepts and ideas are also mentioned.
  • #1
LValfre
5
0
Hello all,

I devised a concept a couple years ago for renewable energy and hit a stumbling block: How do you insert a buoyant object into the bottom of a water column with the smallest amount of energy?

Recently I revisited the concept and found some similar work online including this youtube video.


From my understanding, this concept is flawed because he's pumping the water in some fashion which uses energy. Otherwise the water would spill out of the bottom side opening.

I've attached an image to help visualize my question.

My question is this: Like holding your hand on a straw to hold water in place when lifted out of the water ... Can you seal the top of the water chamber (Valve A) to insert an object into the unsealed bottom of the water column (Valve B)? Likewise, can you seal the bottom of the chamber (Valve B) and unseal the top chamber (Valve A) to allow the object to float to the top?

Any thoughts are appreciated.
 

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  • #2
Since this would appear to give you an endless supply of energy, the answer has to be no.
 
  • #3
I don't know about endless but it would be a start. I'm trying to sort out whether there is any way to insert a buoyant object at the bottom of a container of water. Gravity would of course have to supply the energy ... but if we can use buoyancy to raise our objects to that height in an efficient manner than we can maybe get close to it.

Either way I'd like to learn the physics behind the question and whether sealing the valves would allow such a process to be feasible.
 
  • #4
When you insert the object at B you need to do work. Work to raise the water level if valve A is open, or work compressing the air pocket if valve A is closed. In either case the work you do pushing the object into the column would be less than the work you can extract by letting the object float to the top. TANSTAFL
 
  • #5
That's been my stumbling block ... finding a way to push the object in with less work than it takes to get it to the top. If the object was not buoyant could it be dropped into B granted valve A is sealed? Say a rock or a metal object.

I'm under the impression if Valve A is sealed it won't rush out of an open Valve B. This is following the hand over the straw example where you pull the straw out and the water doesn't flow because the top is sealed creating a suction.

Appreciate the comments.
 
  • #6
ModusPwnd said:
When you insert object B you need to do work. Work to raise the water level if valve A is open, or work compressing the air pocket if valve A is closed. In either case the work you do pushing object B into the column would be less than the work you can extract by letting the object float to the top. TANSTAFL

If valve A is closed, you don't need to compress anything to get the object in. You simply need to allow the water level at point B to rise when you put it in.

However, here's the problem I see with the idea in the first post: When you insert the object at point B, you'll displace a certain mass of water. After you allow the object to float to point A, the object has replaced the water at the top of the column and the water would then be located down at point B. I made an illustration to help visualize what I'm describing:
CT7WG.png


This mass of water that ends up at point B is what allows the object to float in the first place. In order to pull the object out of the tube at point A, you must replace the water at the top of the tube again, reversing the process that allowed the object to float to the top. Since you're moving the water back to the same energy state it started at, you would need to put in more energy to remove the object from the top than any machine could produce from the motion of the object.
 
  • #7
Ahh I see the flaw with this.

Originally I had envisioned an insertion chamber underneath the water column where the object could be inserted and then the water could be 'dropped' on it. The same issue would rise though ... how to get the displaced water to the top of the column again without wasting more energy than you gain from gravitational potential energy.

I want to experiment with it but I can't solve whether it's even possible and all odds are against it. If the weight reached the top and made it out of the column would the GPE ever be enough to surpass the energy needed to 'pump' the water back to the top?

Outside of this question, here's a man in India's idea that is roughly similar.
http://powerfromgravity.com/invention.aspx

Thought it was an interesting approach but it's lax in details so I can only speculate the majority of what's going on.

Thanks again for the insights guys!
 
  • #8
LValfre said:
If the weight reached the top and made it out of the column would the GPE ever be enough to surpass the energy needed to 'pump' the water back to the top?
No. This is a clear contradiction of the law of conservation of energy. And we don't discuss perpetual motion machines here. Thread locked.
 

1. What is buoyancy and how does it work?

Buoyancy is the upward force exerted by a fluid on an object immersed in it. It is caused by the difference in pressure between the top and bottom of the object, with the higher pressure at the bottom pushing up on the object. This force is equal to the weight of the fluid that the object displaces, and it allows objects to float or sink in a fluid.

2. How do you calculate buoyant force?

The buoyant force is calculated by multiplying the density of the fluid by the volume of the displaced fluid and the acceleration due to gravity. This can be represented by the equation Fb = ρVg, where Fb is the buoyant force, ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

3. How does the density of an object affect its buoyancy?

The density of an object plays a crucial role in its buoyancy. Objects with a higher density than the fluid they are in will sink, as they displace less fluid and therefore experience a lower buoyant force. On the other hand, objects with a lower density than the fluid will float, as they displace more fluid and experience a higher buoyant force.

4. Can an object be buoyant in one fluid but not in another?

Yes, an object's buoyancy can vary depending on the fluid it is placed in. For example, a metal object may sink in water but float in mercury due to the difference in densities of these fluids. This is why some objects, such as boats, are designed to be buoyant in one specific type of fluid.

5. How can you use buoyant weights to control the position of an object in a water column?

Buoyant weights can be used to control the position of an object in a water column by adjusting the weight and volume of the weights. By adding or removing weight, the buoyant force acting on the object can be adjusted, allowing it to float at a specific depth in the water column. Additionally, changing the shape or volume of the weights can also affect the position of the object in the water column.

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