Thought experiment on buoyant forces on a system

[y2jayy]

I'm a former physics student and I've been thinking about an interesting problem that eventually led me to the following thought experiment that I'm having trouble resolving.

Imagine a two-compartment system, where one compartment is filled with He gas, and the other is filled with standard atmospheric gas. Both compartments have the same volume, both are cubes, and the heights of both compartments are small enough so that changes in g are negligible between the top and bottom.

The second compartment with the atmospheric gas has

(1) a tiny valve that can be opened and closed located on the face (recall the compartment is cubic) farthest away from the other compartment containing the He gas

Also,
(2) the walls of the second compartment can be fully folded in and collapsed (and attached to the first compartment containing the He gas) so that the total volume of the system in its collapsed/folded state is only non-negligibly greater than the volume of just the first compartment filled with the He gas.

Initially, the (1) valve is closed, and (2) the walls are not collapsed/folded in, and the relative ratio of He to atmospheric gas is such that the system begins to rise because the buoyant force exceeds the weight of the system.

Let's consider two scenarios.
In scenario #1, only the valve is opened. Recalling that the valve is extremely small, with itswidth <<<< size of the container, does the buoyant force get halved (the two compartments are of equal volume, as stated in the beginning) even though the valve's size is <<< size of container? I would think no, since the hole is too small for the second compartment to suddenly "not count" toward the total volume of the air displaced by the overall structure. But on the flip side, I tend to think yes, since if you made the hole incrementally larger until it became as large as the entire side of the face it's situated on, the second compartment with the now giant hole couldn't really be said to be displacing the surrounding air anymore.

In scenario #2, the walls of the second compartment are collapsed and folded in, and then attached to the second compartment so that the total volume is basically halved. This would in turn halve the mass and hence weight of the air displaced by the structure, thus halving the buoyant force.

Making the hole larger and larger, as in scenario #1, seems to make scenario #1 effectively indistinguishable from scenario #2. In scenario #2, the buoyant force is clearly (I think?) halved. How large does the hole have to be to eliminate that compartment from contributing to the buoyant force? Is this reduction of buoyancy discontinuous or continuous?

What errors have I made in this thought experiment given the assumptions stated?

 

[Charles Link]

One thing you need to consider as part of the weight of the system is the weight of the gas that is in the containers. I think you may have forgotten to include that. It doesn't matter which compartment this gas is in, or even what the temperature or pressure is inside the containers. The total weight of the gas inside the system does not change. ## \\ ## The buoyant force is caused by volumes that are assumed to be sealed. The volume displaces the outside air.## \\ ## I had a little trouble following your logic of putting a hole in the container, and trying to then compute a buoyant force. When you make a hole in the second container, the air inside the container is no longer part of the weight of the system. There is no air displaced by the second container in that case either. For the case where you just start making a hole, particularly with inside pressure that could be different from the outside pressure, then the problem becomes a dynamic one with pressure differentials and the forces would be more difficult to compute.

 

[y2jayy]

One thing you need to consider as part of the weight of the system is the weight of the gas that is in the containers. I think you may have forgotten to include that. It doesn't matter which compartment this gas is in, or even what the temperature or pressure is inside the containers. The total weight of the gas inside the system does not change. ## \\ ## The buoyant force is caused by volumes that are assumed to be sealed. The volume displaces the outside air.## \\ ## I had a little trouble following your logic of putting a hole in the container, and trying to then compute a buoyant force. When you make a hole in the second container, the air inside the container is no longer part of the weight of the system. There is no air displaced by the second container in that case either. For the case where you just start making a hole, particularly with inside pressure that could be different from the outside pressure, then the problem becomes a dynamic one with pressure differentials and the forces would be more difficult to compute.

Would the net change be zero, then, with the decrease in the buoyant force caused by the reduction of the air displaced being exactly offset by the decrease in the weight of the system?

 

[Charles Link]

Would the net change be zero, then, with the decrease in the buoyant force caused by the reduction of the air displaced being exactly offset by the decrease in the weight of the system?

Yes, if the pressures are equal outside and inside. If not, it gets much more complicated.

 

[y2jayy]

If the pressures were equal outside and inside. If not, it gets much more complicated.

Okay, got it. In my original problem that motivated the thought experiment, I was trying to get the buoyancy to oscillate. I think the following approach would work, but I'm trying to find a way to achieve this without a periodic input of energy. In this situation, the hole isn't being made.

Imagine there's a square piston (with negligible thickness and mass) inside the second container that can be depressed a max distance of half the side length of the cubical compartment. Let's say that I press down on the piston by this much, and then collapse down/fold the remaining half-volume of the second container so that the volume of the second compartment is reduced by half. This would keep the mass of the system constant since no gas is being expelled. The volume of this second compartment would be halved as described above, and hence the volume of the overall system would be reduced by a quarter. When the piston bounces back to its original position, assume that the compartment is reconfigured into its original cubical shape. The mass again would remain the same while the volume would be increased back to its original amount.

In this scenario, the buoyancy oscillates in a cycle, decreasing as the volume of the structure decreases, and vice versa. But there needs to be a periodic input of energy to depress the piston each cycle (is this correct?)

Is there a way to get the buoyancy to oscillate, assuming a frictionless system, without having to add energy into the system on a periodic basis while maintaining the same amplitude (in this case, decreasing by a quarter each cycle and then returning to its original value)?

 

[Charles Link]

You could simply attach a piston to a spring system, without any gas in the cylinder. Dissipative forces would damp the system rather quickly though.

 

[y2jayy]

Thanks. Can you think of any other systems that would have minimal dissipative forces given the constraints of the problem?

 

[LURCH]

NASA was considering a similar contraption for a rover on Mars. It would be an airborne rover that flies to a new location every day. Their solution was to use expansion and contraction of a balloon caused by temperature changes at sunrise and sunset. That is almost like not having an external energy supply to change buoyancy.