Why the potential energy is unlimited?

[Zhang Yalin]

Do you mean the following ideas:?
1. In the device
W=G(water)H(10m) ; W(negative)= 2G(water)h(1m)
2. The result is just the same by using a ball or a balloon.
W= G(water)H(10m) ; W(negative)= G(water)H(10m)
If so, the first one is correct, the second is wrong.

[TinselKoala]

Perhaps we are encountering a language barrier.
I certainly do not understand how my questions are answered, or even addressed, by your cryptic post, Yalin.
Maybe we need an interpreter...

I ask you again, very plainly:

How do you intend to deflate or withdraw the balloon, which must inevitably raise up an equal volume of water?
How do you overcome the suction of the water column, that I have very clearly demonstrated for you in my video?

Yes, allowing the balloon to ascend in the column "releases" energy which can do work. But it takes exactly this much energy to reset the system to its initial state, and there isn't any left over. Remember, you have to raise that same volume of water back up, that the balloon displaced downward in its rise to the top.
No matter how you deflate or withdraw the balloon, this fact remains.

[Zhang Yalin]

Maybe my device will not do, but the theory is correct. The way to use the potential energy without doing work will be found in the future.

[FreeEnergy]

what if you have the basin sealed as well? wouldn't that eliminate leaks?

[TinselKoala]

No, Yalin, your theory is not correct, sorry. There will not be found any way to use potential energy without doing work, and potential energy isn't unlimited.
And, FreeEnergy, what does sealing the basin accomplish, except make it harder to introduce the balloon? Leaks don't keep this system from working. What keeps it from working is the mass of the water that must be raised back up to reset the system to its initial state (a requirement for any kind of cycle). Sealing the basin won't affect that.