Quote from: phwest on July 27, 2012, 10:16:57 AM
Great.
Next we take this frictionless piston/cylinder, and we fully submerge it in water.

We position the cylinder vertically with the piston/rod assembly extending downwards and the cylinder on top. We keep some air in the cylinder (air pocket), and we make the whole thing buoyancy neutral. We can do it as a whole unit or we can make the piston rod with an effective density equal to water, so it is buoyancy neutral and we have enough air  inside the cylinder so it exactly counterbalances its own weight. (cylinder also buoyancy neutral)

Furthermore we make the rod same diameter as the piston, and now it looks like a medical syringe (like the old all glass ones â€" keep it still frictionless)
So the whole assembly as it stands now is buoyancy neutral, and has some air in it. And lets say is at level1 inside the water.

Do we agree on the following?
a.   If we compress the assembly at that level1 and let go, there is zero net energy for the cycle same as per previous post.
b.   If we move the assembly from level1 to level2 and back to level1 there is no energy gain or loss. The assembly is weightless inside the water. And actually even moving from level1 to level2 requires no net energy (weightless-buoyancy neutral)
c.   Now, lets do both. Compress at level 1 move (translate) to level 2 decompress at level2 and move back to level 1. Net energy is zero, the translation contributes/subtracts nothing, the compression/decompression cycle is also net zero.

Please don't rush through this step so fast. Compress at level 1 --performing work -- this means your formerly neutrally buoyant assembly will now sink, if you are doing it like a syringe or Cartesian diver, since the total water volume displaced by the assembly decreases but its mass is still the same. The external pressure rises as the assembly sinks. Decompress back to neutral buoyancy at level 2 -- that is, back to the same volume it had at level 1 before compression --- against a higher external pressure --  using only the stored energy in the compressed air? Please explain.

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Agreed?
Let me know yes or no, so we’ll continue next with the simplified  pod/riser model as you asked.

Quote from: TinselKoala on July 27, 2012, 11:28:53 AM
Please don't rush through this step so fast. Compress at level 1 --performing work -- this means your formerly neutrally buoyant assembly will now sink, if you are doing it like a syringe or Cartesian diver, since the total water volume displaced by the assembly decreases but its mass is still the same. The external pressure rises as the assembly sinks. Decompress back to neutral buoyancy at level 2 -- that is, back to the same volume it had at level 1 before compression --- against a higher external pressure --  using only the stored energy in the compressed air? Please explain.

Ok for clarity, none of the scenarios a,b,c show any net energy gain or loss after one cycle. Agreed?

Are you are talking about a?
The cylinder can be held in place at level1 and the piston is compressed and let go as previous post net gain/loss is zero. Same situation at level 2 and so on. I’m only describing compression/decompression after submersion at some depth, (level1) for scenario a.

Scenario b?
We can lock the piston/cylinder relative to each other in place, move from level1 to level2, the air pocket is the same (locked no relative movement) so no net energy gain or loss there either.
The piston/cylinder remains neutrally buoyant throughout the motion. Translation of a buoyancy neutral body within the medium requires no net energy.  Agreed?

When the piston is allowed to move freely relative to the cylinder, then I take it you are talking about c?
Glad you brought it up. It equals out at the end of the cycle. As it moves up the outside pressure to the piston/cylinder is also reduced, so when it expands at the level 2 the expansion is more than level 1. Any added energy during raise up (or sink)  is given back by the “extra” expansion at the top. I left that aspect out to keep it simple. The end result is the same, net gain/loss from level1 to level 2 and back to level1 is zero.

Cartesian diver side note,
True, the Cartesian diver, we can take it down to a certain depth and as the “bubble” collapses it will sink to the bottom. (friction excluded it will keep on accelerating as the buoyancy becomes less and less.)It woluld take an ever increasing force as it sinks lower to stop the fall and reverse up Then unless the liquid is decompressed (air removed vacuum etc) to allow re-expansion of the “bubble” and ascension of the diver. Again just a side note here.

Now we can also take the Cartesian diver at level 1. Let’s say neutrally buoyant. Nudge it up and let it move to level 2. As it moves up it can increasingly lift more weight until we stop it at lets say level2.
To move it back down to 1 is the reverse process (weight substation to stabilize it back to level1) and from 1 to 2 to 1 the net gain/loss is zero,(integral dw*dx for each path and so on) right?

Ok back to the scenarios a,b,c, no net energy gain/loss for the cycle.
Agreed?

Quote from: phwest on July 27, 2012, 12:48:48 PM
Ok for clarity, none of the scenarios a,b,c show any net energy gain or loss after one cycle. Agreed?

For clarity? You are kidding me now, I know.

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Are you are talking about a?
The cylinder can be held in place at level1 and the piston is compressed and let go as previous post net gain/loss is zero. Same situation at level 2 and so on. I’m only describing compression/decompression after submersion at some depth, (level1) for scenario a.

Scenario b?
We can lock the piston/cylinder relative to each other in place, move from level1 to level2, the air pocket is the same (locked no relative movement) so no net energy gain or loss there either.
The piston/cylinder remains neutrally buoyant throughout the motion. Translation of a buoyancy neutral body within the medium requires no net energy.  Agreed?

When the piston is allowed to move freely relative to the cylinder, then I take it you are talking about c?

I am talking about the same cases that you are talking about. Since you started at Level 1 and compressed, in your first setup, then your level 2 must be lower than level 1. Let's stick with that convention please, since we are trying to be clear. Now... is the piston FREE, or is it LOCKED? In case A the piston must be free and frictionless. Right? The work you put in is returned instantly when you stop compressing the piston.  In case C..... you first have the piston FREE, to compress at level 1 (call this C1). Then you must LOCK the piston otherwise it will expand back out to neutral buoyancy. Then you sink to level 2, with the piston LOCKED (C2). This sinking comes "for free" since it is a result of reduced buoyancy; in fact you will have to do work to make it stop sinking at level 2. At level 2, you unlock the piston (C3). What happens? The piston expands until the pressures on both sides are equal.... that is, to lesser volume than it was at level 1 because the external pressure is greater, and you haven't changed the mass of air, just its volume. Now you lock the piston again, but your apparatus is still just a little negatively buoyant, since it is not expanded back to the same volume as it had at level 1 but still masses the same (C4). Hence it will require work to get it back up to level 1. Now, at level 1 again, you can unlock the piston, let the chamber expand, reestablish the correct volume for neutral buoyancy (C1) and  NOW get that work back.
But where did the work come from, that raised the negatively buoyant apparatus from C4 to C1? 

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Glad you brought it up. It equals out at the end of the cycle. As it moves up the outside pressure to the piston/cylinder is also reduced, so when it expands at the level 2 the expansion is more than level 1. Any added energy during raise up (or sink)  is given back by the “extra” expansion at the top. I left that aspect out to keep it simple. The end result is the same, net gain/loss from level1 to level 2 and back to level1 is zero.

Did you not just reverse the order of level 1 and level 2? In your first setup and in my discussion, level 1 is above level 2, is it not? Let's please remain clear about this point.

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Cartesian diver side note,
True, the Cartesian diver, we can take it down to a certain depth and as the “bubble” collapses it will sink to the bottom. (friction excluded it will keep on accelerating as the buoyancy becomes less and less.)It woluld take an ever increasing force as it sinks lower to stop the fall and reverse up Then unless the liquid is decompressed (air removed vacuum etc) to allow re-expansion of the “bubble” and ascension of the diver. Again just a side note here.

Now we can also take the Cartesian diver at level 1. Let’s say neutrally buoyant. Nudge it up and let it move to level 2. As it moves up it can increasingly lift more weight until we stop it at lets say level2.
To move it back down to 1 is the reverse process (weight substation to stabilize it back to level1) and from 1 to 2 to 1 the net gain/loss is zero,(integral dw*dx for each path and so on) right?

Let's point out that the Cartesian diver is a case where the "piston" is free to move, that is, unlocked, so that it is the water pressure only that changes the diver's volume. There is no "Mr. Hand" doing the initial compression of the internal volume, then locking it against expansion regardless of the external water pressure, as there must be in your a-b-c scenario. In your "b" case, the piston could be locked, as you postulate, or free to move, as in the Cartesian diver. But not both, one on the way up and the other on the way down.

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Ok back to the scenarios a,b,c, no net energy gain/loss for the cycle.
Agreed?

Are we agreed? Where did the work come from that raises your negatively buoyant apparatus from C4, piston locked, back up to C1? In order for things to equal out, for you to get this work back at the top, you first have to put it in from the bottom.

Are you going to tell me that this work comes from the other Zed, at the top of its cycle?

Nice job keeping this focused and detailed TK.
I think he's saying that the external input from gravity that is causing the compressed device to sink is countered by another external input ( maybe a wind powered sail... ) bringing it back up.

OR maybe that IF the energy from gravity were to be returned into this perfect model and applied to lift the device that at the end of the cycle all forces would be equal, energy expended = energy captured.

At any rate, again, great detail. Next?
Dale

I took it that @phwest always meant level2 was ABOVE level1.  It was my mistake also at first to assume it was below.

At no point in the examples is the piston assembly moving up or down on its own due to a change in buoyancy I think.  I believe he is stating that it must be transferred between levels 1 and 2 by an applied force (work) and that the net gain once returned to the start condition, again by an applied force (work), is zero.

M.