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Very close
- the only real difference in my idea was separate systems for generation and return - but I think your way is probably better. I don't think we need the side and bottom conveyors - those could be replaced with a slide for the shuttles to follow and there should be enough kinetic energy to slide it right into the airlock in one single drop. So when it goes over the edge, the airlock should be open and ready to receive it.
We could calculate the gravitational potential energy involved and see if a gain can be produced for a single bottle path.
When the airlock is flooded, the Gravitational Potential Energy (EGP) of the shuttle becomes negative according to the displacement weight of the water minus the EGP of the bottle in air at that height (using EGP = mgh where g = 9.81).
I just weighed a 2L bottle full of air @ 300 feet above sea level. Here it weighs 57g. So that is the mass of the bottle. The same bottle filled with tap water weighs 2182g. I am neglecting the thickness of the bottle wall which will undoubtedly displace some volume itself - so these calculations will be on the conservative side.
So our displacement weight will be 2182 - 57 = 2125g
The overall height of a 2L bottle is exactly 12 inches or 0.3048m.
The diameter is 4 inches or 0.1016m
Let us suppose we have a tank that is 2.5m deep. And let us suppose it stands on a support frame 0.5m tall. So all together, the overall height is 3m tall. When our shuttle (a 2L bottle as described) is at the top of the tank, floating on the water it will have a center height of 3.1016m minus that amount that is submerged which will equate to the 57g of water displaced. For ease of calculations and to be on the conservative side we can reduce the height to 3.1m
On the outside of the tank then, our shuttle would have a EGPEGP of the water that replaces the shuttle has that value. Where do we get the 0.6524 value from? That is the height off the ground, 0.5m plus the center of the bottle height in the airlock which is (0.3048 / 2). So that is the mass center of the shuttle at rest in the airlock. I used a negative in this case for the -2.125 to indicate that the force vector on the shuttle was away from the Earth. From these calculations, we would expect the shuttle to obtain a kinetic energy of 51.02J - 1.73J = 49.29J by the time it reaches the surface. If this were true, we would have plenty of energy left to open the interior gate and move the shuttle over the edge of the tank.
But there is a serious issue here - drag:Use of a Drag Coefficient to Calculate Drag Force due to Fluid Flow past an Immersed Solid
We would have to accurately model the drag on the shuttle as it moves through the water. This would no doubt lead us to apply methods of reducing that drag, which by far is greatly associated with waters adhesive characteristics. We would want our shuttle to be a slippery as possible in the fluid and the same with the submerged conveyor. Otherwise, the greatest amount of energy would be expended in friction with the water resulting in heating the water with nothing left for us to use.
This calculation used EGP as the means of evaluating the energy involved - but the more common approach would be to use fluid dynamics and pressures. It would be interesting to compare the two results and see if mistakes have been made in the above calculations and if so where.
- the only real difference in my idea was separate systems for generation and return - but I think your way is probably better. I don't think we need the side and bottom conveyors - those could be replaced with a slide for the shuttles to follow and there should be enough kinetic energy to slide it right into the airlock in one single drop. So when it goes over the edge, the airlock should be open and ready to receive it.We could calculate the gravitational potential energy involved and see if a gain can be produced for a single bottle path.
When the airlock is flooded, the Gravitational Potential Energy (EGP) of the shuttle becomes negative according to the displacement weight of the water minus the EGP of the bottle in air at that height (using EGP = mgh where g = 9.81).
I just weighed a 2L bottle full of air @ 300 feet above sea level. Here it weighs 57g. So that is the mass of the bottle. The same bottle filled with tap water weighs 2182g. I am neglecting the thickness of the bottle wall which will undoubtedly displace some volume itself - so these calculations will be on the conservative side.
So our displacement weight will be 2182 - 57 = 2125g
The overall height of a 2L bottle is exactly 12 inches or 0.3048m.
The diameter is 4 inches or 0.1016m
Let us suppose we have a tank that is 2.5m deep. And let us suppose it stands on a support frame 0.5m tall. So all together, the overall height is 3m tall. When our shuttle (a 2L bottle as described) is at the top of the tank, floating on the water it will have a center height of 3.1016m minus that amount that is submerged which will equate to the 57g of water displaced. For ease of calculations and to be on the conservative side we can reduce the height to 3.1m
On the outside of the tank then, our shuttle would have a EGPEGP of the water that replaces the shuttle has that value. Where do we get the 0.6524 value from? That is the height off the ground, 0.5m plus the center of the bottle height in the airlock which is (0.3048 / 2). So that is the mass center of the shuttle at rest in the airlock. I used a negative in this case for the -2.125 to indicate that the force vector on the shuttle was away from the Earth. From these calculations, we would expect the shuttle to obtain a kinetic energy of 51.02J - 1.73J = 49.29J by the time it reaches the surface. If this were true, we would have plenty of energy left to open the interior gate and move the shuttle over the edge of the tank.
But there is a serious issue here - drag:Use of a Drag Coefficient to Calculate Drag Force due to Fluid Flow past an Immersed Solid
We would have to accurately model the drag on the shuttle as it moves through the water. This would no doubt lead us to apply methods of reducing that drag, which by far is greatly associated with waters adhesive characteristics. We would want our shuttle to be a slippery as possible in the fluid and the same with the submerged conveyor. Otherwise, the greatest amount of energy would be expended in friction with the water resulting in heating the water with nothing left for us to use.
This calculation used EGP as the means of evaluating the energy involved - but the more common approach would be to use fluid dynamics and pressures. It would be interesting to compare the two results and see if mistakes have been made in the above calculations and if so where.
. I just didn't think that through well enough 
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