# Can anyone help me figure out the net work/energy of this system?

I was never terribly good at physics (although I wish I was), and I had this thought experiment. I was wondering if anyone could help me out with it. It goes as follows:

We have a square vertical shaft 3 m by 3 m and 100 m tall filled with water. At the base on one side there is a door. This door is attached to a basin that is 3 m by 3 m and 3 m tall. The top of this basin is covered by an identical door. Assuming the doors can withstand the pressure, either door could be open without the tank draining so long as at least one door is always closed. The adjacent basin will be full of water.

Suppose I run a line up the inside of the water tank out the top and back down the outside to the basin, through both doors and connected it to itself. I then attach standard steel drums full of isopentane or perhaps some safer liquid with low density. The drums are spaced every 5 m along the line. When the drums are in the water tank, they float up. When they are outside the water tank, they fall down. Both motions pull the line in the same direction. The water lock on the basin should allow the drums to pass from air to water. Assuming I pump the excess water (the drums will displace water when they enter the basin of water meaning each time the lock is used, the basin gains a drum’s volume of water) how well does this work if I use the movement of the line to generate electricity?

Like all perpetual motion ideas, this won’t work. The energy needed to insert the drum into the bottom of the water column is at least as large as the energy you get from the drum falling down from the top of the water column. In your specific scenario, this energy comes from pumping the water back from the basin into the shaft - to do this, you must overcome the water pressure at the bottom of the shaft.

p.s. To elaborate, once the “basin” is filled, the 100 m shaft is filled up to 97 m level. To pump the water back from the basin to the shaft, you are essentially lifting the 97 m column of water up by 3 m to squeeze the water from the basin at the bottom of the shaft. Or you can think of it as lifting the 3x3x3 m of water up by 97 m and filling the 3 m gap at the top. Either way, the energy you need is 3 m x 3 m x 3 m x 97 m x 1000 kg/m^3 x 9.8 m/s/s.

You get the most energy from the “barrel” if it fits the basin exactly (i.e. 3x3x3 m), and is only fractionally lighter than water. Even then, the energy you get is 3 m x 3 m x 3 m x 97m x 1000 kg/m^3 * 9.8 m/s/s.

(p.s. I was about to refer to this “basin” as an “airlock for water” and then realized that would just be a “lock”. I’d never though about where the term “airlock” came from…)

Alternately, you could fairly easily dunk the barrel into the lower basin, if you didn’t mind that water would overflow out of the top and spill onto the ground. Then when you close the outer door, open the inner door, and cycle the barrel into the tall column, water from the column would flow into the lower basin, and the water level in the column would drop. This process would repeat every time you cycled a barrel through, and each time, you would lose a barrel’s volume of water. You could extract energy from the system, but only until you ran out of water: The energy you are extracting is the potential energy from having a 100-m-high column full of water.

The reason your barrels float up in the first place is because an equivalent volume of water, which weighs more, is falling down to take the place of the barrel. So when you re-insert the barrel into the base of the column, you have to push that water back up again, to make room for the barrel. By definition you will get less energy out of the barrel falling in air than it will take to push the water back up the same distance to re-insert the barrel – because the barrel weighs less than the amount of water it displaces, by design.

The flaw is very common to perpetual motion machines. You tend to forget the energy cost of restoring your machine to its initial state.