12 ms·

> At some point in the chain, you need to consume energy from the environment to drive the system. Agreed, that's where spilling water comes into play:

> At some point in the chain, you need to consume energy from the environment to drive the system. Agreed, that's where spilling water comes into play: ┌───┐ ┃ A ┊┃ │ ┗━━━━━┛ │ │ │┃┈┈┈┈┈┈┈┈┃ └┨ B ┃ ┗━━━━━┳━━┛ ┇ ┃ C ┇ ┃ ┗━━━━━━┛ Here, B->A is driven by B->C spillage. The critical point here is that B is a low quality source, but plentiful. rocqua's point was likely similar, if you replace height with thermal energy. You can't use the energy you put in A to move it from B to A, because it costs you at least as much to move it up. However, we're not bringing it down to B. We're bringing it down to C (ambient 21°C in their example). There will be losses, but shouldn't it be doable, depending on the B-C potential? Wouldn't Peltier modules be 100% efficient in theory for rising B to A (as long as thermal losses go to the hot side, that is: tey keep a side cooler), for instance? If so, the generated energy is the one stored between A and C, minus conversion losses, minus A-B. That is, B-C minus conversions losses from A-C. Probably not very efficient at the end, though worth it if you have plenty of B (and a excellent "C" sink).

jjk166 6y ago
Draining B into C is consuming energy from the environment. You most certainly can drain some of the water from B to C so as to move other water from B to A. But you will always do so at a net loss. The unit of water moved from B=60 to A=61 gains A-B=1 unit of potential energy, in addition to the B-C=40 units of potential energy it already had for a total of A-C=41 units. However, without a 100% efficient (read impossible) pump and turbine combo, it takes more than 1 unit of energy to raise up that unit of mass. This means you need to spill more than 1/40 units of water from B to C to get that 1 unit from B to A. Now dropping that water from A to C would get you 41 units of energy, 1 more than if you had spilled it straight from B to C, but it cost you more than 1 unit of energy to get into this situation. A is indeed a higher quality source than B, but you are using that low quality source of B to generate your energy, and the efficiency losses there will always be greater than the gains you make on the higher one. To break even the initial height/temperature differential would have to be infinitely high. Now if for some reason you could not harvest energy directly from B-C, yes you could use water going from A to C to power the pump from B to A, it would just be less efficient. This would be the equivalent of a siphon.
- MayeulC 6y ago
  I think we agree on everything, thank you for expanding your reasoning. > Now if for some reason you could not harvest energy directly from B-C, yes you could use water going from A to C to power the pump from B to A, it would just be less efficient. I also think that was the initial point: some higher-efficiency generators ("steam turbine") are only available for high potential differentials. Whether raising the potential offsets the gains likely depends on the specific setup and energy source.

jjk166 6y ago

Draining B into C is consuming energy from the environment. You most certainly can drain some of the water from B to C so as to move other water from B to A. But you will always do so at a net loss. The unit of water moved from B=60 to A=61 gains A-B=1 unit of potential energy, in addition to the B-C=40 units of potential energy it already had for a total of A-C=41 units. However, without a 100% efficient (read impossible) pump and turbine combo, it takes more than 1 unit of energy to raise up that unit of mass. This means you need to spill more than 1/40 units of water from B to C to get that 1 unit from B to A. Now dropping that water from A to C would get you 41 units of energy, 1 more than if you had spilled it straight from B to C, but it cost you more than 1 unit of energy to get into this situation. A is indeed a higher quality source than B, but you are using that low quality source of B to generate your energy, and the efficiency losses there will always be greater than the gains you make on the higher one. To break even the initial height/temperature differential would have to be infinitely high. Now if for some reason you could not harvest energy directly from B-C, yes you could use water going from A to C to power the pump from B to A, it would just be less efficient. This would be the equivalent of a siphon.

MayeulC 6y ago
I think we agree on everything, thank you for expanding your reasoning. > Now if for some reason you could not harvest energy directly from B-C, yes you could use water going from A to C to power the pump from B to A, it would just be less efficient. I also think that was the initial point: some higher-efficiency generators ("steam turbine") are only available for high potential differentials. Whether raising the potential offsets the gains likely depends on the specific setup and energy source.