5 ms·
The water's not there to directly store gravitational potential energy, it's there to provide a variable sized containment vessel under nearly constant pressure
by taneq 3y ago
The water's not there to directly store gravitational potential energy, it's there to provide a variable sized containment vessel under nearly constant pressure conditions. This has two advantages:
1) You're not losing thermal energy from the main body of air (assuming their thermal recovery at the compressor works well enough) because the change in air pressure only occurs at the compressor.
2) You get constant pressure and constant power output for the entire volume of your compressed air storage, instead of both dropping rapidly as air is released. This gives you far better bang for buck per unit volume of pressurized air.
I haven't done the calcs but I'd guess that if they're bothering doing all this extra stuff the energy stored in a cubic meter of compressed air at these pressures is significantly higher than the energy you'd get from just lowering a cubic meter of water down to the reservoir.
- usrusr 3y agoCompressed air and water head are in a balance, my physics instincts strongly suggest that the hypothetical sum is 2x the amount of energy stored by raising the water alone. But just like you I do hope that my instincts are missing something! (see nephew comment https://news.ycombinator.com/item?id=40273030 https://news.ycombinator.com/item?id=40273030 )
- taneq 3y agoHmm.. the more I think about it the more I agree with your intuition. If the air's at constant pressure then the work it's doing as it exits the chamber is coming directly from the water entering the chamber. There's no difference (in terms of energy) between having a turbine at the bottom of the column of water vs. having the water push the air and then the air spin a turbine. The whole thing seems like they started with "let's do compressed air energy storage" and realised half way through that the best way to do compressed air energy storage is for it to actually be pumped hydro.
- usrusr 3y agoIt's more capacity than just pumped hydro of identical head x volume: even if we ignore the "compressed spring" of the air involved, charging would require at least the energy for lifting the water plus the energy for heating up their heat source. If you close a valve in the water connection, then pop open the air pressure vessel letting all energy still in the compressed gas you to waste, you'd still have the heat storage, and you could let the water rush into the lower reservoir through a turbine, harvesting the entire energy contained in the water (minus inefficiency). So it's definitely more, the heat and the gas discharge we just wasted. The question is how much more, whether the symmetry suggested by the balanced state implies 2x energy storage or not. Another mental model to further separate the two storage media (water lift and air compression) is this: you could charge separately, first raising the water with a pump, with the dry cavity at ambient pressure via an open connection to daylight, then seal the cavity an charge it as a compression vessel in the non-isobaric way. I still can't decide between trusting the "balance implies 2x" intuition or not, but it should be easy enough to calculate. At least the thought experiments show that it can't be 1x the energy stored by lifting water alone. I consider starting with "let's do compressed air energy storage" as a given, the "do it by pumped hydro" does not come into play because pressure vessels are hard, it comes into play because building compressors and turbines that have good efficiency over a wide range of pressure is hard.