3 ms·

I'm confused why this technique is better than just compressing air?

schraeds 14y ago
I took a tour of LightSail's operations when I was on a trip to the valley a couple months ago. They had just moved to a building that used to be a firehouse so it had beefed up power systems which they needed to test their products, a lot of neat equipment, and really nice people. From my understanding, they are using compressed air to store the energy but are also using water to cool the system and reclaim the heat energy created during the process.
throwit1979 14y ago
schraeds - You may be interested in knowing that nobody will see your useful answer to the question posed, because you were capriciously hellbanned 218 days ago due to your unpopular opinion of the baby boomer generation.
newman314 14y ago
schraeds, FYI, it appears that you are hellbanned.
Gravityloss 14y ago
Air heats as it compresses. If you cool it, you lose pressure and thus energy. So you would have to keep it hot. It's problematic because it's less dense then and because you'd have to insulate it. Water stores much much more heat per volume and you don't have to handle the pressure if you keep the heat modest. This way your air pressure vessel can be smaller and probably uninsulated.
- D_Alex 14y ago
  As I read it, in the proposed system the heat generated during compression is stored, and then recovered, hence the advantage over the plain old air compression. But I'm going to be that guy who says it will not work, and cannot work, based on fundamental thermodynamic theory. Here goes: 1. The best efficiency of a thermodynamic cycle is 1-TL/TH, where TL is the temperature of heat rejection from the cycle, and TH is the temperature of heat addition to the cycle. TL and TH are ABSOLUTE scale temperatures. 2. In the proposed system, TH is necessarily low, no higher than the temperature generated during compression. Efficient compressors work at low temperatures, usually no higher than 450 K (certain INEFFICIENT compressors, eg gas turbine compressors go as high as 700 K, the mechanical inefficiency gets converted into heat). 3. If <450 K assumption is correct... and if TL is the local ambient temperature (what else?) of around 300 K the MAXIMUM efficiency of the heat recovery cycle is 1-300/450, or about 33%. At least 67% of the heat energy would be lost, leading to a very low overall cycle efficiency, far less than you would get with say pumped water storage, or batteries. Dani Fong, please comment on the above. I would be most happy if there was something wrong in my analysis, and the new technology was a success.
  - DaniFong 14y ago
    Consider a big Carnot cycle. It's 100% reversible -- as efficient as you can get. No losses. And it's 100% reversible no matter what temperature it operates at. The Carnot cycle is: 1. Isothermal compression (T_c) 2. Adiabatic compression (T_c -> T_h) 3. Isothermal expansion (T_h) 4. Adiabatic expansion. (T_h -> T_c) The adiabatic compression and expansion processes are just to get between the two temperatures, T_hot and T_cold. Suppose T_hot = T_cold. Then there's no adiabatic section, and it's just 1. Isothermal compression (T) 2. Isothermal expansion (T) The amount of energy out/energy in is T_h/T_c = 1. This is a 100% efficient energy storage system. It is also our idealized cycle -- a gigantic, single Carnot cycle, storing energy in heat and in low entropy, highly compressed air. Now, the Carnot efficiency, or the efficiency of a heat engine, is a completely different kind of efficiency. It's the TOTAL energy out divided by the heat in. This is (W_out-W_in)/Q_in In our case, the Carnot efficiency is zero, even though the cycle is reversible, and the energy storage process is completely reversible. Interesting, no?
    - D_Alex 14y ago
      Interesting indeed, and thanks for the reply. I almost understand it, unfortunately I just returned from our Melbourne Cup party, where lots of wine was consumed. For now, I concede that your concept seems theoretically possible, best of luck!

I took a tour of LightSail's operations when I was on a trip to the valley a couple months ago. They had just moved to a building that used to be a firehouse so it had beefed up power systems which they needed to test their products, a lot of neat equipment, and really nice people. From my understanding, they are using compressed air to store the energy but are also using water to cool the system and reclaim the heat energy created during the process.

throwit1979 14y ago

schraeds - You may be interested in knowing that nobody will see your useful answer to the question posed, because you were capriciously hellbanned 218 days ago due to your unpopular opinion of the baby boomer generation.

newman314 14y ago

schraeds, FYI, it appears that you are hellbanned.

Gravityloss 14y ago

Air heats as it compresses. If you cool it, you lose pressure and thus energy. So you would have to keep it hot. It's problematic because it's less dense then and because you'd have to insulate it. Water stores much much more heat per volume and you don't have to handle the pressure if you keep the heat modest. This way your air pressure vessel can be smaller and probably uninsulated.

D_Alex 14y ago
As I read it, in the proposed system the heat generated during compression is stored, and then recovered, hence the advantage over the plain old air compression. But I'm going to be that guy who says it will not work, and cannot work, based on fundamental thermodynamic theory. Here goes: 1. The best efficiency of a thermodynamic cycle is 1-TL/TH, where TL is the temperature of heat rejection from the cycle, and TH is the temperature of heat addition to the cycle. TL and TH are ABSOLUTE scale temperatures. 2. In the proposed system, TH is necessarily low, no higher than the temperature generated during compression. Efficient compressors work at low temperatures, usually no higher than 450 K (certain INEFFICIENT compressors, eg gas turbine compressors go as high as 700 K, the mechanical inefficiency gets converted into heat). 3. If <450 K assumption is correct... and if TL is the local ambient temperature (what else?) of around 300 K the MAXIMUM efficiency of the heat recovery cycle is 1-300/450, or about 33%. At least 67% of the heat energy would be lost, leading to a very low overall cycle efficiency, far less than you would get with say pumped water storage, or batteries. Dani Fong, please comment on the above. I would be most happy if there was something wrong in my analysis, and the new technology was a success.
- DaniFong 14y ago
  Consider a big Carnot cycle. It's 100% reversible -- as efficient as you can get. No losses. And it's 100% reversible no matter what temperature it operates at. The Carnot cycle is: 1. Isothermal compression (T_c) 2. Adiabatic compression (T_c -> T_h) 3. Isothermal expansion (T_h) 4. Adiabatic expansion. (T_h -> T_c) The adiabatic compression and expansion processes are just to get between the two temperatures, T_hot and T_cold. Suppose T_hot = T_cold. Then there's no adiabatic section, and it's just 1. Isothermal compression (T) 2. Isothermal expansion (T) The amount of energy out/energy in is T_h/T_c = 1. This is a 100% efficient energy storage system. It is also our idealized cycle -- a gigantic, single Carnot cycle, storing energy in heat and in low entropy, highly compressed air. Now, the Carnot efficiency, or the efficiency of a heat engine, is a completely different kind of efficiency. It's the TOTAL energy out divided by the heat in. This is (W_out-W_in)/Q_in In our case, the Carnot efficiency is zero, even though the cycle is reversible, and the energy storage process is completely reversible. Interesting, no?
  - D_Alex 14y ago
    Interesting indeed, and thanks for the reply. I almost understand it, unfortunately I just returned from our Melbourne Cup party, where lots of wine was consumed. For now, I concede that your concept seems theoretically possible, best of luck!