5 ms·

Also I wonder if, using similar principles to a heat pump's operation, you could still get juice out of stored heat at lower temperatures. Surely you can have t

Also I wonder if, using similar principles to a heat pump's operation, you could still get juice out of stored heat at lower temperatures. Surely you can have this hot graphite sitting at under 2000°C, heat some fluid/gas to say 1000°C, and then compress the gas to increase its temperature? Surely that would be the ideal solution anyway, since you don't want your hot graphite to become a chunk of useless heat simply for dropping below temp briefly. On the topic of heat pumps, you could also use a TPV for geothermal power. Since there are no moving parts and presumably no huge steam engine installation, it would be more feasible to have one of these in your back yard. The grid powers a heat pump, you compress the fluid till it hits 2000°C, and your TPV extracts power. The heat pump itself is >100% efficient, so overall you can steal a fair bit of electricity from the ground. Right?

ok_computer 5y ago
I just want to point out that compressing gas takes a non-zero amount of work. Compressing gas to increase temperature to use temperature to make energy would be better spent using the compressing work directly to make electricity.
Szpadel 5y ago
I'm not sure if there is any way to make heat pump operating in this scale of temperatures, and keep in mind that current heat pumps can only "pump" over about 50°C max until it's no longer profitable.
- elil17 5y ago
  That’s not really true at all. Home heat pumps are stuck there because of the refrigerants used. Ammonia heat pumps can go way hotter (though still nowhere close to those temps - more like 150c iirc), it’s just that ammonia requires safety controls that are impractical on a small scale.
qsdf38100 5y ago
The best theoretical efficiency of a heat pump is 1/CarnotEfficency. So the best you can do is to have a round trip efficiency of 1. See Carnot batteries. So using a heat pump to increase the temperature so that a thermal engine is more efficient can’t possibly increase the overall efficiency.
- mcbishop 5y ago
  Heat pumps are different than heat engines... I don't think Carnot efficiency applies. A heat pump can move several units of heat energy per input unit of electrical energy (thanks to awesome compressor / refrigerant / phase-change technology). ...They're >100% efficient.
  - hinkley 5y ago
    > The best theoretical efficiency of a heat pump is 1/CarnotEfficency. The facts have already been stated that refute your what if. Whether you missed it or not, you're arguing for a perpetual motion machine. The Layman's version of the 3 Laws of Thermodynamics are: 1) You can't win 2) You can't cheat 3) You can't even choose not to play
  - qsdf38100 5y ago
    Heat pumps really are reversed heat engines. For instance, take a Brayton thermodynamic cycle. It's motor cycle, but you can literally just reverse the arrows on this cycle, it becomes a reversed Brayton cycle, which corresponds to a heat pump (or a refrigeration machine, depending on which inlet of the machine you're interested in). In practice, the reversing of the arrow means the reversing of the fluid flow, which means compression processes are now expansion processes and vice versa. Carnot theory does still apply. The COP (Coefficient Of Performance) of a Carnot heat pump is 1/n, where n is the usual Carnot efficiency of a heat engine. With n=1/3, you get a COP of 1/1/3 = 3 = 300% Carnot efficiencies consider ideal machines. So ideal Carnot batteries have a n * 1/n = 100% efficiencies. It means you can't increase the efficiency of a heat engine by putting a heat pump in front of it. In an ideal case (ignoring all losses and inefficiencies), you would get the same efficiency. In practice, you can only be worse.
bumby 5y ago
>The heat pump itself is >100% efficient Are you sure you aren't confusing the coefficient of performance with the efficiency? If a heat pump was >100% efficient, I could cool my house by leaving my refrigerator door open.
- hnuser123456 5y ago
  If you pipe the heat from the compressor directly outside instead of venting it into the same room, that would work.
  - bumby 5y ago
    Correct, that's what refrigerated air conditioners do. But that's really just playing a trick with the system boundary in the problem; it doesn't mean the global efficiency has exceeded 100%.
    - cormacrelf 5y ago
      Yes, it's an imprecise term.
      - bumby 4y ago
        To be clear, I don't think the term is imprecise. Efficiency is always (work out)/(work in). The numbers get wonky when we aren't fully accounting for the "work in". In the case of refrigeration, we have to account for the work it takes to compress the fluid and the resulting waste heat.

ok_computer 5y ago

I just want to point out that compressing gas takes a non-zero amount of work. Compressing gas to increase temperature to use temperature to make energy would be better spent using the compressing work directly to make electricity.

Szpadel 5y ago

I'm not sure if there is any way to make heat pump operating in this scale of temperatures, and keep in mind that current heat pumps can only "pump" over about 50°C max until it's no longer profitable.

elil17 5y ago
That’s not really true at all. Home heat pumps are stuck there because of the refrigerants used. Ammonia heat pumps can go way hotter (though still nowhere close to those temps - more like 150c iirc), it’s just that ammonia requires safety controls that are impractical on a small scale.

qsdf38100 5y ago

The best theoretical efficiency of a heat pump is 1/CarnotEfficency. So the best you can do is to have a round trip efficiency of 1. See Carnot batteries. So using a heat pump to increase the temperature so that a thermal engine is more efficient can’t possibly increase the overall efficiency.

mcbishop 5y ago
Heat pumps are different than heat engines... I don't think Carnot efficiency applies. A heat pump can move several units of heat energy per input unit of electrical energy (thanks to awesome compressor / refrigerant / phase-change technology). ...They're >100% efficient.
- hinkley 5y ago
  > The best theoretical efficiency of a heat pump is 1/CarnotEfficency. The facts have already been stated that refute your what if. Whether you missed it or not, you're arguing for a perpetual motion machine. The Layman's version of the 3 Laws of Thermodynamics are: 1) You can't win 2) You can't cheat 3) You can't even choose not to play
- qsdf38100 5y ago
  Heat pumps really are reversed heat engines. For instance, take a Brayton thermodynamic cycle. It's motor cycle, but you can literally just reverse the arrows on this cycle, it becomes a reversed Brayton cycle, which corresponds to a heat pump (or a refrigeration machine, depending on which inlet of the machine you're interested in). In practice, the reversing of the arrow means the reversing of the fluid flow, which means compression processes are now expansion processes and vice versa. Carnot theory does still apply. The COP (Coefficient Of Performance) of a Carnot heat pump is 1/n, where n is the usual Carnot efficiency of a heat engine. With n=1/3, you get a COP of 1/1/3 = 3 = 300% Carnot efficiencies consider ideal machines. So ideal Carnot batteries have a n * 1/n = 100% efficiencies. It means you can't increase the efficiency of a heat engine by putting a heat pump in front of it. In an ideal case (ignoring all losses and inefficiencies), you would get the same efficiency. In practice, you can only be worse.

bumby 5y ago

>The heat pump itself is >100% efficient Are you sure you aren't confusing the coefficient of performance with the efficiency? If a heat pump was >100% efficient, I could cool my house by leaving my refrigerator door open.

hnuser123456 5y ago
If you pipe the heat from the compressor directly outside instead of venting it into the same room, that would work.
- bumby 5y ago
  Correct, that's what refrigerated air conditioners do. But that's really just playing a trick with the system boundary in the problem; it doesn't mean the global efficiency has exceeded 100%.
  - cormacrelf 5y ago
    Yes, it's an imprecise term.
    - bumby 4y ago
      To be clear, I don't think the term is imprecise. Efficiency is always (work out)/(work in). The numbers get wonky when we aren't fully accounting for the "work in". In the case of refrigeration, we have to account for the work it takes to compress the fluid and the resulting waste heat.