6 ms·

Source temperature of waste heat is an important factor. Heat radiating from a 2000C source is a lot more valuable & useful than that from some warm water at 60

Source temperature of waste heat is an important factor. Heat radiating from a 2000C source is a lot more valuable & useful than that from some warm water at 60C. You can't use that heat to raise the temperature of something to 61C, for example.

rocqua 6y ago
But you could use a heat pump and 60C water to raise the temperature of something to 61C. And would take a lot less energy to run that heat-pump than to actually heat that something to 61C. What I've become interested in, is how easy it is to 'abuse' heat pumps like this. Could you take a heat-pump and this 60C water to heat a boiler to the point you can run a steam-turbine that produces more energy than it takes to run the heat-pump? If not at 60C water, does this work at any temperature? There remains free energy in the temperature differential between an ambient 21C and the temperature of warmer water. It seems to me this should be harvest able. There are significant limits here from the Carnot theory, but they don't say this is impossible.
- jjk166 6y ago
  No that explicitly violates the second law of thermodynamics. A heat pump is like a water pump: it's a lot easier to move water to a tank 61 feet off the ground from another tank 60 feet off the ground than from ground level, but you can't power the pump producing 1 foot of pressure head with a turbine being driven by 1 foot of pressure head.
  - JoeAltmaier 6y ago
    Well, not without losing some water in the process. If you spill 10 gallons of water through the pump to pump 1 upward, sure that works.
    - MayeulC 6y ago
      And this is the exact idea behind ram pumps: https://en.wikipedia.org/wiki/Hydraulic_ram https://en.wikipedia.org/wiki/Hydraulic_ram
    - jjk166 6y ago
      Well in this case you need to pump 10 gallons up to get those 10 gallons to spill down. At some point in the chain, you need to consume energy from the environment to drive the system.
      - MayeulC 6y ago
        > 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.
  - rocqua 6y ago
    > Not if the total free energy in the system goes down right? The heat pump requires power to run, the question is whether the power to run the heap pump is smaller than the power required to just generate the same heat differential directly.
    - jjk166 6y ago
      No, the question is whether the power to run the heat pump is smaller than the additional power you'd gain by increasing the differential, which is always no. You would always get more energy out of draining the whole 60 foot water tank directly to ground than you would by lifting the water to 61 feet and then draining it to ground.
  - rocqua 6y ago
    So, say you have a tank at 61 feet, one at 60 feet, and a desire to move the water up. Just let some of the water out of the 60 foot tank through a turbine. Put a shaft on that turbine, and put a pump on the other end. That pump should be able to pump water from the 60 foot tank up to the 61 foot tank. You would get less water in the 61 foot tank than in the 60 foot tank. Quite a bit less actually, but you could still harness the energy in the 60 ft tank to get some water to the 61 ft tank.
    - jjk166 6y ago
      That is an external energy input. No one is saying it's impossible to lift water up 1 foot (that's incredibly easy), the problem is it always takes more energy to lift water 1 foot than you can ever get from lowering it 1 foot. Likewise a heat pump can easily and efficiently create a heat reservoir, but you need to drain some other, larger heat reservoir to do so.
      - rocqua 6y ago
        Yes, we are explicitly losing free energy from this. Thing is, the free energy in 60C wastewater is hard to use, and the free energy in 200C steam is much easier to use (through steam-turbines). Hence, it might be worth it to sacrifice a large part of the free energy in the 60C wastewater to get it to a useable form. What are the alternatives to use 60C wastewater to generate electricity? I can think of a sterling engine or a thermoelectric material. These would probably be more efficient, but they are new technologies, with practical limitations and development limitations. Hence, if my idea could generate any form of electricity, it might be more feasible than the other ideas. From what I know about heat engines, I know it might not be possible to generate electricity this way.

rocqua 6y ago

But you could use a heat pump and 60C water to raise the temperature of something to 61C. And would take a lot less energy to run that heat-pump than to actually heat that something to 61C. What I've become interested in, is how easy it is to 'abuse' heat pumps like this. Could you take a heat-pump and this 60C water to heat a boiler to the point you can run a steam-turbine that produces more energy than it takes to run the heat-pump? If not at 60C water, does this work at any temperature? There remains free energy in the temperature differential between an ambient 21C and the temperature of warmer water. It seems to me this should be harvest able. There are significant limits here from the Carnot theory, but they don't say this is impossible.

jjk166 6y ago
No that explicitly violates the second law of thermodynamics. A heat pump is like a water pump: it's a lot easier to move water to a tank 61 feet off the ground from another tank 60 feet off the ground than from ground level, but you can't power the pump producing 1 foot of pressure head with a turbine being driven by 1 foot of pressure head.
- JoeAltmaier 6y ago
  Well, not without losing some water in the process. If you spill 10 gallons of water through the pump to pump 1 upward, sure that works.
  - MayeulC 6y ago
    And this is the exact idea behind ram pumps: https://en.wikipedia.org/wiki/Hydraulic_ram https://en.wikipedia.org/wiki/Hydraulic_ram
  - jjk166 6y ago
    Well in this case you need to pump 10 gallons up to get those 10 gallons to spill down. At some point in the chain, you need to consume energy from the environment to drive the system.
    - MayeulC 6y ago
      > 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.
- rocqua 6y ago
  > Not if the total free energy in the system goes down right? The heat pump requires power to run, the question is whether the power to run the heap pump is smaller than the power required to just generate the same heat differential directly.
  - jjk166 6y ago
    No, the question is whether the power to run the heat pump is smaller than the additional power you'd gain by increasing the differential, which is always no. You would always get more energy out of draining the whole 60 foot water tank directly to ground than you would by lifting the water to 61 feet and then draining it to ground.
- rocqua 6y ago
  So, say you have a tank at 61 feet, one at 60 feet, and a desire to move the water up. Just let some of the water out of the 60 foot tank through a turbine. Put a shaft on that turbine, and put a pump on the other end. That pump should be able to pump water from the 60 foot tank up to the 61 foot tank. You would get less water in the 61 foot tank than in the 60 foot tank. Quite a bit less actually, but you could still harness the energy in the 60 ft tank to get some water to the 61 ft tank.
  - jjk166 6y ago
    That is an external energy input. No one is saying it's impossible to lift water up 1 foot (that's incredibly easy), the problem is it always takes more energy to lift water 1 foot than you can ever get from lowering it 1 foot. Likewise a heat pump can easily and efficiently create a heat reservoir, but you need to drain some other, larger heat reservoir to do so.
    - rocqua 6y ago
      Yes, we are explicitly losing free energy from this. Thing is, the free energy in 60C wastewater is hard to use, and the free energy in 200C steam is much easier to use (through steam-turbines). Hence, it might be worth it to sacrifice a large part of the free energy in the 60C wastewater to get it to a useable form. What are the alternatives to use 60C wastewater to generate electricity? I can think of a sterling engine or a thermoelectric material. These would probably be more efficient, but they are new technologies, with practical limitations and development limitations. Hence, if my idea could generate any form of electricity, it might be more feasible than the other ideas. From what I know about heat engines, I know it might not be possible to generate electricity this way.