6 ms·

> The temperature of a natural gas flame is 2770C. Using it to heat a house to 30C is ridiculous. Why?

cjfd 3y ago
Entropy. Delta S = Delta Q / T. An efficient use of energy produces as little entropy as possible. Here the loss is Delta Q / 303 - Delta Q / 3043. T in Kelvin.
kragen 3y ago
the carnot limit for 2770°/0° is 1 - 273/3043 = 91% the carnot limit for 30°/0° is 1 - 273/303 = 9.9% so if your gas flame is really 2770° (which i doubt, the adiabatic flame temperature for methane in air is only 1963° and actual flames are even colder than that) you're wasting 90% of your exergy by diluting it down to 30° also tho i do not want to go to your house if you heat it to 30°
- pi-e-sigma 3y ago
  Carnot limit has nothing to do with heating efficiency, only with converting heat into mechanical work
  - kragen 3y ago
    incorrect, the carnot cycle is reversible (like everything in physics up to cp reversal, except for entropy increase), so it also has to do with converting mechanical work into heat (differences) specifically, in this case, if you convert a hypothetical 2770° flame into mechanical work in a hypothetical heat engine capable of withstanding it, you can use that mechanical work to drive a heat pump to pump ten times as much heat into the house to heat it up to 30°. or more, if the temperature outside is higher than 0°
    - pi-e-sigma 3y ago
      very correct, because we are talking about using high temperature flame as somehow wasteful when used for heating. How does converting heat into mechanical work even enter equation?
      - kragen 3y ago
        sorry, i was still editing as you replied
        pi-e-sigma 3y ago
        Even in this revised scenario it's not clear if you were able to get ahead. Because you are converting heat to mechanical work and back to heat in the heat pump so you are subject to Carnot limit _twice_
        kragen 3y ago
        yes, but the second time it works backwards: for each joule of mechanical work you put in, you pump 10.1 joules of heat. this is called the 'coefficient of performance' of a heat pump and it is almost always more than 1, even with the losses present in any real system. values of 2 or 3 are typical, though some mass-produced systems reach 6. sometimes this is quoted as an 'efficiency' of 200% or 300% or 600% which is of course impossible, but it's as if you had 200% efficiency. that's why using heat pumps saves energy
- out_of_protocol 3y ago
  You get exactly 100% efficiency (as heating) at any temperature, since anything will be converted to heat anyway. Well, except in real systems you have exhaust temperature for CO2 etc going out of house, the lower the temperature the lower your heat loses. If you add some internal channels to chimney, you can absorb most of the heat back, making exhaust temperature very low therefore have low efficiency loss
  - kragen 3y ago
    yes, that's correct, but heat pumps do much better than that, as explained downthread
hasmanean 3y ago
Because if you have a natural gas flame in air (lets say 2000C, my previous number was wrong as someone pointed out), then you can do useful work with it. A carnot heat engine with a hot side of 2000C(Th=2273K) and a cold side of 200C(Tc=473K), has an efficiency of 1-Tc/Th or 79%. So you can do useful work (hopefully inside the house, where it too will eventually end up as waste heat) AND recover energy from the exhaust gas of the carnot engine--which is 200C! Imagine burning natural gas in your furnace, vs using your natural gas to run a electricity generator and mining bitcoin with it inside your bedroom. A Bitcoin rig makes a great space heater.
- pi-e-sigma 3y ago
  Or you can just burn it at 2000C and convert all that heat into... well, heating. The initial claim was about using 2000C flame for heating as somehow wasteful, which it clear is not.

cjfd 3y ago

Entropy. Delta S = Delta Q / T. An efficient use of energy produces as little entropy as possible. Here the loss is Delta Q / 303 - Delta Q / 3043. T in Kelvin.

kragen 3y ago

the carnot limit for 2770°/0° is 1 - 273/3043 = 91% the carnot limit for 30°/0° is 1 - 273/303 = 9.9% so if your gas flame is really 2770° (which i doubt, the adiabatic flame temperature for methane in air is only 1963° and actual flames are even colder than that) you're wasting 90% of your exergy by diluting it down to 30° also tho i do not want to go to your house if you heat it to 30°

pi-e-sigma 3y ago
Carnot limit has nothing to do with heating efficiency, only with converting heat into mechanical work
- kragen 3y ago
  incorrect, the carnot cycle is reversible (like everything in physics up to cp reversal, except for entropy increase), so it also has to do with converting mechanical work into heat (differences) specifically, in this case, if you convert a hypothetical 2770° flame into mechanical work in a hypothetical heat engine capable of withstanding it, you can use that mechanical work to drive a heat pump to pump ten times as much heat into the house to heat it up to 30°. or more, if the temperature outside is higher than 0°
  - pi-e-sigma 3y ago
    very correct, because we are talking about using high temperature flame as somehow wasteful when used for heating. How does converting heat into mechanical work even enter equation?
    - kragen 3y ago
      sorry, i was still editing as you replied
      - pi-e-sigma 3y ago
        Even in this revised scenario it's not clear if you were able to get ahead. Because you are converting heat to mechanical work and back to heat in the heat pump so you are subject to Carnot limit _twice_
        kragen 3y ago
        yes, but the second time it works backwards: for each joule of mechanical work you put in, you pump 10.1 joules of heat. this is called the 'coefficient of performance' of a heat pump and it is almost always more than 1, even with the losses present in any real system. values of 2 or 3 are typical, though some mass-produced systems reach 6. sometimes this is quoted as an 'efficiency' of 200% or 300% or 600% which is of course impossible, but it's as if you had 200% efficiency. that's why using heat pumps saves energy
out_of_protocol 3y ago
You get exactly 100% efficiency (as heating) at any temperature, since anything will be converted to heat anyway. Well, except in real systems you have exhaust temperature for CO2 etc going out of house, the lower the temperature the lower your heat loses. If you add some internal channels to chimney, you can absorb most of the heat back, making exhaust temperature very low therefore have low efficiency loss
- kragen 3y ago
  yes, that's correct, but heat pumps do much better than that, as explained downthread

hasmanean 3y ago

Because if you have a natural gas flame in air (lets say 2000C, my previous number was wrong as someone pointed out), then you can do useful work with it. A carnot heat engine with a hot side of 2000C(Th=2273K) and a cold side of 200C(Tc=473K), has an efficiency of 1-Tc/Th or 79%. So you can do useful work (hopefully inside the house, where it too will eventually end up as waste heat) AND recover energy from the exhaust gas of the carnot engine--which is 200C! Imagine burning natural gas in your furnace, vs using your natural gas to run a electricity generator and mining bitcoin with it inside your bedroom. A Bitcoin rig makes a great space heater.

pi-e-sigma 3y ago
Or you can just burn it at 2000C and convert all that heat into... well, heating. The initial claim was about using 2000C flame for heating as somehow wasteful, which it clear is not.