How "efficient" is electron transport as an energy medium?

[PlasmaBomb]

Originally posted by: Jeff7

Originally posted by: So

Originally posted by: sao123
I think his point was why arent we collecting these high energy photons and convert them to a useable form, thus being slightly more efficient. After all, its only the infrared radiation being used to express heat. We should be able to collect the full spectrum of radiation inside a reactor as such and use it.

We *are* collecting these photons! They collide with the moderator and the vessel and heat it up!

<bangs head on desk>

Google was not fruitful; I'm likely not searching for the right terms.

About how much energy is released by uranium that's just undergoing radioactive decay? The closest I could find was info on plutonium decay in RTEGs, but the power figures given were of the electrical output, after the thermal energy worked its way through the thermopiles.


I'd still guess that it's almost nothing when compared to what you get from the fission reaction.

Your googlefu is weak old man...

 

[SunnyD]

Originally posted by: PlasmaBomb
People need to do basic research...
wiki

Instantaneously released energy...............MeV
Kinetic energy of fission fragments...............169.1
Kinetic energy of prompt neutrons...................4.8
Energy carried by prompt ?-rays.....................7.0

Energy from decaying fission products
Energy of ß--particles....................................6.5
Energy of anti-neutrinos..................................8.8
Energy of delayed ?-rays................................6.3
Sum.........................................................202.5




Energy converted into heat in an operating thermal nuclear reactor 202.5 MeV

Efficiency at producing heat in an operating nuclear reactor = (Used energy/Released energy)*100

(202.5/202.5)*100 = 100% at generating heat.

Gee how can we get more than 100% efficiency :roll:

The losses occur elsewhere, where efficiency is limited by Carnot's Rule, which has been linked before.

See, I am infinitely skeptical that all 202.5 MeV of the reactions products are directly converted into heat even under our best technology and conditions. I understand losses occur elsewhere (which is a big part of my original question), but there will also be losses due to non-reacting radiation, etc.

Let's get away from the nuclear reactor example anyway and move back to the question at hand, and observe Carnot's rule as the main basis of what I'm talking about. It observes heat as the principle form of energy - and thus thermodynamics indicates there will essentially always be a loss due to the fact that we have to convert energy mediums (principally heat).

Going back to the reactor - okay, say I grant you all 202.5MeV of energy is being transmitted as heat. You are still going to incur a loss (minimal) transporting the energy to the turbine, and a further loss converting that energy into electrical energy via the turbine. My conjecture is then how much of the raw energy output (yes, I said it again) of that 202.5MeV actually is converted into electricity? A quick search of Google indicates we're at best around 50%. So even if we're talking about 100% efficiency of the transfer of energy to heat, we still suck at converting it to electricity.

And with that, I got sidetracked at work and lost my train of thought. God I hate it when that happens!

 

[Fenixgoon]

Originally posted by: SunnyD

Originally posted by: PlasmaBomb
People need to do basic research...
wiki

Instantaneously released energy...............MeV
Kinetic energy of fission fragments...............169.1
Kinetic energy of prompt neutrons...................4.8
Energy carried by prompt ?-rays.....................7.0

Energy from decaying fission products
Energy of ß--particles....................................6.5
Energy of anti-neutrinos..................................8.8
Energy of delayed ?-rays................................6.3
Sum.........................................................202.5




Energy converted into heat in an operating thermal nuclear reactor 202.5 MeV

Efficiency at producing heat in an operating nuclear reactor = (Used energy/Released energy)*100

(202.5/202.5)*100 = 100% at generating heat.

Gee how can we get more than 100% efficiency :roll:

The losses occur elsewhere, where efficiency is limited by Carnot's Rule, which has been linked before.

See, I am infinitely skeptical that all 202.5 MeV of the reactions products are directly converted into heat even under our best technology and conditions. I understand losses occur elsewhere (which is a big part of my original question), but there will also be losses due to non-reacting radiation, etc.

Let's get away from the nuclear reactor example anyway and move back to the question at hand, and observe Carnot's rule as the main basis of what I'm talking about. It observes heat as the principle form of energy - and thus thermodynamics indicates there will essentially always be a loss due to the fact that we have to convert energy mediums (principally heat).

Going back to the reactor - okay, say I grant you all 202.5MeV of energy is being transmitted as heat. You are still going to incur a loss (minimal) transporting the energy to the turbine, and a further loss converting that energy into electrical energy via the turbine. My conjecture is then how much of the raw energy output (yes, I said it again) of that 202.5MeV actually is converted into electricity? A quick search of Google indicates we're at best around 50%. So even if we're talking about 100% efficiency of the transfer of energy to heat, we still suck at converting it to electricity.

And with that, I got sidetracked at work and lost my train of thought. God I hate it when that happens!

you mean how much *heat* is converted to electricity?

large power plants are around 50%.

for what its worth, energy can take the following forms

kinetic energy (motion)
potential energy (gravity, usually)
electromagnetic energy (electricity, magnetism, and EM radiation)
(if i missed any, feel free to add, scientists of ATOT)

heat is the most "lossy" form of energy transfer - kind of like a crappy audio encoding
electricity is more like a good rip from a vinyl - some loss, but you get most of what you put in.

 

[BrownTown]

Originally posted by: Analog
Utilities lose about 40% of all generated electricity from the powerplant to the user. So its not really that good. However, if you use superconductors, the losses would be theoretically zero. That doesn't include transformers, interconnects and the like.

this is completely untrue, losses in the transmission grid are less than 10%.



As for the point at hand, it really is something to understand that when all is sand and done only like 25% of the chemical or nuclear energy in a source (oil, coal, natural gas, uranium) actually does useful work whereas 75% is lost. However, it is not nescecarrily fair to assume that there is some way which we could easily do it much better. For example efficiency is in large part determined by how cold your heat sink is and how hot you heat source is. Well, the temperature of your heat sink can't be any colder than what it is outside, and the temperature of your heat source can't be any hotter than the temperature at which the materials sued would lose their strength. Just those two factors kill you alot. Also there is a factor of cost, being more efficient isn't always better. I work at a nuclear plant and the equipment required to go from like 25% efficiency to 30% probably costs an extra billion dollars. At some point it jsut gets cheaper to burn a little more fuel instead of trying to get into diminishing returns of efficiency. you might note that the power plants with the highest fuel prices (natural gas) are the most efficient, and those with the cheapest fuel (nuclear) are the least efficient.