PIDEC: Nuclear-Photoelectric Powerplant

[Morbius]

And why would the coolant be hotter than the spheres? Just because of gamma-heating?
I would expect the coolant to be cooler than the spheres, and thus to absorb their heat by conduction.
Furthermore, the small size of the spheres should also improve the radiative heat transfer as well.
I don't see why the coolant in a nuclear lightbulb would be so much hotter than the coolant in a regular thermal reactor, that its temperature should pose a problem.

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sanman,

First - I didn't say that the coolant would be hotter than the spheres. I was attempting to take your
argument that because the spheres only received 20% of the energy deposition that they would be
cooler than the coolant - to its logical conclusion. I'm glad you now agree that the spheres will be
hotter than the ooolant.

The coolant in the nuclear "lightbulb" doesn't have to be hottter than a regular reactor - the PVs
wouldn't work very well at normal reactor temperatures. Typical PWR coolant outlet temperatures
are in excess of 300 C; and PVs essentially don't work above temperatures of 200 C.

So even normal reactor temperatures are too high for PVs.

Dr. Gregory Greenman
Physicist

 

[sanman]

You're essentially saying that radiative energy transfer and collection cannot escape the effects of thermal heat transfer and collection. I don't see that this is inherently the case.

The smaller the fuel elements are and th efaster the coolant flow is, then the lower the temperature under which conductive heat transfer takes place. The more robust the PV (eg. diamond) the more heat it can withstand.

 

[Morbius]

You're essentially saying that radiative energy transfer and collection cannot escape the effects of thermal heat transfer and collection. I don't see that this is inherently the case.

The smaller the fuel elements are and th efaster the coolant flow is, then the lower the temperature under which conductive heat transfer takes place. The more robust the PV (eg. diamond) the more heat it can withstand.

sanman,

I don't see why you are having such a difficult time with this.

There are essentially two issues here. One is which way the heat flows. The other is how fast
it flows. Your arguments about surface area, how conductive the material is - only affect the latter -
how fast the heat flows.

The former is determined by the relative temperatures. The spheres have a heat source in them and
only one way for the heat to get out - conduction to the coolant. It doesn't matter what the
strength of the heat source is in the spheres. Because the spheres have a heat source - they need
to be cooled - and there has to be a path for heat to get out. Because the only path for heat to escape
is to the coolant - the spheres HAVE to be higher in temperature than the coolant.

QED.

Both conduction and radiative transfer require that the spheres be hotter than the material to which
they are transferring heat.

You can NOT say - "The heat transfer path is efficient with low heat transfer resistance; therefore
the temperature will be low." That does NOT follow. If there is only one path for heat to flow out -
then the temperature of the spheres HAS TO BE GREATER than the material to which it is giving
up its heat - the coolant. Heat doesn't flow "uphill".

The fact that the heat transfer will be efficient because of the high surface area only means that the
temperature of the spheres will only be slightly higher than the coolant. But the temperature of the
spheres will be higher than the coolant, none the less.

If the spheres are to be cooled by the coolant - then they have to be hotter than the coolant; and the
spheres need to have a mechanism to reject heat because they have an internal source - it doesn't
matter how small. [ The magnitude of the heat source will only determine the magnitude of the
temperature differential needed to drive the heat transfer - but it will not affect the sign { plus or minus ]
of the temperature difference. ]

Even if your PVs are only slightly hotter than the reactor coolant - they are essentially too hot to work.

As I said before; there's no way to turn this sow's ear into a silk purse.

Dr. Gregory Greenman
Physicist

 

[Morbius]

The smaller the fuel elements are and th efaster the coolant flow is, then the lower the temperature under which conductive heat transfer takes place. The more robust the PV (eg. diamond) the more heat it can withstand.

sanman,

No matter how fast the heat transfer - the spheres can NEVER get cooler than the coolant that is
cooling them.

No matter how fast your car goes; no matter how fast the air flows through the radiator; no matter
how fast the conduction between the tubes of the radiator to the air - the car's coolant will NEVER
get cooler than the air that is dissipating the heat.

Why is this so difficult?

Dr. Gregory Greenman
Physicist

 

[sanman]

I wasn't the one claiming that the spheres will be cooler than the coolant. I was merely saying that if the heat is drawn off fast enough, then this would prevent excessive buildup of heat or high temperatures.

So therefore, I don't see why heat has to necessarily be completely disruptive to the PVs, so as to make their operation impossible or impractical.

 

[Morbius]

I wasn't the one claiming that the spheres will be cooler than the coolant. I was merely saying that if the heat is drawn off fast enough, then this would prevent excessive buildup of heat or high temperatures.

So therefore, I don't see why heat has to necessarily be completely disruptive to the PVs, so as to make their operation impossible or impractical.

sanman,

As I stated before, PVs typically will not work above 200 C; and the reactor coolant will be about 300 C.

Since the coolant is going to be the coolest part [ otherwise it can't cool ] and the coolant is at 300 C;
then your PVs are above 300 C and they don't work above 200 C.

Dr. Gregory Greenman
Physicist

 

[mheslep]

First, as posted above here, common photovoltaics materials such as gallium arsenide are not proposed. Ultraviolet PVs made from, e.g. diamond are proposed. Second, as posted above, only reflective materials need be in contact with the excited coolant. Third, since some kind of heat transfer system is required regardless, the temperature of the coolant depends on it flow rate and and temperature of the down stream heat sink. We can not just say 'its 300C'.

 

[Morbius]

First, as posted above here, common photovoltaics materials such as gallium arsenide are not proposed. Ultraviolet PVs made from, e.g. diamond are proposed. Second, as posted above, only reflective materials need be in contact with the excited coolant.

mheslep,

Doesn't matter whether the material is reflective or not when it comes to heat transfer - if it needs to be
cooled - which it will, it will be in the the gamma field, regardless - it will have to be hotter than the coolant.

All this debate to make an energy conversion system that will be both less efficient and more costly
than the energy conversion systems we have now. Your ultraviolet PVs are not going to match a
conventional Rankine cycle in efficiency.

Again - there's no way to turn this sow's ear into a silk purse.

Dr. Gregory Greenman
Physicist

 

[Astronuc]

I expect that the temperature will be about 300C, based on the fact that even with 40% efficiency, one is going to have to dump 60% of the energy somewhere outside the core, and that will take a primary cooling system. The size of the core will be dictated by requirements of power (and power density) and enrichment/criticality. The more non-fuel stuff (fluorescent medium, PV, conductor, insulator) that is added to the system besides the fuel (U/Pu compound), cladding (isolates fuel from coolant), and coolant (pressurized water most likely), the greater the enrichment requirement, or the larger the volume, which decreases the specific energy.

It comes down to $$\dot{Q}\,=\dot{m}c_p\Delta{T}$$ and T_cold among a few other technical issues.

Perhaps they propose a highly enriched system that is sealed and basically it operates like a naval reactor and power plant, i.e. the whole core is discharged. I don't see that as practical for a commercial utility, unless they are doing a small reactor like the Toshiba4S.

Again, the electrical penetrations through the pressure vessel would be rather complicated, and perhaps make the cooling problematic from an FIV perspective.

 

[vanesch]

Third, since some kind of heat transfer system is required regardless, the temperature of the coolant depends on it flow rate and and temperature of the down stream heat sink. We can not just say 'its 300C'.

That's true, but the idea was that we were going to ADD a rankine cycle to the original light bulb, because otherwise we would just have (even in the optimistic scenario) just 40% efficiency with the light, and that's not any better than the classical cycle. So the 60% heat must also be converted partially to electricity, and to have some bit of efficiency there, you'd need a high temperature, like in a current reactor, of the order of 300 degrees centigrade. Otherwise, what you win with the bulb, you loose in the thermodynamic cycle (ok, you might still get 10% out of it or something...)

 

[mheslep]

That's true, but the idea was that we were going to ADD a rankine cycle to the original light bulb, because otherwise we would just have (even in the optimistic scenario) just 40% efficiency with the light, and that's not any better than the classical cycle. So the 60% heat must also be converted partially to electricity, and to have some bit of efficiency there, you'd need a high temperature, like in a current reactor, of the order of 300 degrees centigrade. Otherwise, what you win with the bulb, you loose in the thermodynamic cycle (ok, you might still get 10% out of it or something...)

Good point, that helps bound the problem of identifying overall system efficiency.

 

[DaemonStudent]

I know this is a slightly old thread... I'm a high school student and I'm going to admit that I probably don't know much in the grand scheme of physics but why do you need florescence as a "middle man"? Gamma rays ionize in three different ways into matter: photoelectric, compton scattering and pair production. All of which eject an electron at a velocity. An electron with a velocity is current, right? I know the material, that would be creating this current, would have to be extremely dense because of the penetration of the gamma ray but it seems feasible. Unless the electrons are ejected in some non coherent way. I doubt that would be usable electricity. I just wanted to know if this is possible or already thought of...

 

[mheslep]

I know this is a slightly old thread... I'm a high school student and I'm going to admit that I probably don't know much in the grand scheme of physics but why do you need florescence as a "middle man"? Gamma rays ionize in three different ways into matter: photoelectric, compton scattering and pair production. All of which eject an electron at a velocity. An electron with a velocity is current, right? I know the material, that would be creating this current, would have to be extremely dense because of the penetration of the gamma ray but it seems feasible. Unless the electrons are ejected in some non coherent way. I doubt that would be usable electricity. I just wanted to know if this is possible or already thought of...

That is a very good question coming out of high school. I think I can only answer crudely here, hopefully someone with a better background in solid state physics offer a better explanation.

The high energy, 50keV and up electrons released by photoelectric or other effects due to gamma interactions would not be efficiently used to make electricity. In a non-fluorescing material, all that extra energy is going to be wasted as heat in the lattice of the material or the electron may simply escape the surface. Another way to see this is to recall that a current is defined simply as the amount of charge passing a point in a given time. Thus one very high speed electron passing a point per minute provides no more current than one slow one per minute. What we want here is a valence electron that jumps up in energy just enough to escape to the conduction band of the material (where it will not leave the surface), around 3- 10eV depending on the material. An incoming visible or UV photon from fluorescing material will do just that.

 

[DaemonStudent]

Have they started using new reactors like this? From what I can understand from the thread above is that the gamma rays that are used to create visible light lowers the amount of gamma rays used to heat the reactor. Thus lowering the amount of power for heat transfer/carnot cycle. Meaning that the efficiency of this PIDEC reactor is probably not what we would need to breathe life into the direct energy field in nuclear engineering, right? Isn't there a "better" technology that uses particle emission for energy? Do any of these actually exist in industry?

 

[mheslep]

Have they started using new reactors like this?

No, not in power fission reactors. This was purely a research initiative, and one that apparently never quite reached the prototyping stage.

Isn't there a "better" technology that uses particle emission for energy? Do any of these actually exist in industry?

Depending on what you mean by 'particle', http://en.wikipedia.org/wiki/Nuclear_fission" , powered by decay not fission, produces energy in the form of either charged particles (beta, alpha) or electromagnetic (gamma).