PIDEC: Nuclear-Photoelectric Powerplant

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Main Question or Discussion Point

Nuclear power generator uses radioactive particle emission to stimulate fluorescent material which then lights up photoelectric cell. It's called PIDEC:

http://www.physorg.com/news158848950.html

http://nextbigfuture.com/2009/04/direct-conversion-of-nuclear-power-to.html

They're claiming upto 40% conversion efficiency, which is pretty good.

They're also saying it could make for a much more compact and portable nuclear powerplant.
I'm wondering if that could make it suitable for spacecraft. For instance, it could be used to power an ion-engine, or a VASIMR type of rocket.
This could make SSTO practical.

At the least, it should be suitable for powering space probes (eg. rovers)

Can anyone comment on the feasibility of this design?
 

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  • #3
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  • #4
mgb_phys
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It doesn't say what kind of isotope you need or how much it costs to generate the isotope.
A beta emitter and a wire gives you 100% efficiency but it's not exactly practical.
 
  • #5
This wouldn't allow an ion based engine to achieve SSTO because ion engines aren't used to lift off Earth. Their thrust to weight ratio is too low to be useful there, though it could be useful once the craft is in space.
 
  • #6
Andrew Mason
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Nuclear power generator uses radioactive particle emission to stimulate fluorescent material which then lights up photoelectric cell. It's called PIDEC:
...

They're claiming upto 40% conversion efficiency, which is pretty good.
40% is not that good. You can get 40% efficiency from some designs of nuclear power plants. Besides, how do you get 40% efficiency when the photo-electric cell alone is about 20% efficient?

AM
 
  • #7
vanesch
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I'm extremely sceptical about this. First off, most energy in a nuclear reactor is deposited by fission products in the fuel, so I don't see how you can make a good scintillator out of the fuel and worse, how to couple that to some kind of photocel :bugeye:

But next, we all know that solar cells are not extremely efficient in converting light into electricity. So trying to convert radiation into scintillation light, and then trying to convert that scintillation light into electricity sounds like about the worse idea one could have to make an efficient plant. So I don't know what that 40% means.
 
  • #8
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Here is a slide presentation on PIDEC from U Missouri:
https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/769/NuclearEnergyConversion.ppt?sequence=5#256,1,A Research Program: Nuclear Energy Conversion
They are proposing using either krypton 85, strontium 90, or plutonium 238 as the radioactive isotope. On page 17 they calculate useful-power-to-raw-power efficiencies in the range of 10% to 31%, so their 40% efficiency number apparently does not include this. The fluorescence apparently pumps an eximer laser which is converted in a photocell.
 
  • #9
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It's one thing to talk about solar photovoltaic efficiencies, since solar photovoltaics are designed to work with the Sun's spectrum.

But it's another thing to talk about photovoltaic efficiencies for other photon sources, like from a fluorescer, since their emission spectrum and bandwidth are different.

Nextly, 40% may be low in comparison to steam turbines, however a steam turbine system is not necessarily the lightest in weight or the most durable/reliable system out there. These things are factors in choice of nuclear conversion setup too.

A simpler nuclear photovoltaic system could be simpler, more compact, and more physically robust. For applications like space propulsion, it might be much more practical.
The fact that the fluorescer surrounds and envelopes the nuclear radiation source to absorb and harvest its radiation energy as much as possible, while still being a fluid, means reduction in radiation damage, in contrast to nuclear thermal systems which expose components to radiation damage over time.

Unlike the thermal fluid in the nuclear thermal reactor, the scintillator is not kept under extreme conditions that could lead to containment breach, etc. It seems to be a much more comfortable and reliable setup.
 
  • #10
vanesch
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Here is a slide presentation on PIDEC from U Missouri:
https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/769/NuclearEnergyConversion.ppt?sequence=5#256,1,A Research Program: Nuclear Energy Conversion
They are proposing using either krypton 85, strontium 90, or plutonium 238 as the radioactive isotope. On page 17 they calculate useful-power-to-raw-power efficiencies in the range of 10% to 31%, so their 40% efficiency number apparently does not include this. The fluorescence apparently pumps an eximer laser which is converted in a photocell.
It is not clear on these transparencies how they hope to recover the essence of the kinetic energy of the fission products in a nuclear fission reactor. As the path of these things is already a few microns, this means that the uranium fuel must be part itself of the fluorescent medium, right ? So you have a solution of uranium in a fluorescent medium and that medium must convert most of the kinetic energy in photons, which must then escape the fluorescent medium and impact upon photo-electric cells ?
So the 160 MeV something get directly converted into light ? And that light gets converted into electricity ?
 
  • #11
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Yes, that's right, the fissile material must be in closest proximity to the scintillator material, so that the two are mixed. The fluorescent scintillator can have a low opacity, so that its photons escape it to reach the surrounding photocell, to be converted to electricity.
 
  • #12
Astronuc
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Yes, that's right, the fissile material must be in closest proximity to the scintillator material, so that the two are mixed. The fluorescent scintillator can have a low opacity, so that its photons escape it to reach the surrounding photocell, to be converted to electricity.
Depending on the medium, the liquid in which the scintillator is suspended is likely very hot, and hot enough that it will change phase (from liquid to vapor) if not under high pressure. If scintillator is in water, then it would require high pressure, e.g. 7 - 15.5 MPa depending on the temperature.

I've heard of incore thermionics, which are only practical in small systems, but I share vanesch's skepticism about PIDEC. It requires too many penetrations from the containment/pressure vessel, and even with 40% efficiency, the system still need to transfer 60% of the energy to the environment.

Further problem/complication with the fissile material in contact with the sintillator is the fact that fission products accumulate in the scintillator. How does PIDEC handle fission products and their accumulation, particularly volatiles and gases?
 
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  • #13
vanesch
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Like Astro says, this looks like sheer horror from the side of a reactor core designer, compared to which any gen-iv design is a child's game. In fact, it seems to come rather close to the molten-salt reactor concept where the fuel is a liquid and is in a continuous reprocessing loop (except for the photocells in the core, uh... wonder how they do under intense neutron and gamma bombardment btw).
 
  • #14
vanesch
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Yes, that's right, the fissile material must be in closest proximity to the scintillator material, so that the two are mixed. The fluorescent scintillator can have a low opacity, so that its photons escape it to reach the surrounding photocell, to be converted to electricity.
Maybe there are scintillators that do this, but several I know of absorb their own scintillation light, so that you need wavelength shifters. All of those need to be simple inorganic molecules, because no covalent bounds are going to survive a long time in a reactor core - which excludes delicate organic molecules.
 
  • #15
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I think quantum dots have shown themselves to be radiation-hardened photovoltaic elements, which can stand upto heavy radiation bombardment. As I recall, there are even liquid bandgap materials, such as liquid Gallium which has been researched with Am-241.

It seems like liquid, or "healable"/ductile materials are being seen to have some promise for coping with structural damage from heavy radiation bombardment.
 
  • #16
Astronuc
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I think quantum dots have shown themselves to be radiation-hardened photovoltaic elements, which can stand upto heavy radiation bombardment. As I recall, there are even liquid bandgap materials, such as liquid Gallium which has been researched with Am-241.
There needs to be a containment/pressure vessel and piping to remove heat. Gallium embrittlement would be a concern. There is also the matter of strings of electrical elements penetrating the containment vessel, which would have to be electrically insulated from the vessel. The insulation and other structures have to be chemically compatible with the coolant, gallium or otherwise.

It seems like liquid, or "healable"/ductile materials are being seen to have some promise for coping with structural damage from heavy radiation bombardment.
Certainly at some temperature, annealing of radiation damage is possible.

The core and coolant must be contained by something solid with strength, and the electrical lines have to penetrate the coolant boundary.

Then let's add reactivity control. Perhaps they envision external reflectors/absorbers.

I see numerous problems with this concept.
 
  • #17
mheslep
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I'm extremely sceptical about this. First off, most energy in a nuclear reactor is deposited by fission products in the fuel, so I don't see how you can make a good scintillator out of the fuel and worse, how to couple that to some kind of photocel :bugeye:
This appears to be mostly a geometry problem, and that sandwiches of thin film materials would be a start towards doing away with the internals of the fuel, where most of the material is exposed on a surface.

But next, we all know that solar cells are not extremely efficient in converting light into electricity. So trying to convert radiation into scintillation light, and then trying to convert that scintillation light into electricity sounds like about the worse idea one could have to make an efficient plant. So I don't know what that 40% means.
PVs achieve over http://www.nanowerk.com/news/newsid=10511.php" [Broken] now. They are quite expensive, but if one has an extremely high energy density source they can pay off.
 
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Astronuc
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  • #20
vanesch
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This appears to be mostly a geometry problem, and that sandwiches of thin film materials would be a start towards doing away with the internals of the fuel, where most of the material is exposed on a surface.
The path of the fission products in the fuel is *a few micrometer*. The energy (the bulk of the fission energy) is dissipated over this length. If you want to recuperate most of this energy, this path has to be mainly in the scintillating material. You cannot reasonably think of solid fuel elements within a confinement of less than a micron thick, perfectly light reflecting (otherwise you'd loose half of the scintillation light), but withstanding for an extended period of time, the radiation and thermal conditions, can you ?

No, the only way to hope to recuperate a serious part of the fission energy is to have it intimately mixed with the scintillator, hence as a fluid (or a transparant solid mixture).

PVs achieve over http://www.nanowerk.com/news/newsid=10511.php" [Broken] now. They are quite expensive, but if one has an extremely high energy density source they can pay off.
Yes, but you need 40% efficiency for the overall process: the ionisation/scintillation (that's quite some difficulty: we make detectors with scintillator, and I can assure you that this step is not 100% efficient!) ; the recuperation of the light on the active elements without light absorption, and finally the conversion of the captured light into current.
We do the same things in detectors (but on a much less energetic scale), and if you end up with a few tens of photons converted, you're happy!
 
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  • #21
Astronuc
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PV's normally don't operate in a gamma/neutron radiation field (~1014 n/cm2-s), as well as ~280-300 C. Then there is the problem of activation (and transmutation) over time.

Then there is the complication of stringing the PV together to form a conductor, and the coolant compatibility.
 
  • #22
Morbius
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This appears to be mostly a geometry problem, and that sandwiches of thin film materials would be a start towards doing away with the internals of the fuel, where most of the material is exposed on a surface.
mheslep,

As vanesch correctly points out - the fission energy is dissipated over length scales that are smaller
than the thinnest thin film PVs. Unless one does as Astronuc suggests and have the fuel and PV
intimately mixed; one isn't going to recover the bulk of the fission energy.

However, as Astronuc also points out; that puts the PV in a radiation field regime where PVs don't
work.

There's really not much that can be done to make a silk purse out of this sow's ear.

Dr. Gregory Greenman
Physicist
 
  • #23
Astronuc
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I forgot to mention that I think the Compton effect would overwhelm the photoelectric effect, not to mention the atoms being knocked out their lattice positions.

I'm not sure how the properties of PV change with dpa.


I once did a calculation on small high power density core which would operate over several months. In that period, I was getting about 3 dpa, which is quite low considering we push structural materials into the 50 dpa range over 30-40 years, and higher.
 
  • #24
mheslep
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PV's normally don't operate in a gamma/neutron radiation field (~1014 n/cm2-s), as well as ~280-300 C. Then there is the problem of activation (and transmutation) over time.

Then there is the complication of stringing the PV together to form a conductor, and the coolant compatibility.
1) Prelas intends, rather requires, UV widebandgap PVs like diamond which I imagine hold up better under radiation, 2) In Prelas's two step process its the excimers that need proximity to the fuel, they fluoresce UV which may be directed to the more remote PVs.
 
  • #25
mheslep
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The path of the fission products in the fuel is *a few micrometer*. The energy (the bulk of the fission energy) is dissipated over this length. If you want to recuperate most of this energy, this path has to be mainly in the scintillating material. You cannot reasonably think of solid fuel elements within a confinement of less than a micron thick, perfectly light reflecting (otherwise you'd loose half of the scintillation light), but withstanding for an extended period of time, the radiation and thermal conditions, can you ?

No, the only way to hope to recuperate a serious part of the fission energy is to have it intimately mixed with the scintillator, hence as a fluid (or a transparant solid mixture).



Yes, but you need 40% efficiency for the overall process: the ionisation/scintillation (that's quite some difficulty: we make detectors with scintillator, and I can assure you that this step is not 100% efficient!) ; the recuperation of the light on the active elements without light absorption, and finally the conversion of the captured light into current.
We do the same things in detectors (but on a much less energetic scale), and if you end up with a few tens of photons converted, you're happy!
Here we go: With UO2 microsphere's Prelas theorizes 50 to 80% of the kinetic energy can escape the sphere before thermalizing, figure attached.
http://prelas.nuclear.missouri.edu/Publications/NDF.pdf
Prelas et al 1987 said:
...for scaling nuclear energy conversion systems with reasonable efficiencies. One of the reasons for this optimism is that theoretical estimates suggest that 50 to 80% of the energy generated in micropellet will be available to [pump a lasing material, excite an excimer, etc]
Brackets are mine.Then there's the efficiency of the Fluorescer (30-50%) and the efficiency of the PV to consider, which apparently can reach 97% with monochromatic light at these wavelengths. https://mospace.umsystem.edu/xmlui/bitstream/handle/10355/769/NuclearEnergyConversion.ppt?sequence=5". Thus Prelas claims a 30-40% over all direct conversion efficiency from a fission reactor (much higher from simpler radioisotope devices), which then can be supplemented with a traditional heat cycle the overall system efficiency can reach "50-70%".
 

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