PIDEC: Nuclear-Photoelectric Powerplant

[sanman]

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?

 

[sanman]

Here's more:

http://isjaee.hydrogen.ru/pdf/AEE04-07_Prelas.pdf

 

[sanman]

Here's more on this:

http://nuclearstreet.com/blogs/nuclear_power_news/archive/2009/04/15/etf.aspx

So is PIDEC an idea whose time has come? What particularly could it have uses for?
Perhaps for nuclear propulsion systems? Maybe a nuclear-electric power source for a VASIMR type of rocket?

 

[mgb_phys]

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.

 

[sour_kremlin]

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.

 

[Andrew Mason]

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

 

[vanesch]

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

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.

 

[Bob S]

Here is a slide presentation on PIDEC from U Missouri:
https://mospace.umsystem.edu/xmlui/...uclearEnergyConversion.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.

 

[sanman]

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.

 

[vanesch]

Here is a slide presentation on PIDEC from U Missouri:
https://mospace.umsystem.edu/xmlui/...uclearEnergyConversion.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 ?

 

[sanman]

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.

 

[Astronuc]

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?

 

[vanesch]

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).

 

[vanesch]

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.

 

[sanman]

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.

 

[Astronuc]

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.

 

[mheslep]

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

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" now. They are quite expensive, but if one has an extremely high energy density source they can pay off.

 

[mheslep]

Astronuc - Your colleague George Miley did some of the earlier work on this topic in 1990:
http://prelas.nuclear.missouri.edu/Publications/PIDEC Fission Conversion.pdf

 

[Astronuc]

Astronuc - Your colleague George Miley did some of the earlier work on this topic in 1990:
http://prelas.nuclear.missouri.edu/Publications/PIDEC Fission Conversion.pdf

I'll bet he doesn't remember that. He's kind of stepped back from a lot of activity, and he's slowed way down since then.

 

[vanesch]

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" 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!

 

[Astronuc]

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.

 

[Morbius]

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

 

[Astronuc]

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.

 

[mheslep]

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.

 

[mheslep]

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

...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/...5/769/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%".

 

[vanesch]

Ok, so if I have it right, we have a scintillating fluid (and coolant ?) in which there are suspended microspheres of fuel. This scintillating fluid emits light, and this is then captured in a 4-pi sphere of PV cells.

Now, this implies that the spheres are so thinly distributed that they hardly absorb any light (you don't want to have some dark black muddy stuff, right ?).

Next, this implies that you have or your PV into the reactor fluid (and hence exposed to very high radiation levels), or it implies that you have a transparant reactor vessel and radiation protection (uh!). Ok, this last bit could be achieved by a large swimming pool. But now, the surface to be covered by PV is rather large.

So the final design looks like a big light bulb: a transparant reactor vessel with a fluid inside with some suspended UO2 microspheres, suspended in the middle of a large swimming pool on the walls of which the PV are placed. I don't know if this is workable. The transparent reactor vessel sounds scary to me :-)

 

[Astronuc]

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

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/...5/769/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%".

The abscissa in the cited figure, 0-6 microns, is on the order of typical grain size in UO₂ ceramic pellets. Sphere-pac fuel uses microspheres of larger size - 50-200 microns diameter - and these produce a pretty tightly packed column. Fast reactor fuel pins are on the order of 6 mm, and pellet OD of standard 17x17 PWR fuel is about 8.2 mm. At normal fuel power densities, the fuel surface is about 400°C, and futher in, near the center of the pellet, the temperature is in the range of 900-1200°C depending on the power level.

Add to that the high gamma field, I can't imagine that PV's function well. The Compton and photoelectric effects would result in considerable ionization in the PV material. I wonder if Prelas has placed a PV in a high rad field at 300°C and measured it's performance.

I've read about nuclear lightbulb concepts. Some are pretty wacky, and conveniently ignore reality.

 

[theCandyman]

It helps get you funding if you make ideal assumptions.

 

[mheslep]

... Prelas has placed a PV in a high rad field at 300°C and measured it's performance...

Recall that optimistically 80% of the energy escapes the Prelas fuel spheres into the flourescer (sp? his term) material so we would not expect high temperatures in the fuel.

 

[mheslep]

Ok, so if I have it right, we have a scintillating fluid (and coolant ?) in which there are suspended microspheres of fuel. This scintillating fluid emits light, and this is then captured in a 4-pi sphere of PV cells.

Now, this implies that the spheres are so thinly distributed that they hardly absorb any light (you don't want to have some dark black muddy stuff, right ?).

Next, this implies that you have or your PV into the reactor fluid (and hence exposed to very high radiation levels), or it implies that you have a transparant reactor vessel and radiation protection (uh!). Ok, this last bit could be achieved by a large swimming pool. But now, the surface to be covered by PV is rather large.

The surface covered by UV mirrors is rather large. The PVs can be down at the end of some wave guide. Same trick commercial solar PV concentrators use.

 

[Morbius]

Recall that optimistically 80% of the energy escapes the Prelas fuel spheres into the flourescer (sp? his term) material so we would not expect high temperatures in the fuel.

mheslep,

NOT TRUE AT ALL! Just because you have less than 80% of the energy being deposited there
does NOT mean that the temperature will not be high.

It reminds me of a question that one of my professors at MIT posed. You have a liquid fuel reactor -
the liquid serves as both fuel and coolant. The reactor is a thermal reactor - so this liquid combination
of fuel / coolant is pumped through channels in a big block of graphite. Where are the hottest locations
in the reactor?

Of course most of the energy is deposited in the fuel / coolant material and not in the graphite - but there
is a little energy deposited in the graphite by gamma heating. Because of that gamma heating - the
graphite has to have a cooling mechanism to maintain steady state. The only cooling mechanism is
conduction to the coolant. Therefore, even though the % of heat energy deposited in the graphite is
low - the graphite has to heat up HOTTER than the fuel / coolant in order to transfer heat to the coolant.

If the only way for the fuel spheres to get rid of that 20% of the power that they do get is to transfer
that energy to the scintillator / coolant - then they will heat up until they ARE hotter than the coolant
in order to get rid of the energy.

You've made the mistake of equating energy deposition and temperature - and they do NOT
necessarily go together.

Dr. Gregory Greenman
Physicist

 

[sanman]

But these spheres are supposed to be very small in size for scintillation capture and transparency, as per everyone's discussions thusfar. The smaller the sphere, the more easily it cools via conductive heat transfer.

 

[Morbius]

But these spheres are supposed to be very small in size for scintillation capture and transparency, as per everyone's discussions thusfar. The smaller the sphere, the more easily it cools via conductive heat transfer.

sanman,

But the only way a sphere can cool by conduction to the coolant is if it is hotter than the coolant!

The spheres - no matter how small are not going to cool by heat conduction to a coolant
that is hotter than the spheres.

The small size only means that your surface area to volume ratio will be high.

But the sphere still has to be hotter than the coolant in order for heat to flow from
sphere to coolant.

Dr. Gregory Greenman
Physicist

 

[theCandyman]

But these spheres are supposed to be very small in size for scintillation capture and transparency, as per everyone's discussions thusfar. The smaller the sphere, the more easily it cools via conductive heat transfer.

But heat flows in the direction where the temperature gradient is negative. For that to happen, the spheres would need to be hotter.

Edt: I see Morbius already answered that.

 

[sanman]

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.

 

[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.

===============================

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).