# Breeder fuel efficiency question

## Summary:

What types of materials and how much mass is left behind if say 1kg of "natural" Uranium (99.3% U238, 0.7% U235) or Thorium (99.98% Th232) is consumed in a breeder capable of consuming 100% of the breed-able/fissionable material? (Maximum energy extraction would be the goal, not burning existing transunranic waste)

## Main Question or Discussion Point

I've looked reasonably to see if I could find this answer, I'm not a nuke engineer so perhaps my search terms are not correct. I've seen the fuel cycle charts (eg:https://en.wikipedia.org/wiki/Thorium_fuel_cycle, but not clear how to work from this to a percentage mass consumed)

I'm interested to know more details of the claims made in wiki (https://en.wikipedia.org/wiki/Breeder_reactor):
"Breeder reactors could, in principle, extract almost all of the energy contained in uranium or thorium, decreasing fuel requirements by a factor of 100 compared to widely used once-through light water reactors, which extract less than 1% of the energy in the uranium mined from the earth."

Basically if 100% of the fuel is consumed, how much mass is actually consumed (of the theoretical 1kg fuel), what mass of fission products is left, and what types of fission products?

Then more practically is this realizable in 4th gen reactors, eg molten salt reactors? If not how close?

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anorlunda
Mentor
Basically if 100% of the fuel is consumed, how much mass is actually consumed (of the theoretical 1kg fuel), what mass of fission products is left, and what types of fission products?
Are you sure you mean mass? It is true that nuclear energy converts mass to energy, but the 100% consumed number refers to 100% of the fissionable atoms transmuted to lighter atoms. But weighed on a scale, the before/after mass difference is trivial.

It can be confusing. Matter-antimatter annihilation, converts 100% of the mass to energy. That is not what we do in reactors.

But it sounds like you are asking about the following which compares U and thorium cycles.
From https://en.wikipedia.org/wiki/Spent_nuclear_fuel#Uranium
96% of the mass is the remaining uranium: most of the original 238U and a little 235U. Usually 235U would be less than 0.83% of the mass along with 0.4% 236U.

Reprocessed uranium will contain 236U, which is not found in nature; this is one isotope that can be used as a fingerprint for spent reactor fuel.

If using a thorium fuel to produce fissile U-233, the SNF (Spent Nuclear Fuel) will have U-233, with a half-life of 159,200 years (unless this uranium is removed from the spent fuel by a chemical process). The presence of U-233 will affect the long-term radioactive decay of the spent fuel. If compared with MOX fuel, the activity around one million years in the cycles with thorium will be higher due to the presence of the not fully decayed U-233.

For natural uranium fuel: Fissile component starts at 0.71% 235U concentration in natural uranium. At discharge, total fissile component is still 0.50% (0.23% 235U, 0.27% fissile 239Pu, 241Pu) Fuel is discharged not because fissile material is fully used-up, but because the neutron-absorbing fission products have built up and the fuel becomes significantly less able to sustain a nuclear reaction.

Some natural uranium fuels use chemically active cladding, such as Magnox, and need to be reprocessed because long-term storage and disposal is difficult.

DEvens
Gold Member
You need reprocessing to get anything like 100% use of the fuel in a breeder. The various isotopes produced have a strong tendency to be neutron suckers. This involves both the fission products, and the isotopes that result from neutrons getting captured by various isotopes.

Some of those will, eventually, be potentially useful. U238 captures a neutron and becomes U239, and then enters into your breeding chain. A beta to Np239, then another beta to Pu239, which is fissile. But you still need to account for the neutron used to do that. It affects your neutron economy.

Typical light water reactors start with their fuel enriched to something in the neighborhood of 5% U235. And they don't burn all of that, since below some degree of enrichment (I mis-remember the exact number, it's about 2%) the process stops. So your spent fuel has about that much of the 5% still left. And the depleted U238 that is the waste product of the enrichment process isn't absolutely cleansed of U235, just very close. So you get something like 60% of the U235, and it's only 0.7% of the original Uranium. But it's complicated, because most Uranium-based reactors breed a little Plutonium. A CANDU reactor, for example, gets something like half its total energy from Plutonium. So that brings you pretty close to 1% of the original Uranium.

However, with reprocessing, you could extract the isotopes that get in the road of fission continuing. You could, for example, extract the Np and let it decay to Pu outside the core. And you could extract all the lower mass fission products so they don't suck up neutrons. This is pretty much required to keep a molten salt type reactor operating. It's annoying for solid fuel reactors.

But given reprocessing, you should be able to extract pretty close to the full available fission energy.

As to how much energy that is, I will now give you homework. The exact amount of energy a fission gives is an average. That's because the specific fission products that are produced are selected through the quantum mechanics of the process. So that's an average. The most probable masses for fission products from Uranium are round about 100 and round about 130. But it's statistical.

Also, the energy comes out in several forms. The immediate nuclei that are produced come out with kinetic energy. The process releases some neutrons, and they have energy. And the fission products are often radioactive, sometimes with very short half lives, and that releases some energy. So you would need to run around and total all of that. And you would need a time scale.

But, as a working number, a reasonable value for the energy per fission is 180 MeV. I use a number not too far from that in my work.

Here's your homework. Assume 180 MeV. And assume 1 kg of pure natural uranium. So count up the number of nuclei in there. Then assume they all fission. And total up the energy released. Then use the e=mc^2 equation and estimate the mass converted to energy by the process. To make it easier, the 180 MeV isn't very accurate. So you can neglect the difference between U238 and U235, assume it's all U238. The error will be less than the error in the 180 MeV.

• essenmein
Are you sure you mean mass? It is true that nuclear energy converts mass to energy, but the 100% consumed number refers to 100% of the fissionable atoms transmuted to lighter atoms. But weighed on a scale, the before/after mass difference is trivial.

It can be confusing. Matter-antimatter annihilation, converts 100% of the mass to energy. That is not what we do in reactors.

But it sounds like you are asking about the following which compares U and thorium cycles.
From https://en.wikipedia.org/wiki/Spent_nuclear_fuel#Uranium
Exactly, it is precisely the realization that 100% consumed fuel does not mean there is nothing left in the end that bred the question. So if 100% of the fuel is used, what is left, and how much mass is converted, and therefore how much energy. This is kind of a follow up of the nuke waste discussion, assuming 4th gen breeders, and given that there will be waste produced, how much is it, what is it (ultimately interested in half lives and storage times), and how much energy per unit of waste.

It may well be that starting with 1kg you end up with 0.95Kg of waste (made up number) x bazillion joules of energy, and , n %Strontium, m % cesium, z % iodine etc.

russ_watters
Mentor
It may well be that starting with 1kg you end up with 0.95Kg of waste (made up number) x bazillion joules of energy, and , n %Strontium, m % cesium, z % iodine etc.
It's closer to 0.999kg, so it isn't something worth bothering with.

DEvens
Gold Member
It's closer to 0.999kg, so it isn't something worth bothering with.
If I did the arithmetic correctly, it's 0.9997 kg.

• • anorlunda and russ_watters
russ_watters
Mentor
If I did the arithmetic correctly, it's 0.9997 kg.
Yeah, I just googled for mass loss in a fission reaction and that's what came back (0.1%). There's no correction for % fissible fuel, differences between fuels or any other considerations.

<snip.

Here's your homework. Assume 180 MeV. And assume 1 kg of pure natural uranium. So count up the number of nuclei in there. Then assume they all fission. And total up the energy released. Then use the e=mc^2 equation and estimate the mass converted to energy by the process. To make it easier, the 180 MeV isn't very accurate. So you can neglect the difference between U238 and U235, assume it's all U238. The error will be less than the error in the 180 MeV.
Interesting: Interestingly, assuming basically the lost mass is a rounding error, 1kg of Uranium (~19g/cm3) = ~52cm3 of volume, and the fission by products say Zirconium (6.5g/cm3) and Cesium (1.93g/cm3) would imply the fuel would expand in volume by ~4-6x as the Uranium is consumed. Does this cause problems?!

anorlunda
Mentor
That's far too simplified. Most of the fission products are radioactive isotopes. Therefore they decay into other species other isotopes, that may in turn decay, and so on and on. Each isotope has properties, gas, metal, oxide, and each has its own half life.

But looking at it pragmatically, since nuclear fuel is canned in metal rods, and since the rods do not burst during operation, the total volume can not change significantly.

From the same Wikipedia article on spent fuel:

Table of chemical data
The chemical forms of fission products in uranium dioxide
ElementGasMetalOxideSolid solution
Br KrYes---
RbYes-Yes-
Sr--YesYes
Y---Yes
Zr--YesYes
Nb--Yes-
Mo-YesYes-

Tc Ru Rh Pd Ag Cd In Sb-Yes--
TeYesYesYesYes
I XeYes---
CsYes-Yes-
Ba--YesYes
La Ce Pr Nd Pm Sm Eu---Yes

Simplified yes! It was more an observation, that splitting a dense atom results in two far less dense materials, which would imply the volume would increase as the reaction progresses. Its something I had not considered before.

Now maybe this is not a problem in current designs because the amount of fuel consumed is small, and thus the volume change is also small, likely swamped by things like CTE of the materials themselves, or the fact that some of the by products are gasses.

To me at least it was just an interesting tidbit.

I get a similar result, 0.9992 kg.

Interestingly, assuming basically the lost mass is a rounding error, 1kg of Uranium (~19g/cm3) = ~52cm3 of volume, and the fission by products say Zirconium (6.5g/cm3) and Cesium (1.93g/cm3) would imply the fuel would expand in volume by ~4-6x as the Uranium is consumed. Does this cause problems?!
But that 4 to 6 x is assuming 100 percent fission, where the real value is more like 1 percent.

Now maybe this is not a problem in current designs because the amount of fuel consumed is small, and thus the volume change is also small, likely swamped by things like CTE of the materials themselves, or the fact that some of the by products are gasses.
Beat me to it!

Even so, fuel in current reactors does swell with burnup. The gap between the fuel pellets and the inside of the cladding can close, and the internal rod pressure can increase.

• essenmein
DEvens
Gold Member
Fuel does swell during operation. Primarily this is due to gaseous fission products. But it is also due to the fact there are more atoms after fission.

Some of the extra material can get impacted in the grain structure of the uranium oxide pellets. They are not made at theoretical maximum density, partly because of this exact issue.

Some of the gaseous products escape the fuel and go into the space between the fuel and the cladding. CANDU fuel consists of pellets, and the pellets have some space deliberately left at the end of the pellet in a structure called "the dish" for obvious reasons.

• essenmein
Interesting:
View attachment 253063
So been doing more reading and noodling on this, and seems there are some discrepancies...

First (https://en.wikipedia.org/wiki/Uranium) :
One kilogram of uranium-235 can theoretically produce about 20 terajoules of energy (2×1013 joules), assuming complete fission; as much energy as 1.5 million kilograms (1,500 tonnes) of coal..

This number is about 4x lower than that I calculated above using 180MeV given by @DEvens

Then (https://en.wikipedia.org/wiki/Plutonium-239) :
The fission of one atom of 239Pu generates 207.1 MeV = 3.318 × 10−11 J, i.e. 19.98 TJ/mol = 83.61 TJ/kg, or about 23,222,915 kilowatt hours/kg.

And (https://en.wikipedia.org/wiki/Uranium-235):
The fission of one atom of uranium-235 releases 202.5 MeV = 3.24 × 10−11 J inside the reactor. That corresponds to 19.54 TJ/mol, or 83.14 TJ/kg.

So why do these numbers all not add up? If I take the 202/207MeV number, the energy release I calculated would increase simply because the MeV number went up from 180.

Is it because when they talk about nuclear fuel they are talking about the metal oxide, ie not all of the atoms in the 1kg of fuel are actually available for fission, hence the lower number. The number I calculated would be for a block of metallic U23x, not oxide/salts.

Then I assume none of these consider decay heat from all the by products. Regarding the breeding of Pu239 from U238: This would mean the U239 would be mostly Np239 after a few hrs, and the Np239 would be mostly Pu239 after about 10 days (I'm using 4 half lives to mean "mostly", ie about 93% decayed).

Would you really need to remove the Np239 from the reactor to let it decay to Pu239 externally? Seems to me since fuel likely stays in there for longer than a couple of weeks, then the first few U238's that caught a neutron would already be fissile Pu239 and participate in fission?

With molten salt reactors could a basic level of reprocessing be done "live", ie since the salt is moving anyway as a mechanism of pulling heat out, could you run this through some sort of cyclone like device (for lack of better description) to partially separate the heavy stuff from the lighter stuff and then put the slightly cleaned up fuel back in?

cmb
Simplified yes! It was more an observation, that splitting a dense atom results in two far less dense materials, which would imply the volume would increase
Just to add something interesting to this point, bear in mind that for fission reactors, of the mass lost (i.e. energy output) about 4.5% of it is lost as neutrinos. So 1/20th of that mass change simply boils away into outer space!

Astronuc
Staff Emeritus
"Breeder reactors could, in principle, extract almost all of the energy contained in uranium or thorium, decreasing fuel requirements by a factor of 100 compared to widely used once-through light water reactors, which extract less than 1% of the energy in the uranium mined from the earth."
That's an awkward statement, and not very helpful in understanding the concept of breeding fissionable material.

For a given system, one must determine the breeding ratio, and doubling time (if doubling is the goal). See the following example:
https://www.osti.gov/servlets/purl/4257307

The thorium cycle is interesting because it is based on thermal flux rather than fast flux. Molten salt reactors (MSRs) are of interest because in a fluid (molten salt) filled core, one eliminates much of the structural material and one can extract fission products online, in theory. On the other hand, one must consider that absorption of neutrons by components of the salt, e.g., Li, Be, F, Cl, besides the fissionable and fertile isotopes of Th, U, Pu, or other actinides.

The chemical forms of fission products in uranium dioxide
That table is a bit misleading IMO. While Xe and Kr are noble gases, elements like Br and I, and Se, Te, Rb and Cs are considered volatiles due to low melting points. The decay chain Te -> I -> Xe -> Cs is of interest because of their abundance and mobility. In the past decade, there has been studies (by French CEA) of Te behavior in conjunction with I and Cs. Te forms numerous compounds with I and Cs. The complementary fission product of Te (Z = 52) is Zr (Z = 40); the Z's of both elements must add to 92 in binary fission. There is also a small fraction (on order of 10-4) of ternary fissions in which T or He are produced in addition to two larger fission products.

A better representation of fission products if found in Figure 3.1 of http://abulafia.mt.ic.ac.uk/publications/theses/stanek/solutioninuo2.pdf.

• Asymptotic and anorlunda
DEvens