# Liquid Fluoride Thorium Reactor

by gcarlin
Tags: fluoride, liquid, reactor, thorium
PF Gold
P: 3,077
 Quote by mesa ... With this data we can simply calculate the accumulation of this element of the course of say a 30 year life cycle based off of anticipated (MWt energy of a reactor)/(energy per fission)*time for a rough estimate. ...

PWR typical burnup is around 50 GWdays/ton, or 5% of the fuel. Up to 500 GWdays/ton is expected in an experimental reactor, says the wiki. LFTR supposedly will have very high burnup, so optimistically assume 500 GWdays/ton, or ~120GWdays per 1000 moles of U, or given a 33% efficient reactor, 40GWe-days/1000 moles, or ~11GWe-years/1e5 moles U.

So for every 11 years of operation, and again following the fission products curve, a 1GWe reactor produces 7e3 moles of Zr, 6e3 moles of Cs, etc, for the high probability products. Or, all products with amu's from 82 to 105, and 127 to 150 would accumulate 5e2 moles, or higher, in 11 years. Those concentrations will change through decay or neutron capture.

The consequence of the result would depend on chemistry of the particular element in contact with the alloy which is beyond me.
P: 509
 Quote by Astronuc Note that MSRE was only 8 MW (without the electrical generation system, and with off-line batch processing), while the MSBR was planned for 2250 MWt (1000 MWe) and use of on-line continuous processing. See Table III, p. 21 of WASH-1222.
 Quote by mheslep PWR typical burnup is around 50 GWdays/ton, or 5% of the fuel. Up to 500 GWdays/ton is expected in an experimental reactor, says the wiki. LFTR supposedly will have very high burnup, so optimistically assume 500 GWdays/ton, or ~120GWdays per 1000 moles of U, or given a 33% efficient reactor, 40GWe-days/1000 moles, or ~11GWe-years/1e5 moles U. So for every 11 years of operation, and again following the fission products curve, a 1GWe reactor produces 7e3 moles of Zr, 6e3 moles of Cs, etc, for the high probability products. Or, all products with amu's from 82 to 105, and 127 to 150 would accumulate 5e2 moles, or higher, in 11 years. Those concentrations will change through decay or neutron capture. The consequence of the result would depend on chemistry of the particular element in contact with the alloy which is beyond me.
Interesting approach, I did it this way using Astronuc's thermal value above for a commercial generating facility of 2250MWt:
2250MWtx24hoursx365daysx30years/((MeV per fission)x(4.4504902416667x10^(-17))) = total number of fissions for the life cycle of the reactor. From here we can just multiply by the Cumulative Fission Yields to get:

4.9x10^21 Ga atoms produced, or .0081mols
Using your method I get .0025mols Ga for the same time frame.

If we are correct Gallium will not be an issue. Granted we could also account for Ga production from U235 since small amounts will also appear in this reactor but that lowers our values since they are an order of magnitude less in production of Ga in the thermal spectrum. Also, as Astronuc pointed out in the other thread, 8-10% of fission in LFTR will be fast neutrons, however this value is comparitively insignificant as well for this particular case.
P: 509
 Quote by mheslep PWR typical burnup is around 50 GWdays/ton, or 5% of the fuel. Up to 500 GWdays/ton is expected in an experimental reactor, says the wiki. LFTR supposedly will have very high burnup, so optimistically assume 500 GWdays/ton, or ~120GWdays per 1000 moles of U, or given a 33% efficient reactor, 40GWe-days/1000 moles, or ~11GWe-years/1e5 moles U. So for every 11 years of operation, and again following the fission products curve, a 1GWe reactor produces 7e3 moles of Zr, 6e3 moles of Cs, etc, for the high probability products. Or, all products with amu's from 82 to 105, and 127 to 150 would accumulate 5e2 moles, or higher, in 11 years. Those concentrations will change through decay or neutron capture. The consequence of the result would depend on chemistry of the particular element in contact with the alloy which is beyond me.
We should go visit Borek in the Chemistry section and see what his thoughts are on this.

As for the remainder, calculations for the rest of the elements produced along with their constituent isotopes (and variations) would be helpful but improvement is needed on how calculations are performed to get decent sig figs.

Any thoughts?
 Admin P: 21,827 * Independent fission yield (%): number of atoms of a specific nuclide produced directly (not via radioactive decay of precursors) in 100 fission reactions * Cumulative fission yield (%): total number of atoms of a specific nuclide produced (directly and via decay of precursors) in 100 fission reactions From http://www-nds.iaea.org/publications...ecdoc-1168.pdf These may not include activation (n-capture). -------------------------------------------------- Fission product pairs for U (Z, 92-Z; A, 234-A for U235 or 232-A for U233), assuming 2 neutrons released per fission. The neutrons affect A, not Z. Z A 92-Z 234-A for U-235; 232-A for U-233 63 Eu 29 Cu 62 Sm 30 Zn 61 Pm 31 Ga 60 Nd 32 Ge 59 Pr 33 As 58 Ce 34 Se 57 La 35 Br 56 Ba 36 Kr 55 Cs 37 Rb 54 Xe 38 Sr 53 I 39 Y 52 Te 40 Zr 51 Sb 41 Nb 50 Sn 42 Mo 49 In 43 Tc 48 Cd 44 Ru 47 Ag 45 Rh 46 Pd 46 Pd -------------------------------------------------- Another factor to consider is the delayed neutron precusors that leave the core. Delayed neutrons are important with respect to control the reactor as well as irradiating the structure and piping outside the core. Reactivity control is another consideration, so a large MSBR may require use of control elements. The graphite must be supported, so there is a core support plate (not graphite), which will receive a neutron flux.Differences in thermal expansion between graphite and the structural alloy will have to be investigated. Hideout of the molten salt could be an issue. Note the MSRE operated 4 years and surface defects of 7 mils were found. Larger defects may propagate. Also, a 40 to 60 year lifetime is preferable. The numerous technical issues should be listed and discussed separately.
P: 509
 Quote by Astronuc * Independent fission yield (%): number of atoms of a specific nuclide produced directly (not via radioactive decay of precursors) in 100 fission reactions * Cumulative fission yield (%): total number of atoms of a specific nuclide produced (directly and via decay of precursors) in 100 fission reactions From http://www-nds.iaea.org/publications...ecdoc-1168.pdf These may not include activation (n-capture). -------------------------------------------------- Another factor to consider is the delayed neutron precusors that leave the core. Delayed neutrons are important with respect to control the reactor as well as irradiating the structure and piping outside the core. Reactivity control is another consideration, so a large MSBR may require use of control elements. The graphite must be supported, so there is a core support plate (not graphite), which will receive a neutron flux.Differences in thermal expansion between graphite and the structural alloy will have to be investigated. Hideout of the molten salt could be an issue. Note the MSRE operated 4 years and surface defects of 7 mils were found. Larger defects may propagate. Also, a 40 to 60 year lifetime is preferable. The numerous technical issues should be listed and discussed separately.
Agreed.

I recieved an email from FliBe energy giving a link to the pdf files of the ORNL research program on the MSR. There is a substantial amount of information:

http://energyfromthorium.com/pdf/

PF Gold
P: 3,077
 Quote by Astronuc Reactivity control is another consideration, so a large MSBR may require use of control elements.
The MSRe had a *negative* temperature reactivity coefficient. The salt expands with temperature, density falls, reactivity falls. Is there some reason that control method must change with large reactor?
P: 509
 Quote by mheslep The MSRe had a *negative* temperature reactivity coefficient. The salt expands with temperature, density falls, reactivity falls. Is there some reason that control method must change with large reactor?
Here is Chris Holdens reason for it @6:16 in his presentation for his reactor design, here is a link:

Calculating for if they are neccessary would be good, however there are many things Astronuc suggested that seem like viable avenues to look at. This is already a proven technology and it would seem the question is whether it is needed or not; it is reasonable to assume regulatory agencies could insist on such measures as they are a standard today even if shown to be unneccesary for LFTR.
P: 509
 Quote by Astronuc The graphite must be supported, so there is a core support plate (not graphite), which will receive a neutron flux.Differences in thermal expansion between graphite and the structural alloy will have to be investigated. Hideout of the molten salt could be an issue. Note the MSRE operated 4 years and surface defects of 7 mils were found. Larger defects may propagate. Also, a 40 to 60 year lifetime is preferable. The numerous technical issues should be listed and discussed separately.
Rusty Holden had an interesting idea about a different moderator @ 3:12:

What is 'hideout'? Are you referring to areas in the reactor where flow rates of the salt drop significantly?

"Also, a 40 to 60 year lifetime is preferable."
That would seem reasonable.
PF Gold
P: 3,077
 Quote by mesa Here is Chris Holdens reason for it @6:16 in his presentation for his reactor design, here is a link: http://www.youtube.com/watch?v=ZbtVk8r6-3U ... This is already a proven technology and it would seem the question is whether it is needed or not; it is reasonable to assume regulatory agencies could insist on such measures as they are a standard today even if shown to be unneccesary for LFTR.
Regulatory agencies could insist on anything they like, just because that's the way it has been done. But that's not technically relevant. No MSR is going to see approval in the US by the NRC for decades to come. The design will have to be built abroad, so I don't see tailoring a design to NRC inertia without valid technical reasons, driving up cost, as particularly wise.
P: 1,042
 Quote by mesa So now come the questions, how hard would it be? how much energy does it use? what is the cost of a system like this? From my own experience in refrigeration I don't think this would be difficult to add on. What is your opinion?
It sounds stupid, wasteful in terms of energy on the one hand and on the other I do not see how you could control the quality of the salt layer. The interface would surely see a lot of stress and cracks and whatnot. Would they propagate to the walls? How could you tell if they did? And so on.
P: 21,827
 Quote by mheslep The MSRe had a *negative* temperature reactivity coefficient. The salt expands with temperature, density falls, reactivity falls. Is there some reason that control method must change with large reactor?
The negative temperature and void coefficients are useful for limiting a reactivity excursion, which is the case in LWRs. However, they are not suitable for power maneuvering a reactor. The delayed neutrons determine the period or rate at which power increases for a given insertion of positive reactivity (e.g., increase in fuel enrichment or removal of a neutron poison). The objective is to maintain control of the power level, and to avoid a rapid increase in reactor power.

Another matter to consider is the guide structure in the core. Control rods are positioned at the edge of the core for rapid insertion. The control rod and guide structure materials must be able to resist the high fluence and fluoride salt interaction.

A lot of the issues mentioned in this thread are also being explored in the Gen-IV MSR program.
 PF Gold P: 3,077 As I recall the ONR MSR ~7MWth experiment mainly used load following to control the reactor. Increase the load which removes heat faster, the salt cools, reactivity increases to meet the load.
P: 509
 Quote by zapperzero It sounds stupid, wasteful in terms of energy on the one hand and on the other I do not see how you could control the quality of the salt layer. The interface would surely see a lot of stress and cracks and whatnot. Would they propagate to the walls? How could you tell if they did? And so on.
One of the big issues with this type of reactor is the materials reacting with the salt and byproducts of fission; keep in mind that rates of reaction go up drastically with temperature and solids are no where near as reactive as liquids so this idea, (that came from the scientists at ORNL/MSR), seems to have some validity.

Either way we should look through the documents first to see what their proposed approach was before attempting to invalidate/validate this idea with arguement. Here is the link if you missed it:

http://energyfromthorium.com/pdf/
P: 509
 Quote by Astronuc * Independent fission yield (%): number of atoms of a specific nuclide produced directly (not via radioactive decay of precursors) in 100 fission reactions * Cumulative fission yield (%): total number of atoms of a specific nuclide produced (directly and via decay of precursors) in 100 fission reactions
Okay, thank you.

 Quote by Astronuc These may not include activation (n-capture). -------------------------------------------------- Fission product pairs for U (Z, 92-Z; A, 234-A for U235 or 232-A for U233), assuming 2 neutrons released per fission. The neutrons affect A, not Z. Z A 92-Z 234-A for U-235; 232-A for U-233 63 Eu 29 Cu 62 Sm 30 Zn 61 Pm 31 Ga 60 Nd 32 Ge 59 Pr 33 As 58 Ce 34 Se 57 La 35 Br 56 Ba 36 Kr 55 Cs 37 Rb 54 Xe 38 Sr 53 I 39 Y 52 Te 40 Zr 51 Sb 41 Nb 50 Sn 42 Mo 49 In 43 Tc 48 Cd 44 Ru 47 Ag 45 Rh 46 Pd 46 Pd --------------------------------------------------
This information is very useful but just for clarification what column is Z and which is A, or are the columns just not lined up?
P: 21,827
 Quote by mesa This information is very useful but just for clarification what column is Z and which is A, or are the columns just not lined up?
The Z is over the atomic number (number of protons in the nucleus). The A and 234-A are over the letters designating the element (nuclide) corresponding to the Z.

If one fission produces Eu (Z=63, A=158) then the other fission product is necessarily Cu (Z=29, A = 234-158 = 76) + 2 neutrons. If Eu-159 was the fission product, then Cu-75 would be the other fission product + 2 neutrons. If 3 neutrons are released during fission, then the pair would be Eu-158, Cu-75 or Eu-159, Cu-74.

When U-233/U-235 absorbs a neutron and becomes an excited U-234/U-236 nucleus and fissions, the atomic numbers of the fission products, Z1 and Z2 must sum to 92 (or Z, 92-Z). The atomic numbers, A1 and A2, sum to 232/234 if 2 fission (prompt) neutrons are released (or A2 = 232-A1, or 234-A1), or 231/233 if 3 fission (prompt) neutrons are released. Some fission products release 'delayed' neutrons as well - usually fractions of a second up to 60 to 80 seconds later. The fraction of delayed neutrons with U-233 is less than for U-235.
P: 509
 Quote by Astronuc The Z is over the atomic number (number of protons in the nucleus). The A and 234-A are over the letters designating the element (nuclide) corresponding to the Z. If one fission produces Eu (Z=63, A=158) then the other fission product is necessarily Cu (Z=29, A = 234-158 = 76) + 2 neutrons. If Eu-159 was the fission product, then Cu-75 would be the other fission product + 2 neutrons. If 3 neutrons are released during fission, then the pair would be Eu-158, Cu-75 or Eu-159, Cu-74. When U-233/U-235 absorbs a neutron and becomes an excited U-234/U-236 nucleus and fissions, the atomic numbers of the fission products, Z1 and Z2 must sum to 92 (or Z, 92-Z). The atomic numbers, A1 and A2, sum to 232/234 if 2 fission (prompt) neutrons are released (or A2 = 232-A1, or 234-A1), or 231/233 if 3 fission (prompt) neutrons are released. Some fission products release 'delayed' neutrons as well - usually fractions of a second up to 60 to 80 seconds later. The fraction of delayed neutrons with U-233 is less than for U-235.
Okay, I understand; I thought your chart represented something else, but it is still good for quick reference.

I would like to put together a data table on fission products that have high cross sectional areas for capturing thermal neutrons in the Th/U233 breeder cycle and see which are of biggest concern (like zenon 135).

It would also be good to run through the fission products and see which will have a high likelyhood for rate of reactivity/concentration (like tellurium) with the Hastelloy N. This part will likely prove difficult to compute without experimentation; hopefully there is sufficient information in the ORNL documents.
 Admin P: 21,827 One would have to do some calculations based on flux and fuel composition, or find detailed tables that list specific nuclides and their decay chains, for example - Ba147 -> La147 -> Ce147 -> Pr147 -> Nd147 -> Pm147 -> Sm147 (stable), but each nuclide can absorb a neutron (but with different cross sections). Sm is a moderate neutron poison. And there are heavier nuclides, e.g., Pm155 -> Sm155 -> Eu155 -> Gd155, where Eu and Gd are stronger neutron poisons, but their fractional yields are quite low. Meanwhile, these can provide some idea of the FP vector. http://www.doitpoms.ac.uk/tlplib/nuc..._processes.php http://en.wikipedia.org/wiki/File:Th...ssionYield.svg http://en.wikipedia.org/wiki/Fission...ts_(by_element) http://en.wikipedia.org/wiki/Fission....2C_127_to_132 http://en.wikipedia.org/wiki/Fission..._152.2C_154.29 http://upload.wikimedia.org/wikipedi...nYield.svg.png There are preferred nuclides, i.e., those with high yield fractions. Also of interest - http://en.wikipedia.org/wiki/Fluoride_volatility
P: 509
 Quote by Astronuc One would have to do some calculations based on flux and fuel composition, or find detailed tables that list specific nuclides and their decay chains, for example - Ba147 -> La147 -> Ce147 -> Pr147 -> Nd147 -> Pm147 -> Sm147 (stable), but each nuclide can absorb a neutron (but with different cross sections). Sm is a moderate neutron poison. And there are heavier nuclides, e.g., Pm155 -> Sm155 -> Eu155 -> Gd155, where Eu and Gd are stronger neutron poisons, but their fractional yields are quite low. Meanwhile, these can provide some idea of the FP vector. http://www.doitpoms.ac.uk/tlplib/nuc..._processes.php http://en.wikipedia.org/wiki/File:Th...ssionYield.svg http://en.wikipedia.org/wiki/Fission...ts_(by_element) http://en.wikipedia.org/wiki/Fission....2C_127_to_132 http://en.wikipedia.org/wiki/Fission..._152.2C_154.29 http://upload.wikimedia.org/wikipedi...nYield.svg.png There are preferred nuclides, i.e., those with high yield fractions. Also of interest - http://en.wikipedia.org/wiki/Fluoride_volatility
Yes, this will take some time.

It shold be fairly straightforward to find the products that need the most attention, we need to set up a formula to account for concentration (based on fission products/decay chains) and 'poisoning/absorbance' via cross secions, pretty straight forward.

To keep things simple a strictly Th232/U233 breeder cycle should be considered including U235 and other fissile isotopes formed in meaningful concenrations for calculations.
Any thoughts?