Tokamak produces radioisotopes

[Enthalpy]

Hello dear friends!

I'd like to propose to produce radioisotopes using the D-D reaction in miniature Tokamaks, especially for medicine.

 

Tokamaks (including stellarators) top the rate of permanent nuclear fusion reactions for a given size and input power
https://en.wikipedia.org/wiki/Tokamak
https://en.wikipedia.org/wiki/Stellarator
so big machines fed with D-T claim to produce net energy (present) at affordable cost (uncertain future)
https://en.wikipedia.org/wiki/ITER

As a neutron source instead, the machines would

• Not try to produce any energy, even less net energy;
• Receive only deuterium (²H or D) without the scarce 50% tritium (³H or T);
• Be 10*10*10 times smaller than Iter with the same operating conditions:
  D=1.2m and 50kW input and 20M€ (...err);
• Emit neutrons to irradiate fertile material like ⁹⁸Mo.

Their activity or misuse would produce little plutonium, tritium and radioactive waste. From my estimates, the isotopes production would be naturally good - maybe at a lower cost than the other alternatives to fission reactors.

 

---------- Figures

 

Welcome to double-checkers, even more as usually, as a 3.7*10¹⁰ factor may well lack somewhere!

 

Iter is to produce 500MW heat (over 400s, let's forget that) from a 17.6MeV reaction, that's 1.8*10²⁰/s. At the same induction, density and 150MK, the D-D reaction is 0.012* as frequent as D-T and the machine is 1000* smaller, for a reaction rate of 2.1*10¹⁵/s. Every second D+D reaction produces ³He+n, the other T+p, but T is consumed 80* faster in a D+T reaction that produces one neutron too: ⁴He+n. So 2.1*10¹⁵/s neutrons as well.

 

The target shall catch all neutrons (how?) and consist of pure ⁹⁸Mo (that's more expensive than presently) in the example I choose. Something (Nitrogen behind graphite and molybdenum? Heavy methane?) shall thermalize the 4kW neutron flux to 77K=6.6meV:

http://www.nndc.bnl.gov/sigma/index.jsp thank you!
(n, total) 6.07b http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=15091&mf=3&mt=1⊄=10
(n, elastic) 5.79b http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=15091&mf=3&mt=2⊄=10
(n, gamma) 0.26b http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=15091&mf=3&mt=102⊄=10
I heavily overinterpretate curves made by models and don't integrate over the energy distribution. Then, the inelastic collisions section is 0.28b and (n, gamma) make 90% of these or 1.9*10¹⁵/s. Still 60% at 300K so money shall decide.

 

Over a 5*24h week, the tokamak produces 8.3*10²⁰ atoms of ⁹⁹Mo. 2.75 days half-life = 343ks exponential decay mean 2.4*10¹⁵Bq=65 000 Ci produced per week.

 

⁹⁹Mo decays fully to ^99mTc used for medical imaging. The worldwide demand is 12 000 Ci per week according to Aiea
https://www.iaea.org/About/Policy/GC/GC54/GC54InfDocuments/English/gc54inf-3-att7_en.pdf
satisfied by one mini-tokamak - rather several ones, since ⁹⁹Mo must be transported swiftly. This allows for:

• Correction of limited errors in my estimate;
• Account for limited design constraints;
• A smaller machine, or if possible less strong fields;
• Production of other radioisotopes;
• Work during daytime.

Build and sell two mini-tokamaks per continent for redundancy.

Marc Schaefer, aka Enthalpy

[swansont]

Over a 5*24h week, the tokamak produces 8.3*10²⁰ atoms of ⁹⁹Mo. 2.75 days half-life = 343ks exponential decay mean 2.4*10¹⁵Bq=65 000 Ci produced per week.

You would not be shipping this continuously, so it is decaying away as you accumulate more. Amount produced ≠ amount avalable

[John Cuthber]

How do they compare (dollars per neutron) with the other sources?
https://en.wikipedia.org/wiki/Neutron_source

[swansont]

 

(n, gamma) 0.26b http://www.nndc.bnl.gov/sigma/getPlot.jsp?evalid=15091&mf=3&mt=102⊄=10

I heavily overinterpretate curves made by models and don't integrate over the energy distribution. Then, the inelastic collisions section is 0.28b and (n, gamma) make 90% of these or 1.9*10¹⁵/s. Still 60% at 300K so money shall decide.

 

Over a 5*24h week, the tokamak produces 8.3*10²⁰ atoms of ⁹⁹Mo. 2.75 days half-life = 343ks exponential decay mean 2.4*10¹⁵Bq=65 000 Ci produced per week.

 

 

How are you getting this number? You appear to assume every neutron is absorbed, and ignore the cross section.

[Enthalpy]

Thanks both for your interest!

You would not be shipping this continuously, so it is decaying away as you accumulate more. Amount produced ≠ amount available

 

Yes. The present chain for ⁹⁹Mo is organized to minimize this loss. Accumulation and shipment are done over few hours. This is an excellent reason to have one or more production sites per continent.

 

It reminds me of how the ancient Romans had ice in their fridges: it was brought down from mountains by slaves.

How do they compare (dollars per neutron) with the other sources?
https://en.wikipedia.org/wiki/Neutron_source

 

If only I knew that! But what I have read is that the other alternatives to fission reactors produce too few neutrons, fusors being worse than accelerators, and that's a strong limitation. Apparently, a small tokamak has at least this argument in favour.

[John Cuthber]

Thanks both for your interest!

 

Yes. The present chain for ⁹⁹Mo is organized to minimize this loss. Accumulation and shipment are done over few hours. This is an excellent reason to have one or more production sites per continent.

 

It reminds me of how the ancient Romans had ice in their fridges: it was brought down from mountains by slaves.

 

If only I knew that! But what I have read is that the other alternatives to fission reactors produce too few neutrons, fusors being worse than accelerators, and that's a strong limitation. Apparently, a small tokamak has at least this argument in favour.

And if the slaves dug a ton of ice a week out of the mountain snow, could Caesar have added a ton of ice to his drinks every week?

[Enthalpy]

 

How are you getting this number? You appear to assume every neutron is absorbed, and ignore the cross section.

 

Yes, that's it. Thick graphite is required, but that's affordable. Cooling it and placing the target needs some engineering. Then, I only compare the relative cross-sections to know what proportion of neutrons is absorbed by ⁹⁸Mo to produce ⁹⁹Mo.

How do they compare (dollars per neutron) with the other sources?

https://en.wikipedia.org/wiki/Neutron_source

 

This paper of 2016 describes their accelerator that produces 10¹²n/s

http://onlinelibrary.wiley.com/doi/10.1002/er.3572/full

and cites existing machines that produce up to 10¹³n/s and announce a goal of 10¹⁴ to 10¹⁵n/s for their future machine. Compare with 2*10¹⁵n/s estimated here.

 

That's a partial picture. All cited machines consume D+T, not D+D as I assumed. Feeding the 0.001*Iter machine with D+T would make 100* more neutrons, or 2*10¹⁷n/s.

It's the basic reason favouring tokamaks to produce energy too. The same energy expense is reused in many collisions in a plasma. An accelerator uses the expended energy in few collision.

 

An other, qualitative comparison: if the operator of a fusor or accelerator has tritium available, which is produced by a fission reactor, he can obtain ⁹⁹Mo from that same fission reactor, more easily and efficiently than by consuming the tritium. The D+D tokamak, as opposed, works without the uranium cycle. This gives more flexibility against politics and ageing reactors.

Edited July 9, 2017 by Enthalpy

[John Cuthber]

 

It's the basic reason favouring tokamaks to produce energy too. The same energy expense is reused in many collisions in a plasma. An accelerator uses the expended energy in few collision.

How do the colliding nuclei know whether they are in a solid target, or in a plasma?

At these sorts of energies, the atoms are not held in place by a lattice.

If a fast deuteron hits another deuteron it's going to lose a lot of energy whether the second one is in a solid target or in a gas.

[Enthalpy]

How do the colliding nuclei know whether they are in a solid target, or in a plasma?

At these sorts of energies, the atoms are not held in place by a lattice.

If a fast deuteron hits another deuteron it's going to lose a lot of energy whether the second one is in a solid target or in a gas.

 

Yes. But in a plasma, that energy is not lost through the elastic collision. If one D or T has more energy than average, it will lose energy over the collisions in average, but the other ions receive this energy, which remains available for future collisions. If, ideally, the plasma is perfectly contained, the only energy loss is by X-ray emission, due to electrons deflected by ions (a good reason to avoid heavier ions in a tokamak).

 

As opposed, with an accelerator, the target nuclei are cold, and the energy transferred by the impinging ion is just spread over the cold matter which can't make successful collisions from that: this energy is lost.

 

The storage ring of an accelerator wouldn't be good neither. Here all ions are "hot" but collisions give them a transverse energy which isn't useable in the future.

 

I can only think of a plasma where collisions keep the energy reusable. Among the devices, fusors and polywells give a small reaction rate, tokamaks a big one.

[John Cuthber]

With an accelerator you take some ions and hurl them at a target with a fairly well defined energy of- say - 100KV.
They hit the target and, if you are lucky, they react and you get neutrons.

With a tokomak, you heat some gas to say 150 million degrees until some of the ions are fast enough to react when they collide.

But most are not moving that fast. The average energy is only something like 15KeV. My old colour telly did better job of accelerating things than that.
So, the difference is that with an accelerator, all those primary collisions actually have enough energy to produce fusion. But with a tokomak nearly all of them don't.
It's far from clear which system is more efficient.
Something like this might give you the best of both worlds
https://en.wikipedia.org/wiki/Inertial_electrostatic_confinement

[swansont]

Yes, that's it. Thick graphite is required, but that's affordable. Cooling it and placing the target needs some engineering. Then, I only compare the relative cross-sections to know what proportion of neutrons is absorbed by ⁹⁸Mo to produce ⁹⁹Mo.

 

More pertinent is how thick would the Mo have to be to capture a large fraction of the neutrons?

 

Thanks both for your interest!

 

 

Yes. The present chain for ⁹⁹Mo is organized to minimize this loss. Accumulation and shipment are done over few hours. This is an excellent reason to have one or more production sites per continent.

 

It reminds me of how the ancient Romans had ice in their fridges: it was brought down from mountains by slaves.

 

Owing to the large amount of energy needed to melt ice, the analogy works if this were a long-lived isotope.

 

 

A quick calculation. The capture cross section for a macroscopic chunk of material is the atom density x individual cross section. So that's something like 10^22 atoms per cm^3 and 10^-24 cm^2. So it's 0.01 per cm.

 

Absorption is an exponential, so you need a meter to absorb a reasonable fraction and get the answer you gave before. How do you extract the radioisotopes? How do you put this material in the tokamak and make it function?

[John Cuthber]

Is it going to be cheaper to get many tonnes of isotopically enriched 98Mo than to get a few tonnes of 235U and build a fission reactor?

How are you going to shred the massive blocks of metal to dissolve them and extract the product?

[Enthalpy]

With an accelerator you take some ions and hurl them at a target with a fairly well defined energy of- say - 100KV.

They hit the target and, if you are lucky, they react and you get neutrons.

 

With a tokamak, you heat some gas to say 150 million degrees until some of the ions are fast enough to react when they collide.

But most are not moving that fast. The average energy is only something like 15KeV. My old colour telly did better job of accelerating things than that.

So, the difference is that with an accelerator, all those primary collisions actually have enough energy to produce fusion. But with a tokamak nearly all of them don't.

It's far from clear which system is more efficient.

Something like this might give you the best of both worlds

https://en.wikipedia.org/wiki/Inertial_electrostatic_confinement

 

Even when the ions have enough energy, which happens at an accelerator, the lucky collisions are rare. Most events are quasi-collisions that dissipate the energy without result. Because the target is cold, the energy lost by the impinging ion isn't available any more for future collisions. In a plasma as opposed, where all ions are hot (if not all hot enough to react), ion collisions don't waste energy: all the heat energy remains available. Only light radiation (and plasma leaks) let the plasma loose energy, and this is a much slower process.

 

The net result is that you get more reactions with the same energy input from a plasma. This is the very reason why attempts to obtain net energy from fusion reactions use a plasma rather than an accelerator.

 

Fusors and polywells (electrostatic confinement) are much worse than tokamaks at obtaining many fusion reactions and at obtaining them with a small power input. The fusion experts' consensus is that they can't produce net energy, for instance. Independently of any rationale, if you check the measured neutron production rate and the reaction rate per power input unit, fusors are very week, much weaker than particle accelerators.

[...] A quick calculation. The capture cross section for a macroscopic chunk of material is the atom density x individual cross section. So that's something like 10^22 atoms per cm^3 and 10^-24 cm^2. So it's 0.01 per cm.

Absorption is an exponential, so you need a meter to absorb a reasonable fraction and get the answer you gave before. [...]

 

Thermalized neutrons are more favourable than that. As the cross-section for elastic collisions exceeds the absorption cross-section by far, they change their direction many times before getting absorbed, instead of flying through in a right path. As soon as the thickness produces many direction changes, the flux decrease isn't any more an exponential depending just on the absorption cross-section.

 

I must have a look first, but my gut feeling is that the flux decrease with thickness resembles a diffusion equation then, with a faster decrease. Interleaving ⁹⁸Mo with elements that deflect neutrons better would improve that effect.

 

That's for material properties. There are engineering means too. At 4K instead of 77K, the exponential decrease length resulting only from the absorption drops from 0.6m to 0.14m. The production of ⁹⁹Mo doesn't need to catch all neutrons neither: I estimated one single 0.001*Iter tokamak to cover ideally twice the worldwide demand, but several machines are needed for short delivery routes and for redundancy, so using 10% or 5% of the neutrons would suffice, even if not intellectually pleasant. Then we're far under 7mm thickness at 4K.

 

Neutrons can also be channelled - and many more subtle operations - but that's too widely out of my knowledge.

Is it going to be cheaper to get many tonnes of isotopically enriched 98Mo than to get a few tonnes of 235U and build a fission reactor?

How are you going to shred the massive blocks of metal to dissolve them and extract the product?

 

⁹⁹Mo doesn't have to be separated from ⁹⁸Mo if I understand properly - but I could be horribly wrong.

⁹⁸Mo is stable, while the decay of ⁹⁹Mo produces the useful ^99mTc which is separated at the hospital

https://en.wikipedia.org/wiki/Technetium-99m_generator

At least, the activity of the presently delivered "⁹⁹Mo" is much lower than if it were pure.

Edited July 10, 2017 by Enthalpy

[swansont]

Thermalized neutrons are more favourable than that. As the cross-section for elastic collisions exceeds the absorption cross-section by far, they change their direction many times before getting absorbed, instead of flying through in a right path. As soon as the thickness produces many direction changes, the flux decrease isn't any more an exponential depending just on the absorption cross-section.

It's less than 10b, so the mean free path for scattering is of order 10cm.

 

And if you're going to use Nitrogen, note that it has an absorption cross section more than an order of magnitude bigger than Mo.

 

I must have a look first, but my gut feeling is that the flux decrease with thickness resembles a diffusion equation then, with a faster decrease. Interleaving ⁹⁸Mo with elements that deflect neutrons better would improve that effect.

 

I don't think that buys you much, and adding a reflector just makes the whole thing bigger.

 

That's for material properties. There are engineering means too. At 4K instead of 77K, the exponential decrease length resulting only from the absorption drops from 0.6m to 0.14m. The production of ⁹⁹Mo doesn't need to catch all neutrons neither: I estimated one single 0.001*Iter tokamak to cover ideally twice the worldwide demand

But your calculation is suspect.

 

You don't know that you can make to tokamak small and have it work as advertised. Did you scale the cost with the size? That's not right. The fixed costs don't scale. Sometimes making it small costs more.

 

Your target has to be outside of whatever containment you have, and whatever moderator you use. (Iron, by the way, also has an absorption cross section 10x bigger than Mo. You've ignored these parasitic losses)

 

Your neutron production is isotropic, so to use them all, you need to surround the reaction chamber completely. After all of the mechanical/structural material, let's say you can get your Mo within 0.75m of the target. Let's make it 0.25m thick. That's more than 2 cubic meters of material. 20 metric tons of the stuff. More if it's not isotopically pure, or otherwise you need to scale by the purity (i.e. lose a factor of 4 for naturally occurring).

 

All of that material is going to take time to put assemble/dismantle. So you have down time to factor is. Another loss for which you must account.

 

 

 

Addendum:

One of the reasons that reactors are a good source of Mo-99 is that the net fission yield is 6.1% via direct production and neutron capture in Mo-98. This means the effective cross section for production is 37 barns — more than two orders of magnitude larger than the capture cross section.

 

https://www.ncbi.nlm.nih.gov/books/NBK215146/

 

Note that the listed neutron fluxes for reactors producing Mo-99 are 10^14 -10^15 cm^-2sec^-1, whereas having a total production of 10^15 sec^-1 means that the flux at a ~1 meter radius is around 10^10 cm^-2sec^-1

 

So you have 100x larger production cross section and 10^4-10^5 larger flux.

[Enthalpy]

[Molybdenum's absorption cross section] is less than 10b, so the mean free path for scattering is of order 10cm.

 

If computing with more details than the exponent, you get figures smaller than that.

I don't think that buys you much, and adding a reflector just makes the whole thing bigger.

 

Deflecting (you write "reflector", I din't) the neutrons by an other material that thermalizes them makes a huge difference, just as I had expected. It gives the neutrons a Brownian motion that lets them pass many times through the molybdenum. Under such conditions, the absorption is much more favourable than in the "straight path" model.

 

Very roughly, if neutrons diffuse from a neutron-rich side to a neutron-poor one through a material that is N times thicker than their mean free path there, they make mean N² collisions during the diffusion, and their Brownian motion is N times longer than the material thickness, giving many more opportunities for the absorbing material to act.

You don't know that you can make to tokamak small and have it work as advertised. Did you scale the cost with the size? That's not right. The fixed costs don't scale. Sometimes making it small costs more.

Your target has to be outside of whatever containment you have, and whatever moderator you use. (Iron, by the way, also has an absorption cross section 10x bigger than Mo. You've ignored these parasitic losses)

 

Your neutron production is isotropic, so to use them all, you need to surround the reaction chamber completely. After all of the mechanical/structural material, let's say you can get your Mo within 0.75m of the target. Let's make it 0.25m thick. That's more than 2 cubic meters of material. 20 metric tons of the stuff. More if it's not isotopically pure, or otherwise you need to scale by the purity (i.e. lose a factor of 4 for naturally occurring).

 

 

A D=2m machine won't cost more than a D=20m one. But yes, I did scale the cost as the volume, which is wrong (as I has suggested). Can anybody do it better at this point?

 

No, the target doesn't have to be outside the containment. Why do you suggest that, and are affirmative? At Iter, the tritium-breeding blankets will be inside.

 

By taking arbitrary and uniform thicknesses, implicitly relying on a wrong neutron straight-path model, you get wrong figures.

Edited July 10, 2017 by Enthalpy

[swansont]

If computing with more details than the exponent, you get figures smaller than that.

I'm using definition of mean free path. The inverse of the macroscopic cross section.

 

Deflecting (you write "reflector", I din't) the neutrons by an other material that thermalizes them makes a huge difference, just as I had expected. It gives the neutrons a Brownian motion that lets them pass many times through the molybdenum. Under such conditions, the absorption is much more favourable than in the "straight path" model.

"Reflector" is what these objects are called in fast reactors.

 

Very roughly, if neutrons diffuse from a neutron-rich side to a neutron-poor one through a material that is N times thicker than their mean free path there, they make mean N² collisions during the diffusion, and their Brownian motion is N times longer than the material thickness, giving many more opportunities for the absorbing material to act.

 

To apply, you need the material to be several times thicker than the mean free path. Which accentuates my other objections (see: 20000 kg of Mo required)

[Enthalpy]

One of the reasons that reactors are a good source of Mo-99 is that the net fission yield is 6.1% via direct production and neutron capture in Mo-98. This means the effective cross section for production is 37 barns — more than two orders of magnitude larger than the capture cross section.

 

https://www.ncbi.nlm.nih.gov/books/NBK215146/

 

Note that the listed neutron fluxes for reactors producing Mo-99 are 10^14 -10^15 cm^-2sec^-1, whereas having a total production of 10^15 sec^-1 means that the flux at a ~1 meter radius is around 10^10 cm^-2sec^-1

 

So you have 100x larger production cross section and 10^4-10^5 larger flux.

 

The question is whether the small tokamak suffices. If a fission reactor can produce more ⁹⁹Mo than necessary, it doesn't make it better.

 

The huge neutron flux in a fission reactor is an equally huge drawback. This, and the fission itself, produces huge amounts of radioisotopes that pollute the desired ⁹⁹Mo, lead to proliferation, and are unwanted independently of any production.

I'm using definition of mean free path. The inverse of the macroscopic cross section.

 

"Reflector" is what these objects are called in fast reactors.

 

 

To apply, you need the material to be several times thicker than the mean free path. Which accentuates my other objections (see: 20000 kg of Mo required)

 

Computing the mean free path gives figures of few cm, not 10cm.

 

"Moderator". Nobody wants to keep fast neutrons.

 

I didn't write "moderator made of molybdenum". Why do you suggest this bad idea? Any reactor designer and design uses several materials.

 

----------

 

I'm a bit uncomfortable by the many times you introduce arguments based on implicit and false assumptions: straight neutron path, only molybdenum, target outside the containment, neutron reflector, and so on. I estimate that your knowledge for neutronics is better than that. In case this is deliberate, it resembles more propaganda than scientific arguments.

[swansont]

 

The question is whether the small tokamak suffices. If a fission reactor can produce more ⁹⁹Mo than necessary, it doesn't make it better.

 

The huge neutron flux in a fission reactor is an equally huge drawback. This, and the fission itself, produces huge amounts of radioisotopes that pollute the desired ⁹⁹Mo, lead to proliferation, and are unwanted independently of any production.

 

 

Reactors already exist. You have to make the case that either you have a cheaper alternative, or that paying a premium is worthwhile.

 

But also, I posted those numbers because you concluded that you could produce a certain amount, and your proposal is about a millionth as efficient as existing technology (in raw production), and yet you claim that you would be able to produce ~5x more than current demand. Something does not add up. And that's before all of the other issues that have been raised.

Here's another problem: For every ~100 g (10 cm^3) of Mo, you have a mole of Mo and a quarter of that is Mo-98. Let's say that presents a 1 cm x 1 cm cross section and is 10 cm thick. Magically, every neutron is captured, and none lost to parasitic events.

 

You are exposing that to 10^10 neutrons/s (as per above). After 5 days, you have somewhere around 2 x 10^15 atoms of Mo-99 (you produced more, but some decayed away) But your sample contains 6 x 10^23 atoms. IOW, your product will have a much smaller fraction of the desired isotope than from sources with a higher neutron flux. "Milking" the system for Tc-99m is going to be that much harder.

 

So that's another advantage the existing systems have: their sample has a much larger density of the desired product.

Computing the mean free path gives figures of few cm, not 10cm.

 

I see no actual calculation from you. I have presented my "back of the envelope" numbers, which were rough estimates (and literally done on an envelope). Easily off by a factor of 2 here or there.

 

Basically you're agreeing with me.

 

"Moderator". Nobody wants to keep fast neutrons.

Reflectors are used in fast reactors; you were objecting to my use of terminology. They serve a different function than the moderator does. But the same purpose as what you have proposed: to scatter neutrons back into the reaction area. Why? Because the neutrons tend to forward-scatter. So what you have proposed is a reflector, in addition to having a moderator.

 

I didn't write "moderator made of molybdenum". Why do you suggest this bad idea? Any reactor designer and design uses several materials.

Good. Neither did I.

 

I pointed out that the moderator will add thickness to the design and reduce your flux (from increasing the surface area and also parasitic losses) I would think these issues would not require explanation. Was I wrong?

 

I'm a bit uncomfortable by the many times you introduce arguments based on implicit and false assumptions: straight neutron path, only molybdenum, target outside the containment, neutron reflector, and so on. I estimate that your knowledge for neutronics is better than that. In case this is deliberate, it resembles more propaganda than scientific arguments.

 

 

The mean free path is standard reactor physics. It is the inverse of the macroscopic cross section.

http://www.nuclear-power.net/nuclear-power/reactor-physics/nuclear-engineering-fundamentals/neutron-nuclear-reactions/mean-free-path/

 

It is, in fact, the average distance a neutron travels before undergoing a reaction.

 

I have not assumed only molybdenum. I have raised several issues that these other materials make your estimations even less reasonable.

 

You were the one who introduced the idea of a neutron reflector. "Interleaving ⁹⁸Mo with elements that deflect neutrons better would improve that effect." In reality I doubt that this would do much since Mo is already much heavier than a neutron. (IOW whatever backscattering you can get isn't going to increase much by going heavier, so it's pointless unless you find something with a much larger scatter cross-section, i.e. it presents a bigger target)

 

If you can put the target inside the containment by all means explain how this is done and who has done it. This is one case where I have assumed something, but you have not presented anything to suggest that a tokamak will operate stuffed full of molybdenum.

[John Cuthber]

 

Even when the ions have enough energy, which happens at an accelerator, the lucky collisions are rare. Most events are quasi-collisions that dissipate the energy without result. Because the target is cold, the energy lost by the impinging ion isn't available any more for future collisions. In a plasma as opposed, where all ions are hot (if not all hot enough to react), ion collisions don't waste energy: all the heat energy remains available. Only light radiation (and plasma leaks) let the plasma loose energy, and this is a much slower process.

 

The net result is that you get more reactions with the same energy input from a plasma. This is the very reason why attempts to obtain net energy from fusion reactions use a plasma rather than an accelerator.

 

Fusors and polywells (electrostatic confinement) are much worse than tokamaks at obtaining many fusion reactions and at obtaining them with a small power input. The fusion experts' consensus is that they can't produce net energy, for instance. Independently of any rationale, if you check the measured neutron production rate and the reaction rate per power input unit, fusors are very week, much weaker than particle accelerators.

 

 

⁹⁹Mo doesn't have to be separated from ⁹⁸Mo if I understand properly - but I could be horribly wrong.

⁹⁸Mo is stable, while the decay of ⁹⁹Mo produces the useful ^99mTc which is separated at the hospital

https://en.wikipedia.org/wiki/Technetium-99m_generator

At least, the activity of the presently delivered "⁹⁹Mo" is much lower than if it were pure.

"Only light radiation (and plasma leaks) let the plasma loose energy, "

You have forgotten bremsstrahlung which gives every "near miss" or collision a chance to waste energy.

 

"The net result is that you get more reactions with the same energy input from a plasma. "

Not all plasmas are tokomaks.

"This is the very reason why attempts to obtain net energy from fusion reactions use a plasma rather than an accelerator."

You are looking for cheap neutrons. That may or may not be the same as looking for energy production.

 

"Fusors and polywells (electrostatic confinement) are much worse than tokamaks at obtaining many fusion reactions and at obtaining them with a small power input. "

Nonsense.

Here's a fusor that runs a kilowatt or so, and produces neturons.

http://makezine.com/projects/make-36-boards/nuclear-fusor/

Please show me the kilowatt tokamak.

"The fusion experts' consensus is that they can't produce net energy, "

You need better experts. Any fusion reactor must release energy.

Even if it only fuses 2 deuterons a week, that's a net release of energy.

They may never be economical but, again, energy production isn't the goal here.

You want neutrons.

 

"⁹⁹Mo doesn't have to be separated from ⁹⁸Mo if I understand properly - but I could be horribly wrong."

I think you are horribly wrong.

If I am reading this

http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/28/060/28060364.pdf

correctly, the other isotopes have higher neutron cross sections so nearly all your precious neutrons would be eaten up by the "wrong" Mo isotopes and in the end, just produce heat.

[swansont]

The fusion experts' consensus is that they can't produce net energy, "

You need better experts. Any fusion reactor must release energy.

Even if it only fuses 2 deuterons a week, that's a net release of energy.

Um, no. Net energy accounts for running the equipment that enables the fusion. Fusion (human-driven, on earth) rarely produces net energy. Just a few bursts here and there.

[John Cuthber]

 

Um, no. Net energy accounts for running the equipment that enables the fusion. Fusion (human-driven, on earth) rarely produces net energy. Just a few bursts here and there.

Imagine I have a fusor in my cellar and I want to heat the house.

I can fill it with air or hydrogen and get zero fusion.

The heat generated as "waste" from the pumps, power supply and so on will still heat the house.

If I fill it with deuterium I will still get that heat (which I pay for as my electricity bill) and I will get the additional heat from fusion.

So fusion's a net benefit, even if it's only one reaction a week.

The problem is that it's a very expensive investment to get very little heat, but that's not the same thing. It's economics, not physics.

 

The same logic applies if I want to heat water to generate steam and thus make electricity.

Any fusion is a net energy gain.

Edited July 15, 2017 by John Cuthber

[Sensei]

net = output - input

It can be positive value when output > input,

or negative value when output < input.

 

In currently existing fusion reactors, a lot of energy is wasted on cooling superconducting materials, creating superconducting magnets, which squeeze plasma.

Discovery of high temperature superconducting materials, or other ways to create powerful magnets, would help decreasing energy needed by fusion reactors.

 

[swansont]

Imagine I have a fusor in my cellar and I want to heat the house.

I can fill it with air or hydrogen and get zero fusion.

The heat generated as "waste" from the pumps, power supply and so on will still heat the house.

If I fill it with deuterium I will still get that heat (which I pay for as my electricity bill) and I will get the additional heat from fusion.

So fusion's a net benefit, even if it's only one reaction a week.

The problem is that it's a very expensive investment to get very little heat, but that's not the same thing. It's economics, not physics.

 

The same logic applies if I want to heat water to generate steam and thus make electricity.

Any fusion is a net energy gain.

If you can't unplug it from the wall and have it run itself, it's not generating net energy. You don't need to plug in a nuclear reactor, or any other device a fusion reactor would replace, to make it work.

[John Cuthber]

If you can't unplug it from the wall and have it run itself, it's not generating net energy. You don't need to plug in a nuclear reactor, or any other device a fusion reactor would replace, to make it work.

If you can't unplug it from the wall and have it run itself, it's not reached break-even.

Every fusion produces a few MeV of energy.

It's a net gain- albeit a tiny one.

I can buy a KWHr hour of electricity and warm my house with 1.0000000000000001 KWhr of energy.

That's a net gain.

A tiny, pointless one; but a net gain.

[swansont]

If you can't unplug it from the wall and have it run itself, it's not reached break-even.

Every fusion produces a few MeV of energy.

It's a net gain- albeit a tiny one.

I can buy a KWHr hour of electricity and warm my house with 1.0000000000000001 KWhr of energy.

That's a net gain.

A tiny, pointless one; but a net gain.

Break-even and net energy are talking about useful energy generation, not heat.