p + 15N

[kunkmiester]

It would even be a perfect candidate for SSTO, since the reason it carries an enormous load of fuel is because it needs to produce a truckload of thrust to counteract air resistance along with gravity.

Hey, wait, what was this thing using as fuel again? Wink

I can already imagine the coolness of having launch vehicles with massive jet intakes. Given the fact that you need to carry only a tiny bit of fuel (if any at all), this would make spacecraft almost featherlight. The only thing that would add weight is payload (and some fuel supply that you'd use when you're clear of atmo). The word 'holy grail' comes to mind.

Problem is, the .37% number. You have to get a LOT of air to get that .37% of the nitrogen you want, when you have to get several grams a minute probably. And is your isotope separator going to weigh less than the relatively small tank of fuel you'd be carrying instead?

[icarus]

15N gas for sale .... 250ml, 500ml, 1L and 5L quantities.

~ USD500 for 1litre (~1.6 grams)

http://www.sigmaaldrich.com/chemistry/s ... e=11594607

... and to put things in perspective, there is almost 10 times more 15N in the atmosphere than CO2 ...

[happyjack27]

happyjack:

you really need to consider the rate of energy dissipation (loss) rather than the amount of energy stored in the system (conserved).

Since you choose to interject why don't you just tell us the answer you are alluding to know .... ?

Cryptic half-statements just smell like BS .... surely not.

i don't know what the answer is. but its mathematics tells you that, well, you're really talking about how much energy you have to put into the system, right? and by conservation of energy, when the system reaches equilibrium (i.e. the total energy & mass in the system stays constant) that would have to be exactly equal to how much energy goes out of the system. because at equilibrium, energy in minus energy out = change of energy in system = 0 (by definition of "equilibrium"). that's all i'm saying. i don't know the math, but presumably rate of energy out is proportional to amount of energy inside the system, anyways. (like the heat equation - transfer of heat across a membrane is proportional to the difference in temperature between the two sides - or the equation for current and voltage, which is the same thing - current is proportional to difference in voltage). so i guess it's the "same difference", as they say around here.

[D Tibbets]

P-15N fusion, along with P-O17 fusion seems to be the fastest of the CNO fusion cycles. This reference shows the reaction rate of various reactions. Toggle the radiobuttons to compare all of the posibilities listed. The graph only goes to 100 KeV, but by looking at the slopes of the graph, it is probable that the two above reactions catches, or even passes D-D fusion at perhaps ~ 150- 200 KeV. The overall CNO cycle is ~ 1% of the D-D fusion rate at these levels (from another source) and this would fit the extrapolated graphs.

http://www.astrophysicsspectator.com/to ... nRate.html

Two problems for 15N or 17O is that they are heavier elements, and I understand the isotopic purification becomes more difficult as you go up the Periodic table..

The other problem (as mentioned above) is Bremsstrulung radiation. This goes up logrithmically as the Z of the fuel increases. There are some who argue that P-B11 reactions cannot reach breakeven because of this. 15N or 17O would only be worse.

Bussard did mention in one paper the minimum temperature needed for the fusion rate to overcome the Bremsstrulung rate. D-T was ~ 5 KeV, D-3He was ~ 15 KeV, and P-11B was ~ 35-50 KeV (if I remember correctly). Strangely, D-D was not listed in the brief table. Bussard also said that the minimum Barns needed was ~ 0.01. I think 15N and 17O would both qualify, but at temperatures higher than P-11B. Theoretically, they are possible, but are greater engineering challenges, larger machines, etc.
Considering all of the tradeoffs, I don't know which would be best for particular situations, but if the system works, you are presented with several fuel combinations, within your engineering limits.

Dan Tibbets

[icarus]

Oh, I see, you made out like you knew something specific about chris's magnetron device ... you were just spouting off in general, carry on.

[happyjack27]

Oh, I see, you made out like you knew something specific about chris's magnetron device ... you were just spouting off in general, carry on.

yes, i was just spouting off in general. i didn't mean to sound like i knew anything specific. i don't. sorry for any confusion.

[D Tibbets]

Concerning the penetrating radiation threat from P-11B. I have heard two numbers.
Neutrons making up ~ one 60 millionth (compared to one half for D-D) of the energy output.
This assumes isotopicallly pure hydrogen.

Gamma radiation of ~ 10^-4. I don't know the source of this estimate, just that it has been mentioned. The only reference I've seen is the side reation of 12C * 12C making a gamma and Magnesium isotope. This had a probability of 10^-10, so it would be much lower than the quoted estimate, but there may be some other side reaction ( perhaps involving the intermediate berylium that I do not know about) or perhaps the excited 12C isomer that is formed is not the same as the ground state 12C's that I mentioned.

In any case, I don't know the energy of the gamma, so I don't know it's penatrability.
Assumeing it is similar to D-D fusion neutrons and causes similar biological damage, then the comparative shielding needed for P-11B compared to D-D would be 0.5 / 0.0001 or a difference of ~ 2000.
Using the 90% blockage rule, if 1 inch of material (equivalent weight) is needed to shield the boron reactor, then ~ 3.2 inches would be needed to stop the deuterium.

A further consideration is the direct conversion. If this allows you to retrieve ~ three times as much useful electricity per unit of fusion, then the reactor could be run at 1/3 the levels to get the job done. This would push the shielding advantage closer to 4:1.

Another advantage is the charged particle from these advanced fuels lend themselves to magnetic steering and can thus be more easily used for direct thrust.

Also, consider that the reactor (if on a space ship) only needs to be shielded on one side, assuming the reactor is on the back of the ship.

Finally, the weight dedicated to shielding may be overshadowed by the weight needed for heat dissipation/ radiators. Reactors that utilize direct conversion has a distinct advantage here. Assume the advantage is ~ 3X. That combined with the decreased shielding, should result in a weight saving in shielding and radiation systems of ~ 10X. This has to be counterbalanced against the presumably smaller and possibly lighter D-D reactor.

Dan Tibbets

[chrismb]

Concerning the penetrating radiation threat from P-11B. I have heard ...

What is your source?

Wiki says;

"Detailed calculations show that at least 0.1% of the reactions in a thermal p–11B plasma would produce neutrons, and the energy of these neutrons would account for less than 0.2% of the total energy released.[28]"

{28: # ^ Heindler and Kernbichler, Proc. 5th Intl. Conf. on Emerging Nuclear Energy Systems, 1989, pp. 177-82. Even though 0.1% is a small fraction, the dose is rate still high enough to require very good shielding, as illustrated by the following calculation. Assume we have a very small reactor producing 30 kW of total fusion power (a full-scale power reactor might produce 100,000 times more than this) and 30 W in the form of neutrons. If there is no significant shielding, a worker in the next room, 10 m away, might intercept (0.5 m²)/(4 pi (10 m)2) = 4×10−4 of this power, i.e., 0.012 W. With 70 kg body mass and the definition 1 gray = 1 J/kg, we find a dose rate of 0.00017 Gy/s. Using a quality factor of 20 for fast neutrons, this is equivalent to 3.4 millisieverts. The maximum yearly occupational dose of 50 mSv will be reached in 15 s, the fatal (LD50) dose of 5 Sv will be reached in half an hour. If very effective precautions are not taken, the neutrons would also activate the structure so that remote maintenance and radioactive waste disposal would be necessary.}

[Stoney3K]

Assume we have a very small reactor producing 30 kW of total fusion power (a full-scale power reactor might produce 100,000 times more than this) and 30 W in the form of neutrons. If there is no significant shielding, a worker in the next room, 10 m away, might intercept (0.5 m²)/(4 pi (10 m)2) = 4×10−4 of this power, i.e., 0.012 W. With 70 kg body mass and the definition 1 gray = 1 J/kg, we find a dose rate of 0.00017 Gy/s. Using a quality factor of 20 for fast neutrons, this is equivalent to 3.4 millisieverts.

You're blindly applying the inverse square law to determine the dose received by said worker.

That would work, but only if there were no walls between the reactor and said worker, or even air! I guess the presence of sheer vacuum would pose a more significant and direct threat to the worker in question than neutron radiation coming from the reactor.

I guess even a 4" wall would have some effects on the dose received. I have no idea why though, it had something to do with particles and their properties when running into solid brick walls....

[D Tibbets]

Concerning the penetrating radiation threat from P-11B. I have heard ...

What is your source?

Wiki says;

"Detailed calculations show that at least 0.1% of the reactions in a thermal p–11B plasma would produce neutrons, and the energy of these neutrons would account for less than 0.2% of the total energy released.[28]"....

My source is hear-say from 23617 (actually made up name, but someone with a similar name posted it here I believe). [EDIT] Actually it was 93143. The 7th post on this page might be the most definative:

viewtopic.php?t=2561&highlight=neutrons+boron

He states that the 12C- 4He side reaction is the most significant neutron producing reaction and it is supressed in large part because the alphas do not hang around in the Polywell.

The above claimed 0.1% neutrogenicity is based on a thermalized plasma. I don't know the average assumed temperature or the high energy tail distribution and what resultant temperatures are being used to calculate the rate of possible side reactions. But, I think everyone agrees that , a thermalized plasma cannot reach breakeven for P-11B fusion anyway, so the point would be mostly moot. They need to modify their assumptions for the numbers to have any real significance. Also, I don't know if they included the natural deuterium in the hydrogen mix (whether it was isotropically pure hydrogen/ protons).

Concerning shielding by air. I don't know the value for neutrons, but for gamma rays it is ~ 50% reduction for 1.5 inches of steel, ~ 6 inches for concrete, ~ 24 inches for wood. I suspect it is at least several hundred or more meters for air. ie: air shielding would only be a small correction to the inverse square law mediated decrease in neutron radiation.

Dan Tibbets

[ladajo]

Time, Distance and Shielding baby!

[chrismb]

You're blindly applying the inverse square law to determine the dose received by said worker.

a) inverse square law is the one that applies, and is correctly applied blindly (why not? and if why not, do you have a better one to apply!?)
b) this was a direct quote from the wiki page. I wasn't applying anything....

[chrismb]

He states that the 12C- 4He side reaction is the most significant neutron producing reaction and it is supressed in large part because the alphas do not hang around in the Polywell.

I think the correct term is 'he speculates that...'.

Let's be very clear here that when we blow hot-air about p-11B this-or-that, in the first instance we have to at least presume the worst case for radiation. Worst case, or otherwise, the only one we've got right now is for a thermal plasma. I recommend you presume to need was someone has already recommended in a published work, and if you find polywell runs cleaner, great... if you find polywell doesn't run, well.... y'know, let's just remember that we're not talking about something 'known' here. You may well find that polywell p-11b runs more aneutronic than expected. No-one knows, so let's stick to calculations from known circumstances in the first instance, eh?

[chrismb]

~ USD500 for 1litre (~1.6 grams)

Thanks for the look-up.

OK, so 1.6 g = 6e22 15N atoms, at 5MeV per reaction (per atom) = 5e10J

Let's say 10kWh = (36MJ) == $1, so that 1 litre 15N represents $1380.

If that $500 is a representative cost of the effort of getting 15N, it looks like cost/effort will be against this, because even if 100% of the fusion energy could be recovered to heat, then the conversion to power (30%) would mean there is only $400 worth of energy to get out of it. So the whole plan would end up about how efficiently that 15N could be extracted from the atmosphere, because the numbers don't work out for p15N at the moment with current isotope extraction....

[Stoney3K]

You're blindly applying the inverse square law to determine the dose received by said worker.

a) inverse square law is the one that applies, and is correctly applied blindly (why not? and if why not, do you have a better one to apply!?)

I wasn't implying the law wouldn't apply. I was only trying to make a point that your calculated dosage would only be valid as an absolute worst-case scenario with nothing in the way, e.g. a 10m vacuum chamber with both the reactor AND the worker inside. Neutron radiation + solid object = less intense neutron radiation. (And a rather hot solid object)

I hope you can understand such a thing would be impractical, both from the worker's POV, as well as from the reactor's, which is going to need something solid around it anyway to keep all of the air out. (or the vacuum in, whatever way you prefer)