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[KitemanSA]

Also, almost all tritium nuclei will react with deuterium giving 17.6MeV per event.

This is an interesting thought.

For many years this forum assumed that the alphas from the pB&J would exit the MaGrid immediately. The same assumption was held, AFAIK, regarding the fusion products from the WB6 D-D reaction. A comment by Dr Nebel lead to further examination regarding the alphas in a 7T level field and lo-and-behold, they DIDN'T exit the MaGrid immediately but bounced around a bit and left only nearer the cusps.

Has anyone done an equivalent assessment of the products of the WB6 level fields with its products? Are we WRONG to assume each D-D reaction yields HALF a neutron when it might also result in a D-T reaction that produces one?

Hmmm, maybe the D-D reaction rate is half what we've assumed by neutron count, but it is accompanied by a D-T half the time which makes a neutron ALL the time.

More math to do boys! Git bizzay!

[ladajo]

So, from first type of reaction 4.03 MeV= 4.03E6 eV
While from the second - 0.82+2.45=3.27 MeV = 3.27E6 eV
Equal probability means that if e.g. million events occured, there will be aprox 500000 of first and 500000 of the second type. Not million times of both types so not 7.3E6 eV/event.
But average energy gain from each DD event 7.3/2=3.65E6 eV if to be correct. And not aprox value 4E6 eV as I typed for simplicity

Thanks Joseph for that.

To walk the dog one more time and correct my math error:

With 2.5E9 Fusions/sec
and 50% D+D=T+P and 50% D+D=3He+N
therefore 50% 1.01MeV + 3.02MeV
and 50% .82MeV + 2.45Mev

For stated fusion rate, that means:
1.25E9 (T+P) * (1.01E6 + 3.02E6) = 5.04E15
and the other half of the products
1.25 (3He+N) * (.82E6 + 2.45E6) = 4.09E15
and, adding the two reactions for Total sum eV energy gives:
5.04E15+4.09E15 = 9.13E15
and then converting to Wattsecs:
9.13E15 * 1.6eV/Joule = 1.46 mW

So you are right. Thanks or helping clarify.

[Joseph Chikva]

For many years this forum assumed that the alphas from the pB&J would exit the MaGrid immediately.

Yes, they will escape zone immediately if their kinetic energy will exceed the depth of potential well.
Similarity.
This is like to that if the bode has energy exceeded potential well of Earth (body has velocity exceeded the first space velocity - this is right term in English?) only in this case that can escape the gravitational field.

And if the well is deeper than kinetic energy, that high energy particle will take participation in thermalization.

[Joseph Chikva]

So, from first type of reaction 4.03 MeV= 4.03E6 eV
While from the second - 0.82+2.45=3.27 MeV = 3.27E6 eV
Equal probability means that if e.g. million events occured, there will be aprox 500000 of first and 500000 of the second type. Not million times of both types so not 7.3E6 eV/event.
But average energy gain from each DD event 7.3/2=3.65E6 eV if to be correct. And not aprox value 4E6 eV as I typed for simplicity

Thanks Joseph for that.

To walk the dog one more time and correct my math error:

With 2.5E9 Fusions/sec
and 50% D+D=T+P and 50% D+D=3He+N
therefore 50% 1.01MeV + 3.02MeV
and 50% .82MeV + 2.45Mev

For stated fusion rate, that means:
1.25E9 (T+P) * (1.01E6 + 3.02E6) = 5.04E15
and the other half of the products
1.25 (3He+N) * (.82E6 + 2.45E6) = 4.09E15
and, adding the two reactions for Total sum eV energy gives:
5.04E15+4.09E15 = 9.13E15
and then converting to Wattsecs:
9.13E15 * 1.6eV/Joule = 1.46 mW

So you are right. Thanks or helping clarify.

You are welcome any time.

[KitemanSA]

For many years this forum assumed that the alphas from the pB&J would exit the MaGrid immediately.

Yes, they will escape zone immediately if their kinetic energy will exceed the depth of potential well.

That was the question I was raising. FUSORS will do this because there is no magnetic field.
Dr N suggested that this WOULDN'T be the case in a full size Polywell with a 7T field and the alphas from a pB&J reaction (p-¹¹B if you are not up on our slang). Someone checked and at those conditions, the gyro radius of the alphas was about 5cm, IIRC, so they would be turned by the MaGrid. My only remaining question is would the .1T field of the WB6 restrain the T and allow it to react with the remaining D? I suspect not, but....

This is like to that if the bode has energy exceeded potential well of Earth (body has velocity exceeded the first space velocity - this is right term in English?) only in this case that can escape the gravitational field.

And if the well is deeper than kinetic energy, that high energy particle will take participation in thermalization.

To me, "first space" would seem to indicate LEO insertion (Low Earth Orbit) but I think the actual term you are looking for is "Escape Velocity" or more specifically "Earth Escape Velocity".

[KitemanSA]

With 2.5E9 Fusions/sec
.....
9.13E15 * 1.6eV/Joule = 1.46 mW

Funny, the only published value for fusion rate I can find is the Valencia paper at 1E9 fusions/sec which results in a power level of ~ 0.6mW as shown in the Polywell Wiki FAQ.

Source for 2.5E9?

[Joseph Chikva]

Dr N suggested that this WOULDN'T be the case in a full size Polywell with a 7T field

I think at 7 T some charged particles (not only alphas) will be confined. As 7 T is really high field. If I recall correctly ITER has about 5.5 T of axial and I think that about 0.5 T poloidal. And there for ITER is meant that charged products will be confined.
But for magnetic traps there is the losses cone and and all depends on angle. So, some particles will confine and some escape. I think that if thermalization is problem you prefer that all would escape.

[Joseph Chikva]

To me, "first space" would seem to indicate LEO insertion (Low Earth Orbit) but I think the actual term you are looking for is "Escape Velocity" or more specifically "Earth Escape Velocity".

You are right. Sorry. The second - not first.
But you understood what I meant.

[ladajo]

With 2.5E9 Fusions/sec
.....
9.13E15 * 1.6eV/Joule = 1.46 mW

Funny, the only published value for fusion rate I can find is the Valencia paper at 1E9 fusions/sec which results in a power level of ~ 0.6mW as shown in the Polywell Wiki FAQ.

Source for 2.5E9?

I normally go with 1E9, but the quoted 2.5E9 is from the "Quick History" Document posted on the EMC2Fusion homepage.

WB-6 showed 1/10 of loss coefficient of WB-4, and ran as a deep well Polywell at 10-12keV, producing DD fusions at 2.5E9 fus/sec. This is 200,000 times higher than the early work of Hirsch/Farnsworth and a world’s record for such IEF devices at same conditions.

So we can say that for DD that is about 1.46mW, and say for arguments sake that they have gotten another magnitude out of WB8 so far, then it would be 14.6mW for 2.5E10 fus/sec, and they would need 2.5E11 fus/sec for 145mW, and 2.5E12 fus/sec to get 1.45W.

The $64 question is what did they get out of WB8 so far, and how does it compare to scaling law? We know that they went for 10X on B-Field, (but maybe got 8X, and we think we know that it is bigger in diameter. From the drawings, we have guessed 1.5 to 2X diameter. So if we go conservative and say 1.5X on diameter, how does that stack up for power scaling? 8X B-Field and 1.5X on coil diameter with 1.46mW for WB6, means what target for WB8?

[KitemanSA]

So if we go conservative and say 1.5X on diameter, how does that stack up for power scaling? 8X B-Field and 1.5X on coil diameter with 1.46mW for WB6, means what target for WB8?

(your words - WB8 two orders more powerful than WB6)

Assuming scaling is as expected (and hoped) WB-8 would be 5 orders more powerful than WB-6. This of course assumes going from .1 T to .8 T for B field and radius increased 1.5 times (estimated from the EMC2 website jpg labeled WB-8 ). This says nothing about confinement voltage changes which might also be increased, but that was not mentioned in the contract.
Best regards

8^4*1.5^3=13824 ~ 4 orders, but who's counting

So 13824*1.46mW ~ 20W.

[ladajo]

Also, almost all tritium nuclei will react with deuterium giving 17.6MeV per event.

This is an interesting thought.

For many years this forum assumed that the alphas from the pB&J would exit the MaGrid immediately. The same assumption was held, AFAIK, regarding the fusion products from the WB6 D-D reaction. A comment by Dr Nebel lead to further examination regarding the alphas in a 7T level field and lo-and-behold, they DIDN'T exit the MaGrid immediately but bounced around a bit and left only nearer the cusps.

Has anyone done an equivalent assessment of the products of the WB6 level fields with its products? Are we WRONG to assume each D-D reaction yields HALF a neutron when it might also result in a D-T reaction that produces one?

Hmmm, maybe the D-D reaction rate is half what we've assumed by neutron count, but it is accompanied by a D-T half the time which makes a neutron ALL the time.

More math to do boys! Git bizzay!

So...

(15KeV)D+D = 50% T(1.01MeV) +P(3.02MeV) and 50% 3He(.82MeV) + N(2.45MeV)

and...the New T's AND P's running around at 50% the fusion rate means possibly...

(13.6KeV)D+T = 4He (3.517MeV) = n (14.069MeV)
P+P = D + (e+) + (v sub e) + .42MeV (But this chain produces about nil for power density...think our sun...so we can ignore it)

and possibly a little...but not likely at 58KeV, so we can discount it...
D+He = He4(3.6MeV) + p(14.7MeV)

and note the D from P + P, which in turn can chain some more...

(15KeV)D+D = negligiable power
and
(13.6KeV)D+T = negligiable power due to low input D count from P+P reaction

So overall, it would seem we can get a decent secondary chain from the initial T products acting within the D fuel construct.

In summary:

(15KeV)D+D = 50% T(1.01MeV) +P(3.02MeV) and 50% He3(.82MeV) + N(2.45MeV)

and a viable secondary chain of:
(13.6KeV)D+T = He4 (3.517MeV) + n (14.069MeV)

means that (detector dependant) for every neutron we see on a counter, it could be a D+D neutron OR a D+T neutron.

So, now we need to look at the cross sections and see how the barns stack up for the distribution probability. But without number crunching, the fact that D+T is easier than D+D, should mean to me that we would not see much of a T population accruing in a D+D machine, as it would most certainly burn off. So for arguments sake, we should see 2 neutrons for every one D+D event. But that also means that the power produced must also account for the double burn of the D+D and the D+T secondary.
On simple terms: 2.5E9 DD fus/sec should be about 5mW, not counting unburned fuel that escapes confinement.

Following on, this line of thought would argue that in a machine counting x neutrons, 50% of DD produces a n count, and the other 50% produces a T which then (assumed 100% burn) produces another n. So, that would say that every D+D makes 2 neutrons. Thus, the initial DD fusion rate is 50% of the neutron count, but then one must also account for additional energy produced by the T secondary.

In our case we thought initially that the WB6 machine produced 1.46mW, whereas looking at it this way, it produced 5mW.
2.5E9 fus/sec
1.25 T+p events/sec for a total of 5.04E15 in energy
1.25 He3 + n events/sec for a total of 4.09E15 in energy
and secondary
1.25E9 He4 + n events/sec for a total of 2.2E16 in energy
which gives
3.11E16 in total energy, converted to Watts gives
.005 Wattsecs
or
5mW

Thoughts?

This, if I am not mistaken, also means that Joel Rogers did his math wrong in his latest paper at IEC 2011.

[ladajo]

And, using the B^4R^3 rule...

We should expect no less than 69W out of WB8...

[ladajo]

And as my brain wanders a little more, it begs the question of using seed fuel in a PB&J reactor of D to "goose" the process, like adding a little nitrus. Of ocurse you would get some pesky neutrons zipping around...but hey, properly managed during start up, could be a good warm-up kick if need be to put some extra protons in the mix.

[mvanwink5]

Ladajo,
EMC2's latest recovery.gov report said

During 4Q of 2011, EMC2 has modified the electron injectors to increase the plasma heating. The higher plasma density in WB-8 prompted the need for higher heating power. We plan to operate WB-8 in high beta regime with the modified electron injectors during 1Q of 2012.

So, one might take that report to imply higher than expected densities and maybe they are running at higher drive voltages too (effect on barnes?).

[Joseph Chikva]

We know that they went for 10X on B-Field, (but maybe got 8X, and we think we know that it is bigger in diameter. From the drawings, we have guessed 1.5 to 2X diameter. So if we go conservative and say 1.5X on diameter, how does that stack up for power scaling?

Ladajo,
Scaling B^4*R^3 implies from one assumption: the same behavior of plasma at different densities.
But the following parameters change with the changing of plasma density:
-free pass of particles
-Debye length
-plasma frequency
So, you will have another properties of plasma and as a rule some instabilities not occurring at lower densities take place at higher.
And, so, I am not sure in that scaling law. To be correct sure that this law is not applicable at all.
As fusion researches have not less than 60 years of history and in case of absence of instabilities that law would be correct for any device and there is not a problem to create even 20 Tesla for short solenoids of B-field and and by increasing of B-field in any fusion device from 1 T to 20 you will get in 20^4=160000 higher fusion rate.

For example I saw the single investigation of instabilities (quoted above), in which is shown that 2-stream instability will be damped with the help of large angular momentums of background electrons.
This regards to electron-electron 2-stream.
But at orders of magnitude higher densities also electron-ion 2-stream may take place. And ions before thermalization will not have large angular momentums.