How Will LPP Get There From Here?


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Posted by Rezwan on Jun 25, 2010 at 04:32 PM
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FFI (Focus Fusion 1), LPP’s experimental device, has achieved higher fusion yields than have been achieved with any other DPF at the same peak current. 

Though remarkable, this yield is still 5 orders of magnitude short of the fusion yield required to prove scientific feasibility of focus fusion.  How does the team hope to make up the difference? 

Step by experimental step:  stairway to fusion

How will LPP go from their current 1/12 of a Joule of fusion energy to 33,000 Joules of fusion energy? 

Figure 1 depicts LPP’s past and planned fusion yields per shot in Joules.  The team will need to get over 10,000 Joules per shot to demonstrate scientific feasibility of net energy production.  Their theory predicts that they may ultimately get as high as ~33,000 Joules per shot.  (”~” means “approximately”.  Pointing this out because if the font is small, it looks like a minus sign.)

x-axis represents stages of research Figure 1.  LPP’s past and planned fusion energy yields per shot in Joules

The pink points in the chart above correspond to yields actually achieved so far.

The blue points correspond to LPP’s goals based on the theories they are testing. 

This is a simplified representation of what LPP plans to do, and should give a rough idea of the jump in yield for each experimental stage.  We used to have dates on the x-axis, but this confused people into thinking there is a set schedule.  Ideally there would be, but contingency issues emerge.  Things don’t always go smoothly as we know from the switch delays.  As such, we’ll take it one stage at a time and not fantasize about the end until all the conditions of the present stage are set for optimum research results.  After the stage is completed, we can enter the time it took.

Research parameters and anticipated yields

Each of the blue points in Figure 1 is plotted based on the theories being tested by the LPP experiment.  Figure 2 below shows the variables that are expected to cause an increase in the fusion yield (left column), and the factors by which the yield is expected to increase (right column).

Variable/causeFactor of increase
Scaling with increased current, I5 scaling to 1.4 MA55
Scaling with increased current, I4 scaling from 1.4 MA to 2.8 MA16
Optimization of axial magnetic field3
Subtotal (55 x 16 x 3)2640
(Change in fuel to pB11) 
Increase in energy yield per reaction pB11 vs. DD3.6
Increase in reaction rate pB11@600keV vs. DD@100 Kev12
Additional compression for pB113.7
Subtotal (3.6 x 12 x 3.7) 160
Total increase factor expected (2640 x 160)422,000
Ultimate fusion yield (~1/12 Joules x 422,000)33,000 Joules
Figure 2.  Variable/cause and corresponding factor of increase

 

The points in Figure 1 were obtained by taking LPP’s recent yield and multiplying by the factors at each step of the way.  Multiplying the factors gives an expected increase of 422,000 times. Taking the current level of slightly less than 1/12 of a Joule that LPP has achieved and multiplying it by 422,000 yield gives us ~33,000.  If all goes well, the experiment will validate this theory and follow the points.

Theoretical basis for anticipated yield

The first two variables in Figure 2 above (increasing current) are based on LPP’s theory, but they are backed up by extensive experiment [links needed]. So far, LPP has been achieving much faster scaling, almost I7

The third item (optimization of axial magnetic field) is also based on LPP’s theory, but requires experimental verification. 

For changing the fuel to pB11, the first two variables LPP is certain of, and are based on well-established measurements by others.  [links needed] 

The third item (additional compression for pB11 with a DPF) is also based on LPP’s theory, which has to be experimentally verified.

Why are 10,000 Joules required for scientific feasibilitiy?

As noted, we need at least 10,000 Joules per shot.  The team hopes that they will ultimately get ~33,000 Joules per shot and that this will demonstrate scientific feasibility of net energy production with this device and pB11 fuel. 

The 33,000 Joule yield was derived based on the idea of firing at full capacity for a capacitor bank of 100,000 Joules. 

Some of you may be wondering why a yield of 33,000 Joules from a shot of 100,000 Joules represents scientific feasibility.  Doesn’t that indicate a loss of 67,000 Joules?

33,000 Joules is the “fusion energy yield”.  This is how much additional energy comes into the system from fusion reactions.  This means you start with 100,000 Joules and you get a yield of ~33,000 Joules of fusion energy.  Well then – that means you now have 133,000 Joules, right?  Sounds like a 33% increase in energy!  Net energy and beyond!  Sounds like you can afford to lose an order of magnitude.  After all, 103,000 Joules would be 3,000 Joules of net energy, no?

Sadly, no.  Energy is lost to inefficiencies.  The goal for fusion yield has to be high enough to make up for losses of the system.  Assuming 80% efficiency, (80% x 133,000 Joules) gives you 106,400 Joules – 6,400 Joules of net energy.  Electric energy recovery efficiency is a variable that can be increased to a certain extent by more careful engineering. 

It was stated above that scientific feasibility could be had with a minimum of 10,000 Joules of fusion yield.  10,000 joules would require a system efficiency of at least 91%. 

Here is the hypothetical sequence

[Volunteers required to animate this]

  • A shot is fired.
  •  
  • An initial current of 100,000 Joules enters the system. 
  •  
  • About 70,000 go toward generating the “pinch” and making fusion happen.
  •  
  • The other 30,000 are not lost.  They are recovered/recycled – stored in a second capacitor bank called the mirror capacitors that is charging up for the next shot. 
  •  
  • The 70,000 Joules in the pinch will theoretically yield 33,000 Joules of fusion-generated energy. 
  •  
  • 70,000 + 33,000 gives us 103,000 Joules of energy to be recovered in the ion beam conversion device.  103,000 Joules from the ion beam conversion device + 30,000 recovered from the shot gives us 133,000 Joules. 
  •  
  • Less ~20% energy lost by inefficiencies and you end up with the 106,400 Joules. 
  •  
  • And, of course, “your mileage may vary”.
  •  

“Scientific feasibility of net energy production” vs. “net energy”

This phase of the LPP experiment measures “fusion yields,” not “net energy”. 

We speak of “scientific feasibility of net energy production” and not “net energy” because the experiment will demonstrate if net energy is feasible, without actually generating net energy. 

The shots LPP fires do not need to produce net energy.  All they need is fusion energy that is a big enough fraction of our total input – over 10% and probably 30%, depending on recovery efficiency. 

“Net energy” itself awaits Phase II, the prototype reactor phase, in which a team of engineers determines how to recover energy and crank up the efficiency.  And of course, this all depends on the success of this Phase I – proof of concept.  Stay tuned!

 


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What a fantastically informative post!  Thank you very much for laying out things so clearly.

This is a hugely exciting time!


Brian H's avatar

I take it that recovery from the X-Ray shell would be in addition to this? That was to be about 40%, last I heard. That would make for a much more productive device.


This is a really excellent post!  Thanks!


Brian H's avatar

If that scaling (I^7) holds up on both factors, it looks like the 55 becomes 275, and the 16 becomes 128. The net result would be an additional factor of 5x8=40. 40x433,000 = 17,320,000. Can you handle that?  wink cheese


That’s a lot of heat.  Following the hypothetical scenario, for each shot you get 6400 Joules of output and you get 26600 Joules of heat.

I suppose you could always use the heat for district heating (and cooling?).


Breakable's avatar

Thermal management will probably be one of the biggest challenges in engineering phase


Brian H,

Actually, recovery from the x-Ray shell would be included in the amount we said for the ion beam conversion device.  The plasmoid energy comes out as a lot of ion beam and a little x-rays.  So we could get by without an x-ray shell for recoveries around 80%, but to get to 91% would require an x-ray shell.  I guess that’s a detail that got glossed over in the article.


DerekShannon's avatar

where is @novapbs special/ @discovery chnl series documenting NJ’s @focusfusion, most exciting science project out there? Even if it fails!


Brian H's avatar

Derek;
?? Who mentioned such a program? First I’ve heard of it.


DerekShannon's avatar

that was rhetorical, 140 character tweet limit required change from “why isn’t there”” to “where is”.  7 characters saved!  At least 1 person confused!


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