Eight (8) Goals of LPP's Experiment

[Brian H]

Tulse wrote:

The real “net energy” is usable energy, I think, which is either captured heat or electric current (the ideal end result, which FF gets to far more directly than D-D fusion or D-T fusion can). Energy in the form of neutron emissions etc. is pretty much “lost”.

Is there a standard definition of “over unity” for fusion projects? The impression I have always gotten is that the way “breakeven” is typically used is very theoretical, and involves all released energy, whether it would be practical or not to harvest it (and that such things like the Carnot limit and other parasitic energy losses are not considered). It seems to me that the real beauty of the FF approach is that since most of the energy gets captured directly as electricity, it will be very easy to demonstrate practical breakeven, that is, breakeven with all those losses factored in. (By contrast, even if ITER or NIF produce theoretical breakeven, that tells us very little about whether it can be practically turned into more usable power than it consumes, since the actual generation of electricity requires so much additional engineering.)
As I understand it, the ability of the pulsed FF system to recharge its own capacitors and refire indefinitely means it has infinite Q, the ideal. Perhaps I’m wrong about that, but that’s how I understand it so far.

[Tulse]

Brian H wrote: As I understand it, the ability of the pulsed FF system to recharge its own capacitors and refire indefinitely means it has infinite Q, the ideal. Perhaps I’m wrong about that, but that’s how I understand it so far.

But that would be true of any over-unity fusion device — surely for FF one would determine Q on the basis of the energy balance for an individual shot.

[Brian H]

Tulse wrote:

As I understand it, the ability of the pulsed FF system to recharge its own capacitors and refire indefinitely means it has infinite Q, the ideal. Perhaps I’m wrong about that, but that’s how I understand it so far.

But that would be true of any over-unity fusion device — surely for FF one would determine Q on the basis of the energy balance for an individual shot.
My understanding is “ignition refers to an infinite Q, that is, a self-sustaining plasma where the losses are made up for by fusion power without any external input“, but I’m not sure what constitutes ignition in a pulsed system. However, once the initial blip is fed into the electrodes by the capacitors, there is no external power required from the grid.

[Tulse]

Brian H wrote: once the initial blip is fed into the electrodes by the capacitors, there is no external power required from the grid.

Right, but once a commercial tokamak is running, it will not require grid power either (just like a conventional generating plant doesn’t). That doesn’t tell you how efficient the device is, how much extra energy is produced above what is needed to sustain the reaction. I may be misunderstanding how one calculates Q, however.

[Brian H]

Tulse wrote:

once the initial blip is fed into the electrodes by the capacitors, there is no external power required from the grid.

Right, but once a commercial tokamak is running, it will not require grid power either (just like a conventional generating plant doesn’t). That doesn’t tell you how efficient the device is, how much extra energy is produced above what is needed to sustain the reaction. I may be misunderstanding how one calculates Q, however.
See how you do with this: http://en.wikipedia.org/wiki/Fusion_energy_gain_factor. That’s what I was grappling with.

[Phil’s Dad]

Brian H wrote:
As I understand it, the ability of the pulsed FF system to recharge its own capacitors and refire indefinitely means it has infinite Q, the ideal. Perhaps I’m wrong about that, but that’s how I understand it so far.

That’s a good way of putting it. Even if it is only outputting say 1.5 times its input, it recycles the original charge. The rest is coming at you 300 times a second. That’s why it doesn’t need anything like Q=30-50 per cycle to be viable, as some commentators have suggested.

In the end though it comes down to bang for buck. How much net electricity per dollar (including capital payback, fuel sourcing and processing, decommissioning costs, waste disposal, security costs, tea, coffee and so on over the expected lifetime of the machine). Let’s call it $Q.

[Brian H]

Phil’s Dad wrote:

As I understand it, the ability of the pulsed FF system to recharge its own capacitors and refire indefinitely means it has infinite Q, the ideal. Perhaps I’m wrong about that, but that’s how I understand it so far.

That’s a good way of putting it. Even if it is only outputting say 1.5 times its input, it recycles the original charge. The rest is coming at you 300 times a second. That’s why it doesn’t need anything like Q=30-50 per cycle to be viable, as some commentators have suggested.

In the end though it comes down to bang for buck. How much net electricity per dollar (including capital payback, fuel sourcing and processing, decommissioning costs, waste disposal, security costs, tea, coffee and so on over the expected lifetime of the machine). Let’s call it $Q.
Yeah; here’s another definition I found: “Q-factor: Ratio of power produced by fusion to power put into the reactor to heat the plasma and drive the magnetic fields. “
By that measure, the Q for pulse 1 is 1.5. For pulse 2 et seq. it is ∞.
But $Q is the overhead costs allocated on a per-pulse basis; lessee:
330 p/s x 60 sec. x 60 min x 24 hr. x 180 days-between-servicings = ~5 billion pulses. Each pulse is worth (¼¢x(5,000kw/pulse)/(330x60x60) = $0.00001. 5bn pulses are worth $50,000. Estimated overheads&amortization per half year are ~ $25,000. $Q is ~2.0. But the “background” power price is around 9¢/kwh (what FF would be replacing), so that 2.0 can be multiplied by ((9 / ¼)=36) = 72. So until “costly” power is entirely supplanted, that would be the payoff ratio ($Q) for putting up new FF generators.

:cheese: :coolgrin: 😆

[Tulse]

Phil’s Dad wrote: In the end though it comes down to bang for buck. How much net electricity per dollar (including capital payback, fuel sourcing and processing, decommissioning costs, waste disposal, security costs, tea, coffee and so on over the expected lifetime of the machine). Let’s call it $Q.

And the thing I find hugely exciting about FF is that it’s perhaps the only fusion technique that even offers a ballpark estimate of this — with tokamaks and ICF, who the hell knows how expensive the actual physical plant will be, or how much it will cost to decommission the radioactive gear? Big Fusion is so far away from engineering an actual plant that we don’t have a clue whether it will be economical, even if it reaches theoretical breakeven. By contrast, it’s very clear that if FF actually produces fusion, it will be extremely cheap to engineer it to start pumping out electricity.

[Phil’s Dad]

Brian H wrote: $Q is the overhead costs allocated on a per-pulse basis; lessee:
330 p/s x 60 sec. x 60 min x 24 hr. x 180 days-between-servicings = ~5 billion pulses. Each pulse is worth (¼¢x(5,000kw/pulse)/(330x60x60) = $0.00001. 5bn pulses are worth $50,000. Estimated overheads&amortization per half year are ~ $25,000. $Q is ~2.0. But the “background” power price is around 9¢/kwh (what FF would be replacing), so that 2.0 can be multiplied by ((9 / ¼)=36) = 72. So until “costly” power is entirely supplanted, that would be the payoff ratio ($Q) for putting up new FF generators.

:cheese: :coolgrin: 😆

So, on the assumption that to begin with electricity would be charged at the current rate, early adopters would see break even in 2.5 days! They’ll be $Qing round the block. 🙂

[Brian H]

Phil’s Dad wrote:

$Q is the overhead costs allocated on a per-pulse basis; lessee:
330 p/s x 60 sec. x 60 min x 24 hr. x 180 days-between-servicings = ~5 billion pulses. Each pulse is worth (¼¢x(5,000kw/pulse)/(330x60x60) = $0.00001. 5bn pulses are worth $50,000. Estimated overheads&amortization per half year are ~ $25,000. $Q is ~2.0. But the “background” power price is around 9¢/kwh (what FF would be replacing), so that 2.0 can be multiplied by ((9 / ¼)=36) = 72. So until “costly” power is entirely supplanted, that would be the payoff ratio ($Q) for putting up new FF generators.

:cheese: :coolgrin: 😆

So, on the assumption that to begin with electricity would be charged at the current rate, early adopters would see break even in 2.5 days! They’ll be $Qing round the block. 🙂
Nah; I get 60 hrs X 5,000kw X $0.09 = $27,000 gross receipts. You’re off by an order of magnitude, or more. About 1 month for break-even. :down:

[Phil’s Dad]

Brian H wrote:
Nah; I get 60 hrs X 5,000kw X $0.09 = $27,000 gross receipts. You’re off by an order of magnitude, or more. About 1 month for break-even. :down:

Mea Culpa. :red: Still; a one month payback in the energy industry should get a fair uptake.

[Brian H]

Phil’s Dad wrote:

Nah; I get 60 hrs X 5,000kw X $0.09 = $27,000 gross receipts. You’re off by an order of magnitude, or more. About 1 month for break-even. :down:

Mea Culpa. :red: Still; a one month payback in the energy industry should get a fair uptake.
More like a mad sharp-elbowed and knives-drawn stampede! (After a couple of first-adopters are past that payoff month.)

[Augustine]

Rezwan wrote: For more detail on the 8 goals, read this post on 8 goals of LPP’s experiment

First goal: machine assembled, and a pinch. Check!

Four goals are supposed to be wrapped up in just two months. Is this realistic?
2. At 25kV: Produce 1 MA, determine optiumum gas pressure
3. Test theory of axial magnetic field
4. Move to 45kV, 2MA, with Deuterium
5. Confirm Texas results, with better instruments

OK. Discuss.

This will be a success if you can pass step 5. I think you really only need three things:

1. Show that fusion is really happening (neutron count)
2. Empirically measure how the potential and amperage of each attempt influences the rate of fusion
3. Have the empirical data from step 2 verify the theoretical model

[Phil’s Dad]

Augustine wrote:

This will be a success if you can pass step 5. I think you really only need three things:

1. Show that fusion is really happening (neutron count)
2. Empirically measure how the potential and amperage of each attempt influences the rate of fusion
3. Have the empirical data from step 2 verify the theoretical model

Welcome Augustine. 🙂

I hope however that a neutron count will not figure significantly as a sucess criteria. :sick: Ideal result? 0
(Although I understand the inevitability of it with Deuterium)

[Augustine]

Phil’s Dad wrote:

Welcome Augustine. 🙂

I hope however that a neutron count will not figure significantly as a sucess criteria. :sick: Ideal result? 0
(Although I understand the inevitability of it with Deuterium)

The neutron count seems to be more of a smoking gun. Of course we all want p-B for our reaction but i’ll take the neutron count of a Deuterium reaction. Once you show that you can fuse Deuterium all you need to do is establish how the machine scales. The rest is engineering.

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