Direct energy conversion - alpha particles to electricity

[Doug]

Several recent articles on proton-boron fusion state: "The energy is given by the number
of alpha particles times 2.8 MV which energy can be converted into three phase
ac-electricity by using the HVDC (high voltage direct current) transmission technology
used for electricity power transmission over 1000 km or much higher distance with minimum
of losses."

The method of conversion is referenced in 'Kanngiesser Karl-Werner., D. Hartmut Huang,
Hans Peter Lips and Georg Wild. HVDC Systems and their Planning Siemens EV HA 7, Siemens
Monographien, München (1994)'.

I have been unsuccessful in finding a detailed description of how the alpha particles are
'converted' into a discharge current via direct energy conversion or the Siemens
reference.

Appreciate any suggested references or detailed explanation of how the conversion process
is accomplished.

Thank you.

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[Libor 'Poutnik' Stříž]

Dne 03/02/2018 v 19:44 Doug napsal(a):

Energy 2.8 MeV means one need a voltage 2.8 MV
to accelerate alpha particles to such an energy.

It does not mean 2/8 MeV particles
would be charging eventual capacitor to 2.8 MV.

Alpha particles are not used as the direct source
of electricity. Their huge energy is used
when converted to thermal energy.

In small power sources, there is used 238Pu -> 234U + alpha
in space probes and pacemakers.

In the former, there is usually a thermocouple DC source,
but there are functional models with Stirling engines as well.
The engine has about 4 times higher efficiency,
but it has supposed lower reliability due moving parts.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

[Doug]

Following from Laser Focus World, ref. https://tinyurl.com/yawnfsgr

Direct electricity
Hydrogen-boron fusion produces no neutrons and, therefore, no radioactivity in its primary
reaction. And unlike most other sources of power production, like coal, gas and nuclear,
which rely on heating liquids like water to drive turbines, the energy generated by
hydrogen-boron fusion converts directly into electricity.

The mechanism of the referenced means of direct energy conversion is what I am looking
for.

Thanks.

[Libor 'Poutnik' Stříž]

Dne 03/02/2018 v 21:15 Doug napsal(a):

> Following from Laser Focus World, ref. https://tinyurl.com/yawnfsgr
>
> Direct electricity
> Hydrogen-boron fusion produces no neutrons and, therefore, no radioactivity in its primary
> reaction. And unlike most other sources of power production, like coal, gas and nuclear,
> which rely on heating liquids like water to drive turbines, the energy generated by
> hydrogen-boron fusion converts directly into electricity.
>
> The mechanism of the referenced means of direct energy conversion is what I am looking
> for.

Alpha particles are charged particles.
There is no needed mechanism to convert the charge into the charge.

There is only the question of energy efficiency
of collecting the opposite charges.

[Jos Bergervoet]

Yes but those alphas have even more energy, 5.593 MeV.

The ones Doug refers to are apparently those from Boron
fusion (where you get three alphas with together 8.7MeV,
so the separate alphas have some distribution of energies
with on average 2.9MeV, but a broad spread around that
value, wich may complicate matters even further.)

What "high voltage direct current" transmission technology
(HVDC) could do to convert alphas to "three phase ac" is
rather unclear. In fact it seems completely unrelated:
<https://en.wikipedia.org/wiki/High-voltage_direct_current

> In the former, there is usually a thermocouple DC source,
> but there are functional models with Stirling engines as well.
> The engine has about 4 times higher efficiency,
> but it has supposed lower reliability due moving parts.

The direct conversion to electricity for nuclear reactions
is probably based on MHD generator concepts:
<https://en.wikipedia.org/wiki/Magnetohydrodynamic_generator>

--
Jos

[Libor 'Poutnik' Stříž]

Dne 03/02/2018 v 23:00 Jos Bergervoet napsal(a):

> On 2/3/2018 8:44 PM, Libor 'Poutnik' Stříž wrote:
>>
>> Energy 2.8 MeV means one need a voltage 2.8 MV
>> to accelerate alpha particles to such an energy.
>>
>> It does not mean 2/8 MeV particles
>> would be charging eventual capacitor to 2.8 MV.
>>
>> Alpha particles are not used as the direct source
>> of electricity. Their huge energy is used
>> when converted to thermal energy.
>>
>> In small power sources, there is used 238Pu -> 234U + alpha
>> in space probes and pacemakers.
>
> Yes but those alphas have even more energy, 5.593 MeV.

Sure.

>
> The ones Doug refers to are apparently those from Boron
> fusion (where you get three alphas with together 8.7MeV,
> so the separate alphas have some distribution of energies
> with on average 2.9MeV, but a broad spread around that
> value, wich may complicate matters even further.)

Yes, I am aware of this.
It is 11B + p --> 12C* --> 8Be + 4He --> 3 4He
with the spread alpha particle energy spectrum.

>
> The direct conversion to electricity for nuclear reactions
> is probably based on MHD generator concepts:
> <https://en.wikipedia.org/wiki/Magnetohydrodynamic_generator>

I am rather sceptical about the efficiency and cost.

[Doug]

Thanks Jos.

In patent application US 2017/0125129 A1, 'Method for generating electrical energy by
laser-based nuclear fusion and laser reactor', section [0065] states:

'At the potential of -1.4MV the energy of the alpha particles is available, and at a
charge of -1.4MV the energy is supplied as a seconds long discharge current of 714 Amperes
by means of the high voltage direct current transmission technique known from [9].'

The patent further notes the HVDC is converted into conventional 3Ph. AC in a known
manner, which I understand to be similar to the inverter section of a VFD. Reference [9]
is Kanngiesser K.W., Huang D.H. & Lips H. (1994). Highvoltage direct current transmission
– Systems and Planning. Siemens monographs Munich, EV HA 7.

As there is no mention of capturing thermal energy to generate electricity, the fusion
reactor must rely solely on the alpha particles. Is not clear the usion reactor's means
of generating the current is the same as that of an MHD generator. Perhaps I am not
seeing the obvious.

Thanks.



Jos Bergervoet <jos.ber...@xs4all.nl> wrote:

[Libor 'Poutnik' Stříž]

Dne 04/02/2018 v 15:57 Doug napsal(a):

> Thanks Jos.
>
> In patent application US 2017/0125129 A1, 'Method for generating electrical energy by
> laser-based nuclear fusion and laser reactor', section [0065] states:
>
> 'At the potential of -1.4MV the energy of the alpha particles is available, and at a
> charge of -1.4MV the energy is supplied as a seconds long discharge current of 714 Amperes
> by means of the high voltage direct current transmission technique known from [9].'
>

> [...]

>
> As there is no mention of capturing thermal energy to generate electricity, the fusion
> reactor must rely solely on the alpha particles. Is not clear the usion reactor's means
> of generating the current is the same as that of an MHD generator. Perhaps I am not
> seeing the obvious.
>

It looks like there is supposed to be an electrode
kept at the highly positive potential,
but not high enough to repulse energetic positive alpha particles.

They would then bite 2 electrons from the electrode surface,
becoming neutral helium, charging it as even more positive.

[Jos Bergervoet]

On 2/4/2018 4:43 PM, Libor 'Poutnik' Stříž wrote:
> Dne 04/02/2018 v 15:57 Doug napsal(a):
>> Thanks Jos.
>>
>> In patent application US 2017/0125129 A1, 'Method for generating electrical energy by
>> laser-based nuclear fusion and laser reactor', section [0065] states:
>>
>> 'At the potential of -1.4MV the energy of the alpha particles is available, and at a
>> charge of -1.4MV the energy is supplied as a seconds long discharge current of 714 Amperes
>> by means of the high voltage direct current transmission technique known from [9].'
>>
>> [...]
>>
>> As there is no mention of capturing thermal energy to generate electricity, the fusion
>> reactor must rely solely on the alpha particles. Is not clear the usion reactor's means
>> of generating the current is the same as that of an MHD generator. Perhaps I am not
>> seeing the obvious.
>>
>
> It looks like there is supposed to be an electrode
> kept at the highly positive potential,
> but not high enough to repulse energetic positive alpha particles.

And that probably is different from MHD generators. It would
perhaps look more like a Farnsworth Fusor, but then working
the opposite way (not providing the energy for the reaction
but harvesting it.)

How they handle the negative electrode in this case is not clear,
"rely solely on the alpha particles" is of course nonsense, you
must close the circuit somehow.

And the fact that harvested DC power can be converted to AC
power (as might be, but need not, be useful for distribution)
remains totally unrelated.

In fact it all doesn't contribute much to solving the real
problem: you first have to use enough of the alphas' energy
for collisions with protons to give them a high energy, in
order to create enough chance to have (on average) one new
reaction! That must happen before they escape the reaction
site and before this harvesting could take place.

The patent seems not so much concerned with the real problem,
but more with some engineering optimizations afterwards..

--
Jos

[Libor 'Poutnik' Stříž]

Dne 04/02/2018 v 17:21 Jos Bergervoet napsal(a):

> On 2/4/2018 4:43 PM, Libor 'Poutnik' Stříž wrote:

>>
>> It looks like there is supposed to be an electrode
>> kept at the highly positive potential,
>> but not high enough to repulse energetic positive alpha particles.
>
> And that probably is different from MHD generators. It would
> perhaps look more like a Farnsworth Fusor, but then working
> the opposite way (not providing the energy for the reaction
> but harvesting it.)
>
> How they handle the negative electrode in this case is not clear,
> "rely solely on the alpha particles" is of course nonsense, you
> must close the circuit somehow.

Sure. It may be possible that the geometry arrangement
and the dynamics of the fusion just distribute alpha particles
and electrons to the different electrodes,
that get their potential just from the charges.

>
> In fact it all doesn't contribute much to solving the real
> problem: you first have to use enough of the alphas' energy
> for collisions with protons to give them a high energy, in
> order to create enough chance to have (on average) one new
> reaction! That must happen before they escape the reaction
> site and before this harvesting could take place.

The needed proton energy at the cross-section resonance peak
of proton 11B fusion is near 150 keV,
about 2% of the fusion reaction yield.
http://www.fusor.eu/Images/crossSections.jpg

The idea is somewhat similar to 232Th fission reactors,
that need support from a proton accelerator.

[Jos Bergervoet]

On 2/4/2018 5:56 PM, Libor 'Poutnik' Stříž wrote:
> Dne 04/02/2018 v 17:21 Jos Bergervoet napsal(a):
>> On 2/4/2018 4:43 PM, Libor 'Poutnik' Stříž wrote:

..
...

>> In fact it all doesn't contribute much to solving the real
>> problem: you first have to use enough of the alphas' energy
>> for collisions with protons to give them a high energy, in
>> order to create enough chance to have (on average) one new
>> reaction! That must happen before they escape the reaction
>> site and before this harvesting could take place.
>
> The needed proton energy at the cross-section resonance peak
> of proton 11B fusion is near 150 keV,
> about 2% of the fusion reaction yield.
> http://www.fusor.eu/Images/crossSections.jpg

Yes, the problem is not that this amount of energy would be
a burden on the reaction's output! But that this is much too
high to be available from heating the plasma (i.e. you can't
give all particles on average this amount of energy!)

So you rely on giving only a few protons temporarily this
high energy, but they still should have enough chance to
cause a new reaction, before they lose it. So you probably
need to give much more than one proton this high energy to
have just one (on average) succeed. And *then* it actually
does become a burden on the output energy in the alphas.

Of course if you just want to use the old-fashioned thermal
energy generation (and not the patented current harvesting)
then this is no problem. The protons that fail give of
their energy as heat anyway.

> The idea is somewhat similar to 232Th fission reactors,
> that need support from a proton accelerator.

A proton accelerator would provide the energy from an
electric field. But that also happens in the Farnsworth
Fusor. The idea discussed here would in addition extract
the output energy into an electric field. You can imagine
doing both by electric fields. (But if that were easy then
Farnsworth would have done it, after all he also was the
first to create fully electronic television!)

--
Jos

[bob.mont...@gmail.com]

> The method of conversion is referenced in 'Kanngiesser Karl-Werner., D. Hartmut Huang,
> Hans Peter Lips and Georg Wild. HVDC Systems and their Planning Siemens EV HA 7, Siemens
> Monographien, München (1994)'.

In the proposed spherical reactor, the shell is charged at -1.4MV while the central HB11 target is oppositely charged. "For conversion of most of the 2.99 MeV α-particle energy into electrostatic energy, the particles have to be slowed down by an electric field of −1.4 MV."; "On the basis of an avalanche process during up to 1 ns duration, the energy in the α-particles is more than 1 GJ (277 kWh)."

(https://www.physics.unsw.edu.au/sites/default/files/LPB%202015%20Fusion%20Energy%20Laser%20Pulses.pdf )

Generated alphas have ( H +11 B = 3^4 He + 8.7MeV) 8.7/3 = 2.9MeV of kinetic energy "slowed" ?contained? by the spherical -1.4MV E field, initially the alhpas/electrons are confined by the 10kT 'cylindrical', really somewhat of a stretched prolate toroidal field between two amperian loops, and would be 'shot' out one end of it as a dispersing cloud of electric charges in a plasma of Helium and unspent Boron fragments. The charged plasma might be somewhat contained by the spherical electrostatic counter-field.

Using the provided 277kWH figure, (Also 1GJ = 0.28 MWH ) and iterating reactions at 1Hz intervals, the alpha particles initially at least have 2.99MeV of kinetic energy, and since KE=qV, energy of electrons in electron-volts numerically the same as the voltage between the plates of a capacitor, and W/V=A the instantaneous (per second) wattage of 277 KWH in ~1 sec is about 76 KW, divided by 2.99MeV = 25 milliAmps of current at 3MV would have to be drained from the sphere, and quickly, as fig 7. in the paper illustrates a rapidly-dissipating longitudinal E field, presumably accompanied by linearly inverse increases in current as voltage drops.

The "714 Amp averaged between each fusion reaction" or 714 Coulombs/sec, or 714 * 6.2415 × 10E+18, or 4.45E+21 electrons somewhere between the outer reactor shell and the HB11 reaction assembly at it's center would likely have to be coaxed about magnetically in order for an initially anisotropic charge cloud to be confined, at least long enough to drain it's electric potential. The reaction vessel and external power-conversion electronics would need to be designed as a 714*whateverVoltage Farad capacitor to directly electroscatically capture the charge of the plasma.

There are half-dozen ways of direct energy conversion from plasma, conveniently described in Wikipedia. ( https://en.wikipedia.org/wiki/Direct_energy_conversion ).

Combined megavoltage potentials and high, greater than nanofarad, electrostatic capacitance is somwehat of an oxymoron implying large plate surface areas which have to be in close proximity yet not so close as to counter the tendancy of corona arcs across a conductive plasma. An ignited plasma at human time-scales is essentially a direct short, but again the W. article shows the high-energy physics crowd has tricks for charge extraction from plasmas.

Whatever means used to drain the charge it seems would have to be effective in about half-cycle at about one or two GHz, before the E-field dissipates, to feed a switched inductor which in turn would feed a switched series-connected string of capacitors. Plasma+confinement vessel, external inductor and capacitor string all would have carefull matched impedances at the operating frequency to facilitate charge transfer e.g. of a few MV and a few 10s of mA of current within ns-scale frequency cycles (=GHz)

Though there doesn't appear to be a source for the Kanngiesser Siemens reference, there appears to be a related patent

( https://patentimages.storage.googleapis.com/5f/3a/1f/b606f81a0c97a0/US4429357.pdf )

Which just describes one aspect of industry-standard switched electronic power conversion, e.g. downconversion or cross-conversion of high-tension high-frequency power to mains poly(3)phase frequency.

Essentially switches in the series-connected string of capacitors disconnect the capacitors from the plasma generator, and from each other,and each capcaitor, now at some megavoltage divided by numbers, perhaps hundreds or thousands of capacitors, and/or groups of capacitors are reconnected shorter strings and all strings in parallel, effectively dividing the voltage and multiplying the amperage, and that potential can be then switched in sync with mains waveforms to deliver power to the grid. 3MV@25ma = 75KVA = 0.68A @110KV = 6,250A@110V

High frequency switching power converters are everywhere - in power cubes, PCs, handhelds, EV powertrains, and are scaling up to megawatt utiltiy distribution and transmission, soon will be replacing pole and EHV distribution transmission and regional-grid interconnect transformers. Google "Electronic power conversion" or "buck-boost" to open the door to further references.