I see no way a multi kA beam of alphas will collimate into a very small beam. They'll surely spread out very willingly, before neutralising and going every-which-way.EricF wrote:
just spitting out ideas here,and assuming a good confined jet of alphas through the cusps, but would it be realistic to allow the alpha particle to travel far enough to unload its voltage into the collection grid, but neutralize the particle while it is still in motion with enough inertia left over to allow it to travel the remaining distance of the chamber and exit via a one way valve. That would be one small darn valve though. Carbon fiber nanovalve?
Polywell Electrical System
In a turbo pump the rpm is determined by the tip speed so the larger the pump the lower the rpm.
==
I also think gas puffing would not be used in an operational machine. Puffing is useful to get a range of pressures in a pulsed machine.
Even in a pulsed machine gas flows could be used.
BTW it is only in table top fusors that controlling the flows would be hard. And I have even figured out how to do that - if it was useful.
==
I also think gas puffing would not be used in an operational machine. Puffing is useful to get a range of pressures in a pulsed machine.
Even in a pulsed machine gas flows could be used.
BTW it is only in table top fusors that controlling the flows would be hard. And I have even figured out how to do that - if it was useful.
Engineering is the art of making what you want from what you can get at a profit.
chrismb wrote:I see no way a multi kA beam of alphas will collimate into a very small beam. They'll surely spread out very willingly, before neutralising and going every-which-way.EricF wrote:
just spitting out ideas here,and assuming a good confined jet of alphas through the cusps, but would it be realistic to allow the alpha particle to travel far enough to unload its voltage into the collection grid, but neutralize the particle while it is still in motion with enough inertia left over to allow it to travel the remaining distance of the chamber and exit via a one way valve. That would be one small darn valve though. Carbon fiber nanovalve?
As chrismb said the alpha beam would be dispersing as it left the cusp. A math type could probably calculate the dispersion angle. I'm guessing it would not be great because of the speed of the particles. I suspect some collimation/focusing could be done by a conversion grid (at the cost of some energy), but probably only to limit the rate of dispersion. To neutralize without hitting a grounded surface (or at least create a neutral plasma of alphas and electrons), you would need electron guns shooting electrons into the stream of the mostly decelerated alphas. I'm guessing this would be a process similar to how neutral beams are shot into tokamacs to help heat the working plasma (except in this case you would want the resulting stream of plasma to be at low energy (perhaps a few hundred eV). This would cost energy and add complexity, though if the vacuum pumping was the limiting factor, it could make the difference. How much of a thermal load this would put on the pumps may be a concideration. Of course with a neutral plasma streaming into the pump the turbine blades could be magnetically shielded (?)- more complexity.
[EDIT] Though wait!- spinning magnetic fields, charged particles moving at possibly high speeds- sounds suspisously like a motor/generater. Is it possible that a pump could double as a direct energy converter?
I once suggested in another thread something similar to the tiny valve except on a more grand scale - having the alpha stream focused to such a small beam that they would pass through a small magnetically shielded orifice at such a high density and speed that you wouldn't need a pump and you could use the stream as direct thrust for a rocket in the atmosphere. That idea did not survive long.
In a sense a one way valve is how the turbomolecular pump (or turbines in general) work. Due to the tilted spinning blades the particles tend to bounce from one layer of blades to the next, but have a much more difficult time bouncing backwards. The specifics change depending on the density, but I believe that is the basic premise.
Dan Tibbets
To error is human... and I'm very human.
Back to the reason I posted in the first place, I think I stumbled on an explanation for the direct conversion to electricity that worked for me.
It is here –
viewtopic.php?t=541&postdays=0&postorder=asc&start=15.
The key phrase from Carlos, “You could say: Hey, if you just take electrons from the inside, electrons that you had to put there in the first place, to later on take them away, then there is no gain. Well, there is a BIG difference between the electrons that you inject in a Polywell and the electrons that go from M to G, and that is that you inject them with, lets say, 50 keV of energy, while the ones that go from M to G gain no less than 2.5 MeV. They are the same in number, but 50 times more energetic, and that's where the gain comes from.” Thanks Carlos.
I was confused because it looked like the approximately the same amount of current was entering and leaving the reactor, so I didn’t see where the gain was.
Viewed from the outside of the reactor,
Electrons are removed from the H and B11 then the ions fired into polywell. The ion guns are a kV load.
Electrons are returned to the alpha particles. The alpha particle collectors are an MV source.
What I failed to see was that current enters the reactor in the kV range and leaves in the MV range.
P = V*I
Pin = kV*I and Pout = MV*I so Pin<<Pout
For anyone new to the polywell the answers to all their questions are probably already in this forum. There are so many threads it’s difficult to find them.
It is here –
viewtopic.php?t=541&postdays=0&postorder=asc&start=15.
The key phrase from Carlos, “You could say: Hey, if you just take electrons from the inside, electrons that you had to put there in the first place, to later on take them away, then there is no gain. Well, there is a BIG difference between the electrons that you inject in a Polywell and the electrons that go from M to G, and that is that you inject them with, lets say, 50 keV of energy, while the ones that go from M to G gain no less than 2.5 MeV. They are the same in number, but 50 times more energetic, and that's where the gain comes from.” Thanks Carlos.
I was confused because it looked like the approximately the same amount of current was entering and leaving the reactor, so I didn’t see where the gain was.
Viewed from the outside of the reactor,
Electrons are removed from the H and B11 then the ions fired into polywell. The ion guns are a kV load.
Electrons are returned to the alpha particles. The alpha particle collectors are an MV source.
What I failed to see was that current enters the reactor in the kV range and leaves in the MV range.
P = V*I
Pin = kV*I and Pout = MV*I so Pin<<Pout
For anyone new to the polywell the answers to all their questions are probably already in this forum. There are so many threads it’s difficult to find them.
the future is near.
Think grounded grid power amplifier. The current gain is 1 (or very close). It is the voltage gain that gives the power amplification.zimdlg wrote:Back to the reason I posted in the first place, I think I stumbled on an explanation for the direct conversion to electricity that worked for me.
It is here –
viewtopic.php?t=541&postdays=0&postorder=asc&start=15.
The key phrase from Carlos, “You could say: Hey, if you just take electrons from the inside, electrons that you had to put there in the first place, to later on take them away, then there is no gain. Well, there is a BIG difference between the electrons that you inject in a Polywell and the electrons that go from M to G, and that is that you inject them with, lets say, 50 keV of energy, while the ones that go from M to G gain no less than 2.5 MeV. They are the same in number, but 50 times more energetic, and that's where the gain comes from.” Thanks Carlos.
I was confused because it looked like the approximately the same amount of current was entering and leaving the reactor, so I didn’t see where the gain was.
Viewed from the outside of the reactor,
Electrons are removed from the H and B11 then the ions fired into polywell. The ion guns are a kV load.
Electrons are returned to the alpha particles. The alpha particle collectors are an MV source.
What I failed to see was that current enters the reactor in the kV range and leaves in the MV range.
P = V*I
Pin = kV*I and Pout = MV*I so Pin<<Pout
For anyone new to the polywell the answers to all their questions are probably already in this forum. There are so many threads it’s difficult to find them.
Or if you are too young for space bottles. Think grounded base amplifier.
Engineering is the art of making what you want from what you can get at a profit.
Apologies if this has been discussed on some other thread - I have not found it and am surprised.
Direct energy conversion is held out as desirable on grounds of:
a) cost
b) higher conversion efficiency (=> lower Q needed).
Given that p-B11 produces 3 alphas & 13Mev (I think) I am wondering how easy is direct conversion?
There are two problems - doing the conversion anyhow, and, given that is possible, matching efficiently to the energy spectrum of the reaction product charged particles
two methods:
electrostatic
inductive
Electrostatic is deceptively simple but one is left with power in the form of a current at a very high voltage (2MV average, since 3 alpha have change of 6e). Semiconductors to switch this do not to my knowledge exist directly, but of course 10kV can easily be switched and so 200 stages in series could presumably deal with the problem. I am not sure about cost and efficiency. For max efficiency you need a whole load of collector grids at different potentials, and then DC->DC conversion from all of these down to something sensible.
Inductive conversion is probably easier, especially from a pulsed system, but I am not sure how feasible.
Is the assumption of 80% conversion efficiency justified, or is this to a DC power source at unusably high voltage, with extra loss (and cost) to come from DC-DC conversion?
I have not found much discussion of whether electric power at 2MV can sensibly be used for anything other than generating heat, but many people discussing direct conversion designs seem to view high voltage DC electrical power as end product:
http://www.askmar.com/Fusion_files/Dire ... actors.pdf
Best wishes, Tom
Direct energy conversion is held out as desirable on grounds of:
a) cost
b) higher conversion efficiency (=> lower Q needed).
Given that p-B11 produces 3 alphas & 13Mev (I think) I am wondering how easy is direct conversion?
There are two problems - doing the conversion anyhow, and, given that is possible, matching efficiently to the energy spectrum of the reaction product charged particles
two methods:
electrostatic
inductive
Electrostatic is deceptively simple but one is left with power in the form of a current at a very high voltage (2MV average, since 3 alpha have change of 6e). Semiconductors to switch this do not to my knowledge exist directly, but of course 10kV can easily be switched and so 200 stages in series could presumably deal with the problem. I am not sure about cost and efficiency. For max efficiency you need a whole load of collector grids at different potentials, and then DC->DC conversion from all of these down to something sensible.
Inductive conversion is probably easier, especially from a pulsed system, but I am not sure how feasible.
Is the assumption of 80% conversion efficiency justified, or is this to a DC power source at unusably high voltage, with extra loss (and cost) to come from DC-DC conversion?
I have not found much discussion of whether electric power at 2MV can sensibly be used for anything other than generating heat, but many people discussing direct conversion designs seem to view high voltage DC electrical power as end product:
http://www.askmar.com/Fusion_files/Dire ... actors.pdf
Best wishes, Tom
Switching is regularly done with 500 KV DC lines. Going up to 2 MV is not a big stretch given advances in semiconductors.
DC-AC conversion efficiencies can be very good. Losses under 5% of power in.
With SiC based high power semiconductors (not yet in mass production for power devices) losses can be reduced by factors of 2 to 5.
Carbon based semiconductors (lab experiments for now) will have even lower losses.
===
Currently the highest voltage rating for IGBTs is around 6,500 volts. In a practical circuit you could use them at 4,000 volts. That says 500 devices per phase for 2 MV.
Figure around $20 million for a 500 MW converter. BOE And with higher production rates costs will come down.
DC-AC conversion efficiencies can be very good. Losses under 5% of power in.
With SiC based high power semiconductors (not yet in mass production for power devices) losses can be reduced by factors of 2 to 5.
Carbon based semiconductors (lab experiments for now) will have even lower losses.
===
Currently the highest voltage rating for IGBTs is around 6,500 volts. In a practical circuit you could use them at 4,000 volts. That says 500 devices per phase for 2 MV.
Figure around $20 million for a 500 MW converter. BOE And with higher production rates costs will come down.
Engineering is the art of making what you want from what you can get at a profit.
Yes, that sounds right to me.
I just have not seen circuits for these EHT converters with v large numbers of IGBTs and (presumably) protection against catastrophic failure should any one stage fail.
But (having done belatedly a bit of internet research on HVDC conversion) 500kV bipolar systems are standard so the app here is only X2 voltage.
And since HVDC cables are used as part of transmission systems I suppose 2MV DC is quite usable - only double what is needed to interface directly with HVDC cable.
There is still the issue of how to interface lots of collectors each at different voltages (needed to collect at high efficiency). The HVDC convertors seem to be implemented as single very high voltage switching stages, with air-cored transformers. So something different is needed to interface economically with a number of collectors.
Best wishes, Tom
I just have not seen circuits for these EHT converters with v large numbers of IGBTs and (presumably) protection against catastrophic failure should any one stage fail.
But (having done belatedly a bit of internet research on HVDC conversion) 500kV bipolar systems are standard so the app here is only X2 voltage.
And since HVDC cables are used as part of transmission systems I suppose 2MV DC is quite usable - only double what is needed to interface directly with HVDC cable.
There is still the issue of how to interface lots of collectors each at different voltages (needed to collect at high efficiency). The HVDC convertors seem to be implemented as single very high voltage switching stages, with air-cored transformers. So something different is needed to interface economically with a number of collectors.
Best wishes, Tom
Is there any reason BIG synchronous buck converters (http://en.wikipedia.org/wiki/Buck_converter) can't be used to step the two different high voltage levels down to something more manageable like 500 kV. Surely they could be constructed out of existing high power IGBTs in series.
See http://www.ptd.siemens.de/HVDC_Solution ... 7_V_1b.pdf for 800 kV equipment under development, it’s all huge. It makes me think transmission at 2MV levels may not be practical.
Has the question of whether or not alpha particles shooting towards the spherical deceleration grid from the inside would be slowed down by the +ve charge on the grid been settled. Surely an alpha particle would feel repulsive forces from the grid from all sides which would then cancel each other out. The net effect would then be no force on the alpha particle. Am I wrong?
From the discussion I’ve seen in other threads it seems to be accepted that it will work. Anybody know what the expected current would be flowing to the deceleration grid of a 100 MW reactor?
I = P/V = 100*10^6 / 2*10^6 = 50 A
Is it correct to calculate it like that?
See http://www.ptd.siemens.de/HVDC_Solution ... 7_V_1b.pdf for 800 kV equipment under development, it’s all huge. It makes me think transmission at 2MV levels may not be practical.
Has the question of whether or not alpha particles shooting towards the spherical deceleration grid from the inside would be slowed down by the +ve charge on the grid been settled. Surely an alpha particle would feel repulsive forces from the grid from all sides which would then cancel each other out. The net effect would then be no force on the alpha particle. Am I wrong?
From the discussion I’ve seen in other threads it seems to be accepted that it will work. Anybody know what the expected current would be flowing to the deceleration grid of a 100 MW reactor?
I = P/V = 100*10^6 / 2*10^6 = 50 A
Is it correct to calculate it like that?
the future is near.
No reason I can see. I just don't know what is the best type of converter for this application given you need 500 or so switching elements in series.
But if there are say 22 distinct collector grids that would be 22 Buck converters, as far as I can see.
Probably, however, there is some cleverer way to deal with this problem. For example neighbouring grids could be averaged with a switched inductor (effectively a Buck), and in this process the switching voltage is only the difference between the two grid voltages. A binary tree of similar converters can be used to combine all the grids. Given N grids at the nth level there are N.2**-n Bucks needed at a voltage of V0*2**n/N where V0 is the max grid voltage (this assumes voltages are linear, not true but OK for ball-park calculation).
The number of switch transitors thus scales as log2(N)/N times the number needed for N independent Bucks, or log2(N) more than would be needed for a single grid converter.
How many collector grids do you need? The reference above suggests 22 - for efficiency of 86%. This is from D/D fusion however. The reason for the large number is that reaction product energy distribution spreads out over a large energy range.
Tom
PS - this is quite a clever configuration - unfortunately I have just put it into public domain so can't now patent it! (Simon, if you feel strongly, you can always delete this post).
But if there are say 22 distinct collector grids that would be 22 Buck converters, as far as I can see.
Probably, however, there is some cleverer way to deal with this problem. For example neighbouring grids could be averaged with a switched inductor (effectively a Buck), and in this process the switching voltage is only the difference between the two grid voltages. A binary tree of similar converters can be used to combine all the grids. Given N grids at the nth level there are N.2**-n Bucks needed at a voltage of V0*2**n/N where V0 is the max grid voltage (this assumes voltages are linear, not true but OK for ball-park calculation).
The number of switch transitors thus scales as log2(N)/N times the number needed for N independent Bucks, or log2(N) more than would be needed for a single grid converter.
How many collector grids do you need? The reference above suggests 22 - for efficiency of 86%. This is from D/D fusion however. The reason for the large number is that reaction product energy distribution spreads out over a large energy range.
Tom
PS - this is quite a clever configuration - unfortunately I have just put it into public domain so can't now patent it! (Simon, if you feel strongly, you can always delete this post).
Last edited by tomclarke on Mon Jul 20, 2009 10:08 pm, edited 2 times in total.
Zim -
Sorry: to answer your questions -
Calculation about current is correct.
The way to analyse electrostatic deceleration is simple. The collector grid must have a potential relative to that where the fusion occurs of the required particle energy divided by particle's charge. Using eV for energy this is just voltage = energy/ atomic number.
Ideally the particles are slowed down to near zero energy at the collection grid, but a small positive energy simply means power lost to heat. Of course that is a LOT of power, and cooling these collector grids is an issue. See my reference above for a design which has been simulated and tested for D/D fusion products.
Sorry: to answer your questions -
Calculation about current is correct.
The way to analyse electrostatic deceleration is simple. The collector grid must have a potential relative to that where the fusion occurs of the required particle energy divided by particle's charge. Using eV for energy this is just voltage = energy/ atomic number.
Ideally the particles are slowed down to near zero energy at the collection grid, but a small positive energy simply means power lost to heat. Of course that is a LOT of power, and cooling these collector grids is an issue. See my reference above for a design which has been simulated and tested for D/D fusion products.
You're right, but you've visualized the system wrong. The trap grid doesn't take effect until the alphas pass it. After that, they see the negative charge on the trap grid and slow down to a stop just as they reach a collector plate. Approximately.zimdlg wrote:Has the question of whether or not alpha particles shooting towards the spherical deceleration grid from the inside would be slowed down by the +ve charge on the grid been settled. Surely an alpha particle would feel repulsive forces from the grid from all sides which would then cancel each other out. The net effect would then be no force on the alpha particle. Am I wrong?
Really it's about potential difference, not charge; charge takes care of itself. You could theoretically not bother with the trap grid, and just have the magrid be a couple million volts lower than the collector plates - the magrid would wind up with a huge negative charge, and this setup would stop alphas fine, but then you wouldn't get electron recirculation and the wiffleball wouldn't form properly, so there wouldn't be a whole lot of alphas to stop...
One way you might accomplish the multi-voltage is to use diodes in series with each voltage in a tapped inductor.tomclarke wrote:No reason I can see. I just don't know what is the best type of converter for this application given you need 500 or so switching elements in series.
But if there are say 22 distinct collector grids that would be 22 Buck converters, as far as I can see.
Probably, however, there is some cleverer way to deal with this problem. For example neighbouring grids could be averaged with a switched inductor (effectively a Buck), and in this process the switching voltage is only the difference between the two grid voltages. A binary tree of similar converters can be used to combine all the grids. Given N grids at the nth level there are N.2**-n Bucks needed at a voltage of V0*2**n/N where V0 is the max grid voltage (this assumes voltages are linear, not true but OK for ball-park calculation).
The number of switch transitors thus scales as log2(N)/N times the number needed for N independent Bucks, or log2(N) more than would be needed for a single grid converter.
How many collector grids do you need? The reference above suggests 22 - for efficiency of 86%. This is from D/D fusion however. The reason for the large number is that reaction product energy distribution spreads out over a large energy range.
Tom
PS - this is quite a clever configuration - unfortunately I have just put it into public domain so can't now patent it! (Simon, if you feel strongly, you can always delete this post).
I'm sure if we took the design constraints to a company that already produces very high power inverters something cost effective could be worked out.
To make the unit more compact - oil cooling and insulation. This is probably not done where the inverter hall is relatively low cost for ease of maintenance.
If individual switches fail shorted - no problem. Have enough extra switches to give a reasonable MTBF. If they fail open - a parallel SCR may do the trick.
Engineering is the art of making what you want from what you can get at a profit.
I'm well aware of that. That's why I said "approximately"... Back before we knew the alphas would be coming out in beams, I tried to conceptualize a plausible collector system, and multiple collectors at a range of voltages - perhaps a significant number of them, fairly closely spaced - was considered to be a baseline requirement.tomclarke wrote:You absolutely need multiple collector grids - otherwise the cooling requirements are horrendous (and also you lose output power).
The point I was making in my post above is that, to a first-order approximation, it's not the charge on the collectors that decelerates the alphas; it's the charge on everything else. Gauss' Law is not disrespected. zimdlg seemed to be a bit fuzzy on that.