## Why is ignition required for a fusion power plant

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
james6742
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### Why is ignition required for a fusion power plant

I'm new to this area, so I'm sure this is a newbie question, but one I haven't found answered in several days of research. Why is it assumed that ignition is necessary for a break-even fusion reactor design? According to the figures I've found here and elsewhere (please correct me if I'm wrong), a D-T fusion event requires 3.8 keV input and outputs 17 MeV.

Why wouldn't a simple design of a Deuterium ion beam firing into a pressurized Tritium gas (or vica versa) achieve breakeven energy production? According to the reaction cross section measurements at http://home.earthlink.net/~jimlux/nuc/sigma.htm, I calculate that a 106 keV Deuterium beam firing into a 10 cm long pressurized Tritium gas cylinder at 3000 PSI would experience a 25% reaction rate (that is, 1 in 4 ions would experience a fusion event), resulting in a theoretical gain of 40X (17 MeV output for one in four input ions at 106 keV).

I'm sure I'm missing something here but I haven't been able to find what it is, as all the literature talks about fusion assuming that you need a plasma that is at temperature, not just harvesting the energy output from isolated fusion events.

Thanks!
-James

james6742
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### Re: Why is ignition required for a fusion power plant

Here's my math, by the way, derived from: http://home.earthlink.net/~jimlux/nuc/fusormath.htm

Tritium beam: 100 keV
D-T cross section @ 100 keV: 4.937 Barns ~= 5e-24 cm^2 (http://home.earthlink.net/~jimlux/nuc/tdn.txt)
To find atoms per cc in Deuterium gas at 3000 psi:
n/V = P / RT
P = 3000psi = 20,684,271 pascal
R = 8314 J/kmol*K
T = 293 K
n/V = 20,684,271 pascal / ( 8314 J/kmol*K * 293 K ) = 8.49 pascal * kmol / J = 8.49 kmol/m^3
multiplying to convert units:
8.49 kmol/m^3 * 1e3 mol/kmol * 1e-6 m^3/cm^3 * 6.02e23 atoms/mol = 5.11e21 atoms/cm^3

Thus the reaction rate as a fraction of incident ions should be (for a 10 cm long chamber):
reaction rate = atoms/cm^3 * cross section (cm^2) * depth (cm) = 5.11e21 * 5e-24 * 10 = 0.25

With a 170:1 output to input energy ratio on the D-T fusion events (17 MeV to 100 keV), that would mean a theoretical maximum gain of 40x, more than enough to overcome inefficiencies in the ion beam generation and power harvesting.

I'd appreciate if anyone can point out what I'm obviously missing, or at least point me to better resources. Thanks!

Rich Feldman
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### Re: Why is ignition required for a fusion power plant

Hi James.

You put the question well. I'm sure there are regular readers here who can give you a good answer.

Maybe they are waiting to see your personal introduction. Check rules 1 and 2 in the fusor.net registration dialog. I missed them myself, first time.

-Rich
Mike echo oscar whisky! I repeat! Mike echo oscar whisky, how do you copy? Over.

Richard Hull
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### Re: Why is ignition required for a fusion power plant

Calculations can be deceiving if mis-applied or applied without a real knowledge of what goes on in the real doing of fusion.

1. Not even in idealized real world situations will even a tiny fraction of the fuel burn, (Fuse). However all that input energy will be needed to make anything at all happen. (Get all the atoms up to fusion energy.)

2. If fusion is a 170:1 gain per unit fusion and only 1 in every 100 million atoms at fusion energy fuse, that is a net loss of between 100,000:1 and 1,000,000:1.....Now that is what you call a real loser!

3. Stars are the best examples of real ignition fusion but, per unit mass and volume are among the worst fusion energy producers in the universe. Considering they are almost 100% pure fusion fuel, their burn rates related to the mass and volume make them volumetrically grossly inefficient fusion engines. Good thing, too! Thanks to their gross inefficiency they last for many Billions of years if on the main sequence.

Math is one thing, the reality of fusion in the doing is another. Fusion is not an easy process, like fission, yet it is fully understood and has been for many, many years. It just shows that not all things that are calculable and fully understood are necessarily readily doable.

The reason ignition is desirable is that fusion would become self-sustaining in that no input energy would be needed once the fusion engine or reactor is running. Like a diesel, just keep the fuel coming and I'll keep running.

Without ignition, but with well above unity fusion, (net gain), you might need a 20 megawatt coal powered plant to start and keep a fusion plant of 50 megawatts on line. Not a realistic solution, but still a net energy gain.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

prestonbarrows
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### Re: Why is ignition required for a fusion power plant

You have the right idea, but you are missing the Achilles's heel of beam-on-target fusion; scattering collisions.

Each time a D or T ion in your beam encounters a target nucleus, it can generally do one of two things; either fuse and release energy or simply scatter and 'bounce off' like a billiard ball with out a nuclear reaction. Just like there is a fusion cross-section as you mentioned, there is also a scattering cross-section. The problem is the scattering cross-section is always on the order of 100's or 1000's times larger than that for fusion. This means that statistically, your ion will bounce off something like 100 targets before fusing.

Now comes the big problem, your target nuclei are also D or T just like your beam. This means each time a beam ion scatters off a stationary target it will transfer about half of its momentum to the target. That 100 keV beam quickly becomes 50 keV then 25 keV and so on. If you look at the fusion cross-sections, you can see that after only a few scattering events your beam is no longer energetic enough to ever produce a fusion reaction.

This means that the great majority of ions you accelerate into your target just bounce around and deposit their energy into the target as heat and never produce fusion; much more than if you simply consider the fusion cross-section. You will never get more energy out from fusion than you put in to accelerate the beam. World-class beam-on-target devices take on the order of 100,000 W to power the beam and produce on the order of 1 W of fusion yield.

That is why any scheme geared towards net power needs some type of confinement. That way, the 'background' is just as hot as the 'beam' and during a scattering collision there is no momentum lost on average. Things just keep rattling around until that 1-in-100 collision where it actually fuses.

You need to:
1) be hot enough that the fusion cross-section is significant
2) keep the plasma around long enough that those 1-in-100 fusions can happen
3) have enough volume/density that the overall output is meaningful

Doing all those things at once is exceedingly difficult. (see Lawson criterion)

For example, a fusor will generally have much higher temperatures (>20 keV) than something like a tokamak (~10 keV), but have a small size and most of the accelerated ions don't stay around very long before impacting the grid and running into the whole beam-on-target limitations above. On the other hand, the sun has a rather cold temperature (<1.5 keV !) but makes up for it with an enormous volume and confinement time.

For your other question, controlled ignition is not required for a fusion power plant and will quite likely never happen in our lifetimes for a non-pulsed device on the current schedule. Even the most optimistic plans call for gains on the order of Q=10 for the next generation devices within the next few decades. (ignition is essentially Q=infinity)

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### Re: Why is ignition required for a fusion power plant

1. Beam interactions:- In regards the question of a beam passing through a gas, Preston has provided a sound description.

However, it is more dire than that still, because in a gas you have electrons and these sap the beam too.

I have written at length on this subject here because this is the very issue my patent intends to overcome. If you want further reading then you might like to see my first ever post here:
viewtopic.php?f=14&t=6798&p=42481#p42481

and a further discussion:
viewtopic.php?f=15&t=7183

or you could wade through my patent for that discussion:
http://patft.uspto.gov/netacgi/nph-Pars ... icyclotron

do a word search (ctl-F) in the page for "beam interactions" to take you to the relevant point.

2. Thermalisation and thermonuclear energy:-

The issue of the beam passing into a gas is the subject of 'thermalisation' of the beam energy. Its energy is dissipated into thermal energy pretty much in the first half dozen collisions. This will heat the gas, but do no more than that.

But imagine instead that you keep on pumping the gas with the beam particles. Eventually, the gas particles may become so excited that they will have a similar energy as the beam particles. When they are this excited, they will bounce around and off each other until either a) they cool down, because each collision causes the emission of EM energy that takes some of the energy away, or b) they react. In a way, you could look at a thermal plasma as if it were zillions of beams all bouncing off each other, continuously. Because they are giving or taking energy off each other all the time, there is no 'thermalisation' process as the gas is already 'thermalised'. Instead, if the heat is kept in then the nucleii will just keep on bouncing off each other until they fuse.

For these reasons, a thermal plasma really is the only way to go to get some real, practical specific power out of a nuclear reaction. It's a pretty poor power density as it is in a DT reactor (~0.5W/cc) but in other beam devices it will simply never be enough to generate real grid power.

Fusion is a beguiling, but ultimately demanding and unfulfilling mistress. The returns are so poor but the temptation of 'free energy' is so great. There is nothing more expensive than free energy. You would generate more energy from an acre of land with solar panels all over it than from a thermonuclear fusion power plant taking up the same space. If you tried to do it with beam-fusion processes, you're looking at a few orders of magnitude less than that, even.

Maybe tech will one day overtake my dire prediction, but that's how it looks today.

3: 'Ignition':-

Ignition is where the reactor is capable of self-powering itself. There is a power loss in a thermal plasma - it's from the EM energy from all those continual collisions. This is in the X-ray spectrum for a thermal-fusion capable plasma. It cools very rapidly. It can't afford to do that because if the plasma cools the fusion reactions will snuff out.

The idea with DT is that a quarter of the total fusion energy comes out as alpha particles, and as these are strongly decelerated and held in the plasma that it will be this that heats and drives the plasma to continue in its reactions. So, for example, 1GW of DT fusion reactions would have 800MW of neutron power that is drawn off by interaction with a reactor blanket and the heat taken away by a cooling system, whilst the remaining 200MW of alpha energy makes up for the thermal losses from the plasma. The plasma has to be large enough that this energy balance is matched - the bigger it is the higher is the ratio of the alpha heating energy versus the radiative losses.

The ratio of power loss to energy contained is called the 'confinement time'. For example, if the plasma contains 1GJ and the radiative losses are 200MW, then the confinement time is 5 seconds. The two are related in that the more energy is in the plasma the quicker the radiative losses are, hence these things have been getting bigger and bigger as that trade-off is reached. The higher the confinement time, the less power needed to keep the reactor going.

I do not recall if ITER is meant to achieve 'ignition' or 'burning'. So far, reactors have been 'burning' reactors in that they have had to be fed by an external power source. Still, the total energy out exceeds the energy in, in a burning reactor, but that it will not self-sustain. As far as I can see, a 'burning'
reactor is a better bet, because you can control precisely the power output, and turn it off instantly, though it will be a lesser power. Anyhow, that's what ignition is, and it's not strictly essential to achieve so long as 'burning' is.

Addendum: I am still hopeful that someone may develop my ideas of recirculating beam energy to create an over-unity beam device, and I do think that is possible and may have certain niche applications, but I now believe it will never reach a sufficient power level to be practical as an energy source, and I don't believe any beam device will, of any description. I think thermal plasma fusion devices will become viable on energy (and if you look deeper at my patent you might see I suggest the epicyclotron principle might aid toroidal plasma confinement) but such thermal reactors will then not be practical on cost. Having spend a long time studying all this, it is my considered conclusion that beam energy amplifier systems activating fertile elements is the long term future for man's energy needs into the 10's of millennia ahead.... if they survive that long.

Dan Tibbets
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### Re: Why is ignition required for a fusion power plant

All reasonable posts. But, just to add my two cents worth again.

Ignition refers to keeping the plasma hot enough only from the energy from the fusion products thermalizing (transferring heat) with the fuel plasma. This is a seemingly attractive way to keep the plasma hot, despite a number of processes that quickly cool the plasma. There is enough excess energy in the fusion products, that there is enough to go around, to both keep the plasma hot, and to deliver useful energy to the electrical grid. Actually, achieving this state has proven to be very difficult though. And, ignition is not necessarily needed. There are schemes that do not depend on this heating method. The Fusor is not a ignition machine. It is an amplifier. A very inefficient amplifier. Can a simple fuser reach breakeven? It depends on definitions. Can you get as much fusion derived heat out as fuel heating put in? No. But can you get out more useful energy out than you put in? Yes, provided you make some ridiculous assumptions. Generally all of the energy comes out as heat, from energy input plus any fusion output. If you have a magic steam turbine that can convert heat to electricity at an efficiency of perhaps 99.9999999%, then the heat input plus the tiny amount of fusion output could result in a net electrical energy gain. Perhaps even enough to light a single very dim light bulb. But, as steam thermal conversion to electricity is more in the region of 25-40%, you have to come very close to breakeven to even begin to think about making a profit. Actually Q would need to be ~ 2-3 for you to make a few Watts of electricity for your multibillion dollar investment.

I actually think the Sun is a pretty good fusion reactor, considering the fusion cross section of protons- Hydrogen. Actually any deuterium present in the core of the Sun is quickly consumed, even before the Sun settled down into the main sequence hydrogen burning. Brown dwarf stars are stars that fuse the ~ 1 part per 2000 of deuterium , and the small amounts of lithium. Their cores do not get hot enough to fuse hydrogen in appreciable quantities. If the hydrogen in the Suns core was magically transformed into deuterium, it would burn ~ a billion billion times faster. It would blow its self apart within hours, if not seconds. It is not that the Sun (and especially stars with a little more mass than the Sun) are terrible fusion machines, it is that hydrogen is fortunately a very poor fuel, but not too poor- it is just right, or we wouldn't be here.

My appreciation for D-T fusion is that the best temperature is a little less then 50,000 eV, just below the fusion cross section peak. Thermalization concerns may push the ideal average temperature more towards ~ 30,000 eV. D-T fusion collisions at the cross section peak will occur about once for every 10 Coulomb scattering collisions.For D-D the best that can be achieved is ~ 1 fusion collision for every ~ 1000 Coulomb collisions. Thus the efforts for making a fusion reactor that releases more energy than it consumes is to figure out ways to prevent particle or energy loss from the Coulomb collisions or non colliding escapes from overwhelming the energy output from fusion. Then there is the useful fusion output problem. Even if you could breakeven with a dilute plasma (say a density of ~ 1 trillionth of an atmosphere), you would only be producing a few Watts of fusion power. While this is a physics success, it is useless. It is not even enough to power the vacuum pumps or lights in your billion dollar plant. As you increase the density to levels where useful fusion output might happen (say densities of ~ 1 millionth of an atmosphere), the various losses are multiplied exponentially- losses through the magnetic bottle, edge instabilities, Bremsstruhlung radiation losses, etc. all become much more challenging to control. It is essentially a process of taking one step forward and two steps back. But physicists have been very successful despite this- successful at maintaining their careers anyway. Whether a solution will eventually be reached is open to debate. I personally think the tokamak may eventually be a success from a physics standpoint. But, making it economically viable, even if you are a green energy fanatic, would still be a pessimistic possibility. While approaches like the Polywell, or Dense Plasma Focus , or General Fusion, or Lockheed, or Field Reversed Configurations are further back in demonstrating viability, at least they are much cheaper to research and if successful offer a real economic opportunity.

PS: We already have access to a very successful fusion reactor. It is called Solar Energy. We can utilize it directly via photovoltaics and wind, or indirectly through biomass or fossil fuels.

Dan Tibbets

james6742
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### Re: Why is ignition required for a fusion power plant

Thanks everyone for such great responses! Scattering was obviously the elephant in the room that I wasn't aware of.

Knowing what to look for now, I found a good set of lecture slides that goes into the math behind scattering cross sections, both on cold atoms and cold electrons. According to his math, the D-T scattering only degrades the gain by about a factor of 2 and it is the electron scattering that is the real problem:
http://www.lehigh.edu/~eus204/Teaching/ ... ture03.pdf
http://www.lehigh.edu/~eus204/Teaching/ ... ture04.pdf

Richard Hull
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### Re: Why is ignition required for a fusion power plant

it must be remembered that for every deuteron and every triton at fusion energy there is a corresponding rather worthless electron at the same or nearly the same energy. So some decent fraction of any input energy is lost or effectively lost as heat. Plus, in anything less than ideal plasma, (never truly obtained in pure form), there are recombination losses. Lotsa' flys in the fusion ointment.

A COP of 2-3 would not be usable in a non-ignition fusion reactor. The power companies would like to see a COP of 10 or better to get serious about fusion. This assumes the COP of 10 comes from a very inexpensive to construct and smallish fusion reactor. Regardless of COP, you would need a coal fired power plant or small fission reactor to power a successful, non-ignition, fusion reactor 24-7-365.

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
Retired now...Doing only what I want and not what I should...every day is a saturday.

Bob Reite
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### Re: Why is ignition required for a fusion power plant

Once you get the hypothetical 800 MW (electrical) fusion reactor started with the 200 MWe coal plant, what's to stop you from diverting 200 MW from the fusion reactor to sustain the reaction, leaving a net output of 600 MW electrical?
The more reactive the materials, the more spectacular the failures.
The testing isn't over until the prototype is destroyed.