Public: Ecology
EMP Fusion   (+4, -5)  [vote for, against]
Compact fusion device using EMP to heat a ring of LiD.

Posted as a separate idea per advice:

Here are the steps for EMP Fusion:

1. Take Lithium Deuteride (LiD)

2. Make it into a ring shape

3. Coat it with a Tantalum (high melting point)

4. Heat the whole thing to 1000 C (LiD melts)

5. Since LiD is a salt, it conducts electricity when has melted

6. Now take an Explosively Pumped Flux Compression Generator (EPFCG) or another EMP (electromagnetic pulse) device

7. Put the LiD ring at the end of the EMP device

8. Hit it with EMP. It will heat up very fast!

9. The LiD fuel goes thru fusion and releases heat

10. Collect the heat to generate power

Tantalum could be replaced with cheaper metal. Since there is Lithium, there is very little neutron emmission. The blast is small so could be easily digested. Bon apetit.
-- xkuntay, Feb 02 2009

Z-pinch fusion future direction http://fire.pppl.go...load configurations
The summary of research on z-pinch at Sandia National Ls. (to show the magnitude of investment on such a program!) [xkuntay, Feb 12 2009]

Article Link
Shows formulas used in calculations.. heat loss omitted. [xkuntay, Jul 28 2009]

What force do you need to confine the plasma tightly enough for fusion? What are the dimensions of the ring? What's the conductivity of molten LiD? What will the energy yield be? How does the EMP device survive the blast? Will LiD boil (and therefore stop conducting) before it fuses? If tantalum can be replaced with a cheaper metal, why not start out that way?

Sorry, but this seems lacking in details. Without them, it's effectively saying "squash light atoms and heat them until they fuse", which people are currently trying.
-- MaxwellBuchanan, Feb 02 2009

Well answers come as it is tried. I can answer those questions with some 100 gran. For free I can offer some projections: Plasma will be confined by Lorenz forces. Why, it works in z-pinch, why shouldn't it in this case? Dimensions of the ring: could be anything, as small as a wedding ring. Energy yield: each mg yields same as 5 barrels of oil. Will the LiD boil? Yes, but even if it does the tantalum continues to conduct, until LiD becomes a plasma. Will EMP device survive the blast? If the blast is small enough. For instance 0.1 mg of LiD. Why don't we use the alternative of Tantalum to start with? It's not commercial yet. We can use whatever we want. Tantalum sounds okay for now.

//squash light atoms and heat them until they fuse// It is really strange that something not more complicated than making a bread dough could actually win the Nobel prize. But then it's good to have a plaything for all those lazy dreamers.
-- xkuntay, Feb 03 2009

//answers come as it is tried.// No, for something like this, answers come from quantitative modelling, before someone spends the 100K. (We usually get those 'lazy dreamers' to do the maths.)

I'm not saying that it's a silly idea, but it might be silly - there's no indication as to whether it's remotely feasible or impossible by six orders of magnitude.
-- MaxwellBuchanan, Feb 03 2009

so Tantalum is transparent to the spectra of the EMP blast whereas LiD isn't ?
-- FlyingToaster, Feb 03 2009

[F-T] No, Tantalum absorbs EMP too, but it is very thin. Tantalum has high melting point. Its function is to confine LiD when it is melted. Another choice would be...Tungsten. Sounds more familiar?

If this idea fails, it was [Custardguts]' idea. He proposed the LiD. If it doesn't, nevermind.

Also, [M-B], I don't know for a fact but highly suspect that LiD will conduct electricity even when vaporized. Since the current is induced in circles, the circuit need not be complete. The ions can simple be accelerated in vacuum.
-- xkuntay, Feb 03 2009

Well, OK. However, most of the questions here are answerable by calculation, pre-experiment.
-- MaxwellBuchanan, Feb 03 2009

Ah, fusion without the fusion. Well, at least we know for sure it's not baked.
-- colorclocks, Feb 04 2009

You really have to give me discomfort don't you. Okay, I will get my hands dirty:

Ref: id=DygveTQU4YkC& pg=PA315&lpg=PA31 5&dq=salt+melt+conduc tivity&source=web&ots= PFu-ubzLXJ&sig=aTAMv 595jA82Ox_ZjVn6k4-1dw A&hl=en&sa=X&oi=book_ result&resnum=3&ct=result

c (LiI) = 10^-2 ro = 10^4 ohm.m

Lorentz Force = q*[E+(vxB)]

Ignoring electric field (repulsion)

L.F. = q.v.B

v = velocity of particle

B = magnetic field

Assuming two current carrying wires (divide the ring into two):

Lorentz Force = Laplace Force = I.LXB =I.L.B (for parallel wires)

Biot-Savart law:

dB = mu/4pi*I*dLxr/r2

where r represents the distance between wires..

since r and L are always perpendicular and r does not change:

B = mu/4pi*I*L/r2

mu for vacuum is = 1.2566 x 10^-06 T.m/A

ring perimeter L = 2 pi 0.01m = 0.0628 m

r = 0.000001m

Faraday's law of induction: EMF (V) = A.dB/dt

dB/dt = dB / 1 nanosecond

EMP Internal Current = 100 M Amp

Distance of target from EMP = 0.01 m

EMP B = 6283 T

ring area A = pi (0.01)^2 = 0.000314 m2

EMF = 1,972,862,000 Volts

Ref: http://www.geocities. com/capecanaveral /5971/emp.html

I = V/R

ring crossection S = 0.000001 m2

R = = 628000000

I = 3.14 amps

Ohmic heating:

P = I2.R = 6191828800 J/s

Going back to the Lorentz Force:

B = 19,728 T

L.F. = L.I.B = 3890 Newtons

so, pressure-wise:

area of ring = 0.0000628 m2

Pressure = 61,942,675 Pascals = 62,000 kPa

So, there is a force of at least 62 atmospheres on the molten salt. Not much, but enough to keep it together, until it become a plasma.

PV = NRT, so if T goes 100 times (which is 30,000 K) and you want V to stay same, P must go up 100 times. By 30,000 I'm sure it is a plasma. The plasma will have a conductivity much higher than the molten salt, so Lorentz force would be much higher. And I can prove that some other time. Now you must be one important man because I don't do this for everyone. So now, all yours! Please prove me wrong (Austin Powers laugh).
-- xkuntay, Feb 04 2009

ok, so I cheated on r. r was supposed to be 0.0005, making B 250,000 times less. But if I is 1000 times more it is balanced. Cmon, with 1,972,862,000 Volts you must have at least 3140 amps! All metals have resistivities less than 10^-6. Why can't I hope for 10^1?
-- xkuntay, Feb 05 2009

going back to the plasma, which I promised to prove:

the initial plasma resisitivity (ro) is around 1 ohm-m and goes down with temperature. ref: http://www.oldcitypub Pfulltext/HTHP37.3fullte xt/Khomkin.pdf

So, the current is around 31400 amps and that takes our Laplace force to 6200 atm. Still not much, but we can afford a temperature of 3.000,000 K with this! At that temperature the plasma resistivity will be around 0.00001 ohm-m and our Laplace force is 62,000,000,000,000 atm (yes, something like the economic stimulus package). QED.
-- xkuntay, Feb 06 2009

calculation of heating: assuming 0.01 grams of fuel + tungston

specific heat of 0.385 J/g.K

initial heating rate: 6191828800 J/s

if conductivity increased 10 folds, the heating would increase 100 folds, and temp. rise would be:

160,826,722,077,000 K/s or 160,826 K/ns

a 1000 fold increase in conductivity would result in 1M increase in heating rate, making it:

160,826,000,000 K/ns

that temperature coupled with Lorentz pressure would be sufficient for a thermonuclear explosion.
-- xkuntay, Feb 12 2009

I would like to let you know that my paper named EMP Induced Fusion was finally published in a scientific journal. Thanks to [MB] for challenging my calculations, I coded a VB program that simulated the heat accumulation in the target. The target, a lithium wire-tube of 0.1 cm diameter and filled with liquid deuterium, reaches 100 million K in about 1500 ns. Well the effect of Bremhauslung radiation was ignored in this calculation, but when you look at it, it is pretty viable. There are a number of advantages not mentioned earlier: 1.The magnetic field fixes the ions in place so that they can only move along the axis. This preserves the ring shape of the target. 2. There is good reason to expect the initial shock to squeeze the liquid deuterium to the point where it is metallized, thus quadrupling the conductivity. 3. There is good chance that lithium will stay in metal form for some 500 ns. It may not and it may evaporate, but then you can keep it conductive using UV lamps. No croissants until Nobel prize??
-- xkuntay, Jul 27 2009

[marked-for-deletion] - unnacceptable ammount of annotation deletion. I know I, and many other people had discussed this idea in detail earlier this year. I"ve got no problem with deleting defunct, or obviously trollish annos, but this is rediculous. Is it entirely appropriate for [x] to use this kind of scorched-earth editing?
-- Custardguts, Jul 27 2009

what issue of what journal?
-- FlyingToaster, Jul 27 2009

[C] Jeez, it's like I destroyed some important evidence or erased all traces that I stole the idea from somewhere. If you remember the anno, please do post again. Maybe it wasn't me who deleted it. I just deleted two of my own annos which I felt redundant. I didn't know they were so important for you. This is a democratic environment and every comment is appreciated. Why would I post this idea if I didn't care for comments? Same with publishing the paper. The purpose is to share info and ideas. So cheer up fella. I didn't forget you suggested removing the bullet from the pictures. I am not hiding anything here, and I can send a copy of my paper to anyone who might be interested. And I was joking about the Nobel, you figured I hope.

[FT] G.U. Journal of Science 21(2): 33-36 (2008) I can send the file if you like.
-- xkuntay, Jul 27 2009

If I could even pretend to understand it I'd say yes :)
-- FlyingToaster, Jul 27 2009

My technical writing skills can't be that bad!
-- xkuntay, Jul 28 2009

ok, article me.
-- FlyingToaster, Jul 28 2009

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