Tuesday, May 29, 2007

Deriving Operating Voltage For An IEC Reactor

I have had some complaints (justifiable in my opinion) on how I derived the operating voltage for a given fusion reactor to within a few percent. I dont't take into account Einstein since the particle energies we are talking about for a proton are on the order of .6 Mev maximum and a proton "weighs" around 900 Mev. It might matter in the fifth place decimal (I estimate about a 1 part in 300,000 difference) but we don't really need to know the voltage that close since creation of the polywell will cost us around 10% of the drive voltage, indicating that any reasonable supply should be capable of about the calculated voltage +20% or maybe +25% to give a little more margin.

Well off to the races. First let me define the terms.

m1 = mass of the lowest mass particle amu (protons)
m2 = the other mass in amu
c1 = the charge of m1 in electron charge units
c2 = the charge of m2 in electron charge units
AV = accelerator voltage in KV
TV = the voltage from a chart or graph where one particle is stationary
V1 = the velocity of m1 when accelerated by AV
V2 = the velocity of m2 when accelerated by AV
V1tv = the velocity of m1 when accelerated by TV

V1 + V2 = V1tv

When deriving the equations the = sign from now on will mean exactly proportional to.

m1 * V12 = c1 * AV
m2 * V22 = c2 * AV

m1 * V1tv2 = c1 * TV

Since we are using Newtonian mechanics velocities add:

V12 = c1 * AV / m1
V1 = √( c1 * AV / m1 )

V22 = c2 * AV / m2
V2 = √( c2 * AV / m2 )

V1tv = √( c1 * TV / m1 )

I'm going to skip a few obvious steps here for brevity.

√( c1 * TV / m1 ) =

√ AV * ( √( c1 / m1 ) + √( c2 / m2 ))

√ AV = (√( c1 * TV / m1 )) / ( √( c1 / m1 ) + √( c2 / m2 ))

Which is in a spread sheet availabe at the IEC Fusion Newsgroup as an attachment.


Anonymous said...

I don't agree with this step:

m1 * V1tv2 = c1 * TV

- In order to be consistent with the way the energy calculations are done in your fusion-reaction reference, this calculation has to be done in the center-of-mass frame.

- accordingly, it should not have m1, but rather mr, where
mr = m1*m2/(m1+m2)

- and TV should be defined in the CoM frame.

Check out Section 1.2 of your reference. All of his energy calculations are done in the CoM frame.

Anonymous said...

A further problem, which is either typographical or possibly consequential from the first: Your final results don't depend at all on m2.

But that's impossible. In the limit that m1/m2 => 0, it must be that

Vo = epsilon-max/e

M. Simon said...

I agree that I made a typo.

Corrected I think.

I'll look up the center of mass bit.

M. Simon said...


I'm looking at the velocities prior to collision where the velocities add.

In addition the center of mass is only important in determing where the collision will take place. It does not determine collision velocity.

The center of mass will determine if there is a velocity bias in the collision. Which will bias the results for a particular collision. However due to the design of the reactor that should average out.

M. Simon said...

Actually, the above definition applies in general
to particles with relative velocity v, and is therefore symmetric in the two
particles, since we have �12(v) = �21(v).

From the reference.

Anonymous said...

M. Simon,

You are correct about the relative velocity: It is the same in all frames.

The point I'm making is that when you set the equation:
epsilon-max = total kinetic energy,

this calculation has to be done in the CoM frame, and there are two separate velocities:
KE = 2 * (m1*u1**2 + m2*u2**2)

where u1 and u2 are seen in the CoM frame; or equivalently,

KE = 2 * (mr * v-rel**2)

where mr = m1*m2/(m1+m2) ,

and v-rel = u1 + u2
= v1 + v2

Look at eqn 1.7 on page 3 of the fusion-reaction pdf.

Anonymous said...

Since you don't seem to believe me, I'll do it again, re-using the equations from the reference more closely.

a) Because both nuclei are experiencing the same voltage well, of magnitude Vo, their kinetic energies (when at the bottom of the well) are related:

- for the proton, for charge = +e:

kinetic energy = (Mp*vp**2)/2
potential energy = e*Vo
so e*Vo = (Mp*vp**2)/2

and so
vp = sqrt(2*e*Vo/Mp)

- for the B-11, for charge = +Ze:

kinetic energy = (Mb*vb**2)/2
potential energy = Z*e*Vo
so Z*e*Vo = (Mb*vb**2)/2

and so
vb = sqrt(2*Z*e*Vo/Mb)

b) The relative velocity is the sum of vp and vb, because we assume they are in opposite direction. Therefore
v-rel = vp + vb
= sqrt(2*e*Vo) [sqrt(Z/Mb) + sqrt(1/Mp)]

c) Looking at eqn. 1.7, the center-of-mass energy, epsilon, is

epsilon = ((M-reduced)*(v-rel)**2)/2

M-reduced = Mp*Mb/(Mp+Mb)

(v-rel)**2 = (vp + vb)**2
= (2*e*Vo)*[Z/Mb + 1/Mp + 2sqrt(Z/(Mb*Mp))]

epsilon = [e*Vo/(Mp+Mb)]*[Z*Mp + Mb + 2*sqrt(Z*Mp*Mb)]

If now we set Mp/Mb = a, we get

epsilon = [e*Vo/(a+1)]*[Z*a + 1 + 2*sqrt(Z*a)]

= [e*Vo/(1+a)]*(1+sqrt(Z*a))**2

Turning this around, we get:
Vo = (epsilon/e)(1+a)/[1+sqrt(Z*a)]**2

which is exactly what I had before.

In the particular case of interest,
a = Mp/Mb = 1/11

Z = 5

Note: This calculation assumes that the proton and the B11 are headed directly at each other. If you expect some divergence in the beam, you'll need more voltage so that there will be enough energy even when they come at each other with a slight angle.

Anonymous said...

Final nail in the coffin: on Section 1.2, page 4, paragraph 2, your fusion-reaction reference states:

"Cross sections can also be expressed in terms of the centre-of-mass energy 1.7, and we have σ12(epsilon) = σ21(epsilon). In most cases, however, the
cross sections are measured in experiments in which a beam of particles with energy epsilon-1 measured in the laboratory frame, hits a target at rest.
The corresponding beam-target cross-section σbt12(epsilon-1) is related to the centre-of-mass cross-section σ12(epsilon) by
σ12(epsilon) = σbt(epsilon-1) , 1.9
with epsilon-1 =  epsilon ·(m1 + m2)/m2. From now on, we shall refer to centre-of-mass cross-sections and omit the indices 1 and 2.