Monday, May 21, 2007

Operating Voltage For B11

I want to work out the accelerating voltage for the B11 proton reaction.

From this excellent discussion of fusion reactions [pdf] we find that the optimum voltage for the p-B11 reaction when accelerating protons into B11 is 550 KV. Representing a proton energy of 550 Kev. However because we are accelerating protons and B11 from the same virtual grid voltage some adjustments must be made.

First off the reaction runs based on velocity not energy so we have to adjust the accelerating voltage to account for that.

The energy of a non-relativistic particle = 1/2 mV2 However to keep from cluttering up the calculation we will throw out the 1/2 factor since we are not concerned with actual energy but just input voltage and for that the 1/2 cancels out. Masses will be given in Atomic Mass Units where the proton or neutron = 1. Also since we are not concerned with actual values but values relative to 550 Kev, energy will be given as Kev.

So let us start with proton velocity.

mV2=550Kev

Since the mass of a proton = 1 that simplifies to the required velocity = √550,000 = 741.6198 in arbitrary units.

For a given accelerating voltage the energy of the B11 will be 5X the energy of the proton because B11 has 5X the charge of a proton.

Ep=Vp2

EB11=11*VB2

Vp + VB = √550,000

Since the energy of the B11 is 5X the energy of the p then

5*Vp2 = 11*VB2

Vp2 = (11/5)*VB2

Vp = (√(11/5))*VB

Vp + VB = √550,000

substituting:

(√(11/5))*VB + VB = √550,000

calculating the result

VB = 298.650

Vp = 442.970

The energy of the proton will be:

442.9702 = 196,222.175 or an accelerating voltage of about 200KV

The energy of the boron will be:

11 * 298.6502 = 981,110.048

which when divided by 5 = 196,222.010 which checks within the accuracy of the numbers used.

Total energy required will be 1,177,332 ev or about 1.177 Mev

10 comments:

Dave said...

So this would be a 200 KV well?

José Luis said...

Hi, I'm joseluis at fusor.net.

Inspired by dave's post, I applied your calculations to the resonance of the p-B11 reaction at a energy of 148keV, cross-section of about 0.1 barn and obtained the following energies:

E(p) = 52.801 keV
E(B11) = 264.008 keV
Etotal = 316.809 keV

Looks like one can find something interesting with a well of only 55kV, giving the high collision efficiency or a MaGrid, but depending on the width of the peak.

Neal J. King said...

M. Simon,

I'm a little puzzled by your calculation:

- I think epsilon-max is a center-of-mass energy. Even though the units are in keV, that doesn't convert to an accelerating voltage right away.

- Likewise, the relevant mass is the reduced mass
= mp*mB11/(mp+mB11)
if you want to get relative velocity calculations.

Neal J. King said...

M. Simon,

Essentially repeating your calculation, my assumptions are:
- Bare proton, mass = m1 = 1; charge = +1
- Bare boron-11, mass = m2 = 11; charge = Z = +5

Then the lab-frame speeds are both
defined by the same accelerating voltage, so
(m1*v1**2)/2 = e*Voltage

(m2*v2**2)/2 = Z*e*Voltage

Transforming to the center-of-mass frame,
u1 = v1 + x
u2 = v2 - x

where m1*u1 = m2*u2,

we eventually find that:
u1 = (v1 + v2)/(1+a)

u2 = (v1 + v2)* (a/(1+a)),

where a = m1/m2

At the maximum cross-section, epsilon-max =
(m1*u1**2 + m2*u2**2)/2,

the accelerating voltage can be expressed as:

Vo = {(1+a)/[(1+sqrt(Za))**2]}*(epsilon-max/e)

When epsilon-max = 550 keV
a = m1/m2 = 1/11 = 0.09
Z = 5

I find Vo = 214 kV, pretty close to your 196; and hence
Energy-p = 214 keV
Energy-B11 = 1070 keV
Total energy = 1284 keV

(As a check, I can put in a = 1 and Z = 1; my general formula then gives e*Vo/epsilon-max = 1/2, which is what you would expect to get.)

For epsilon-max = 148 keV, I find:
Voltage = 57.6 kV
Energy-p = 57.6 keV
Energy-B11 = 288 keV
Total energy = 345.6 keV


The differences come from taking into account the center-of-mass motion.

Neal J. King said...

As another check on the formula, try Z = 0 and m2 = m1, so a = 1. This corresponds to a proton (m1) hitting a neutron (m2 = m1). It's easy to see that the voltage has to produce twice the energy of the center-of-mass frame. The formula does indeed give you that:

Vo = 2*epsilon-max/e

M. Simon said...

Neal,

My calculations are not rigorous in that I mix units in a realy horrible way to simlify the caculations.

The deal is that the velocity of m1 plus the velocity of m2 must equal the velocity that produces fusion. i.e. the velocity of m1 when accelerated through the required potential where the velocity of m2 is zero. Which is how the energies are given.

Actually the velocity is given as an energy of m1 in electron volts.

The cross section of p-B11 peaks at around 560 to 580 Kev (of the proton where the velocity of B11 is zero) . The peak cross section is about 1 to 1.5 barns.

I'll see if I can dig up a reference.

M. Simon said...

Here is a good semi-techical primer on fusion reactions:

Fusion reactions [pdf]

It is also on the sidebar.

It has graphs on fusion cross sections vs energy.

M. Simon said...

I have redone my calculations in as more general way at:

General Calculation.

Neal J. King said...

M. Simon,

From my point of view, the problem remains.

If you check Section 1.2 of your reference on nuclear-fusion reactions, you will see that everything is defined in the CoM frame:
- all energies (epsilon-whatever)
- the coefficient of the relative velocity is neither m1 nor m2, but the reduced mass

Look at equations 1.7 - 1.8.

This implies that the issue of "what is the right energy" has to be decided in the CoM frame.

The issue of "what voltage does this correspond to" is established in the lab frame.

Raelik said...

So, given that according to Bussard, the well depth is ~80% of the drive voltage, then 250kV is going to be required for optimal p-B11 fusion, which I believe the good Dr. has stated before.