Proton Boron Consortium

There is a new kid on the block when it comes to Polywell Fusion. The Proton Boron Consortium. It is working on an open source Polywell type design. You can also find the group on Facebook Proton Boron Consortium - Facebook. I am personally associated with the effort. They haven't started serious fund raising yet but if you would like to donate to the electronics part of the work you can use this PayPal button.

Please note if it is for the PB (Proton Boron) Project. Or the KW (kilowatt hour) offer.

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.

E_p=V_p²

E_B11=11*V_B²

V_p + V_B = √550,000

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

5*V_p² = 11*V_B²

V_p² = (11/5)*V_B²

V_p = (√(11/5))*V_B

V_p + V_B = √550,000

substituting:

(√(11/5))*V_B + V_B = √550,000

calculating the result

V_B = 298.650

V_p = 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