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XYLENE POWER LTD.

**NUCLEAR FUSION:**

Plasma Impact Fusion (PIF) is essentially a pulse based energy generation technology. An issue of great importance is the fusion energy output per pulse.

**DESIGN CHOICES:**

Nda = Nta = initial numbers of deuterium and tritium ions

Choose theoretical maximum fusion pulse energy at 1200 MJ. Then:

**Ndd = Ntd** = (1200 MJ / 17.59 MeV) X (1 eV / 1.602 X 10^-19 J)

**= 42.586 X 10^19 ions**

From the energy position of the relative maximum in D-T fusion cross section the theoretical minimum required initial liquid lead kinetic energy Ekld is given by:

Ekld = 2 (2) (Ndd + Ntd) X 120 keV

= 8 X 4.2586 X 10^20 partices X 1.2 X 10^5 eV / particle X 1.602 X 10^-19 J / eV

= 65.492 X 10^6 J

**= 65.492 MJ**

**DEFINITIONS:**

Ri = plasma radius

T = time

T = Tf at Ri = Rif

Ekld = initial liquid lead kinetic energy at Ri = Rid;

T = Tg at Ri = Rig

Rig = plasma radius at which plasma kinetic energy = Ekld / 4

Eklg = (3 / 4) Ekld = liquid lead kinetic energy at Ri = Rig

T = Ti at Ri = Rii

Rii = Rig / 2 = minimum plasma radius

Epi = 4 Epg = plasma thermal energy at Ri = Rii

T = Tk at Ri = Rik = Rig

T = Tl at Ri = Ril = Rif

Nd = number of deuterium ions in plasma

Nd = Ndd at time T = Td

Nd = Ndg at time T = Tg

Nd = Ndi at time T = Ti

Nd = Ndl at time T = Tl

Vd = thermal velocity of deuterium ions

Nt = number of tritium ions in plasma

Vt = thermal velocity of tritium ions

Sigma = average deuterium-tritium fusion cross section for Rii < Ri < Rig

T = time

Vol = plasma volume

**Ri BEHAVIOR:**

In the PIF process for Ri starts at its initial value of:

Ri = Rif,

continues decreasing through:

Ri = Rig

to its minimum value at:

Ri = Rii.

Ri = Rik = Rig

and eventally reaches:

Ri = Ril = Rif.

In the region:BR>
Ri > Rig

both the ion thermal velocities Vd and Vt and the fusion cross section Sigma decrease with increasing Ri, so from the perspective of calculating fusion rates fusion can be ignored in the region:

Ri > Rig

and

Ri > Rik

Hence:

Ndd = Ndg

and

Ndl = Ndk

**FUSION RATE:**

Deuterium and tritium fuse together in accordance with the equation:

H-2 + H-3 = He-4 + n + 17.59 MeV

Note that in these interactions:

d(Nd) = d(Nt)

For deuterium-tritium fusion interactions we can write:

d(Nd) = - Nd (Nt / vol) Sigma Vd dT - Nt (Nd / Vol) Sigma Vt dT
= - Nd Nt Sigma (Vd + Vt) dT / Vol

For properly mixed (H-2 + H-3) fuel:

Nd = Nt

so this equation simplifies to:

**d(Nd) / Nd^2** = - Sigma (Vd + Vt) dT / Vol

**= - [Sigma (Vd + Vt) / Vol] [dT / dRi] dRi**

**SIGMA:**

**In order to solve the fusion rate equation we will assume that in the ranges:
Rig > Ri > Rii
and
Rik > Ri > Rii
Sigma = 3.0 barns
and in the ranges
Ri > Rig
and
Ri > Rik
Sigma = 0.**

Data published by Kaye & Laby indicates that in actual fact:

At Ri = Rig:

Sigma = 2.0 barns

and then as Ri decreases from Rig to Rii Sigma increases from 2.0 barns to 5.0 barns before decreasing to 2.0 barns at:

Ri = Rii.
Similarly as Ri increases from Rii to Rik Sigma increases from 2.0 barnes back up to 5.0 barns and then decreases back to 2.0 barns at:

Ri = Rik.

For:

Ri > Rig

and

Ri > Rik

Sigma decreases rapidly to zero with increasing Ri.

**ASSUMPTIONS:**

In order to allow closed form calculation of fusion rate a number of assumptions are made. To the extent that these assumptions do not reflect physical reality the calculation of fusion rate will be in error.

1. As shown above:

for:

Rii < Ri < Rig

Sigma = 3.0 barns

= 3.0 X 10^-28 m^2

and for

Ri > Rig

Sigma = 0.0

2. The liquid lead is assumed to be an incompressible fluid. We know that this assumption is not correct at very high pressures, but this assumption is made anyway to allow a closed form solution.

3. During the adiabatic compression of the plasma there is no energy loss in the liquid lead. Since the liquid lead is assumed to be incompressible it cannot convert energy of compression into heat.

4. In the plasma the relationship between contained energy and volume is of the form:

Ep = Epb(Rig / Ri)^2

This relationship is valid for adiabatic compression where:

[Rig / (2)^0.5] < Ri < Rif

but is not valid for:

Rii < Ri < Rig / (2)^0.5

but we will still use it in the region:

(Rig / 2) = Rii < Ri

5. Once fusion starts energy feedback from the fusion will increase the thermal velocites Vb and Vt which will tend to increase the fusion rate. However the increasing particle energies will reduce Sigma which will tend to decrease the fusion rate. For now we will assume that these two effects cancel each other out.

Thus the fusion rate calculation made using these assumptions is fairly crude.

**MAXIMUM ION CONCENTRATION:**

The concentration of lead atoms in the liquid lead is:

10.66 X 10^3 kg / m^3 X 1 mole / 207.2 g X 1000 g / kg X 6.023 X 10^23 atoms / mole

= 0.30987 X 10^29 atoms / m^3

= 0.30987 X 10^20 atoms / mm^3

Hence a liquid lead sphere with a radius of 2 mm contains:

0.30987 X 10^20 atoms / mm^3 X (4/3) Pi (2 mm)^3 = 10.3838 X 10^20 lead atoms

This number of lead atoms is close to the number of deuterium and tritium ions. When the ion concentration exceeds the lead atom concentration the deuterium and tritium mix with the liquid lead, which sharply slows D-T reactions. In order for the deuterium plus tritium ion concentration to not exceed the lead atom concentration:

**Ric > 2 mm**

Hence fusion only occurs in the range:

Rig > Ri > Rii

where:

Rii > 2 mm

and

Rii = Rig / 2

To ensure that this condition is met the PIF system design parameters are chosen so that:

Rii ~ 3.5 mm

**FIND Vd AND Vt**

The PIF equipment is designed to operate at particle energies where Sigma is close to its physical maximum. Specifically, for:

Ri = Rig / (2)^0.5

Ekdh = Ekth = 120 keV.

Define:

Md = mass of deuterium ion;

Mt = mass of tritium ion.

In general:

Ekd = (Md / 2) Vd^2

and

Ekt = (Mt / 2) Vt^2

or

Vd = [2 Ekd / Md]^0.5

and

Vt = [2 Ekt / Mt]^0.5

However:

Ekt = Ekd

because all ions tend to adopt the same thermal energy. Hence:

Vt = [2 Ekd / Mt]^0.5

From the properties of adiabatic compression:

Ekd = [Ekdg] [Rig^2 / Ri^2]

At Ri = Rih = (Rig / (2)^0.5):

Ekdh = 2 Ekdg

or

120 keV = 2 Ekdg

or

Ekdg = 120 keV / 2

= 60 keV

In general:

Ekd = Ekdg (Rig / Ri)^2

Thus:

**Vd** = [2 Ekd / Md]^0.5

= [2 Ekdg (Rig / Ri)^2 / Md]^0.5

**= [2 Ekdg / Md]^0.5 (Rig / Ri)**

= Vdg (Rig / Ri)

Similarly:

**Vt = [2 Ektg / Md]^0.5 (Rig / Ri)**

= Vtg (Rig / Ri)

**PLASMA VOLUME:**

**Vol = (4 / 3) Pi Ri^3**

**LIQUID LEAD INITIAL KINETIC ENERGY:**

Recall that:

Ekdg = 60 keV.

Epg = (4 Ndd X 60 keV).

The liquid lead kinetic energy at Ri = Rig is given by: Eklg = (3 / 4) Ekld

Conservation of energy gives:

Epg = Ekld / 4.

Thus we have:

Ekld / 4 = (4 Ndd X 60 keV)

or

**Ekld = 16 Ndd X 60 keV**

= 16 X 4.2584 X 10^20 X 60 keV

= 16 X 4.2584 X 10^20 X 60 keV X 1000 eV / keV X 1.602 X 10^-19 J / eV

= 6549.07 X 10^4 J

**= 65.4907 MJ**

**FIND Rif AND Ekdf:**

(Ekdg / Ekdf) = (Rif / Rig)^2

or

Ekdg Rig^2 = Ekdf Rif^2

and

**Choose Rif = 1.000 m**

as being a reasonable value which has some room for compromise via an increase in spheromak energy if it should turn out to be necessary to reduce Rif due to a requirement for more time for the fuel injection and for the plasma to come to equilibrium.

Then:

**Ekdf** = Ekdg (Rig / Rif)^2

= 60,000 eV (.0075419 m / 1.0 m)^2

**= 3.413 eV**

**PLASMA POTENTIAL ENERGY:**

Assume adiabatic compression conditions which give the plasma energy Ep as:

Ep = Epg(Rig / Ri)^2

**LIQUID LEAD KINETIC ENERGY:**

On the web page titled SPHERICAL COMPRESSION A
it is shown that for Ri << Ro, as pertains near fusion conditions:

Ekl = Rhol 2 Pi Ri^3 (dRi / dT)^2

where:

Ekl = kinetic energy in liquid lead;

Rhol = density of liquid lead = 10.66 X 10^3 kg / m^3

Thus:

(dRi / dT)^2 = Ekl / (Rhol 2 Pi Ri^3)

However conservation of energy gives:

Ekl = Ekld - Ep

= Ekld - Epg(Rig / Ri)^2

Recall that:

Epg = Ekld / 4

Thus:

Ekl = Ekld [1 - (1 / 4)(Rig / Ri)^2]

Hence:

(dRi / dT)^2 = Ekl / (Rhol 2 Pi Ri^3)

= Ekla [1 - (1 / 4)(Rig / Ri)^2] / (Rhol 2 Pi Ri^3)

or

**(dRi / dT) = - {Ekld [1 - (1 / 4)(Rig / Ri)^2] / (Rhol 2 Pi Ri^3)}^0.5**

or

**(dT / dRi) = - {(Rhol 2 Pi Ri^3) / (Ekld [1 - (1 / 4)(Rig / Ri)^2])}^0.5**

**SOLVE THE ORIGINAL DIFFERENTIAL EQUATION:**

Recall that:

**d(Nd) / Nd^2 = - [Sigma (Vd + Vt) / Vol] [dT / dRi] dRi**

Substitution into this equation of previously developed expressions for Sigma, Vd, Vt, Vol, (dRi / dT) gives:

**d(Nd) / Nd^2** = - [Sigma (Vd + Vt) / Vol] [dT / dRi] dRi

= - {Sigma [2 Ekdg]^0.5 (Rig / Ri)[(1 / Md)^0.5 + (1 / Mt)^0.5] / [(4 / 3) Pi Ri^3]}

X [- {(Rhol 2 Pi Ri^3) / Ekld [1 - (1 / 4)(Rig / Ri)^2]}^0.5] dRi

= {Sigma [2 Ekdg]^0.5 [(1 / Md)^0.5 + (1 / Mt)^0.5] / (4 / 3) Pi}{(Rig Ri^1.5 / Ri^4) dRi}

X {(Rhol 2 Pi) / (Ekld [1 - (1 / 4)(Rig / Ri)^2])}^0.5

{Sigma [2 Ekdg]^0.5 3 [(1 / Md)^0.5 + (1 / Mt)^0.5] / (4 Pi)}{(Rig^2.5 / Ri^2.5) d(Ri / Rig)}

X {(Rhol 2 Pi) / (Rig Ekld [1 - (1 / 4)(Rig / Ri)^2])}^0.5

Z = (Ri / Rig)

Hence:

**d(Nd) / Nd^2**

= {Sigma [2 Ekdg]^0.5 3 [(1 / Md)^0.5 + (1 / Mt)^0.5] / 4 }{(1 / Z)^2.5 dZ}

X {(Rhol 2) / (Rig Ekld Pi[1 - (1 / 4)(1 / Z)^2])}^0.5

= {Sigma [2 Ekdg]^0.5 6 [(1 / Md)^0.5 + (1 / Mt)^0.5] / 4}{(1 / Z)^1.5 dZ}

X {(Rhol 2) / (Rig Pi Ekld [4 Z^2 - 1])}^0.5

= {Sigma 3 [(1 / Md)^0.5 + (1 / Mt)^0.5]}{dZ / (Z^1.5 [4 Z^2 - 1]^0.5 }

X {(Rhol Ekdg) /(Rig Pi Ekld)}^0.5

Thus:

**[(-1 / Ndi) + (1 / Ndg)] = {3 Sigma [(1 / Md)^0.5 + (1 / Mt)^0.5]}
X {(Rhol Ekdg) / (Rig Pi Ekld)}^0.5
X Integral from Z= 1 to Z = 0.5 of:
{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}**

or

X {(Rhol Ekdg) / (Rig Pi Ekld)}^0.5

X Integral from Z= 0.5 to Z = 1.0 of:

{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}

Similarly:

**[(1 / Ndk) - (1 / Ndi)] = {3 Sigma [(1 / Md)^0.5 + (1 / Mt)^0.5]}
X {(Rhol Ekdg) / (Rig Pi Ekld)}^0.5
X Integral from Z= 0.5 to Z = 1.0 of:
{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}**

Adding these two equations gives:

**[(1 / Ndk) - (1 / Ndg)]** = {6 Sigma [(1 / Md)^0.5 + (1 / Mt)^0.5]}

X {(Rhol Ekdg) / (Rig Pi Ekld)}^0.5

X Integral from Z= 0.5 to Z = 1.0 of:

{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}

or

**[(Ndg / Ndk) - 1] = {6 Ndg Sigma [(1 / Md)^0.5 + (1 / Mt)^0.5]}
X {(Rhol Ekdg) /(Rig Pi Ekld)}^0.5
X Integral from Z= 0.5 to Z = 1.0 of:
{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}**

**FUSION PARAMETER Fr:**

Define:

Ndd = initial number of deuterium ions in the confined plasma. Hence the number of injected deuterium molecules is:

**(Ndd / 2)**

Fr = fraction of Ndd deuterium ions that have participated in fusion reaction up to time T;

Then:

Frg = value of Fr at Ri = Rig

Fri = value of Fr at Ri = Rii

Frk = value of Fr at Ri = Rik

Frl = value of Fr at Ri = Ril

Since there is negligible fusion in the region:BR>
Rig < Ri < Rif

hence:

Frg = 0

Since there is negligible fusion in the region

Rik < Ri < Ril

hence:

Frl = Frk

Hence:

(Frk Ndd) = number of deuterium atoms that fuse during each fusion pulse

where:

Frk = [1 - (Ndk / Ndd)]

**COEFFICIENT EVALUATION:**

**{6 Ndg Sigma [(1 / Md)^0.5 + (1 / Mt)^0.5]}
X {(Rhol Ekdg) / (Rig Pi Ekld)}^0.5**

= {6 X 4.2584 X 10^20 ions X 3.0 X 10^-28 m^2 [(1/ 2)^0.5 + (1 / 3)^0.5]

X [1 / (1.67 X 10^-27 kg)]^0.5}

X {(10.66 X 10^3 kg / m^3 X 60,000 eV X 1.602 X 10^-19 J / eV) / (0.01621 m X Pi X 65.4907 X 10^6 J)}^0.5

= {76.6512 X 10^-8 m^2 [.70710 + .57734] [2.447 X 10^14] kg^-0.5 X [30.722 X 10^-18 X kg J/ m^4 J]^0.5

=240.91 X 10^6 m^2 kg^-0.5 X 5.5427 X 10^-9 kg^0.5 / m^2

= 1335.30 X 10^-3

= **1.3335**

**INTEGRAL EVALUATION:**

Integral from Z= 0.5 to Z = 1.0 of:

{dZ / (Z^1.5 [4 Z^2 - 1]^0.5)}

~ Integral from Z= 0.5 to Z = 1.0 of:

{dZ / (Z [4 Z^2 - 1]^0.5)}

= arc sin{(-2) / (|Z| 4)}| Z = 1

- arc sin{(-2) / (|Z| 4)| z= 0.5

= arc sin{(-0.5} - arc sin{(-1}

= (- Pi / 6) - (- Pi / 2)

= **Pi / 3**

**FIND Fr:**

[(Ndg / Ndk) - 1 = 1.3335 (Pi / 3)

or

**(Ndd / Ndk)** = (Ndg / Ndk)

**= 2.3964**

Thus:

**Frk** = [1 - (Ndk / Ndd)]

= 1 - (1 / 2.3964)

**= 0.5827**

Thus from a theoretical perspective the PIF process appears to be viable with (H-2 + H-3) fuel at an initial liquid lead radial velocity of -300 m /s. Note that for (H-2 + He-3) fuel the value of Sigma is two orders of magnitude smaller which reduces Fr below the minimum value necessary for net power generation. Hence the fuel (H-2 + He-3) will not provide net power generation via the PIF process.

This web page last updated January 17, 2015.

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