XYLENE POWER LTD.

FUSION OUTPUT

By Charles Rhodes, P.Eng., Ph.D.

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.

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 for lack of a better expression.

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.

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

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

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
[(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= 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.