XYLENE POWER LTD.

D-T FUSION FUEL

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

FUEL CHOICE:
This web page assumes the use of Deuterium-Tritium (D-T) or (H-2 - H-3)) fusion fuel.

Deuterium-tritium (D-T) or (H-2 + H-3) has a low reaction temperature, a high fusion cross section and a high energy release. However, (H-2 + H-3) has low energy feedback and poor availability.

The H-3 must be produced via a dedicated breeding program.
 

MAIN NUCLEAR REACTIONS:
The nuclear reaction for (H-2 + H-3) fuel is:
H-2 + H-3 = He-4 + n + 17.59 MeV
 

BREEDING REACTIONS:
The H-3 that is required is obtained by utilization of at least half of the energetic fast neutrons emitted by the main reaction using the reactions:
n + Li-7 = 2 n + Li-6
followed by:
n + Li-6 = He-4 + H-3
Note that in storage H-3 decays with a half life of 12.6 years to less desirable He-3.
 

(H-2 + H-3) FUEL CONCEPT DETAIL:
The plasma consists of an integral multiple of a particle group consisting of 1 H-2 ion plus 1 H-3 ion plus 2 associated electrons. There are 4 particles in this particle group (2 electrons + 2 ions). An electrical apparatus produces a random plasma with an average particle kinetic energy of 3.5 eV in addition to the ionization energies. The plasma is adiabatically compressed to raise the average energy per particle at
Epg = Ekld / 4
to
60 keV.

This choice results in a peak particle energy, independent of fusion heating of 240 keV and a particle energy at:
Eph = Ekld / 2
of
Ekdh = 120 keV
= 0.120 MeV.

Note that Ekdh is the energy position of the peak in the fusion cross section Sigma.

Thus the maximum possible potential fusion heating energy release is:
17.59 MeV per particle group.

The chosen number of such particle groups contained within a single fusion pulse is:
[1200 MJ / (17.59 MeV / particle group)] X [1 eV / 1.602 X 10^-19 J]
= 42.5846 X 10^19 particle groups

The energy required to ionize each particle group is:
[(1 H-2 ion X 13.6 eV / ion) + (1 H-3 ion X 13.6 eV / ion) + (4 particles X 3.5 eV / particle)]
= 13.6 eV + 13.6 eV + 14.0 eV
= 41.2 eV / particle group

The minimum amount of energy that must be delivered to the random plasma by the spheromaks to fully ionize the fuel is:
(42.5846 X 10^19 particle groups / fusion pulse) X (41.2 eV / particle group)
= 1754.48 X 10^19 eV / fusion pulse X 1.602 X 10^-19 J / eV
= 2810.6 J / fusion pulse

If as a result of adiabatic compression, at
Ri = Rii
the plasma contains 240 keV per particle the total plasma energy at Ri = Rii is:
240 X 10^3 eV / particle X 4 particles / particle group X 42.5846 X 10^19 particle groups
X 1.602 X 10^-19 J / eV
= 240 X 10^3 X 4 X 42.5846 X 1.602 J
= 65.492 MJ
~ 65.5 MJ

In order to provide this amount of energy to the plasma the liquid lead needs an initial kinetic energy of at least:
Ekld = 65.5 MJ.
 

PLASMA VOLUME:
As shown on the web page titled ADIABATIC COMPRESSION, the condition for for adiabatic compression is:
Ekl = (dRi / dT)^2 [Rhol 2 Pi Ri^3]
> (2 / Ml)^2 (Ekt Mt) [Rhol 2 Pi Ri^3]
or at the threshold of being adiabatic:
Pi Ri^3 = Ekl (Ml / 2)^2 / 2 Ekt Mt Rhol

We want to maintain adiabatic compression until Ri = Rih.

Note that:
Rih = Rig / (2)^0.5
and
Rii = (Rig / 2)

Thus at Rih = Rig / (2)^0.5 the compressed plasma volume Vol is given by:
Vol = (4 / 3) Pi Rih^3
= (4 / 3) (Ekld / 2) / {(2 / Ml)^2 (Ekt Mt) (Rhol 2)}
= (4 / 3) ((65.5 / 2) X 10^6 J) / {[2 / (207 X 1.67 X 10^-27 kg)]^2 (120 X 10^3 eV X 1.602 X 10^-19 J / eV X 3 X 1.67 X 10^-27 kg) X (10.66 X 10^3 kg / m^3 X 2)}
= (1 / 3) (32.75 X 10^6 J) (1.67 X 10^-27 kg)(207)^2 / {(120 X 10^3 eV X 1.602 X 10^-19 J / eV X 3) X (10.66 X 10^3 kg / m^3 X 2)}
= 63.532 X 10-8 J kg m^3 / J kg
= 0.63532 X 10^-6 m^3

The correspondinng plasma radius Rih is given by:
Rih^3 = 0.63532 X 10^-6 m^3 X (3 / 4 Pi)
or
Rih = 10^-2 m [0.63532 X 3 / 4 Pi]^0.333
= 10^-2 m [.15167]^0.333
= 10^-3 m [151.67]^0.333
= 5.3329 mm

Thus:
Rig = (2)^0.5 X 5.3329 mm
= 7.5419 mm
and
Rii = 5.3329 mm / (2)^0.5
= 3.771 mm

The number of deuterium ions in the plasma volume is:
42.5846 X 10^19 particle groups X 1 deuterium ions / particle group
= 4.25846 X 10^20 deuterium ions

Thus at Ri = Rih the deuterium ion density (Nd / Vol) is:
Nd / Vol = (4.25846 X 10^20 ions) / (0.63532 X 10^-6 m^3)
= 6.70285 X 10^26 deuterium ions / m^3

At Ri = Rih the H-3 ion velocity Vth is given by:
Mt Vth^2 / 2 = 120 X 10^3 eV X 1.602 X 10^-19 J / eV
or
Vth^2 = [2 / (3 X 1.67 X 10^-27 kg)] (120 X 10^3 eV X 1.602 X 10^-19 J / eV
)
= 76.743 X 10^11 J / kg = 7.6743 X 10^12 (m /s)^2

Hence:
vth = 2.77022 X 10^6 m / s
 

COMPRESSION RADIUS:
(7.5 eV)(Rif / Rih)^2 > 120,000 eV
or
Rif > [120,00 / 7.5]^0.5 Rio
= 126.49 (.0053329 m)
= 0.674 m
Thus there is adequate margin in Rif to allow fuel injection and plasma randomization.
 

BLOCK DIAGRAM:
A block diagram of a fusion power plant is shown below. The energy amounts shown correspond to the fusion of one particle group.

A Plasma Impact Fusion (PIF) power plant has four major sub components:

1)PLASMA GENERATOR:
The plasma generator uses electricity to produce 2 X 3 kJ plasma spheromaks. These spheromaks heat the deuterium-tritium fuel gas to form a plasma with an average particle kinetic energy ~ 3.5 eV.
 

2)ADIABATIC PLASMA COMPRESSOR:
The plasma compressor uses spherically convergent liquid lead with an initial radial velocity of - 300 m / s to adiabatically compress the plasma to a temperature of 120 keV / plasma particle at Ri = Rih. The fraction of the plasma particles that react is Fr. Thus for a 17.591 MeV per particle group fusion energy output the required input energy is:
4 (120 keV) / Fr = 0.48 MeV / Fr.

The source of energy for implementing the adiabatic compression is the kinetic energy of the high velocity liquid lead. At Ri= Rih half of the initial kinetic energy in the liquid lead has coupled to the plasma. Hence the initial kinetic energy of the liquid lead per particle group in the plasma must be:
2 (0.48 MeV) / Fr = 0.96 MeV / Fr
 

3) LIQUID LEAD ACCELERATOR:
Assume that the combined efficiency of converting electrical energy to initial liquid lead kinetic energy and then converting initial liquid lead kinetic energy into plasma energy is Fc. Then the electricity input Ef to the liquid lead acceleration system per particle group is:
Ef = 0.96 MeV / Fr Fc
 

4)HEAT TO ELECTRICITY CONVERTER:
The heat to electricity converter is assumed to be a two stage steam turbogenerator with an efficiency of:
Fg = Ee / Et
= 0.3333.
 

MATHEMATICAL ANALYSIS:
For each reacting particle group the system's gross thermal output Et is:
Et = 17.59 MeV + (0.96 MeV / Fr)

Assuming a turbogenerator efficiency Fg of:
Fg = (electricity output Ee) / (thermal power input Et)
= (1 / 3)
then the net electricity output from the fusion power system is:
Ee - Ef
= (Et / 3) - 0.96 MeV / Fr Fc
= [17.59 MeV + (.96 MeV / Fr)] / 3 - (.96 MeV / Fr Fc)
= 5.8633 MeV + (0.32 MeV / Fr) - (0.32 MeV / Fr)(3 / Fc)
= 5.8633 MeV - (0.32 MeV / Fr)[(3 / Fc) - 1]

The reject heat flow is:
(2 Et / 3) + (1 - Fc)Ef
= (2 / 3)[[17.59 MeV + (.96 MeV / Fr)] + (1 - Fc) [.96 MeV / Fr Fc]
= 11.7266 MeV + (0.64 MeV / Fr) + (.96 MeV / Fr Fc) - .96 MeV / Fr
= 11.7266 MeV - (0.32 MeV / Fr) + (.96 MeV / Fr Fc)
= 11.7266 MeV + [(3 / Fc) - 1] (0..32 MeV / Fr)

Thus the ratio of net electricity output to heat rejection is:
{5.8633 MeV - (0..32 MeV / Fr)[(3 / Fc) - 1]} / {11.7266 MeV + [(3 / Fc) - 1] (0.32 MeV / Fr)}

For Fr = 0.5 and Fc = 0.7 the ratio of net electricity output to heat rejection is:
{5.8633 MeV - (0.32 MeV / Fr)[(3 / Fc) - 1]} / {11.7266 MeV + [(3 / Fc) - 1] (0.32 MeV / Fr)}
= {5.8633 MeV - (0.32 MeV / 0.5)[(3 / 0.7) - 1]} / {11.7266 MeV + [(3 / 0.7) - 1] (0.32 MeV / 0.5)}
= {5.8633 MeV - (0.64 MeV)[23 / 7]} / {11.7266 MeV + [23 / 7] (0.64 MeV)}
= {5.8633 MeV - 2.1029 MeV} / {11.7266 MeV + 2.1029 MeV}
= 3.0604 / 13.8295
= 0.2213

The practical success of fusion power generation is entirely dependent on achieving:
Fr > 0.5
and
Fc > 0.7
Otherwise the ratio of:
(net electricity output) / (reject heat)
becomes impractically small.
 

This web page last updated January 28, 2015.