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**INTRODUCTION:**

This web page identifies liquid lead related constraints that must be satisfied by the Plama Impact Fusion (PIF) process.

**SPHEROMAK LIFETIME CONSTRAINT:**

The muzzle velocity of the liquid lead guns must be sufficiently high that the spherical liquid lead shell forms around the injected spheromak before the spheromak spontaneously randomizes.

The spheromak is injected at time:

T = Tc.

The liquid lead shell forms at time:

T = Td.

In time (Td - Tc) the liquid lead must move from:

Ric = 2.0 m

to

Rid = 1.45 m.

Hence we have relationship:

(dRi / dT)|Ric > (2.0 m - 1.45 m ) / (Td - Tc)

Assume that:

(Td - Tc) = 1.9 ms

then:

(dRi / dT) > 0.55 m / 1.9 ms

= 289 m / s.

Thus if the minimum spheromak lifetime is 2.0 ms then:

**(dRi / dT)|Ric ~ 300 m / s**

implying that the muzzle velocity of the liquid lead guns must be at least:

**300 m / s**.

**ADIABATIC COMPRESSION CONSTRAINT AT Rid:**

As shown on the web page titled:ADIABATIC COMPRESSION the requirement for adiabatic compression at Ri = Rid gives:

**(-dRid / dT)** > (2 / Ml)(Ektd Mt)^0.5

= (2 Mt / Ml) (Ektd / Mt)^0.5

= (2 X 3 / 207.2) [(3.5 eV X 1.602 X 10^-19 J / eV) / (3 X 1.67 X 10^-27 kg)]^0.5

= (.028957) [1.0579 X 10^4 m / s]

= **306 m / s**

**ADIABATIC COMPRESSION CONSTRAINT AT Rif:**

As Ri shrinks from Rid to Rif the LHS of the above inequality increases whereas the RHS remains nearly constant, so the inequality continues to be met.

**GAIN CONSTRAINT AT Rif:**

As Ri shrinks from Rif to Rih the deuterium ion energy must climb from Ekdf to Ekdh.

For:

Rif = 1.0 m

Rih = 5.3329 mm

Ekdf = 3.5 eV

then:

**Ekdh** = (Rif / Rih)^2 Ekdf

= (1.0 m / .0053329 m)^2 (3.5 eV)

**= 123,066 eV**

which value is close to the target value of 120,000 eV.

**ADIABATIC COMPRESSION NEAR FUSION IGNITION:**

From web page titled:SPHERICAL COMPRESSION A

**Ekl = Rhol 2 Pi Ri^3 (dRi / dT)^2 [(Ro - Ri) / Ro]**

or

|(dRi / dT)| = [Ekl / {Rhol 2 Pi Ri^3 [(Ro - Ri) / Ro]}]^0.5

Near fusion ignition:

Ri << Ro

giving:

|(dRi / dT)| = [Ekl / {Rhol 2 Pi Ri^3}]^0.5

Recall that the requirement for adiabatic compression is:

(-dRi / dT) > (2 / Ml)(Ekt Mt)^0.5

Hence the compression will be adiabatic as long as:

[Ekl / {Rhol 2 Pi Ri^3}]^0.5 > (2 / Ml)(Ekt Mt)^0.5

or

Ekl > (2 / Ml)^2 (Ekt Mt){Rhol 2 Pi Ri^3}

= 8 (Mt / Ml) Ekt (Rho Pi Ri^3) / Ml

= 8 (3 / 207.2) (Ektf) (Rif / Ri)^2 (Rho Pi Ri^3 / Ml)

= 8 (3 / 207.2) (Ektf) Rif^2 (Rho Pi Ri / Ml)

Howevver, Ekl goes to zero at Rii = 3.77 mm, so this inequality fails at small Ri.

At Ri = Rih:

** Ekl** = (Ekld / 2)
** = 32.75 MJ**.

At Ri = Rih:

8 (3 / 207.2) (Ektf) Rif^2 (Rho Pi Rih / Ml)

= 8 (3 / 207.2) (3.5 eV X 1.602 X 10^-19 J / eV) (1.0 m)^2 (10.66 X 10^3 kg / m^3 X Pi X .0053329 m / (207.2 X 1.67 X 10^-27 kg)

= 3.35209 X 10^-4 X 10^11 J m^2 m kg / m^3 kg

= 3.35209 X 10^7 J

**= 33.5209 MJ**

Thus the compression becomes non-adiabatic at a radius slightly greater than Rih.

**LIQUID LEAD VOLUME Voll:**

Define:

Voll = liquid lead volume

Rhol = liquid lead density

The kinetic energy Ekla of the lead at the start of the plasma compression is:

Ekld = (Rhol Voll / 2) [(-dRi / dT)^2]|Rid

or

**Voll** = 2 Ekld / [Rhol (-dRid / dT)^2]

= 2 (65.5 MJ) / [10.66 X 10^3 Kg / m^3 X (3.00)^2 X 10^4 m^2 / s^2]

= **0.13654 m^3**

**LIQUID LEAD RADIUS Roi:**

The final liquid lead sphere outside radius Roi is given by:

Voll = (4/ 3) Pi Roi^3

or

**Roi ** = [3 Voll / 4 Pi]^0.333

= [3 (0.13654 m^3) / 4 Pi]^0.333

= [.0325974283 m^3]^0.333

= [32.5974 X 10^-3 m^3]^0.333
= **0.31944 m**

At this lead thickness there may be significant fast neutron wear on the reaction chamber pressure vessel and the liquid lead guns.

There must be enough liquid lead present in the pressure vessel to absorb fast neutrons and to carry away the heat released. Hence, in addition to the gun accelerated liquid lead the system needs additional liquid lead circulating over the inner surface of the spherical pressure vessel for heat removal and to attenuate impacts from fast neutrons and high radial velocity liquid lead droplets.

**INITIAL LIQUID LEAD THICKNESS:**

The compressing liquid lead volume Voll is given by:

Voll ~ 4 Pi Rid^2 (Rod - Rid)

Hence:

(Rod - Rid) = Voll / (4 Pi Rid^2)

= 0.13654 m^3 / [4 Pi (1.45 m)^2]

= 0.0051680 m

= **5.168 mm**

Thus the initial liquid lead sphere wall thickness is **5.168 mm**

**COMPRESSING LIQUID LEAD MASS:**

The lead sphere mass is given by:

Rhol Voll = 10.66 X 10^3 kg / m^3 X 0.13654 m^3

= **1455.52 kg**

**PROPERTIES OF LEAD:**

Liquid density at m.p. = 10.66 g / cm-3

Melting point = 327.46 C

Boiling point = 1749 C

Heat of vaporization = 179.5 kJ mol-1

Molar heat capacity = 26.650 J mol-1 K-1

**HEAT ABSORPTION:**

The liberated nuclear energy appears primarily as kinetic energy of neutrons and He-4 nuclei. The neutrons penetrate the liquid lead and their kinetic energy becomes heat deep in the liquid lead and heat in the Li based neutron absorption blanket.

The He-4 nuclei bounce off the lead atoms and add their kinetic energy to heating of the plasma. The plasma electrons emit x-rays that are in absorbed near the inner surface of the liquid lead creating high pressure lead vapor.

The heat capacity of lead is 26.650 J / mole-K .The liquid lead temperature rise (Delta T) between the melting point and boiling point is 1421 K. The maximum energy absorption per fusion pulse by compressing liquid lead heating is:

1455.52 kg X 1 mole / 207.2 g X 1000 g / kg X 26.650 J / mole-K x (1421 K)

= **266.023 MJ**

so that the pressure vessel must be engineered to safely absorb repeated vapor pressure pulses of:

(600 MJ + 65.5 MJ - 266.023 MJ = **399.477 MJ**

**AMOUNT OF LIQUID LEAD VAPORIZED:**

[400 MJ / (179.5 kJ mol-1)] X .2072 kg / mole

= 2228.4 mole X .2072 kg / mole

**= 461.73 kg**

**BINDING ENERGY OF LIQUID LEAD:**

Assume that the liquid lead is initially at 400 degrees C. The binding energy per lead atom **Ebl** is given by:

**Ebl = {[(1749 -400) deg C X 26.650 J / (mol deg C)] + [179.5 X 10^3 J / mol]} X [1 mol / 6.023 X 10^23 atoms]
X [1 eV / 1.602 X 10^-19 J]
= {[35950 J / mol] + [179,500 J / mole]} [(1 / 9.649) X 10^-4 mole eV / atom J]
= 2.23 eV / atom**

This low binding energy as compared to the need for a ~ 300 keV deuterium ion kinetic energy is clearly a major issue.

**LEAD SPUTTERING / EVAPORATION THRESHOLD:**

The evaporation / sputtering threshold is reached when an impacting deuterium ion imparts sufficient kinetic energy to a liquid lead atom at the liquid lead wall surface to change the lead atom from its liquid state to it gaseous state.
During the adiabatic compression the liquid lead near the liquid lead inner wall will increase in temperature due to energy absorption from deuterium ion impacts. Liquid lead sputtering and/or evaporation starts when the random kinetic energy imparted by a deuterium ion to a lead atom exceeds the liquid lead atom's binding energy, or:

(Ml / 2) Vl^2 > 2.23 eV

or

(Ml / 2)(Md Vtw / Ml)^2 > 2.23 eV

or

(1 / 2) (Md / Ml)(Md Vtw^2) > 2.23 eV

or

** 2 (Md / Ml) Ekd > 2.23 eV**

or

**Ekd > (Ml / 2 Md) 2.23 eV = (208 / 4) 2.23 eV = 116.0 eV**

However, when a lead atom jumps off the surface of the liquid lead wall it has only the radial velocity that it had at the instant of jump off plus the energy that it gains from the impacting deuterium or tritium ion less its binding energy of 2.23 eV. Meanwhile the liquid lead surface behind it is constantly accelerating and will soon catch up with the stray liquid lead atom and reabsorb it. Hence, provided that the liquid lead wall has sufficient acceleration this liquid lead adiabatic compression system will tolerate deuterium and tritium ion induced liquid lead sputtering.

This web page last updated January 23, 2015

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