INTRODUCTION TO INERTIAL CONFINEMENT FUSION

R. Betti

Lecture 5

Ignition of a simple DT plasma

Recap from previous lecture • Achieving energy gain from inertial fusion requires extreme power sources if the fuel is directly heated and free to expand (100PW for 1g of DT) • To reduce these extreme requirement, we will:

(a) use implosions and deliver energy “slowly” to accelerate a shell (piston) that in turn will heat up a plasma to thermonuclear temperatures (several keV) (b) we will consider a very small amount of fuel (hot spot) heated to keV (<<1mg) (c) we will surround this hot fuel with a dense a cold fuel (the shell or piston) to slow down the expansion (d) we will use alpha heating and ignition (to be defined) of this tiny heated plasma (hot spot) as a spark lighting up the rest of the surrounding dense fuel (if the piston is made of fusion fuel) by propagating the burn (e) for high gains G=Energy-Out>>Energy-In using lasers (poor efficiency) one needs ignition and burn propagation into the cold fuel

MHS THS VHS

Hot spot

Cold fuel (piston)

Vf

Hot spot ignition

(simplest quasi-static model)

The rate of change of the plasma energy is determined by the energy sources and energy losses

• Alpha heating is an energy source

• Work from the spherical piston on the hot spot is an energy source when the piston is converging (pdV work) • If the spherical piston is on its way out than energy is transferred back from the hot spot to the piston and therefore it is a loss (expansion losses) • At the exact time of stagnation, the piston is at rest and no compression work is exchanged with the hot spot

• Bremmstrahlung radiation emission is a loss • Thermal conduction “could” be a loss. This is a bit complicated and will be addressed later

MHS THS VHS

Hot spot

Cold fuel (piston)

Vf

α ν

DT Plasma Pressure P Volume V 3

2pE PV=

• Plasma thermal energy (neglect kinetic energy and approximate plasma as Ideal Gas). Think of this plasma as the hot spot plasma

ppdV rad cond

dEW W W W

dt α= + − −

• Energy balance equation

α-heating>0 pdV piston work ><=

-Radiation losses <0

-Conduction losses < = 0

Rate of change of plasma energy

Simplest ignition model • Consider a DT plasma at stagnation

• Assume that negligible alpha heating occurred before

stagnation (inaccurate! To be corrected later)

• Assume that the energy and pressure of the plasma at stagnation are only determined by the compression of the

piston without any “help” from alphas • We call these stagnation conditions as no-alpha

• Assumes that the alpha heating becomes important starting from the stagnation conditions (not quite right but we will do better later) • Define as t=0 the stagnation condition and Pnoα as the pressure

at t=0

P(0)=Pnoα V(0)=V0

32 pdV rad cond

d PV W W W Wdt α = + − −

3 32 2pdV rad cond

dP dVV W W W W Pdt dtα= + − − −

After stagnation this is a loss

loss loss Loss of pressure because dV>0 after stagnation

32

PV

τ−

Combine all losses together into a heuristic loss term with a characteristic loss time τ (Confinement time)

0 after stagnationdV >

23 32 4 2

dP n PVV v Vdt αε σ

τ= −

23 32 4 2

dP n Pvdt αε σ

τ= −

2

2

3 32 16 2

vdP P Pdt Tα

σε

τ= −

2

224vdP P P

dt Tα

σε

τ= −

Simplified Energy Balance equation

Divide by V

Use P=2nT

Final form

2

2

ˆ ˆ ˆˆ 24 no

vdP P P Pdt Tα α

σε τ= −

ˆno

PPP α

≡

ˆ ttτ

≡

2ˆ ˆ ˆˆ

dP P Pdt

χ= − 224noP

Tv

α

α

τχ

ε σ

≡

Dimensionless form using P(0)=Pnoα

Final dimensionless form

The parameter χ is a function of T and therefore a function of time through the dependence of <σv>/T2

Restricting the temperature range between 8 and 23keV around the maximum of <σv>/T2 (where <σv>∼T2 ) then <σv>/T2 ∼constant

2

vTσ

( )T keV

2 2424no no

f

P P constT

Sv

α α

αα

τ τχ

εε σ

≡ = ≈

28 23

fT

vS const

Tσ

< <

= ≈

24no

no

f

P const

S

αα

α

τχ χ

ε

= ≡ ≈

Note that the denominator has the dimensions of Pressure×time (same as the numerator) because χ is dimensionless

If one can neglect the time dependence of the confinement time and treat it as a constant, then the energy balance is reduced to a single ODE

2ˆ ˆ ˆˆ no

dP P Pdt αχ= − ˆ(0) 1P =

2

ˆ1 1 1ˆ ˆ ˆˆ ˆ no

dP ddt dtP P Pαχ = − = −

( )ˆ noddt αχΦ = − +Φ

ˆ

1 0ˆt

no

d dtαχ

Φ Φ=

Φ −∫ ∫ ˆ1

no

no

Ln tα

α

χχ

Φ −=

−

( ) ˆ1 tno no eα αχ χΦ = + −

1P̂

Φ ≡

( )1ˆ

1 tno no

Peα αχ χ

=+ −

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0

1

2

3

4

5

1noαχ =

1.5noαχ =

0.5noαχ =

ˆ ˆ( )P t

t̂

ignition

no-ignition

Singular for χnoα>1

Singular solution (ignition) for χnoα>1

124 /

nono

f

PSα

αα

τχε

≡ =

[ ] 24no ignition

f

PSα

α

τε

≡24no

f

PSα

α

τε

=

Ignition condition

[ ] 61.1 10 11 11no ignitionP Pa s atm s Gbar nsατ ≡ × ≈ ≈

(0) 8noT T keVα=

2P nTτ τ= Is the Lawson parameter or Fusion Triple Product (Lawson 1956). It determines the ignition conditions of any plasma (ICF, MFE and other fusion concepts)

General results for ignition of a DT plasma regardless of the fusion concept

• χ is the normalized Lawson parameter (PY Chang et al PRL 2010, Betti et al, PoP 2010)

2

0

222

20 0

( )2 2

0 0

4

ˆ ˆ16 16

3 ˆ ˆˆ ˆ2

r

f no

p nono no no

nYield N v Vdt

V S PvP Vdt P dtT

EP V P dt P dt

α

αα α α

α α

σ

τσ

χ χε ε

∞

∞ ∞

∞ ∞

= = =

= ≈ =

=

∫

∫ ∫

∫ ∫

Neutron Yield

( ) 2

0ˆ ˆp no

no

EYield P dtα

αα

χε

∞= ∫

( ) 2

0ˆ ˆp no

no

EYield P dtα

αα

χε

∞= ∫

( ) ˆ1ˆ

1 tno no

Peα αχ χ

=+ −

( ) 1 11

p nono

no no

EYield Lnα

αα α α

χε χ χ

= − −

( )

1 11 no

p no no no

Yieldf LnEα

α αα α α

ε χχ χ

≡ = − −

Carry out time integration

Fractional alpha heating: Energy from alphas over hot spot energy without alphas depends on Lawson parameter

Energy in plasma without alphas

Function of Lawson parameter. Contribution from alpha heatin

No-alpha yield If the alpha does not contribute or if we wish to assess what yield we would be getting if the alpha heating is somehow turned off then take limit for χnoα0

( )0

1 11no

p nono no

no no

EYield Lim Ln

α

αα αχ

α α α

χε χ χ→

= − −

( )

2p no no

no

EYield α α

αα

χε

=

Same answer if setting χnoα=0 in energy balance, solve and perform integration

( )ˆ ˆexpP t= −

( ) ( )2

0ˆ ˆ

2p no p no no

no no

E EYield P dtα α α

α αα α

χχε ε

∞= =∫

2ˆ ˆ ˆˆ no

dP P Pdt αχ= −

Measure of alpha heating Yield amplification due to alphas: it is only a function of the Lawson parameter

0.0 0.2 0.4 0.6 0.8 1.0 1

2

3

4

5

6

7

noαχ

[ ]minno

nono ign

PP

αα

α

τχτ

≡

Singular

2

2 11 no

no no no

Yield LnYield α

α α α

χχ χ

= − −

no

YieldYield α

Yield amplification due to alphas is often used as measure of progress in laser ICF

Yield amplification from alpha heating to date on NIF is 2-3X

Hot spot ignition: dynamic model

Hot spot

Cold fuel (piston)

α ν

Vi

• The shell has been accelerated (by the laser) to a peak implosion velocity.

• The shell is compressing the hot spot and raising its pressure by converting its kinetic energy into hot spot internal energy. • The hot spot is gaining energy from the shell and from alpha heating while loosing energy through radiation emission and loosing heat through heat conduction

Mshell

• The simplest shell model is the one of a thin dense shell (high aspect ratio) whose motion (after being accelerated by the laser) is governed by Newton’s law.

Hot spot

R(t)

24shM R PRπ=

*(0)R R=(0) iR V= −

• t=0 is now the time of peak implosion velocity or beginning of the coasting/deceleration phase

• P(t) is the hot spot pressure pushing on the shell surface 4πR2 P

• Need to couple Newton’s law for the shell to the energy equation and momentum equation of the hot spot because there are two unknowns: P(r,t) and R(t). Note that in general the pressure inside the hot spot is a function

of radius and time

• Start with the hot spot momentum equation (Euler equation)

U U PUt r r

ρ ∂ ∂ ∂ + = − ∂ ∂ ∂

• Scale each term. Here ↔ means “compare”

U U PUt r r

ρ + ⇔ −

• Time scale: t ~ R/Vi; velocity: U ~ Vi; Spatial scale r ~ R

2 2i iV V PR R R

ρ

+ ⇔

2 2 22

2~ ~/ /i i i

i th

V V VMachP T mρ υ

=

• Define the Mach number of the hot spot

• Therefore the LHS is of order Mach2 with respect to the RHS

22~iV PMach

R Rρ

• Hot spot is multi keV and the thermal velocity of its plasma is >> implosion velocity of the shell, therefore Mach<1: SUBSONIC HOT SPOT

22

2~ 1i

th

VMachυ

<<

• Therefore neglect LHS of momentum equation leads to P=P(t): ISOBARIC HOT SPOT

0Pr

∂≈

∂( )P P t≈