who will save the tokamak – harry potter, arnold schwarzenegger, shaquille o’neal or donald...

1

Who will save the tokamak – Harry Potter, Arnold Schwarzenegger,

Shaquille O’Neal or Donald Trump?

J. P. Freidberg, F. Mangiarotti, J. MinerviniMIT Plasma Science and Fusion Center

PPPL Jan. 20 , 2015

2

Why does the tokamak need saving?

• Standard tokamak does not scale to a reactor

• Design determined almost entirely by

nuclear physics and engineering constraints

• One piece of plasma physics enters the design –

empirical

• The plasma physics is not up to the task

E

3

How can I prove this?

• Design a tokamak reactor from an engineering view

• Use only plasma physics

• Obtain reasonable values for all engineering parameters

• Obtain reasonable values for all plasma parameters

• More important – give a non-plasma physics reason for choosing

each parameter

• Test design against known tokamak operational limits

• The design FAILS

E

4

The approach• Use simple models

• No pretense of high precision

• Show there is one possible show stopper

• Discuss 4 strategies to solve the problem

• Many of you are aware of the key results

• How we get there is what is interesting

• Let the design begin

5

ARIES Tokamak Reactor

6

Theoretical approximation

0R

baa

c

TF magnet

Plasma

Blanket

7

Goals of the designQuantity Symbol

Minor radius of the plasma

Major radius of the plasma

Elongation

Thickness of the blanket region

Thickness of the TF magnets

Average plasma temperature

Average plasma pressure

Average plasma density

Energy confinement time

Magnetic field at

Normalized plasma pressure

Plasma current

Bootstrap fraction

0R R

a

c

b

0R

0BE

npT

Bf

I

8

Nuclear Physics Constraints

• Neutron slowing down:

• Tritium breeding:

• D-T Fusion:

2 barnssd

1/2( / )

960 barns

0.025 eV

BR B T

B

T

E E

E

2

max/ at 14 keVv T T

9

Engineering constraints 1

Steady state ignited reactor

• Electric power out:

• Neutron wall loading:

• Recirculating power fraction:

• Thermal conversion efficiency:

1000 MWEP

24 MW/mWP

0.4T

0.1RPf

10

Engineering constraints 2

Superconducting niobium-tin magnets

• Maximum field at the coil:

• Maximum mechanical stress:

• Maximum current density:

• Wall to RF conversion efficiency:

maxB 13 T

max 600 MPa

0.4T

2max 20 MA/mJ

11

Analysis strategy• Use the constraints to express all parameters in

terms of

• Define a reference case:

• No justification for reference case – just an example

• Examine design as a function of

• Show no value of works: identify show-stopper• Examine 4 possible solutions

a

0 / 4R a

a

a

12

Ellipticity

• Not very controversial• Good for plasma physics: increased and • Good for engineering: lower • But there are limits• Plasma limit – vertical instabilities• Engineering limit – feedback control design• We choose

Icost

1.7

13

First wall, Blanket, Shield, Vac. Ch.

First wall Blanket Shield

Plasma

b

Vac. Ch.

14

Focus on the Blanket• Blanket thickness from nuclear physics constraints

• Use conservation of mass and energy

7

0 6

(0) 14.1 MeV 1 / 0.1 m

(0) 1 / 0.003 m at 0.025 eV

S Li SS

BR Li BRBR

dE EE N

dx

dN

dx

15

Blanket (cont.)

• Calculate to reduce neutron flux to

• Add some shield, first wall, and vacuum chamber, improve optimization

• We choose 1.2 mb

0ln 1 ln 0.9 S Bb

x

1/2

710B FB

S T

EE

16

Neutron Wall LoadingRelation between R0 and a

• Wall loading relation

1/222

0

14.1

22.4

1 4

2

n W

n En F

F T

P P S

E PP P

E

S R a

17

Solve for R0

• Solution

• Reference case:• Then

1/22

0 2

1 1 1 7.14 2

n E

F T W

E PR

E P a a

0 / 4R a

01.3 m 5.3 ma R

18

Magnetic field at R = R0

• A good approximation:

• Then

• For the reference case

00

RB B

R

0 max0

(1 ) 0.14 ( )B B

a bB B a a b

R

0 6.8 TB

0.47B

19

TF Coil DesignRelation between c and a

• Coil thickness = structure + superconductor

• Structure - Magnet forces

• Tensile forces

• Centering forces

• Overturning forces

• Overturning forces small - neglect

20

Tensile force balance• Tensile forces

• Solve for the tensile stress

2 20 0

0

2 2 20

0

1 ln

1

(1 ) (1 )

/ stress thickness tensile e s

2

str s

BZ

B

T

Z

T T T B B M

M M

M

T

T

F B R

F A R

c

F

c R

F

20

0

ln[(1 ) / (1 )]2 (2 2 )

B BT

M B M

B

ZF

TF TF

21

Centering force balance• Centering forces

• Solve for the compression stress

20 0

0

20 0

0

2 ( )

2 2

2 ( )

2 2 2 ( ) compression stress

R in out

inB M

o

R

utB M

C C C c M

C

C

F F F

B a b RF

B a b RF

F A c

F

a

F

b

20

0

2(2 2 ) 1

BC

M B M B

B

22

Stress thickness

• Tresca Stress = Maximum stress

• Solve for stress thickness

maxT C

2 1/2

20

0 max

0

2 11 ln

1 2 1

1 [(1 ) ] 0.39 mM

BM

B

B B

B

B

M

B

c R

23

SC current density thickness• How much SC is needed to carry TF current?

• Solve for

2 2

0 0max

2

0

0

0

[(1 ) (1 ) ]

2

/

J

J B B J

J J

B

A Rc R

RI J A

Jc

2 1/2

x

0

0

0 0 ma

1 [(1 ) ] 0

2

.58 mJ

J

B B Jc R

BR J

24

Total coil thickness

• Total = sum of tensile plus centering


M Jc c c

0.97 mc

25

Plasma temperature

• Assume

• Fusion power density

• Choose to maximize

e iT T T

22

1( )

16p n

vS E E p

T

T 2/T

v T

14 keVT

26

Plasma pressure

• Thermal power must produce desired output power

• Simple “vanilla” profiles

2216

T FE

vEp d P

T

r

20

2 1/2

2 3/2

2 (1 ) cos

1.5 (1 ) sin

2.5 (1 )

T T R R a

n n Z a

p p

27

Plasma Pressure (cont.)

• Cross section

• Volume element

• Solve for


4

0

exp (ln )mmv k T

2 204d a R d r

p1/2

2

12 2 1/220

0

64 8.8 atm

25 (1 )E

T F

P Tp

E R a av d

7.6 atmp

28

Plasma beta

• From the definition


02 1/2 20

2 1.3 %

(1 )B

pB a

4.1 %

29

Plasma density

• Density from


2p nT

20 31/2

5 1.7 (10 )

12p

n mT a

20 31.4 10 mn

30

Energy confinement time

• Ignited operation: alpha power = thermal conduction losses

• Solve for


32 E

P pd r

E2 2 1/2

03 0.8F TE

E

ER a p a

E P

0.94 secE

31

The plasma current• Here is where plasma physics enters

• Empirical ELMy H-mode scaling

• Solve for the current


0.93 1.39 0.58 0.78 0.41 0.15 0.190

0.690.145 secE

I R a n B AH

P

0.741.08 1.63

1.08 1.49 0.62 0.84 0.44 0.16 0.20 0.160 0 0

7.9812E E

F T

E P aI MA

H R a n B A E B

14 MAI

32

The kink safety factor

• From the definition


2 21.37 1.160

* 00 0

2 10.11

2a B

q a BR I

* 1.6q

33

Bootstrap fraction

• From recirculating power fraction and CD efficiency

• Assume Lower Hybrid current drive – outside launch

• Recirculating power fraction

• Electrical conversion efficiency

(0.1) 100 MWR RF EF RP EP P PP

(0.5)(0.8) 40 MWCD R CD RFF RF PP P P

34

Bootstrap fraction (cont.)• Current drive


• The bootstrap fraction

1/2 1/22 2

2 2

20 0

( ) 1 1

2 ( )

1

0 8

.

.

2

pe pe LHm

e e

CD CDCD C

LH m

D

m

P PI

R n R n

n

n

2.3 MACDI

1 0.84CDB

If

I

35

Two additional figures of merit• Cost:

• Reference case

• Heat flux:

• Reference case

2 20

0

2

Volume of materialElectric power

( )(

ou

)

4 (2 2 2 )[(

tput

1 ) 2 ]

B

I B TF

E E

TF

V R

V V V

a b a b a

V c R a b

P P

c a b c

3/ 1.0 /I EV P m MW

0

0 02 2 SOL

LP B PQ

R a R

500 MW-T/mQ

36

How well does the plasma shape up?

• Test the plasma against 4 plasma limits

• Greenwald density limit

• Troyon beta limit

• Kink safety factor limit

• Achievable bootstrap fraction

37

Greenwald density limit

• Density limit

• Reference case

2G

In n

a

1.43 2.56

38

Troyon beta limit

• Beta limit

• Reference case0

2.8%T N N

IaB

4.13 4.39

39

Kink safety factor limit

• Safety factor limit

• Reference case

* 2Kq q

1.56 2

40

Bootstrap fraction limit• The maximum bootstrap fraction:

• Model for the total

1/2

0

3/2 2 1/2

1/4 1/2 1/20

1 1( ) 2.44 0.055

(1 )6.8

B

r p n TJ

R B n r T r

pa R B

B

1/20

4/9

( )( )

/ 2

1 (1 ) 1 2.53

1

x

Bb

I a

x ex

e

41

Total current density profile

42

Bootstrap fraction limit (cont.)• The neoclassical bootstrap fraction

• The bootstrap limit


5/2 5/4 5/2 2 1/21

1/2 2 00 0

(1 )268B

NC

I a pf d

I R I b

B NCff

0.84 0.44

43

The basic problem

• Too much current is needed for ignition• Yes but we assumed that• Maybe there is a better value for ?• Let’s see!• Plot

• All must simultaneously be less than 1 for success

0 / 4R a a

*

/ vs. / vs. /q vs. / vs.

G

T

K

B NC

n n aa

q aff a

44

No value of works!! a

45

What’s the consequence?

• Without steady-state the tokamak is on the path to nowhere

SHOWSTOPPER

46

What should we do about it?

Four possible solutions

• The Harry Potter solution

• The Arnold Schwarzenegger solution

• The Shaquille O’Neal solution

• The Donald Trump solution

47

The Harry Potter SolutionAdvanced Plasma Physics

• Raise , set to fix

• Lowers the required I

• Lowers the achievable n

• Lowers the achievable

• Still violates limit

• Success requires

1 1.3 (ACT1 = 1.65)2.8 3.6 (ACT1 = 4.7)

Robust, disruption free operationN N

H H

B NCff aH

Plasma Physics Strategy - Magic

0

0

0

*

1.3 m5.7 m

/ 4.57.4 T

10 MAq 2

0.78B

aRR aBI

f

1.33.4N

H

49

The Arnold Schwarzenegger SolutionAdvanced Magnet Technology


• Improves plasma physics

• Raises

• Success requires

maxB

/I EV P

max max13 17.6HTS already exist (YBCO)This is a game changer

B T B T

B NCff a

50

Engineering Strategy – Strong B

0

0

0

*

3

12.4 T0.97 m7.4 m

/ 7.77.6 MA1.5 %

q 20.69

/ 1.87 m /MWQ 660 MW-T/m

B

I E

BaRR aI

f

V P

max 17.6 TB

51

The Shaquille O’Neal SolutionEconomy of Scale


• Keep standard

• Economy of scale benefits

• Leads to a larger plant

• About the same• Success requires

EP

max, ,NH B

1000 1530Improved grid capacity

E EP MW P MW

B NCff a

/I EV P

52

Utility Risk – Large Power Plant

0

0

0

*

3

8.5 T1.44 m7.7 m

/ 5.311.2 MA

2.5 %q 2

0.70

/ 1.2 m /MWQ 670 MW-T/m

B

I E

BaRR aI

f

V P

1550 MWEP

53

The Donald Trump SolutionEasier Engineering

• Lower , set to fix• Keep standard• Eases engineering• A large, expensive plant • Much larger • Success requires

WP

max, ,NH B

/I EV P

2 24 MW/m 2.1MW/mA big wallet

W wP P

B NCff aa

54

Economic Risk – Large $/W

3

0

0

0

*

/ 2.6 m /MWQ 370 MW-T/m

9.7 T1.35 m10.7 m

/ 7.58.5 MA1.5 %

q 20.66

I E

B

V P

BaRR aI

f

22.1 MW/mWP

55

What if the tokamak doesn’t work?There is always the stellarator

56

Our opinion

• Donald Trump: Not attractive economically

• Shaquille O’Neal: Not attractive utility-wise

• Harry Potter: May work, but risky if only option

• Arnold Schwarzenegger: Best new hope

because of game changing technology

57

Some observations

• Some say plasma physics is done – we should focus on engineering

• We do not agree!• Do not know how to make a steady state tokamak at

high• No solution yet for heat load problem – not primarily

a materials problem, but a divertor problem• Designs tend towards larger• Plasma performance improves at high

0, ,n T B

0 0/ ,R a B

0B

58

The US Fusion Program

• What elements should be included?

• How well are they supported on a scale of 10?

• Advanced tokamak physics 8

• Steady state high performance plasma 6

• Stellarator research 3

• Advanced magnet technology 1

• Fusion-fission hybrids 1

59

Can we improve the US Program?

• We have a chance

• Community input mandated by Congress

• Charge and procedures very important

• Don’t need a charge by Machiavelli and Madoff

They can’t be trusted!

• Do need a charge by Maxwell and Newton

They can’t be fooled!

who will save the tokamak – harry potter, arnold schwarzenegger, shaquille o’neal or donald...

Documents