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Numerical Solutions

of Tokamak Equilibria-(1/2)

- 핵융합 플라즈마 특강 2 - 2009. 2nd Semester

YoungMu Jeon

(2)

Example of Tokamak Equilibrium Analysis

(3)

Physical Understanding of Tokamak Equilibrium

How much plasma can be supported by B

and B

p

? 0 B

J B

B

J       

 p , μ

0

,

 

0 '

     

2 ' 2

r F p F

r        

Bp p B

* Image fetched from IOP-physicsweb

(4)

Physical Understanding of Tokamak Equilibrium

βp=0.088, βN[%]=0.185 βp=0.484, βN[%]=0.944 βp=1.129, βN[%]=1.991 βp=2.098, βN[%]=3.113

“how much plasma”  plasma beta

0

2 2

B

p

High beta plasmalarge shafranov shift & J() profile change J() profile changesafety factor q 2 & inductance li()

(5)

Numerical/Mathematical Understanding of Tokamak Equilibrium

         

r F rp F

J r

r

0 ' '

2 2

* ,

where

 ( , ) 

0

*     rJ   r z

Current Sources Magnetic Flux

Non-linear, elliptic PDE (partial differential equation)

A sort of non-linear Poisson equation in toroidal geometry

Unknown: 

(r,z),

Known(given):

J(

(r,z))

Solve mag. flux, 

(r,z),

with given source,

J(

(r,z))
(6)

(Ref.) Green Function Method In General

A linear differential equation in a general form can be expressed as follows

Its solution can be written as

More specifically, define the inverse operator as

Recall the properties of the Dirac delta function

term s

homogeneou -

non known :

) (

function unknown

: ) (

operator al

differenti adjoint

- self linear, a

: ) (

) ( )

( ) (

x f

x u

x L

x f x

u x L

operator Identity

where ,

) ( ) ( )

(x L1 x f x LL1 L1L I u

L x

x G

dx x f x x G f

L

operator of

Function s

Green' the

is ) '

; ( kernel the

where

) ( )

;

( ' ' '

1

) (

x

1 )

( ),

( )

( )

(  ' ' '

' '

dx x x

f dx

x f x

x

* Reference : http://www.boulder.nist.gov/div853/greenfn/tutorial.html

(7)

(Ref.) Green Function Method – Cont.

The Green‟s function G(x;x) then satisfies

The solution can then be written directly in terms of the Green‟s function as

From basic physics, the Green‟s function gives the potential at the point x due to a point charge at the point x‟ the source point. And the Green‟s function only depends on the distance between the source and field points.

) (

)

; ( )

(

x G x x' x x'

L

( ;

'

) (

'

)

'

)

(

x G x x f x dx u
(8)

Green Function for Toroidal Ring Current

   



 

coils from

) (

plasma from

) (

) (

) (

) (

0

0 0

*

*

c c

G

I

dS G

J

r G

rJ

r r,

r r, r

r r r

r,

   





 

kind second

the of integral elliptic

complete )

(

kind first

the of integral elliptic

complete )

(

) (

) (

4

) ( 2 ) ( ) 2

2 ( )

(

2 0 2

0 2 0

2 2 1 0 0

k E

k K

z z r

r k rr

k E k

K k

k rr

G

r

r,

Z

Y

r0

In cylindrical (r, , z) coordinates

Howard S. Cohl et. al.,“A Compact Cylindrical Green‟s Function Expansion for the Solution of Potential Problems”,The Astrophysical Jornal, Vol. 527, p86-101 (1999)

B.J. Braams, “The interpretation of tokamak magnetic diagonostics”, Plasma Physics and Controlled Fusion, Vol. 33, No. 7, p715~748 (1991)

(9)

Remark On Use Of Green Function

If there is a toroidal ring current source such as PF coil, then one can calculate the magnetic flux at the position, (r,z), from the source, (r

0

,z

0

), directly using Green function.

In addition, one can calculate the magnetic field strength easily as follows with

B=0

Basically one could solve the GS equation using Green function by assuming the plasma as a set of toroidal ring current elements. However it requires rather large computational costs.

) ,

; , ( )

,

( r z  I

c

G r z r

0

z

0

r G r

I r

B r

z G r

I z

B r

c z

c r

 

 

 

 

 

 

) (

1

) (

1

' '

r r,

r r,

(10)

Direct Solution By Green Function

 

    

c

c c

l

l l

l

J G I G

S

dS G

J J

z r dS z

r z r G z r J

z r

) (

) (

) (

) ( )

( ) ( )

(

) , ( )

,

; , ( ) , ( )

, (

' '

' plasma ,

external ,

plasma ,

' ' '

' '

' total ,

r r;

r r;

r

r r

r;

r

r

' ' ' '

 ( , ) 

total , 0

*

    rJ

 r z

By assuming the plasma as a set of current elements, the magnetic flux can be calculated by Green function directly as follows

Slow computation speed

Not enough smoothness of solution

Drawbacks
(11)

Various Numerical Approaches

For Tokamak Equilibria

(12)

Numerical Methods For Tokamak Equilibria

Real space solver Inverse equilibrium solver

Iterative method

- SOR, ADI, or MGM with FDM/FEM

Direct method

- DCR (double cyclic reduction) or FACR (fourier analysis cyclic reduction)

Iterative metric method

Direct inverse solution method

Poloidal angle expansion method

n m

R

J    ( r )

0

Others

Green function method

Orthogonal function expansion method

Conformal mapping method

(13)

Types Of Equilibrium Solutions

Real space equilibrium With free boundary

Image fetched from http://www-fusion.ciemat.es

Inverse equilibrium With fixed boundary

(14)

Approximation Using Finite Difference Method

   

 

 

 

 

 

1 2 1

' 2 1 '

2 2 1

2

1 2 1

1 2

1

1 2 1

2

k k

k k

k k

X k X

k k k

k k k

X X

h h

X

h h

X

k k

 

 

 

Using Centered Difference Scheme

By applying it to GS eq.

l l j l

j Z l

j R l R

l j Z R

l j R l R

l j Z

J h R

h R h

h h

h R h

h2 1, 2 , 1 2 2 , 2 , 1 2 1, 0 ,

1 2

1 1 1

1 2 1

2 1 1 1

1

























l l j l

j Z l

j R l R

l j R l R

l j Z Z

R l

j R J

h h

R h

h R h

h h

h 2 1, 2 , 1 2 , 1 2 1, 0 ,

1

2 , 2

1 2

1 1 1

2 1 1 1

1 1

2 1

Let‟s consider a simple discretization based on rectangular grids

(15)

Approximation Using Finite Difference Method

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

l l j

Z R

Z R

l j l

R Z

R Z l

j l

R Z

R Z

l j Z

R R l

j Z

R R l

j

J h R

h h h

R h h

h h R

h h

h h

h h

h h

h h

0 , 2

2 2 2

1 2 ,

2 2 1

2 , 2

2

, 2 1

2 2 ,

2 1 2

2 ,

1 2 1 2

2 1

Discretized Grad-Shafranov Eq.

(j,l)th element can be updated/adjusted with adjacent 4 elements and its current source on the same grid point

Now it‟s a boundary value problem (B.C. required ).

This linearized equation can be solved using various numerical algorithms such as SOR and MGM so on.

(16)

Boundary Condition

(j,l)

(j+1,l)

(j-1,l)

(j,l-1) (j,l+1)

l=1 l=NR

J=NZ

j=1

(NR-2)x(NZ-2) eqs. Obtained from the previous page

- (j,l)th element is calculated with adjacent 4 elements

2(NR-2) + 2(NZ-2) – 4 eqs. required to close the linear equations boundary condition

For free boundary problems, B.C. can be given from external current sources and plasma currents using „free-space Green function‟

For fixed boundary problems, the B.C. can be specified by user. Normally zero flux.

(17)

How To Handle Non-linearity ??

“Picard Iteration”

In the initial value problem as follows

Suppose is Lipschitz continuous in and continuous in . Then, for some value , there exists a unique solution to the initial value problem within the range .

     

0 0 0 0

'

( t ) f ( t , y ( t )), y ( t ) y , t t , t y

f y t

 0

y(t)

t0 ,t0

Successive Approximation By Picard Iteration

Existence/Uniqueness Of Solution : Picard-Lindelof Theorem

converged is

until 1

with (2)

iterate )

3 (

)) ( ,

( )

(

from

solve (2)

guess (1)

1 '

1 1

0

(t) y

, k

t y t f t

y (t)

y (t) y

k

k k

k k

Refer to the chapter 2 of„First-order differential equations‟ in „Advanced Engineering Mathematics‟

(18)

Homework

Make two contour plots of

(r,z) with a computational region

(1.1m r

2.4m, -1.5m

z

1.5). Here

(r,z) is solved with

following information. One contour is without PF0 and the other one is with PF0. Compare these two plots and understand the difference. Probably 3D contour might be better than 2D contour.

Current

Sources R [m] Z [m] # of turn

Current [kA]

PF1U/PF1L 0.57 0.25 180 -8.85 PF2U/PF2L 0.57 0.70 144 -1.43 PF3U/PF3L 0.57 1.00 72 9.49 PF4U/PF4L 0.57 1.26 108 10.17 PF5U/PF5L 1.09 2.30 208 10.66 PF6U/PF6L 3.09 1.92 128 -2.07 PF7U/PF7L 3.73 0.98 72 -17.20

PF0 1.80 0.00 1 2000

Tips

Use Green function formula as a direct method

One can solve this in an iterative

way studied above if one want.

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