Page 398 - Fundamentals of Magnetic Thermonuclear Reactor Design
P. 398
376 Fundamentals of Magnetic Thermonuclear Reactor Design
Magnetic stiffness may be positive or negative depending on the coils’ ori-
entation relative to one another and the currents’ direction. For example, mutual
magnetic stiffness of two coils lying in the same plate is negative if currents
are running in the same direction. Otherwise, the magnetic stiffness is positive.
Hence, electromagnetic forces caused by relative displacements of the coils
may either destabilise the coil position or, vice versa, play a stabilising role.
We introduce the magnetomechanical interaction potential
1 2
U ( PF) (uu, ) = c ( PF) (u − u ) .
2
Uij(PF)ui,uj=12cij(PF)ui−uj . ij i j 2 ij i j
The overall potential of N PFCs interaction is the sum of potentials
PF
() 1 N PF N PF
m
U PF (u ,..., u N PF ) = ∑ ∑ j , ≠ i.
i
UPFmui,...,uNPF=14∑j=1NPF∑i 4 = j 1 = i 1
=1NPFcijmui−uj2,j≠i.
We convert this expression to a standard form:
1 N PF N PF ) 1
()
m
m
()
U PF = ∑ ∑ a (PF u u = u Τ ⋅ A PF ⋅u,
ij
ij
UPF(m)=12∑i=1NPF∑j=1NPFaij 2 = i 1 = j 1 2
PFuiuj=12uT⋅APF(m)⋅u, m
APFm where the components of the A magnetic stiffness matrix are
PF
N PF
a ij ( PF) = δ ij ∑ ij ( PF) − C ij ( PF) ,with ij
δ being the delta-function.
c
aij(PF)=δij∑j=1NPFcij(PF)− = j 1
Cij(PF),with δij being the delta- Next, we consider the interaction of the poloidal field coils with the CS.
function.
The CS parameters are full current (I ), inner radius (R ), wall thickness (t )
in
0
w
and height (H). Replacing the CS with a current-carrying shell with equivalent
radius
+ t ) 3 0.5
= w 2 ,
R( in
R 0 − R in
3
R =(Rin+tw) 3tw−Rin20.5, t 3 w
0
we make linear magnetic energy (and magnetic flux) of the discussed current-
carrying shell equal to the energy of the solenoid of a given thickness in case of
an infinite length.
To determine the force acting on the PFCs, we isolate an elementary CS
ingredient, that is, current-carrying band dz, and estimate the force of its interac-
tion with the i-th PFC. To this end, we use the initial expression for F and take
ij
the integral of the solenoid height
Fi(CS)=ci(CS)⋅ui, F i ( CS) = c i ( CS) ⋅u ,
i
µ 1
c ( i CS) = 0 I IR R i ⋅
0
i 0
ci(CS)=µ 2I IiR Ri1H⋅ 2 H
0
0
0