Page 403 - Fundamentals of Magnetic Thermonuclear Reactor Design

Mechanics of Magnetic Fusion Reactors Chapter | 12 381

q = I(r ′ × B+ r ′ × ).B q~e=I(r~’×B+r’×B~).
e
Since

r ′ = ′ u, Bu ′ r′=u′,B~=u⋅∇B and r′=t,
×⋅∇B and r= t,
we obtain
t
q 1 = q e = I [ ′ ×u B + × (u ⋅∇B ] ) . q1=qe=Iu′×B+t×(u⋅∇B).
Let us introduce unit vector n = e . Since t’ = kn and n’ = −kt (curvature
r
−1
k = R ), we obtain
u ′ = ′ + u(k t ′ − ku )t + u n t ) .n u′=u′zk+(u′t−kun)t+u′n+kutn.
( ′ + ku
u
n
z
Since the field gradient
t
k
B
∇ = e ∂ + ke ϕ ∂ +∂ z ) B r()e ϕ =− ′ B r()nt + kB , ∇B=er∂r+ke∂+k∂zB(r)e=−B′(r
n
ϕ
( r
r
)nt+kBtn,
the equation for the linear load takes the form
q 1 = IBu nu k). q =IB(u′zn−u′nk).
(
z
′ − ′ n
1
In reality, there are quite a lot of elastic supports. Therefore, their distributed
reaction can be written as
u
q 2 =−c . q =−cu.
2
By summing the two effective distributed forces, we obtain the total load on
the coil q = q + q .
1
2
The initial equations can be written by component as
= 0 = 0,
Q t ′ − kQ n − cu t Q n ′ − kQ t + IBu z ′ − cu t
= 0 = 0,
Q z ′ − IBu n ′ − cu z M t ′ − kM n
M n ′ + kM t − Q z = 0 M z ′ + Q n = 0,
θ ) θ )
t
n
M t = a t θ ( ′ − k n M n = a n θ ( ′ + k t
= 0
M z = a z θ′ z u t ′ − ku n
= θ .
u n ′ + ku t θ z u z ′ =− n Q′t−kQn−cut=0Q′n−kQt+IB
u′z−cut=0,Q′z−IBu′n−cuz=0
The general solution of this system of 12 equations is a linear combination of
Mt’−kMn=0,Mn’+kMt−Qz=0-
exponents of the form = e ,u u λs …,Q t = Q t e . By applying transformations to Mz’+Qn=0,Mt=atθ′t−kθnMn=anθ′n+kθt
λs
u=ueλs,…,Qt=Qteλs
the amplitudes, we get the following linear system of four algebraic equations: Mz=azθ′zut’−kun=0un’+kut=θzuz’=−θn.

2
λ ( a 3 ( +k λ k −1 − IB λ u − kQ = 0,
z ) ) + ut z t

−1
2
2
2 2
4
IB λ ku t + (a n λ − a t λ k + ) c u z − k λ (a t + a n θ ) t = 0,

2
2
( ka z λ ( + λ k −1 ) − ) cut + λQ t = 0,
k

2
2
k λ (a t + au + λ ( 2 a t − k a n θ ) t = 0, azλ k+λ k−1+ut−IBλuz−kQt=0,IB
2
3
) z
n
2 2
4
2
2
λ k−1ut+anλ −atλ k +cuz−kλ at+
2
2
anθt=0,kazλ k+λ k−1−cut+λQt=0,
2
2
2
kλ at+anuz+λ at−k anθt=0.