22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8: Effect of a Vertical Field on Tokamak Equilibrium
Toroidal Force Balance by Means of a Vertical Field
1. Let us review why the vertical field is important
2.
3. For very short times, the vacuum chamber acts like a perfectly conducting shell:
t ∼ 1 msec.
4. On a longer time scale, the fields diffuse through the shell and a vertical field is
required for equilibrium.
5. Analytic procedure: Assume an external vertical field slowly penetrates a highly
conducting shell. The shell then becomes superconducting. We take the limit as the
shell moves to infinity: b → ∞ .
6. The limit b → ∞ is nontrivial. To begin, we leave the shell in place.
Influence of the Vertical Field
1. Bv causes a shift of the plasma surface with respect to R0 , the center of the shell.
2. The applied vertical field is given by B= BV eZ
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 1 of 9
3. Assume BV scales with the shift Δ . BV is clearly a component of the poloidal field.
Consider
Bv
∼ ∈,
Bθ
Bv
∼ ∈2
B0
4. Write BV in terms of ψv , the vacuum vertical field flux function
a. Bv = Bv e z
= BV ⎣⎡er sin θ + eθ cos θ ⎦⎤
≡
1 ⎡ ∂ψv
1 ∂ψv
⎤
er ⎥
eθ −
⎢
R0 ⎣ ∂r
r ∂θ
⎦
b. ψv = R0 Bv r cos θ
5. The new boundary condition including the vertical field is given by
Solve the Grad–Shafranov Equation Using the Tokamak Expansion to Account for
Bv
1. Zero order: same as before: radial pressure balance
2. First order: same equation as before: ∇2 ψ1 = ...
note cos θ DEPENDENCE
but with a new boundary condition: ψ1 ( b, θ ) = R0 Bv b cos θ
3. Let ψ ( r, θ ) = ψ0 ( r ) + ψ1 ( r ) cos θ
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
ψ1 ( b ) = R0 Bv b
Lecture 8
Page 2 of 9
4. The solution for ψ1 is found as follows:
a.
d ⎛ 2 d ψ1 ⎞
2
2 dp
⎜ rBθ
⎟ = rBθ − 2μ0 r
dr ⎝
dr Βθ ⎠
dr
b. ψ1 = ψ1old + ψ1hom
ψ1old satisfies the equation with B.C. ψ1old ( b ) = 0
ψ1hom satisfies the homogeneous equation with B.C. ψ1hom ( b ) = R0 Bv b
5. Homogeneous solution
'
⎛ψ ⎞
c
a. ⎜ 1 ⎟ = 12
B
rBθ
⎝ θ ⎠
r
dr
0
rBθ2
b. ψ1 = c2 Bθ + c1Bθ ∫
c. Choose c1 = 0 for regularity
⎡ R bB ⎤
d. ψ1hom = c2 Bθ ( r ) = ⎢ 0 v ⎥ Bθ ( r )
⎣⎢ Bθ ( b ) ⎦⎥
6. The full solution can be written as
ψ1 ( r ) = −Bθ ∫
dx
b
r
∫
R0 Bv b
⎛ 2
2 dp ⎞
Bθ ( r )
⎜ yBθ − 2y
⎟ dy +
dy ⎠
Bθ ( b )
⎝
x
xB2 0
θ
7. The new Shafranov shift is given by
a.
Δa = −
ψ1 ( a)
ψ'0 ( a)
=−
ψ1old ( a)
ψ'0 ( a )
−
ψ1hom ( a)
ψ'0 ( a)
⎡R B b
⎤
1
b. Δ a = Δold − ⎢ 0 v Bθ ( a) ⎥
B
b
R
B
⎥⎦ 0 θ ( a)
⎣⎢ θ ( )
= Δ old −
bBv
Bθ ( b )
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 3 of 9
8. Thus
Δa
b
=
2R0
b
⎡⎛
l
BV
1⎞⎛
a2 ⎞
b⎤
⎢⎜ β p + i − ⎟ ⎜⎜ 1 − 2 ⎟⎟ + ln ⎥ −
2 2⎠⎝
a ⎥⎦ Bθ ( b )
b ⎠
⎢⎣⎝
9. a. How much vertical field do we need to keep the plasma centered?
(
b. Set Δ a = 0, Bθ ( b ) = μ0 I p 2πb
c. Bv =
)
μ0 I ⎡⎛
l
1⎞⎛
a2 ⎞
b⎤
⎢⎜ β p + i − ⎟ ⎜⎜1 − 2 ⎟⎟ + ln ⎥
4πR0 ⎣⎢⎝
2 2⎠⎝
a ⎦⎥
b ⎠
The Limit b → ∞
1. How much field is required for Δ a = 0 if the shell is not present?
2. Imagine the shell receding further and further away so that b a → ∞
3. Take this limit in the expression for Bv
1−
ln
a2
b2
→1
b
→?
a
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 4 of 9
4. What is the difficulty?
a. Physically b < R0. Also we assumed b R0
1
b. Here is a simple approximation: ln ( b a) → ln ( R0 a)
Electrical Engineering Derivation of the ln b/a Limit
1. ln b/a represents the force due to the change in magnetic energy between the
plasma and the wall as the plasma shifts outward by an amount dR0 = Δ
2. It is the analog of the li term except applied to the external field
3. External field changes:
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 5 of 9
4. The force = −∇ (potential energy) = −∇ (magnetic energy)
as the plasma is displaced by an amount dR0. Note that since the plasma is a perfect
conductor, the flux linking the plasma remains fixed during the displacement
a.
F =−
d ⎡ 1
⎢
dR0 ⎣ 2μ0
2
⎤
d ⎛1
∫ Bθ dr ⎥⎦ = − dR0 ⎜⎝ 2 Le I
2
⎞
⎟
⎠
external inductance
b. Constant flux implies Le I = ψ e = const.
F =−
c.
L2e I 2 d 1
I 2 dLe
=
2 dR Le
2 dR0
5. If the plasma is surrounded by a shell
Le = μ0 R0 ln b a
F =
μ0 I 2
2
ln
b
a
6. For a plasma without a shell (homework problem)
⎡ 8R
⎤
Le = μ0 R0 ⎢ln 0 − 2⎥
a
⎣
⎦
F =
μ0 I 2 ⎡
a
8R0
⎤
⎢ln a − 1⎥
⎣
⎦
7. Therefore, the proper limit is
ln
8R
R
b
→ ln 0 − 1 = ln 0 + 1.08
a
a
a
8. Substitute into the Bv formula
Bv =
μ0 I ⎡
8R ⎤
l
3
β p + i − + ln 0 ⎥
⎢
4πR0 ⎣
2 2
a ⎦
9. This is a widely used formula in the design of circular tokamaks
Summary of Ohmically Heated Tokamak Equilibria
1. low β : βt ∼ ∈2 , β p ∼ 1
2. q ∼ 1 : required for stability
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 6 of 9
3. radial pressure balance: poloidal field (Z pinch)
4. toroidal force balance: vertical field Bv
5. toroidal field: needed only for stability to keep q ∼ 1
6. Ordering
βt ∼ ∈2
βp ∼ 1
q ∼1
Bθ B0 ∼ ∈
δ Bφ B0 ∼ ∈2
Intuitive Form of Toroidal Force Balance
1. Multiply the Bv equation by 2πR0 I
2. Then
μ0 I 2 ⎡
2πR0 IBv =
2
8R0 ⎤
li 3
⎢ β p + 2 − 2 + ln a ⎥
⎣
⎦
3. T1 = 2πR0 IBv = Bv IL force due to the vertical field acting on the current I
4. T2 =
5. T3 =
μ0 I 2 ⎡
2
8R0
⎤ I 2 dLe
ln
1
−
⎢
⎥ = 2 dR force due to the change in the external field
a
⎣
⎦
μ0 I 2 li
2
2
=
μ0 I 2
4
Li 4π I 2 Li
=
2πR0 μ0
2 R0
a. but
1
Li I 2 =
2
Bθ2
2π2 R0
d
=
r
∫ 2μ0
μ0
⎡ 4π2
Li = ⎢
⎢⎣ μ0
2
∫ Bθ rdr
2
⎤
⎛ Bθ ⎞
⎥ × R0
r
dr
⎜
⎟
∫⎝ Ι ⎠
⎥⎦
independent of R0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 7 of 9
b. Thus Li ∝ R0 and
c.
T3 =
6. T4 =
μ I 2 li
Li
dLi
I 2 dLi
=
=
so that 0
2 2
2 dR0
R0 dR0
I 2 dLi
2 dR0
μ0 I 2 ⎡
1⎤
⎢βp − 2 ⎥ =
2
⎣
⎦
μ0 I 2 ⎡16π2
⎢
2
⎢⎣ μ0 I
2
1⎤
∫ pr dr − 2 ⎥⎥
= 8π2 ∫ pr dr −
⎦
μ0 I 2
4
a. Recall the general pressure balance relation
2π ∫ pr dr =
μ0 I 2
8π
+ 2π ∫ r
B02 − Bφ2
2μ0
dr
let Bφ ( r ) = B0 + B2 ( ψ0 ) ≡ B0 + δBφ ( r )
μ0 I 2
4
= 4π2 ∫ pr dr + 4π2 ∫ r
B0 δBφ
μ0
dr
b. T4 = 8π2 ∫ pr dr − 4π2 ∫ pr dr − 4π2 ∫ r dr
B0 δBφ
⎛
= 4π2 ∫ ⎜⎜ p −
μ0
⎝
B0 δBφ
μ0
⎞
⎟⎟ r dr
⎠
7. Summary
2πR0 IBv =
B δB
⎛
I2 d
( Le + Li ) + 4π2 ∫ ⎜⎜ p − 0 φ
2 dR0
μ0
⎝
⎞
⎟⎟ r dr
⎠
vertical field force hoop force tire tube force
1/R force
8. Proof of the tire tube and 1/R force
a. Ftt :
Ftt = − ∫ eR ⋅ ∇p dr
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 8
Page 8 of 9
1
R
⎡
⎤
= − ∫ ⎢∇ ⋅ ( eR p ) − p∇ ⋅ eR ⎥ dr
⎢
⎥
⎣
⎦
integrates to zero
=
p
∫ R Rrdφ
dr dθ
= 4π2 ∫ pr dr
b. F1 R
Jp × Bφ ⋅ eR =
B
1
⎡∇RBφ × eφ ⎤ × Bφ eφ ⋅ eR = − φ eR ⋅ ∇RBφ
⎣
⎦
Rμ0
Rμ0
=−
=−
=−
c.
F1 R =
1
R2
1
R2
eR ⋅ ∇
eR ⋅ ∇
R02
μ0 R2
R2 Bφ2
2μ0
R2
1
F2
= − 2 eR ⋅ ∇ 0 B02 + 2B0δBφ
2μ0
2μ0
R
(
)
eR ⋅ ∇B0 δBφ
∫ ( Jp × Bφ ⋅ eR ) dr
=−
R02
1
e ⋅ ∇B0 δBφ dr
μ0 ∫ R2 R
−
=−
R02
=−
R02
⎡
⎛ B0 δBφ
⎣
2
⎝ R
⎢∇ ⋅ ⎜
μ0 ∫ ⎢ ⎜
μ0 ∫
≈ −4π2 ∫
B0 δBφ
3
R
B0 δBφ
μ0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
1
R3
⎞
e ⎤
eR ⎟⎟ − B0 δBφ ∇ ⋅ R2 ⎥ dr
R ⎥⎦
⎠
dr ≈ −
R02
μ0 ∫
B0 δBφ
R3
Rrdφ dθ dr
r dr
Lecture 8
Page 9 of 9