22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Resistive Wall Mode
1. We have seen that a perfectly conducting wall, placed in close proximity to
the plasma can have a strong stabilizing effect on external kink modes.
2. In actual experiments, the metallic vacuum chamber surrounding the plasma
is a good approximation to a perfectly conducting wall.
3. However, its conductivity is not infinite but is finite.
4. In fact we do not want the conductivity too high and/or, too thick because it
would take too long externally applied feedback fields to penetrate the shell
and interact with the plasma.
5.
Also, higher resistivity, smaller currents are induced in the chamber during
transients, alleviating power supply requirements.
6. The question raised here concerns the effect of finite resistivity of the wall on
external kink stability.
7. There are three possible situations and only one is really interesting.
8. In the first case the plasma is stable to external kinks with the wall at ∞
Here, since the plasma is already stable, a wall, either ideal or resistive does
not affect stability. This case is uninteresting.
9. In the second case, the plasma is unstable with the wall at ∞ and with the
wall at its actual position, assuming the wall is perfectly conducting. Since the
plasma is unstable with a perfectly conducting wall as r=b, making the wall
resistive does not help. This case is also uninteresting.
10. The interesting case is when the plasma is unstable with the wall at ∞ , but
stable with a perfectly conducting wall at r=b. Does the resistivity of the wall
destroy wall stabilization?
11. To address this issue we investigate the problem in a straight cylindrical
geometry. However, the results are valid for a general toroidal geometry as
well.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 1 of 10
Plan of attack
1. The analysis of the resistive wall mode is carried out in four steps.
2. First, reference values of δW are calculated for an ideal wall located as
∞ ( δW∞ ) and at r=b ( δWb )
3. The full eigenvalue problem is solved region by region assuming slow growing
modes - on the scale of the wall diffusion time.
4. Third, the fields within the resistive wall are calculated using the then wall
approximation. This gives rise to a set of jump conditions across the wall.
5. The resulting set of coupled equation and boundary conditions are solved
yielding the dispersion relation.
The Reference Cases
1. Recall that δW for a general screw pinch surrounded by a perfectly
conducting wall is given by
δW
2π2R 0 μ0
=
∫
a
0
2
2
⎡ fξ'2 + gξ2 ⎤ dr + ⎡ FF + r ΛF ⎤ ξ2
⎢
⎥ a
2
⎢⎣
⎥⎦
m ⎥⎦
⎣⎢ k 0
a
where F = kBz +
mBθ
mBθ
, F = kBz −
r
r
1 + (a b)
2m
Λ≈
1 − (a b)
2m
k20 = k2 +
m2
r2
2. The exact minimizing ξ satisfies
(Fξ ) − gξ = 0
'
'
ξ ( a) = ξa
ξ ( 0 ) regular
rF2
and for an external mode with a resonant surface outside
k 20
the plasma this implies that F ≠ 0 in the plasma.
3. Recall that F =
4. Thus the variational equation for ξ is non-singular. Its solution is important,
but boring.
5.
Assume the solution for ξ is known, either analytically or computationally.
6. If we multiply the equation for ξ by
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
∫
a
0
( ) ξ dr
we find that
Lecture 20
Page 2 of 10
∫
a
0
⎛ Fξ'2 + gξ2 ⎞ dr = Fξξ'
⎜
⎟
⎝
⎠
a
7. This allows us to write
δW
2
2π R 0
⎡ FF r2 ΛF2 F2
=⎢ 2 +
+ 2
m
μ0 ⎣⎢ k 0
k0
(
8. Note that rξ' ξ
)
a
⎛ rξ' ⎞ ⎤ 2
⎜⎜
⎟⎟ ⎥ ξa
⎝ ξ ⎠ ⎦⎥ a
is a known quantity from the solution for ξ .
9. The first reference case corresponds to the wall at ∞ : Λ = Λ ∞ = 1
For this case
δW∞
2π2R 0
⎡ FF r2F2 Λ ∞ F2 ⎛ rξ' ⎞ ⎤ 2
=⎢ 2 +
+ 2⎜
⎟ ⎥ ξa
m
μ0 ⎢⎣ k 0
k 0 ⎜⎝ ξ ⎟⎠ ⎥⎦ a
10. The second reference case corresponds to the wall at b
2m
2m
Λ = Λb = ⎡1 + ( a b ) ⎤ ⎡1 − ( a b ) ⎤
⎥⎦
⎣⎢
⎦⎥ ⎢⎣
11. Keep in mind that Λb > Λ ∞ (well stabilization)
(
12. For both reference cases rξ' ξ
)
a
is the same. It is unaffected by the wall.
13. These relations allow us to write
δWb
2
2π R 0 μ0
=
δW∞
2
2π R 0
⎡ r2F2 ⎤
2
+⎢
⎥ ( Λb − Λ ∞ ) ξa
m
μ0 ⎣⎢
⎦⎥ a
14. The interesting case under consideration corresponds to
δW∞ < 0 unstable with the wall at ∞
δWb > 0 stable with perfect wall at r=b
The eigenvalue problem with a resistive wall
1. We solve the full eigenvalue problem with the resistive wall
2. However, we can make use of much of what we have already done by
assuming slow growing modes - resistive wall diffusion term.
3. Example: a = .3 m, R 0 = 1 m, Tc = Tc − 2 keV, b ≈ a
4. Then τMHD = R 0 ν Tc = 2.3 × 10−6 sec.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 3 of 10
5. Consider a stainless steel vacuum chamber of thickness d = 1 mm . Then, with
η = 11 × 10−8 Ωm
τD = μ0bd η = 3.4 × 10−3 sec.
6. For a thick copper wall d=1 cm, η = 1.7 × 10−8 Ωm .
τD = μ0bd η = .22 sec.
7. Clearly τD
τMHD for either case.
8. The implication is that in the plasma eigenfunction equation,
ω2
k2 ν2a , ω2
k2 ν2τι and k ≠ 0 for external mode. Therefore we can ignore
ω2 in the plasma region.
9. The resulting equation for ξ thus corresponds to the ideal marginal stability
equation which is our old friend.
( fξ ) − g ξ = 0
'
'
10. The ω ' s will appear where we discuss the wall.
11. The region between the plasma and the wall satisfies
BI = ∇φI , ∇2 φI = 0
(rφ ) − (k
'
I
'
2
)
+ m2 r2 φI = 0
12. The solution, neglecting k2 for simplicity (to have polynomials rather than
Bessel functions) is given by
⎛r ⎞
φI = c1 ⎜ ⎟
⎝b ⎠
m
⎛b⎞
+ c2 ⎜ ⎟
⎝r ⎠
m
13. We will find c1 and c2 shortly by matching jump conditions
14. A similar analysis holds for the outer vacuum region where
BII = ∇φII
∇2 φII = 0
15. The solution here has only a decaying solution since the fields must be
regular as ∞ . Thus
⎛b ⎞
φII = c3 ⎜ ⎟
⎝r ⎠
m
The wall solution
1. Now lets look within the wall
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 4 of 10
2. Assume the wall is then d
b . The wall looks rectangular
3. Let r = b + x, θ = y b
4. The equation for B in the wall is obtained as follows
η 2
∂B
η
∇×B =
∇ ×B
= −∇ × E = −∇ × ηJ = −∇ ×
μ0
∂t
μ0
5. Focus on the r (i.e. x component), and assume B ∝ e−ιωt ∝ eωιt
ωι = growth rate.
∂Bx
∂2x
⎛
μ ω
m2 ⎞
− ⎜ k 2 + 2 ⎟ B x = 0 t Bx
⎜
⎟
η
b ⎠
⎝
6. By and Bz are found from ∇ ⋅ B = 0 and the assumption Jx=0 (then wall
approx - all current flows parallel to the surface): ex ⋅ ∇ × B = 0
7. We do not need By and Bz so we will not calculate them.
8. Then wall ordering: Assume ωι ∼
9. Then
μ0 ωιb2
ηm2
∼
μ0b2 η b
∼
η bd
d
10. Also Bx μ0 ωι Bx ( η) ∼
η
J
∼
μ0bd τD
1
η
1 ημ0bd b
= 2
∼
d
d μ0 ωι
d μ0 η
1
2
1
⎛
m2 ⎞
11. This implies that the ⎜ k2 + 2 ⎟ Bx can be neglected and that Bx = Bx0 + Bx1 ( x )
⎜
b ⎠⎟
⎝
where BX0 = const, Bx1 Bx0 ∼ d b 1
12. The equation and solution for Bx1 are given by
∂2Bx1
∂x
2
=
μ0 ωι
Bx0
η
Bx = Bx0 +
μ0 ωι
x2
Bx0
η
2
13. For a thin wall d b → 0, this solution translates into the following two jump
conditions
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 5 of 10
b+
Bx b− = Bx0 +
B'x
b+
=
b−
μ0 ωι
d2
− Bx0 ≈ 0
Bx0
η
2
μ0 ωιBX0d μ0 ωιB
≈
×d
η
η
Br' =
14. Or Br = 0
μ0 ωιdBr
η
The jump condition and dispersion relation
1. There are four unknowns in the problem c1, c2, c3, ω
2. There are four jump conditions. 2 at the wall given above, and 2 on the
plasma we must now determine
3. The first is the usual n ⋅ B condition
n⋅B
B1r
a
a
(
= n⋅∇× ξ×B
)a
= ι Fa ξa
4. The second is the pressure balance jump condition (lots of work)
2
2
⎛
B ⎞⎟
B
μ0p1 + B ⋅ B1 + ξ ⋅ ∇ ⎜ μ0p +
= B ⋅ B1 + ξ ⋅ ∇
⎜
2 ⎟
2
⎝
⎠
5. For no surface currents and p, p’ as r=0 vanishing this reduces to
(
B⋅∇× ξ×B
) a = B ⋅ B1 a
6. Vacuum part B ⋅ B1 = B ⋅ ∇φI = ιFφI
(
)
(
=0 at the edge
)
7. Plasma part B ⋅ ∇ × ξ × B = ∇ ⋅ ξ × B × B − ξ × B ⋅ ∇ × B
= −∇ ⋅ ξ ⊥B2 = −ξ ⊥ ⋅ ∇B2 − B2∇ ⋅ ξ ⊥
8. Now B2 = B2z + B2θ . Near the edge Bz = const and B0 ∼
∇B2 : −
2B2θ
r
er and ξ ⊥ ⋅ ∇B2 = −
a
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
k
. Therefore
r
2ξB2θ
r
Lecture 20
Page 6 of 10
9. The last term is B2∇ ⋅ ξ⊥ where
∇ ⋅ ξ⊥ =
=
G=
B e ⎞
⎛ B e
1
1
(rξ )' + ∇ ⋅ η = r (rξ )' + ∇ ⋅ ⎜ η zB 0 − θB z ⎟
r
⎝
⎠
(
⎞ 1
1
η ⎛ ιmBz
ιG ι
'
'
'
− ιkBθ ⎟ = (rξ ) +
rξ ) + ⎜
G (rξ ) − 2k0Bθ ξ
(
2
r
B⎝ r
r
B
rk
B
⎠
0
mBz
− kBθ
r
10. Note
1
k 20B2
(
ι
η=
)
⎡G (rξ )' + 2kB ξ ⎤
θ ⎥
⎢⎣
⎦
rk 20B
k20B2 − G2 =
)
F2
k20B2
11. Combine term
∇ ⋅ ξ⊥ =
F2
(rξ )
k20B2
r
'
−
2kGBθ
rk20B2
ξ
12. Collect term
B ⋅ ∇ × ( ξ × B) =
2Bθ
2kGBθ
F2 (rξ )
ξ− 2
+
ξ
r
r
k0
rk20
=−
=−
'
⎛ 2B2θ
2kGBθ ⎞
F2
'
ξ
+
−
+
⎜
⎟ξ
2
2
⎜
k0
rk0
rk20 ⎟⎠
⎝ r
F2
F2
k20
ξ' −
FF
rk20
ξ
13. Pressure balance boundary condition
ιFφI a = −
F2
k 20
ξ' −
FF
rk20
ξ
a
or
φI a =
rξ'F ⎞
ι ⎛
F
+
⎜
⎟ξ
ξ ⎟⎠
rk20 ⎜⎝
a
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 7 of 10
Summary of where we are
⎛r ⎞
φI = c1 ⎜ ⎟
⎝b⎠
m
⎛b⎞
ϕII = c3 ⎜ ⎟
⎝r ⎠
m
m
'
As r = b
As r = a
⎛b⎞
+ c2 ⎜ ⎟
⎝r ⎠
Br = 0, Br =
B1r = ιFξa , φI =
μ0 ωι dBr
η
r ξ'F ⎞
ι ⎛
⎜F +
⎟ ξa
2 ⎜
ξ ⎟⎠
rk 0 ⎝
Apply B.C (note: Br = ϕ' )
As r = b
B1r = 0
As r = b
Br =
'
m
b
m
( c1 − c2 ) =
b
( −c3 )
(
)
m m +1
m
μ0 ωιdBr
− 2 ⎡ m − 1 c1 + m + 1 c2 ⎤ +
c3 =
⎦
η
b ⎣
b2
(
)
(
)
+
'
As r = a
B1r = ιFξ
As r = a
φ=
m ⎡ c1
⎤
− Wm +1c2 ⎥ = ιFξa
2 ⎢
m −1
b ⎣W
⎦
ι ⎛
rξ'F ⎞
F
+
⎜
⎟ξ
ξ ⎟⎠
rk20 ⎜⎝
c1
Wm
+ c2Wm =
W=
m
b
( c1 − c2 )
μ0 ωι d
η
b
a
ι ⎛
rξ'F ⎞
F
+
⎟
⎜
ξ ⎟⎠
rk20 ⎜⎝
Solve for c1 , c2 , c3 from 3 equations
c1 − c2 + c3 = 0
c1 − W2mc2 =
ιaFWmξ
m
c1 + W2mc2 =
rξ'F ⎞
ιaWm ⎛
F
+
⎜
⎟ξ
ξ ⎟⎠
a2k2a ⎜⎝
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 8 of 10
Solution
⎡F
aξ' F ⎤
F
⎢ + 2 2 + a 2 2⎥ξ
ξa k aa ⎥⎦
⎢⎣ m k aa
c1 =
ιaWm
2
c2 =
aξ'a F' ⎤
F
ιa ⎡ F
−
+
+
⎢
⎥ξ
ξa k2aa2 ⎥⎦
2Wm ⎢⎣ m k2aa2
c3 = c2 − c1 = −
'
ιa ⎡ 2m
F
F
F2 aξa ⎤
2m
2m
+
+
−
+
−
W
1
W
1
W
1
⎢
⎥
m
2Wm ⎣⎢
k2aa2
k2aa2 ξa ⎦⎥
(
)
(
)
(
)
Dispersion Relation (last equation)
− (m − 1) c1 − (m + 1) c2 + (m + 1) c3 = ( c1 − c2 )
μ0 ωι db
η
Define τD = μ0 db n, set c3 = c2 − c1
Then ωι τD =
2c1
c3
Simplify
⎡F
F
2Wm ⎢ + 2 2 +
m
k
a
⎣⎢
ωι τD = −
⎡ F
1
F
W2m − 1 ⎢ 2 2 + 2 2
W
k a
⎣⎢ k a
(
)
aξ'a F ⎤
⎥
ξa k2a2 ⎦⎥
aξ'a W2m + 1 F ⎤
+ 2m
⎥
ξa
W − 1 m ⎦⎥
Recall
δW∞
2π2R 0 μ0
δWb
2π2R 0 μ0
'
⎡ F
F
F aξa ⎤
= Fa2 ⎢ 2 2 +
+ 2 2
⎥
m k aa ξa ⎥⎦
⎢⎣ k aa
'
⎡ F
F ⎛ W2m + 1 ⎞
F aξa ⎤
= Fa2 ⎢ 2 2 +
⎥
⎜⎜ 2m
⎟⎟ + 2 2
m ⎝ W − 1 ⎠ k aa ξa ⎥⎦
⎢⎣ k aa
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 20
Page 9 of 10
Therefore
ωι τD = −
2W2m δW∞
W2m − 1 δWb
Resistive wall mode is unstable!! δW∞ < 0
Growth rate ∼ 1 τD
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
δWh > 0
Lecture 20
Page 10 of 10