22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9: The High Beta Tokamak
Summary of the Properties of an Ohmic Tokamak
1. Advantages:
a. good equilibrium (small shift)
b. good stability ( q ∼ 1)
(
c. good confinement τ ∼ naR2
)
d. good ohmic heating (Te ∼ Ti ∼ 2keV )
2. Disadvantages:
a. low β : β ∼ ∈2 , βt =
∈2 β p
qa2
∼
0.32 × 0.5
32
= 1 2%
b. external heating is required: joule heating is not adequate. External heating is
expensive, but also raises β
c. tight aspect ratio is required to raise β : this is technologically difficult
d. pulsed operation is required unless current drive works efficiently
3. The high β tokamak resolves the problem of low β
a. It allows a tokamak to operate at higher β ∼ 5 − 10% . This leads to more
economic devices
b. How do we achieve higher β ? We apply additional auxiliary heating, keeping
Bφ fixed. This raises p relative to Bφ2 2μ .
0
High β Tokamak Expansion
1. We again assume large aspect ratio: a R0
1 ∈ ≡ a R0
2. Stability is produced by a large toroidal field: q ∼ 1
3. Thus, as in the ohmic tokamak q ∼ rBφ RBθ , implying that Bθ Bφ ∼ ∈
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 1 of 12
4. Radial pressure balance, however is produced by the toroidal field
Bφ = B0 + δBφ
diamagnetic
'
⎛
B2 ⎞
B
'
⎜ p + φ ⎟ + θ ( rBθ ) = 0
⎜
2μ0 ⎟ μ0 r
⎝
⎠
'
⎛
B δB2 ⎞
B
'
⎜ p + 0 φ ⎟ + θ ( rBθ ) = 0
⎜
⎟
μ0 r
2μ0
⎝
⎠
neglect, confines only β ∼ ∈2
5. To improve β over that achievable in the ohmic tokamak we need
p∼
B0 δBφ
μ0
Bθ2
μ0
6. Or in terms of β
β ∼
δBφ
Bθ2
B0
Bφ2
∼ ∈2
7. How large can β and δBφ B0 get?
8. The limiting condition is determined by toroidal force balance which is still
accomplished by a combination of I and Bv
9. Increasing β increases the toroidal shift. The largest possible β occurs when the
shift becomes of order unity Δ a ∼ 1. Recall that in an ohmic tokamak Δ a ∼ a R0 .
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 2 of 12
10. Let us estimate the shift using the small shift relation
ψ1 = −Bθ ∫
b
r
dr '
r'
∫
r 'B2 0
θ
⎛ 2
2 dp ⎞
⎜ yBθ − 2μ0 y
⎟ dy
dy ⎠
⎝
neglect since μ0 p
ψ1 ∼
Bθ2
μ0 a2 p
Bθ
11. Therefore
2
β q2
ψ
1 ⎛ μ0 a p ⎞ a ⎛ μ0 p ⎞
Δ
∼ − 1' ∼
⎜
⎟ ∼ ⎜ 2 ⎟ ∼ ∈ βp ∼ t
∈
a
aRBθ ⎝⎜ Bθ ⎠⎟ R ⎝⎜ Bθ ⎠⎟
aψ0
12. For q ∼ 1, then Δ a ∼ 1 when βt ∈ ∼ 1
13. This suggests the following ordering for the high β tokamak
β ∼ ∈,
δBφ
Bφ
∼ ∈, Δ a ∼ 1
Comparison of Expansions
Ohmic Tokamak
High Beta Tokamak
q ∼1
q ∼1
Bθ Bφ ∼ ∈
Bθ Bφ ∼ ∈
β ∼ 2μ0 p B02 ∼ ∈2
β ∼ 2μ0 p B02 ∼ ∈2
δ Bφ B0 ∼ ∈2 (para)
δBφ B0 ∼ ∈ (dia)
β p ∼ 2 μ0 p Bθ2 ∼ 1
β p ∼ 2 μ0 p Bθ2 ∼ 1 ∈
Expansion of Grad-Shafranov Equation
1. Since Δ a ∼ 1 , toroidal force balance and radial pressure balance enter together
in zeroth order.
2. Good news: we need only the zeroth order equations. No first order corrections
are required.
3. Bad news: The zero other equations are still nonlinear partial differential
equations.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 3 of 12
4. Expansion:
a. ψ = ψ0 ( r , θ ) + ..., ψ ∼ rRBθ ∼ a2 B0
b.
p ( ψ ) = p ( ψ 0 ) + ..., μ0 p ∼ ∈ B02
c.
F ≡ RBφ = F ( ψ )
new free function
F
2
=
R02
⎡ B02 − 2μ0 p ( ψ ) + 2B0 B2 ( ψ ) ⎤
⎣
⎦
∼1
∼ ∈2
∼∈
B2 B0 ∼ ∈2
d. This automatically produces a θ pinch pressure balance
p+
Bφ2
2μ0
≈ const
5. Substitute the expansion into the Grad-Shafranov equation
∇2 ψ = − μ0 ( R0 + r cos θ )
2
T1 = ∇2 ψ ∼
T3 =
ψ
a2
dp
d F 2 1 ⎛ ∂ψ
1 ∂ψ
⎞
cos θ −
sin θ ⎟
−
+ ⎜
dψ dψ 2
R ⎝ ∂r
r ∂θ
⎠
∼ B0
1 ⎛ ∂ψ
1 ∂ψ
ψ
⎞
∼ ∈ B0 neglect
cos θ −
sin θ ⎟ ∼
R ⎜⎝ ∂r
r ∂θ
aR
⎠
T2 = − μ0 ( R0 + r cos θ )
2
≈−
dp
d F2
−
dψ dψ 2
⎞
d ⎛ F2
dp
+ μ0R02 p ⎟ − 2μ0 R0 r cos θ
+ ...
⎜⎜
⎟
dψ ⎝ 2
d
ψ
⎠
=0
2
⎡
⎤
⎛
⎞
B
d
dp
= −R02
⎢⎜ 0 − μ0 p + B0 B2 ⎟ + μ0 p ⎥ − 2μ0 R0 r cos θ
⎟
d ψ ⎣⎢⎜⎝ 2
d
ψ
⎠
⎦⎥
= −R02
d
( B0B2 )
dψ
−2μ0 R0 r cos θ
∼
dp
dψ
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
∼
R02 B0 B2
∼ B0
ψ
μ0 pB0 a
ψ
∼ B0
Lecture 9
Page 4 of 12
6. Therefore, to leading order the Grad-Shafranov equation reduces to
∇2 ψ0 = −R02
dp ( ψ0 )
d
B0 B2 ( ψ0 ) − 2μ0 R0 r cos θ
d ψ0
d ψ0
7. Note that μ0 RJφ ≈ −∇2 ψ , so that on a circular plasma flux surface
μ0 RJφ = R02
d
B0 B2
dψ
average over θ
We see that dB2 dψ is proportional to the average toroidal current within a given
flux surface.
8. Even though the equation is simpler, it is still a nonlinear PDE.
9. In general, it must be solved numerically.
10. The difficulty arises because the shifts are finite and cannot be treated
perturbatively.
11. We shall determine general features of high β tokamak by examining a special
case.
Special Case
1. Choose
2μ0 R0
R02 B0
dp
= −C
d ψ0
dB2
= −A
d ψ0
C = const
A = const
2. This implies p ∼ −C ψ (assume ψ (boundary)=0) and Jφ ∼ − A
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 5 of 12
3. Solution: with these choices the Grad-Shafranov equation becomes
2
1 ∂ ∂ψ0
1 ∂ ψ0
r
+ 2
= A + Cr cos θ
r ∂r
∂r
r ∂θ 2
4. Boundary conditions: We assume a circular plasma of radius r = a
ψ ( a, θ ) = 0 (normalization of flux function is arbitrary)
ψ ( r,θ )
regular for r ≤ a
The circular assumption is made for simplicity and can be generalized to other
cross sections.
5. Solution: (We need only cos nθ terms because of up-down symmetry.)
ψ part = A
r2
r3
+C
cos θ
4
8
ψ hom = k1 + k2 ln r + k3r cos θ +
not regular
∞
k4 cos θ
+ ∑ an r n + bn r −n cos nθ
r
n =2
(
not regular
)
not regular
= k3r cos θ + k1
only terms which are regular and required
to balance ψ part on the boundary r = a
For all n ≥ 2 , it follows that an = 0
6. Choose k1 and k3 to make ψ ( a, θ ) = 0
ψ (r, θ) =
(
)
(
)
A 2
C 3
r − a2 +
r − a2 r cos θ
4
8
7. Then
Bθ =
1 ∂ψ0
1
=
R0 ∂r
2R0
μ0 p = −
(
)
C
⎡
⎤
2
2
⎢ Ar + 4 3r − a cos θ ⎥
⎣
⎦
C
C
ψ0 =
2R0
8R0
(
)
(
)
C 2
⎡
⎤
2
2
3
⎢ A a − r + 2 a r − r cos θ ⎥
⎣
⎦
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 6 of 12
Physical Interpretation
Let us express A and C in terms of more physical quantities: βt , I or equivalently
βt , q* ∝ 1 I
1. q* is a parameter related to kink stability and the surface MHD safety factor qa
q* ≡
=
2 Ap B0
μ0R0 I
2πa2 B0
μ0 R0 I
∫ Bθ ( a, θ )adθ
2. μ0 I = ∫ B p ⋅ dl =
ON A CIRCLE
=
3.
a
2R0
2π
∫0
⎡
⎤
πAa2
Ca2
Aa2
cos θ ⎥ dθ =
⋅ 2π =
⎢ Aa +
2
2R0
R0
⎢⎣
⎥⎦
μ0 R0 πAa2
1
A
=
=
2
q* 2πa B0 R0 μ0
2B0
4. βt =
=
2μ0
B02
p =
2μ0 1
B02 πa2
2μ0
C
8
μ
πa B0
0 R0
2
∫ pr
a 2π
∫0 ∫0
dr dθ
(
)
(
)
C 2
⎡
⎤
r dr dθ ⎢ A a2 − r 2 +
a r − r 3 cos θ ⎥
2
⎣
⎦
2πa4
4
=
a2
8R0 B02
AC =
a2 C
4R0 B0 q*
5. Thus
A
1
=
2B0
q*
a2C
= q* βt
4R0 B0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 7 of 12
6. General equilibrium relation for a high β circular tokamak (minor degression)
∫ pr
βp =
a.
dr dθ
μ0 I 2 8π
=
8π πa2 B02
βt
μ0 I 2 2μ0
2
⎛ 2πaB0 ⎞
=⎜
⎟ βt
⎝ μ0 I ⎠
=
q*2 βt
∈2
, ∈≡ a R0
b. The general equilibrium relation is given by
βt =
∈2 β p
q*2
7. Substitute A and C back into the solutions. Define ν =
ψ0
2
a B0
=
(
βt q*2
∈
=∈ β p ∼ 1, ρ = r a
)
1 ⎡ 2
ρ − 1 + ν ρ 3 − ρ cos θ ⎤
⎦
2q* ⎣
Bθ
1
=
∈ B0
q*
ν
⎡
⎤
2
⎢ ρ + 2 3 ρ − 1 cos θ ⎥
⎣
⎦
(
)
Br
ν
=−
ρ 2 − 1 sin θ
∈ B0
2q*
(
2 μ0 p
B02
Bφ
B0
)
= 2 βt 1 − ρ 2 (1 + νρ cos θ )
μ0 R0 Jφ
B0
(
)
=−
2
(1 + 2νρ cos θ )
q*
(
)
= 1− ∈ ρ cos θ - βt 1 − ρ 2 (1 + νρ cos θ )
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 8 of 12
Properties of High Beta Tokamak Equilibria
1. Sketched below are typical midplane profiles (Z=0) showing radial pressure
balance.
2. Shown here are p and Jφ profiles along the midplane for different ν . Increasing
ν implies higher β .
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 9 of 12
a. Observe the increased shift of the magnetic axis as β increases.
b. Observe the buildup of current on the outside of the torus to produce toroidal
force balance at higher β .
c. Observe Jφ reversing on the inside of the torus when ν > 1 2 .
3. The flux surfaces are “round”, but are not circles except for the boundary
(
)
ρ 2 − 1 + ν ρ 3 − ρ cos θ = const
4. Let us calculate the magnetic axis shift Δ0 by finding the value of r where
∂ψ
∂p
∝
= 0 . By symmetry this occurs when θ = 0 ( or π ) .
∂r
∂r
a. At θ = 0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 10 of 12
(
ψ ∝ ρ2 − 1 + ν ρ3 − ρ
)
b. Set
Δ0
⎞
∂ψ ⎛
,θ = 0 ⎟ = 0
⎜ρ =
∂ρ ⎝
a
⎠
c. This yields
2Δ 0
+ν
a
⎛ Δ2
⎞
⎜ 3 20 − 1 ⎟ = 0
⎜ a
⎟
⎝
⎠
d. The value of Δ0 is given by
Δ0
ν
=
a
1 + 1 + 3ν 2
(
)
12
∼1
Δ0
∼1
a
Δ
ν
Ohmic Tν ∼ ∈ → 0 ∼ ∼ ∈
2
a
e. For the HBT ν ∼ 1 →
1
5. Find the shape of the flux surfaces near the magnetic axis.
a. Let x = ρ cos θ
y = ρ sin θ
(
)
b. Then ψ ∝ x 2 + y 2 − 1 ⎡⎣1 + ν x ⎤⎦
c. Expand x =
Δ0
+ δx
a
y = δy
δx, δy
1
and define x0 = Δ0 a
d. Substitute
2
2
ψ ∝ ⎡⎢ x02 + 2 x0 δx + ( δx ) + ( δy ) − 1⎤⎥ ⎣⎡1 + ν x0 + νδx ⎦⎤ = const
⎣
⎦
(
)
= x02 − 1 (1 + ν x0 ) + δx ⎡2 x0 + 2ν x02 + ν x02 − ν ⎤
⎣
⎦
0 definition of x0
+ ( δy ) (1 + ν x0 ) + ( δx ) ⎡⎣1 + ν x0 + 2ν x0 ⎤⎦
2
2
or
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 11 of 12
(1 + 3ν x0 ) ( δx )2 + (1 + ν x0 ) ( δy )2
= const
e. This is the equation of an ellipse
elongated flux surfaces, squashed near the outside
f.
The elongation κ0 is defined by
κ20 =
b2
2
a
=
1 + 3ν x0
1 + ν x0
(
= 1 + 3ν 2
κ20 ≈ 1 for ν
)
(
⎡
2
⎢⎣1 + 1 + 3ν
12
(
)
12⎤
⎡
2
2
⎢⎣1 + ν + 1 + 3ν
)
⎥⎦
12⎤
⎥⎦
1 and κ20 = 3 2 for ν = 1 .
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 9
Page 12 of 12