22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7: The First Order Grad–Shafranov Equation
First Order Equation
The first order Grad–Shafranov equation is given by
2
⎡
⎤
d2 p
dp
1 d ψ0
2 d
cos θ
∇2 ψ1 + ⎢ μ0 R02
+
R
B B ⎥ ψ1 = −2μ0 R0
r cos θ +
0
2
2 0 2
d ψ0
R0 dr
d ψ0
d ψ0
⎢⎣
⎥⎦
Simplify
This equation is simplified as follows:
1. Bθ =
2. R0
1 d ψ0
, ψ0 = ψ0 ( r )
R0 dr
dp
=
d ψ0
R0 dp
1 dp
=
' dr
Bθ dr
ψ0
⎡
2μ0 dp ⎤
r
3. RHS = ⎢ Bθ −
⎥ cos θ
Bθ
dr ⎦
⎣
⎤
1 d ⎡ 1 d
4. ⎡⎣ ⎤⎦ = R02
( μ0 p + B0B2 )⎥
⎢
R0 Bθ dr ⎣ R0 Bθ dr
⎦
=−
5.
⎞
1 d ⎛ 1 Bθ d
1 d ⎛1 d
⎞
rBθ ⎟ = −
⎜
⎜ r dr rBθ ⎟
Bθ dr ⎝ Bθ r dr
B
dr
⎝
⎠
θ
⎠
∇2 ψ1 −
⎡ 2μ r dp
⎤
1 d ⎛1 d
⎞
rBθ ⎟ ψ1 = ⎢ − 0
+ Bθ ⎥ cos θ
⎜
Bθ dr ⎝ r dr
⎠
⎣ Bθ dr
⎦
Solve
This equation is solved as follows:
1. Note that all the forcing terms are proportional to cos θ
2. a. The boundary conditions for a circle of radius b are given by
ψ ( b, θ ) = const = ψ0 ( b ) + ψ1 ( b, θ )
ψ1 ( b, θ ) = 0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 1 of 7
b. For an ellipse r = b ⎡⎣1 + δ cos 2θ ⎤⎦ . Assume δ ∼∈
1.
ψ ⎡⎣b + bδ cos 2θ, θ ⎤⎦ ≈ ψ0 ( b ) + bψ'0 δ cos 2θ + ψ1 ( b, θ ) + ...
ψ1 ( b, θ ) = −bψ'0 δ cos 2θ
a second harmonic is required in the solution
3. Thus, for a circular boundary with cos θ driving terms we can write
ψ ( r, θ ) = ψ0 ( r ) + ψ1 ( r ) cos θ
explicit cos θ dependence
ψ1 ( b ) = 0
4. Simplify the ψ1 equation
ψ1
'
2μ0 rp'
1 ⎡ '' Bθ Bθ ⎤
− 2 ⎥ ψ1 = Bθ −
⎢Bθ +
Bθ ⎣⎢
r
Bθ
r ⎥⎦
( )
−
( )
( rBθ ) ψ
−
1
r ψ1'
r
1
r ψ1'
r
'
'
r2
−
'
'
1
rBθ
= Bθ −
2μ0 r '
p
Bθ
5. Note
'
'
⎡
⎛ψ ⎞ ⎤
⎢ rBθ2 ⎜ 1 ⎟ ⎥ = rBθ ψ1' − r ψ1Bθ'
⎢
⎝ Bθ ⎠ ⎥⎦
⎣
(
)
'
( ) + rBθ ψ
= Bθ r ψ1'
'
'
'
1
( ) − r ψ Bθ
− ψ1 rBθ'
'
' '
1
⎛ 2 d ψ1 ⎞
2
2 dp
⎜ rBθ
⎟ = rBθ − 2μ0 r
dr
B
dr
θ ⎠
⎝
6.
d
dr
7.
d ψ1
1
=
dr Bθ
rBθ2
r
⎛
2
∫0 ⎜⎝ yBθ
− 2μ0 y 2
dp ⎞
⎟ dy
dy ⎠
regularity at r = 0
8. ψ1 = −Bθ ∫
b
r
dx
∫
x
xB2 0
θ
⎛ 2
2 dp ⎞
⎜ yBθ − 2μ0 y
⎟ dy
dy ⎠
⎝
9. This expression for ψ1 represents the toroidal correction to the equilibrium
solution.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 2 of 7
Consequences of Toroidicity
1. The main consequence is an outward shift of the flux surfaces.
2. From ψ1 it is straightforward to calculate Δ ( r ) the flux surface shift.
3. a. ψ ( r, θ ) = ψ0 ( r ) + ψ1 ( r ) cos θ = const.
b. Assume the flux surfaces are of the form r = r0 + r1 ( r0 , θ ) .
c. Then ψ0 ( r0 ) + ψ'0 ( r0 ) r1 + ψ1 ( r0 ) cos θ... = const.
d. Solve for r1
r1 ( r, θ ) = −
Δ (r ) ≡ −
ψ1 ( r ) cos θ
ψ'0 ( r )
= Δ ( r ) cos θ
ψ1 ( r )
ψ'0 ( r )
4. The equation for the flux surfaces is given by r = r0 + Δ ( r0 ) cos θ ,
assuming Δ
r0
5. a. Note that ( x − Δ ) + ( y ) = r02 is the equation of a shifted circle,
2
2
equivalent to the equation for r.
b. Let
x = r cos θ y = r sin θ
r 2 − 2r Δ cos θ ≈ r02
Δ
⎡
⎤
r ⎢1 − cos θ ⎥ ≈ r0
r
⎣
⎦
r ≈ r0 + Δ cos θ
6. The flux surfaces are shifted circles, with shift Δ = − ψ1 ψ'0
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 3 of 7
The Shafranov Shift
1. Calculate the Shafranov shift Δ a ≡ Δ ( a) where a is the last surface to
carry current, i.e., the edge of the plasma.
2. Simplify ψ1 ( a) given by
ψ1 ( a) = −Bθ ( a) ∫
b
a
dx
∫
x
xB2 0
θ
⎛ 2
2 dp ⎞
⎜ yBθ − 2μ0 y
⎟ dy
dy ⎠
⎝
3. Consider the term T1 , noting that a < x < b
a.
x
a
0
0
− μ0 ∫ 2y 2 p'dy = − μ0 ∫ 2y 2 p'dy
= − 2 μ0 y 2 p
=
μ02 I 2
4π2
a
0
a
+ 4∫ μ0 py dy
0
βp
= a2 Bθ2 ( a) β p
b
dx
a
xBθ2
b. T1 = − a2 Bθ3 ( a ) β p ∫
c. For x > a Bθ = Bθ ( a )
a
x
d. Therefore
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 4 of 7
b
T1 = −Bθ ( a) β p ∫ x dx
a
Bθ ( a) β p b2 ⎛
a2 ⎞
=−
⎜⎜ 1 − 2 ⎟⎟
2
b ⎠
⎝
4. Consider the term T2
a. Separate the integral into two parts
b
T2 = −Bθ ( a )
∫a
= −Bθ ( a )
∫a
b. T2b :
b
dx
x
2
∫0 yBθ dy
2
xBθ
dx ⎡ a 2
∫ yBθ dy +
xB2 ⎣⎢ 0
θ
x
x
2
2
2
∫a yBθ dy = Bθ ( a) a ∫a
x
2
⎤
∫0 yBθ dy ⎦⎥
dy
= a2 Bθ2 ( a) ⎡⎣ln x − ln a⎤⎦
y
c. Substitute
T2b = −Bθ ( a) x ⎡⎣ln x − ln a⎤⎦
⎡ ⎛ b2 − a2 ⎞
b2
b2 a2
a2 ⎤
= −Bθ ( a) ⎢ − ⎜
ln a +
ln b −
−
ln a +
⎥
⎟
⎜
⎟
2
2
4
2
4 ⎥⎦
⎢⎣ ⎝
⎠
= Bθ ( a )
d.
T2a =
b2
4
⎡
a2
b⎤
⎢1 − 2 − 2 ln ⎥
a ⎦⎥
b
⎣⎢
− Bθ ( a ) ∫
dx
b
2
a
xBθ
a
2
∫0 yBθ dy
e. Introduce the normalized internal inductance per unir length li.
f.
This follows from
1
Li I 2 =
2
Bθ2
∫p 2μ0 dr
= (2πR0 ) (2π )
=
2π2 R0
μ0
a
1
2μ0
2
∫ Bθ r
dr
2
∫0 Bθ r dr
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 5 of 7
g. Define li ≡
μ0
Li
2πR0
4π
the internal inductance per unit length normalized
to μ0 4π
h. Then
a
∫0
⎛ μ
Bθ2 r dr = ⎜ 20
⎜ 2π R
0
⎝
=
⎞1
μ ⎡ 2πaBθ ( a ) ⎤
⎟ (2πR0 li ) 0 ⎢
⎥
⎟2
μ0
4π ⎣⎢
⎦⎥
⎠
2
1 2 2
a Bθ ( a ) li
2
i.
li is a profile parameter related to the width of the Jφ profile
j.
Then
T2a = −Bθ ( a)
= −Bθ ( a)
li
2
b
∫a x
dx
li 2 ⎛
a2 ⎞
b ⎜1 − 2 ⎟
⎜
4
b ⎟⎠
⎝
5. Combine these terms to evaluate Δ a
a.
Δa = −
ψ1 ( a )
ψ'0
( a)
=−
ψ1 ( a )
R0 Bθ ( a )
b. Then
Δa = −
⎡ Bθ ( a) β p b2
1
⎢−
2
R0 Bθ ( a) ⎢
⎣
⎛
b2 li
a2 ⎞
b2 ⎛
a2
b⎞
⎜⎜1 − 2 ⎟⎟ + Bθ ( a)
⎜⎜1 − 2 − 2 ln ⎟⎟ − Bθ ( a)
4 ⎝
4
a⎠
b ⎠
b
⎝
c. The Shafranov shift is given by
Δa
b
=
2R0
b
⎡⎛
l
1⎞⎛
b⎤
a2 ⎞
⎢⎜ β p + i − ⎟ ⎜⎜1 − 2 ⎟⎟ + ln ⎥
2 2⎠⎝
a ⎦⎥
b ⎠
⎣⎢⎝
Properties of the Shafranov Shift
1.
Δa
b
∼
∼∈
b
R0
1. The shift is small, implying that our approximations are
consistent.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 6 of 7
⎛
a2 ⎞ ⎤
⎜⎜1 − 2 ⎟⎟ ⎥
b ⎠ ⎥⎦
⎝
Δ(a )
∝ β p. This is the outward shift due to the tire tube force and the 1/R
b
force.
1
2.
3.
2
Δ(a )
l ⎛
a2 ⎞
b 1⎛
a2 ⎞
∝ i ⎜1 + 2 ⎟ + ln − ⎜1 − 2 ⎟
2 ⎜⎝
b
a 2 ⎜⎝
b ⎟⎠
b ⎟⎠
internal field
external field
hoop force
This is the shift due to the hoop force.
22.615, MHD Theory of Fusion Systems
Prof. Freidberg
Lecture 7
Page 7 of 7