Equilibrium Reconstruction with Grad-Shafranov Equation

Transcription

1 Equilibium Reconstuction with Gad-Shafanov Equation Qian Peng (qp215) Apil 14, Gad-Shafanov equation [1]Fo axial symmetic case, the tooidal B eld is B to = F φ. Whee 2π F (, θ) is the total poloidal cuent though the cicle with adius at(, θ). The total B eld is given by B = F φ φ ψ p notice that hee 2π ψ p is the polodial ux, but dened slightly dieent fom that in Plasma II. ψ p is at magnetic axis and is the integation of counte-clockwise B eld fom magnetic axis to the suface. Then we have thus µ j = B = d F d ψ ψ p φ (R p R ( 1 R = d F d ψ p ψ p φ ψp φ j toodial = ψ p µ R ) Z ) ψ p φ It should be pointed out that cuently is dened in cylindical coodinates while othe functions ae dened in ux coodinates. Fo MHD model: µ j B = µ p d F ( F d ψ ψp )( φ) 2 p = d(µ p) d ψ p ψp = 2 d(µ p) d ψ p = 2 P T F d F d ψ p 1

2 Gad-Shafonov equation is ψp = 2 P ( ψ p ) T ( ψ p ) (1) and we have µ j toodial = P + T 1.1 Geen's function Seach fo Geens function G(, z,, z ) such that[2],z G(, z,, z ) = δ(, z z ) (2) G,z In fact the poloidal component of B eld: φ ψ p = ( ψ p φ) and ψ p is thus the φ component of magnetic potential A φ coesponding to the B eld. The physical meaning of the RHS of (2) is a ing cuent (1/µ ) locating at (, z ) and its magnectic potential is A( ) = µ 4π = ( φ) 4π j( ) d3 π π cos(θ) cos(θ) dθ thus π G(, z,, z ) = 4π cos(θ) dθ (3) π cos(θ) 1.2 Solution of GSh equation We only want to solve GSh equation within a domain: ψ p = f(, z) (, z) Ω (4) we can extend Ω to the whole space to include the extenal coils and have ψ p (, z) = f(, z )G(, z,, z )d dz (5) +Σ i I i G(, z, i, z i) whee I i denote extenal coils. This is not vey numeically ecient because if Ω is a N by N space and f(, z) is non-zeo almost evey whee in the 2

3 domain then this would equie O(N 4 ) opeations to obtain the value of ψ p at evey point. Anothe way to solve this is to st constuct the bounday value ψ p (, z) (,z) Ω using (5) (Which will take O(N 3 ) opeations) and then discetize (4) and solve it numeically (Which is expected to take O(N 3 ) opeations). [4] To constuct ψ p on the bounday, we still need to calculate Geen's function O(N 3 ) times befoehand and stoe them in the memoy. We can impove it futhe moe.[3] u = ( (1 = ( 1 u) ) z )u using Geen's identity G(, z,, z ) ψ p (, z) ψ p (, z) G(, z,, z )ddz = G(, z,, z )f(, z) ψ p (, z)δ(, z z )ddz = G(, z,, z )f(, z)ddz ψ p (, z ) since = = we have Ω Ω Ω ψ p (, z ) = G(, z,, z ) ψ p (, z) ψ p (, z) G(, z,, z )ddz G(, z,, z ) ( 1 ψ p (, z)) ψ p (, z) ( 1 G(, z,, z ))ddz G(, z,, z ) + Ω Ω n ψ p (, z) ψ p (, z) n G(, z,, z ) ds,z G(, z,, z )f(, z)ddz ψ p (, z) n G(, z,, z ) G(, z,, z ) n ψ p (, z) ds,z (6) Notice that the st tem of RHS of (6) is what we need in (5), we can obtain the value of this tem by the following pocedue: Solve the equation numeically by discetizing the opeato ψ p = f(, z) (, z) Ω ψ p Ω = 3

4 whee the domain Ω is lage than Ω. Then calculate ψ n p on the bounday and since ψ p is zeo on the bounday, we have G(, z,, z )f(, z)ddz = G(, z,, z )f(, z)ddz Ω Ω = ψ p (, z G(, z,, z ) ) + Ω n ψ p (, z) ds,z and we only need to know the value of G(, z,, z )with (, z) Ω, (, z ) Ω. Substitute this into (5) we have ψ p (, z) = ψ G(, z,, z) p (, z) + Ω n ψ p (, z ) ds,z (7) +ΣI i G(, z, i i, z i) Use (7) to calculate ψ p (, z) on the bounday and then solve it numeically. 1.3 Non-lineaity of GSh equation. In the pevious section we ae assuming the RHS of GSh equation is f(, z) while in eality the RHS of (1) is f( ψ p ) which depends on the function of p( ψ p ) and F ( ψ p ). Even assuming that p and F is given. f is a function of (, z) in a sense that we aleady know the function ψ p (, z) and f(, z) = f( ψ p (, z)). But ψ p (, z) is the function that we ae solving fo! One way to teat this poblem is to solve this equation iteatively. Stat with an initial guess of ψ p, (, z) and use this to calculate f n+1 (, z) = f( ψ p,n (, z)) and then solve the equation to obtain ψ p,n+1, iteate until ψp conveges. In fact, the plasma bounday can also change duing each iteation. Plasma bounday is detemined by a set of limites located at ( s, z s ) in this case and the limite with lowest ψ p,n,min = ψ p,n ( s, z s ) detemines the bounday of plasma. The function of p and F ae actually functions of nomalized poloidal ux p( ψ p (, z)/ ψ p,min (, z)), F ( ψp (, z)/ ψ p,min (, z)) because they ae dened on the plasma egion. 1.4 GSh equation on ux coodinates [1]Solving GSh equation on (, z) coodinates gives simplicity in discetizing the opeato, but it it may not be the most ecient way. Use geneal coodinates (a, θ) to descibe the (, z) plane, whee a can be teated as a suface label and θ the poloidal angle. Let D(, z) D = D(a, θ) 4

5 then whee f = ( 1 f) = 1 D a (DG a,a a f) + 1 D a (DG a,θ θ f) + 1 D θ (DG θ,a a f) + 1 D θ (DG θ,θ θ f) G a,a = x a x G a,θ = G θ,a = x a a x θ The GSh equation is thus D ψ p = DP D T G θ,θ = x θ x θ = V P LT whee V D, L = D If (a, θ) is aleady the ux coodinate (a, θ) that ψ p = ψ p (a ), p = p(a ), F = F (a ) then we shall have a (DG a,a aveaging ove θ we have ψ p) + θ (DG θ,a ψ p) = V P LT (8) (( DG a,a ) ψ p ) = T L P V, whee (f) = 1 2π Dene the total coss-section cuent J(a ) = 1 2π = 1 2π = Compae it with (9) we have: and since we have Which is an 1D equation. a 2π 2π dθ dθ a a V P + L T da µ j tooidal D(, z) D(a, θ) da DP + DT da (( DG a,a ) ψ p ) a = J(a ) B θ (a = ) = ψ p a= = fdθ (9) ( DG a,a ) ψ p = J (1) 5

6 1.5 Petubed GSh equation [1]When solving the GSh equation, we don't eally have the ux coodinate (a, θ) until we get the tue solution. Howeve, we can use the coodinate (a, θ) geneated fom each appoximation of ψp and assume that (a, θ) is close to (a, θ) in the sense that a = a ξ(a, θ) then (, z) which is initially a function of (a, θ) can be expessed as = (a, θ) = (a + ξ, θ) (a, θ) + aξ z = z(a, θ) = z(a + ξ, θ) z(a, θ) + z aξ the suface of (a, θ) has to satisfy that ψ p (a, θ) ψ p + ψ(a, θ) = const whee ψ p is the main ode appoximation and can be obtained by solving the equation (1), in the coodinate (a, θ) but taken as (a, θ). Thus: ψ pξ + ψ = const Notice that ξ have the pat ξ and oscillatoy pat ξ thus ξ = ψ ψ p ξ = ψ p (a, θ) ψ p ψ ψ p chose ψ p (a, θ) = ψ p then ξ = ψ ψ p Now, GSh equation can be lineaized: ψp = 2 P (a ) T (a ) with P (a ) = P (a) P ξ + δp T (a ) = T (a) T ξ + δt (δp and δt denote the vaiations of the given function pole which will be used late.) such that L ψ p = V P LT + V P ξ + LT ξ V δp LδT (11) Fo the pevious algoithms, at each step, when ψ p n n+1 ψ p, only the change to the LHS of the above equation is consideed. If one can also calculated the elated ξ due to the change in ψ p, and use both of them in the above equation, a bette conveging ate should be achieved. 6

7 2 Equilibium econstuction Solving GSh equation is one majo pat of econstucting the equilibium pole. Yet, one also needs to have the pole of P ( ψ p ) and T ( ψ p ) to solve the equation. These poles ae the input paametes fo GSh equation solve and ae often detemined by the measuement fom pobes. We stat with an initial guess of the poles with paametes P ( ψ p ) = P (α 1, α 2,..., α n, ψ p ) T ( ψ p ) = T (β 1, β 2,..., β n, ψ p ) and solve fo ψ p (, z). Then we have P (α 1, α 2,..., α n,, z) and T (β 1, β 2,..., β n,, z), fom which we can get the expect eading fom the ith pobe E i (α 1, α 2,..., α n, β 1, β 2,..., β n ) Since we also have the actually eadings, the paametes can be adjusted to bette t the eading. Update the paametes and solve the GSh equation again until it conveges. [4] One can see that the convegence ate does not tend to be vey fast, because the tting pocedue does not take into account that ψ p (, z) would be changed when the paametes change. A bette way to do it might be using δp and δt in the equation (11) and obtain an estimated esponse of ψp. Taking that into account when doing the tting pocesue may esult in a bette convegence ate. 2.1 Eects of measuing techniques on econstuction. It is a commen sense that the moe accuate you measuements ae, the bette econstuction you will get. But, it would be bette if one can tell quantitatively how does the accuacy of the pobe eadings aects the cedibility of one's econstuction esult. In L.E. Zakhaov's pape[5], he gives the elations quantitatively. The idea is to view eos as petubations and see how lage the magnitude of the petubation could be as long as they ae undetectable by the device. When we change the P and T pole by δp and δt, it will esult in a change of δ ψ p and eventually a change of signals we ead, δs. Fo smal petubation, we can lineaize the elationship and have δs = AδX whee δs is a N dimensional vecto of all the signals fom pobes and δx ae a M dimensional discete vesion of the change of δp and δt. Since all pobes have cetain degees of eo allowance, we use δɛ to denote that. Nomalize each ow of A(i, :) with δɛ i and we get Ã. Thus, δ S = ÃδX 7

8 and each enty of δ S has to have a magnitude less than 1 so that the petubation is within the eo ange of pobes. One can single value decompse à = UW V T and fo δx k = γv k we have δ S k = γw k U k. Set cetain constains fo the nom of δs and we'll get the eo ange δx fo the econstuction. Refeences [1] `Theoy of petubed equilibia fo solving the GadShafanov equation.' Zakhaov, L E Pletze, A (1999). [2] `Numeical Solution of Gad-Shafanov Equation fo the Distibution of Magnetic Flux in Nuclea Fusion Devices.' Aydin, Selcuk Han, Teze-Sezgin Munevve. (28). [3] `Computation of ideal MHD equilibia.' Lackne K, (1976). [4] `Numeical detemination of axisymmetic tooidal magnetohydodynamic equilibia.' Johnson, J L Dalhed, H E Geene, J M Gimm, R C Hsieh, Y Y Jadin, S C Manickam, J Okabayashi, M Stoe, R G Todd, A M M Voss, D E Weime, K E. (1979). [5] `Reconstuction of the p and q poles in ITER fom extenal and intenal measuements.' Zakhaov. L, Foley. E, Levinton. F, Yuh. H. (28). 8