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Simulation of Electromagnetic Fields  Chapter | 4    81


                We pick out field H  = H − H . The vector H  describes the field due
                                          ext
                                M
                                                        M
             to magnetised structures and satisfies equation ∇ × H  = ∇×(H − H ). We
                                                         M
                                                                      ext
             further introduce scalar potential  : H  = –∇ . From the equations for the
                                             M
                                         M
                                                    M
             divergence of B and µ H  it follows that ∇·(B − µ H ) = 0, or that
                                                         ext
                                                      0
                                 ext
                               0
                                                                    =
                  ∇⋅[ µµ( H M  +  H ext  ) −  µ 0 H ext ]  =∇ ⋅[ µ µ H M  −  µ µ −1) H ext ]0.                       ∇⋅[µ µ(HM+Hext)−µ Hext]=∇⋅[µ 0
                                                          (
                      0
                                                         0
                                                 0
                                                                                                                                      0
                                                                                                                        0
                                                                                                                            µHM−µ (µ−1)Hext]=0.
                Finally, to determine   the following equation is used:                                                           0
                                  M
                                     ∇
                                                 (
                               ∇⋅( µµ ϕ )  = ∇⋅ µ µ −1) H ext  .        (4.7)                                       ∇⋅(µ µ∇M)=∇⋅µ (µ−1)Hext.
                                   0
                                                0
                                        M
                                                                                                                                 0
                                                                                                                        0
                The algorithm for solving the problem is similar to that used for the earlier
             electrical vector potential P. The multiplier (µ − 1) in Eq. (4.7) limits the com-
             putation of H  to the subdomain with µ ≠ 1 (i.e. within the ferromagnetics).
                        ext
             Note, for this problem the field H  is known at least in quadratures and spends
                                        ext
             a lot of computation cost compared to the calculation of P.
                With respect to a magnet system (MS), all other current sources may be
             divided conventionally into ‘external’ j  and ‘internal’ j . This is driven by
                                                            int
                                             ext
             the search for the most effective approach to finding a solution. One can in-
             troduce two vector fields, H  and P, simultaneously. For these fields, equa-
                                    ext
             tions
                  ∇×  H ext  =  j ,  ∇ ×  P =  j ,  ∇⋅ µ 0 H ext  = 0,  H int  = −∇ ϕ +  P,
                                        int
                            ext
                  H =  H ext +  H int ,  ∇⋅ B = 0,  B =  µµ H =  µµ( H ext +  H int )                                ∇×Hext=jext,  ∇×P=jint,  ∇⋅µ H
                                               0
                                                       0
                                                                                                                                              0
                                                                                                                     ext=0,  Hint=−∇+P,  H=Hext+
             are valid.
                From the last two equations, it follows that                                                         Hint,  ∇⋅B=0,  B=µ µH=µ µ(He
                                                                                                                                      0
                                                                                                                                           0
                                                                                                                                         xt+Hint)
                                    ∇
                                      )
                             ∇⋅( µµ ϕ = ∇⋅ µ µ P+ ∇⋅ µ µ H .            (4.8)                                       ∇⋅(µ µ∇)=∇⋅µ µP+∇⋅µ µHext.
                                                          ext
                                  0
                                                      0
                                             0
                                                                                                                                      0
                                                                                                                               0
                                                                                                                        0
                Eq.  (4.8) must account for the magnetic properties of materials and the
             boundary conditions. As earlier, to avoid many computations, Eq. (4.8) can be
             transformed into
                                 ∇
                           ∇⋅( µµ ϕ = ∇⋅ µ µ P+ ∇⋅  µ µ −1) H ext .     (4.9)                                       ∇⋅(µ µ∇)=∇⋅µ µP+∇⋅µ (µ−1)Hext.
                                    )
                                                     (
                                           0
                                                    0
                               0
                                                                                                                               0
                                                                                                                       0
                                                                                                                                      0
                Among the possible external current sources, first, we consider current coils.
             For a wide range of applications, the coils can be modelled as a combination
             of ring, rectangular, or arc-shaped conductors with rectangular or polygonal
             cross-sections, where current densities are uniformly distributed. The coils are
             complemented with ring-shaped current filaments to describe a plasma cur-
             rent and with uniformly magnetised volume elements bounded by flat facets to
             simulate ferromagnetic structures such as the tokamak building reinforced with
             steel rebar and inserts. For a given magnetisation, a field due to such elements
             can be calculated analytically or using known quadratures that enables control
             of the computation accuracy.
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