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users_manual [2020/10/20 21:48]
stanzurek [A.3.2 Asymptotic Boundary Conditions] 		users_manual [2021/11/22 21:25] (current)
stanzurek [A.1 Modeling Permanent Magnets]
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 (A.1)  $  Hc = \frac{5·10^5· \sqrt{E}}{π}  $ (A.1)  $  Hc = \frac{5·10^5· \sqrt{E}}{π}  $
  
-where //E// is the energy product in MGOe and the resulting //Hc// is in units of A/m (e.g. 40 MGOe ≈ 106 A/m).+where //E// is the energy product in MGOe and the resulting //Hc// is in units of A/m (e.g. 40 MGOe ≈ 10<sup>6</sup> A/m).
  
 {{fig_a-3.png}} {{fig_a-3.png}}
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 === A.3.3 Kelvin Transformation === === A.3.3 Kelvin Transformation ===
  
-A particularly good approach to “open boundary” problems is the Kelvin Transformation, a technique +A particularly good approach to “open boundary” problems is the Kelvin Transformation, a technique first discussed in the context of computational magnetics in [18] and [19]. The strengths of this technique are: 
-first discussed in the context of computational magnetics in [18] and [19]. The strengths of +  * the effects of the exterior region are, in theory, exactly modeled by this approach; 
-this technique are: +  * a sparse matrix representation of the problem is retained (unlike FEM-BEM methods, which give the same “exact solution” but densely couples together the boundary nodes). 
-• the effects of the exterior region are, in theory, exactly modeled by this approach; +  * requires no “special” features in the finite element solver to implement the technique, other than the ability to apply periodic boundary conditions. 
-• a sparse matrix representation of the problem is retained (unlike FEM-BEM methods, which + 
-give the same “exact solution” but densely couples together the boundary nodes). +The purposes of this note are to explain what the Kelvin transformation is derived and to show how it is implemented in the context of the FEMM finite element program. 
-• requires no “special” features in the finite element solver to implement the technique, other + 
-than the ability to apply periodic boundary conditions. +** Derivation ** 
-The purposes of this note are to explain what the Kelvin transformation is derived and to show how + 
-it is implemented in the context of the FEMM finite element program. +In the “far field” region, the material is typically homogeneous (e.g. air and free of sources. In this case, the differential equation that describes vector potential A is the Laplace equation: 
-Derivation + 
-In the “far field” region, the material is typically homogeneous (e.g. air and free of sources. In this +(A.19) $ \nabla^2 A = 0  $ 
-case, the differential equation that describes vector potential A is the Laplace equation: + 
-Ñ2A = 0 (A.19) +If we write (A.19) in polar notation, //A// is described by: 
-If we write (A.19) in polar notation, A is described by: + 
-1 +(A.20) $ \frac{1}{r} \left( r \frac{\partial A}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 A}{\partial θ^2} = 0  $ 
-r + 
-¶ +Assume that the “near field” region of the problem can be contained in a circle of radius //ro// centered at the origin. The far-field region is then everything outside the circle. 
-¶r + 
- +One approach to unbounded problems is to attempt to map the unbounded region onto a bounded region, wherein problems can more easilby be solved. Specifically, we desire a way to transform the unbounded region outside the circle into a bounded region. One simple way to make such a mapping is to define another variable, //R//, that is related to //r// by: 
-r + 
-¶A +(A.21) $ R = \frac{r_0^2}{r} $ 
-¶r + 
- +By inspecting (A.21), it can be seen that this relationship maps the exterior region onto a circle of radius //ro//. 
-+ + 
-1 +The next step is to transform (A.19), the differential equation that the field must satisfy, into the mapped space. That is, (A.19) must be written in terms of //R// and //θ// rather than //r// and //θ//. We can evaluate derivatives in terms of //R// instead of //r// by employing the chain rule: 
-r2 + 
-¶2A +(A.22) $ \frac{\partial}{\partial r} = \frac{\partial}{\partial R} \left( \frac{\partial R}{\partial r} \right) = - \frac{\partial}{\partial R} \left( \frac{R}{r_0} \right)^2  $ 
-¶q2 = 0 (A.20) + 
-Assume that the “near field” region of the problem can be contained in a circle of radius ro centered +Now, we can note that at //r// = //R// = //ro//, 
-at the origin. The far-field region is then everything outside the circle. + 
-One approach to unbounded problems is to attempt to map the unbounded region onto a +(A.23)  $ \frac{\partial A}{\partial r} = -\frac{\partial A}{\partial R} $ 
-bounded region, wherein problems can more easilby be solved. Specifically, we desire a way +
-to transform the unbounded region outside the circle into a bounded region. One simple way to +
-make such a mapping is to define another variable, R, that is related to r by: +
-R = +
-r2 +
-o +
-r +
-(A.21) +
-By inspecting (A.21), it can be seen that this relationship maps the exterior region onto a circle of +
-radius ro. +
-158 +
-The next step is to transform (A.19), the differential equation that the field must satisfy, into +
-the mapped space. That is, (A.19) must be written in terms of R and q rather than r and q. We can +
-evaluate derivatives in terms of R instead of r by employing the chain rule: +
-¶ +
-¶r +
-= +
-¶ +
-¶R +
- +
-dR +
-dr +
- +
-= − +
-¶ +
-¶R +
- +
-R +
-ro +
-2 +
-(A.22) +
-Now, we can note that at r = R = ro, +
-¶A +
-¶r +
-= − +
-¶A +
-¶R +
-(A.23)+
 and we can substitute (A.22) into (A.19) to yield, after some algebraic manipulation: and we can substitute (A.22) into (A.19) to yield, after some algebraic manipulation:
-1 
-R 
- 
-¶R 
- 
-R 
-¶A 
-¶R 
- 
-+ 
-1 
-R2 
-¶2A 
-¶q2 = 0 (A.24) 
  
-Eq. (A.24), the transformed equation for the outer region, has exactly the same form as inner +(A.24) $  \frac{1}{R} \frac{\partial}{\partial R} \left(R \frac{\partial A}{\partial R} \right) + \frac{1}{R^2}  \frac{\partial^2 A}{\partial θ^2}  = 0 $
-region, only in terms of R rather than r. The implication is that for the 2-D planar problem, the +
-exterior can be modeled simply by creating a problem domain consisting of two circular regions: +
-on circular region containing the items of interest, and an additional circular region to represent +
-the “far field.” Then, periodic boundary conditions must be applied to corresponding edges of the +
-circle to enforce the continuity of A at the edges of the two regions. The is continuity of A at the +
-boundary between the exterior and interior regions. For a finite element formulation consisting +
-of first-order triangles, (A.23) is enforced automatically at the boundaries of the two regions. The +
-second circular region exactly models the infinite space solution, but does it on a bounded domain– +
-one could always back out the field for any point in space by applying the inverse of (A.21). +
-Kelvin Transformation Example – open1.fem+
  
-As an example, consider an E-core lamination stack with a winding around it. Suppose that the +Eq. (A.24), the transformed equation for the outer region, has exactly the same form as inner region, only in terms of //R// rather than //r//. The implication is that for the 2-D planar problem, the exterior can be modeled simply by creating a problem domain consisting of two circular regions: on circular region containing the items of interest, and an additional circular region to represent the “far field.” Then, periodic boundary conditions must be applied to corresponding edges of the circle to enforce the continuity of //A// at the edges of the two regions. The is continuity of //A// at the boundary between the exterior and interior regions. For a finite element formulation consisting of first-order triangles, (A.23) is enforced automatically at the boundaries of the two regions. The second circular region exactly models the infinite space solution, but does it on a bounded domain - one could always back out the field for any point in space by applying the inverse of (A.21). 
-objective is to determine the field around the E-core in the absence of any flux return path (i.e. + 
-when the magnetic circuit is open). In this case, the flux is not constrained to flow in a path that is +** Kelvin Transformation Example ** – ''open1.fem'' 
-a priori well defined, because the laminations that complete the flux path have been removed. + 
-The geometry was chosen arbitrarily, the purpose here being more the procedure than the actual +As an example, consider an E-core lamination stack with a winding around it. Suppose that the objective is to determine the field around the E-core in the absence of any flux return path (i.e. when the magnetic circuit is open). In this case, the flux is not constrained to flow in a path that is a priori well defined, because the laminations that complete the flux path have been removed.  
-problem. The E-core was chosen to have a 0.5” thick center leg, 0.25” thick outer legs, and a slot + 
-depth of 0.75”. The material for the core is linear with a relative permeability of 2500. The coil +The geometry was chosen arbitrarily, the purpose here being more the procedure than the actual problem. The E-core was chosen to have a 0.5” thick center leg, 0.25” thick outer legs, and a slot depth of 0.75”. The material for the core is linear with a relative permeability of 2500. The coil carries a bulk current density of 2 MA/m2. The input geometry is picture in Figure A.8. 
-carries a bulk current density of 2 MA/m2. The input geometry is picture in Figure A.8. + 
-In Figure A.8, the core is placed within a circular region, and a second circular region is drawn +In Figure A.8, the core is placed within a circular region, and a second circular region is drawn next to the region containing the core. Periodic boundary conditions are applied to the arcs that define the boundaries as shown in Figure A.8. The way that periodic boundary conditions are implemented in FEMM, each periodic boundary condition defined for the problem is to be applied to two and only two corresponding entities. In this case, each boundary circle is composed of two arcs, so two periodic boundary conditions must be defined to link together each arc with in the domain with the core to its corresponding arc in the domain representing the exterior region. 
-next to the region containing the core. Periodic boundary conditions are applied to the arcs that + 
-define the boundaries as shown in Figure A.8. The way that periodic boundary conditions are +Also notice that a point has been drawn in the center of the exterior region. A point property has been applied to this point that specifies that //A// = 0 at this reference point. The center of the circle maps to infinity in the analogous open problem, so it makes sense to define, in effect, //A// = 0 at infinity. If no reference point is defined, it is fairly easy to see that the solution is only unique to within a constant. The situation is analogous to a situation where Neumann boundary conditions have been defined on all boundaries, resulting in a non-unique solution for A. Due to the type of solver that FEMM employs, the problem can most likely be solved even if a reference point is not defined. However, defining a reference point eliminates the possibility of numerical difficulties due to uniqueness issues.
-implemented in FEMM, each periodic boundary condition defined for the problem is to be applied +
-to two and only two corresponding entities. In this case, each boundary circle is composed of two +
-arcs, so two periodic boundary conditions must be defined to link together each arc with in the +
-domain with the core to its corresponding arc in the domain representing the exterior region. +
-Also notice that a point has been drawn in the center of the exterior region. A point property +
-has been applied to this point that specifies that A = 0 at this reference point. The center of the +
-circle maps to infinity in the analogous open problem, so it makes sense to define, in effect, A = 0+
  
 {{fig_a-8.png}} {{fig_a-8.png}}
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 Figure A.8: Example input geometry. Figure A.8: Example input geometry.
  
-at infinity. If no reference point is defined, it is fairly easy to see that the solution is only unique to + 
-within a constant. The situation is analogous to a situation where Neumann boundary conditions +The resulting solution is shown in Figure A.9. As is the intention, the flux lines appear to cross out of the of the region containing the core as if unaffected by the presence of the boundary. The flux lines reappear in the domain representing the exterior region, completing their flux paths through the exterior region.
-have been defined on all boundaries, resulting in a non-unique solution for A. Due to the type of +
-solver that FEMM employs, the problem can most likely be solved even if a reference point is not +
-defined. However, defining a reference point eliminates the possibility of numerical difficulties +
-due to uniqueness issues. +
-The resulting solution is shown in Figure A.9. As is the intention, the flux lines appear to +
-cross out of the of the region containing the core as if unaffected by the presence of the boundary. +
-The flux lines reappear in the domain representing the exterior region, completing their flux paths +
-through the exterior region.+
  
  
 ==== A.4 Nonlinear Time Harmonic Formulation ==== ==== A.4 Nonlinear Time Harmonic Formulation ====
  
-Starting with the the 3.3 version of FEMM, the program includes a “nonlinear time harmonic” +Starting with the the 3.3 version of FEMM, the program includes a “nonlinear time harmonic” solver. In general, the notion of a “nonlinear time harmonic” analysis is something of a kludge. To obtain a purely sinusoidal response when a system is driven with a sinusoidal input, the system must, by definition, be linear. The nonlinear time harmonic analysis seeks to include the effects of nonlinearities like saturation and hysteresis on the fundamental of the response, while ignoring higher harmonic content. This is a notion similar to “describing function analysis,” a widely used tool in the analysis of nonlinear control systems. There are several subtly different variations of the formulation that can yield slightly different results, so documentation of what has actually been implement is important to the correct interpretation of the results from this solver. 
-solver. In general, the notion of a “nonlinear time harmonic” analysis is something of a kludge. + 
-To obtain a purely sinusoidal response when a system is driven with a sinusoidal input, the system +An excellent description of this formulation is contained in [21]. FEMM formulates the nonlin-ear time harmonic problem as described in this paper. Similar to Jack and Mecrow, FEMM derives an apparent BH curve by taking H to be the sinusoidally varying quantity. The amplitude of B is obtained by taking the first coefficient in a Fourier series representation of the resulting B. For the purposes of this Fourier series computation, FEMM interpolates linearly between the user-defined points on the BH curve to get a set of points with the same H values as the input set, but with an adjusted B level. The rationale for choosing H to be the sinusoidal quantity (rather than B) is that choosing B to be sinusoidal shrinks the defined BH curve–the B values stay fixed while the H values become smaller. It then becomes hard to define a BH curve that does not get interpolated. In contrast, with H sinusoidal, the B points are typically larger than the DC flux density levels, creating a curve with an expanded range.
-must, by definition, be linear. The nonlinear time harmonic analysis seeks to include the effects +
-of nonlinearities like saturation and hysteresis on the fundamental of the response, while ignoring +
-higher harmonic content. This is a notion similar to “describing function analysis,” a widely used +
-tool in the analysis of nonlinear control systems. There are several subtly different variations of +
-the formulation that can yield slightly different results, so documentation of what has actually been +
-implement is important to the correct interpretation of the results from this solver. +
-An excellent description of this formulation is contained in [21]. FEMMformulates the nonlin-+
  
 {{fig_a-9.png}} {{fig_a-9.png}}
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 Figure A.9: Solved problem. Figure A.9: Solved problem.
  
-ear time harmonic problem as described in this paper. Similar to Jack and Mecrow, FEMM derives +A “nonlinear hysteresis lag” parameter is then applied to the effective BH curve. The lag is assumed to be proportional to the permeability, which gives a hysteresis loss that is always proportional to |B|<sup>2</sup>. This form was suggested by O’Kelly [22]. It has been suggested that that the Steinmetz equation could be used to specify hysteresis lag, but the Steinmetz equation is badly behaved at low flux levels (i.e. one can’t solve for a hysteresis lag that produces the Steinmetz |B|<sup>1.6</sup> form for the loss as B goes to zero.)
-an apparent BH curve by taking H to be the sinusoidally varying quantity. The amplitude of B is +
-obtained by taking the first coefficient in a Fourier series representation of the resulting B. For the +
-purposes of this Fourier series computation, FEMM interpolates linearly between the user-defined +
-points on the BH curve to get a set of points with the same H values as the input set, but with +
-an adjusted B level. The rationale for choosing H to be the sinusoidal quantity (rather than B) is +
-that choosing B to be sinusoidal shrinks the defined BH curve–the B values stay fixed while the H +
-values become smaller. It then becomes hard to define a BH curve that does not get interpolated. +
-In contrast, with H sinusoidal, the B points are typically larger than the DC flux density levels, +
-creating a curve with an expanded range. +
- +
-A “nonlinear hysteresis lag” parameter is then applied to the effective BH curve. The lag +
-is assumed to be proportional to the permeability, which gives a hysteresis loss that is always +
-proportional to |B|2. This form was suggested by O’Kelly [22]. It has been suggested that that +
-the Steinmetz equation could be used to specify hysteresis lag, but the Steinmetz equation is badly +
-behaved at low flux levels (i.e. one can’t solve for a hysteresis lag that produces the Steinmetz +
-|B|1.6 form for the loss as B goes to zero.)+
  
-For nonlinear in-plane laminations, an additional step is taken to obtain an effective BH curve +For nonlinear in-plane laminations, an additional step is taken to obtain an effective BH curve that also includes eddy current effects. At each H level on the user-defined BH curve, a 1D nonlinear time harmonic finite element problem is solved to obtain the total flux that flows in the lamination as a function of the H applied at the edge of the lamination. Then dividing by the lamination thickness and accounting for fill factor, and effective B that takes into account saturation, hysteresis, and eddy currents in the lamination is obtained for each H.
-that also includes eddy current effects. At each H level on the user-defined BH curve, a 1D nonlinear +
-time harmonic finite element problem is solved to obtain the total flux that flows in the +
-lamination as a function of the H applied at the edge of the lamination. Then dividing by the lamination +
-thickness and accounting for fill factor, and effective B that takes into account saturation, +
-hysteresis, and eddy currents in the lamination is obtained for each H.+
users_manual.1603223292.txt.gz · Last modified: 2020/10/20 21:48 by stanzurek

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