III-2.4. Comparison Between Sinusoidal and Trapezoidal Waveforms 411 References [1] M.R.Dubois,“Review of Electromechanical ConversioninWindTurbines”,Report EPP00.R03, TU Delft, ITS Faculty, The Netherlands, April 2000. [2] M.R. Dubois, H. Polinder, J.A. Ferreira, “Comparison of Generator Topologies for Direct- Drive Wind Turbines”, Proceedings of the Nordic Countries Power and Industrial Electronics Conference (NORPIE), Denmark, 2000, p. 22–26. [3] J.F.Gieras, M. Wing, Permanent Magnet Motor Technology, 2nd edition, Revisedand Expanded Marcel Dekker Inc., New York. [4] J.R. Henderson Jr., T.J.E. Miller, Design of Brushless Permanent-Magnet Motors, Oxford: Magna Physics Publishing and Calderon Press, 1994. III-3.1. EQUIVALENT THERMAL CONDUCTIVITY OF INSULATING MATERIALS FOR HIGH VOLTAGE BARS IN SLOTS OF ELECTRICAL MACHINES P.G. Pereirinha 1,2 and Carlos Lemos Antunes 1 1 ISR-Lab. CAD/CAE, University of Coimbra, 3030-290 Coimbra, Portugal ppereiri@isr.uc.pt 2 Inst. Sup. Engenharia de Coimbra. Rua Pedro Nunes, 3030-199 Coimbra, Portugal lemos.antunes@deec.uc.pt Abstract. The equivalent thermal conductivity of insulating materials for a high voltage bar in slots of electrical machines is calculated using the finite element method. This allows the use of much coarser meshes with an equivalent thermal conductivity k e , without accuracy loss in the hot spot temperature calculation. It is shown the dependency of k e value on the equivalent mesh used. Some considerations are also presented on the heat flux finite element calculation. Introduction One of the major thermal problems in electrical machines is the steady-state hot spot temperatures in the windings, which are responsible for its thermal aging and degrada- tion. So it is important to correctly determine those hot spots in the thermal design of the machine. In the thermal finite element (FE) modeling of electrical machines, all the different ma- terials to be crossed by the heat flux should be considered, namely the windings insulations. For a bundle of conductors a simple explicit formula for the thermal conductivity [1] and a statistical approach for the temperature calculation [2] were presented elsewhere. An- other study was presented [3] for a voice coil loudspeaker motor in which the real coil was replaced by an equivalent bulk coil. A different problem arises in electrical machine slots with high voltage bars containing several different narrow insulation materials. Despite the amazing development of the com- puter and FE software capabilities it is still a problem to model the different materials when analyzing the thermal problem for a full machine due to the different dimensions involved and the consequent huge number of nodes required for the mesh discretization. The aim of this paper is to present a method to replace the several insulation layers by a single layer with equivalent thermal conductivity allowing the use of much less number of FE elements without significant accuracy loss, namely in the bar maximum temperatures. S. Wiak, M. Dems, K. Kom ˛ eza (eds.), Recent Developments of Electrical Drives, 413–422. C 2006 Springer. 414 Pereirinha and Antunes Formulation The classic heat diffusion model is [4,5] ρc dT dt +∇(−k∇T) = q (1) where ρ is the density [kg/m 3 ], c the specific heat capacity (also called specific heat) [J/kgK], T thetemperature [K], ∇ the nabla differential operator, k the thermal conductivity [W/(Km)], and q the thermal sources [W/m 3 ]. The boundary conditions depend on the problem type. The heat flow due to conduction is given by the Fourier’s law, F =−k∇T (2) where F is the heat flux vector [W/m 2 ], and the heat flux φ h [W] crossing a surface, closed or not, is then given by φ h = F ˆ ndS (3) where ˆ n is the unit outer normal vector to the boundary. In a Cartesian coordinate system (3) becomes φ h = yz F x dydz + xz F y dxdz+ xy F z dxdz (4) where F x , F y , and F z are the components of the heat flux density vector F in the x, y, and z directions respectively. The case of the steady-state heat transfer problem, is described by the following partial differential equation [4,5]: ∇(−k∇T) = q (5) where the thermal sources q for the present problem are only the Joule losses in the bar copper which can be given by q = ρ 0 J 2 (6) where ρ 0 is the electric resistivity [m] at a reference temperature T 0 [K] and J the current density [A/m 2 ], or by q = ρ 0 (1 + α(T − T 0 ))J 2 (7) where α is the linear expansion coefficient [K −1 ], if it is necessary to consider the resistivity variation with the temperature rise [6]. The thermal problem was solved using first order FE thermal processor [6] of our finite element package CADdyMAG. Case study and methodology As the case study it was considered the bar presented in Fig. 1(a), of a three phase high voltage synchronous generator (1 MVA, 6.3 kV) driven by a hydraulic turbine of 250 III-3.1. Equivalent Thermal Conductivity of Insulating Materials 415 Figure 1. (a) Bar with 9 × 2 wires; (b) original FE mesh (2,035 nodes/3,920 elements) for 1/4 of the bar in a slot. rpm. Each side of the winding (bar) consists of 9 × 2 copper wires (“1” in Fig. 1a) each one individually insulated with 0.2 mm layers of paper and cotton (“2” and “3” in Fig. 1a) packed together with a 0.4 mm “bitumen” layer and a final 2.5 mm layer of molded micanite (“4” and “5” in the same figure). Finally a 0.25 mm impregnation resin layer between the bar and the slot was also considered (“6” in Fig. 1a). As in each slot there are two bars with an additional spacer between them and supposing the analysis of half machine with 48 slots, this would lead to nearly 400,000 nodes only to model the stator slots in a 2D problem. The idea is thento replace the different insulator materials by only onebulk materialwith a considered equivalent thermal conductivity k and significantly lower number of elements and nodes of the corresponding finite element mesh. Due to symmetry it can be analyzed only 1/4 of the bar, and it was used for the original bar a mesh with 2,055 nodes and 3,920 first order finite elements, as shown in Fig. 1(b). Tothe nominal currentof 91.64 A, flowing in two wires in parallel, correspondsacurrent density J = 3.916 A/mm 2 . In the original bar the insulators thermal conductivities vary from 0.17 to 0.7 W/(Km), and to solve the original bar problemit was considered a boundary condition corresponding to a temperature in the slot border T ref =373.15 K (100 ◦ C). It was used (5) for the thermal sources. The steady-state thermal conduction problem was solved using first order FEthermal processor of ourFE packageCADdyMAG [6]. The temperature results obtained for the original bar are presented in Fig. 2(a) and in Fig. 2(b) the heat flux density vector distribution can be seen. The hot spot temperature rise for the bar T ref = T max − T ref (8) was T ref = 10.103 K. 416 Pereirinha and Antunes Figure 2. Original bar temperature; and (b) heat flux density vector F distribution. To easily calculate the heat flux leaving the bar one can see that, for a 2D problem in the xy plane, like the presented in Figs. 1–5, F z in (4) is zero. So (4) can be simplified to φ h l = y F x dy + x F y dx =− y k ∂T ∂x dy − x k ∂T ∂y dx (9) where l is the bar length in the z direction. The thermal flux φ h /l [W/m] leaving the cable crossing lines 1, 2, or 3 (red lines in Fig. 3), as the integration path only crosses one material with constant thermal conductivity, Figure 3. (a) Integration lines for the bar; (b) relative positions of lines 1 to 3 to the mesh elements. III-3.1. Equivalent Thermal Conductivity of Insulating Materials 417 is given by φ h l = y=a F x dy + x=b F y dx = k − y=a ∂T ∂x dy − x=a ∂T ∂y dx (10) As the lines chosen for the integral calculation are parallel to the Cartesian axes, it is interesting to note that for the vertical part of the integration lines (line “a” in Fig. 3a) only F x is to be considered and for the horizontal part of the integration lines (line “b” in Fig. 3a) only F y is to be considered. As the lines chosen for the integral calculation are parallel to the Cartesian axes, it is interesting to note that for the vertical part of the integration lines (line “a” in Fig. 3a) only F x is to be considered and for the horizontal part of the integration lines (line “b” in Fig. 3a) only F y is to be considered. The thermal sources q [W/m] for 1/4 of the bar are q = 13.91942 W/m. The heat flux crossing lines 1, 2, and 3 in Fig. 3 was calculated by using (10) for the original bar and it was confir med that it is equal to the thermal sources q (with an error of 0.839%, 0.256%, and −0,083%, respectively). So it was checked that (5) is verified and the finite element solution is validated. As a methodology we have chosen to replace the original bar by one composed by copper, where the current losses are produced, surrounded by a b ulk insulation material. Two different approaches were considered: first, replace the original bar by an equivalent one with the same copper distribution (“Equal,” Fig. 4) and second, replace it by a bulk bar with all the copper area concentrated (“Conc.,” Fig. 5) in only one bigger wire. The idea of keeping the same copper area as in the original bar is to use the same current density J values for both the thermal and magnetic problems, although other possibilities may be considered. For both cases, the five different insulating materials were replaced by only one equivalent material. Two different meshes for each case were also considered: fine meshes (“Fine”) and coarse meshes (“Coarse”). For the models with the same copper Figure 4. Models with same copper distribution, “Equal”: (a) “Fine” mesh (2,035 nodes/3,920 elements); (b) “Coarse” mesh (33 nodes/49 elements). 418 Pereirinha and Antunes Figure 5. Models with different copper distribution, concentrated “Conc.”: (a) “Fine” mesh (1,089 nodes/2,048 elements); (b) “Coarse” mesh (nine nodes/eight elements). distribution, “Equal,” thefine mesh (Fig. 4a) hasthe same geometry as theoriginal bar mesh (Fig. 1b), and the coarse mesh (Fig. 4b) has the minimum number of nodes and elements required for the analysis. For the models with concentrated copper, “Conc.,” the fine mesh (Fig. 5a) has about half of the nodes of the original bar (Fig. 1b), and the coarse mesh (Fig. 5b) has also the minimum number of nodes and elements required for the analysis. As in the proposed methodology the heat does not have to cross the whole bar from side to side (what would probably lead to the consideration of two different thermal con- ductivities, one for the horizontal and another for the vertical directions of the bar), but instead it is generated inside the bar, the equivalent thermal conductivity was considered isotropic. The issue here is how to calculate the value of the equivalent thermal conductivity. To calculate this, the steady-state thermal problem for the four cases mentioned before (con- centrated and equal copper distribution, for both a very coarse and a fine mesh) were solved with thermal conductivities k ranging from 0.2 to 0.6 W/(Km) and the hot spot temperature rise were calculated (Fig. 6). An equivalent thermal conductivity k e was calculated as will be seen in more detail in the next section “Results and Validation.” Results and validation The steady-state thermal problem (5) for the four equivalent meshes in Figs. 4 and 5 was solved for several thermal conductivities ranging from 0.2 to 0.6 W/(Km), as in the bar the real insulators thermal conductivities vary from 0.17 to 0.7 W/(Km). The hot spot temperature raise T is given by T = T max − T ref (11) where T max is the maximum temperature for each mesh, and is presented in Fig. 6 with thick color lines. III-3.1. Equivalent Thermal Conductivity of Insulating Materials 419 T = 2.57098 k –0.99889 T = 2.40877 k –0.99861 T = 3.03175 k –0.99778 T = 3.47952 k –0.99875 4 6 8 10 12 14 16 18 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 k [W//K.m] T [k] or[ ° C] Equal / Coarse Equal / Fine Conc. / Coarse Conc. / Fine Equiv. Bar // Mesh Figure 6. Hot spot temperature rise as a function of thermal conductivity for the different meshes. It is seen that the curves presented can be perfectly fitted by power functions (black thin lines) in the form T = ak −b (12) where“a” and “b” are coefficients given inFig. 6 using the “trendline” function of Microsoft Excel c . Toobtain the samehotspot rise ofthe original bar andmesh, T ref ,theequivalentthermal conductivity k e for each particular FE mesh (as the temperature solution will depend on the mesh) can then be very easily calculated by k e = T ref a −1/b (13) The values of the “a” and “b” coefficients as well as the resulting k e are presented in Table 1. At this point, it should be mentioned that for the thermal sources, it should not be used the more accurate expression (7) which accommodates the resistivity variation with the temperature. Indeed, although (7) is the expression that should be used to solve the global thermal problem, for the particular case presented in this communication, it is mandatory to use (6). The reason is that if (7) is used the thermal sources will not remain constant and Table 1. Coefficients and equivalent conductivities k e Bar/mesh “a” coefficient “b” coefficient k e , W/(Km) Equal/coarse 2.40877 0.99861 0.237946 Equal/fine 2.57098 0.99889 0.25409 Conc./coarse 3.03175 0.99778 0.299282 Conc./fine 3.47952 0.99875 0.343945 420 Pereirinha and Antunes Figure 7. Temperature distribution for equal copper distribution, “Equal”: (a) Fine mesh (2,035 nodes/3,920 elements), k e = 0.254090; (b) Coarse mesh (33 nodes/49 elements), k e = 0.237946. consequently the results of the hot spot temperature rise T would not be only a function of the thermal conductivity (as are those presented in Fig. 6) but also of the cur rent density and of the boundary condition T ref . Using the equivalent k e calculated by (13) and presented in Table 1, the thermal con- duction problem was solved and the results for the fine and coarse meshes are presented in Figs. 7 and 8, for the “Equal” and “Conc.” bars respectively. Comparing these results with the reference ones in Fig. 2(a) it can be seen that the reference hot spot temperature, T ref = 383.253 K (T ref = 10.103 K), is obtained with the very small errors presented in Table 2. Figure 8. Temperature distribution for concentrated copper distribution, “Conc.”: (a) Fine mesh (1,089 nodes/2,048 elements), k e =0.343945; (b) Coarse mesh (9 nodes/8 elements), k e =0.299282. III-3.1. Equivalent Thermal Conductivity of Insulating Materials 421 Table 2. Errors in hot spot temperature Bar/mesh Nodes/elements k e , W/(Km) T(K) Error (%) Equal/coarse 33/49 0.237946 10.1031 0.0010 Equal/fine 2,035/3,920 0.254090 10.1024 −0.0059 Conc./coarse 9/8 0.299282 10.1014 −0.0158 Conc./fine 1,089/2,048 0.343945 10.1020 −0.0099 Table 3. Errors in hot spot temperature for slightly less coarse meshes Bar/mesh Nodes/elements k e , W/(Km) T (K) Error (%) Equal/coarse2 35/53 0.241042 10.1031 0.0010 Conc./coarse2 10/10 0.319620 10.1017 −0.0129 It can also be seen that for the equivalent concentrated bar “Conc.” the temperature distribution as well as the average copper temperature has some differences to the original bar. However, for the “Equal” mesh (Figs. 4b and 7b), which has 33 nodes, i.e. 62 times less nodes than the original one (Figs. 1b and 2a, 2,035 nodes), a very similar temperature distribution is obtained. The temperature distribution in the coarse meshes can be further improved by simply adding two or one more nodes in the line from the upper right corner of the copper to the upper right corner of the FE mesh in Figs. 4(b) and 5(b), respectively, as presented in Fig. 9, where the thermal solution is plotted along with the new meshes “Coarse2.” The k e for these two new meshes are presented in Table 3, with the resulting hot spot temperature rise and the correspondent errors. Figure 9. Temperature distribution for concentrated and equal copper distribution, with slightly less coarse meshes “Coarse 2”: (a) “Conc.” (10 nodes/10 elements), k e = 0.319620; (b) “Equal” (35 nodes /53 elements), k e = 0.241042. . use of much less number of FE elements without significant accuracy loss, namely in the bar maximum temperatures. S. Wiak, M. Dems, K. Kom ˛ eza (eds.), Recent Developments of Electrical Drives, . INSULATING MATERIALS FOR HIGH VOLTAGE BARS IN SLOTS OF ELECTRICAL MACHINES P.G. Pereirinha 1,2 and Carlos Lemos Antunes 1 1 ISR-Lab. CAD/CAE, University of Coimbra, 303 0-2 90 Coimbra, Portugal ppereiri@isr.uc.pt 2 Inst Rua Pedro Nunes, 303 0-1 99 Coimbra, Portugal lemos.antunes@deec.uc.pt Abstract. The equivalent thermal conductivity of insulating materials for a high voltage bar in slots of electrical machines