Recent Developments of Electrical Drives - Part 42 ppsx

This allows the use of much coarser meshes with an equivalent thermal conductivity k e, without accuracy loss in the hot spot temperature calculation.. The case of the steady-state heat

III-2.4 Comparison Between Sinusoidal and Trapezoidal Waveforms 411

References

[1] M.R Dubois, “Review of Electromechanical Conversion in Wind Turbines”, 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, Revised and 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 Pereirinha1,2and Carlos Lemos Antunes1

1ISR-Lab CAD/CAE, University of Coimbra, 3030-290 Coimbra, Portugal

ppereiri@isr.uc.pt

2Inst 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 evalue 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.

2006 Springer.

414 Pereirinha and Antunes

Formulation

The classic heat diffusion model is [4,5]

ρc d T

where ρ is the density [kg/m3], c the specific heat capacity (also called specific heat) [J/kgK], T the temperature [K], ∇ the nabla differential operator, k the thermal conductivity [W/(Km)], and q the thermal sources [W/m3]

The boundary conditions depend on the problem type The heat flow due to conduction

is given by the Fourier’s law,



where F is the heat flux vector [W/m2], and the heat fluxφ h[W] crossing a surface, closed

or not, is then given by

φ h =





where ˆn is the unit outer normal vector to the boundary In a Cartesian coordinate system

(3) becomes

φ h=



yz

F x d yd z+



x z

F y d xd z+



x y

where F x , F y , and F zare 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]:

where the thermal sources q for the present problem are only the Joule losses in the bar

copper which can be given by

whereρ0is the electric resistivity [m] at a reference temperature T0[K] and J the current

density [A/m2], or by

whereα 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

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 then to replace the different insulator materials by only one bulk material with

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)

To the nominal current of 91.64 A, flowing in two wires in parallel, corresponds a current

density J = 3.916 A/mm2 In the original bar the insulators thermal conductivities vary from 0.17 to 0.7 W/(Km), and to solve the original bar problem it 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 FE thermal processor of our FE package CADdyMAG [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

wasT = 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 zin (4) is zero So (4) can be simplified to

φ h

l =



y

F x d y+



x

F y d x= −



y

k ∂T

∂x d y−



x

k ∂T

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.

is given by

φ h

l =



y =a F x d y+



x =b F y d x = k



−



y =a

∂T

∂x d y−



x =a

∂T

∂y d x



(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 yis 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 yis 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 confirmed 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 bulk 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,” the fine mesh (Fig 4a) has the same geometry as the original 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 ewas 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 raiseT is given by

where T maxis the maximum temperature for each mesh, and is presented in Fig 6 with thick color lines

T = 2.57098k –0.99889

4

6

8

10

12

14

16

18

k [W//K.m]

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

where “a” and “b” are coefficients given in Fig 6 using the “trendline” function of Microsoft

Excelc.

To obtain the same hot spot rise of the original bar and mesh,T ref, the equivalent thermal

conductivity k efor 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)

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 riseT would not be only a function

of the thermal conductivity (as are those presented in Fig 6) but also of the current density

and of the boundary condition T ref

Using the equivalent k ecalculated 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 = 0.343945; (b) Coarse mesh (9 nodes/8 elements), k = 0.299282

Table 2 Errors in hot spot temperature

Table 3 Errors in hot spot temperature for slightly less coarse meshes

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 efor 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 = 0.241042