Development of Thermal Models for PM Tractions Motors

Development of Thermal Models forPermanent-Magnet Traction Motors

BO YANG

Master of Science Thesis in Electrical Machines and Power Electronicsat the School of Electrical Engineering

Royal Institute of TechnologyStockholm, Sweden, July 2009

Supervisor: Juliette Soulard, Florence Meier

XR-EE-EME 2009:005

AbstractThis thesis deals with the thermal behavior of permanent magnet trac-tion motors. Finite element (FE) and lump parameter (LP) thermalmodels are developed. Temperature distribution in the motor is calcu-lated both by analytical methods and FE simulations

The first part of the thesis consists of a literature study on principles ofheat transfer in electrical machines. Convection heat transfer is furtherdiscussed in the following part. After that, 2D-FEM thermal simula-tions are carried out using FLUX. Temperatures at different conditionsare compared. Since it is of great importance to predict the temperatureof permanent magnets, a LP thermal model of the rotor is developed.The temperatures given by the LP model are verified by FE simulations.

Keywords: PM Motor, Heat Transfer, Finite Element, LumpedParameter Thermal Model

Acknowledgment

This master thesis was carried out at the Department of Electrical Machines andPower Electronics (EME), KTH. This project was performed in cooperation withBombardier Transportation AB, Sweden.

First of all, I would like to express my deepest appreciation to my supervisorsJuliette Soulard and Florence Meier for all their assistance during the whole work.A special thank to my room mate, Andreas Kings, for his help with softwares.

I would like to thank all the personal of EME of helping me in any way. It hasbeen a pleasure to work here.

Finally, I would like to thank my parents and my girl friend, for their endlesslove.

Yang Bo

2009-07

Contents

Contents

1 Introduction 1

2 Literature study 32.1 Principle of conduction and convection heat transfer . . . . . . . . . 3

2.1.1 Conduction heat transfer . . . . . . . . . . . . . . . . . . . . 32.1.2 Convection heat transfer . . . . . . . . . . . . . . . . . . . . . 4

2.2 Lumped parameter (LP) thermal model . . . . . . . . . . . . . . . . 5

3 Modelling of convection heat transfer 73.1 Useful definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Convection in airgap . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.3 Convection in air ducts . . . . . . . . . . . . . . . . . . . . . . . . . 113.4 Correlation of natural convection . . . . . . . . . . . . . . . . . . . . 13

4 FEM thermal models and simulations 154.1 FEM thermal models in FLUX . . . . . . . . . . . . . . . . . . . . . 15

4.1.1 Face and line regions in FLUX thermal application . . . . . . 154.1.2 Conduction and convection heat transfer in FLUX thermal

model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.2 Losses at different conditions . . . . . . . . . . . . . . . . . . . . . . 164.3 FEM thermal simulations . . . . . . . . . . . . . . . . . . . . . . . . 18

4.3.1 Thermal simulation at normal condition . . . . . . . . . . . . 204.3.2 FEM thermal simulation at turn-to-turn fault condition . . . 224.3.3 FEM thermal simulation at three-phase short-circuit condition 244.3.4 Comparison of different fault conductions . . . . . . . . . . . 26

4.4 FEM thermal model of the rotor . . . . . . . . . . . . . . . . . . . . 26

5 LP thermal model for the rotor 295.1 The T-equivalent circuit . . . . . . . . . . . . . . . . . . . . . . . . . 295.2 Geometry of the LP thermal model . . . . . . . . . . . . . . . . . . . 325.3 Thermal resistances . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.4 Verification of the rotor LP thermal model . . . . . . . . . . . . . . . 34

5.4.1 Verification in Case 1 . . . . . . . . . . . . . . . . . . . . . . 355.4.2 Verification in Case 2 . . . . . . . . . . . . . . . . . . . . . . 355.4.3 Verification in Case 3 . . . . . . . . . . . . . . . . . . . . . . 38

5.5 Discussion and conclusion . . . . . . . . . . . . . . . . . . . . . . . . 39

6 Conclusions and future works 41

Bibliography 43

List of symbols 45

A Fan characteristics 47

B Rotor eddy current losses and magnet losses 49

C Calculation of thermal resistances 51C.1 Rotor dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51C.2 Thermal resistances . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

Chapter 1

Introduction

Thermal analysis is of equal importance as electromagnetic analysis in electricalmachines. The life of insulation materials is strongly temperature dependent [1]. InPM machines, the coercivity and remanence of permanent magnets decrease withan increasing temperature [2]. This thesis focuses on the thermal behavior of PMtraction motors.

The motor investigated in this thesis is a four-pole V-shaped interior permanentmachine. The radial cross-section of the motor is shown in Fig 1.1. Air ducts areadopted to cool the motor effectively. Heat transfer in electrical machines is in facta three-dimensional problem. The focus of this work is on the radial cross-section,which means the three-dimensional problem is simplified as two-dimensional.

Figure 1.1. Radial cross-section of the motor.

1

CHAPTER 1. INTRODUCTION

Thermal modelling and analysis can be classified into two categories: lumped pa-rameter (LP) thermal model and numerical method. In this thesis, 2D finite elementanalysis is performed and a LP thermal model for the motor’s rotor is developed.The following softwares are used:

- FLUX

FLUX is a finite element software for electromagnetic and thermal simula-tions, which is particularly suited to electric motor analysis [3].

- PORTUNUS

PORTUNUS is a system simulation software. Besides the simulation for drivesystems, its application also includes the thermal analysis of complex systems[4]. The rotor LP thermal network is solved by PORTUNUS in this work.

- Motor-CAD

MOTOR-CAD is aimed to simulate the heat transfer in electrical machinesusing the LP thermal network [5]. It also provides the correlations of con-vection heat transfer coefficients. However, Motor-CAD needs complete anddetailed information about the geometry and materials of the motor. In thiswork, only the correlations are used.

Outline of the thesis• Chapter 2 gives a literature review of the principle of conduction and convec-

tion heat transfer. The LP thermal network is then introduced.

• Chapter 3 focuses on the modelling of the convection heat transfer in electricalmotors. Correlations of convection heat transfer coefficients for natural andforced convection are introduced.

• Chapter 4 describes the FEM thermal model in FLUX. Thermal simulationsat different conditions are carried out and the results are analyzed.

• Chapter 5 covers the development of the LP thermal model for the motorrotor, aiming to predict the temperature of the permanent magnets.

• Chapter 6 summarizes the conclusions and provides some suggestions for pos-sible future work.

2

Chapter 2

Literature study

In this chapter, the principles of conduction and convection heat transfer are in-troduced. Also the analogy between thermal and electrical circuit is presented. Atlast, the lumped parameter thermal model is described.

2.1 Principle of conduction and convection heat transfer

2.1.1 Conduction heat transferConduction heat transfer is due to a temperature gradient in a solid material [6].By means of molecular thermal activity within the material, energy is transportedwithout motion of the material. The basic relation of conduction heat transfer isdefined by Fourier’s law as shown in Fig 2.1 and Equ (2.1): [6],

AT1

T2

Q

∆T=T1-T2

∆x

Figure 2.1. Conduction heat transfer.

Q = −λA∆T∆x (2.1)

where ∆T is the temperature difference in ◦C, Q is the heat flow in W , λ is thethermal conductivity in W/(m◦C), A represents the cross-sectional area in m2 and

3

CHAPTER 2. LITERATURE STUDY

∆x is the distance along the heat flow in m.

The unknown in system with conduction heat transfer is usually the temperaturedistribution. This can be determined by solving the "heat diffusion equation" [6]with boundary conditions. The equation is derived from Fourier’s law and theconservation of energy [7]:

∇ · (∇λT ) + qg = ρcpdT

dt(2.2)

where qg represents the heat generation rate inW/m3, ρ is the density in kg/m3 andcp is the specific heat in J/(kg ◦C). In case of complex geometries, finite elementis a powerful method to solve the equation. At steady state, Equ (2.2) is simplifiedas:

∇ · (∇λT ) = −qg . (2.3)

In case of no heat generation i.e. qg = 0, Laplace’s equation is obtained.

∇2T = 0 . (2.4)

2.1.2 Convection heat transferConvection is the heat transfer between a solid and fluid medium due to the motionof the fluid [6]. In electrical motors, convection heat transfer happens on the motorouter surface, in the airgap and in the cooling ducts. In general, there are two typesof convection:

1. Natural convection, which is a result of the buoyancy force caused by thevariation of the medium’s density [8].

2. Forced convection. In this case, the movement of the medium is caused byan external force, such as fan or pump [8]. The external force enhances themixing of the hot and cool medium, which makes the heat exchange moreefficient.

The basic relation of convection heat transfer is described by Newton’s law:

Q = hcAs(Ts − Tf ), (2.5)

where Q is the heat transfer rate in W , hc is the convection heat transfer coefficientin W/(m2 ◦C), As is the surface area in m2, Ts and Tf are the temperatures of solidand fluid medium in ◦C, respectively. It is clear that the major difficulty in systemwith convection is to use the right value of convection heat transfer coefficient, hc.The state of the fluid flow, i.e. laminar, vortex or turbulent, has a great influenceon convection heat transfer. To correlate hc, the first step is to determine the stateof the fluid flow.

4

2.2. LUMPED PARAMETER (LP) THERMAL MODEL

In MOTOR-CAD, the "Airgap Model" gives the correlation of hc in the motor’sairgap [9]. In [10] and [11] other methods to calculate hc of airgap are given.Forced convection happens in the motor’s air cooling ducts. The "Enclosed Chan-nel Convection Correlation" is used in MOTOR-CAD [9]. Correlations of convectioncoefficients are further described in Chapter 3.

2.2 Lumped parameter (LP) thermal model

Comparing Fourier’s (Equ (2.1)) and Ohm’s law (Equ (2.6)), the electrical analogyfor heat conduction is observed.

I = ∆VR

. (2.6)

The analogies are listed in Table 2.1.

Table 2.1. Analogy between thermal and electrical circuits.

Thermal ElectricalFlow Thermal flow Q [W ] Current I [A]Potential Temperature T [◦C] Voltage V [V ]Resistance Rth = ∆T

Q [◦C/W ] R = ∆VI [V/A]

Conductivity λ [W/(m ◦C)] σ [S/m]

Because of the analogy, it is natural to use a lumped parameter thermal network,which is similar to electrical network, to predict the temperature distribution inelectrical machines [12]. In a thermal network, the parts assumed to be at an ho-mogeneous temperature are lumped together and represented as one node. Differentnodes are separated by the thermal resistances, which represent the heat transferpath [8]. The heat power sources which are like current sources in electrical circuitrepresent the losses in the machine. Then the temperatures at different nodes canbe solved. In steady state, the LP model only consists of thermal resistances andheat power sources. For transient analysis, the thermal capacitances are also in-cluded accounting for the storage of heat in different parts.

Thermal resistances are determined by the geometric parameters, thermal conduc-tivities of the materials and convection heat transfer coefficients of the fluid medium[12]. For conduction heat transfer, to calculate the thermal resistances, the first stepis to solve Laplace’s equation (Equ (2.4)) with boundary conditions. Once the tem-perature distribution is determined, the heat flow can be calculated from Fourier’slaw. The thermal resistance is the quotient of the temperature difference and heatflow.

5

CHAPTER 2. LITERATURE STUDY

The thermal resistance of a rectangular component is [1]:

Rth = ∆TQ

= l

λA, (2.7)

where l is length of the component along the heat flow in m and A the cross sec-tional area perpendicular to the heat flow in m2 and λ is the thermal conductivityin W/(m ◦C).

Thermal resistances of a cylindrical component in axial and radial directions are[1]:

Rth,ax = l

λA, (2.8)

Rth,r = ln(ro/ri)λ · 2πl , (2.9)

where ro and ri are the inner and outer radius of the cylinder, l is the length of thecylinder in axial direction, and A is the radial cross sectional area.

Calculations of thermal resistances for arc segments and trapezoidal shapes can befound in [13]. However, if the internal heat generation is not zero, the T-equivalentcircuit is used to represent the component, instead of only one thermal resistance[1], [14]. The T-equivalent circuit is discussed in Chapter 4.

For convection heat transfer, the thermal resistance is directly derived from Fourier’slaw:

Rthconv = ∆TQ

= 1hcA

. (2.10)

6

Chapter 3

Modelling of convection heat transfer

In this chapter, several useful definitions in fluid and thermal dynamics are intro-duced at first. After that, correlations of convection heat transfer coefficients inmotor airgap, air ducts and outer surface are discussed.

3.1 Useful definitionsIn analysis of convection heat transfer, several definitions in thermal and fluid dy-namics are used to determine the state of the fluid medium and correlate the heattransfer coefficients. These definitions are not extensively discussed in textbooks ofelectrical engineering. Therefore, it is necessary to introduce them at first.

- Dynamic viscosity, µ

Dynamic viscosity defines the tangential force required to move one horizon-tal plane with respect to another when they are maintained a distance apartby the fluid [15]. The effect of temperature on dynamic viscosity of gas isdescribed by Sutherland’s formula [16]:

µ = µ0T0 + C

T + C

(T

T0

)3/2, (3.1)

where µ0 is the reference viscosity in kg/(m ·s) at reference temperature, C isthe Sutherland’s constant, T and T0 are the reference and input temperaturein K, respectively.

- Reynolds number, Re

Reynolds number determines if the fluid flow is laminar, vortex or turbulent.It gives a measure of the ratio of inertial forces to viscous forces [16]:

Re = ρvl

µ, (3.2)

7

CHAPTER 3. MODELLING OF CONVECTION HEAT TRANSFER

where ρ is the density of the fluid in kg/m3, v is the velocity of the fluid inm/s, and l is the characteristic length in m.

- Taylor number, Ta

Instead of Reynolds number, Taylor number is often used to determine thestate of the fluid flow [10] [11]. There are various calculations of Taylor num-ber which are not the same. The relationship between Taylor and Reynoldsnumber is given in Equ (3.3) [9].

Ta = Re1√dc, (3.3)

where dc is the dimensionless curvature defined by:

dc = l

R, (3.4)

where R is the radius of the surface and l is the characteristic length in m.

- Nusselt number, Nu

Nusselt number is defined as the ratio of convective to conductive heat transfer[7]:

Nu = hcl

λ, (3.5)

where hc is convection heat transfer coefficient inW/(m2 ◦C). From Equ (3.5),the convection heat transfer coefficient is correlated as:

hc = Nuλ

l, (3.6)

- Prandtl number, Pr

Prandtl number defines the ratio of viscous diffusion to thermal diffusion:

Pr = cpµ

λ, (3.7)

where cp is the specific heat in J/(kg ◦C).

- Grashof number, Gr.

Together with Prandtl number, Grashof number is used to determine the fluidflow state in natural convection. It is a ratio of buoyancy force to viscous forceand calculated as [9]:

Gr = βgθρ2l3

µ2 , (3.8)

8

3.2. CONVECTION IN AIRGAP

where β is the volumetric thermal expansion coefficient in 1/K and approxi-mately equal to the reciprocal of absolute temperature. The absolute temper-ature can be calculated as an average of the ambient and surface temperature.g is the gravity acceleration in m/s2, θ is the temperature difference betweensurface and fluid flow in ◦C. In case of horizontal cylinder, the characteristiclength l equals the cylinder diameter [9].

3.2 Convection in airgapThe "Airgap Model" in MOTOR-CAD describes how the heat transfer coefficient iscorrelated in the airgap of electrical machines [9]. The method is used in this thesis.In general, the first step is to determine the state of the air flow in the airgap. TheReynolds number is calculated from Equ (3.2):

Re = ρωrrsδagµ

, (3.9)

where ω is the rotor speed in rad/s, rrs is the rotor surface radius in m and δag isthe airgap length in m. The air flow state is determined as:

laminar flow if Re < Revotvortex flow if Revot < Re < Returturbulent flow if Re > Retur

(3.10)

where

- Revot = 41√dc, the critical vortex air flow Reynolds number,

- Revot = 100√dc, the critical turbulent air flow Reynolds number,

- dc = rrot sur/δag.

Based on the air flow state, different Nusselt numbers are selected to correlate theheat transfer coefficient as shown in Equ (3.11) and (3.12).

hc ag = Nu · λair2δag, (3.11)

where λair is the thermal conductivity of air.

Nu =

2 if laminar air flow

0.212[Re(

1dc

)0.5]0.63

Pr0.27 if vortex air flow

0.386[Re(

1dc

)0.5]0.5

Pr0.27 if turbulent air flow

(3.12)

where the Prandtl number is given as a constant equal to 0.0739 in MOTOR-CAD.

9


Seen from Equ (3.1), (3.2), (3.9) and (3.11), the convection heat transfer coeffi-cient of airgap (hc ag) is a function of temperature and rotor speed, when assumingthe density and thermal conductivity of air are constant. Fig 3.1 shows variationof hc,ag with different temperatures and rotor speeds. It can be seen that hc ag

0 1000 2000 3000 4000 5000 600010

20

30

40

50

60

70

80

90

Speed [rpm]

Con

vect

ion

heat

tran

sfer

coe

ffici

ent [

W/m

2K]

T = 40 degT = 140 degT = 240 deg

turbulent

Increasingtemperature

vortex

laminar

Figure 3.1. hc ag varies with temperature and rotor speed.

increases with an increasing rotor speed and decreasing temperature. With thegiven rotor diameter and airgap thickness, the air flow changes from laminar tovortex flow when the rotor speed is above 300 rpm. When the rotor speed is above600 rpm, the air changes to turbulent flow. The convection heat transfer coefficientrises sharply when the air flow state changes from laminar to turbulent flow. Atthe rotor speed of 3000 rpm, hc,ag reduces 5 units in average when the temperaturerises every 100 ◦C.

The airgap has to be modeled as a thermal conducting region when using FLUX.The principle of equivalence is that amount of heat transfer through the airgap iskept. As a result, the thermal resistances of conduction and convection must be thesame, which leads to:

1hc agA

= δagλeqA

, (3.13)

λeq = hc agδ = Nuλair2 , (3.14)

where λeq is the equivalent thermal conductivity of the airgap.

10

3.3. CONVECTION IN AIR DUCTS

3.3 Convection in air ductsForced convection happens in the air ducts. Fig 3.2 illustrates the heat transfer inair ducts in radial and axial directions. The cooling air enters the motor from thefan. It is warmed by the heat from the motor and takes the heat out of the motor.

Stator yoke

Air channel T0TdumTend

(a)

Stator yoke

Housing

Air duct

Tdu m

Ts

Air duct wallsH

W

(b)

Figure 3.2. Geometry of the air ducts in (a) axial direction, (b) radial direction.

In axial direction, the relationship between the extracted heat, air flow rate andtemperature rise is:

q = cpρQ(Tend − T0) = cpρQ∆T, (3.15)

where q is the amount of heat taken out by the cooling air in W , cp is the specificheat in J/(kg ◦C), Q is the air flow rate in m3/s, T0 and Tend are the air enteringand exiting temperatures in ◦C, respectively.

The air flow rate at different rotor speeds is shown in App.A. Seen from (3.15),if the air flow rate Q is high enough, the temperature rise ∆T can be neglected,which means T0 ≈ Tend. Therefore, in LP thermal model, the air cooling ducts canbe modeled as a constant temperature reference [10]. On the other hand, if Q is notlarge enough, a linear temperature rise along axial direction needs to be assumed[17]. In MOTOR-CAD, the air ducts are modeled as a heat sink. In order to getthe mean temperature in the air ducts, an iterative process is needed. With thelinear temperature rise, the mean temperature (Tdum) in the air ducts is:

Tdum = 12(T0 + Tend) = T0 + 1

2∆T (3.16)

11


The air entering temperature T0 is usually 40 ◦C according to [1]. The temperaturerise ∆T is between 15 ◦C and 40 ◦C [1].

In radial direction, from Newton’s law (see Equ (2.5)):

q = hc duAs(Tdum − Ts), (3.17)

where hc du is the convection heat transfer coefficient of the cooling air inW/(m2 ◦C),As is the surface area inm2, and Ts is the surface temperature of the air ducts in ◦C.

Correlation of hc du is given by the "Enclosed Channel Convection Correlation" inMOTOR-CAD [18]. Again, the first step is to determine the state of the air flow.In this case, the Reynold number is calculated as:

Re = DhV

ν, (3.18)

where Dh is the air duct hydraulic diameter in m, V is the air flow speed in m/s,and ν is the air flow kinematic viscosity in Pa s. In MOTOR-CAD, ν equals 1.689 ·10−5 Pa s. And in case of rectangular air ducts, the hydraulic diameter is obtainedby:

Dh = 4 · AchSch

, (3.19)

where Ach and Sch are the cross sectional area and perimeter of the air duct, re-spectively.

The state of the cooling air is determined as:laminar flow if Re < 2800transition if 2800 < Re < 10000turbulent flow if Re > 10000

(3.20)

Once the air flow state is determined, hc du is calculated as shown in Equ (3.21) and(3.22).

hc du = Nu · λairDh

, (3.21)

Nu =

7.49− 17.02 · HW + 22.43(HW

)2− 9.94

(HW

)3+ 0.065(Dh/l)·Re·Pr

1+0.04[(Dh/l)·Re·Pr]2/3 if laminar air flowf8 ·

(Re−1000)Pr1+12.7·(f/8)0.5·(Pr2/3−1) if turbulent air flow

(3.22)where l is the characteristic length of the surface which equals the active lengthof the motor in this case, H and W are the height and width of the air duct,respectively, and f is the friction factor and equals:

f = [0.79 · ln(Re)− 1.64]−2 (3.23)

Within the range of the given fan characteristic (see App.A), the air flow is turbu-lent. Fig 3.3 shows the convection heat transfer coefficient as a function of rotorspeed. At 3000 rpm, hc du equals 104.4W/(m2 ◦C).

12

3.4. CORRELATION OF NATURAL CONVECTION

1000 2000 3000 4000 5000 6000

40

60

80

100

120

140

160

180

Motor speed [rpm]

heat

tran

sfer

coe

ffici

ent

[W/m

2K]

Figure 3.3. variation of hc du with rotor speed.

3.4 Correlation of natural convectionNatural convection happens at the motor outer surface. Reference [9] gives themethod of correlation of natural convection for horizontal cylinder, which is suitablefor electrical motors. The air flow state is determined by the product of Prandtl(Equ 3.7) and Grashof number (Equ 3.8), as shown in Equ (3.24).{

laminar if 104 < GrPr < 109

turbulent if 109 < GrPr < 1012 (3.24)

Natural convection heat transfer coefficient hc na is calculated as:

hc na = Nu · λairD

, (3.25)

where D is the outer diameter of the motor.

Nu ={

0.525(Gr · Pr)0.25 if laminar air flow0.129(Gr · Pr)0.33 if turbulent air flow (3.26)

Since the Grashof number is temperature dependent, for precise calculation of hc na,an iterative process is needed.

13

Chapter 4

FEM thermal models and simulations

In this chapter, the FEM thermal models in FLUX are introduced. Then the FEMthermal simulations at different conditions, i.e. normal, turn-to-turn fault andthree-phase short-circuit are discussed. The temperatures in different parts of themotor are compared. At last, the FEM model for the rotor is developed to check theeffect of the magnet insulation in order to verify the LP thermal model in Chapter5.

4.1 FEM thermal models in FLUX

4.1.1 Face and line regions in FLUX thermal applicationBesides magnetic and electric simulations, FLUX acn also be used for thermal appli-cations [3]. Face regions in FLUX are used to model a group of components whichhave the same thermal properties. There are two different types of face regions inthermal application [3]:

1. Thermal conducting region with heat thermal source, in which material andpossible thermal source need to be specified. The thermal conductivity isdefined in material’s properties.

2. Inactive region, which means this region is disregarded during simulation.

Face regions can only be used to model the conduction heat transfer.

Line regions in FLUX are used to provide the boundary conditions or to char-acterize the convection and radiation heat transfer [3]. Four different types of lineregions are defined in FLUX thermal application [3]:

1. Thermal conducting region with heat thermal source. In this type of lineregion, the thermal conductivity and thermal source need to be defined. Itis supposed to be efficient to model thin layers with this type of line region.However, the thermal conductivity is ignored during simulation (reported buggin the software).

15

CHAPTER 4. FEM THERMAL MODELS AND SIMULATIONS

2. Region with surface thermal exchanges and heat source. With this type ofline region, the convction and radiation thermal heat transfer can be modeled.The coefficients of convection and radiation transfer need to be specified.Adiabatic boundary conditions can also be realized using this type of lineregion by allying zero heat transfer coefficients.

3. Imposed temperature on the boundary. This type of line region characterizesthe boundary conditions with fixed temperatures.

4. Inactive region.

4.1.2 Conduction and convection heat transfer in FLUX thermal modelFinite element analysis is a powerful method to solve conduction heat transfer prob-lems in solid components, especially for complex geometries. In FLUX, the thermalconducting parts of the motor i.e. magnets, windings, iron, etc. are modeled as"thermal conducting face regions" in which the thermal conductivities and thermalsources are taken into account.

Convection heat transfer happens in the airgap, air ducts and outer surface of themotor. However, the convection heat transfer can not be directly modeled in FLUX.Only "line region with surface of thermal exchange" can be applied to boundariesto characterize the convection heat transfer. The solution is to use conducting faceregions with an equivalent thermal conductivities or line regions instead:

- Airgap: Thermal conducting region with an equivalent thermal conductivity(λag eq);

- Air ducts: The air-ducts walls are modeled as line regions with a heat transfercoefficient of hc du. The air ducts are modeled as "inactive face regions";

- Motor outer surface: A line region with heat transfer coefficient for naturalconvection (hc na) is applied to the motor outer surface.

Calculations of λag eq, hc du and hc na can be found in Chapter 3. Fig 4.1 shows thegeometry of the thermal model in FLUX.

4.2 Losses at different conditionsAn electromagnetic FEMmodel for the investigated motor was developed by JulietteSoulard [19]. In this model, the electrical circuit associated to the geometry modelis shown in Fig 4.2 [19]. By opening or closing the switches appropriately, the turn-to-turn and three-phase fault conditions can be simulated. A detailed descriptionof the model can be found in [19].

In case of turn-to-turn fault, the current supplied to the stator winding is 200A

16

4.2. LOSSES AT DIFFERENT CONDITIONS

Motor outer surface

Boundary condition -natural convection.

Frame, thermal conducting region,

steel.

Air-d

uct walls.

Boundary condition -forced convection.

Air ducts, inactive region.

Stator yoke, thermal conducting region with thermal source,

iron lim

itation.

Stator to

oth, thermal conducting region with thermal source,

iron lamination.

Winding, thermal conducting region with thermal source,

copper.

Coil in

sulation and spacer, th

ermal conducting region ,

mica.

Magnet, th

ermal conducting region with thermal source,

NdFeB.

Wedge, thermal conducting region , epoxy.

Rotor iro

n, thermal conducting region with thermal source,

iron lamination.

Rotor slot, th

ermal conducting region ,

epoxy.

Airgap, equivalent thermal conducting region.

Shaft.

Boundary condition -imposed temperature 100 °C or 150 °C

Symmetrical lines: adiabatic boundary conditions

Figure 4.1. Geometry of the FLUX thermal model.

17


Short-circuit turns

Phase A

Phase B

Phase C

Figure 4.2. Electrical circuit in FLUX [19].

and the rotor speed is 3000 rpm, respectively. The windings are disconnected fromthe current sources one period after the short circuit happens. Fig 4.3 and 4.4 showthe currents in the short-circuit and healthy turns in different three phases. It canbe seen that, in steady state, a small amount of current goes through the healthyturns and almost all the current goes through the short-circuit turns. Therefore thecopper losses in the faulted slots are much largerin the healthy ones. The losses atnormal conditions are also obtained from the same simulation.

Using the same model, the FE electromagnetic simulation of a three-phase short-circuit condition was carried out by Florence Meier [20]. The same current androtor speed as the case of turn-to-turn fault condition are used. Table 4.1 gives thelosses at different conditions.

In turn-to-turn fault condition, although large amount of losses is generated in theshort-circuit turns, the difference in total losses are not large compared with normalcondition. The three-phase fault gives much more losses than normal condition. Themagnet losses are not included in Table 4.1. The magnet losses are obtained from"active power" in FEM simulations. It is observed that the magnet losses are muchsmaller compared with other losses. The total losses in all magnets are 42W atnormal condition.

4.3 FEM thermal simulations

The loss powers are obtained from of finite element electromagnetic simulations andare applied to the corresponding face regions. At the rotor speed of 3000 rpm, anequivalent thermal conductivity (λag eq) of 0.12W/(m ◦C) is applied to the air gap.The heat transfer coefficient for natural convection (hnu = 5.68W/(◦Cm2)) is ap-

18

4.3. FEM THERMAL SIMULATIONS

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−1000

−500

0

500

1000

1500

2000

Time [s]

Cur

rent

[A]

I_RSC

Normal

Short−circuit

disconnected from the current source

(a)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−200

−150

−100

−50

0

50

100

150

Time [s]

Cur

rent

[A]

I_RA1I_RA2

(b)

Figure 4.3. Currents at turn-to-turn fault in phase A (a) short-circuit turns,(b) healthy turns.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−150

−100

−50

0

50

100

150

200

Time [s]

Cur

rent

[A]

I_RB1I_RB2

(a)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−150

−100

−50

0

50

100

150

200

Time [s]

Cur

rent

[A]

I_RC1I_RC2

(b)

Figure 4.4. Currents at turn-to-turn fault in (a) phase B, (b) phase C.

19


Table 4.1. Losses at different conditions.

.

Losses [W ] Copperlosses

Stator ironlosses

Rotor ironlosses

total

Normal condition 4800 1562 466 6792Turn-to-turnfault

3659 1185 307 5151

Three-phaseshort-circuit

15644 573 115 16332

plied to the motor outer surface. The heat transfer coefficient for forced convection(hdu) of 104W/(◦Cm2) is applied to the air-duct walls. The ambient tempera-ture is 40 ◦C. Two different temperatures, i.e. 100 ◦C and 150 ◦C are applied tothe shaft [20]. The mean temperature of air ducts are obtained from Equ (3.17).Temperatures at the points shown in Fig 4.5 are checked.

1

2

3

4

5

6

7

8

9

10

11

12

1 Shaft

2 Magnet (central)

3 Magnet (peripheral)

4 Rotor iron (mid)

5 Rotor iron (top)

6 Winding (bottom)

7 Winding (top)

8 Teeth (mid)

9 Teeth (top)

10 Yoke (mid)

11 Yoke (top)

12 Frame

Figure 4.5. Temperature check points.

4.3.1 Thermal simulation at normal conditionIn this case, because of the symmetry, only one quarter of the motor is simulated.As a result, the losses in Table 4.1 are divided by 4 and applied to the model. Themean temperature of the air ducts is 75 ◦C [20]. Temperatures at points shown inFig 4.5 are checked and illustrated in Fig 4.6.

It can be seen the highest temperature of 199 ◦C appears in the windings near theairgap, since they are far from the cooling ducts. The temperature drops greatlyacross the airgap. As a result, although the rotor temperatures varies greatly with

20


(a)

(b)

Figure 4.6. Temperature distribution at normal condition (a) color shadedtemperature distribution, Tshaft = 150 ◦C, (b) temperatures at check points,Tshaft = 100 ◦C and 150 ◦C.

different shaft temperatures, the stator temperature is slightly influenced. Thetemperature of the motor frame is 138 ◦C and 142 ◦C with the shaft temperatureof 100 ◦C and 150 ◦C, respectively.

With the simulation results, the amount of heat that transfers through the air-gap can be calculated. Equ(4.1) gives the heat transfer rate through a cylinder in

21


radial direction:Q = 2πLλ∆T

ln(Ro/Ri)(4.1)

where Q is the heat transfer rate in W , L is the length of the cylinder in m, λ isthe thermal conductivity in W/(m◦C), ∆T is the temperature difference in ◦C, Roand Ri are the outer and inner radius of the cylinder in m, respectively.

Fig 4.7 shows the temperature difference along the airgap. The airgap is dividedinto 200 segments in the circumferential direction. The heat transfer rate is:

qr = 4200

2πLλag eqln(Ro/Ri)

200∑i=1

∆Ti = 1179W , (4.2)

about 18.5% of the total stator losses.

0 50 100 150 20010

12

14

16

18

20

22

distance along the rotor surface [mm]

tem

pera

ture

diff

eren

ce [K

]

Figure 4.7. Temperature difference across the airgap.

4.3.2 FEM thermal simulation at turn-to-turn fault condition

In this case, the complete motor is simulated since the symmetry is broken by thefault. The rotor speed is 3000 rpm. Because of the asymmetry, the face region ofstator windings is further divided into short-circuit turns and different phases asshown in Table 4.2. Assuming all the stator losses are extracted by the cooling air,the mean temperature of 66 ◦C in the air ducts is obtained from Equ (3.17).

Temperatures at points (see Fig 4.5) near the faulted slots shown in Fig 4.8 arechecked. It can be seen from Fig 4.9 that an extremely high temperature (820 ◦C)

22


Table 4.2. Winding face regions in turn-to-turn condition.

.

Face regions Losses [W ]Short-circuitturns

3492

Phase A1 51.85Phase A2 59.62Phase B1 23.48Phase B2 23.48Phase C1 4.6Phase C2 4.6

total 3659

Fault slot 1

Fault slot 2

Area 1

Area 2

Figure 4.8. Fault areas.

would appear if a thermal steady-state would be reached without changing condi-tions in the faulted slot due to the large short-circuit current. As a result, the statoryoke and teeth near the faulted slots are exposed to a high temperature (300 ◦C).The highest frame temperature is 174 ◦C, about 30 ◦C higher than normal condi-tion. Thanks to the low thermal conductivity of the airgap, the temperatures inthe middle of the magnets do not increase very much.

23


(a)

(b)

Figure 4.9. Temperature distribution at turn-to-turn fault condition, (a) colorshaded temperature distribution, Tshaft = 150 ◦C (b) temperatures at checkpoints, Tshaft = 150 ◦C.

4.3.3 FEM thermal simulation at three-phase short-circuit condition

In this case, only one quarter of the motor is simulated and the losses in Table 4.2are divided by 4 when applied to the FEM thermal model. The mean temperaturein the air ducts is 124 ◦C from Equ (3.17). Fig 4.10 shows the temperature distri-bution.

24


(a)

(b)

Figure 4.10. Temperature distribution at three-phase fault condition (a) colorshaded temperature distribution, Tshaft = 150 ◦C, (b) temperatures at checkpoints, Tshaft = 100 ◦C and 150 ◦C.

The highest temperature of 427 ◦C would appear if a thermal steady-state would bereached without changing conditions in the windings of bottom slot. Temperaturerise in magnet is 30 ◦C comparing with normal condition. Average temperature ofrotor frame is 280 ◦C, about 140 ◦C higher than that of normal condition.

25


4.3.4 Comparison of different fault conductions

It is of interest to compare the temperature in different parts of the motor at turn-to-turn and three-phase short-circuit conditions. Generally, the turn-to-turn faultgives rise to a local hot area near the fault slots. In three-phase symmetrical fault,since the total losses are much higher than that for a turn-to-turn fault, the averagetemperatures in different parts of the motor are higher.

Table 4.3 gives the temperature comparison in different parts of the motor withthe shaft temperature of 150 ◦C. It can be seen that three-phase short circuit giveshigher temperatures in magnets, rotor iron, stator teeth, yoke and frame. Sincealmost all the copper losses are concentrated in the faulted windings when a turn-to-turn fault happens, the temperatures in the faulted windings are much higher.

Table 4.3. Temperature comparison between three-phase short-circuit and turn-to-turn fault conditions, Tshaft = 150 ◦C.

.

Temperature [◦C] Three-phase short-circuit Turn-to-turn faultArea 1 Area 2

Shaft 150 150 150Magnets central 193 166 164

peripheral 219 174 174Rotor iron mid 183 168 167

top 205 185 180Winding bottom 427 826 325

top 411 322 731Teeth mid 386 373 307

top 370 298 294Yoke mid 325 223 230

top 307 200 203Frame 282 176 179

4.4 FEM thermal model of the rotorTwo FEM thermal models for the motor rotor are build in FLUX as shown Fig4.11. Magnet insulation is included in Model 2. The material of magnet insulationis Kapton whose thermal conductivity is 0.385W/(m ◦C). The magnet insulationis modeled as thermal conducting region. The rotor iron is divided into four parts

26

4.4. FEM THERMAL MODEL OF THE ROTOR

as shown in Fig 4.11. Adiabatic boundary conditions are applied to both sides todefine the symmetry.

Rotor iron , q-axis

Rotor iron , middle

Rotor iron , d-axis

Rotor surface

PM_C

PM_P

Rotor slot

(a)

Rotor iron , q-axis

Rotor iron , middle

Rotor iron , d-axis

Rotor surface

Central magnet

Rotor slot

PM_C

PM_P

Magnet insulation

(b)

Figure 4.11. Half pole of the motor rotor, (a) Model 1 without magnet insu-lation, (b) Model 2 with magnet insulation.

In order to investigate the influence of magnet insulation, temperatures in the centerof magnets are checked at the following two conditions:

- Case 1: Temperature difference is introduced between the rotor surface (Trot suf =200 ◦C) and shaft (Tsh = 100 ◦C). The rotor iron and magnet losses are nottaken into account.

- Case 2: Temperatures at rotor surface and shaft are the same, Trot suf = Tsh =100 ◦C. Losses are introduced in the rotor iron regions and magnets. FromFEM electromagnetic simulations, the losses in magnets are quite small (seeApp. B). In order to make the temperature rise more obvious, a "fictional"large amount of losses are assigned to the magnets and rotor iron regions. Thelosses in the central and peripheral magnets are 1 and 0.5 kW , respectively.The losses input to the four rotor iron regions are 0.5 kW , respectively.

Table 4.4 shows the comparison of the magnets temperatures in these two cases.It can be seen that, if there are no losses in the magnets and rotor iron, the dif-ference is not obvious. Once the losses are introduced in the rotor iron, especiallyin the magnets, the temperatures of magnets with insulations are 8.5% higher inaverage. This is reasonable, since the heat generated in the magnets is blocked bythe insulation material whose thermal conductivity is low.

27


Table 4.4. Influence of the magnet insulation.

.

Case 1 Case 2Temperature [◦C] PMC PMP PMC PMP

Without magnet insulation 156 181 207 197With magnet insulation 159 184 230 212

difference [%] 1.88 1.63 10 7.08

28

Chapter 5

LP thermal model for the rotor

In this chapter, the T-equivalent circuit used to model the heat transfer with internalheat generation is firstly introduced. Then the lumped parameter (LP) thermalmodel for the rotor at steady state is described. Temperature predictions from thethermal network are compared with FEM thermal simulations at different lossesand boundary conditions.

5.1 The T-equivalent circuit

The T-equivalent circuit used to analyze the relation between heat flow and themean component temperature is extensively used for induction motors [1] [14]. Theprinciple is explained with the example of heat transfer in a rectangular plate asshown in Fig 5.1.

x

T1 T2

q

Figure 5.1. One dimensional heat transfer in rectangular plate.

From Equ (2.3), Poisson’s equation for simple one-dimensional temperature distri-

29

CHAPTER 5. LP THERMAL MODEL FOR THE ROTOR

bution is:λd2T (x)dx2 = −qg (5.1)

where T is the temperature in ◦C, λ is the thermal conductivity in W/(m ◦C) andqg is the heat density in W/m3. Integrating Equ (5.1) and applying the boundaryconditions, the solution is:

T (x) = −qgx2

2λ + T2 − T1l

x+ qgl

2λx+ T1 (5.2)

where l is the length of the plate.

Therefore, the mean temperature Tm of the rectangular plate is calculated as:

Tm = 1l

∫ l0T (x)dx , (5.3)

Tm = qgl2

12λ + T1 + T22 . (5.4)

Seen from Equ (5.4), the mean temperature consists of two parts: the averagetemperature rise due to the internal heat generation (first term on the right side)and the mean value of the boundary temperatures (second term on the right side).Recalling Equ (2.7), the thermal resistance of rectangular plate is defined as:

R = l

λA(5.5)

Inserting Equ (5.5) into Equ (5.4) leads to,

Tm = qglAl

λA+ T1 + T2

2 = QR

12 + T1 + T22 (5.6)

where Q = qglA is the total heat dissipated in. the component

Based on Equ (5.6), an equivalent circuit shown in Fig 5.2 [1],[14] is developed.A negative thermal resistance (Rm = −R/6) is introduced to keep Tm with thesame value as Equ (5.6).

The T-equivalent circuit can also be used when there is no internal heat generation,i.e. Q equals zero. In some other conditions, if the internal heat generation Q orthe thermal resistance R are small, Rm can be approximated to zero, which leadsto:

Tm = QR

4 + T1 + T22 . (5.7)

In this case, the error is QR6 .

30

5.1. THE T-EQUIVALENT CIRCUIT

x

T1 T2

6m

RR = −

2

R

2

R

Tm

Q

Figure 5.2. T-equivalent circuit for one-dimensional heat transfer.

In case of a two-dimensional heat transfer problem, for example, the heat trans-fer in permanent magnets, it is assumed that the heat transfer in horizontal andvertical directions are independent [1]. Two T-equivalent circuits are paralleled con-nected as shown in Fig 5.3.

x

T1 T2

6

x

mx

RR = −

2

xR

2

xR

TmQ

T3

T4

6

y

my

RR = −

2

yR

2

yR

y

Figure 5.3. T-equivalent circuit for two-dimensional heat transfer.

In [1] and [14], the analytical calculation of Rm for rectangular and cylindricalshapes are given. In [13], the calculation and approximation of Rm for trapezoidalshape and arc segment are discussed.

31


5.2 Geometry of the LP thermal modelBecause of the symmetry, only 1/8 (half pole) of the complete rotor is modeled. Asshown in Fig 5.4, the rotor is generally divided into three parts:

- Permanent magnets: central and peripheral magnets;

- Magnet slots;

- Rotor iron which is further divided into four regions: rotor iron in the middle,along d and q axis and the rotor surface.

Rotor iron , q-axis

Rotor iron , middle

Rotor iron , d-axis

Rotor surface

Central magnet

Peripheral magnet

Rotor slot

Figure 5.4. Half pole of the rotor.

The effect of magnet insulation is not considered.

Fig 5.5 shows the LP thermal model of the rotor. Since the thermal network isonly used to predict the temperature at steady state, the thermal capacitances arenot included. The loss sources are placed at the middle of the magnets and ironparts where the losses are generated. The thermal resistance of the iron near ro-tor surface is neglected and only the losses are introduced. Two parallel-connectedT-equivalent circuits are used to model the magnets both in radial and tangentialdirections. In the d axis and middle parts of rotor iron, the T-equivalent circuit isonly applied in the radial direction.

Simplifying the thermal network in Fig 5.5 by series and parallel connections of

32

5.3. THERMAL RESISTANCES

P

N2

N3

N4

N5

N6

N7

N8

N9

N10

N1

Rpm_c_rad

Rpm_c_rad

Rpm_c_tg

Rpm_c_tg_m

Rpm_c_rad_m

Rpm_p_tg

R pm_p_rad

R pm_p_tg_c

Rpm_p_rad_c

Rpm_p_rad

Rpsl_c_rad

Rpsl_c_tg

R psl_p_rad

Rpsl_p_tg

Rcsl_c_tgRcsl_c_tgRcsl_c_tgRcsl_c_tg

P

R rot_q_p1

R rot_q_c

Rrot_sh

Rrot_mid_p

Rrot_mid_c

Rrot_d_p

Rrot_d_c1

Rrot_d_c2

Rrot_d_c3Rrot_d_c3

Rrot_d_c4Rrot_d_c4

Rrot_d_c3Rrot_d_c3

R rot_q_p2

Rrot_q_m

P

P

P

P

Rrot_mid_m

Figure 5.5. LP model of the rotor half pole.

the resistances (see App.C), a thermal circuit with 11 nodes and 22 resistances isobtained as illustrated in Fig 5.6. This circuit is solved by Portunus and the nodetemperatures are obtained.

5.3 Thermal resistancesThe thermal resistances of magnets are easily calculated from Equ (5.5). However,in the other parts, because of the irregular shapes, exact analytical calculationsof the the thermal resistances [13] are complex. The solution is to simplify theseirregular shapes as rectangles with mean widths and heights. Take the rotor ironalong the q axis for example, as shown in Fig 5.7:

Rrot q p1 = W782λferL(H78 +H77)/2 (5.8)

Rrot q c = W782λferL(H78 +H88)/2 (5.9)

Rrot q m = −Rrot q p1 +Rrot q c2 · 3 (5.10)

33


1

2

Rpm_c_rad_p

Rpm_c_rad_c

Rpm_c_tg_mid

43

Rrot_q_p

Rrot_q_c

Rpm_c_tg_q

R_rot_mid_p

Rrot_mid_c

Rpm_p_tg_mid

Rpm_p_rad_p

Rpm_p_rad_c

Rpm_p_tg_d

5

Rrot_d_c

Rrot_cbri_p

6

Rrot_cbri_c

T T_shaft

8910 7

Rrot_d_p

Rrot_cbri_m

11

TM1

TM4TM6

TM7TM8TM9

Rpm_c_tg_c

Rpm_c_rad_mPpm_c

TM3

Rpm_p_tg_c

Rpm_p_rad_mPpm_p

TM5

T

T_rot_surf

Pri_q Pri_d

Pri_rot_surf

Rsh

Rrot_q_m

TM2

Rrot_mid_m

TM10

Pri_mid

WMWM

Figure 5.6. Simplified thermal network for the rotor.

where L is the active length of the motor and λfer is the thermal conductivity ofthe rotor iron.

The calculation of all the resistances in Fig 5.5 and 5.6 can be found in App.C.It should be noticed that all the resistances should be divided by 8 when this modelis extended to the complete rotor.

5.4 Verification of the rotor LP thermal modelTo verify the accuracy of the LP model, the results from Portunus are compared withthe ones from FEM thermal simulations at different boundary and losses conditions:

- Case 1: temperature difference is introduced between shaft and rotor surfaceand there is no magnet and iron losses;

- Case 2: same temperature is applied to shaft and rotor surface and magnet

34

5.4. VERIFICATION OF THE ROTOR LP THERMAL MODEL

W78

H77

H78

H88

Figure 5.7. Rotor iron along the q axis.

and iron losses are considered;

- Case 3: Both temperature difference between shaft and rotor surface andmagnet and iron losses are taken into account.

The FEM model of the rotor is described in Section 4.4. To make the temperaturerise more obvious, the relatively large amount of "fictional" losses are put into themagnets and rotor iron.

5.4.1 Verification in Case 1In this case, the temperature of shaft (Tshaft) and rotor surface (Trot sur) are set to100◦C and 200◦C, respectively. Neither magnet nor iron losses are introduced. Fig5.8 shows the temperature distribution from FEM thermal simulation. Table 5.1lists the comparison of the results.

The error at central and peripheral magnets are −2.04% and −0.33%, respectively.

5.4.2 Verification in Case 2In this case, the same temperature of 100◦C is applied to the shaft and rotor surface.The losses in the central and peripheral magnets are 1kW and 0.5kW , respectively.The losses applied to the middle, d and q axis, rotor surface iron regions are 0.5kW ,respectively. Fig 5.9 shows the temperature distribution from FEM thermal simu-lation. Table 5.2 lists the comparison of the results.

The temperatures in central and peripheral magnets from the LP model are 3.47%and 2.68% lower than those from the FEM simulation, respectively. The largest

35


(a) (b)

Figure 5.8. Case 1 (a) Temperature distribution, (b) Equi-temperature lines.

Table 5.1. Temperature comparison in Case 1.

.

Node Circuit FEM Error %1 200 200 –2 148,8 152,4 -2,363 153,5 156,7 -2,044 178,2 176,2 1,145 181,3 181,9 -0,336 195,5 196,5 -0,517 183,5 176,7 3,858 166 163,2 1,729 137,2 135,7 1,1110 120,6 117,7 2,4611 100 100 –

36

5.4. VERIFICATION OF THE ROTOR LP THERMAL MODEL

(a) (b)



.

Node Circuit FEM Error %1 100 100 –2 184,8 184,5 0,163 200 207,2 -3,474 181 187,5 -3,475 192,1 197,4 -2,686 133,3 129,5 2,937 162,9 174 -6,388 194,2 190,9 1,739 169,4 160,7 5,4110 152 138,2 9,9911 100 100 –

37


error appears at point near the shaft (Node 10), where the temperature predictedby the LP model is 10% higher than given by the FEM simulations..

5.4.3 Verification in Case 3

Case 3 combines Case 1 and Case 2. Tshaft equals 100◦C and Trot sur equals 200◦C.The losses in the central and peripheral magnets are 1kW and 0.5kW , respectively.The losses applied to the middle, d and q axis, rotor surface iron regions are 0.5kW ,respectively. Fig 5.10 shows the temperature distribution from FEM thermal sim-ulation. Table 5.3 lists the comparison of the results.

(a) (b)



.

Node Circuit FEM Error %1 200 200 –2 233,6 235,9 -0,973 253,2 261,4 -3,144 259,2 269,5 -3,825 273,4 280,2 -2,436 228,9 223,2 2,557 246,4 251 -1,838 260,2 253,9 2,489 206,7 195,1 5,9510 172,7 155,3 11,2011 100 100 –

38

5.5. DISCUSSION AND CONCLUSION

The temperature at central and peripheral magnets from the LP model are 3.14%and 2.43% lower than that from the FEM simulation, respectively. The biggesterror appears at point near the shaft (Node 10), which is 11.2% higher.

5.5 Discussion and conclusionUsing the T-equivalent circuits based LP thermal model, a relatively accurate tem-perature prediction for the magnets can be achieved in different losses and boundaryconditions. However, large errors (around 10%) appears at the point near the shaftwhen losses are introduced.

On the other hand, "fictional" losses and boundary temperature have been usedto verify the thermal model. Since there is no relative movement between the rotorand fundamental flux field, the losses of the rotor are small in PM synchronousmachines. This means that Case 1 is closest to the real conditions, and in that casereally small differences between FEM and LP results have been obtained.

39

Chapter 6

Conclusions and future works

The presented work in this thesis has mainly revolved around the thermal behaviorof permanent magnet motors in traction application. FEM thermal simulations givethe temperature distribution of the motor at different conditions in steady state.FEM analysis provides details of the temperature distribution. It is a powerfultool to solve the thermal problem of complex geometry in steady state. A lumpedparameter thermal model is developed for the rotor and gives the temperature pre-diction of the permanent magnets.

Suggestions for the future work are summarized in the following:

• The work in this thesis focuses in the radial cross section. The thermal modelscan be extended to three dimensions including the heat transfer in axial di-rection. The temperatures of the end windings and bearing can be predicted.

• Thermal transient analysis can be carried out both with FEM and LP thermalmodels.

• The temperature prediction is strongly dependent on the losses prediction ofthe motor. Therefore, one possible future work is to investigate the iron lossesand magnet losses of the motor and to predict them with accuracy.

• The results obtained from FEM and LP thermal models in this thesis shouldbe validated with measurements.

41

Bibliography

[1] T.A. Lipo. Introduction to AC Machine Design. University of Wisconsin, 2ndedition, 2004.

[2] Chandur Sadarangani. Electrical Machines, Design and Analysis of Inductionand Permanent Magnet Motors. KTH, 2006.

[3] Flux 9.30 user’s guide, April 2006.

[4] Portunus, April 2009. http://www.motor-design.com/partners.php.

[5] Motor-cad software, April 2009. Motor Design Ltd.

[6] Frank P. Incropera and David P. DeWitt. Fundamentals of Heat and MassTransfer . Wiley, 6th edition, 2006.

[7] Heat transfer. http://en.wikipedia.org/.

[8] AE Bergles. Evolution of cooling technology for electrical, electronic, andmicroelectronic equipment. IEEE Transactions on Components and PackagingTechnologies, 26(1):6–15, 2003.

[9] DA Staton. Thermal Analysis of Electric Motors and Generators, TrainingCourse. Motor Design Ltd, 2007.

[10] Lindstrom J. Thermal Model of a Permanent-Magnet Motor for a Hybrid Elec-tric Vehicle. Licentiate thesis, Chalmers University of Technology, Goteborg,Sweden, 1999.

[11] N. Bianchi, S. Bolognani, and F. Tonel. Thermal analysis of a run-capacitorsingle-phase induction motor. IEEE Transactions on Industry Applications,39(2):457–465, 2003.

[12] Y.R. Chin. A permanent magnet traction motor for electric forklifts–Designand iron loss analysis with experimental verifications. PhD thesis, KTH, Swe-den, 2006.

[13] ZJ Liu, D. Howe, PH Mellor, and MK Jenkins. Thermal analysis of permanentmagnet machines. In Electrical Machines and Drives, 1993. Sixth InternationalConference on (Conf. Publ. No. 376), pages 359–364, 1993.

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[14] PH Mellor, D. Roberts, and DR Turner. Lumped parameter thermalmodel for electrical machines of TEFC design. IEE Proceedings B [see alsoIEE Proceedings-Electric Power Applications] Electric Power Applications,138(5):205–218, 1991.

[15] Dynamic viscosity. http://www.engineeringtoolbox.com/.

[16] Fluid dynamics. http://en.wikipedia.org/.

[17] Y. Lee, S. Hahn, and S.K. Kauh. Thermal analysis of induction motor withforced cooling channels. IEEE Transactions on Magnetics, 36(4), 2000.

[18] DA Staton and A. Cavagnino. Convection heat transfer and flow calculationssuitable for analytical modeling of electric machines. In 32nd Annual Confer-ence of the IEEE Industrial Electronics Society (IECON), pages 7–10.

[19] J. Smeets. Study of Turn-to-turn Failure in Permanent Magnet Traction Motorfor Railway Applications. Internal Report, Electrical Machines and PowerElectronics division, KTH, 2008.

[20] Discussion with F. Meier. Bombardier Transportation, Sweden, 2009.

44

List of Symbols

Symbols

A cross sectional area m2

As stator surface area m2

cp specific heat J/(kg ◦C)dc dimensionless curvatureD diameter mGr Grashof numberhc convection heat transfer coefficient W/(m2) ◦Cl length of component mL active length of the machine mNu Nusselt numberPr Prandtl numberqr heat density W/m3

Q heat flow / flow rate W m3/sRe Reynolds numberSch air channel perimeter mT temperature ◦CTa Taylor numberV velocity m/sλ thermal conductivity W/(m ◦C)ρ density kg/(m3)β load angle ◦

µ Dynamic viscosity P

Acronyms and abbreviations

FE finite elementLP lumped parameter

45

Appendix A

Fan characteristics

The fan characteristics are given by Florence Meier, Bombardier Transportation[20]. Table A.1 lists the air flow rate, air speed and pressure drop at different rotorspeeds. Fig A.1 shows air flow rate variation with the rotor speed.

Table A.1. Fan characteristics.

.

Motor speed [rpm] Air flow [m3/s] Air speed [m/s] Pressure fall [Pa]1000 0.091 8.6 80.22000 0.186 17.5 313.52500 0.234 22 486.33000 0.282 26.5 696.33500 0.33 31 943.54000 0.378 35,5 1227.24500 0.426 40 1547.85000 0.474 44.5 1905.15500 0.522 49.1 2298.96000 0.571 53.6 2729.3

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APPENDIX A. FAN CHARACTERISTICS

Figure A.1. Flow rate variation with rotor speed.

48

Appendix B

Rotor eddy current losses and magnetlosses

Table B.1 lists FEM simulation results of the rotor eddy current losses and magnetlosses with different current angles (β). The magnet losses are obtained from the"active power" in FLUX. The magnitude of the stator current (Is) is 200A and therotor speed (ω) is 3000 rpm.

The rotor iron is divided into several regions shown in Fig B.1.

PM_pos_c

PM_pos_p

PM_neg_c

PM_neg_p

Rotor iron_pos_d

Rootr iron_pos_mid

Rotor iron_pos_q

Rotor iron_neg_q

Rotor rion neg_mid

Rotor iron_ neg _d

Rotor iron_neg_sur

Rotor iron_pos_sur

Figure B.1. Rotor regions.

49

APPENDIX B. ROTOR EDDY CURRENT LOSSES AND MAGNET LOSSES

Table B.1. Rotor eddy current and magnet losses, Is = 200A, ω = 3000 rpm.

.

Rotor iron eddy current losses [W ]regions noload beta=126 beta=170neg q 1.42 14.04 30.18negmid 7.65 21.25 28.82neg d 14.58 35.08 17.88neg sur 25.25 136.33 42.39pos q 1.45 7.66 25.47posmid 6.53 25.34 35.89pos d 14.19 63.52 28.07pos sur 24.5 146.14 36.3Total 95.57 449.36 245

Magnets active power [W ]PM neg, p 1.09 6.58 1.88PM pos p 1.08 6.31 1.98PM neg c 5.01 24.36 4.08PM pos c 5.01 5.15 3.75Total 12.19 42.4 11.69

50

Appendix C

Calculation of thermal resistances

C.1 Rotor dimensions

The geometrical parameters used to calculate the thermal resistances of the rotorare shown in Fig C.1 and Table C.1.

Wpm_c

W78

H88

H77

H78

Wpsl_c

Wpm_p

Hc

Wpsl_p

Hpsl_p

Wcsl_c

Hpsl_c

Hcsl_c

Hself_p

Hp

Wp

Hpm_c

Hpb2

Hpb1

Wpb1Wpb2

Ri

Ro

Hslgap

α_pm_p

Figure C.1. Dimensions of the rotor (half pole).

51

APPENDIX C. CALCULATION OF THERMAL RESISTANCES

Table C.1. Dimensions of the rotor (half pole).

Dimensions Value [mm] Dimensions Value [mm]Permanent magnetsHpmp – Wpmp –Hpmc – Wpmc –Magnet slotsWpsl c – Wsl c –Hpsl c – Hcsl c –Wpsl p – Hslgap –Hpsl p – Hself –Wcsl c –Rotor ironWp – Ri –Hp – Ro –Hc – H88 –Hpb1 – H77 –Hpb2 – H78 –Wpb1 – W78 –Wpb2 – H37 –αpmp –

C.2 Thermal resistancesThermal resistances of magnets:central magnets

Rpmc rad = Wpmc2λpm LHpmc

(C.1)

Rpmc radm = −13Rpmc rad (C.2)

Rpmc tg = Hpmc2λpm LWpmc

(C.3)

Rpmc tgm = −13Rpmc tg (C.4)

peripheral magnetsRpmp rad = Wpmp

2λpm LHpmp(C.5)

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C.2. THERMAL RESISTANCES

P

N2

N3

N4

N5

N6

N7

N8

N9

N10

N1

Rpm_c_rad

Rpm_c_rad

Rpm_c_tg

Rpm_c_tg_m

Rpm_c_rad_m

Rpm_p_tg

R pm_p_rad

R pm_p_tg_c

Rpm_p_rad_c

Rpm_p_rad

Rpsl_c_rad

Rpsl_c_tg

R psl_p_rad

Rpsl_p_tg

Rcsl_c_tgRcsl_c_tgRcsl_c_tgRcsl_c_tg

P

R rot_q_p1

R rot_q_c

Rrot_sh

Rrot_mid_p

Rrot_mid_c

Rrot_d_p

Rrot_d_c1

Rrot_d_c2

Rrot_d_c3Rrot_d_c3

Rrot_d_c4Rrot_d_c4

Rrot_d_c3Rrot_d_c3

R rot_q_p2

Rrot_q_m

P

P

P

P

Rrot_mid_m

Figure C.2. Thermal circuit of the rotor (half pole)

Rpmp radm = −13Rpmp rad (C.6)

Rpmp tg = Hpmp2λpm LWpmp

(C.7)

Rpmp tgm = −13Rpmp tg (C.8)

Thermal resistances of the magnet slots:

Rpsl c rad = Wpsl cλsl LHpsl c

(C.9)

Rpsl c tg = Hsl gapλsl LWpmc

(C.10)

Rpsl p rad = Wpsl pλsl LHpsl p

(C.11)

53

APPENDIX C. CALCULATION OF THERMAL RESISTANCES

1

2

Rpm_c_rad_p

Rpm_c_rad_c

Rpm_c_tg_mid

43

Rrot_q_p

Rrot_q_c

Rpm_c_tg_q

R_rot_mid_p

Rrot_mid_c

Rpm_p_tg_mid

Rpm_p_rad_p

Rpm_p_rad_c

Rpm_p_tg_d

5

Rrot_d_c

Rrot_cbri_p

6

Rrot_cbri_c

T T_shaft

8910 7

Rrot_d_p

Rrot_cbri_m

11

TM1

TM4TM6

TM7TM8TM9

Rpm_c_tg_c

Rpm_c_rad_mPpm_c

TM3

Rpm_p_tg_c

Rpm_p_rad_mPpm_p

TM5

T

T_rot_surf

Pri_q Pri_d

Pri_rot_surf

Rsh

Rrot_q_m

TM2

Rrot_mid_m

TM10

Pri_mid

WMWM

Figure C.3. Simplified thermal circuit of the rotor (half pole).

Rpsl p tg = Hsl gapλsl LWpmp

(C.12)

Rcsl c tg = Hcsl c2λsl LWcsl csin(αpmp)

(C.13)

Thermal resistances of the rotor iron:

Rrot q c = W78/2λfer L (H88 +H78)/2 (C.14)

Rrot q p1 = W78/2λfer L (H78 +H77)/2 (C.15)

Rrot q m = −Rrot q c +Rrot q p12 · 3 (C.16)

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C.2. THERMAL RESISTANCES

Rrot q p2 = Wcsl cλferLH77

(C.17)

Rrotmid p = (Wpmc/2 +Wpmp/2 +Wpsl,c +Wpsl p)/2λfer L (Hc +Hself )

(C.18)

Rrotmid c = (Wpmc +Wpmp +Wcsl,c)/22λfer L (Hc +Hself )

(C.19)

Rrotmidm = −Rrotmid c +Rrotmid p2 · 3 (C.20)

Rrot d p = Wp2λfer LHp

(C.21)

Rrot d c1 = Rrot d p (C.22)

Rrot d c2 = Wp b12λfer LHp b1

(C.23)

Rrot d c3 = Hcsl c2λfer LHp b2

(C.24)

Rrot d c4 = Wp b22λfer LHp b1

(C.25)

Rrot sh = 8 · ln(Ro/Ri)2πLλfer

(C.26)

Serial and parallel connections:

Rrot q p = Rrot q p1 +Rrot q p2 (C.27)

Rpmc rad p = Rpmc rad +Rpsl c rad (C.28)

Rpmc tgmid = Rpmc tg +Rpsl c tg (C.29)

Rpmp rad p = Rpmp rad +Rpsl p rad (C.30)

Rpmp tg d = Rpmp tg +Rpsl p tg (C.31)

Rrot d c = Rrot d c1 +Rrot d c2 +Rcsl c tg||Rrot d c3 (C.32)

Rrot cbri c = Rrot cbri p = Rrot d c4 +Rcsl c tg||Rrot d c3 (C.33)

Rrot cbrim = Rcsl c tg||Rrot d c3 (C.34)

55

Development of Thermal Models for PM Tractions Motors

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